From 415fbe5d9615cc99ceaa20e90436a1a99f452110 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Thu, 25 Dec 2025 23:41:30 +0800 Subject: [PATCH 001/274] vault backup: 2025-12-25 23:41:30 --- 编写小组/课后测/课后测4.md | 15 ++++++++++- 编写小组/课后测/课后测解析版4.md | 26 ++++++++++++++++++- 2 files changed, 39 insertions(+), 2 deletions(-) diff --git a/编写小组/课后测/课后测4.md b/编写小组/课后测/课后测4.md index 7a1e9fe..7f19bc6 100644 --- a/编写小组/课后测/课后测4.md +++ b/编写小组/课后测/课后测4.md @@ -2,4 +2,17 @@ -2.设数列 $\{a_n\}$ 的三个子列 $\{a_{2n}\}$、$\{a_{2n+1}\}$、$\{a_{3n+1}\}$ 均收敛，那么 $\{a_n\}$ 是否一定收敛？说明理由。 \ No newline at end of file +2.设数列 $\{a_n\}$ 的三个子列 $\{a_{2n}\}$、$\{a_{2n+1}\}$、$\{a_{3n+1}\}$ 均收敛，那么 $\{a_n\}$ 是否一定收敛？说明理由。 + + + + + + +3.设$\{x_n\}$是数列，下列命题中不正确的是( ) +A. 若$\lim\limits_{n \to \infty} x_n = a$，则$\lim\limits_{n \to \infty} x_{2n} = \lim\limits_{n \to \infty} x_{2n+1} = a$ +B. 若$\lim\limits_{n \to \infty} x_{2n} = \lim\limits_{n \to \infty} x_{2n+1} = a$，则$\lim\limits_{n \to \infty} x_n = a$ +C. 若$\lim\limits_{n \to \infty} x_n = a$，则$\lim\limits_{n \to \infty} x_{3n} = \lim\limits_{n \to \infty} x_{3n+1} = a$ +D. 若$\lim\limits_{n \to \infty} x_{3n} = \lim\limits_{n \to \infty} x_{3n+1} = a$，则$\lim\limits_{n \to \infty} x_n = a$ + + diff --git a/编写小组/课后测/课后测解析版4.md b/编写小组/课后测/课后测解析版4.md index 90424ab..13fdb7a 100644 --- a/编写小组/课后测/课后测解析版4.md +++ b/编写小组/课后测/课后测解析版4.md @@ -22,4 +22,28 @@ $\{a_n\}$ **一定收敛**，理由如下： - $\{a_{3n+1}\}$ 中的奇数项子列（属于 $\{a_{2n+1}\}$）的极限 = $\{a_{3n+1}\}$ 的极限。 - 故 $\{a_{2n}\}$ 与 $\{a_{2n+1}\}$ 的极限相等。 -综上，数列 $\{a_n\}$ 的偶数项子列和奇数项子列极限相同，因此 $\{a_n\}$ 收敛。 \ No newline at end of file +综上，数列 $\{a_n\}$ 的偶数项子列和奇数项子列极限相同，因此 $\{a_n\}$ 收敛。 + +3.设$\{x_n\}$是数列，下列命题中不正确的是( ) +A. 若$\lim\limits_{n \to \infty} x_n = a$，则$\lim\limits_{n \to \infty} x_{2n} = \lim\limits_{n \to \infty} x_{2n+1} = a$ +B. 若$\lim\limits_{n \to \infty} x_{2n} = \lim\limits_{n \to \infty} x_{2n+1} = a$，则$\lim\limits_{n \to \infty} x_n = a$ +C. 若$\lim\limits_{n \to \infty} x_n = a$，则$\lim\limits_{n \to \infty} x_{3n} = \lim\limits_{n \to \infty} x_{3n+1} = a$ +D. 若$\lim\limits_{n \to \infty} x_{3n} = \lim\limits_{n \to \infty} x_{3n+1} = a$，则$\lim\limits_{n \to \infty} x_n = a$ + +**解析** +选项A：✅ 正确 +若$\lim\limits_{n \to \infty} x_n = a$，$\{x_{2n}\}和\{x_{2n+1}\}是\{x_n\}$的子列，根据子列性质，子列必收敛于a。 +- 选项B：✅ 正确 +数列$\{x_n\}$的项由偶数项$\{x_{2n}\}$和奇数项$\{x_{2n+1}\}$全部组成，二者都收敛于a，则对任意$\varepsilon>0$，存在N，当n>N时，偶数项和奇数项都满足$|x_n - a| < \varepsilon$，故$\lim\limits_{n \to \infty} x_n = a$。 +- 选项C：✅ 正确 +$\{x_{3n}\}$和$\{x_{3n+1}\}$是$\{x_n\}$的子列，由子列性质，原数列收敛则子列必收敛于同一值a。 +- 选项D：❌ 不正确 +$\{x_{3n}\}$（3的倍数项）和$\{x_{3n+1}\}$（3的倍数加1项）只是$\{x_n\}$的部分子列，遗漏了$\{x_{3n+2}\}$（3的倍数加2项）。 +反例：构造数列 +$$x_n = \begin{cases} +a, & n=3k \text{ 或 } n=3k+1 \ +b \ (b\neq a), & n=3k+2 +\end{cases}$$ +此时$\lim\limits_{n \to \infty} x_{3n} = a$，$\lim\limits_{n \to \infty} x_{3n+1} = a$，但$\lim\limits_{n \to \infty} x_{3n+2} = b \neq a$，故$\{x_n\}$的极限不存在，无法推出$\lim\limits_{n \to \infty} x_n = a$。 + +答案：$\boldsymbol{D}$ -- 2.34.1 From 4c44b50e161eaa7abba136d3cdf48e02ec86ad11 Mon Sep 17 00:00:00 2001 From: Cym10x Date: Fri, 26 Dec 2025 07:03:23 +0800 Subject: [PATCH 002/274] minor edit --- 编写小组/讲义/子数列问题&考试易错点汇总.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总.md b/编写小组/讲义/子数列问题&考试易错点汇总.md index 338c02d..a28402c 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总.md @@ -466,6 +466,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f > [!example] 例2 > 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ +这个也是无界但不是无穷大，请自行证明。 总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” -- 2.34.1 From e2c78a458196d292cd8a5af955bb2472a26dcd8b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Fri, 26 Dec 2025 07:04:13 +0800 Subject: [PATCH 003/274] vault backup: 2025-12-26 07:04:13 --- .../子数列问题&考试易错点汇总（解析版）.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index bf45c72..2517960 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -243,9 +243,9 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 **解**： 对方程 $y = \ln(1 + \sin^2 x)$ 两边关于 $x$ 求导。 -$$ y‘ = \frac{1}{1+\sin^2 x} \cdot (2\sin x \cos x) = \frac{\sin 2x}{1+\sin^2 x} $$ +$$ y'= \frac{1}{1+\sin^2 x} \cdot (2\sin x \cos x) = \frac{\sin 2x}{1+\sin^2 x} $$ -根据微分的定义，$dy = y’ dx$，故 +根据微分的定义，$dy = A dx$，而当$y$可导时，$A=y'$，故 $$ dy = \frac{\sin 2x}{1+\sin^2 x} dx $$ 注意：不要漏写 $dx$ ！ @@ -343,7 +343,7 @@ $$\lim_{n \to \infty} \frac{ \frac{1}{n^{1 + \frac{1}{n}}} }{ \frac{1}{n} } = \l #### ❌ 经典错误思路 -无法直接套用p级数，因多了一个 `ln n` 因子。错误地认为 `1/(n ln n) < 1/n`，认为原级数收敛 +无法直接套用p级数，因多了一个 `ln n` 因子。错误地认为 `1/(nln n) < 1/n`，认为原级数收敛 #### ✅ 正确分析与解法(超纲，仅供拓展) **正确解法**（积分判别法）： -- 2.34.1 From 94192852ed1871de6a8ab4bc742b371a257a7edf Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 10:39:53 +0800 Subject: [PATCH 004/274] vault backup: 2025-12-26 10:39:53 --- .../证明题方法：单调有界定理，介值定理.md | 4 ++++ ...法：单调有界定理，介值定理（解析版）.md | 4 ++++ 2 files changed, 8 insertions(+) diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md index 98d80bd..3eb43b3 100644 --- a/编写小组/讲义/证明题方法：单调有界定理，介值定理.md +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md @@ -16,6 +16,10 @@ tags: - **函数单调有界性质**： 若函数 $f(x)$ 在区间 $I$ 上单调且有界，则 $f(x)$ 在 $I$ 的端点或无穷远处存在单侧极限。 +- **这类题的基本思路**： + 先看有界性：尝试用不等式或者使用数学归纳法 + 再看单调性：尝试作差、作商或者使用数学归纳法 + ## 适用情况 适用于证明数列或函数极限的存在性，尤其是： diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md index 4bc4449..d85674a 100644 --- a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md @@ -15,6 +15,10 @@ tags: - **函数单调有界性质**： 若函数 $f(x)$ 在区间 $I$ 上单调且有界，则 $f(x)$ 在 $I$ 的端点或无穷远处存在单侧极限。 +- **这类题的基本思路**： + 先看有界性：尝试用不等式或者使用数学归纳法 + 再看单调性：尝试作差、作商或者使用数学归纳法 + ## 适用情况 适用于证明数列或函数极限的存在性，尤其是： -- 2.34.1 From f70a26d769997e0b2da4a9a509aafa17f5c7a61e Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 10:43:18 +0800 Subject: [PATCH 005/274] vault backup: 2025-12-26 10:43:18 --- .../证明题方法：单调有界定理，介值定理.md | 4 +++- ...法：单调有界定理，介值定理（解析版）.md | 4 +++- 2 files changed, 6 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md index 3eb43b3..8c4903b 100644 --- a/编写小组/讲义/证明题方法：单调有界定理，介值定理.md +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理.md @@ -17,7 +17,9 @@ tags: 若函数 $f(x)$ 在区间 $I$ 上单调且有界，则 $f(x)$ 在 $I$ 的端点或无穷远处存在单侧极限。 - **这类题的基本思路**： - 先看有界性：尝试用不等式或者使用数学归纳法 + + 先看有界性：尝试用不等式或者使用数学归纳法，注意在上下界不好找的时候，可以先取极限，来确定一个大致范围，再对猜测的有界性尝试证明 + 再看单调性：尝试作差、作商或者使用数学归纳法 ## 适用情况 diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md index d85674a..e0467e9 100644 --- a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md @@ -16,7 +16,9 @@ tags: 若函数 $f(x)$ 在区间 $I$ 上单调且有界，则 $f(x)$ 在 $I$ 的端点或无穷远处存在单侧极限。 - **这类题的基本思路**： - 先看有界性：尝试用不等式或者使用数学归纳法 + + 先看有界性：尝试用不等式或者使用数学归纳法，注意在上下界不好找的时候，可以先取极限，来确定一个大致范围，再对猜测的有界性尝试证明 + 再看单调性：尝试作差、作商或者使用数学归纳法 ## 适用情况 -- 2.34.1 From 411a1b268a57c468e040191737fcf3957724358d Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 10:57:15 +0800 Subject: [PATCH 006/274] vault backup: 2025-12-26 10:57:15 --- ...方法：单调有界定理，介值定理（解析版）.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md index e0467e9..8c868c6 100644 --- a/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md +++ b/编写小组/讲义/证明题方法：单调有界定理，介值定理（解析版）.md @@ -4,6 +4,7 @@ tags: --- **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 + # 单调有界准则 ## 原理 -- 2.34.1 From cfc14f1ec859826b4334f9fd8be9a7a971974623 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:01:14 +0800 Subject: [PATCH 007/274] vault backup: 2025-12-26 11:01:14 --- .../讲义/极限计算的基本方法（解析版2）.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/编写小组/讲义/极限计算的基本方法（解析版2）.md b/编写小组/讲义/极限计算的基本方法（解析版2）.md index e91db4c..f63d7d9 100644 --- a/编写小组/讲义/极限计算的基本方法（解析版2）.md +++ b/编写小组/讲义/极限计算的基本方法（解析版2）.md @@ -343,7 +343,8 @@ $$ ### 4.3 使用原则 1. **乘除运算可直接替换** 2. **加减运算需谨慎**：只有同阶无穷小相加减时，不能随意替换 -3. **复合函数可整体替换** +3. **无穷乘方运算一般也不行**（即乘方上下均有变量时）：一对等价量可以写成差一个无穷小的形式，经过无穷乘方后无法得知其情况 +4. **复合函数可整体替换** ### 4.4 典型例题 -- 2.34.1 From 0018f177c28ce101275de1ad7f6cec2cdf7dca33 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:03:45 +0800 Subject: [PATCH 008/274] vault backup: 2025-12-26 11:03:45 --- .../讲义/极限计算的基本方法.md | 22 +++++++++- ...限计算的基本方法（解析版）.md | 42 ++++++++++++++++++- 2 files changed, 61 insertions(+), 3 deletions(-) diff --git a/编写小组/讲义/极限计算的基本方法.md b/编写小组/讲义/极限计算的基本方法.md index 86a0f37..e227386 100644 --- a/编写小组/讲义/极限计算的基本方法.md +++ b/编写小组/讲义/极限计算的基本方法.md @@ -192,8 +192,26 @@ $$\lim\limits_{x \to \infty} x(e^\tfrac{1}{x}-1)=\lim\limits_{x \to \infty } \tf # 等价无穷小求极限法 - -本质是利用已知公式进行等价无穷小因子的代换将复杂函数化简 +## 4.1 基本原理 + +在乘积和商的极限运算中，可以将复杂的无穷小量替换为等价的简单无穷小量，简化计算。 + +## 4.2 常用等价无穷小（$x\to 0$） +| 等价形式 | 条件 | +|---------|------| +| $x \sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x$ | 基础等价 | +| $x \sim e^x-1 \sim \ln(1+x)$ | 指数对数等价 | +| $1-\cos x \sim \frac{1}{2}x^2$ | 三角函数等价 | +| $(1+x)^a-1 \sim ax$ | 幂函数等价 | +| $a^x-1 \sim x\ln a\ (a>0)$ | 指数函数等价 | +| $x - \sin x \sim \frac{1}{6}x^3$ | 高阶等价 | +| $\tan x - x \sim \frac{1}{3}x^3$ | 高阶等价 | + +## 4.3 使用原则 +1. **乘除运算可直接替换** +2. **加减运算需谨慎**：只有同阶无穷小相加减时，不能随意替换 +3. **无穷乘方运算一般也不行**（即乘方上下均有变量时）：一对等价量可以写成差一个无穷小的形式，经过无穷乘方后无法得知其情况 +4. **复合函数可整体替换** > [!example] 例1 diff --git a/编写小组/讲义/极限计算的基本方法（解析版）.md b/编写小组/讲义/极限计算的基本方法（解析版）.md index b49dde7..d8e0b7e 100644 --- a/编写小组/讲义/极限计算的基本方法（解析版）.md +++ b/编写小组/讲义/极限计算的基本方法（解析版）.md @@ -371,7 +371,47 @@ $$= \frac{2\tan t}{1 - \tan t}.$$ # 等价无穷小求极限法 -本质是利用已知公式进行等价无穷小因子的代换将复杂函数化简 +## 4.1 基本原理 + +在乘积和商的极限运算中，可以将复杂的无穷小量替换为等价的简单无穷小量，简化计算。 + +## 4.2 常用等价无穷小（$x\to 0$） +| 等价形式 | 条件 | +|---------|------| +| $x \sim \sin x \sim \tan x \sim \arcsin x \sim \arctan x$ | 基础等价 | +| $x \sim e^x-1 \sim \ln(1+x)$ | 指数对数等价 | +| $1-\cos x \sim \frac{1}{2}x^2$ | 三角函数等价 | +| $(1+x)^a-1 \sim ax$ | 幂函数等价 | +| $a^x-1 \sim x\ln a\ (a>0)$ | 指数函数等价 | +| $x - \sin x \sim \frac{1}{6}x^3$ | 高阶等价 | +| $\tan x - x \sim \frac{1}{3}x^3$ | 高阶等价 | + +## 4.3 使用原则 +1. **乘除运算可直接替换** +2. **加减运算需谨慎**：只有同阶无穷小相加减时，不能随意替换 +3. **无穷乘方运算一般也不行**（即乘方上下均有变量时）：一对等价量可以写成差一个无穷小的形式，经过无穷乘方后无法得知其情况 +4. **复合函数可整体替换** + + +> [!example] 例1 +> 计算 $\displaystyle \lim_{x\to 0}\frac{\sin mx}{\sin nx}$ + + + + + + + +> [!example] 例2 +> 计算 $$\displaystyle \lim_{x\to 0}(\cos x)^{\frac{1+x}{\sin^2 x}}$$ + + + + + + +熟悉常见的极限，灵活运用好极限的四则运算法则和等价无穷小，已经能够解决绝大多数的极限计算题了，无论是简单题还是难题。很多看上去很复杂的所谓难题，无非是在四则运算和等价无穷小之间反复套娃而已，如果可以熟练的分离这些特征，其实不需要泰勒公式等“高级工具”就能快速准确地得出极限值。当然，不是贬低其他的工具，只是说不要学习了所谓的一些高级工具之后，就不重视这些初级的工具和结论。 + 引入练习：计算 $$\lim\limits_{x\to0}\frac{\sqrt{1+2tanx}-\sqrt{1+2sinx}}{xln(1+x)-x^2}$$     -- 2.34.1 From a40ea30f91218c6ee54bdc1ac4d9352d54083029 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:10:21 +0800 Subject: [PATCH 009/274] vault backup: 2025-12-26 11:10:21 --- .../讲义/极限计算的基本方法（解析版）.md | 6 +----- 1 file changed, 1 insertion(+), 5 deletions(-) diff --git a/编写小组/讲义/极限计算的基本方法（解析版）.md b/编写小组/讲义/极限计算的基本方法（解析版）.md index d8e0b7e..534af70 100644 --- a/编写小组/讲义/极限计算的基本方法（解析版）.md +++ b/编写小组/讲义/极限计算的基本方法（解析版）.md @@ -350,9 +350,7 @@ $$t = x - \frac{\pi}{4} \quad \Rightarrow \quad x = t + \frac{\pi}{4},\ t \to 0. $$= \frac{2\tan t}{1 - \tan t}.$$ 代入原极限：$$\lim\limits_{x\to\pi/4} \frac{\tan x - 1}{x - \pi/4} - = \lim\limits_{t\to 0} \frac{\frac{2\tan t}{1 - \tan t}}{t} - = \lim\limits_{t\to 0} \frac{2\tan t}{t(1 - \tan t)}.$$ 利用重要极限 $\displaystyle \lim_{u\to 0} \frac{\tan u}{u} = 1:\frac{2\tan t}{t(1 - \tan t)} = 2 \cdot \frac{\tan t}{t} \cdot \frac{1}{1 - \tan t}$ @@ -360,7 +358,6 @@ $$= \frac{2\tan t}{1 - \tan t}.$$ 当 $t \to 0$ 时，$\tan t \to 0，\lim\limits_{t\to 0} \frac{\tan t}{t} = 1,\quad \lim\limits_{t\to 0} (1 - \tan t) = 1.$ 所以$$\lim\limits_{t\to 0} 2 \cdot \frac{\tan t}{t} \cdot \frac{1}{1 - \tan t} - = 2 \cdot 1 \cdot \frac{1}{1 - 0} = 2$$ （4） @@ -558,5 +555,4 @@ $$\alpha n + \ln \left( a + \frac{\cdots}{e^{\alpha n}} \right)$$ 小练习： - -$$\lim_{x \to 0} \frac{\csc(x) - \cot(x)}{x}=$$ \ No newline at end of file +$$\lim_{x \to 0} \frac{\csc(x) - \cot(x)}{x}= $$ -- 2.34.1 From d0bf5aedb11e08aa82732f0fe8980acbb6a97b9e Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:23:54 +0800 Subject: [PATCH 010/274] vault backup: 2025-12-26 11:23:54 --- ...&考试易错点汇总（解析版）.md | 29 +++++++++++++++---- 1 file changed, 23 insertions(+), 6 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index bf45c72..2d80338 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -226,12 +226,29 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 **解析** -1. 构造第一个数列$\{x_n^{(1)}\}$： - 取$x_n^{(1)} = x_0 + \frac{1}{n}$（有理数），则$$\lim\limits_{n \to \infty} x_n^{(1)} = x_0$$由于$x_n^{(1)}$是有理数，所以$D(x_n^{(1)}) = 1$，因此$$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ -2. 构造第二个数列$\{x_n^{(2)}\}$： - 取$x_n^{(2)} = x_0 + \frac{\sqrt{2}}{n}$（无理数），则$$\lim\limits_{n \to \infty} x_n^{(2)} = x_0$$由于x_n^{(2)}是无理数，所以$D(x_n^{(2)}) = 0$，因此$$\lim\limits_{n \to \infty} D(x_n^{(2)}) = 0$$由于$$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1 \neq 0 = \lim\limits_{n \to \infty} D(x_n^{(2)})$$根据海涅定理，$\lim\limits_{x \to x_0} D(x)$不存在。 - 由于$x_0$是任意一点，所以狄利克雷函数在任何点处都不存在极限。 +1. 构造第一个数列 $\{x_n^{(1)}\}$： + + 取 $x_n^{(1)} = x_0 + \frac{1}{n}$（有理数），则 + $$\lim\limits_{n \to \infty} x_n^{(1)} = x_0$$ + + 由于 $x_n^{(1)}$ 是有理数，所以 $D(x_n^{(1)}) = 1$，因此 + $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ + +2. 构造第二个数列 $\{x_n^{(2)}\}$： + + 取 $x_n^{(2)} = x_0 + \frac{\sqrt{2}}{n}$（无理数），则 + $$\lim\limits_{n \to \infty} x_n^{(2)} = x_0$$ + + 由于 $x_n^{(2)}$ 是无理数，所以 $D(x_n^{(2)}) = 0$，因此 + $$\lim\limits_{n \to \infty} D(x_n^{(2)}) = 0$$ + + 由于 + $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1 \neq 0 = \lim\limits_{n \to \infty} D(x_n^{(2)})$$ + + 根据海涅定理，$\lim\limits_{x \to x_0} D(x)$ 不存在。 + + 由于 $x_0$ 是任意一点，所以狄利克雷函数在任何点处都不存在极限。 # 考试易错点总结 @@ -298,7 +315,7 @@ $$ \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{1 + \frac{2}{x} - \frac{3}{ **错误做法示范**：观察级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n} n^2}{3^n}$，若直接对其使用比值判别法： - $$ \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right| = \lim_{n \to \infty} \frac{(n+1)^2 / 3^{n+1}}{n^2 / 3^n} = \frac{1}{3} < 1 $$ +$$ \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right| = \lim_{n \to \infty} \frac{(n+1)^2 / 3^{n+1}}{n^2 / 3^n} = \frac{1}{3} < 1 $$ 若由此断言“原级数收敛”，则犯了**滥用判别法**的错误。因为比值判别法（及比较、根值判别法）仅 -- 2.34.1 From 0ab5ce154b9649911ff53dbeef1359528f809819 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:24:29 +0800 Subject: [PATCH 011/274] vault backup: 2025-12-26 11:24:29 --- .../子数列问题&考试易错点汇总（解析版）.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 2d80338..1b46175 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -268,6 +268,7 @@ $$ dy = \frac{\sin 2x}{1+\sin^2 x} dx $$ 注意：不要漏写 $dx$ ！ ## **Vol.2等价无穷小问题** + 注意，等价无穷小只能用于乘除，用于加减虽然有时也会得到正确的答案，但这并不是有保证的。归根到底这是因为等价无穷小是一种**近似**，在乘除中它的近似程度还可以用，但在加减中就未必了，加减中我们需要更精确的近似方法：泰勒展开。另外重要极限也是等价无穷小的两种特殊情况。 >[!example] 例题 >求极限$$\lim\limits_{x\to0}\frac{tanx-sinx}{x^3}$$. -- 2.34.1 From de5bd95f9d5cc8ce488c2ae970062a385acf1c40 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:24:33 +0800 Subject: [PATCH 012/274] vault backup: 2025-12-26 11:24:33 --- .../子数列问题&考试易错点汇总（解析版）.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 1b46175..db5fe03 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -311,6 +311,7 @@ $$ \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{1 + \frac{2}{x} - \frac{3}{ **结论**：曲线的渐近线为 $x = 2$ 和 $y = 1$。 ## Vol. 4:正项级数的判别法勿滥用 + >[!example] 例3 >判断级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n} n^2}{3^n}$ 的敛散性（绝对收敛、条件收敛或发散）。 -- 2.34.1 From 841d7aeb986ec77cdba8ad5d746bee1f0a86ec5a Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:30:33 +0800 Subject: [PATCH 013/274] vault backup: 2025-12-26 11:30:33 --- .../子数列问题&考试易错点汇总.md | 9 +++++---- ...&考试易错点汇总（解析版）.md | 20 +++++++++++++------ 2 files changed, 19 insertions(+), 10 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总.md b/编写小组/讲义/子数列问题&考试易错点汇总.md index a28402c..e176882 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总.md @@ -232,7 +232,7 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 ## Vol. 1:补药漏写dx口牙！ ->[!example] 例1 +>[!example] 例题 >设函数 $y = \ln(1 + \sin^2 x)$，求其微分 $dy$。 ``` @@ -268,7 +268,7 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 ## Vol. 3:可去间断点的说明 ->[!example] 例2 +>[!example] 例题 >求曲线 $y = \frac{x^2 + 2x - 3}{x^2 - 3x + 2}$ 的所有渐近线。 ``` @@ -284,7 +284,8 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 ## Vol. 4:正项级数的判别法勿滥用 ->[!example] 例3 + +>[!example] 例题 >判断级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n} n^2}{3^n}$ 的敛散性（绝对收敛、条件收敛或发散）。 ``` @@ -434,7 +435,7 @@ $y=f^{-1}(x)$，即$x = f(y)$，求$f^{-1'}$就是在求$\frac{dy}{dx}$，而$\f 所以，我们还需要将$y = f^{-1}(x)$代入，即： $\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$ -> [!example] 例1 +> [!example] 例题 > 求$d(\arcsin x)$ ``` diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index db5fe03..01f067c 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -254,7 +254,7 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 ## Vol. 1:补药漏写dx口牙！ ->[!example] 例1 +>[!example] 例题 >设函数 $y = \ln(1 + \sin^2 x)$，求其微分 $dy$。 **解**： @@ -270,8 +270,9 @@ $$ dy = \frac{\sin 2x}{1+\sin^2 x} dx $$ ## **Vol.2等价无穷小问题** 注意，等价无穷小只能用于乘除，用于加减虽然有时也会得到正确的答案，但这并不是有保证的。归根到底这是因为等价无穷小是一种**近似**，在乘除中它的近似程度还可以用，但在加减中就未必了，加减中我们需要更精确的近似方法：泰勒展开。另外重要极限也是等价无穷小的两种特殊情况。 + >[!example] 例题 ->求极限$$\lim\limits_{x\to0}\frac{tanx-sinx}{x^3}$$. +>求极限$$\lim\limits_{x\to0}\frac{tanx-sinx}{x^3}$$ **解**：如果直接用$tanx\sim x,sinx\sim x(x\to0)$的话，分子就会变成$0$，从而极限为$0$. 然而从另一个角度看，$$\lim\limits_{x\to0}\frac{tanx-sinx}{x^3}=\lim\limits_{x\to0}\frac{sinx-sinxcosx}{x^3cosx}=\lim\limits_{x\to0}\frac{sinx(1-cosx)}{x^3}\cdot \frac{1}{cosx}=\lim\limits_{x\to0}\frac{x\cdot\frac{1}{2}x^2}{x^3}=\frac{1}{2}.$$ @@ -281,7 +282,7 @@ $$ dy = \frac{\sin 2x}{1+\sin^2 x} dx $$ ## Vol. 3:可去间断点的说明 ->[!example] 例2 +>[!example] 例题 >求曲线 $y = \frac{x^2 + 2x - 3}{x^2 - 3x + 2}$ 的所有渐近线。 **解**： @@ -312,7 +313,7 @@ $$ \lim_{x \to \infty} y = \lim_{x \to \infty} \frac{1 + \frac{2}{x} - \frac{3}{ ## Vol. 4:正项级数的判别法勿滥用 ->[!example] 例3 +>[!example] 例题 >判断级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n} n^2}{3^n}$ 的敛散性（绝对收敛、条件收敛或发散）。 **错误做法示范**：观察级数 $\sum_{n=1}^{\infty} \frac{(-1)^{n} n^2}{3^n}$，若直接对其使用比值判别法： @@ -451,7 +452,9 @@ $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$ - 所以$a_n$ 发散。 ## **Vol.7 分段函数分段点处求导问题** + 分段函数分段点处无论是求导还是判断连续都必须从**左右两边的极限**分别去算，而且计算的时候一定只能**用定义**。 + >[!example] 例题 >设$f(x)=\begin{cases} \frac{2}{3}x ,\ \ x\le1 \\ x^2, \ \ x>1,\end{cases}$则$f(x)$在$x=1$处的\[ \]. >（A）左右导数都存在 （B）左导数存在，右导数不存在 @@ -472,7 +475,7 @@ $y=f^{-1}(x)$，即$x = f(y)$，求$f^{-1'}$就是在求$\frac{dy}{dx}$，而$\f 所以，我们还需要将$y = f^{-1}(x)$代入，即： $\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$ -> [!example] 例1 +> [!example] 例题 > 求$d(\arcsin x)$ 解：设$y=\arcsin x$，即$x=\sin y$，$dx=\cos y\ dy$，即$dy=\frac{dx}{\cos y}$ @@ -482,14 +485,19 @@ $\frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}$ 得$\cos y=\sqrt{1-x^2}$，综上，$dy=\frac{dx}{\sqrt{1-x^2}}$，即$d(\arcsin x)=\frac{dx}{\sqrt{1-x^2}}$ ## Vol. 10: 无界与无穷大的辨析 + 很多人都觉得无界和无穷大是同一个概念，因为它们的实在是太像了：画在坐标系上都是“直指苍穹🚀”或者“飞流直下三千尺”嘛！但是，“无界”准确来说不完全是这样。要准确辨析它们，需要回到它们的**定义**上： + 无穷大的定义：$\forall M > 0, \exists \delta>0$，当$0<|x-x_0|<\delta$时有$|f(x)|>M$，称$f(x)$是当$x\rightarrow x_0$时的无穷大量 + 无界量的定义：由有界的定义($\exists M > 0, \forall x \in D_f,|f(x)| 0$, $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f(x)$是当$x\rightarrow x_0$时的无界量 + **核心区别**：无穷大是**存在某**去心邻域内**任意**$x$都大于$M$，无界是需要对**任意**邻域**存在**一个$x$使得$|f(x)|>M$ + **联系**：无穷大一定是无界量，但是无界量不一定是无穷大。 -> [!example] 例1 +> [!example] 例题 > 无穷震荡$\lim\limits_{x\rightarrow 0}{\frac{1}{x}\sin\frac{1}{x}}$ ![[易错点10-1.png]] -- 2.34.1 From e611c1a03436a04c29b716cee5c0db97ebcdd47b Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 11:32:39 +0800 Subject: [PATCH 014/274] vault backup: 2025-12-26 11:32:39 --- .../子数列问题&考试易错点汇总（解析版）.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 01f067c..0510944 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -463,6 +463,7 @@ $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$ **解：**$f(1)=\frac{2}{3},f'_-(1)=\lim\limits_{x\to1^-}\frac{\frac{2}{3}x-\frac{2}{3}}{x-1}=\frac{2}{3},f'_+(1)=\lim\limits_{x\to1^+}\frac{x^2-\frac{2}{3}}{x-1}=+\infty$,故左导数存在，右导数不存在，选$B$. ## **Vol.8绝对收敛级数** + 绝对收敛的级数满足加法交换律，也就是说，交换各项的顺序不会导致最后结果的改变。但条件收敛的级数是不满足交换律的，改变加法的顺序可能会导致最后结果的改变，甚至可能使原本收敛的级数变成发散级数。这一点了解就行，不会出题目给大家考。 ## Vol. 9: 反函数求导 -- 2.34.1 From 600d4500fac754d08634f8b08932e29508b2146e Mon Sep 17 00:00:00 2001 From: Cym10x Date: Fri, 26 Dec 2025 11:45:56 +0800 Subject: [PATCH 015/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=EF=BC=9A=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0plot=E5=9B=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/讲义/图片/易错点10-2.png | Bin 0 -> 142949 bytes .../子数列问题&考试易错点汇总.md | 22 ++++++++++++++++-- ...&考试易错点汇总（解析版）.md | 3 ++- 3 files changed, 22 insertions(+), 3 deletions(-) create mode 100644 编写小组/讲义/图片/易错点10-2.png diff --git a/编写小组/讲义/图片/易错点10-2.png b/编写小组/讲义/图片/易错点10-2.png new file mode 100644 index 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z4y$E7T_%F;{j`AymOeWCGYYl)Ic#*dc!IBfvZN>}=Tg*cxrq6K%b~D#gO1L+(2-@Y z6l@+XpvKJuwwOHqtY9ssJbZfilwiYPSeCYpt``5KeI&y^Fu#$1V3OhDs?Uxq^*+dR z94*As!g{6(s5}jl=J~M!R2u>hLcuer%2TSJcbyJ(Q>@YoB!1kiUFB`GCP3J;3;s11|$or z#ha6w;s#arlBEZl2#gOuKB7Po!^px(1%HFQ6(-XX z2>n+-X5?NdK9z2iMu~2G&ZGXCGJ=x;dIkve7k~QJ>IhGsXT;d6d-K~2@N3;JV)pg& z>UZI37_zErj^ZUh;p~4CXja#Zvr2*GXY`g_y=gReg*U(R3>cnQ`HVj(JjgO(PY+85Xx~ff5chA+qy|T zjaZUkJND!DHYIXdpxaoy?|yvuJ2~f&btotUVI2TfPzI6aFZJS>jKpYhm`SZ$ zYy{V;EQd+so9%h(;af@$i{m1o5a4=-J<&%lzd7eUP-eHU3^&$a4n-g5QWRFdrGc;K zLJE|ts|=jpLi&^@1KAThzmSj-m*q#T*`$(i5aRdMhgX%kRD!Rc-$B{fq>d3WWB+_duspD^ND^gGzy0e@j?yj18>@>CsGH`oGn{_=!Uq?i6PG=eVm6d5hAyHL00+wTFlAdDZ7bDFu z14VdVkPrjFCb>jh`w}=pD4&;a)B2VTPh0ef<>_++nApyD!1UVK(h`d8aHw={GfpCq oQ_soCL4S!fNy={m6HM5=xM0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f 那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 > [!example] 例2 -> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ +> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos \frac{n\pi}{2}}$ +![[易错点10-2.png]] 这个也是无界但不是无穷大，请自行证明。 -总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” +``` + + + + + + + + + + + + + + + +``` +总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” \ No newline at end of file diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 0510944..87a184a 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -506,6 +506,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f 那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 > [!example] 例2 -> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ +> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos \frac{n\pi}{2}}$ +![[易错点10-2.png]] 总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” -- 2.34.1 From 2361f5cab251caa6f88eef5d2d0c786903ede217 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Fri, 26 Dec 2025 12:49:34 +0800 Subject: [PATCH 016/274] =?UTF-8?q?=E6=8F=92=E5=85=A5=E4=BA=86=E4=B8=80?= =?UTF-8?q?=E5=BC=A0=E5=9B=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 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zdTeFe05|a1CD4@@+0t_O`z|+AImd|hzc&HW+KZ=X9&6WMS5O^Q>Zk3K@{%Gt(wFAR zV>f&KEbQ)4<0X%IQ(<8a+AGV)ac5P7!Xt(p2c|m*z)I{`Vo_V|zUHSB&V1#a13A{S zHb~aH${2d_zr2!}pydNDTLD1a6V<sR{XgT~6O4rYCnKP<$E%KSGO2HXV*ldoUKBb( zHwh$F|C6-qFJOX_spg-><aFvBov?e(UUVivt{PQ8-wb$lSP6RKL`_d!x!%e!!{uA> z#bb0qUxLzfW420qVDm<{95KN+{0g*BYUrMp+{%2m4B&{?k)vDnu|GY@cX2lvwBxNo z2Qo9k2HVYk_f-7@5Joa}!3HTT9vF91FHmz@i$3fFO3NW~{_6lVO-i3GaCTLk78!;} zsh!h8GN-Bhq+3avUNcH|lWhrMvJfepeGq<4qH&U6XpKhAhTF;hn}JJu7;#MXdBCMh nm(I_XoC*qo4_7ai3E?HeJSw>UHE<g6H<y$Z)a6TLO@seG)n8iK literal 0 HcmV?d00001 diff --git a/编写小组/讲义/子数列问题&考试易错点汇总.md b/编写小组/讲义/子数列问题&考试易错点汇总.md index 338c02d..2c4f0c4 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总.md @@ -465,6 +465,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f > [!example] 例2 > 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ +![[xcospix.png]] 总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 2517960..47a94f7 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -479,5 +479,6 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f > [!example] 例2 > 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ +![[xcospix.png]] 总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” -- 2.34.1 From f4df976cac966f6654e026121b4ebb8d581da0ed Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 26 Dec 2025 12:56:23 +0800 Subject: [PATCH 017/274] =?UTF-8?q?=E4=BF=AE=E6=94=B910-2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../子数列问题&考试易错点汇总（解析版）.md | 5 ----- 1 file changed, 5 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 73fe855..6c00582 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -506,12 +506,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f 那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 > [!example] 例2 -<<<<<<< HEAD -> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ -![[xcospix.png]] -======= > 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos \frac{n\pi}{2}}$ ->>>>>>> origin/develop ![[易错点10-2.png]] 总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” -- 2.34.1 From 9968f26b686736f87f4e165b88ac6ef08b5455e0 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 26 Dec 2025 12:59:10 +0800 Subject: [PATCH 018/274] =?UTF-8?q?=E4=BF=AE=E6=94=B910-2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/讲义/子数列问题&考试易错点汇总.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总.md b/编写小组/讲义/子数列问题&考试易错点汇总.md index 91da17c..e6d099a 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总.md @@ -465,8 +465,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f 那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 > [!example] 例2 -> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos n\pi}$ -![[xcospix.png]] +> 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos \frac{n\pi}{2}}$ -- 2.34.1 From 8b89078d380f69e3a079897e3bdf16130b895411 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 17:37:30 +0800 Subject: [PATCH 019/274] vault backup: 2025-12-26 17:37:30 --- .../子数列问题&考试易错点汇总（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 6c00582..65683d7 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -462,7 +462,7 @@ $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$ **解：**$f(1)=\frac{2}{3},f'_-(1)=\lim\limits_{x\to1^-}\frac{\frac{2}{3}x-\frac{2}{3}}{x-1}=\frac{2}{3},f'_+(1)=\lim\limits_{x\to1^+}\frac{x^2-\frac{2}{3}}{x-1}=+\infty$,故左导数存在，右导数不存在，选$B$. -## **Vol.8绝对收敛级数** +## **Vol.8: 绝对收敛级数** 绝对收敛的级数满足加法交换律，也就是说，交换各项的顺序不会导致最后结果的改变。但条件收敛的级数是不满足交换律的，改变加法的顺序可能会导致最后结果的改变，甚至可能使原本收敛的级数变成发散级数。这一点了解就行，不会出题目给大家考。 -- 2.34.1 From fcdd5e65aea646e658b18d0226f66c7254a692c3 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 26 Dec 2025 19:03:25 +0800 Subject: [PATCH 020/274] =?UTF-8?q?=E8=A1=A5=E5=85=85=E8=AF=A6=E7=BB=86?= =?UTF-8?q?=E8=AF=81=E6=98=8E?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...题&考试易错点汇总（解析版）.md | 17 ++++++++++++++--- 1 file changed, 14 insertions(+), 3 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 65683d7..cae636b 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -504,9 +504,20 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f ![[易错点10-1.png]] 这个并不是无穷大——不管取的邻域有多小，我总能找到一个令$\sin\frac{1}{x}=0$的$x$，此时$\frac{1}{x}\sin\frac{1}{x}=0$。 那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 +详细的证明过程如下： + +>对$f(x)={\frac{1}{x}\sin\frac{1}{x}}$，可以取数列$a_n=\frac{1}{2n\pi+\frac{\pi}{2}}$, +>$f(a_n)=\frac{1}{a_n} \sin\frac{1}{a_n}=2n\pi+\frac{\pi}{2}$，$\lim\limits_{n\rightarrow\infty}{f(a_n)}=+\infty$故极限无界； +>另取数列$b_n=\frac{1}{n\pi}$，$f(b_n)=0$。 +>按照无穷大的定义：$\forall M > 0, \exists \delta>0$，当$0<|x-x_0|<\delta$时有$|f(x)|>M$，称$f(x)$是当$x\rightarrow x_0$时的无穷大量。 +>然而，$\forall\delta>0$，$n>\frac{1}{\delta\pi}, x=b_n$时，$0<|x|<\delta$，但是$f(x)=0$，与无穷大定义不符。 +>故：$\lim\limits_{x\rightarrow 0}{\frac{1}{x}\sin\frac{1}{x}}$无界，但不是无穷大 > [!example] 例2 > 双子数列$\lim\limits_{n\rightarrow \infty}{n\cos \frac{n\pi}{2}}$ - -![[易错点10-2.png]] -总结：无穷大是在邻域内“一直都很大”，无界是邻域内“有很大的” +>如图： +>![[易错点10-2.png]] +>与上一题思路类似，令$x_n={n\cos \frac{n\pi}{2}}$，取子数列$a_n=x_{2n-1},b_n=x_{4n}$， +>$a_n=0$，$\lim\limits_{n\rightarrow \infty}{b_n}=+\infty$，故$x_n$无界； +>按照（数列）无穷大的定义：$\forall M > 0, \exists N>0$，当$n>N$时有$x_n>M$，称$x_n$趋近无穷大 +>然而，$\forall N>0$，取$n=4N$，$x_n=b_N=0$，与无穷大定义不符。故$x_n$无界但不是无穷大。 \ No newline at end of file -- 2.34.1 From 2b1148a0f132d84a46169b0196e9dd5afaf5ab2c Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 26 Dec 2025 19:11:45 +0800 Subject: [PATCH 021/274] =?UTF-8?q?=E5=88=9D=E5=A7=8B=E5=8C=96=E8=AF=95?= =?UTF-8?q?=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 1 + 1 file changed, 1 insertion(+) create mode 100644 编写小组/试卷/期中考前押题卷.md diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md new file mode 100644 index 0000000..05be6fe --- /dev/null +++ b/编写小组/试卷/期中考前押题卷.md @@ -0,0 +1 @@ +时量：60分钟 满分：____ \ No newline at end of file -- 2.34.1 From bbb181e6f9d4d2538ca94a16a335be86d3d6fab3 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 19:19:26 +0800 Subject: [PATCH 022/274] vault backup: 2025-12-26 19:19:26 --- .../子数列问题&考试易错点汇总（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index cae636b..28d7cce 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -464,7 +464,7 @@ $$L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$$ ## **Vol.8: 绝对收敛级数** -绝对收敛的级数满足加法交换律，也就是说，交换各项的顺序不会导致最后结果的改变。但条件收敛的级数是不满足交换律的，改变加法的顺序可能会导致最后结果的改变，甚至可能使原本收敛的级数变成发散级数。这一点了解就行，不会出题目给大家考。 +绝对收敛的级数满足加法交换律，也就是说，交换各项的顺序不会导致最后结果的改变。但条件收敛的级数是不满足交换律的，改变加法的顺序可能会导致最后结果的改变，甚至可能使原本收敛的级数变成发散级数。这一点了解就行，基本不会出题目给大家考。 ## Vol. 9: 反函数求导 易错：变量混淆。反函数的导数 = 原函数导数的倒数，但**自变量和因变量角色互换**。 -- 2.34.1 From 18ae8257050a3bc0a90bd4a25ed327f88ec31f94 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 26 Dec 2025 21:08:47 +0800 Subject: [PATCH 023/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 270 ++++++++++++++++++- 1 file changed, 269 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index 05be6fe..f821012 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -1 +1,269 @@ -时量：60分钟 满分：____ \ No newline at end of file +时量：60分钟 ____ +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + + + + + +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 +（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ + +**解析：** +由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 + +$$ +\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} = \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} = \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \frac{1}{2} f'(1). +$$ + +已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 +由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 +因此，曲线在点 $(5,f(5))$ 处的切线斜率为 $-2$，选项（D）正确。 + +**答案：** (D) + +--- + + + +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 + +(A) 可去间断点 +(B) 跳跃间断点 +(C) 无穷间断点 +(D) 振荡间断点 + +**解析：** +分析函数在 $x=0$ 处的左右极限。 + +1. **第一部分**：$\dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$ + + - 当 $x \to 0^+$ 时，$\frac{1}{x} \to +\infty$，$e^{\frac{1}{x}} \to +\infty$，所以 + + $$ +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} \rightarrow \frac{0 + 1}{0 + 1} = 1. +$$ + - 当 $x \to 0^-$ 时，$\frac{1}{x} \to -\infty$，$e^{\frac{1}{x}} \to 0$，所以 + + $$ +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2. +$$ +1. **第二部分**：$\dfrac{\sin x}{|x|}$ + + - 当 $x \to 0^+$ 时，$|x| = x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{x} \to 1$。 + + - 当 $x \to 0^-$ 时，$|x| = -x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{-x} \to -1$。 + +2. **整体极限**： + + - 右极限：$\lim\limits_{x \to 0^+} f(x) = 1 + 1 = 2$。 + + - 左极限：$\lim\limits_{x \to 0^-} f(x) = 2 + (-1) = 1$。 + + +由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 + +**答案：** (B) +--- +3.（多选）下列级数中收敛的有______。 + +A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + +B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ + +C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ + +D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + +E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + +F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + +**解析：** + +- **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 + +- **B**：由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 + +- **C**：由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 + +- **D**：由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **E**：由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **F**：由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 + + +**答案：** ABCF + + + +4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$。 + +**解析：** + +1. 先求 $x=1$ 时的 $y$ 值：代入方程：$1^y + 2 \times 1^2 - y = 1 \implies 1 + 2 - y = 1 \implies y = 2$。 + +2. 隐函数求导：方程两边对 $x$ 求导，注意 $x^y = e^{y \ln x}$： + $\frac{d}{dx}(x^y) + 4x - y' = 0$， + 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 + 代入 $x=1, y=2$： + $1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6$。 + 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 + + +**答案：** $dy|_{x=1} = 6dx$ + + + +5.求 $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 + +**解析：** +用夹逼准则： + +- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{1}{2}$； + +- 上界：$S_n \leq \sum_{k=1}^n \frac{k}{n^2 - n} = \frac{n(n+1)}{2(n^2-n)} \to \frac{1}{2}$（$n \to \infty$）。 + 故极限为 $\frac{1}{2}$。 + + +**答案：** $\frac{1}{2}$ + + +6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 + +**解析：** +这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： +$\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 +利用等价无穷小和泰勒展开： + +- $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； + +- $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 + 因此： + $\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2)$。 + 取对数： + $\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1)$。 + 所以原极限为 $e^{7/2}$。 + + +**答案：** $e^{7/2}$ + +--- + + + + +7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 + +**解析：** +渐近线分垂直、水平、斜渐近线分析： + +1. **垂直渐近线**： + 分母 $2-x=0 \implies x=2$，计算 $\lim_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 + +2. **水平渐近线**（$x \to -\infty$）： + 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， + 所以 $\lim_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 + +3. **斜渐近线**（$x \to +\infty$）： + + - 斜率 $k = \lim_{x \to +\infty} \frac{y}{x} = \lim_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 + 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 + + - 截距 $b = \lim_{x \to +\infty} (y - x) = \lim_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 + 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 + 故斜渐近线为 $y = x - \frac{\pi}{2}$。 + + +**答案：** +渐近线为：$x=2$，$y=\frac{\pi}{2}$，$y=x-\frac{\pi}{2}$。 + +--- + + +8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ +证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + +**解析：** +**第一步：证明数列有下界。** +由算术-几何平均不等式，对任意正数 $x$，有 +由算术-几何平均不等式，对任意正数 $x$，有 +$$ +\frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. +$$​. +因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 + +**第二步：证明数列单调性。** +考虑差值： +考虑数列的递推式：$a_{n+1} = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right)$，则 +$$ +a_{n+1} - a_n = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right) - a_n = \frac{1}{2} \left( \frac{\sigma}{a_n} - a_n \right) = \frac{\sigma - a_n^2}{2a_n}. +$$ + +由于 $a_n > 0$，差值的符号由 $\sigma - a_n^2$ 决定。 + +- 若 $a_n > \sqrt{\sigma}$，则 $a_{n+1} - a_n < 0$，数列单调递减； + +- 若 $a_n < \sqrt{\sigma}$，则 $a_{n+1} - a_n > 0$，数列单调递增； + +- 若 $a_n = \sqrt{\sigma}$，则 $a_{n+1} = a_n$，数列为常数列。 + + +易证 $a_1 \geq \sqrt{\sigma}$，等号仅当 $a = \sqrt{\sigma}$ 时成立。 +若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，结论成立。 +若 $a > \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$，由归纳法所有 $a_n > \sqrt{\sigma}$，且数列单调递减。 +若 $0 < a < \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$（因为 $a_1 = \frac{1}{2}(a + \sigma/a) \geq \sqrt{\sigma}$ 且等号不成立），此时从 $n=1$ 起 $a_n > \sqrt{\sigma}$，且 $a_2 < a_1$（因 $a_1 > \sqrt{\sigma}$），之后单调递减。 + +因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ 或更大），故数列收敛。 + +**第三步：求极限。** +设 $\lim\limits_{n \to \infty} a_n = L$，则 $L \geq \sqrt{\sigma} > 0$。在递推式两边取极限： +$$ +L = \frac{1}{2} \left( L + \frac{\sigma}{L} \right). +$$ +整理得 $2L = L + \frac{\sigma}{L}$，即 $L = \frac{\sigma}{L}$，从而 $L^2 = \sigma$，故 $L = \sqrt{\sigma}$（正根）。 + +因此，数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + +--- + + + +9.飞机离地面 2 km，以 200 km/h 水平飞行，求飞至目标正上方时摄影机的角速率。 + +**解析：** +几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 + +求导：$\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{dx}{dt}$。 + +代入 $x=0$（正上方）：$\theta = 0 \implies \cos \theta = 1$，$\frac{dx}{dt} = -200$ km/h，得 +$$ +\frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. +$$ +角速率为 $100$ rad/h（速率取绝对值）。 + +**答案：** $100$ rad/h。 + +以上即为整理后的Obsidian笔记，可以直接复制到Obsidia + +--- + + +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 +x1≤x2≤⋯≤xn +证明：存在 $\xi \in [x_1,x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ + +**解析：** +由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 + +对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 +m≤$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$≤M. +由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ +证毕。 + +--- + -- 2.34.1 From 154488ad27122699ff65e45afbb6303695a3db6e Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Fri, 26 Dec 2025 21:20:41 +0800 Subject: [PATCH 024/274] vault backup: 2025-12-26 21:20:41 --- ...数列问题&考试易错点汇总（解析版）.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 28d7cce..cfd242b 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -342,6 +342,7 @@ $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{3} < 1 $$ 根据定义，若一个级数的绝对值级数收敛，则该级数**绝对收敛**。绝对收敛的级数必然收敛。 ## Vol. 5:误用p级数 + **机械地套用p级数结论，而忽视了其应用前提：指数 `p` 必须是与 `n` 无关的常数。** > [!example] 例题1： @@ -352,7 +353,7 @@ $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{3} < 1 $$ 3. 因为 `1/n > 0`，所以 `p = 1 + 1/n > 1` 恒成立，误判为收敛 #### ✅ 正确分析与解法 **错误原因**：`pₙ = 1 + 1/n` 不是常数，其极限为1。 -使用比值审敛法与调和级数 `∑ 1/n` 比较： +使用比较判别法与调和级数 `∑ 1/n` 比较： $$\lim_{n \to \infty} \frac{ \frac{1}{n^{1 + \frac{1}{n}}} }{ \frac{1}{n} } = \lim_{n \to \infty} \frac{1}{n^{1/n}} = 1$$ 可知两级数敛散性相同，且调和级数发散 ⇒ 原级数**发散**。 @@ -376,6 +377,7 @@ $$ ## Vol. 6: 条件收敛、绝对收敛、发散 + **仅当利用比值/根值判别法判断出$\sum |a_n|$发散时$\Rightarrow$$\sum a_n$发散 #### 基本定义 @@ -385,7 +387,9 @@ $$ 3. **条件收敛**：如果 $\sum a_n$ 收敛但 $\sum |a_n|$ 发散，则称 $\sum a_n$ **条件收敛** #### 正确分析： + 仅有$\sum |a_n|$ 收敛 ⇒ $\sum a_n$收敛 + **证明**： - 使用柯西收敛准则。对于任意 $\varepsilon > 0$： 因为 $\sum |a_n|$ 收敛，由柯西准则，存在 $N$，使得当 $m > n \geq N$ 时： @@ -503,7 +507,7 @@ $\forall \delta>0, \exists x \in \mathring{U}(x_0,\delta)$有 $|f(x)|>M$，称$f ![[易错点10-1.png]] 这个并不是无穷大——不管取的邻域有多小，我总能找到一个令$\sin\frac{1}{x}=0$的$x$，此时$\frac{1}{x}\sin\frac{1}{x}=0$。 -那这个是有界的吗？也不是。这就是典型的**不是无界量的无穷大**。 +那这个是有界的吗？也不是。这就是典型的**不是无穷大的无界量**。 详细的证明过程如下： >对$f(x)={\frac{1}{x}\sin\frac{1}{x}}$，可以取数列$a_n=\frac{1}{2n\pi+\frac{\pi}{2}}$, -- 2.34.1 From 70874109e828f798eceb570b20ef7e65d7b180be Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 26 Dec 2025 21:23:54 +0800 Subject: [PATCH 025/274] =?UTF-8?q?=E8=AF=95=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 期中考前押题卷.md | 68 ++++++++++++++++++++++++++++++++++++++++ 1 file changed, 68 insertions(+) create mode 100644 期中考前押题卷.md diff --git a/期中考前押题卷.md b/期中考前押题卷.md new file mode 100644 index 0000000..bc83efb --- /dev/null +++ b/期中考前押题卷.md @@ -0,0 +1,68 @@ +时量：60分钟 ____ +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + + + + + +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为（ ）。 +（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ + + + +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 + +(A) 可去间断点 +(B) 跳跃间断点 +(C) 无穷间断点 +(D) 振荡间断点 + + + +3.（多选）下列级数中收敛的有______。 + +A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + +B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ + +C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ + +D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + +E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + +F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + + + +4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$______。 + + + +5. $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$=______。 + + + +6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$______。 + + + +7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 + + +8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ +证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + + + + +9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 + + + + +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 +x1≤x2≤⋯≤xn +证明：存在 $\xi \in [x_1,x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ \ No newline at end of file -- 2.34.1 From d83db0eb37390fc1bcd436afd3052c1f5c59dd90 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 26 Dec 2025 21:24:42 +0800 Subject: [PATCH 026/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E9=83=A8?= =?UTF-8?q?=E5=88=86=E6=A0=BC=E5=BC=8F=EF=BC=8C=E5=B9=B6=E8=A1=A5=E5=85=85?= =?UTF-8?q?=E5=AE=8C=E6=95=B4=E4=BA=86=E4=B8=80=E9=81=93=E9=A2=98=E7=9B=AE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 56 ++++++++++---------- 1 file changed, 28 insertions(+), 28 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index f821012..054d216 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -1,6 +1,6 @@ 时量：60分钟 ____ -**内部资料，禁止传播** -**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** +内部资料，禁止传播 +编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 @@ -12,9 +12,9 @@ **解析：** 由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 -$$ -\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} = \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} = \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \frac{1}{2} f'(1). -$$ +$$\begin{aligned} +\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} &= \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} \\ &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} \\ &= \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\ &= \frac{1}{2} f'(1). +\end{aligned}$$ 已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 @@ -64,6 +64,7 @@ $$ 由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 **答案：** (B) + --- 3.（多选）下列级数中收敛的有______。 @@ -83,15 +84,15 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ - **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 -- **B**：由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 +- **B**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim\limits_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 -- **C**：由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 +- **C**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 -- **D**：由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 +- **D**：由于 $\lim\limits_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 -- **E**：由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 +- **E**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim\limits_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 -- **F**：由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 +- **F**：由于 $\lim\limits_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim\limits_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 **答案：** ABCF @@ -108,7 +109,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ $\frac{d}{dx}(x^y) + 4x - y' = 0$， 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 代入 $x=1, y=2$： - $1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6$。 + $$1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6.$$ 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 @@ -116,7 +117,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ -5.求 $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 +5.求 $\lim\limits_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 **解析：** 用夹逼准则： @@ -130,20 +131,20 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ **答案：** $\frac{1}{2}$ -6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 +6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 **解析：** 这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： -$\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 +$$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ 利用等价无穷小和泰勒展开： - $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； - $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 因此： - $\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2)$。 + $$\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2).$$ 取对数： - $\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1)$。 + $$\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1).$$ 所以原极限为 $e^{7/2}$。 @@ -164,15 +165,15 @@ $\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 2. **水平渐近线**（$x \to -\infty$）： 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， - 所以 $\lim_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 + 所以 $\lim\limits_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 3. **斜渐近线**（$x \to +\infty$）： - - 斜率 $k = \lim_{x \to +\infty} \frac{y}{x} = \lim_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 + - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - - 截距 $b = \lim_{x \to +\infty} (y - x) = \lim_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 + - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 故斜渐近线为 $y = x - \frac{\pi}{2}$。 @@ -188,11 +189,12 @@ $\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 **解析：** **第一步：证明数列有下界。** -由算术-几何平均不等式，对任意正数 $x$，有 由算术-几何平均不等式，对任意正数 $x$，有 + $$ \frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. -$$​. +$$ + 因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 **第二步：证明数列单调性。** @@ -231,7 +233,7 @@ $$ -9.飞机离地面 2 km，以 200 km/h 水平飞行，求飞至目标正上方时摄影机的角速率。 +9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 **解析：** 几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 @@ -242,17 +244,15 @@ $$ $$ \frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. $$ -角速率为 $100$ rad/h（速率取绝对值）。 +角速率为 $100$ $rad/h$（速率取绝对值）。 -**答案：** $100$ rad/h。 +**答案：** $100$ $rad/h$。 -以上即为整理后的Obsidian笔记，可以直接复制到Obsidia --- -10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 -x1≤x2≤⋯≤xn +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 $$x_1≤x_2≤⋯≤x_n$$ 证明：存在 $\xi \in [x_1,x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ @@ -260,7 +260,7 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -m≤$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$≤M. +$m≤$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ 由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 证毕。 -- 2.34.1 From 33a0325cd970a34704160ae3897e9b7876f07289 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 26 Dec 2025 21:25:29 +0800 Subject: [PATCH 027/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/期中考前押题卷 1.md | 0 .../试卷/期中考前押题卷解析版.md | 269 ++++++++++++++++++ 2 files changed, 269 insertions(+) rename 期中考前押题卷.md => 编写小组/试卷/期中考前押题卷 1.md (100%) create mode 100644 编写小组/试卷/期中考前押题卷解析版.md diff --git a/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷 1.md similarity index 100% rename from 期中考前押题卷.md rename to 编写小组/试卷/期中考前押题卷 1.md diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md new file mode 100644 index 0000000..f821012 --- /dev/null +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -0,0 +1,269 @@ +时量：60分钟 ____ +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + + + + + +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 +（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ + +**解析：** +由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 + +$$ +\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} = \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} = \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \frac{1}{2} f'(1). +$$ + +已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 +由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 +因此，曲线在点 $(5,f(5))$ 处的切线斜率为 $-2$，选项（D）正确。 + +**答案：** (D) + +--- + + + +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 + +(A) 可去间断点 +(B) 跳跃间断点 +(C) 无穷间断点 +(D) 振荡间断点 + +**解析：** +分析函数在 $x=0$ 处的左右极限。 + +1. **第一部分**：$\dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$ + + - 当 $x \to 0^+$ 时，$\frac{1}{x} \to +\infty$，$e^{\frac{1}{x}} \to +\infty$，所以 + + $$ +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} \rightarrow \frac{0 + 1}{0 + 1} = 1. +$$ + - 当 $x \to 0^-$ 时，$\frac{1}{x} \to -\infty$，$e^{\frac{1}{x}} \to 0$，所以 + + $$ +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2. +$$ +1. **第二部分**：$\dfrac{\sin x}{|x|}$ + + - 当 $x \to 0^+$ 时，$|x| = x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{x} \to 1$。 + + - 当 $x \to 0^-$ 时，$|x| = -x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{-x} \to -1$。 + +2. **整体极限**： + + - 右极限：$\lim\limits_{x \to 0^+} f(x) = 1 + 1 = 2$。 + + - 左极限：$\lim\limits_{x \to 0^-} f(x) = 2 + (-1) = 1$。 + + +由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 + +**答案：** (B) +--- +3.（多选）下列级数中收敛的有______。 + +A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + +B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ + +C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ + +D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + +E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + +F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + +**解析：** + +- **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 + +- **B**：由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 + +- **C**：由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 + +- **D**：由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **E**：由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **F**：由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 + + +**答案：** ABCF + + + +4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$。 + +**解析：** + +1. 先求 $x=1$ 时的 $y$ 值：代入方程：$1^y + 2 \times 1^2 - y = 1 \implies 1 + 2 - y = 1 \implies y = 2$。 + +2. 隐函数求导：方程两边对 $x$ 求导，注意 $x^y = e^{y \ln x}$： + $\frac{d}{dx}(x^y) + 4x - y' = 0$， + 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 + 代入 $x=1, y=2$： + $1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6$。 + 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 + + +**答案：** $dy|_{x=1} = 6dx$ + + + +5.求 $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 + +**解析：** +用夹逼准则： + +- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{1}{2}$； + +- 上界：$S_n \leq \sum_{k=1}^n \frac{k}{n^2 - n} = \frac{n(n+1)}{2(n^2-n)} \to \frac{1}{2}$（$n \to \infty$）。 + 故极限为 $\frac{1}{2}$。 + + +**答案：** $\frac{1}{2}$ + + +6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 + +**解析：** +这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： +$\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 +利用等价无穷小和泰勒展开： + +- $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； + +- $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 + 因此： + $\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2)$。 + 取对数： + $\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1)$。 + 所以原极限为 $e^{7/2}$。 + + +**答案：** $e^{7/2}$ + +--- + + + + +7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 + +**解析：** +渐近线分垂直、水平、斜渐近线分析： + +1. **垂直渐近线**： + 分母 $2-x=0 \implies x=2$，计算 $\lim_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 + +2. **水平渐近线**（$x \to -\infty$）： + 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， + 所以 $\lim_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 + +3. **斜渐近线**（$x \to +\infty$）： + + - 斜率 $k = \lim_{x \to +\infty} \frac{y}{x} = \lim_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 + 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 + + - 截距 $b = \lim_{x \to +\infty} (y - x) = \lim_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 + 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 + 故斜渐近线为 $y = x - \frac{\pi}{2}$。 + + +**答案：** +渐近线为：$x=2$，$y=\frac{\pi}{2}$，$y=x-\frac{\pi}{2}$。 + +--- + + +8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ +证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + +**解析：** +**第一步：证明数列有下界。** +由算术-几何平均不等式，对任意正数 $x$，有 +由算术-几何平均不等式，对任意正数 $x$，有 +$$ +\frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. +$$​. +因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 + +**第二步：证明数列单调性。** +考虑差值： +考虑数列的递推式：$a_{n+1} = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right)$，则 +$$ +a_{n+1} - a_n = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right) - a_n = \frac{1}{2} \left( \frac{\sigma}{a_n} - a_n \right) = \frac{\sigma - a_n^2}{2a_n}. +$$ + +由于 $a_n > 0$，差值的符号由 $\sigma - a_n^2$ 决定。 + +- 若 $a_n > \sqrt{\sigma}$，则 $a_{n+1} - a_n < 0$，数列单调递减； + +- 若 $a_n < \sqrt{\sigma}$，则 $a_{n+1} - a_n > 0$，数列单调递增； + +- 若 $a_n = \sqrt{\sigma}$，则 $a_{n+1} = a_n$，数列为常数列。 + + +易证 $a_1 \geq \sqrt{\sigma}$，等号仅当 $a = \sqrt{\sigma}$ 时成立。 +若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，结论成立。 +若 $a > \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$，由归纳法所有 $a_n > \sqrt{\sigma}$，且数列单调递减。 +若 $0 < a < \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$（因为 $a_1 = \frac{1}{2}(a + \sigma/a) \geq \sqrt{\sigma}$ 且等号不成立），此时从 $n=1$ 起 $a_n > \sqrt{\sigma}$，且 $a_2 < a_1$（因 $a_1 > \sqrt{\sigma}$），之后单调递减。 + +因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ 或更大），故数列收敛。 + +**第三步：求极限。** +设 $\lim\limits_{n \to \infty} a_n = L$，则 $L \geq \sqrt{\sigma} > 0$。在递推式两边取极限： +$$ +L = \frac{1}{2} \left( L + \frac{\sigma}{L} \right). +$$ +整理得 $2L = L + \frac{\sigma}{L}$，即 $L = \frac{\sigma}{L}$，从而 $L^2 = \sigma$，故 $L = \sqrt{\sigma}$（正根）。 + +因此，数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + +--- + + + +9.飞机离地面 2 km，以 200 km/h 水平飞行，求飞至目标正上方时摄影机的角速率。 + +**解析：** +几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 + +求导：$\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{dx}{dt}$。 + +代入 $x=0$（正上方）：$\theta = 0 \implies \cos \theta = 1$，$\frac{dx}{dt} = -200$ km/h，得 +$$ +\frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. +$$ +角速率为 $100$ rad/h（速率取绝对值）。 + +**答案：** $100$ rad/h。 + +以上即为整理后的Obsidian笔记，可以直接复制到Obsidia + +--- + + +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 +x1≤x2≤⋯≤xn +证明：存在 $\xi \in [x_1,x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ + +**解析：** +由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 + +对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 +m≤$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$≤M. +由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ +证毕。 + +--- + -- 2.34.1 From b57b731d7a17f6b8179199af25b6008cf7767b69 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 26 Dec 2025 21:26:50 +0800 Subject: [PATCH 028/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 269 ------------------ .../试卷/期中考前押题卷解析版.md | 56 ++-- 2 files changed, 28 insertions(+), 297 deletions(-) delete mode 100644 编写小组/试卷/期中考前押题卷.md diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md deleted file mode 100644 index 054d216..0000000 --- a/编写小组/试卷/期中考前押题卷.md +++ /dev/null @@ -1,269 +0,0 @@ -时量：60分钟 ____ -内部资料，禁止传播 -编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 - - - - - -1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 -（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ - -**解析：** -由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 - -$$\begin{aligned} -\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} &= \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} \\ &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} \\ &= \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\ &= \frac{1}{2} f'(1). -\end{aligned}$$ - -已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 -由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 -因此，曲线在点 $(5,f(5))$ 处的切线斜率为 $-2$，选项（D）正确。 - -**答案：** (D) - ---- - - - -2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 - -(A) 可去间断点 -(B) 跳跃间断点 -(C) 无穷间断点 -(D) 振荡间断点 - -**解析：** -分析函数在 $x=0$ 处的左右极限。 - -1. **第一部分**：$\dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$ - - - 当 $x \to 0^+$ 时，$\frac{1}{x} \to +\infty$，$e^{\frac{1}{x}} \to +\infty$，所以 - - $$ -\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} \rightarrow \frac{0 + 1}{0 + 1} = 1. -$$ - - 当 $x \to 0^-$ 时，$\frac{1}{x} \to -\infty$，$e^{\frac{1}{x}} \to 0$，所以 - - $$ -\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2. -$$ -1. **第二部分**：$\dfrac{\sin x}{|x|}$ - - - 当 $x \to 0^+$ 时，$|x| = x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{x} \to 1$。 - - - 当 $x \to 0^-$ 时，$|x| = -x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{-x} \to -1$。 - -2. **整体极限**： - - - 右极限：$\lim\limits_{x \to 0^+} f(x) = 1 + 1 = 2$。 - - - 左极限：$\lim\limits_{x \to 0^-} f(x) = 2 + (-1) = 1$。 - - -由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 - -**答案：** (B) - ---- -3.（多选）下列级数中收敛的有______。 - -A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ - -B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ - -C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ - -D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ - -E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ - -F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ - -**解析：** - -- **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 - -- **B**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim\limits_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 - -- **C**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 - -- **D**：由于 $\lim\limits_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 - -- **E**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim\limits_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 - -- **F**：由于 $\lim\limits_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim\limits_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 - - -**答案：** ABCF - - - -4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$。 - -**解析：** - -1. 先求 $x=1$ 时的 $y$ 值：代入方程：$1^y + 2 \times 1^2 - y = 1 \implies 1 + 2 - y = 1 \implies y = 2$。 - -2. 隐函数求导：方程两边对 $x$ 求导，注意 $x^y = e^{y \ln x}$： - $\frac{d}{dx}(x^y) + 4x - y' = 0$， - 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 - 代入 $x=1, y=2$： - $$1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6.$$ - 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 - - -**答案：** $dy|_{x=1} = 6dx$ - - - -5.求 $\lim\limits_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 - -**解析：** -用夹逼准则： - -- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{1}{2}$； - -- 上界：$S_n \leq \sum_{k=1}^n \frac{k}{n^2 - n} = \frac{n(n+1)}{2(n^2-n)} \to \frac{1}{2}$（$n \to \infty$）。 - 故极限为 $\frac{1}{2}$。 - - -**答案：** $\frac{1}{2}$ - - -6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 - -**解析：** -这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： -$$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ -利用等价无穷小和泰勒展开： - -- $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； - -- $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 - 因此： - $$\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2).$$ - 取对数： - $$\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1).$$ - 所以原极限为 $e^{7/2}$。 - - -**答案：** $e^{7/2}$ - ---- - - - - -7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 - -**解析：** -渐近线分垂直、水平、斜渐近线分析： - -1. **垂直渐近线**： - 分母 $2-x=0 \implies x=2$，计算 $\lim_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 - -2. **水平渐近线**（$x \to -\infty$）： - 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， - 所以 $\lim\limits_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 - -3. **斜渐近线**（$x \to +\infty$）： - - - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 - 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； - 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - - - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 - 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 - 故斜渐近线为 $y = x - \frac{\pi}{2}$。 - - -**答案：** -渐近线为：$x=2$，$y=\frac{\pi}{2}$，$y=x-\frac{\pi}{2}$。 - ---- - - -8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ -证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 - -**解析：** -**第一步：证明数列有下界。** -由算术-几何平均不等式，对任意正数 $x$，有 - -$$ -\frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. -$$ - -因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 - -**第二步：证明数列单调性。** -考虑差值： -考虑数列的递推式：$a_{n+1} = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right)$，则 -$$ -a_{n+1} - a_n = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right) - a_n = \frac{1}{2} \left( \frac{\sigma}{a_n} - a_n \right) = \frac{\sigma - a_n^2}{2a_n}. -$$ - -由于 $a_n > 0$，差值的符号由 $\sigma - a_n^2$ 决定。 - -- 若 $a_n > \sqrt{\sigma}$，则 $a_{n+1} - a_n < 0$，数列单调递减； - -- 若 $a_n < \sqrt{\sigma}$，则 $a_{n+1} - a_n > 0$，数列单调递增； - -- 若 $a_n = \sqrt{\sigma}$，则 $a_{n+1} = a_n$，数列为常数列。 - - -易证 $a_1 \geq \sqrt{\sigma}$，等号仅当 $a = \sqrt{\sigma}$ 时成立。 -若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，结论成立。 -若 $a > \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$，由归纳法所有 $a_n > \sqrt{\sigma}$，且数列单调递减。 -若 $0 < a < \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$（因为 $a_1 = \frac{1}{2}(a + \sigma/a) \geq \sqrt{\sigma}$ 且等号不成立），此时从 $n=1$ 起 $a_n > \sqrt{\sigma}$，且 $a_2 < a_1$（因 $a_1 > \sqrt{\sigma}$），之后单调递减。 - -因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ 或更大），故数列收敛。 - -**第三步：求极限。** -设 $\lim\limits_{n \to \infty} a_n = L$，则 $L \geq \sqrt{\sigma} > 0$。在递推式两边取极限： -$$ -L = \frac{1}{2} \left( L + \frac{\sigma}{L} \right). -$$ -整理得 $2L = L + \frac{\sigma}{L}$，即 $L = \frac{\sigma}{L}$，从而 $L^2 = \sigma$，故 $L = \sqrt{\sigma}$（正根）。 - -因此，数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 - ---- - - - -9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 - -**解析：** -几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 - -求导：$\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{dx}{dt}$。 - -代入 $x=0$（正上方）：$\theta = 0 \implies \cos \theta = 1$，$\frac{dx}{dt} = -200$ km/h，得 -$$ -\frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. -$$ -角速率为 $100$ $rad/h$（速率取绝对值）。 - -**答案：** $100$ $rad/h$。 - - ---- - - -10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 $$x_1≤x_2≤⋯≤x_n$$ -证明：存在 $\xi \in [x_1,x_n]$，使得 -$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ - -**解析：** -由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 - -对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -$m≤$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ -由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 -$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ -证毕。 - ---- - diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index f821012..054d216 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -1,6 +1,6 @@ 时量：60分钟 ____ -**内部资料，禁止传播** -**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** +内部资料，禁止传播 +编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 @@ -12,9 +12,9 @@ **解析：** 由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 -$$ -\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} = \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} = \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} = \frac{1}{2} f'(1). -$$ +$$\begin{aligned} +\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} &= \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} \\ &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} \\ &= \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\ &= \frac{1}{2} f'(1). +\end{aligned}$$ 已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 @@ -64,6 +64,7 @@ $$ 由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 **答案：** (B) + --- 3.（多选）下列级数中收敛的有______。 @@ -83,15 +84,15 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ - **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 -- **B**：由于 $\lim_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 +- **B**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim\limits_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 -- **C**：由于 $\lim_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 +- **C**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 -- **D**：由于 $\lim_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 +- **D**：由于 $\lim\limits_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 -- **E**：由于 $\lim_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 +- **E**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim\limits_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 -- **F**：由于 $\lim_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 +- **F**：由于 $\lim\limits_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim\limits_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 **答案：** ABCF @@ -108,7 +109,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ $\frac{d}{dx}(x^y) + 4x - y' = 0$， 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 代入 $x=1, y=2$： - $1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6$。 + $$1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6.$$ 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 @@ -116,7 +117,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ -5.求 $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 +5.求 $\lim\limits_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 **解析：** 用夹逼准则： @@ -130,20 +131,20 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ **答案：** $\frac{1}{2}$ -6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 +6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 **解析：** 这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： -$\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 +$$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ 利用等价无穷小和泰勒展开： - $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； - $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 因此： - $\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2)$。 + $$\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2).$$ 取对数： - $\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1)$。 + $$\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1).$$ 所以原极限为 $e^{7/2}$。 @@ -164,15 +165,15 @@ $\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 2. **水平渐近线**（$x \to -\infty$）： 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， - 所以 $\lim_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 + 所以 $\lim\limits_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 3. **斜渐近线**（$x \to +\infty$）： - - 斜率 $k = \lim_{x \to +\infty} \frac{y}{x} = \lim_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 + - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - - 截距 $b = \lim_{x \to +\infty} (y - x) = \lim_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 + - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 故斜渐近线为 $y = x - \frac{\pi}{2}$。 @@ -188,11 +189,12 @@ $\lim_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}$。 **解析：** **第一步：证明数列有下界。** -由算术-几何平均不等式，对任意正数 $x$，有 由算术-几何平均不等式，对任意正数 $x$，有 + $$ \frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. -$$​. +$$ + 因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 **第二步：证明数列单调性。** @@ -231,7 +233,7 @@ $$ -9.飞机离地面 2 km，以 200 km/h 水平飞行，求飞至目标正上方时摄影机的角速率。 +9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 **解析：** 几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 @@ -242,17 +244,15 @@ $$ $$ \frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. $$ -角速率为 $100$ rad/h（速率取绝对值）。 +角速率为 $100$ $rad/h$（速率取绝对值）。 -**答案：** $100$ rad/h。 +**答案：** $100$ $rad/h$。 -以上即为整理后的Obsidian笔记，可以直接复制到Obsidia --- -10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 -x1≤x2≤⋯≤xn +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 $$x_1≤x_2≤⋯≤x_n$$ 证明：存在 $\xi \in [x_1,x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ @@ -260,7 +260,7 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -m≤$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$≤M. +$m≤$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ 由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 证毕。 -- 2.34.1 From 0d37c9be4016a65bb98935345dae3f73e3ca3fca Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 26 Dec 2025 21:27:21 +0800 Subject: [PATCH 029/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../{期中考前押题卷 1.md => 期中考前押题卷.md} | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 编写小组/试卷/{期中考前押题卷 1.md => 期中考前押题卷.md} (100%) diff --git a/编写小组/试卷/期中考前押题卷 1.md b/编写小组/试卷/期中考前押题卷.md similarity index 100% rename from 编写小组/试卷/期中考前押题卷 1.md rename to 编写小组/试卷/期中考前押题卷.md -- 2.34.1 From f84d8c2bea240c5a7baf1a983bcd246e43150bd9 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 06:52:38 +0800 Subject: [PATCH 030/274] vault backup: 2025-12-27 06:52:38 --- ...&考试易错点汇总（解析版）.md | 25 +++++++++++++++++++ 编写小组/试卷/期中考前押题卷.md | 2 +- 2 files changed, 26 insertions(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index cfd242b..81fef13 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -227,6 +227,8 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 **解析** +首先考虑 $x_0$ 为有理数的情况： + 1. 构造第一个数列 $\{x_n^{(1)}\}$： 取 $x_n^{(1)} = x_0 + \frac{1}{n}$（有理数），则 @@ -235,6 +237,29 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 由于 $x_n^{(1)}$ 是有理数，所以 $D(x_n^{(1)}) = 1$，因此 $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ +2. 构造第二个数列 $\{x_n^{(2)}\}$： + + 取 $x_n^{(2)} = x_0 + \frac{\sqrt{2}}{n}$（无理数），则 + $$\lim\limits_{n \to \infty} x_n^{(2)} = x_0$$ + + 由于 $x_n^{(2)}$ 是无理数，所以 $D(x_n^{(2)}) = 0$，因此 + $$\lim\limits_{n \to \infty} D(x_n^{(2)}) = 0$$ + + 由于 + $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1 \neq 0 = \lim\limits_{n \to \infty} D(x_n^{(2)})$$ + + 根据海涅定理，$\lim\limits_{x \to x_0} D(x)$ 不存在。 + +再考虑 $x_0$ 为无理数的情况： + +1. 构造第一个数列 $\{x_n^{(3)}\}$： + + 取 $x_n^{(1)} = x_0 + \frac{1}{n}$（有理数），则 + $$\lim\limits_{n \to \infty} x_n^{(1)} = x_0$$ + + 由于 $x_n^{(1)}$ 是有理数，所以 $D(x_n^{(1)}) = 1$，因此 + $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ + 2. 构造第二个数列 $\{x_n^{(2)}\}$： 取 $x_n^{(2)} = x_0 + \frac{\sqrt{2}}{n}$（无理数），则 diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index bc83efb..2c20845 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -40,7 +40,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ -5. $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$=______。 +5.$\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$=______。 -- 2.34.1 From f19bee098c29b08a8171d437ff8a6fa5cd4bf185 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 07:08:43 +0800 Subject: [PATCH 031/274] vault backup: 2025-12-27 07:08:43 --- 编写小组/试卷/期中考前押题卷解析版.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index 054d216..d88bb3d 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -257,6 +257,7 @@ $$ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ **解析：** + 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -- 2.34.1 From 7ed38450b85fb4540c80bc63db86908fc0ff04ae Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 07:14:04 +0800 Subject: [PATCH 032/274] vault backup: 2025-12-27 07:14:04 --- 编写小组/试卷/期中考前押题卷解析版.md | 8 ++------ 1 file changed, 2 insertions(+), 6 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index d88bb3d..ea19c16 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -3,9 +3,6 @@ 编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 - - - 1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 （A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ @@ -153,8 +150,6 @@ $$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ --- - - 7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 **解析：** @@ -181,6 +176,7 @@ $$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ **答案：** 渐近线为：$x=2$，$y=\frac{\pi}{2}$，$y=x-\frac{\pi}{2}$。 + --- @@ -229,8 +225,8 @@ $$ 因此，数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 ---- +--- 9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 -- 2.34.1 From 9ca2bff4ab23e0d001bb4ddb736e8e7df48acc4f Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 08:05:06 +0800 Subject: [PATCH 033/274] vault backup: 2025-12-27 08:05:06 --- 编写小组/试卷/期中考前押题卷解析版.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index ea19c16..e1e1501 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -257,7 +257,8 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -$m≤$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ +$m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ + 由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 证毕。 -- 2.34.1 From 57dab8600aae0f122911e28613e8fc660320086c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 27 Dec 2025 08:30:03 +0800 Subject: [PATCH 034/274] vault backup: 2025-12-27 08:30:03 --- .../试卷/期中考前押题卷解析版.md | 24 +++++++------------ 1 file changed, 9 insertions(+), 15 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index ea19c16..dc622a1 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -130,22 +130,16 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 -**解析：** -这是 $1^\infty$ 型极限，令 $t = \frac{1}{x}$，则当 $x \to \infty$ 时 $t \to 0^+$，原极限化为： -$$\lim\limits_{t \to 0^+} \left( \tan^2 (2t) + \cos t \right)^{\frac{1}{t^2}}.$$ -利用等价无穷小和泰勒展开： - -- $\tan(2t) \sim 2t$，所以 $\tan^2(2t) \sim 4t^2$； - -- $\cos t = 1 - \frac{t^2}{2} + o(t^2)$。 - 因此： - $$\tan^2(2t) + \cos t = 4t^2 + 1 - \frac{t^2}{2} + o(t^2) = 1 + \frac{7}{2}t^2 + o(t^2).$$ - 取对数： - $$\ln \left[ \left( 1 + \frac{7}{2}t^2 + o(t^2) \right)^{\frac{1}{t^2}} \right] = \frac{1}{t^2} \ln \left( 1 + \frac{7}{2}t^2 + o(t^2) \right) = \frac{1}{t^2} \left( \frac{7}{2}t^2 + o(t^2) \right) = \frac{7}{2} + o(1).$$ - 所以原极限为 $e^{7/2}$。 - +**解析：** $$\begin{aligned} +\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2} +&=\lim\limits_{x\to\infty}(1+\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)^{\frac{1}{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}\cdot x^2\cdot(\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)} +\\&=e^{\lim\limits_{x\to\infty}\frac{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}{\frac{1}{x^2}}} +\\&\overset{t=\frac{1}{x}}{=}e^{\lim\limits_{t\to0}\frac{\tan^2(2t)+\cos t-1}{t^2}} +\\&=e^{\lim\limits_{t\to0}\frac{4t^2}{t^2}-\frac{\frac{1}{2}t^2}{t^2}}(四则运算和等价无穷小) +\\&=e^{\frac{7}{2}} +\end{aligned}$$ -**答案：** $e^{7/2}$ +**答案：** $e^{\frac{7}{2}}$ --- -- 2.34.1 From 8f1b6c864f5c5b104927b0563515db56678c55cf Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 08:33:04 +0800 Subject: [PATCH 035/274] vault backup: 2025-12-27 08:33:04 --- 编写小组/试卷/期中考前押题卷解析版.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index 47fcc7b..a176f02 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -250,8 +250,8 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 -对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -$m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ +对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此可将所有不等式加起来， +从而得到 $m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ 由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ -- 2.34.1 From 0cefd077e8a6756b07b384fc797730df35ea56a4 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 08:53:01 +0800 Subject: [PATCH 036/274] vault backup: 2025-12-27 08:53:01 --- ...&考试易错点汇总（解析版）.md | 49 ++++++++++++++----- 1 file changed, 37 insertions(+), 12 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 81fef13..0e1bc85 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -224,9 +224,10 @@ $$\lim_{k \to \infty}a_{4k+1} = \lim_{k \to \infty}\left(1 + \frac{1}{4k+1}\righ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数时}\end{cases}$$ 在$(-\infty,+\infty)$上每一点都不存在极限。 - **解析** +方法一： + 首先考虑 $x_0$ 为有理数的情况： 1. 构造第一个数列 $\{x_n^{(1)}\}$： @@ -252,28 +253,52 @@ $$D(x)= \begin{cases}1, & x\text{为有理数时}, \\ 0, & x\text{为无理数 再考虑 $x_0$ 为无理数的情况： +设 $[x_0]_n$ 为 $x_0$ 取到 $n$ 位小数后的结果，则有 +$$\lim\limits_{n \to \infty} [x_0]_n = x_0$$ +但是需要注意的是，极限并不是完全相等，其实相差了一个无穷小，也就是要多趋近有多趋近，但是它实质上还是一个有理数，因为它毕竟不是 $x_0$ 本身 + +这种思路的目的是为了找到这样一种表达式，极限是 $x_0$ 的同时，与第一种情况类似，仍是有理数 + 1. 构造第一个数列 $\{x_n^{(3)}\}$： - 取 $x_n^{(1)} = x_0 + \frac{1}{n}$（有理数），则 - $$\lim\limits_{n \to \infty} x_n^{(1)} = x_0$$ + 取 $x_n^{(3)} = [x_0]_n$（有理数），则 + $$\lim\limits_{n \to \infty} x_n^{(3)} = x_0$$ - 由于 $x_n^{(1)}$ 是有理数，所以 $D(x_n^{(1)}) = 1$，因此 + 由于 $x_n^{(3)}$ 是有理数，所以 $D(x_n^{(3)}) = 1$，因此 $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ -2. 构造第二个数列 $\{x_n^{(2)}\}$： +2. 构造第二个数列 $\{x_n^{(4)}\}$： - 取 $x_n^{(2)} = x_0 + \frac{\sqrt{2}}{n}$（无理数），则 - $$\lim\limits_{n \to \infty} x_n^{(2)} = x_0$$ + 取 $x_n^{(4)} = [x_0]_n + \frac{\sqrt{2}}{n}$（无理数），则 + $$\lim\limits_{n \to \infty} x_n^{(4)} = x_0$$ - 由于 $x_n^{(2)}$ 是无理数，所以 $D(x_n^{(2)}) = 0$，因此 - $$\lim\limits_{n \to \infty} D(x_n^{(2)}) = 0$$ + 由于 $x_n^{(4)}$ 是无理数，所以 $D(x_n^{(4)}) = 0$，因此 + $$\lim\limits_{n \to \infty} D(x_n^{(4)}) = 0$$ 由于 - $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1 \neq 0 = \lim\limits_{n \to \infty} D(x_n^{(2)})$$ + $$\lim\limits_{n \to \infty} D(x_n^{(3)}) = 1 \neq 0 = \lim\limits_{n \to \infty} D(x_n^{(4)})$$ 根据海涅定理，$\lim\limits_{x \to x_0} D(x)$ 不存在。 - - 由于 $x_0$ 是任意一点，所以狄利克雷函数在任何点处都不存在极限。 + +综上所述， $x_0$ 是任意一点时，都能得到$\lim\limits_{x \to x_0} D(x)$ 不存在，所以狄利克雷函数在任何点处都不存在极限。 + +方法二：利用实数的稠密性 + +1. **取任意实数 $a$**。 +2. **构造序列**： + - 对每个 $n \in \mathbb{N}^*$，由有理数的稠密性，存在有理数 $r_n$ 满足 $|r_n - a| < \frac{1}{n}$。 + - 对每个 $n \in \mathbb{N}^*$，由无理数的稠密性，存在无理数 $s_n$ 满足 $|s_n - a| < \frac{1}{n}$。 +3. **证明序列收敛**（用 $\varepsilon$-$N$ 语言）： + - 对任意 $\varepsilon > 0$，取 $N = \lfloor 1/\varepsilon \rfloor + 1$，则当 $n > N$ 时，$|r_n - a| < 1/n < \varepsilon$，故 $\lim r_n = a$。 + - 同理 $\lim s_n = a$。 +4. **计算函数值极限**： + - $D(r_n) = 1$，常数序列极限为 $1$。 + - $D(s_n) = 0$，常数序列极限为 $0$。 +5. **应用海涅归结原理**： + - 若 $\lim_{x \to a} D(x)$ 存在，则任何收敛于 $a$ 的序列 $\{x_n\}$ 都应有 $\lim D(x_n)$ 相等。 + - 但 $\{r_n\}$ 和 $\{s_n\}$ 都收敛于 $a$，却得到不同的极限 $1$ 和 $0$，矛盾。 +1. **结论**：$\lim_{x \to a} D(x)$ 不存在，且 $a$ 任意，故 $D(x)$ 在每一点都无极限。 + # 考试易错点总结 -- 2.34.1 From f51ea8b94310626d11880073ba3c9da9c7a126ea Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 08:56:26 +0800 Subject: [PATCH 037/274] vault backup: 2025-12-27 08:56:26 --- .../子数列问题&考试易错点汇总（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 0e1bc85..41ce8dd 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -265,7 +265,7 @@ $$\lim\limits_{n \to \infty} [x_0]_n = x_0$$ $$\lim\limits_{n \to \infty} x_n^{(3)} = x_0$$ 由于 $x_n^{(3)}$ 是有理数，所以 $D(x_n^{(3)}) = 1$，因此 - $$\lim\limits_{n \to \infty} D(x_n^{(1)}) = 1$$ + $$\lim\limits_{n \to \infty} D(x_n^{(3)}) = 1$$ 2. 构造第二个数列 $\{x_n^{(4)}\}$： -- 2.34.1 From df8626db065ed8a2db2ca01000de907ba07aed21 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 27 Dec 2025 09:10:55 +0800 Subject: [PATCH 038/274] vault backup: 2025-12-27 09:10:55 --- .../试卷/期中考前押题卷解析版.md | 37 ++++++++----------- 1 file changed, 15 insertions(+), 22 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index dc622a1..8a3b23a 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -3,7 +3,7 @@ 编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 -1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，**周期为$4$**，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 （A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ **解析：** @@ -14,7 +14,7 @@ $$\begin{aligned} \end{aligned}$$ 已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 -由于 $f(x)$ 是周期函数且可导，其导数 $f'(x)$ 也是周期函数，且周期相同。点 $(5,f(5))$ 处的切线斜率为 $f'(5)$。由周期性，若 $5$ 与 $1$ 相差整数个周期，即存在整数 $k$ 使 $5-1 = kT$，则 $f'(5)=f'(1)$。为使答案确定，可认为 $5$ 与 $1$ 满足周期性条件（否则无法从已知求得 $f'(5)$），故 $f'(5)=f'(1) = -2$。 +由于 $f(x)$ 是周期函数且周期为$4$，则有$$f(x)=f(x+4),$$两边对$x$求导得：$$f'(x)=f'(x+4),$$于是$$f'(5)=f'(1)=-2$$ 因此，曲线在点 $(5,f(5))$ 处的切线斜率为 $-2$，选项（D）正确。 **答案：** (D) @@ -38,12 +38,12 @@ $$\begin{aligned} - 当 $x \to 0^+$ 时，$\frac{1}{x} \to +\infty$，$e^{\frac{1}{x}} \to +\infty$，所以 $$ -\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} \rightarrow \frac{0 + 1}{0 + 1} = 1. +\lim\limits_{x\to0^+}\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \lim\limits_{x\to0^+}\frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} = \frac{0 + 1}{0 + 1} = 1. $$ - 当 $x \to 0^-$ 时，$\frac{1}{x} \to -\infty$，$e^{\frac{1}{x}} \to 0$，所以 $$ -\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2. +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2\ \ (x\to0^-). $$ 1. **第二部分**：$\dfrac{\sin x}{|x|}$ @@ -102,14 +102,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 1. 先求 $x=1$ 时的 $y$ 值：代入方程：$1^y + 2 \times 1^2 - y = 1 \implies 1 + 2 - y = 1 \implies y = 2$。 -2. 隐函数求导：方程两边对 $x$ 求导，注意 $x^y = e^{y \ln x}$： - $\frac{d}{dx}(x^y) + 4x - y' = 0$， - 其中 $\frac{d}{dx}(x^y) = x^y \left( y' \ln x + \frac{y}{x} \right)$。 - 代入 $x=1, y=2$： - $$1^2 \left( y' \ln 1 + \frac{2}{1} \right) + 4 \times 1 - y' = 0 \implies (0 + 2) + 4 - y' = 0 \implies y' = 6.$$ - 所以 $dy|_{x=1} = y'(1)dx = 6dx$。 - - +2. 隐函数求导：由原式得$$x^y=1+y-2x^2,$$取对数得$$y\ln x=\ln(1+y-2x^2)$$两边对$x$求导得$$y'\ln x+\frac{y}{x}=\frac{y'-4x}{1+y-2x^2}$$把$x=1,y=2$带入得$$2=\frac{y'-4}{1+2-2},y'=6$$于是$$dy|_{x=1}=6dx$$ **答案：** $dy|_{x=1} = 6dx$ @@ -119,7 +112,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ **解析：** 用夹逼准则： -- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{1}{2}$； +- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{\frac{1}{2}n(n+1)}{n^2+n}=\frac{1}{2}$； - 上界：$S_n \leq \sum_{k=1}^n \frac{k}{n^2 - n} = \frac{n(n+1)}{2(n^2-n)} \to \frac{1}{2}$（$n \to \infty$）。 故极限为 $\frac{1}{2}$。 @@ -133,10 +126,10 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ **解析：** $$\begin{aligned} \lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2} &=\lim\limits_{x\to\infty}(1+\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)^{\frac{1}{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}\cdot x^2\cdot(\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)} -\\&=e^{\lim\limits_{x\to\infty}\frac{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}{\frac{1}{x^2}}} -\\&\overset{t=\frac{1}{x}}{=}e^{\lim\limits_{t\to0}\frac{\tan^2(2t)+\cos t-1}{t^2}} -\\&=e^{\lim\limits_{t\to0}\frac{4t^2}{t^2}-\frac{\frac{1}{2}t^2}{t^2}}(四则运算和等价无穷小) -\\&=e^{\frac{7}{2}} +\\\\&=e^{\lim\limits_{x\to\infty}\frac{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}{\frac{1}{x^2}}} +\\\\&\overset{t=\frac{1}{x}}{=}e^{\lim\limits_{t\to0}\frac{\tan^2(2t)+\cos t-1}{t^2}} +\\\\&=e^{\lim\limits_{t\to0}\frac{4t^2}{t^2}-\frac{\frac{1}{2}t^2}{t^2}}(四则运算和等价无穷小) +\\\\&=e^{\frac{7}{2}} \end{aligned}$$ **答案：** $e^{\frac{7}{2}}$ @@ -150,7 +143,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 渐近线分垂直、水平、斜渐近线分析： 1. **垂直渐近线**： - 分母 $2-x=0 \implies x=2$，计算 $\lim_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 + 分母 $2-x=0 \implies x=2$，计算 $\lim\limits_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 2. **水平渐近线**（$x \to -\infty$）： 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， @@ -159,7 +152,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 3. **斜渐近线**（$x \to +\infty$）： - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 - 由于 $\ln(1+e^x) = x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 @@ -204,11 +197,11 @@ $$ 易证 $a_1 \geq \sqrt{\sigma}$，等号仅当 $a = \sqrt{\sigma}$ 时成立。 -若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，结论成立。 +若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，极限为$\sqrt{\sigma}$。 若 $a > \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$，由归纳法所有 $a_n > \sqrt{\sigma}$，且数列单调递减。 若 $0 < a < \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$（因为 $a_1 = \frac{1}{2}(a + \sigma/a) \geq \sqrt{\sigma}$ 且等号不成立），此时从 $n=1$ 起 $a_n > \sqrt{\sigma}$，且 $a_2 < a_1$（因 $a_1 > \sqrt{\sigma}$），之后单调递减。 -因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ 或更大），故数列收敛。 +因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ ），故数列收敛。 **第三步：求极限。** 设 $\lim\limits_{n \to \infty} a_n = L$，则 $L \geq \sqrt{\sigma} > 0$。在递推式两边取极限： @@ -251,7 +244,7 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 -$m≤$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ +$m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ 由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ 证毕。 -- 2.34.1 From 44fa1bbb636bdf2f9d7d5845a9c18b16ee081070 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Sat, 27 Dec 2025 09:15:00 +0800 Subject: [PATCH 039/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 1 - 1 file changed, 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index bc83efb..cfb315b 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -39,7 +39,6 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$______。 - 5. $\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$=______。 -- 2.34.1 From 8ac5b1a75f3264da527657c0130525886208a96e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Sat, 27 Dec 2025 09:19:56 +0800 Subject: [PATCH 040/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- conflict-files-obsidian-git.md | 17 ----------------- 1 file changed, 17 deletions(-) delete mode 100644 conflict-files-obsidian-git.md diff --git a/conflict-files-obsidian-git.md b/conflict-files-obsidian-git.md deleted file mode 100644 index a1abbee..0000000 --- a/conflict-files-obsidian-git.md +++ /dev/null @@ -1,17 +0,0 @@ -# Conflicts -Please resolve them and commit them using the commands `Git: Commit all changes` followed by `Git: Push` -(This file will automatically be deleted before commit) -[[#Additional Instructions]] available below file list - -- [[期中考前押题卷]] - -# Additional Instructions -I strongly recommend to use "Source mode" for viewing the conflicted files. For simple conflicts, in each file listed above replace every occurrence of the following text blocks with the desired text. - -```diff -<<<<<<< HEAD - File changes in local repository -======= - File changes in remote repository ->>>>>>> origin/main -``` \ No newline at end of file -- 2.34.1 From 9d3c42350e69fa0a7e968cd5c4006f7e180cad85 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 27 Dec 2025 09:41:48 +0800 Subject: [PATCH 041/274] vault backup: 2025-12-27 09:41:48 --- 编写小组/试卷/期中考前押题卷解析版.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index 0e7160f..d87e214 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -152,7 +152,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 3. **斜渐近线**（$x \to +\infty$）： - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 - 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) =x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 -- 2.34.1 From 9a12a121ed2586a0431ac1aff286e80f8f9b2341 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 27 Dec 2025 09:42:18 +0800 Subject: [PATCH 042/274] vault backup: 2025-12-27 09:42:18 --- 编写小组/试卷/期中考前押题卷解析版.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index d87e214..0e7160f 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -152,7 +152,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 3. **斜渐近线**（$x \to +\infty$）： - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 - 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) =x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 -- 2.34.1 From 9cda46d68fae1b374771b9e3828e7d98d954b085 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sat, 27 Dec 2025 09:44:45 +0800 Subject: [PATCH 043/274] vault backup: 2025-12-27 09:44:45 --- 编写小组/试卷/期中考前押题卷解析版.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index 0e7160f..1786e05 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -152,7 +152,7 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 3. **斜渐近线**（$x \to +\infty$）： - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 - 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x})= x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 -- 2.34.1 From 51f1bc3a0972c882ba3d64fa741e40bbcef32bcb Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Sat, 27 Dec 2025 09:44:57 +0800 Subject: [PATCH 044/274] vault backup: 2025-12-27 09:44:57 --- .../子数列问题&考试易错点汇总（解析版）.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 41ce8dd..d2d2f8e 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -398,9 +398,9 @@ $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{3} < 1 $$ > [!example] 例题1： >$$判定\sum_{n=1}^{\infty} \frac{1}{n^{1 + \frac{1}{n}}}的敛散性$$ #### ❌ 经典错误思路 -1. 形式像 `1/n^p` -2. "指数"是 `1 + 1/n` -3. 因为 `1/n > 0`，所以 `p = 1 + 1/n > 1` 恒成立，误判为收敛 +1. 形式像 $1/n^p$ +2. "指数"是 $1 + 1/n$ +3. 因为 $1/n > 0$，所以 $p = 1 + \frac{1}{n} > 1$ 恒成立，误判为收敛 #### ✅ 正确分析与解法 **错误原因**：`pₙ = 1 + 1/n` 不是常数，其极限为1。 使用比较判别法与调和级数 `∑ 1/n` 比较： -- 2.34.1 From 5a5d13bb09a51ac534d8055470b712a9605efc8b Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Sat, 27 Dec 2025 09:45:50 +0800 Subject: [PATCH 045/274] vault backup: 2025-12-27 09:45:50 --- .../子数列问题&考试易错点汇总（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index d2d2f8e..3812425 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -402,7 +402,7 @@ $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{3} < 1 $$ 2. "指数"是 $1 + 1/n$ 3. 因为 $1/n > 0$，所以 $p = 1 + \frac{1}{n} > 1$ 恒成立，误判为收敛 #### ✅ 正确分析与解法 -**错误原因**：`pₙ = 1 + 1/n` 不是常数，其极限为1。 +**错误原因**：$pₙ = 1 + \frac{1}{n}$ 不是常数，其极限为1。 使用比较判别法与调和级数 `∑ 1/n` 比较： $$\lim_{n \to \infty} \frac{ \frac{1}{n^{1 + \frac{1}{n}}} }{ \frac{1}{n} } = \lim_{n \to \infty} \frac{1}{n^{1/n}} = 1$$ 可知两级数敛散性相同，且调和级数发散 ⇒ 原级数**发散**。 -- 2.34.1 From fb3e6cf440f8c885485a9eaec2e9857e3d097a07 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 09:52:03 +0800 Subject: [PATCH 046/274] vault backup: 2025-12-27 09:52:02 --- 编写小组/试卷/期中考前押题卷.md | 6 +----- 1 file changed, 1 insertion(+), 5 deletions(-) diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index f6af797..6ce7490 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -2,11 +2,7 @@ **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** - - - - -1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为（ ）。 +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，**周期为$4$**，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为（ ）。 （A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ -- 2.34.1 From a336260a3801a9014c81ce886777bf41fa953eca Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 27 Dec 2025 10:21:06 +0800 Subject: [PATCH 047/274] vault backup: 2025-12-27 10:21:06 --- .../子数列问题&考试易错点汇总（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md index 3812425..05a2e7b 100644 --- a/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md +++ b/编写小组/讲义/子数列问题&考试易错点汇总（解析版）.md @@ -3,7 +3,7 @@ tags: - 编写小组 --- **内部资料，禁止传播** -**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 +**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** # 子数列及其相关定理 -- 2.34.1 From 4e00d5d1bb2ac6164256427754a2dc6247261235 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 10:31:33 +0800 Subject: [PATCH 048/274] vault backup: 2025-12-27 10:31:33 --- 编写小组/试卷/期中考前押题卷解析版.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index 0e7160f..f038289 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -23,7 +23,7 @@ $$\begin{aligned} -2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 +2.设 $f(x) = \frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \frac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 (A) 可去间断点 (B) 跳跃间断点 -- 2.34.1 From 6efa7bc8a88baa64056036d91b552949a410e971 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 10:34:05 +0800 Subject: [PATCH 049/274] vault backup: 2025-12-27 10:34:04 --- 编写小组/试卷/期中考前押题卷解析版.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷解析版.md b/编写小组/试卷/期中考前押题卷解析版.md index f038289..0e7160f 100644 --- a/编写小组/试卷/期中考前押题卷解析版.md +++ b/编写小组/试卷/期中考前押题卷解析版.md @@ -23,7 +23,7 @@ $$\begin{aligned} -2.设 $f(x) = \frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \frac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 (A) 可去间断点 (B) 跳跃间断点 -- 2.34.1 From 83a7667e2b49e5aaf37363c4814cc8f178542e16 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 11:38:21 +0800 Subject: [PATCH 050/274] vault backup: 2025-12-27 11:38:21 --- 笔记分享/在这个文件夹分享笔记.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) create mode 100644 笔记分享/在这个文件夹分享笔记.md diff --git a/笔记分享/在这个文件夹分享笔记.md b/笔记分享/在这个文件夹分享笔记.md new file mode 100644 index 0000000..e69de29 -- 2.34.1 From a1ce7e9a9ee097ab3493924e089f5a38bfb939f2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 27 Dec 2025 11:39:36 +0800 Subject: [PATCH 051/274] vault backup: 2025-12-27 11:39:36 --- .../在这个文件夹分享笔记.md | 24 +++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/笔记分享/在这个文件夹分享笔记.md b/笔记分享/在这个文件夹分享笔记.md index e69de29..e130342 100644 --- a/笔记分享/在这个文件夹分享笔记.md +++ b/笔记分享/在这个文件夹分享笔记.md @@ -0,0 +1,24 @@ +## **聚点和上下极限** +此知识点仅作为拓展了解，不要求掌握。 +### 一、聚点 +聚点是研究实数性质的重要概念，这里先下定义： +>[!note] **定义：** +>对于一集合$I\subset\mathbb{R}$，如果存在一点$x_0\in \mathbb{R}$,使得$\forall \delta>0$，有$\overset{\circ}{U}(x_0,\delta)\cap I\neq\varnothing$,则称$x_0$是$I$的一个**聚点**. + +**注：** +1、$x_0$未必是$I$中的一个点； +2、一个集合$I$存在聚点$x_0$与以下两个命题等价： + （1）$\forall \delta>0$,在$U(x_0,\delta)$中有$I$中的无穷多个点； + （2）存在$I$中互异的点组成的数列$\{x_n\}$使得$\lim\limits_{n\to\infty}x_n=x_0$; +3、若$x_0\in I$，但它不是$I$的一个聚点，则称之为孤立点. + +对聚点有如下定理： +>[!note] 聚点原理 +>$\mathbb{R}$的任何一个有界无穷子集至少有一个聚点. + +证明略，有兴趣的可以自己去查阅任何数学分析教材。需要指出的是，它与单调有界原理是等价的。实际上，有四五个与之等价的命题，它们之间互相等价，一般会取一个（一般是戴德金分割定理或者确界原理）作为公理，推出另外几个。它们描述的都是实数的连续性。 + +### **二、上下极限** +我们都知道有左右极限，但其实函数和数列都有一个上下极限。这里给出数列上下极限的定义，有兴趣的也同样可以去找数学分析教材。 +对数列$\{x_n\}$，定义$$l_n=\inf\{x_n,x_{n+1},\cdots\} , h_n=\sup\{x_n,x_{n+1},\cdots\}$$其中$\inf$和$\sup$分别表示下确界和上确界.显然有$$l_1\le l_2\le\cdots l_{n}\le\cdots\le h_n\le\cdots h_2\le\cdots h_1.$$于是数列$\{l_n\}$和$\{h_n\}$是单调有界数列，由单调有界原理，它们都存在极限，且$$\lim\limits_{n\to\infty}l_n=\sup\{l_n\},\lim\limits_{n\to\infty}h_n=\inf\{h_n\}.$$记$$\overset{\_\_\_\_}{\lim\limits_{n\to\infty}}x_n=\lim\limits_{n\to\infty}h_n,\underset{n\to\infty}{\underline{\lim}}x_n=\lim\limits_{n\to\infty}l_n$$ +分别为数列$\{x_n\}$的上、下极限。容易证明，$\{x_n\}$极限存在等价于其上下极限都存在且相等。 -- 2.34.1 From 69082d570c581915df96efebbb558246a085a6b9 Mon Sep 17 00:00:00 2001 From: unknown <2974730459@qq.com> Date: Sat, 27 Dec 2025 11:40:51 +0800 Subject: [PATCH 052/274] vault backup: 2025-12-27 11:40:51 --- ...在这个文件夹分享笔记.md => 聚点和上下极限.md} | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 笔记分享/{在这个文件夹分享笔记.md => 聚点和上下极限.md} (100%) diff --git a/笔记分享/在这个文件夹分享笔记.md b/笔记分享/聚点和上下极限.md similarity index 100% rename from 笔记分享/在这个文件夹分享笔记.md rename to 笔记分享/聚点和上下极限.md -- 2.34.1 From fe210798927a3d5503a6c0a7ad41641d639d814f Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Sat, 27 Dec 2025 11:46:16 +0800 Subject: [PATCH 053/274] =?UTF-8?q?=E7=9F=A9=E9=98=B5=E8=BE=93=E5=87=BA?= =?UTF-8?q?=E6=96=B9=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中考前押题卷.md | 8 +++++++- 1 file changed, 7 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index 6ce7490..c7a4185 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -63,4 +63,10 @@ F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ 10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 x1≤x2≤⋯≤xn 证明：存在 $\xi \in [x_1,x_n]$，使得 -$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ \ No newline at end of file +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ + + + + + +$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ \ No newline at end of file -- 2.34.1 From 0a281290cb073f085a74391e887dce5141dd3075 Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Sat, 27 Dec 2025 11:48:19 +0800 Subject: [PATCH 054/274] vault backup: 2025-12-27 11:48:19 --- 笔记分享/矩阵打印方式.md | 4 ++++ 笔记分享/聚点和上下极限.md | 5 +++++ 编写小组/试卷/期中考前押题卷.md | 1 - 3 files changed, 9 insertions(+), 1 deletion(-) create mode 100644 笔记分享/矩阵打印方式.md diff --git a/笔记分享/矩阵打印方式.md b/笔记分享/矩阵打印方式.md new file mode 100644 index 0000000..819e44e --- /dev/null +++ b/笔记分享/矩阵打印方式.md @@ -0,0 +1,4 @@ + + + +$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ \ No newline at end of file diff --git a/笔记分享/聚点和上下极限.md b/笔记分享/聚点和上下极限.md index e130342..7ab39d9 100644 --- a/笔记分享/聚点和上下极限.md +++ b/笔记分享/聚点和上下极限.md @@ -22,3 +22,8 @@ 我们都知道有左右极限，但其实函数和数列都有一个上下极限。这里给出数列上下极限的定义，有兴趣的也同样可以去找数学分析教材。 对数列$\{x_n\}$，定义$$l_n=\inf\{x_n,x_{n+1},\cdots\} , h_n=\sup\{x_n,x_{n+1},\cdots\}$$其中$\inf$和$\sup$分别表示下确界和上确界.显然有$$l_1\le l_2\le\cdots l_{n}\le\cdots\le h_n\le\cdots h_2\le\cdots h_1.$$于是数列$\{l_n\}$和$\{h_n\}$是单调有界数列，由单调有界原理，它们都存在极限，且$$\lim\limits_{n\to\infty}l_n=\sup\{l_n\},\lim\limits_{n\to\infty}h_n=\inf\{h_n\}.$$记$$\overset{\_\_\_\_}{\lim\limits_{n\to\infty}}x_n=\lim\limits_{n\to\infty}h_n,\underset{n\to\infty}{\underline{\lim}}x_n=\lim\limits_{n\to\infty}l_n$$ 分别为数列$\{x_n\}$的上、下极限。容易证明，$\{x_n\}$极限存在等价于其上下极限都存在且相等。 + + + + + diff --git a/编写小组/试卷/期中考前押题卷.md b/编写小组/试卷/期中考前押题卷.md index c7a4185..bdcbf09 100644 --- a/编写小组/试卷/期中考前押题卷.md +++ b/编写小组/试卷/期中考前押题卷.md @@ -69,4 +69,3 @@ $$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ -$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ \ No newline at end of file -- 2.34.1 From 48daaaf91a54590ddc1f0e9cf648d2d2502d83fb Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 27 Dec 2025 11:50:09 +0800 Subject: [PATCH 055/274] vault backup: 2025-12-27 11:50:09 --- .../试卷/2025秋期中考前押题卷.md | 72 +++++ .../2025秋期中考前押题卷解析版.md | 256 ++++++++++++++++++ 2 files changed, 328 insertions(+) create mode 100644 编写小组/试卷/2025秋期中考前押题卷.md create mode 100644 编写小组/试卷/2025秋期中考前押题卷解析版.md diff --git a/编写小组/试卷/2025秋期中考前押题卷.md b/编写小组/试卷/2025秋期中考前押题卷.md new file mode 100644 index 0000000..56826e6 --- /dev/null +++ b/编写小组/试卷/2025秋期中考前押题卷.md @@ -0,0 +1,72 @@ +时量：60分钟 ____ +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航** + +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，**周期为$4$**，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为（ ）。 +（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ + + + +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 + +(A) 可去间断点 +(B) 跳跃间断点 +(C) 无穷间断点 +(D) 振荡间断点 + + + +3.（多选）下列级数中收敛的有______。 + +A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + +B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ + +C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ + +D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + +E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + +F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + + + +4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$______。 + + + + +5.$\lim_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$=______。 + + + + +6.计算 $\lim_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$______。 + + + +7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 + + +8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ +证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + + + + +9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 + + + + +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 +x1≤x2≤⋯≤xn +证明：存在 $\xi \in [x_1,x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ + + + + + +$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ diff --git a/编写小组/试卷/2025秋期中考前押题卷解析版.md b/编写小组/试卷/2025秋期中考前押题卷解析版.md new file mode 100644 index 0000000..0e7160f --- /dev/null +++ b/编写小组/试卷/2025秋期中考前押题卷解析版.md @@ -0,0 +1,256 @@ +时量：60分钟 ____ +内部资料，禁止传播 +编委会（不分先后，姓氏首字母顺序）：程奕铭 韩魏 刘柯妤 卢吉辚 王轲楠 支宝宁 郑哲航 + + +1.设周期函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导，**周期为$4$**，又 $\lim\limits_{x\to0}\dfrac{f(1)-f(1-x)}{2x}=-1$，则曲线 $y=f(x)$ 在点 $(5,f(5))$ 处切线的斜率为[　]。 +（A）$\dfrac{1}{2}$　　　（B）$0$　　　（C）$-1$　　　（D）$-2$ + +**解析：** +由极限表达式变形：令 $h=-x$，则当 $x\to0$ 时 $h\to0$，于是 + +$$\begin{aligned} +\lim_{x \to 0} \frac{f(1) - f(1 - x)}{2x} &= \lim_{h \to 0} \frac{f(1) - f(1 + h)}{-2h} \\ &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{2h} \\ &= \frac{1}{2} \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\ &= \frac{1}{2} f'(1). +\end{aligned}$$ + +已知该极限值为 $-1$，故 $\dfrac{1}{2}f'(1) = -1$，解得 $f'(1) = -2$。 +由于 $f(x)$ 是周期函数且周期为$4$，则有$$f(x)=f(x+4),$$两边对$x$求导得：$$f'(x)=f'(x+4),$$于是$$f'(5)=f'(1)=-2$$ +因此，曲线在点 $(5,f(5))$ 处的切线斜率为 $-2$，选项（D）正确。 + +**答案：** (D) + +--- + + + +2.设 $f(x) = \dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} + \dfrac{\sin x}{|x|}$，则 $x = 0$ 是 $f(x)$ 的（ ）。 + +(A) 可去间断点 +(B) 跳跃间断点 +(C) 无穷间断点 +(D) 振荡间断点 + +**解析：** +分析函数在 $x=0$ 处的左右极限。 + +1. **第一部分**：$\dfrac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}}$ + + - 当 $x \to 0^+$ 时，$\frac{1}{x} \to +\infty$，$e^{\frac{1}{x}} \to +\infty$，所以 + + $$ +\lim\limits_{x\to0^+}\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} = \lim\limits_{x\to0^+}\frac{\frac{2}{e^{\frac{1}{x}}} + 1}{\frac{1}{e^{\frac{1}{x}}} + 1} = \frac{0 + 1}{0 + 1} = 1. +$$ + - 当 $x \to 0^-$ 时，$\frac{1}{x} \to -\infty$，$e^{\frac{1}{x}} \to 0$，所以 + + $$ +\frac{2 + e^{\frac{1}{x}}}{1 + e^{\frac{1}{x}}} \rightarrow \frac{2 + 0}{1 + 0} = 2\ \ (x\to0^-). +$$ +1. **第二部分**：$\dfrac{\sin x}{|x|}$ + + - 当 $x \to 0^+$ 时，$|x| = x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{x} \to 1$。 + + - 当 $x \to 0^-$ 时，$|x| = -x$，$\dfrac{\sin x}{|x|} = \dfrac{\sin x}{-x} \to -1$。 + +2. **整体极限**： + + - 右极限：$\lim\limits_{x \to 0^+} f(x) = 1 + 1 = 2$。 + + - 左极限：$\lim\limits_{x \to 0^-} f(x) = 2 + (-1) = 1$。 + + +由于左右极限存在但不相等，故 $x=0$ 处为**跳跃间断点**。 + +**答案：** (B) + +--- +3.（多选）下列级数中收敛的有______。 + +A $\sin \frac{\pi}{2} + \sin \frac{\pi}{2^2} + \sin \frac{\pi}{2^3} + \cdots$ + +B $\sum_{n=1}^{\infty} \frac{1}{5^n} \cdot \frac{3n^3+2n^2}{4n^3+1}$ + +C $\sum_{n=1}^{\infty} \frac{1}{(a+n-1)(a+n)(a+n+1)} \quad (a > 0)$ + +D $\sum_{n=1}^{\infty} \frac{1}{n\sqrt[n]{n}}$ + +E $\sum_{n=1}^{\infty} \frac{1}{n} \arctan \frac{n}{n+1}$ + +F $\sum_{n=1}^{\infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1}$ + +**解析：** + +- **A**：由于 $0 < \sin \frac{\pi}{2^n} \leq \frac{\pi}{2^n}$，且几何级数 $\sum_{n=1}^{\infty} \frac{\pi}{2^n}$ 收敛，故由比较判别法知原级数收敛。 + +- **B**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{5^n} \cdot \frac{3n^3 + 2n^2}{4n^3 + 1} \right) / \left( \frac{1}{5^n} \right) = \lim\limits_{n \to \infty} \frac{3n^3 + 2n^2}{4n^3 + 1} = \frac{3}{4}$，又几何级数 $\sum_{n=1}^{\infty} \frac{1}{5^n}$ 收敛，故由极限形式的比较判别法知原级数收敛。 + +- **C**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{(a+n-1)(a+n)(a+n+1)} \right) / \left( \frac{1}{n^3} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^3}$ 收敛（$p=3>1$），故原级数收敛。也可直接放缩：$\frac{1}{(a+n-1)(a+n)(a+n+1)} \leq \frac{1}{(a+n-1)^3} < \frac{1}{(n-1)^3}$（当 $n>1$），由比较判别法知收敛。 + +- **D**：由于 $\lim\limits_{n \to \infty} \frac{1}{n\sqrt[n]{n}} / \frac{1}{n} = \lim\limits_{n \to \infty} \frac{1}{\sqrt[n]{n}} = 1$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **E**：由于 $\lim\limits_{n \to \infty} \left( \frac{1}{n} \cdot \arctan \frac{n}{n+1} \right) / \frac{1}{n} = \lim\limits_{n \to \infty} \arctan \frac{n}{n+1} = \arctan 1 = \frac{\pi}{4}$，而调和级数 $\sum_{n=1}^{\infty} \frac{1}{n}$ 发散，故原级数发散。 + +- **F**：由于 $\lim\limits_{n \to \infty} \frac{\sqrt{n+\sqrt{n}}}{n^2+1} / \frac{1}{n^{3/2}} = \lim\limits_{n \to \infty} \sqrt{1 + \frac{1}{\sqrt{n}}} / \left( 1 + \frac{1}{n^2} \right) = 1$，而 $p$ 级数 $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ 收敛（$p=3/2>1$），故原级数收敛。 + + +**答案：** ABCF + + + +4.设 $y=y(x)$ 由方程 $x^y + 2x^2 - y = 1$ 确定，求 $dy|_{x=1}$。 + +**解析：** + +1. 先求 $x=1$ 时的 $y$ 值：代入方程：$1^y + 2 \times 1^2 - y = 1 \implies 1 + 2 - y = 1 \implies y = 2$。 + +2. 隐函数求导：由原式得$$x^y=1+y-2x^2,$$取对数得$$y\ln x=\ln(1+y-2x^2)$$两边对$x$求导得$$y'\ln x+\frac{y}{x}=\frac{y'-4x}{1+y-2x^2}$$把$x=1,y=2$带入得$$2=\frac{y'-4}{1+2-2},y'=6$$于是$$dy|_{x=1}=6dx$$ +**答案：** $dy|_{x=1} = 6dx$ + + + +5.求 $\lim\limits_{n \to \infty} \left( \frac{1}{n^2 + \sin 1} + \frac{2}{n^2 + 2\sin 2} + \cdots + \frac{n}{n^2 + n \sin n} \right)$。 + +**解析：** +用夹逼准则： + +- 下界：$S_n \geq \sum_{k=1}^n \frac{k}{n^2 + n} = \frac{\frac{1}{2}n(n+1)}{n^2+n}=\frac{1}{2}$； + +- 上界：$S_n \leq \sum_{k=1}^n \frac{k}{n^2 - n} = \frac{n(n+1)}{2(n^2-n)} \to \frac{1}{2}$（$n \to \infty$）。 + 故极限为 $\frac{1}{2}$。 + + +**答案：** $\frac{1}{2}$ + + +6.计算 $\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2}$。 + +**解析：** $$\begin{aligned} +\lim\limits_{x \to \infty} \left( \tan^2 \frac{2}{x} + \cos \frac{1}{x} \right)^{x^2} +&=\lim\limits_{x\to\infty}(1+\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)^{\frac{1}{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}\cdot x^2\cdot(\tan^2\frac{2}{x}+\cos \frac{1}{x}-1)} +\\\\&=e^{\lim\limits_{x\to\infty}\frac{\tan^2\frac{2}{x}+\cos \frac{1}{x}-1}{\frac{1}{x^2}}} +\\\\&\overset{t=\frac{1}{x}}{=}e^{\lim\limits_{t\to0}\frac{\tan^2(2t)+\cos t-1}{t^2}} +\\\\&=e^{\lim\limits_{t\to0}\frac{4t^2}{t^2}-\frac{\frac{1}{2}t^2}{t^2}}(四则运算和等价无穷小) +\\\\&=e^{\frac{7}{2}} +\end{aligned}$$ + +**答案：** $e^{\frac{7}{2}}$ + +--- + + +7.求曲线 $y = \ln(1+e^x) + \frac{2+x}{2-x} \arctan \frac{x}{2}$ 的渐近线方程。 + +**解析：** +渐近线分垂直、水平、斜渐近线分析： + +1. **垂直渐近线**： + 分母 $2-x=0 \implies x=2$，计算 $\lim\limits_{x \to 2} y = \infty$，故垂直渐近线为 $x=2$。 + +2. **水平渐近线**（$x \to -\infty$）： + 当 $x \to -\infty$ 时，$\ln(1+e^x) \to 0$，$\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to -\frac{\pi}{2}$， + 所以 $\lim\limits_{x \to -\infty} y = (-1) \times (-\frac{\pi}{2}) = \frac{\pi}{2}$，水平渐近线为 $y = \frac{\pi}{2}$。 + +3. **斜渐近线**（$x \to +\infty$）： + + - 斜率 $k = \lim\limits_{x \to +\infty} \frac{y}{x} = \lim\limits_{x \to +\infty} \frac{\ln(1+e^x)}{x} + \lim\limits_{x \to +\infty} \frac{\frac{2+x}{2-x} \arctan \frac{x}{2}}{x}$。 + 由于 $\ln(1+e^x) =\ln e^x(1+e^{-x}) x + \ln(1+e^{-x})$，所以 $\frac{\ln(1+e^x)}{x} = 1 + \frac{\ln(1+e^{-x})}{x} \to 1$； + 又 $\frac{2+x}{2-x} \to -1$，$\frac{\arctan \frac{x}{2}}{x} \to 0$，所以第二项趋于0，因此 $k = 1$。 + + - 截距 $b = \lim\limits_{x \to +\infty} (y - x) = \lim\limits_{x \to +\infty} \left[ \ln(1+e^x) - x + \frac{2+x}{2-x} \arctan \frac{x}{2} \right]$。 + 由于 $\ln(1+e^x) - x = \ln(1+e^{-x}) \to 0$，且 $\frac{2+x}{2-x} \to -1$，$\arctan \frac{x}{2} \to \frac{\pi}{2}$，所以 $b = -\frac{\pi}{2}$。 + 故斜渐近线为 $y = x - \frac{\pi}{2}$。 + + +**答案：** +渐近线为：$x=2$，$y=\frac{\pi}{2}$，$y=x-\frac{\pi}{2}$。 + + +--- + + +8.设 $a > 0, \sigma > 0$，定义 $a_1 = \dfrac{1}{2} \left( a + \dfrac{\sigma}{a} \right)$，$a_{n+1} = \dfrac{1}{2} \left( a_n + \dfrac{\sigma}{a_n} \right)$，$n = 1, 2, \ldots$ +证明：数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + +**解析：** +**第一步：证明数列有下界。** +由算术-几何平均不等式，对任意正数 $x$，有 + +$$ +\frac{1}{2} \left( x + \frac{\sigma}{x} \right) \geq \sqrt{x \cdot \frac{\sigma}{x}} = \sqrt{\sigma}. +$$ + +因此，对 $n \geq 1$，有 $a_n \geq \sqrt{\sigma}$，即数列有下界 $\sqrt{\sigma}$。 + +**第二步：证明数列单调性。** +考虑差值： +考虑数列的递推式：$a_{n+1} = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right)$，则 +$$ +a_{n+1} - a_n = \frac{1}{2} \left( a_n + \frac{\sigma}{a_n} \right) - a_n = \frac{1}{2} \left( \frac{\sigma}{a_n} - a_n \right) = \frac{\sigma - a_n^2}{2a_n}. +$$ + +由于 $a_n > 0$，差值的符号由 $\sigma - a_n^2$ 决定。 + +- 若 $a_n > \sqrt{\sigma}$，则 $a_{n+1} - a_n < 0$，数列单调递减； + +- 若 $a_n < \sqrt{\sigma}$，则 $a_{n+1} - a_n > 0$，数列单调递增； + +- 若 $a_n = \sqrt{\sigma}$，则 $a_{n+1} = a_n$，数列为常数列。 + + +易证 $a_1 \geq \sqrt{\sigma}$，等号仅当 $a = \sqrt{\sigma}$ 时成立。 +若 $a = \sqrt{\sigma}$，则数列恒为 $\sqrt{\sigma}$，极限为$\sqrt{\sigma}$。 +若 $a > \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$，由归纳法所有 $a_n > \sqrt{\sigma}$，且数列单调递减。 +若 $0 < a < \sqrt{\sigma}$，则 $a_1 > \sqrt{\sigma}$（因为 $a_1 = \frac{1}{2}(a + \sigma/a) \geq \sqrt{\sigma}$ 且等号不成立），此时从 $n=1$ 起 $a_n > \sqrt{\sigma}$，且 $a_2 < a_1$（因 $a_1 > \sqrt{\sigma}$），之后单调递减。 + +因此，无论哪种情况，数列从某项起单调且有界（下界 $\sqrt{\sigma}$，上界为 $a_1$ ），故数列收敛。 + +**第三步：求极限。** +设 $\lim\limits_{n \to \infty} a_n = L$，则 $L \geq \sqrt{\sigma} > 0$。在递推式两边取极限： +$$ +L = \frac{1}{2} \left( L + \frac{\sigma}{L} \right). +$$ +整理得 $2L = L + \frac{\sigma}{L}$，即 $L = \frac{\sigma}{L}$，从而 $L^2 = \sigma$，故 $L = \sqrt{\sigma}$（正根）。 + +因此，数列 ${a_n}$ 收敛，且极限为 $\sqrt{\sigma}$。 + + +--- + + +9.一飞机在离地面$2 km$的高度，以$200 km/h$的速度水平飞行到某目标上空，以便进行航空摄影。试求飞机飞至该目标正上方时，摄影机转动的角速率。 + +**解析：** +几何关系：$\tan \theta = \frac{x}{2}$（$x$ 为水平距离，$\theta$ 为竖直与摄影机连线的夹角）。 + +求导：$\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{2} \cdot \frac{dx}{dt}$。 + +代入 $x=0$（正上方）：$\theta = 0 \implies \cos \theta = 1$，$\frac{dx}{dt} = -200$ km/h，得 +$$ +\frac{d\theta}{dt} = \frac{1}{2} \times (-200) = -100 \text{ rad/h}. +$$ +角速率为 $100$ $rad/h$（速率取绝对值）。 + +**答案：** $100$ $rad/h$。 + + +--- + + +10.若函数 $f(x)$ 在 $(a,b)$ 内连续，任取 $x_i \in (a,b)$ $(i=1,2,\cdots,n)$ 使 $$x_1≤x_2≤⋯≤x_n$$ +证明：存在 $\xi \in [x_1,x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ + +**解析：** + +由于 $f(x)$ 在 $(a,b)$ 内连续，而 $[x_1, x_n] \subset (a,b)$，所以 $f(x)$ 在闭区间 $[x_1, x_n]$ 上连续。根据闭区间上连续函数的最值定理，$f(x)$ 在 $[x_1, x_n]$ 上能取到最大值 $M$ 和最小值 $m$。 + +对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此 +$m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ +对于任意 $x_i \in [x_1, x_n]$，有 $m \leq f(x_i) \leq M$，因此可将所有不等式加起来， +从而得到 $m≤$$\frac{1}{n} \sum_{i=1}^n f(x_i$)$≤M.$ + +由连续函数的介值定理，存在 $\xi \in [x_1, x_n]$，使得 +$$f(\xi) = \frac{1}{n} \sum_{i=1}^n f(x_i)$$ +证毕。 + +--- + -- 2.34.1 From abec3b1380407e49694b0e2a2285311a4601335c Mon 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\rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 + A. 同阶无穷小 + B. 低阶无穷小 + C. 等价无穷小 + D. 高阶无穷小 +4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 + A. 无穷大量 + B. 无穷小量 + C. 有界变量 + D. 无界变量 +5. 下列四个级数，**发散**的是 + A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ + B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ + C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ + D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ +#### 二、填空题（共5小题，每小题3分，共15分） +6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. +7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. +8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. +9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. +10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. +#### 三、解答与证明题（11~19小题，共70分） +11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. +12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. +13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. +14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. +15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. +16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. +17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. +18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] +19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ + (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） + (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file -- 2.34.1 From f3c7687e21a1665a2f55355dc1b56deafeeedefe Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Sun, 28 Dec 2025 08:31:06 +0800 Subject: [PATCH 059/274] =?UTF-8?q?LaTeX=E8=BE=93=E5=85=A5=E8=A7=84?= =?UTF-8?q?=E8=8C=83=E5=8F=8A=E6=8F=90=E7=A4=BA?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 笔记分享/LaTeX（KaTeX）特殊输入.md | 13 +++++++++++++ 笔记分享/LaTeX（KaTeX）输入规范.md | 4 ++++ 笔记分享/矩阵打印方式.md | 4 ---- 3 files changed, 17 insertions(+), 4 deletions(-) create mode 100644 笔记分享/LaTeX（KaTeX）特殊输入.md create mode 100644 笔记分享/LaTeX（KaTeX）输入规范.md delete mode 100644 笔记分享/矩阵打印方式.md diff --git a/笔记分享/LaTeX（KaTeX）特殊输入.md b/笔记分享/LaTeX（KaTeX）特殊输入.md new file mode 100644 index 0000000..6278761 --- /dev/null +++ b/笔记分享/LaTeX（KaTeX）特殊输入.md @@ -0,0 +1,13 @@ + +矩阵打印方式 +$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ +然而，我们教材上是使用的方括号矩阵，要将pmatrix换成bmatrix： +$\begin{bmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{bmatrix}$ +行列式的输入方法如下：(vmatrix) +$\begin{vmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{vmatrix}$ +求和、求积的上下记号可以使用limits套： +$\sum\limits_{i=2}^{n}a_i, \prod\limits_{i=1}^{n}a_i$ +积分符号的上下标，按照原本的上下标来处理： +$\int_1^5 x\mathrm{d}x$ +积分符号的上下记号，用limits套： +$\int\limits_{L}(x+y)\mathrm{d}s$ \ No newline at end of file diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md new file mode 100644 index 0000000..e9538a1 --- /dev/null +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -0,0 +1,4 @@ +通常的，记号严格按照教材中的规范。 +1. 矩阵使用bmatrix +2. 自然常数或电荷量e、虚数单位i应当为**正体**，需要用mathrm记号包裹，例：$\mathrm{e}^{\mathrm{i}\pi}+1=0$；然而，当e,i作为变量时，应当用正常的斜体。例：$\sum\limits_{i=1}^{n}a_i$ +3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ \ No newline at end of file diff --git a/笔记分享/矩阵打印方式.md b/笔记分享/矩阵打印方式.md deleted file mode 100644 index 819e44e..0000000 --- a/笔记分享/矩阵打印方式.md +++ /dev/null @@ -1,4 +0,0 @@ - - - -$\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ \ No newline at end of file -- 2.34.1 From 7672530bfe62fd2f2ac7567bccd2f626cf242c28 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 29 Dec 2025 10:34:04 +0800 Subject: [PATCH 060/274] =?UTF-8?q?=E6=9C=AA=E5=AE=8C=E6=88=90=E7=9A=84?= =?UTF-8?q?=E8=A7=A3=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/期中试卷解析.md | 62 +++++++++++++++++++++++ 1 file changed, 62 insertions(+) create mode 100644 编写小组/试卷/期中试卷解析.md diff --git a/编写小组/试卷/期中试卷解析.md b/编写小组/试卷/期中试卷解析.md new file mode 100644 index 0000000..4874412 --- /dev/null +++ b/编写小组/试卷/期中试卷解析.md @@ -0,0 +1,62 @@ +#官方试卷 +#民间答案 +#### 一、单选题（共5小题，每小题3分，共15分） +1. 若$\lim\limits_{n\rightarrow\infty}a_n=4$，则当$n$充分大时，恒有 + A. $|a_n|\le 1$ + B. $|a_n|>2$ + C. $|a_n|<2$ + D. $|a_n|>4$ +> **答案：B** +> 解析：保号性 +> 若$\lim\limits_{n\rightarrow\infty}a_n=4$，按照定义，$\forall \epsilon >0,\exists N, n>N\text{时},|a_n-4|<\epsilon$， +> 取$\epsilon=2$，则$|a_n-4|<\epsilon$，则$2<a_n<6$，B选项符合题意 +2. 已知$g(x)=\frac{1}{x^2}$，复合函数$y=f(g(x))$对$x$的导数为$-\frac{1}{2x}$，则$f'(\frac{1}{2})$的值为 + A. $1$ + B. $2$ + C. $\frac{\sqrt 2}{4}$ + D. $\frac{1}{2}$ +>**答案：D** +>解析：复合函数求导法则 +>$y'=f'(g(x))g'(x)=\frac{-2}{x^3}f'(\frac{1}{x^2})=-\frac{1}{2x}\Rightarrow f'(\frac{1}{x^2})=\frac{x^2}{4}$，即$f'(\frac{1}{x})=\frac{x}{4}(x> 0)$ +>代入$x=2$得：$f'(\frac{1}{2})=\frac{1}{2}$ +3. 设$f(x)$可导且$f'(x_0)=\frac{1}{3}$，则当$\Delta x \rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 + A. 同阶无穷小 + B. 低阶无穷小 + C. 等价无穷小 + D. 高阶无穷小 +> **答案：A** +> 解析：同阶无穷小；微分的定义 +> $\Delta x \rightarrow 0$时，$\Delta x \sim \mathrm{d}x$，又$\mathrm{d}y|_{x=x_0}=\frac{1}{3}\mathrm{d}x$，故$\mathrm{d}y$与$\Delta{x}$是同阶不等价的无穷小 +4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 + A. 无穷大量 + B. 无穷小量 + C. 有界变量 + D. 无界变量 +>**答案：D** +>解析：易错点10-无穷大与无界的辨析 +>$\lim\limits_{k\to\infty}a_{2k}=+\infty$，因此$a_n$无界；然而，$\lim\limits_{k\to\infty}a_{2k+1}=0$，因此$a_n(n\to\infty)$不是无穷大 +5. 下列四个级数，**发散**的是 + A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ + B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ + C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ + D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ +>**答案：C** +> +#### 二、填空题（共5小题，每小题3分，共15分） +6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. +7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. +8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. +9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. +10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. +#### 三、解答与证明题（11~19小题，共70分） +11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. +12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. +13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. +14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. +15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. +16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. +17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. +18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] +19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ + (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） + (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file -- 2.34.1 From ae9252b244340e81e2d634a06498b15c30e894a2 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:35:02 +0800 Subject: [PATCH 061/274] vault backup: 2025-12-29 10:35:02 --- 笔记分享/LaTeX（KaTeX）特殊输入.md | 11 +++++++++++ 1 file changed, 11 insertions(+) diff --git a/笔记分享/LaTeX（KaTeX）特殊输入.md b/笔记分享/LaTeX（KaTeX）特殊输入.md index 6278761..413f04d 100644 --- a/笔记分享/LaTeX（KaTeX）特殊输入.md +++ b/笔记分享/LaTeX（KaTeX）特殊输入.md @@ -1,13 +1,24 @@ 矩阵打印方式 + $\begin{pmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{pmatrix}$ + 然而，我们教材上是使用的方括号矩阵，要将pmatrix换成bmatrix： + $\begin{bmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{bmatrix}$ + 行列式的输入方法如下：(vmatrix) + $\begin{vmatrix} 1+x_1 & 1+x_1^2 & ... &1+x_1^n \\ 1+x_2 & 1+x_2^2 & ... &1+x_2^n \\ ... \\ 1+x_n & 1+x_n^2 & ... &1+x_n^n \end{vmatrix}$ + 求和、求积的上下记号可以使用limits套： + $\sum\limits_{i=2}^{n}a_i, \prod\limits_{i=1}^{n}a_i$ + 积分符号的上下标，按照原本的上下标来处理： + $\int_1^5 x\mathrm{d}x$ + 积分符号的上下记号，用limits套： + $\int\limits_{L}(x+y)\mathrm{d}s$ \ No newline at end of file -- 2.34.1 From e7f65ba31b15fb44a8fa18af174623fffb40e4ce Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:47:17 +0800 Subject: [PATCH 062/274] vault backup: 2025-12-29 10:47:17 --- 素材/先化简再用洛必达.md | 4 ++++ 1 file changed, 4 insertions(+) create mode 100644 素材/先化简再用洛必达.md diff --git a/素材/先化简再用洛必达.md b/素材/先化简再用洛必达.md new file mode 100644 index 0000000..1005062 --- /dev/null +++ b/素材/先化简再用洛必达.md @@ -0,0 +1,4 @@ +--- +tags: + - 素材 +--- -- 2.34.1 From 62ba6e5cbbeababeeefb1e24299a70d3e8196424 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 29 Dec 2025 10:47:57 +0800 Subject: [PATCH 063/274] =?UTF-8?q?=E6=B4=9B=E5=BF=85=E8=BE=BE=E6=B3=95?= =?UTF-8?q?=E5=88=99-=E6=B3=A8=E6=84=8F=E4=BA=8B=E9=A1=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/洛必达法则-注意事项.md | 3 +++ 1 file changed, 3 insertions(+) create mode 100644 素材/洛必达法则-注意事项.md diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md new file mode 100644 index 0000000..aa9ab02 --- /dev/null +++ b/素材/洛必达法则-注意事项.md @@ -0,0 +1,3 @@ +1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ + 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ +2. 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便 \ No newline at end of file -- 2.34.1 From e638ba34988cb60acc3ffaad6df0c276de8ec82e Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 29 Dec 2025 10:48:36 +0800 Subject: [PATCH 064/274] =?UTF-8?q?=E5=90=88=E5=B9=B6?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/先化简再用洛必达.md | 4 ---- 1 file changed, 4 deletions(-) delete mode 100644 素材/先化简再用洛必达.md diff --git a/素材/先化简再用洛必达.md b/素材/先化简再用洛必达.md deleted file mode 100644 index 1005062..0000000 --- a/素材/先化简再用洛必达.md +++ /dev/null @@ -1,4 +0,0 @@ ---- -tags: - - 素材 ---- -- 2.34.1 From 5a2d01374791f15fc757830398a458eb44ec26b0 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 29 Dec 2025 10:52:28 +0800 Subject: [PATCH 065/274] =?UTF-8?q?=E5=90=88=E5=B9=B6?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/先化简再用洛必达.md | 4 ---- 素材/洛必达法则-注意事项.md | 5 +++++ 2 files changed, 5 insertions(+), 4 deletions(-) delete mode 100644 素材/先化简再用洛必达.md diff --git a/素材/先化简再用洛必达.md b/素材/先化简再用洛必达.md deleted file mode 100644 index 1005062..0000000 --- a/素材/先化简再用洛必达.md +++ /dev/null @@ -1,4 +0,0 @@ ---- -tags: - - 素材 ---- diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index aa9ab02..e3bd120 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -1,3 +1,8 @@ +--- +tags: + - 素材 +--- + 1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ 2. 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便 \ No newline at end of file -- 2.34.1 From 61eeff537edd09a970512cc49d5b80a9c342ad1e Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:54:44 +0800 Subject: [PATCH 066/274] vault backup: 2025-12-29 10:54:44 --- 素材/洛必达法则-注意事项.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index e3bd120..0fd3caf 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -2,7 +2,7 @@ tags: - 素材 --- - 1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ -2. 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便 \ No newline at end of file +2. 先化简，再使用洛必达 + 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便$$\lim_{x \to 0} \frac{e^x-e^{-x}-2x}{\tan^{3}x}$$先利用等价无穷小，将 $\tan^{3}x$ 转化为 $x^3$ ，然后再使用洛必达 \ No newline at end of file -- 2.34.1 From 19af4adcc56ee95770a94d7fe6c423deb67ba077 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:58:20 +0800 Subject: [PATCH 067/274] vault backup: 2025-12-29 10:58:20 --- 素材/洛必达法则-注意事项.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 0fd3caf..00e2e56 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -5,4 +5,6 @@ tags: 1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ 2. 先化简，再使用洛必达 - 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便$$\lim_{x \to 0} \frac{e^x-e^{-x}-2x}{\tan^{3}x}$$先利用等价无穷小，将 $\tan^{3}x$ 转化为 $x^3$ ，然后再使用洛必达 \ No newline at end of file + 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便$$\lim_{x \to 0} \frac{e^x-e^{-x}-2x}{\tan^{3}x}$$先利用等价无穷小，将 $\tan^{3}x$ 转化为 $x^3$ ，然后再使用洛必达 +3. 在满足定理的某些情况下洛必达法则不能解决计算问题 + $$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ \ No newline at end of file -- 2.34.1 From 63da3644c423dd1b4d848053014df98c04d803d7 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:58:45 +0800 Subject: [PATCH 068/274] vault backup: 2025-12-29 10:58:45 --- 素材/洛必达法则-注意事项.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 00e2e56..24da070 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -4,7 +4,9 @@ tags: --- 1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ + 2. 先化简，再使用洛必达 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便$$\lim_{x \to 0} \frac{e^x-e^{-x}-2x}{\tan^{3}x}$$先利用等价无穷小，将 $\tan^{3}x$ 转化为 $x^3$ ，然后再使用洛必达 + 3. 在满足定理的某些情况下洛必达法则不能解决计算问题 $$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ \ No newline at end of file -- 2.34.1 From 8ddd5fdc48eee0d7cb2717f74f652a2ecb948e7d Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 10:59:32 +0800 Subject: [PATCH 069/274] vault backup: 2025-12-29 10:59:32 --- 素材/洛必达法则-注意事项.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 24da070..200c648 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -9,4 +9,4 @@ tags: 适当使用洛必达，不要一直用洛必达，有时用等价无穷小化简更方便$$\lim_{x \to 0} \frac{e^x-e^{-x}-2x}{\tan^{3}x}$$先利用等价无穷小，将 $\tan^{3}x$ 转化为 $x^3$ ，然后再使用洛必达 3. 在满足定理的某些情况下洛必达法则不能解决计算问题 - $$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ \ No newline at end of file + 下面的例子就会导致反复变为倒数$$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ -- 2.34.1 From ae5aaf1c0395459829e7001e4e78280ad5116768 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 11:01:39 +0800 Subject: [PATCH 070/274] vault backup: 2025-12-29 11:01:39 --- 素材/洛必达法则-注意事项.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 200c648..53a9cc7 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -10,3 +10,5 @@ tags: 3. 在满足定理的某些情况下洛必达法则不能解决计算问题 下面的例子就会导致反复变为倒数$$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ +4. 必须要求分子分母的导数极限都存在 + 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是先分子分母同时除以 $x$ \ No newline at end of file -- 2.34.1 From d1fdd1ed0f17c32344a9872adab4757e3f3fd60f Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 29 Dec 2025 11:02:01 +0800 Subject: [PATCH 071/274] vault backup: 2025-12-29 11:02:01 --- 素材/洛必达法则-注意事项.md | 1 + 1 file changed, 1 insertion(+) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 53a9cc7..c037805 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -10,5 +10,6 @@ tags: 3. 在满足定理的某些情况下洛必达法则不能解决计算问题 下面的例子就会导致反复变为倒数$$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ + 4. 必须要求分子分母的导数极限都存在 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是先分子分母同时除以 $x$ \ No newline at end of file -- 2.34.1 From 601dcecb9b907d4b7b8fe039390fedb1fb17558a Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 12:00:59 +0800 Subject: [PATCH 072/274] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E6=B4=9B?= =?UTF-8?q?=E5=BF=85=E8=BE=BE=E6=B3=95=E5=88=99=E7=9A=84=E8=AF=81=E6=98=8E?= =?UTF-8?q?=E5=92=8C=E6=B3=A8=E6=84=8F=E4=BA=8B=E9=A1=B9=EF=BC=8C=E5=8D=B3?= =?UTF-8?q?=E6=B5=81=E7=A8=8B=E5=9B=BE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 笔记分享/LaTeX（KaTeX）特殊输入.md | 2 +- 素材/洛必达法则-注意事项.md | 15 ++++++++++++++- 2 files changed, 15 insertions(+), 2 deletions(-) diff --git a/笔记分享/LaTeX（KaTeX）特殊输入.md b/笔记分享/LaTeX（KaTeX）特殊输入.md index 413f04d..877d957 100644 --- a/笔记分享/LaTeX（KaTeX）特殊输入.md +++ b/笔记分享/LaTeX（KaTeX）特殊输入.md @@ -21,4 +21,4 @@ $\int_1^5 x\mathrm{d}x$ 积分符号的上下记号，用limits套： -$\int\limits_{L}(x+y)\mathrm{d}s$ \ No newline at end of file +$\int\limits_{L}(x+y)\mathrm{d}s$ diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index c037805..958d58e 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -12,4 +12,17 @@ tags: 下面的例子就会导致反复变为倒数$$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ 4. 必须要求分子分母的导数极限都存在 - 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是先分子分母同时除以 $x$ \ No newline at end of file + 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是先分子分母同时除以 $x$ + +## **洛必达法则证明：** +1.$对于在(a,a+\delta)上连续可导函数f(x),g(x)，且g'(x)\neq0.如果\lim\limits_{x\to a^+}f(x)=0,\lim\limits_{x\to a^+}g(x)=0,$$且\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}存在(或为无穷)，则$$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}(或无穷)$$ +证明：令$$F(x)=\begin{cases}f(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},G(x)=\begin{cases}g(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},$$显然有$F(x),G(x)$在$[a,a+\delta)$上连续可导.$\forall x\in (a,a+\delta)$,由柯西中值定理，$\exists\xi\in(a,x)$，使得$$\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F'(\xi)}{G'(\xi)} \Rightarrow \frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)},$$令$\xi=a+\theta(x-a),\theta\in(0,1)$,有$$\frac{f(x)}{g(x)}=\frac{f'(a+\theta(x-a))}{g'(a+\theta(x-a))},$$左右同时取极限$x\to a^+$得$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}.$$得证. +2.类似地可以证明$x\to a^-,x\to a$的情况.若是$x\to +\infty$,则令$t=\frac{1}{x}$则$t\to0^+$，然后用上述结论；$x\to-\infty和x\to-\infty$类似. + +>[!warning] 不能使用洛必达的的情况： +>1、必须要是$\frac{0}{0}$或$\frac{\infty}{\infty}$形式的，其他形式必须转化成前两种形式才能使用洛必达法则； +>2、必须要$\lim\frac{f'(x)}{g'(x)}$存在才能使用； +>3、一些特殊的函数可能会无法使用洛必达，比如$\lim\limits_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$. + +运用洛必达法则解决其他不定式问题的流程： +$$\boxed{\infty-\infty}\overset{通分}{\Longrightarrow}\boxed{\frac{0}{0}或\frac{\infty}{\infty}}\overset{取倒数}{\Longleftarrow}\boxed{0\cdot\infty}\overset{取对数}{\Longleftarrow}\boxed{0^0或1^{\infty}或\infty^0}$$ -- 2.34.1 From 2531e1864195376e37dd3f3aa2fec48f98ad2d16 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 13:22:53 +0800 Subject: [PATCH 073/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E4=B8=80?= =?UTF-8?q?=E4=BA=9B=E6=8E=92=E7=89=88=EF=BC=8C=E5=B9=B6=E5=A2=9E=E5=8A=A0?= =?UTF-8?q?=E4=BA=86=E4=B8=A4=E9=81=93=E4=BE=8B=E9=A2=98=E5=92=8C=E9=83=A8?= =?UTF-8?q?=E5=88=86=E7=9A=84=E7=AD=94=E6=A1=88=EF=BC=8C=E6=9C=80=E5=90=8E?= =?UTF-8?q?=E7=9D=A1=E4=BA=86=E5=8D=88=E8=A7=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- Notice!!!（develop分支）.md | 2 +- 素材/洛必达法则-注意事项.md | 34 +++++++++++++++++++++----- 2 files changed, 29 insertions(+), 7 deletions(-) diff --git a/Notice!!!（develop分支）.md b/Notice!!!（develop分支）.md index 721f6ff..b9f59c6 100644 --- a/Notice!!!（develop分支）.md +++ b/Notice!!!（develop分支）.md @@ -1,5 +1,5 @@ # Nota Bene!!! 每次修改此仓库（也就是在左侧advanced-math-notes栏中的任何文件）时 -请先pull，commit +请先pull，再commit 确保自己修改的仓库是最新版本 diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 958d58e..9dbabd4 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -2,6 +2,12 @@ tags: - 素材 --- +## **洛必达法则证明：** +1.$对于在(a,a+\delta)上连续可导函数f(x),g(x)，且g'(x)\neq0.如果\lim\limits_{x\to a^+}f(x)=0,\lim\limits_{x\to a^+}g(x)=0,$$且\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}存在(或为无穷)，则$$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}(或无穷)$$ +证明：令$$F(x)=\begin{cases}f(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},G(x)=\begin{cases}g(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},$$显然有$F(x),G(x)$在$[a,a+\delta)$上连续可导.$\forall x\in (a,a+\delta)$,由柯西中值定理，$\exists\xi\in(a,x)$，使得$$\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F'(\xi)}{G'(\xi)} \Rightarrow \frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)},$$令$\xi=a+\theta(x-a),\theta\in(0,1)$,故当$x\to a^+$时，$\xi\to a^+$，从而两边取极限得： +$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{\xi\to a^+}\frac{f'(\xi)}{g'(\xi)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}.$$得证. +2.类似地可以证明$x\to a^-,x\to a$的情况.若是$x\to +\infty$,则令$t=\frac{1}{x}$,则$t\to0^+$，从而$$\lim\limits_{x\to+\infty}\frac{f(x)}{g(x)}=\lim\limits_{t\to0+}\frac{f(1/t)}{g(1/t)}=\lim\limits_{t\to0^+}\frac{f'(1/t)\cdot(-\frac{1}{t^2})}{g'(1/t)\cdot(-\frac{1}{t^2})}=\lim\limits_{t\to0^+}\frac{f'(1/t)}{g'(1/t)}=\lim\limits_{x\to+\infty}\frac{f'(x)}{g'(x)};$$$x\to-\infty和x\to\infty$类似. +## **注意事项：** 1. 使用时需要在等号上方写明是用的什么类型的洛必达，$\frac{0}{0}$？$\frac{\infty}{\infty}$？ 例如：$\lim\limits_{x\to\infty}\frac{\mathrm{e}^{2x}+1}{\mathrm{e}^{2x}-1} \overset{\frac{\infty}{\infty}}{=}\lim\limits_{x\to\infty}\frac{2\mathrm{e}^{2x}}{2\mathrm{e}^{2x}}=1$ @@ -12,17 +18,33 @@ tags: 下面的例子就会导致反复变为倒数$$\lim_{x \to +\infty} \frac{\sqrt{1+x^2}}{x}$$ 4. 必须要求分子分母的导数极限都存在 - 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是先分子分母同时除以 $x$ + 例如$$\lim_{x \to +\infty} \frac{x+\sin x}{x}$$正确的办法是转换成两个极限相加。 -## **洛必达法则证明：** -1.$对于在(a,a+\delta)上连续可导函数f(x),g(x)，且g'(x)\neq0.如果\lim\limits_{x\to a^+}f(x)=0,\lim\limits_{x\to a^+}g(x)=0,$$且\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}存在(或为无穷)，则$$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}(或无穷)$$ -证明：令$$F(x)=\begin{cases}f(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},G(x)=\begin{cases}g(x),x\in(a,a+\delta) \\ 0,x=a\end{cases},$$显然有$F(x),G(x)$在$[a,a+\delta)$上连续可导.$\forall x\in (a,a+\delta)$,由柯西中值定理，$\exists\xi\in(a,x)$，使得$$\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F'(\xi)}{G'(\xi)} \Rightarrow \frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)},$$令$\xi=a+\theta(x-a),\theta\in(0,1)$,有$$\frac{f(x)}{g(x)}=\frac{f'(a+\theta(x-a))}{g'(a+\theta(x-a))},$$左右同时取极限$x\to a^+$得$$\lim\limits_{x\to a^+}\frac{f(x)}{g(x)}=\lim\limits_{x\to a^+}\frac{f'(x)}{g'(x)}.$$得证. -2.类似地可以证明$x\to a^-,x\to a$的情况.若是$x\to +\infty$,则令$t=\frac{1}{x}$则$t\to0^+$，然后用上述结论；$x\to-\infty和x\to-\infty$类似. >[!warning] 不能使用洛必达的的情况： >1、必须要是$\frac{0}{0}$或$\frac{\infty}{\infty}$形式的，其他形式必须转化成前两种形式才能使用洛必达法则； >2、必须要$\lim\frac{f'(x)}{g'(x)}$存在才能使用； >3、一些特殊的函数可能会无法使用洛必达，比如$\lim\limits_{x\to\infty}\frac{x}{\sqrt{x^2+1}}$. -运用洛必达法则解决其他不定式问题的流程： +>[!tips] 运用洛必达法则解决其他不定式问题的流程： $$\boxed{\infty-\infty}\overset{通分}{\Longrightarrow}\boxed{\frac{0}{0}或\frac{\infty}{\infty}}\overset{取倒数}{\Longleftarrow}\boxed{0\cdot\infty}\overset{取对数}{\Longleftarrow}\boxed{0^0或1^{\infty}或\infty^0}$$ + +## **例题** + +>[!example] 求下列极限 +>(1)$\lim\limits_{x\to1}(1-x^2)\tan{\frac{\pi}{2}x}$; (2)$\lim\limits_{x\to0}\frac{1}{x^2}-\frac{1}{x\tan x};(3)\lim\limits_{x\to0}(x^2+2^x)^{\frac{1}{x}}.$ + +解： +(1)用等价无穷小：$$原式\overset{t=x-1}{=}-\lim\limits_{t\to0}t(t+2)\tan{\frac{\pi}{2}(t+1)}=2\lim\limits_{t\to0}\frac{t}{\tan{\frac{\pi}{2}t}}=\frac{4}{\pi}.$$ +用洛必达：$$原式=\lim\limits_{x\to1}\frac{1-x^2}{\cot{\frac{\pi}{2}x}}\overset{\frac{0}{0}}{=}\frac{2}{\pi}\lim\limits_{x\to1}\frac{-2x}{-\csc^2{\frac{\pi}{2}x}}=\frac{4}{\pi}.$$ +(2)$$原式=\lim\limits_{x\to0}\frac{\tan x-1}{x^2\tan x}=\lim\limits_{x\to0}\frac{\tan x-x}{x^3}\overset{\frac{0}{0}}{=}\lim\limits_{x\to0}\frac{\sec^2x-1}{3x^2}=\lim\limits_{x\to0}\frac{\tan^2x}{3x^2}=\frac{1}{3}.$$ +(3)$$\begin{aligned}原式&=\lim\limits_{x\to0}\mathrm{e}^{\frac{1}{x}\ln{(x^2)+2^x}}\\\\&=\mathrm{e}^{\lim\limits_{x\to0}\frac{\ln{(x^2+2^x)}}{x}}\\\\&\overset{\frac{0}{0}}{=}\mathrm{e}^{\lim\limits_{x\to0}\frac{2x+2^x\ln2}{x^2+2^x}}\\\\&=\mathrm{e}^{\ln2}=2.\end{aligned}$$ + +>[!example] 例题 +>设$f(x)$在$[0,1]$内二阶可导，且满足$$f(0)=0,f(1)=1,f(\frac{1}{2})>\frac{1}{4}.$$证明： +>(1)至少存在一点$\xi\in(0,1)$,使得$f''(\xi)<2$; +>(2)若$\forall x\in(0,1)$,有$f''(x)\neq2$,则当$x\in(0,1)$时，恒有$f(x)>x^2$. + +证明： +（1）令$F(x)=f(x)-x^2$,则$F(0)=F(1)=0,F(\frac{1}{2})>0$.显然$F(x)$在$[0,\frac{1}{2}]$和$[\frac{1}{2},1]$上都满足拉格朗日中值定理得条件，所以由拉格朗日中值定理得：$$\exists\xi_1\in(0,\frac{1}{2}),\xi_2\in(\frac{1}{2},1),有\frac{F(\frac{1}{2})-F(0)}{\frac{1}{2}-0}=F'(\xi_1)>0,\frac{F(1)-F(\frac{1}{2})}{1-\frac{1}{2}}=F'(\xi_2)<0,$$于是再对$F(x)$在$[\xi_1,\xi_2]$上用拉格朗日中值定理得：$$\exists\xi\in(\xi_1,\xi_2),有F''(\xi)=\frac{F(\xi_2)-F(\xi_1)}{\xi_2-\xi_1}<0,即f'(\xi)<2.$$ +（2）用反证法，假设$\exists \eta\in(0,1),f(\eta)\le \eta^2,则F(\eta)\le 0$, \ No newline at end of file -- 2.34.1 From d7afc80acca3e7ef53f9750110cf0f33f7d84d26 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:14:31 +0800 Subject: [PATCH 074/274] vault backup: 2025-12-29 20:14:31 --- .../2018-19线性代数期末考试卷.md | 67 +++++++++++++++++++ 1 file changed, 67 insertions(+) create mode 100644 编写小组/试卷/2018-19线性代数期末考试卷.md diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md new file mode 100644 index 0000000..5030791 --- /dev/null +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -0,0 +1,67 @@ +1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 + + (A) $AB - BA$; + (B) $AB + BA$; + (C) $BAB$; + (D) $(AB)^2$. + +2. 设 $A, B$ 是可逆矩阵，且 $A$ 与 $B$ 相似，则下列结论错误的是【 】 + + (A) $A^T$ 与 $B^T$ 相似； + (B) $A^{-1}$ 与 $B^{-1}$ 相似； + (C) $A + A^T$ 与 $B + B^T$ 相似； + (D) $A + A^{-1}$ 与 $B + B^{-1}$ 相似。 + +3. 设向量组 + $$ + \alpha_1 = (0, 0, c_1)^T,\quad + \alpha_2 = (0, 1, c_2)^T,\quad + \alpha_3 = (1, -1, c_3)^T,\quad + \alpha_4 = (-1, 1, c_4)^T, + $$ + 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是【 】 + + (A) $\alpha_1, \alpha_2, \alpha_3$; + (B) $\alpha_1, \alpha_2, \alpha_4$; + (C) $\alpha_1, \alpha_3, \alpha_4$; + (D) $\alpha_2, \alpha_3, \alpha_4$. + +4. 设 $A, B$ 为 $n$ 阶矩阵，则【 】 + + (A) $\text{rank}[A \ AB] = \text{rank} A$; + (B) $\text{rank}[A \ BA] = \text{rank} A$; + (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; + (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. + +5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足【 】 + + (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; + (B) 将 $A^*$ 的第一行加上第二行的 2 倍得到 $B^*$; + (C) 将 $A^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $B^*$; + (D) 将 $A^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $B^*$. + +6. 已知方程组 + $$ + \text{(I)} \quad + \begin{cases} + x_1 + 2x_2 + 3x_3 = 0, \\ + 2x_1 + 3x_2 + 5x_3 = 0, \\ + x_1 + x_2 + ax_3 = 0, + \end{cases} + $$ + 与 + $$ + \text{(II)} \quad + \begin{cases} + x_1 + bx_2 + cx_3 = 0, \\ + 2x_1 + b^2x_2 + (c+1)x_3 = 0 + \end{cases} + $$ + 同解，则【 】 + + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + +--- \ No newline at end of file -- 2.34.1 From 4bd3023f9bb58cfcc03d5ebba4bca2c4b9ffd1ed Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:16:39 +0800 Subject: [PATCH 075/274] vault backup: 2025-12-29 20:16:39 --- .../2018-19线性代数期末考试卷.md | 56 +++++++++++++++++++ 1 file changed, 56 insertions(+) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 5030791..30fc61c 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -64,4 +64,60 @@ (C) $a = 2, b = 0, c = 1$; (D) $a = 2, b = 1, c = 2$. + +7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积 + $$ + \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}. + $$ + +8. 设二阶矩阵 $A$ 有两个相异特征值，$\alpha_1, \alpha_2$ 是 $A$ 的线性无关的特征向量，且 $A^2 (\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2$，则 + $$ + |A| = \underline{\qquad\qquad}. + $$ + +9. 若向量组 + $$ + \alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T + $$ + 不能由向量组 + $$ + \beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 = (3,4,a)^T + $$ + 线性表示，则 + $$ + a = \underline{\qquad\qquad}. + $$ + +10. 设矩阵 + $$ + A = \begin{bmatrix} + 1 & a_1 & a_1^2 & a_1^3 \\ + 1 & a_2 & a_2^2 & a_2^3 \\ + 1 & a_3 & a_3^2 & a_3^3 \\ + 1 & a_4 & a_4^2 & a_4^3 + \end{bmatrix},\quad + x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},\quad + b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, + $$ + 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为 + $$ + \underline{\qquad\qquad\qquad\qquad}. + $$ + +11. 若 $n$ 阶实对称矩阵 $A$ 的特征值为 + $$ + \lambda_i = (-1)^i \quad (i=1,2,\dots,n), + $$ + 则 + $$ + A^{100} = \underline{\qquad\qquad\qquad\qquad}. + $$ + +12. 设 $n$ 阶矩阵 $A = [a_{ij}]_{n \times n}$，则二次型 + $f(x_1, x_2, \dots, x_n) = \sum_{i=1}^n (a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n)^2$ + 的矩阵为 + $$ + \underline{\qquad\qquad\qquad\qquad}. + $$ + --- \ No newline at end of file -- 2.34.1 From 2add3ec9a86fc69fabfccd781cca502b5e099beb Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:19:53 +0800 Subject: [PATCH 076/274] vault backup: 2025-12-29 20:19:53 --- .../2018-19线性代数期末考试卷.md | 53 ++++++++++++++++++- 1 file changed, 52 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 30fc61c..70304e2 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -120,4 +120,55 @@ \underline{\qquad\qquad\qquad\qquad}. $$ ---- \ No newline at end of file +13. （10 分）计算 $n$ 阶行列式 + + $$ + D_n = \begin{vmatrix} + 1 & 2 & 3 & \cdots & n-1 & n \\ + 2 & 1 & 2 & \cdots & n-2 & n-1 \\ + 3 & 2 & 1 & \cdots & n-3 & n-2 \\ + \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ + n-1 & n-2 & n-3 & \cdots & 1 & 2 \\ + n & n-1 & n-2 & \cdots & 2 & 1 + \end{vmatrix}. + $$ + + + + + + + +14. （10 分）设 + $$ + \alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T + $$ + 和 + $$ + \beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T + $$ + 是 $\mathbb{R}^3$ 的两组基，求向量 + $$ + u = \alpha_1 + 2\alpha_2 - 3\alpha_3 + $$ + 在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 + + + +15. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 + + （1）证明 $A - E$ 可逆； + + （2）证明 $AB = BA$； + + （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； + + （4）若矩阵 + $$ + B = \begin{bmatrix} + 1 & -3 & 0 \\ + 2 & 1 & 0 \\ + 0 & 0 & 2 + \end{bmatrix}, + $$ + 求矩阵 $A$。 \ No newline at end of file -- 2.34.1 From e682981283b1726ee610669915be2b5c5210a8ce Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:26:04 +0800 Subject: [PATCH 077/274] vault backup: 2025-12-29 20:26:04 --- .../2018-19线性代数期末考试卷.md | 22 +++++++++++++++++-- 1 file changed, 20 insertions(+), 2 deletions(-) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 70304e2..380ce73 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -120,7 +120,7 @@ \underline{\qquad\qquad\qquad\qquad}. $$ -13. （10 分）计算 $n$ 阶行列式 +13. （20 分）计算 下面两个$n$ 阶行列式 $$ D_n = \begin{vmatrix} @@ -134,7 +134,15 @@ $$ +$$ +\begin{vmatrix} +1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n \\ +1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n \\ +\vdots & \vdots & \ddots & \vdots \\ +1+x_n & 1+x_n^2 & \cdots & 1+x_n^n +\end{vmatrix} +$$ @@ -171,4 +179,14 @@ 0 & 0 & 2 \end{bmatrix}, $$ - 求矩阵 $A$。 \ No newline at end of file + 求矩阵 $A$。 + + + + + + + + + +16. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 -- 2.34.1 From b5b95030d28de4c75dbbce5303a9bb86f926130f Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:41:44 +0800 Subject: [PATCH 078/274] vault backup: 2025-12-29 20:41:44 --- 编写小组/试卷/2018-19线性代数期末考试卷.md | 8 +++----- 1 file changed, 3 insertions(+), 5 deletions(-) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 380ce73..ecaddf3 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -70,10 +70,8 @@ \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}. $$ -8. 设二阶矩阵 $A$ 有两个相异特征值，$\alpha_1, \alpha_2$ 是 $A$ 的线性无关的特征向量，且 $A^2 (\alpha_1 + \alpha_2) = \alpha_1 + \alpha_2$，则 - $$ - |A| = \underline{\qquad\qquad}. - $$ +8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 + 9. 若向量组 $$ @@ -120,7 +118,7 @@ \underline{\qquad\qquad\qquad\qquad}. $$ -13. （20 分）计算 下面两个$n$ 阶行列式 +13. （10 分）计算 下面两个$n$ 阶行列式 $$ D_n = \begin{vmatrix} -- 2.34.1 From a87b56d2b1121792cf5acaf6aa3cbf95fe12d5cd Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:46:52 +0800 Subject: [PATCH 079/274] vault backup: 2025-12-29 20:46:52 --- .../2018-19线性代数期末考试卷.md | 48 ++++++++++++++++--- 1 file changed, 42 insertions(+), 6 deletions(-) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index ecaddf3..00835e7 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -5,13 +5,30 @@ (C) $BAB$; (D) $(AB)^2$. -2. 设 $A, B$ 是可逆矩阵，且 $A$ 与 $B$ 相似，则下列结论错误的是【 】 - - (A) $A^T$ 与 $B^T$ 相似； - (B) $A^{-1}$ 与 $B^{-1}$ 相似； - (C) $A + A^T$ 与 $B + B^T$ 相似； - (D) $A + A^{-1}$ 与 $B + B^{-1}$ 相似。 +2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 +$$ +\varepsilon_1 = e_1 + 5e_2,\quad \varepsilon_2 = e_2, +$$ +则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 +$$ +(A) \begin{bmatrix} +-1 & 0 \\ +5 & -1 +\end{bmatrix} \quad +(B) \begin{bmatrix} +0 & -1 \\ +-6 & 0 +\end{bmatrix} \quad +(C) \begin{bmatrix} +1 & 0 \\ +-5 & -1 +\end{bmatrix} \quad +(D) \begin{bmatrix} +1 & 0 \\ +-5 & 1 +\end{bmatrix}. +$$ 3. 设向量组 $$ \alpha_1 = (0, 0, c_1)^T,\quad @@ -73,6 +90,25 @@ 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 + +解析： +步骤1：分析矩阵A的幂次规律 +先计算$A^2$： + +$$A^2 = \begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix} = \begin{bmatrix}18&-6\\-54&18\end{bmatrix} = 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ + +由此递推： +- $$A^3 = A^2 \cdot A = 6A \cdot A = 6A^2 = 6\times6A = 6^2A$$ +- 归纳可得当$n \geq 1$时，$A^n = 6^{n-1}A$ + +步骤2：写出最终表达式 + +将A代入得： +$$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = \begin{bmatrix}3\times6^{n-1}&-6^{n-1}\\-9\times6^{n-1}&3\times6^{n-1}\end{bmatrix}$$ + +答案：$$\boldsymbol{6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix}}$$ + + 9. 若向量组 $$ \alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T -- 2.34.1 From f4a97f31363b6271d9d858b5a626ada84389f3eb Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:48:10 +0800 Subject: [PATCH 080/274] vault backup: 2025-12-29 20:48:10 --- .../试卷/2018-19线性代数期末考试卷.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 00835e7..f77de62 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -224,3 +224,17 @@ $$ 16. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 + + +解析： +因为n=4，$n-\text{rank}A=2$，所以$\text{rank}A=2$。 +对A施行初等行变换，得 +$$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&t-2&-1&-1\end{bmatrix}$$ + +$$\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}\to\begin{bmatrix}1&0&1-2t&2-2t\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}$$ + + +要使$\text{rank}A=2$，则必有t=1。 +此时，与Ax=0同解的方程组为$\begin{cases}x_1=x_3\\x_2=-x_3-x_4\end{cases}$，得基础解系为 +$$\boldsymbol{\xi}_1=\begin{bmatrix}1\\-1\\1\\0\end{bmatrix},\ \boldsymbol{\xi}_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ +方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ \ No newline at end of file -- 2.34.1 From 165c8c9f430083134a467f95887b7204551dc544 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 20:50:30 +0800 Subject: [PATCH 081/274] vault backup: 2025-12-29 20:50:30 --- ...代数期末考试卷.md => 1231线性代数考试卷.md} | 3 +++ 1 file changed, 3 insertions(+) rename 编写小组/试卷/{2018-19线性代数期末考试卷.md => 1231线性代数考试卷.md} (97%) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/1231线性代数考试卷.md similarity index 97% rename from 编写小组/试卷/2018-19线性代数期末考试卷.md rename to 编写小组/试卷/1231线性代数考试卷.md index f77de62..6b3f91d 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -1,3 +1,5 @@ +## 一、选择题，共六道，每题3分，共18分 + 1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 (A) $AB - BA$; @@ -81,6 +83,7 @@ $$ (C) $a = 2, b = 0, c = 1$; (D) $a = 2, b = 1, c = 2$. +## 二、填空题，共六道，每题3分，共18分 7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积 $$ -- 2.34.1 From e1808b06f038d3b651fecc098307fa3d8ba004c5 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 20:51:48 +0800 Subject: [PATCH 082/274] vault backup: 2025-12-29 20:51:48 --- 编写小组/试卷/2018-19线性代数期末考试卷.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/2018-19线性代数期末考试卷.md b/编写小组/试卷/2018-19线性代数期末考试卷.md index 30fc61c..f027edf 100644 --- a/编写小组/试卷/2018-19线性代数期末考试卷.md +++ b/编写小组/试卷/2018-19线性代数期末考试卷.md @@ -120,4 +120,6 @@ \underline{\qquad\qquad\qquad\qquad}. $$ ---- \ No newline at end of file +设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. +(1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； +(2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. -- 2.34.1 From eee4a072b348380869d03cfd2ba360193ab37c62 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 20:55:52 +0800 Subject: [PATCH 083/274] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E4=B8=80?= =?UTF-8?q?=E9=81=93=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/1231线性代数考试卷.md | 9 ++++++++- 1 file changed, 8 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 6b3f91d..c4f8714 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -240,4 +240,11 @@ $$\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}\to\begi 要使$\text{rank}A=2$，则必有t=1。 此时，与Ax=0同解的方程组为$\begin{cases}x_1=x_3\\x_2=-x_3-x_4\end{cases}$，得基础解系为 $$\boldsymbol{\xi}_1=\begin{bmatrix}1\\-1\\1\\0\end{bmatrix},\ \boldsymbol{\xi}_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ -方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ \ No newline at end of file +方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ +设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. +(1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； +(2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. +解析： +(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. + +(2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. -- 2.34.1 From 7d403c0792941e840fe9a69bf13d7e1daa9dab4e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 20:58:00 +0800 Subject: [PATCH 084/274] vault backup: 2025-12-29 20:57:59 --- 编写小组/试卷/{期中试卷.md => 线代期中试卷.md} | 0 .../试卷/{期中试卷解析.md => 线代期中试卷解析.md} | 0 2 files changed, 0 insertions(+), 0 deletions(-) rename 编写小组/试卷/{期中试卷.md => 线代期中试卷.md} (100%) rename 编写小组/试卷/{期中试卷解析.md => 线代期中试卷解析.md} (100%) diff --git a/编写小组/试卷/期中试卷.md b/编写小组/试卷/线代期中试卷.md similarity index 100% rename from 编写小组/试卷/期中试卷.md rename to 编写小组/试卷/线代期中试卷.md diff --git a/编写小组/试卷/期中试卷解析.md b/编写小组/试卷/线代期中试卷解析.md similarity index 100% rename from 编写小组/试卷/期中试卷解析.md rename to 编写小组/试卷/线代期中试卷解析.md -- 2.34.1 From e6fbae7728617761b363aa4d6ff96c72c61a7126 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 29 Dec 2025 20:59:15 +0800 Subject: [PATCH 085/274] =?UTF-8?q?=E6=94=B9=E4=BA=86=E4=B8=A4=E4=B8=AA?= =?UTF-8?q?=E6=A0=87=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/高数期中试卷.md | 46 ++++++++++++++ .../试卷/高数期中试卷解析.md | 62 +++++++++++++++++++ 2 files changed, 108 insertions(+) create mode 100644 编写小组/试卷/高数期中试卷.md create mode 100644 编写小组/试卷/高数期中试卷解析.md diff --git a/编写小组/试卷/高数期中试卷.md b/编写小组/试卷/高数期中试卷.md new file mode 100644 index 0000000..ffed476 --- /dev/null +++ b/编写小组/试卷/高数期中试卷.md @@ -0,0 +1,46 @@ +#官方试卷 +时量：120分钟 满分：100分 +#### 一、单选题（共5小题，每小题3分，共15分） +1. 若$\lim\limits_{n\rightarrow\infty}a_n=4$，则当$n$充分大时，恒有 + A. $|a_n|\le 1$ + B. $|a_n|>2$ + C. $|a_n|<2$ + D. $|a_n|>4$ +2. 已知$g(x)=\frac{1}{x^2}$，复合函数$y=f(g(x))$对$x$的导数为$-\frac{1}{2x}$，则$f'(\frac{1}{2})$的值为 + A. $1$ + B. $2$ + C. $\frac{\sqrt 2}{4}$ + D. $\frac{1}{2}$ +3. 设$f(x)$可导且$f'(x_0)=\frac{1}{3}$，则当$\Delta x \rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 + A. 同阶无穷小 + B. 低阶无穷小 + C. 等价无穷小 + D. 高阶无穷小 +4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 + A. 无穷大量 + B. 无穷小量 + C. 有界变量 + D. 无界变量 +5. 下列四个级数，**发散**的是 + A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ + B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ + C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ + D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ +#### 二、填空题（共5小题，每小题3分，共15分） +6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. +7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. +8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. +9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. +10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. +#### 三、解答与证明题（11~19小题，共70分） +11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. +12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. +13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. +14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. +15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. +16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. +17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. +18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] +19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ + (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） + (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file diff --git a/编写小组/试卷/高数期中试卷解析.md b/编写小组/试卷/高数期中试卷解析.md new file mode 100644 index 0000000..4874412 --- /dev/null +++ b/编写小组/试卷/高数期中试卷解析.md @@ -0,0 +1,62 @@ +#官方试卷 +#民间答案 +#### 一、单选题（共5小题，每小题3分，共15分） +1. 若$\lim\limits_{n\rightarrow\infty}a_n=4$，则当$n$充分大时，恒有 + A. $|a_n|\le 1$ + B. $|a_n|>2$ + C. $|a_n|<2$ + D. $|a_n|>4$ +> **答案：B** +> 解析：保号性 +> 若$\lim\limits_{n\rightarrow\infty}a_n=4$，按照定义，$\forall \epsilon >0,\exists N, n>N\text{时},|a_n-4|<\epsilon$， +> 取$\epsilon=2$，则$|a_n-4|<\epsilon$，则$2<a_n<6$，B选项符合题意 +2. 已知$g(x)=\frac{1}{x^2}$，复合函数$y=f(g(x))$对$x$的导数为$-\frac{1}{2x}$，则$f'(\frac{1}{2})$的值为 + A. $1$ + B. $2$ + C. $\frac{\sqrt 2}{4}$ + D. $\frac{1}{2}$ +>**答案：D** +>解析：复合函数求导法则 +>$y'=f'(g(x))g'(x)=\frac{-2}{x^3}f'(\frac{1}{x^2})=-\frac{1}{2x}\Rightarrow f'(\frac{1}{x^2})=\frac{x^2}{4}$，即$f'(\frac{1}{x})=\frac{x}{4}(x> 0)$ +>代入$x=2$得：$f'(\frac{1}{2})=\frac{1}{2}$ +3. 设$f(x)$可导且$f'(x_0)=\frac{1}{3}$，则当$\Delta x \rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 + A. 同阶无穷小 + B. 低阶无穷小 + C. 等价无穷小 + D. 高阶无穷小 +> **答案：A** +> 解析：同阶无穷小；微分的定义 +> $\Delta x \rightarrow 0$时，$\Delta x \sim \mathrm{d}x$，又$\mathrm{d}y|_{x=x_0}=\frac{1}{3}\mathrm{d}x$，故$\mathrm{d}y$与$\Delta{x}$是同阶不等价的无穷小 +4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 + A. 无穷大量 + B. 无穷小量 + C. 有界变量 + D. 无界变量 +>**答案：D** +>解析：易错点10-无穷大与无界的辨析 +>$\lim\limits_{k\to\infty}a_{2k}=+\infty$，因此$a_n$无界；然而，$\lim\limits_{k\to\infty}a_{2k+1}=0$，因此$a_n(n\to\infty)$不是无穷大 +5. 下列四个级数，**发散**的是 + A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ + B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ + C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ + D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ +>**答案：C** +> +#### 二、填空题（共5小题，每小题3分，共15分） +6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. +7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. +8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. +9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. +10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. +#### 三、解答与证明题（11~19小题，共70分） +11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. +12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. +13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. +14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. +15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. +16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. +17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. +18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] +19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ + (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） + (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file -- 2.34.1 From 4e3dbbe4ce84574e122e58de315ffdc7b9eed475 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:02:14 +0800 Subject: [PATCH 086/274] vault backup: 2025-12-29 21:02:13 --- .../试卷/1231线性代数考试卷.md | 34 ++++++++++++------- 1 file changed, 22 insertions(+), 12 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index c4f8714..8a34ca1 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -95,21 +95,23 @@ $$ 解析： -步骤1：分析矩阵A的幂次规律 先计算$A^2$： -$$A^2 = \begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix} = \begin{bmatrix}18&-6\\-54&18\end{bmatrix} = 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ +$$A^2 += \begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ +$$= \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix} +$$$$= \begin{bmatrix}18&-6\\-54&18\end{bmatrix} $$ +$$= 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ 由此递推： - $$A^3 = A^2 \cdot A = 6A \cdot A = 6A^2 = 6\times6A = 6^2A$$ - 归纳可得当$n \geq 1$时，$A^n = 6^{n-1}A$ -步骤2：写出最终表达式 + 将A代入得： -$$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = \begin{bmatrix}3\times6^{n-1}&-6^{n-1}\\-9\times6^{n-1}&3\times6^{n-1}\end{bmatrix}$$ +$$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ -答案：$$\boldsymbol{6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix}}$$ 9. 若向量组 @@ -157,7 +159,10 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = \begin{bmatrix}3\times6^ \underline{\qquad\qquad\qquad\qquad}. $$ -13. （10 分）计算 下面两个$n$ 阶行列式 + +--- + +12. （10 分）计算 下面的$n$ 阶行列式 $$ D_n = \begin{vmatrix} @@ -182,6 +187,16 @@ $$ $$ +13. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. +(1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； +(2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. + + + +解析： +(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. + +(2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. 14. （10 分）设 @@ -200,6 +215,7 @@ $$ + 15. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 （1）证明 $A - E$ 可逆； @@ -241,10 +257,4 @@ $$\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}\to\begi 此时，与Ax=0同解的方程组为$\begin{cases}x_1=x_3\\x_2=-x_3-x_4\end{cases}$，得基础解系为 $$\boldsymbol{\xi}_1=\begin{bmatrix}1\\-1\\1\\0\end{bmatrix},\ \boldsymbol{\xi}_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ 方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ -设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. -(1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； -(2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. -解析： -(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. -(2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. -- 2.34.1 From 2e0c402add5d53e2ab44d9fc25c2e3a4aaaca33f Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:08:14 +0800 Subject: [PATCH 087/274] vault backup: 2025-12-29 21:08:14 --- .../试卷/1231线性代数考试卷.md | 108 ++++++++++++++++-- 1 file changed, 100 insertions(+), 8 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 8a34ca1..d618d20 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -185,19 +185,21 @@ $$ \end{vmatrix} $$ - +--- 13. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. -(1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； -(2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. - + + (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； + (2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. +--- 解析： (1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. (2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. +--- 14. （10 分）设 $$ @@ -213,7 +215,97 @@ $$ $$ 在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 +--- + +解析： + +已知： + +$$ +A = (\alpha_1, \alpha_2, \alpha_3) = +\begin{bmatrix} +1 & 2 & 1 \\ +0 & 1 & 1 \\ +-1 & 1 & 1 +\end{bmatrix}, +$$ + +$$ +B = (\beta_1, \beta_2, \beta_3) = +\begin{bmatrix} +0 & -1 & 0 \\ +1 & 1 & 2 \\ +1 & 0 & 1 +\end{bmatrix}. +$$ + +设 $u$ 在基 $\alpha_1, \alpha_2, \alpha_3$ 下的坐标为 $x = (1, 2, -3)^T$，在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 $y$，则 + +$$ +u = (\alpha_1, \alpha_2, \alpha_3) x = (\beta_1, \beta_2, \beta_3) y, +$$ + +即 + +$$ +Ax = By. +$$ + +因为 $B$ 可逆，所以 + +$$ +y = B^{-1} A x. +$$ + +用增广矩阵求解 $y$： + +$$ +(B, Ax) = +\begin{bmatrix} +0 & -1 & 0 & \vert & 2 \\ +1 & 1 & 2 & \vert & -1 \\ +1 & 0 & 1 & \vert & -2 +\end{bmatrix} +$$ + +作行初等变换： + +$$ +\begin{aligned} +&\rightarrow +\begin{bmatrix} +1 & 0 & 1 & \vert & -2 \\ +0 & 1 & 1 & \vert & 1 \\ +0 & -1 & 0 & \vert & 2 +\end{bmatrix} \\ +&\rightarrow +\begin{bmatrix} +1 & 0 & 1 & \vert & -2 \\ +0 & 1 & 1 & \vert & 1 \\ +0 & 0 & 1 & \vert & 3 +\end{bmatrix} \\ +&\rightarrow +\begin{bmatrix} +1 & 0 & 0 & \vert & -5 \\ +0 & 1 & 0 & \vert & -2 \\ +0 & 0 & 1 & \vert & 3 +\end{bmatrix}. +\end{aligned} +$$ + +因此向量 + +$$ +u = \alpha_1 + 2\alpha_2 - 3\alpha_3 +$$ +在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 + +$$ +y = (-5, -2, 3)^T. +$$ + +--- 15. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 @@ -235,15 +327,16 @@ $$ 求矩阵 $A$。 +--- - - +--- 16. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 +--- 解析： 因为n=4，$n-\text{rank}A=2$，所以$\text{rank}A=2$。 @@ -256,5 +349,4 @@ $$\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}\to\begi 要使$\text{rank}A=2$，则必有t=1。 此时，与Ax=0同解的方程组为$\begin{cases}x_1=x_3\\x_2=-x_3-x_4\end{cases}$，得基础解系为 $$\boldsymbol{\xi}_1=\begin{bmatrix}1\\-1\\1\\0\end{bmatrix},\ \boldsymbol{\xi}_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ -方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ - +方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ \ No newline at end of file -- 2.34.1 From 15c2439b3e604098544902e8e20b947ebbcf892d Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:22:08 +0800 Subject: [PATCH 088/274] vault backup: 2025-12-29 21:22:08 --- .../试卷/1231线性代数考试卷.md | 46 ++++++++++++++----- 1 file changed, 35 insertions(+), 11 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index d618d20..e2eb5ce 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -31,6 +31,38 @@ $$ -5 & 1 \end{bmatrix}. $$ + +--- + +解析： + +$$ +\varepsilon_1 = e_1 + 5e_2,\quad \varepsilon_2 = 0e_1 + 1e_2. +$$ + +把它们按列排成矩阵形式： + +$$ +[\varepsilon_1, \varepsilon_2] = [e_1, e_2] +\begin{pmatrix} +1 & 0 \\ +5 & 1 +\end{pmatrix}. +$$ + +基变换矩阵为： + +$$ +T = \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}. +$$ +$$ +\quad +T^{-1} = \begin{pmatrix} 1 & 0 \\ -5 & 1 \end{pmatrix}. +$$ +$\quad T^{-1} = \begin{pmatrix} 1 & 0 \\ -5 & 1 \end{pmatrix}$是坐标变换矩阵，即为过渡矩阵，选D + +--- + 3. 设向量组 $$ \alpha_1 = (0, 0, c_1)^T,\quad @@ -143,18 +175,10 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ \underline{\qquad\qquad\qquad\qquad}. $$ -11. 若 $n$ 阶实对称矩阵 $A$ 的特征值为 - $$ - \lambda_i = (-1)^i \quad (i=1,2,\dots,n), - $$ - 则 - $$ - A^{100} = \underline{\qquad\qquad\qquad\qquad}. - $$ -12. 设 $n$ 阶矩阵 $A = [a_{ij}]_{n \times n}$，则二次型 - $f(x_1, x_2, \dots, x_n) = \sum_{i=1}^n (a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n)^2$ - 的矩阵为 + +11. +12. $$ \underline{\qquad\qquad\qquad\qquad}. $$ -- 2.34.1 From 4e4e32d57ddc8e9dd4e8a54b5e2e219f02d7b88e Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:22:39 +0800 Subject: [PATCH 089/274] vault backup: 2025-12-29 21:22:39 --- ...试卷.md => 1231线性代数考试卷（解析版）.md} | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) rename 编写小组/试卷/{1231线性代数考试卷.md => 1231线性代数考试卷（解析版）.md} (99%) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md similarity index 99% rename from 编写小组/试卷/1231线性代数考试卷.md rename to 编写小组/试卷/1231线性代数考试卷（解析版）.md index e2eb5ce..66b54ff 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -124,7 +124,7 @@ $\quad T^{-1} = \begin{pmatrix} 1 & 0 \\ -5 & 1 \end{pmatrix}$是坐标变换矩 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 - +--- 解析： 先计算$A^2$： @@ -143,7 +143,7 @@ $$= 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ 将A代入得： $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ - + --- 9. 若向量组 -- 2.34.1 From e01b539f7cbbc4c56cf0b8bd020250e08e924a85 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:29:15 +0800 Subject: [PATCH 090/274] vault backup: 2025-12-29 21:29:15 --- .../1231线性代数考试卷（解析版）.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 66b54ff..5c0a0d3 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -174,10 +174,24 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ $$ \underline{\qquad\qquad\qquad\qquad}. $$ +--- +解析： + +一眼顶针，鉴定为： $$ + x = (1,0,0,0)^T$$ +--- 11. + +$$ +A^{k} = 0 +$$ + +$$ +k= \underline{\qquad\qquad\qquad\qquad}. + $$ 12. $$ \underline{\qquad\qquad\qquad\qquad}. -- 2.34.1 From 8b4d3b1c479f80509efe0c8651ad344d33268589 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:42:34 +0800 Subject: [PATCH 091/274] vault backup: 2025-12-29 21:42:33 --- ...231线性代数考试卷（解析版）.md | 149 +++++++++++++++++- 1 file changed, 145 insertions(+), 4 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 5c0a0d3..9f5cde8 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -159,7 +159,9 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ a = \underline{\qquad\qquad}. $$ -10. 设矩阵 +--- + +9. 设矩阵 $$ A = \begin{bmatrix} 1 & a_1 & a_1^2 & a_1^3 \\ @@ -200,10 +202,10 @@ k= \underline{\qquad\qquad\qquad\qquad}. --- -12. （10 分）计算 下面的$n$ 阶行列式 +12. （20 分）计算 下面的两个$n$阶行列式 $$ - D_n = \begin{vmatrix} + K_n = \begin{vmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 1 & 2 & \cdots & n-2 & n-1 \\ 3 & 2 & 1 & \cdots & n-3 & n-2 \\ @@ -215,7 +217,7 @@ k= \underline{\qquad\qquad\qquad\qquad}. $$ -\begin{vmatrix} + M_n =\begin{vmatrix} 1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n \\ 1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n \\ \vdots & \vdots & \ddots & \vdots \\ @@ -224,7 +226,146 @@ $$ $$ --- +--- + +解析 +（1）$K_n$： +从第 $n-1$ 行开始，依次乘以 $(-1)$ 加到下一行，再把第 $n$ 列加到前面各列，得 + +$$ +\begin{aligned} +K_n &= +\begin{vmatrix} +1 & 2 & 3 & \cdots & n-1 & n \\ +2 & 1 & 2 & \cdots & n-2 & n-1 \\ +3 & 2 & 1 & \cdots & n-3 & n-2 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +n-1 & n-2 & n-3 & \cdots & 1 & 2 \\ +n & n-1 & n-2 & \cdots & 2 & 1 +\end{vmatrix} \\[4pt] +&= +\begin{vmatrix} +1 & 2 & 3 & \cdots & n-1 & n \\ +1 & -1 & -1 & \cdots & -1 & -1 \\ +1 & 1 & -1 & \cdots & -1 & -1 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +1 & 1 & 1 & \cdots & -1 & -1 \\ +1 & 1 & 1 & \cdots & 1 & -1 +\end{vmatrix} + +\end{aligned} +$$ + +继续化简： + +$$ +\begin{aligned} +&= +\begin{vmatrix} +n+1 & n+2 & n+3 & \cdots & 2n-1 & n \\ +0 & -2 & -2 & \cdots & -2 & -1 \\ +0 & 0 & -2 & \cdots & -2 & -1 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +0 & 0 & 0 & \cdots & -2 & -1 \\ +0 & 0 & 0 & \cdots & 0 & -1 +\end{vmatrix} + +\end{aligned} +$$ + +这是一个上三角行列式，因此 + +$$ +D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1) + +$$ +--- + +$$ +A = (\alpha_1, \alpha_2, \alpha_3) = +\begin{bmatrix} +1 & 2 & 1 \\ +0 & 1 & 1 \\ +-1 & 1 & 1 +\end{bmatrix}, +$$ + +$$ +B = (\beta_1, \beta_2, \beta_3) = +\begin{bmatrix} +0 & -1 & 0 \\ +1 & 1 & 2 \\ +1 & 0 & 1 +\end{bmatrix}. +$$ + +设 $u$ 在基 $\alpha_1, \alpha_2, \alpha_3$ 下的坐标为 $x = (1, 2, -3)^T$，在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 $y$，则 + +$$ +u = (\alpha_1, \alpha_2, \alpha_3) x = (\beta_1, \beta_2, \beta_3) y, +$$ + +即 + +$$ +Ax = By. +$$ + +因为 $B$ 可逆，所以 + +$$ +y = B^{-1} A x. +$$ +用增广矩阵求解 $y$： + +$$ +(B, Ax) = +\begin{bmatrix} +0 & -1 & 0 & \vert & 2 \\ +1 & 1 & 2 & \vert & -1 \\ +1 & 0 & 1 & \vert & -2 +\end{bmatrix} +$$ + +作行初等变换： + +$$ +\begin{aligned} +&\rightarrow +\begin{bmatrix} +1 & 0 & 1 & \vert & -2 \\ +0 & 1 & 1 & \vert & 1 \\ +0 & -1 & 0 & \vert & 2 +\end{bmatrix} \\ +&\rightarrow +\begin{bmatrix} +1 & 0 & 1 & \vert & -2 \\ +0 & 1 & 1 & \vert & 1 \\ +0 & 0 & 1 & \vert & 3 +\end{bmatrix} \\ +&\rightarrow +\begin{bmatrix} +1 & 0 & 0 & \vert & -5 \\ +0 & 1 & 0 & \vert & -2 \\ +0 & 0 & 1 & \vert & 3 +\end{bmatrix}. +\end{aligned} +$$ + +因此向量 + +$$ +u = \alpha_1 + 2\alpha_2 - 3\alpha_3 +$$ + +在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 + +$$ +y = (-5, -2, 3)^T. +$$ + +--- 13. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； -- 2.34.1 From fb32fd9e2f07fe98420c4e61e28267aa620c4ab7 Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:43:28 +0800 Subject: [PATCH 092/274] vault backup: 2025-12-29 21:43:28 --- ...231线性代数考试卷（解析版）.md | 84 +------------------ 1 file changed, 1 insertion(+), 83 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 9f5cde8..4cc5a95 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -202,7 +202,7 @@ k= \underline{\qquad\qquad\qquad\qquad}. --- -12. （20 分）计算 下面的两个$n$阶行列式 +13. （20 分）计算 下面的两个$n$阶行列式 $$ K_n = \begin{vmatrix} @@ -281,89 +281,7 @@ D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1) $$ --- -$$ -A = (\alpha_1, \alpha_2, \alpha_3) = -\begin{bmatrix} -1 & 2 & 1 \\ -0 & 1 & 1 \\ --1 & 1 & 1 -\end{bmatrix}, -$$ - -$$ -B = (\beta_1, \beta_2, \beta_3) = -\begin{bmatrix} -0 & -1 & 0 \\ -1 & 1 & 2 \\ -1 & 0 & 1 -\end{bmatrix}. -$$ - -设 $u$ 在基 $\alpha_1, \alpha_2, \alpha_3$ 下的坐标为 $x = (1, 2, -3)^T$，在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 $y$，则 - -$$ -u = (\alpha_1, \alpha_2, \alpha_3) x = (\beta_1, \beta_2, \beta_3) y, -$$ - -即 - -$$ -Ax = By. -$$ - -因为 $B$ 可逆，所以 - -$$ -y = B^{-1} A x. -$$ - -用增广矩阵求解 $y$： - -$$ -(B, Ax) = -\begin{bmatrix} -0 & -1 & 0 & \vert & 2 \\ -1 & 1 & 2 & \vert & -1 \\ -1 & 0 & 1 & \vert & -2 -\end{bmatrix} -$$ -作行初等变换： - -$$ -\begin{aligned} -&\rightarrow -\begin{bmatrix} -1 & 0 & 1 & \vert & -2 \\ -0 & 1 & 1 & \vert & 1 \\ -0 & -1 & 0 & \vert & 2 -\end{bmatrix} \\ -&\rightarrow -\begin{bmatrix} -1 & 0 & 1 & \vert & -2 \\ -0 & 1 & 1 & \vert & 1 \\ -0 & 0 & 1 & \vert & 3 -\end{bmatrix} \\ -&\rightarrow -\begin{bmatrix} -1 & 0 & 0 & \vert & -5 \\ -0 & 1 & 0 & \vert & -2 \\ -0 & 0 & 1 & \vert & 3 -\end{bmatrix}. -\end{aligned} -$$ - -因此向量 - -$$ -u = \alpha_1 + 2\alpha_2 - 3\alpha_3 -$$ - -在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 - -$$ -y = (-5, -2, 3)^T. -$$ --- 13. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. -- 2.34.1 From 83e174560691958fd3ad120d487dda2a602486fd Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:43:32 +0800 Subject: [PATCH 093/274] vault backup: 2025-12-29 21:43:32 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 4cc5a95..99cb10f 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -225,7 +225,7 @@ $$ \end{vmatrix} $$ ---- + --- 解析 -- 2.34.1 From 04c9cc12e59f40c0446cfe8da95687d96c34056d Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:44:32 +0800 Subject: [PATCH 094/274] vault backup: 2025-12-29 21:44:32 --- .../试卷/1231线性代数考试卷（解析版）.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 99cb10f..90a74a3 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -199,7 +199,7 @@ k= \underline{\qquad\qquad\qquad\qquad}. \underline{\qquad\qquad\qquad\qquad}. $$ - +## 三、解答题，共五道，共64分 --- 13. （20 分）计算 下面的两个$n$阶行列式 @@ -284,7 +284,7 @@ $$ --- -13. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. +14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； (2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. @@ -298,7 +298,7 @@ $$ --- -14. （10 分）设 +15. （10 分）设 $$ \alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T $$ @@ -405,7 +405,7 @@ $$ --- -15. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 +16. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 （1）证明 $A - E$ 可逆； @@ -431,7 +431,7 @@ $$ --- -16. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 +17. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 --- -- 2.34.1 From 2959befb1fd7f952547ad3f9a3f09cd97d8d640a Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Mon, 29 Dec 2025 21:46:40 +0800 Subject: [PATCH 095/274] vault backup: 2025-12-29 21:46:40 --- ...231线性代数考试卷（解析版）.md | 64 +++++++++++++++++++ 1 file changed, 64 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 90a74a3..e476d63 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -427,7 +427,71 @@ $$ --- +**【解】** +**(1)** +由 $AB = A + B$ 得 $(A - E)(B - E) = E$，因此 $A - E$ 可逆。 +$$\text{……3 分}$$ + +**(2)** +由 $(A - E)(B - E) = E$ 得 $(B - E)(A - E) = E$，因此 $AB = BA$。 +$$\text{……6 分}$$ + +**(3)** +由 $AB = A + B$ 得 $A = (A - E)B$，而 $A - E$ 可逆，故 +$$ +\mathrm{rank}(A) = \mathrm{rank}(B). +$$ +$$\text{……9 分}$$ + +**(4)** +由 $AB = A + B$ 得 $A(B - E) = B$，而 $B - E$ 可逆，故 +$$ +A = B(B - E)^{-1}. +$$ +已知 +$$ +B = \begin{bmatrix} +1 & -3 & 0 \\ +2 & 1 & 0 \\ +0 & 0 & 2 +\end{bmatrix}, +$$ +则 +$$ +B - E = \begin{bmatrix} +0 & -3 & 0 \\ +2 & 0 & 0 \\ +0 & 0 & 1 +\end{bmatrix}. +$$ +求逆得 +$$ +(B - E)^{-1} = \begin{bmatrix} +0 & \frac12 & 0 \\[2pt] +-\frac13 & 0 & 0 \\[2pt] +0 & 0 & 1 +\end{bmatrix}. +$$ +于是 +$$ +A = B(B - E)^{-1} = \begin{bmatrix} +1 & -3 & 0 \\ +2 & 1 & 0 \\ +0 & 0 & 2 +\end{bmatrix} +\begin{bmatrix} +0 & \frac12 & 0 \\[2pt] +-\frac13 & 0 & 0 \\[2pt] +0 & 0 & 1 +\end{bmatrix} += \begin{bmatrix} +1 & \frac12 & 0 \\[2pt] +-\frac13 & 1 & 0 \\[2pt] +0 & 0 & 2 +\end{bmatrix}. +$$ +$$\text{……12 分}$$ --- -- 2.34.1 From d8ebdb28e73cc2bbf9843442090731a9656880c8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Mon, 29 Dec 2025 23:15:33 +0800 Subject: [PATCH 096/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...231线性代数考试卷（解析版）.md | 101 ++++++++++++++++++ 1 file changed, 101 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index e476d63..677326c 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -280,8 +280,109 @@ D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1) $$ --- +（2）加边 +$$ +\tilde{D} = +\begin{vmatrix} +1 & 0 & 0 & \cdots & 0 \\ +1 & 1 + x_1 & 1 + x_1^2 & \cdots & 1 + x_1^n \\ +1 & 1 + x_2 & 1 + x_2^2 & \cdots & 1 + x_2^n \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +1 & 1 + x_n & 1 + x_n^2 & \cdots & 1 + x_n^n +\end{vmatrix} + +$$ + +$$ +\tilde{D} = +\begin{vmatrix} +1 & -1 & -1 & \cdots & -1 \\ +1 & x_1 & x_1^2 & \cdots & x_1^n \\ +1 & x_2 & x_2^2 & \cdots & x_2^n \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +1 & x_n & x_n^2 & \cdots & x_n^n +\end{vmatrix} +$$ + +将第一行拆为 $(2,0,0,\dots,0)$ 与 $(-1,-1,\dots,-1)$ 之和： + +$$ +\tilde{D} = +\begin{vmatrix} +2 & 0 & 0 & \cdots & 0 \\ +1 & x_1 & x_1^2 & \cdots & x_1^n \\ +1 & x_2 & x_2^2 & \cdots & x_2^n \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +1 & x_n & x_n^2 & \cdots & x_n^n +\end{vmatrix} ++ +\begin{vmatrix} +-1 & -1 & -1 & \cdots & -1 \\ +1 & x_1 & x_1^2 & \cdots & x_1^n \\ +1 & x_2 & x_2^2 & \cdots & x_2^n \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +1 & x_n & x_n^2 & \cdots & x_n^n +\end{vmatrix} +$$ + +令左边为 $A$，右边为 $B$。 + +计算 $A$，按第一行展开： + +$$ +A = 2 \cdot +\begin{vmatrix} +x_1 & x_1^2 & \cdots & x_1^n \\ +x_2 & x_2^2 & \cdots & x_2^n \\ +\vdots & \vdots & \ddots & \vdots \\ +x_n & x_n^2 & \cdots & x_n^n +\end{vmatrix} += 2 \cdot \left( \prod_{i=1}^{n} x_i \right) \cdot +\begin{vmatrix} +1 & x_1 & \cdots & x_1^{n-1} \\ +1 & x_2 & \cdots & x_2^{n-1} \\ +\vdots & \vdots & \ddots & \vdots \\ +1 & x_n & \cdots & x_n^{n-1} +\end{vmatrix} +$$ + +右边为范德蒙德行列式： + +$$ +A = 2 \prod_{i=1}^{n} x_i \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) +$$ + +计算 $B$，提出第一行的因子 $-1$： + +$$ +B = (-1) \cdot +\begin{vmatrix} +1 & 1 & 1 & \cdots & 1 \\ +1 & x_1 & x_1^2 & \cdots & x_1^n \\ +1 & x_2 & x_2^2 & \cdots & x_2^n \\ +\vdots & \vdots & \vdots & \ddots & \vdots \\ +1 & x_n & x_n^2 & \cdots & x_n^n +\end{vmatrix} +$$ + +该行列式为 $n+1$ 阶范德蒙德行列式，变量为 $1, x_1, x_2, \dots, x_n$： + +$$ +B = (-1) \cdot \prod_{i=1}^{n} (x_i - 1) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) +$$ + +因此： + +$$ +\tilde{D} = A + B = \left( 2 \prod_{i=1}^{n} x_i - \prod_{i=1}^{n} (x_i - 1) \right) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) +$$ + +$$ +\boxed{\tilde{D} = \left(2\prod\limits_{i=1}^{n}x_i - \prod\limits_{i=1}^{n}(x_i-1)\right) \prod\limits_{1\leq i<j\leq n}(x_j-x_i)} +$$ + --- 14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. -- 2.34.1 From 7ad0f18b5b1029592fa9fc1ef6a38fbf07d099f7 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 03:16:29 +0800 Subject: [PATCH 097/274] vault backup: 2025-12-30 03:16:29 --- ...231线性代数考试卷（解析版）.md | 12 ++-- 编写小组/试卷/线代期中试卷.md | 46 -------------- .../试卷/线代期中试卷解析.md | 62 ------------------- 3 files changed, 5 insertions(+), 115 deletions(-) delete mode 100644 编写小组/试卷/线代期中试卷.md delete mode 100644 编写小组/试卷/线代期中试卷解析.md diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 677326c..cb88fad 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -161,7 +161,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -9. 设矩阵 +10. 设矩阵 $$ A = \begin{bmatrix} 1 & a_1 & a_1^2 & a_1^3 \\ @@ -194,10 +194,8 @@ $$ $$ k= \underline{\qquad\qquad\qquad\qquad}. $$ -12. - $$ - \underline{\qquad\qquad\qquad\qquad}. - $$ +12. + $$\underline{\qquad\qquad\qquad\qquad}$$ ## 三、解答题，共五道，共64分 --- @@ -475,13 +473,13 @@ $$ 1 & 0 & 1 & \vert & -2 \\ 0 & 1 & 1 & \vert & 1 \\ 0 & -1 & 0 & \vert & 2 -\end{bmatrix} \\ +\end{bmatrix} \\[1em] &\rightarrow \begin{bmatrix} 1 & 0 & 1 & \vert & -2 \\ 0 & 1 & 1 & \vert & 1 \\ 0 & 0 & 1 & \vert & 3 -\end{bmatrix} \\ +\end{bmatrix} \\[1em] &\rightarrow \begin{bmatrix} 1 & 0 & 0 & \vert & -5 \\ diff --git a/编写小组/试卷/线代期中试卷.md b/编写小组/试卷/线代期中试卷.md deleted file mode 100644 index ffed476..0000000 --- a/编写小组/试卷/线代期中试卷.md +++ /dev/null @@ -1,46 +0,0 @@ -#官方试卷 -时量：120分钟 满分：100分 -#### 一、单选题（共5小题，每小题3分，共15分） -1. 若$\lim\limits_{n\rightarrow\infty}a_n=4$，则当$n$充分大时，恒有 - A. $|a_n|\le 1$ - B. $|a_n|>2$ - C. $|a_n|<2$ - D. $|a_n|>4$ -2. 已知$g(x)=\frac{1}{x^2}$，复合函数$y=f(g(x))$对$x$的导数为$-\frac{1}{2x}$，则$f'(\frac{1}{2})$的值为 - A. $1$ - B. $2$ - C. $\frac{\sqrt 2}{4}$ - D. $\frac{1}{2}$ -3. 设$f(x)$可导且$f'(x_0)=\frac{1}{3}$，则当$\Delta x \rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 - A. 同阶无穷小 - B. 低阶无穷小 - C. 等价无穷小 - D. 高阶无穷小 -4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 - A. 无穷大量 - B. 无穷小量 - C. 有界变量 - D. 无界变量 -5. 下列四个级数，**发散**的是 - A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ - B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ - C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ - D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ -#### 二、填空题（共5小题，每小题3分，共15分） -6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. -7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. -8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. -9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. -10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. -#### 三、解答与证明题（11~19小题，共70分） -11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. -12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. -13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. -14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. -15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. -16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. -17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. -18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] -19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ - (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） - (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file diff --git a/编写小组/试卷/线代期中试卷解析.md b/编写小组/试卷/线代期中试卷解析.md deleted file mode 100644 index 4874412..0000000 --- a/编写小组/试卷/线代期中试卷解析.md +++ /dev/null @@ -1,62 +0,0 @@ -#官方试卷 -#民间答案 -#### 一、单选题（共5小题，每小题3分，共15分） -1. 若$\lim\limits_{n\rightarrow\infty}a_n=4$，则当$n$充分大时，恒有 - A. $|a_n|\le 1$ - B. $|a_n|>2$ - C. $|a_n|<2$ - D. $|a_n|>4$ -> **答案：B** -> 解析：保号性 -> 若$\lim\limits_{n\rightarrow\infty}a_n=4$，按照定义，$\forall \epsilon >0,\exists N, n>N\text{时},|a_n-4|<\epsilon$， -> 取$\epsilon=2$，则$|a_n-4|<\epsilon$，则$2<a_n<6$，B选项符合题意 -2. 已知$g(x)=\frac{1}{x^2}$，复合函数$y=f(g(x))$对$x$的导数为$-\frac{1}{2x}$，则$f'(\frac{1}{2})$的值为 - A. $1$ - B. $2$ - C. $\frac{\sqrt 2}{4}$ - D. $\frac{1}{2}$ ->**答案：D** ->解析：复合函数求导法则 ->$y'=f'(g(x))g'(x)=\frac{-2}{x^3}f'(\frac{1}{x^2})=-\frac{1}{2x}\Rightarrow f'(\frac{1}{x^2})=\frac{x^2}{4}$，即$f'(\frac{1}{x})=\frac{x}{4}(x> 0)$ ->代入$x=2$得：$f'(\frac{1}{2})=\frac{1}{2}$ -3. 设$f(x)$可导且$f'(x_0)=\frac{1}{3}$，则当$\Delta x \rightarrow 0$时，$f(x)$在$x_0$处的微分$\text{d}y$是$\Delta x$的 - A. 同阶无穷小 - B. 低阶无穷小 - C. 等价无穷小 - D. 高阶无穷小 -> **答案：A** -> 解析：同阶无穷小；微分的定义 -> $\Delta x \rightarrow 0$时，$\Delta x \sim \mathrm{d}x$，又$\mathrm{d}y|_{x=x_0}=\frac{1}{3}\mathrm{d}x$，故$\mathrm{d}y$与$\Delta{x}$是同阶不等价的无穷小 -4. 设数列通项为$x_n=\begin{cases}\frac{n^2-\sqrt{n}}{n},n=2k, \\ \frac{1}{n},n=2k+1 \end{cases}(k\in \mathbb{N}^+)$，则当$n\rightarrow\infty$时，$x_n$是 - A. 无穷大量 - B. 无穷小量 - C. 有界变量 - D. 无界变量 ->**答案：D** ->解析：易错点10-无穷大与无界的辨析 ->$\lim\limits_{k\to\infty}a_{2k}=+\infty$，因此$a_n$无界；然而，$\lim\limits_{k\to\infty}a_{2k+1}=0$，因此$a_n(n\to\infty)$不是无穷大 -5. 下列四个级数，**发散**的是 - A. $\sum\limits_{n=1}^{\infty}{(a^{\frac{1}{n}}+a^{-\frac{1}{n}}-2)}(a>0)$ - B. $\sum\limits_{n=1}^{\infty}\frac{1}{n^{2n\sin{\frac{1}{n}}}}$ - C. $\sum\limits_{n=1}^{\infty}{(\frac{\cos{n!}}{n^2+1}+\frac{1}{\sqrt{n+12}})}$ - D. $\sum\limits_{n=1}^{\infty}\frac{[\sqrt{2}+(-1)^n]^n}{3^n}$ ->**答案：C** -> -#### 二、填空题（共5小题，每小题3分，共15分） -6. 函数$f(x)=\frac{\mathrm{e}^\frac{1}{x-1} \ln{|1+x|}}{(\mathrm{e}^x-1)(x-2)}$的第二类间断点的个数为____. -7. 设函数$y=x-\frac{1}{\sqrt{2}}\arctan{(\sqrt{2}\tan x)}$，则$\text{d}y|_{x=\frac{\pi}{4}}=$\_\_\_\_\_. -8. 已知级数$\sum\limits_{n=2}^{\infty}{(-1)^n\frac{1}{n^a}\ln\frac{n+1}{n-1}}$条件收敛，则常数$a$的取值范围是_____. -9. 已知函数$f(x)$在$x=1$处可导，且$f(1)=0,f'(1)=2$，则$\lim\limits_{x\rightarrow 0}{\frac{f(e^{x^2})}{\sin^2 x}}$=\_\_\_\_\_\_. -10. 曲线$y=\frac{1}{2}x^2$与曲线$y=c\ln x$相切，则常数$c$的值为\_\_\_\_\_\_. -#### 三、解答与证明题（11~19小题，共70分） -11. （6分）求极限$\lim\limits_{x\rightarrow 0}(\cos 2x + x\arcsin x)^\frac{1}{x^2}$. -12. （6分）求极限$\lim\limits_{n\rightarrow\infty}\sqrt[n+2]{2\sin^2 n+\cos^2 n}$. -13. （6分）已知函数$f(x)=\begin{cases}\sin 2x+1, x\le 0\\a^{2x}+b, x>0 \end{cases}$在$x=0$处可导，求常数$a,b$的值. -14. （6分）设$y=y(x)$是由方程$y=x\mathrm{e}^y=1$所确定的函数，求曲线$y=y(x)$在$x=0$对应点处的切线方程. -15. （8分）设函数$y=f(x)$的极坐标式为$\rho=\mathrm{e}^\theta$，求$\frac{\mathrm{d}y}{\mathrm{d}x}|_{\theta=\frac{\pi}{2}}$，$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}|_{\theta=\frac{\pi}{2}}$. -16. （8分）设$f(x)$在$(-\infty,+\infty)$内是以$2T$为周期的连续函数，证明对任一实数$x_0$，方程$f(x)=f(x+T)$在区间$[x_0-\frac{T}{2},x_0+\frac{T}{2}]$上至少有一个根. -17. （10分）求曲线$y=\frac{x^2-3x+2}{x-1}+2\ln|x-2|,(x>0)$的所有渐近线方程. -18. （10分）百米跑道上正在举行百米赛跑，摄影师站在50米处为4号跑道夺冠种子选手A录像，摄像头始终对准选手A，摄影师距离4号跑道5米，若选手A以10m/s的速度从摄影师正前方经过，问此时摄影镜头的角速度是多少？ ![[期中试卷-18.png]] -19. （10分）设$a_1=3,a_{n+1}=\frac{a_n}{2}+\frac{1}{n},n=1,2,\dots,$ - (1) 证明$\lim\limits_{n\rightarrow\infty}a_n$存在，并求其极限值；（5分） - (2) 证明：对于任意实数$p$，级数$\sum\limits_{n=1}{\infty}{n^p(\frac{a_n}{a_{n+1}}-1)}$收敛. （5分） \ No newline at end of file -- 2.34.1 From 76ac4a319d5ede47a647d55968d3565c9692bc5e Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 12:02:52 +0800 Subject: [PATCH 098/274] =?UTF-8?q?=E8=AF=95=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/1231线性代数考试卷.md | 91 +++++++++++++++++++ 1 file changed, 91 insertions(+) create mode 100644 编写小组/试卷/1231线性代数考试卷.md diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md new file mode 100644 index 0000000..5771a41 --- /dev/null +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -0,0 +1,91 @@ +--- +tags: + - 编写小组 +--- +## 一、选择题，共六道，每题3分，共18分 + +1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 + + (A) $AB - BA$; + (B) $AB + BA$; + (C) $BAB$; + (D) $(AB)^2$. + +2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 +(A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ +(B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ +(C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ +(D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ + +3. 设向量组 $\alpha_1 = (0, 0, c_1)^T,\quad \alpha_2 = (0, 1, c_2)^T,\quad \alpha_3 = (1, -1, c_3)^T,\quad \alpha_4 = (-1, 1, c_4)^T,$ + 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是 + (A) $\alpha_1, \alpha_2, \alpha_3$; + (B) $\alpha_1, \alpha_2, \alpha_4$; + (C) $\alpha_1, \alpha_3, \alpha_4$; + (D) $\alpha_2, \alpha_3, \alpha_4$. +4. 设 $A, B$ 为 $n$ 阶矩阵，则 + (A) $\text{rank}[A \ AB] = \text{rank} A$; + (B) $\text{rank}[A \ BA] = \text{rank} A$; + (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; + (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. + +5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 + (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; + (B) 将 $A^*$ 的第一行加上第二行的 2 倍得到 $B^*$; + (C) 将 $A^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $B^*$; + (D) 将 $A^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $B^*$. + +6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + +## 二、填空题，共六道，每题3分，共18分 + +7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积$\langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}.$ + +8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 + +9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ + +10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ + 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ +11. $A^{k} = 0, k=\underline{\qquad\qquad\qquad\qquad}.$ +12. $\underline{\qquad\qquad\qquad\qquad}$ + +## 三、解答题，共五道，共64分 + +13. （20 分）计算 下面的两个$n$阶行列式 + + $$ + K_n = \begin{vmatrix} + 1 & 2 & 3 & \cdots & n-1 & n \\ + 2 & 1 & 2 & \cdots & n-2 & n-1 \\ + 3 & 2 & 1 & \cdots & n-3 & n-2 \\ + \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ + n-1 & n-2 & n-3 & \cdots & 1 & 2 \\ + n & n-1 & n-2 & \cdots & 2 & 1 + \end{vmatrix}. + $$ +$$ + M_n =\begin{vmatrix} +1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n \\ +1+x_2 & 1+x_2^2 & \cdots & 1+x_2^n \\ +\vdots & \vdots & \ddots & \vdots \\ +1+x_n & 1+x_n^2 & \cdots & 1+x_n^n +\end{vmatrix} +$$ +14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. + (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； + (2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. + +15. （10 分）设 $\alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T$和$\beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T$是 $\mathbb{R}^3$ 的两组基，求向量$u = \alpha_1 + 2\alpha_2 - 3\alpha_3$在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 + +16. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 + （1）证明 $A - E$ 可逆； + （2）证明 $AB = BA$； + （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； + （4）若矩阵$B = \begin{bmatrix}1 & -3 & 0 \\2 & 1 & 0 \\0 & 0 & 2\end{bmatrix}$，求矩阵 $A$。 + +17. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 \ No newline at end of file -- 2.34.1 From 9921278820c2f6ad855181a05a3ac90d35ed4868 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 30 Dec 2025 23:01:28 +0800 Subject: [PATCH 099/274] =?UTF-8?q?=E6=8A=8A=E4=B8=80=E4=BA=9B=E5=9C=86?= =?UTF-8?q?=E6=8B=AC=E5=8F=B7=E7=9A=84=E7=9F=A9=E9=98=B5=E6=8D=A2=E6=88=90?= =?UTF-8?q?=E4=BA=86=E6=96=B9=E6=8B=AC=E5=8F=B7=E7=9A=84?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/1231线性代数考试卷（解析版）.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 90a74a3..022adc9 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -44,22 +44,22 @@ $$ $$ [\varepsilon_1, \varepsilon_2] = [e_1, e_2] -\begin{pmatrix} +\begin{bmatrix} 1 & 0 \\ 5 & 1 -\end{pmatrix}. +\end{bmatrix}. $$ 基变换矩阵为： $$ -T = \begin{pmatrix} 1 & 0 \\ 5 & 1 \end{pmatrix}. +T = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}. $$ $$ \quad -T^{-1} = \begin{pmatrix} 1 & 0 \\ -5 & 1 \end{pmatrix}. +T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}. $$ -$\quad T^{-1} = \begin{pmatrix} 1 & 0 \\ -5 & 1 \end{pmatrix}$是坐标变换矩阵，即为过渡矩阵，选D +$\quad T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}$是坐标变换矩阵，即为过渡矩阵，选D --- -- 2.34.1 From d70426fe05311c612e10b0e869489ab632b54698 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:03:40 +0800 Subject: [PATCH 100/274] vault backup: 2025-12-30 23:03:40 --- 编写小组/试卷/1231线性代数考试卷.md | 1 + 1 file changed, 1 insertion(+) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 5771a41..37a36ba 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -23,6 +23,7 @@ tags: (B) $\alpha_1, \alpha_2, \alpha_4$; (C) $\alpha_1, \alpha_3, \alpha_4$; (D) $\alpha_2, \alpha_3, \alpha_4$. + 4. 设 $A, B$ 为 $n$ 阶矩阵，则 (A) $\text{rank}[A \ AB] = \text{rank} A$; (B) $\text{rank}[A \ BA] = \text{rank} A$; -- 2.34.1 From 3cbbb539ae3955342101e9c41d87960d87368cbc Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:05:59 +0800 Subject: [PATCH 101/274] vault backup: 2025-12-30 23:05:59 --- 编写小组/试卷/1231线性代数考试卷.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 37a36ba..1598da1 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -12,10 +12,11 @@ tags: (D) $(AB)^2$. 2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 + (A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ -(B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ -(C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ -(D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ + (B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ + (C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ + (D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ 3. 设向量组 $\alpha_1 = (0, 0, c_1)^T,\quad \alpha_2 = (0, 1, c_2)^T,\quad \alpha_3 = (1, -1, c_3)^T,\quad \alpha_4 = (-1, 1, c_4)^T,$ 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是 -- 2.34.1 From c4d5a80a94c731c4cc36b4d21aa9ffc195c1fb73 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 23:12:33 +0800 Subject: [PATCH 102/274] =?UTF-8?q?0103=E9=AB=98=E6=95=B0=E6=A8=A1?= =?UTF-8?q?=E6=8B=9F=E8=AF=95=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 4 ++++ 1 file changed, 4 insertions(+) create mode 100644 编写小组/试卷/0103高数模拟试卷.md diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md new file mode 100644 index 0000000..1fb86d3 --- /dev/null +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -0,0 +1,4 @@ +--- +tags: + - 编写小组 +--- -- 2.34.1 From 201f38a1a1300a945da3035f774d1abc33062bb2 Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Tue, 30 Dec 2025 23:15:00 +0800 Subject: [PATCH 103/274] vault backup: 2025-12-30 23:15:00 --- 编写小组/试卷/1231线性代数考试卷.md | 11 ++++++++++- 1 file changed, 10 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 1598da1..9983091 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -53,7 +53,16 @@ tags: 10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ -11. $A^{k} = 0, k=\underline{\qquad\qquad\qquad\qquad}.$ +11. 矩阵$$A=\begin{bmatrix} +0 & 0 & \cdots & 1 & 1 & \cdots & 1 & 1 \\ +0 & 0 & \cdots & 0 & 1 & \cdots & 1 & 1 \\ +\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +0 & 0 & \cdots & 0 & 0 & \cdots & 1 & 1 \\ +0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 1 \\ +\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ +0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\ +0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 +\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____.$A^{k} = 0, k=\underline{\qquad\qquad\qquad\qquad}.$ 12. $\underline{\qquad\qquad\qquad\qquad}$ ## 三、解答题，共五道，共64分 -- 2.34.1 From 19e820341d2dcd95b614226ce005cd9cc552c75f Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Tue, 30 Dec 2025 23:16:20 +0800 Subject: [PATCH 104/274] vault backup: 2025-12-30 23:16:19 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 9983091..3a14abc 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -62,7 +62,7 @@ tags: \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____.$A^{k} = 0, k=\underline{\qquad\qquad\qquad\qquad}.$ +\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____. 12. $\underline{\qquad\qquad\qquad\qquad}$ ## 三、解答题，共五道，共64分 -- 2.34.1 From 24c01c7651d3efdc8f4c2709062dbd284c3495a0 Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Tue, 30 Dec 2025 23:21:27 +0800 Subject: [PATCH 105/274] vault backup: 2025-12-30 23:21:27 --- .../试卷/1231线性代数考试卷.md | 19 +++++++++++-------- 1 file changed, 11 insertions(+), 8 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 3a14abc..8b1821a 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -54,14 +54,17 @@ tags: 10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ 11. 矩阵$$A=\begin{bmatrix} -0 & 0 & \cdots & 1 & 1 & \cdots & 1 & 1 \\ -0 & 0 & \cdots & 0 & 1 & \cdots & 1 & 1 \\ -\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -0 & 0 & \cdots & 0 & 0 & \cdots & 1 & 1 \\ -0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 1 \\ -\vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 \\ -0 & 0 & \cdots & 0 & 0 & \cdots & 0 & 0 +0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 1 & 1 & 1 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____. 12. $\underline{\qquad\qquad\qquad\qquad}$ -- 2.34.1 From 9c09d38d2d2a72de19f8b9e5d5f39c747c46a6cb Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:23:20 +0800 Subject: [PATCH 106/274] vault backup: 2025-12-30 23:23:20 --- 编写小组/试卷/1231线性代数考试卷.md | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 8b1821a..62a151d 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -53,7 +53,8 @@ tags: 10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ -11. 矩阵$$A=\begin{bmatrix} + +11. 矩阵(陈峰华原创难题)$$A=\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 1 & 1 & 1 \\ @@ -66,6 +67,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____. + 12. $\underline{\qquad\qquad\qquad\qquad}$ ## 三、解答题，共五道，共64分 -- 2.34.1 From f18de13aca31e3dd07a471f36754a052ad012ab8 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:41:23 +0800 Subject: [PATCH 107/274] vault backup: 2025-12-30 23:41:23 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 62a151d..4674573 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -54,7 +54,7 @@ tags: 10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ -11. 矩阵(陈峰华原创难题)$$A=\begin{bmatrix} +11. 矩阵(陈峰华原创题)$$A=\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 1 & 1 & 1 \\ -- 2.34.1 From dd798a212e8f8c1f6171d614957ac61354856091 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 23:42:25 +0800 Subject: [PATCH 108/274] =?UTF-8?q?=E9=AB=98=E6=95=B0=E6=A8=A1=E6=8B=9F?= =?UTF-8?q?=E8=AF=95=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 8 ++++++++ 1 file changed, 8 insertions(+) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 1fb86d3..ed86bba 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -2,3 +2,11 @@ tags: - 编写小组 --- +## 一、单选题（共5小题，每小题2分，共10分） + + +## 二、填空题（共5小题，每小题2分，共10分） + + +## 三、解答题（共11小题，共80分） + -- 2.34.1 From 55779733a777923be8bc0995c5b085faa7b3e3f9 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 23:50:47 +0800 Subject: [PATCH 109/274] =?UTF-8?q?=E8=B0=83=E6=95=B4=E6=A0=BC=E5=BC=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/1231线性代数考试卷.md | 41 +++++++++---------- 1 file changed, 19 insertions(+), 22 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 4674573..b119ecc 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -4,40 +4,38 @@ tags: --- ## 一、选择题，共六道，每题3分，共18分 -1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 - +1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是 (A) $AB - BA$; (B) $AB + BA$; (C) $BAB$; (D) $(AB)^2$. 2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 - -(A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ - (B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ - (C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ - (D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ + (A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ + (B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ + (C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ + (D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ 3. 设向量组 $\alpha_1 = (0, 0, c_1)^T,\quad \alpha_2 = (0, 1, c_2)^T,\quad \alpha_3 = (1, -1, c_3)^T,\quad \alpha_4 = (-1, 1, c_4)^T,$ 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是 - (A) $\alpha_1, \alpha_2, \alpha_3$; + (A) $\alpha_1, \alpha_2, \alpha_3$; (B) $\alpha_1, \alpha_2, \alpha_4$; (C) $\alpha_1, \alpha_3, \alpha_4$; (D) $\alpha_2, \alpha_3, \alpha_4$. -4. 设 $A, B$ 为 $n$ 阶矩阵，则 +5. 设 $A, B$ 为 $n$ 阶矩阵，则 (A) $\text{rank}[A \ AB] = \text{rank} A$; (B) $\text{rank}[A \ BA] = \text{rank} A$; (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. -5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 +6. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; (B) 将 $A^*$ 的第一行加上第二行的 2 倍得到 $B^*$; (C) 将 $A^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $B^*$; (D) 将 $A^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $B^*$. -6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 +7. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 (A) $a = 1, b = 0, c = 1$; (B) $a = 1, b = 1, c = 2$; (C) $a = 2, b = 0, c = 1$; @@ -49,10 +47,9 @@ tags: 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 -9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ +9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad\beta_2 = (1,2,3)^T,\quad\beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ -10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$ - 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ +10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ 11. 矩阵(陈峰华原创题)$$A=\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ @@ -66,15 +63,14 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=0$,则$k$的最小值为____. +\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为____. 12. $\underline{\qquad\qquad\qquad\qquad}$ ## 三、解答题，共五道，共64分 13. （20 分）计算 下面的两个$n$阶行列式 - - $$ +$$ K_n = \begin{vmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 1 & 2 & \cdots & n-2 & n-1 \\ @@ -92,11 +88,12 @@ $$ 1+x_n & 1+x_n^2 & \cdots & 1+x_n^n \end{vmatrix} $$ -14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. - (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； - (2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. -15. （10 分）设 $\alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T$和$\beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T$是 $\mathbb{R}^3$ 的两组基，求向量$u = \alpha_1 + 2\alpha_2 - 3\alpha_3$在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 +14. （10 分）设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. + (1)证明：方程组 $Ax=\alpha$ 的解均为方程组 $Bx=\beta$ 的解； + (2)若方程组 $Ax=\alpha$ 与方程组 $Bx=\beta$ 不同解，求 $a$ 的值. + +15. （10 分）设 $\alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T$和$\beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T$是 $\mathbb{R}^3$ 的两组基，求向量 $u = \alpha_1 + 2\alpha_2 - 3\alpha_3$在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 16. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 （1）证明 $A - E$ 可逆； @@ -104,4 +101,4 @@ $$ （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； （4）若矩阵$B = \begin{bmatrix}1 & -3 & 0 \\2 & 1 & 0 \\0 & 0 & 2\end{bmatrix}$，求矩阵 $A$。 -17. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 \ No newline at end of file +17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 \ No newline at end of file -- 2.34.1 From 9920f45b02ac444764fc0e595cd2b3d3b14a23e0 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 23:53:33 +0800 Subject: [PATCH 110/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E6=A0=BC=E5=BC=8F2?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index b119ecc..4d47222 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -101,4 +101,4 @@ $$ （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； （4）若矩阵$B = \begin{bmatrix}1 & -3 & 0 \\2 & 1 & 0 \\0 & 0 & 2\end{bmatrix}$，求矩阵 $A$。 -17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 \ No newline at end of file +17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组 $Ax=0$ 的基础解系中含有两个解向量，求 $Ax=0$ 的通解。 \ No newline at end of file -- 2.34.1 From cebc0d6abbc09fc1cd4bee3e0e80f5f4aa1136d5 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 30 Dec 2025 23:56:35 +0800 Subject: [PATCH 111/274] =?UTF-8?q?=E8=A1=A5=E5=85=8512=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 4d47222..b62bfc1 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -65,7 +65,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为____. -12. $\underline{\qquad\qquad\qquad\qquad}$ +12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{2cm}}.$ ## 三、解答题，共五道，共64分 -- 2.34.1 From ead59c5806441ac954060b6cd184d8adce6d3441 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:57:08 +0800 Subject: [PATCH 112/274] vault backup: 2025-12-30 23:57:08 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index b62bfc1..d20890a 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -63,7 +63,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为____. +\end{bmatrix}_{n \times n}$$其中第一行有 $m$ 个$0$.若$A^k=O$,则$k$的最小值为____. 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{2cm}}.$ -- 2.34.1 From c1dfc7c52d3c0ac00aeb5426917ba7fb253b224f Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 30 Dec 2025 23:59:53 +0800 Subject: [PATCH 113/274] vault backup: 2025-12-30 23:59:53 --- 编写小组/试卷/1231线性代数考试卷.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index d20890a..ded34e2 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -63,9 +63,9 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有 $m$ 个$0$.若$A^k=O$,则$k$的最小值为____. +\end{bmatrix}_{n \times n}$$其中第一行有 $m$ 个$0$.若$A^k=O$,则 $k$ 的最小值为____. -12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{2cm}}.$ +12. 设 $A, B$ 均为 $n$ 阶方阵，满足 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B$ ，且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{2cm}}.$ ## 三、解答题，共五道，共64分 -- 2.34.1 From cac456a4c1b9a3f8989ff4dcf5b1f524ffd67eaf Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 31 Dec 2025 00:12:01 +0800 Subject: [PATCH 114/274] =?UTF-8?q?=E7=BA=BF=E6=80=A7=E4=BB=A3=E6=95=B0?= =?UTF-8?q?=E9=80=82=E5=BA=94=E6=80=A7=E8=B0=83=E7=A0=94?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/1231线性代数考试卷.md | 126 +++++++++++++++++- 1 file changed, 119 insertions(+), 7 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index b62bfc1..8998026 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -2,6 +2,7 @@ tags: - 编写小组 --- +# 线性代数适应性调研 ## 一、选择题，共六道，每题3分，共18分 1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是 @@ -23,19 +24,19 @@ tags: (C) $\alpha_1, \alpha_3, \alpha_4$; (D) $\alpha_2, \alpha_3, \alpha_4$. -5. 设 $A, B$ 为 $n$ 阶矩阵，则 +4. 设 $A, B$ 为 $n$ 阶矩阵，则 (A) $\text{rank}[A \ AB] = \text{rank} A$; (B) $\text{rank}[A \ BA] = \text{rank} A$; (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. -6. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 +5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; (B) 将 $A^*$ 的第一行加上第二行的 2 倍得到 $B^*$; (C) 将 $A^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $B^*$; (D) 将 $A^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $B^*$. -7. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 +6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 (A) $a = 1, b = 0, c = 1$; (B) $a = 1, b = 1, c = 2$; (C) $a = 2, b = 0, c = 1$; @@ -45,7 +46,7 @@ tags: 7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积$\langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}.$ -8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 +8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\qquad\qquad\quad\quad}$。 9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad\beta_2 = (1,2,3)^T,\quad\beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ @@ -63,9 +64,9 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为____. +\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ -12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{2cm}}.$ +12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{3cm}}.$ ## 三、解答题，共五道，共64分 @@ -88,17 +89,128 @@ $$ 1+x_n & 1+x_n^2 & \cdots & 1+x_n^n \end{vmatrix} $$ +```text + + + + + + + + + + + + + + + + + + + + +``` 14. （10 分）设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. (1)证明：方程组 $Ax=\alpha$ 的解均为方程组 $Bx=\beta$ 的解； (2)若方程组 $Ax=\alpha$ 与方程组 $Bx=\beta$ 不同解，求 $a$ 的值. +```text + + + + + + + + + + + + + + + + + + + + +``` 15. （10 分）设 $\alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T$和$\beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T$是 $\mathbb{R}^3$ 的两组基，求向量 $u = \alpha_1 + 2\alpha_2 - 3\alpha_3$在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 +```text + + + + + + + + + + + + + + + + + + + + +``` 16. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 （1）证明 $A - E$ 可逆； （2）证明 $AB = BA$； （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； （4）若矩阵$B = \begin{bmatrix}1 & -3 & 0 \\2 & 1 & 0 \\0 & 0 & 2\end{bmatrix}$，求矩阵 $A$。 +```text + + + + + + + + + + + + + + + + + + + + + +``` +17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组 $Ax=0$ 的基础解系中含有两个解向量，求 $Ax=0$ 的通解。 +```text + + + + + + + + + + + + + + + + + + + + -17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组 $Ax=0$ 的基础解系中含有两个解向量，求 $Ax=0$ 的通解。 \ No newline at end of file +``` \ No newline at end of file -- 2.34.1 From 6ebb0889dde4d4d754a5121b9f5f67eb531f40b5 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 00:13:04 +0800 Subject: [PATCH 115/274] vault backup: 2025-12-31 00:13:04 --- .../试卷/1231线性代数考试卷（解析版）.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index a512488..b0e1f21 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -1,3 +1,7 @@ +--- +tags: + - 编写小组 +--- ## 一、选择题，共六道，每题3分，共18分 1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 -- 2.34.1 From 58c776b49f1ebf82fedaf25d4e9ed43ec2eee18b Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Wed, 31 Dec 2025 00:21:35 +0800 Subject: [PATCH 116/274] vault backup: 2025-12-31 00:21:35 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 8998026..f3cbb24 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -64,7 +64,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ +\end{bmatrix}_{(nm) \times (nm)}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -- 2.34.1 From 279dcb03efb8b4c56e444f5abf3dddc777436f6d Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 00:22:50 +0800 Subject: [PATCH 117/274] vault backup: 2025-12-31 00:22:50 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index f3cbb24..46d2817 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -64,7 +64,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{(nm) \times (nm)}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ +\end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -- 2.34.1 From 667b93902dc8bb0d23b46166e887545e4a330142 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 31 Dec 2025 00:34:08 +0800 Subject: [PATCH 118/274] =?UTF-8?q?=E4=BF=AE=E6=94=B912=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 8998026..466c9f4 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -66,7 +66,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{n \times n}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ -12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} A & O \\ B & E \end{bmatrix} =\underline{\hspace{3cm}}.$ +12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ ## 三、解答题，共五道，共64分 -- 2.34.1 From 81f15ba90f21b6a745a1d2d2e34737107639b78c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 31 Dec 2025 00:46:59 +0800 Subject: [PATCH 119/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1231线性代数考试卷（解析版）.md | 11 +++++++---- 1 file changed, 7 insertions(+), 4 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index b0e1f21..a920854 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -194,12 +194,15 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ $$ A^{k} = 0 $$ +观察得每乘一次第一行少m个0 $$ -k= \underline{\qquad\qquad\qquad\qquad}. - $$ -12. - $$\underline{\qquad\qquad\qquad\qquad}$$ +k= + n$$ +12.又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 +$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ + + ## 三、解答题，共五道，共64分 --- -- 2.34.1 From 2afc53d9e14542302ec7f5b38607de0fde5e2e68 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 00:47:40 +0800 Subject: [PATCH 120/274] vault backup: 2025-12-31 00:47:40 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index a920854..b846e7d 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -199,7 +199,7 @@ $$ $$ k= n$$ -12.又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 +12. 又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ -- 2.34.1 From edc9efe7f36692d274dfa9a78471ca3f3c51f5c0 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 00:49:20 +0800 Subject: [PATCH 121/274] vault backup: 2025-12-31 00:49:20 --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index db2878a..2808020 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -64,7 +64,7 @@ tags: 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 -\end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$,则$k$的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ +\end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -- 2.34.1 From c50d0019762c5fe775e2271a290903738538a543 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 31 Dec 2025 19:09:45 +0800 Subject: [PATCH 122/274] =?UTF-8?q?Issue=20resolved:=20=E4=BF=AE=E6=94=B9?= =?UTF-8?q?=E4=BA=86=E7=BA=BF=E4=BB=A3=E8=AF=95=E5=8D=B7=E7=9A=84=E4=B8=80?= =?UTF-8?q?=E4=B8=AA=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/1231线性代数考试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷.md b/编写小组/试卷/1231线性代数考试卷.md index 2808020..ee4547a 100644 --- a/编写小组/试卷/1231线性代数考试卷.md +++ b/编写小组/试卷/1231线性代数考试卷.md @@ -11,7 +11,7 @@ tags: (C) $BAB$; (D) $(AB)^2$. -2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 +2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $\varepsilon_1, \varepsilon_2$ 到基 $e_1, e_2$ 到基的过渡矩阵是 (A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ (B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ (C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ -- 2.34.1 From aeb7b164bd194985e0ed054c3aca3461f78db231 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:14:05 +0800 Subject: [PATCH 123/274] vault backup: 2025-12-31 19:14:05 --- ...231线性代数考试卷（解析版）.md | 28 ++++++++++--------- 1 file changed, 15 insertions(+), 13 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index b846e7d..59fe444 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -189,20 +189,22 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ -11. +11. 矩阵(陈峰华原创题)$$A=\begin{bmatrix} +0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 1 & 1 & 1 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 1 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +\vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \\ +0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 +\end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ + +12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -$$ -A^{k} = 0 -$$ -观察得每乘一次第一行少m个0 - -$$ -k= - n$$ -12. 又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 -$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ - - ## 三、解答题，共五道，共64分 --- -- 2.34.1 From 182aeeddbf1a6beb7b8808c11b97502c42e59c67 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:15:47 +0800 Subject: [PATCH 124/274] vault backup: 2025-12-31 19:15:47 --- .../试卷/1231线性代数考试卷（解析版）.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 59fe444..645fe8b 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -188,7 +188,6 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- - 11. 矩阵(陈峰华原创题)$$A=\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0 & 1 & \cdots & 1 & 1 & 1 \\ @@ -203,6 +202,10 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ +--- +解析： + + 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -- 2.34.1 From c9e1f14b6a1994967fa289eb6ca8b3af124edf0a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:26:18 +0800 Subject: [PATCH 125/274] vault backup: 2025-12-31 19:26:18 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 645fe8b..26a0a51 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -203,7 +203,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ --- -解析： +解析：观察得每乘一次第一行少 $m$ 个0，故最少进行 $n$ 次即可 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ -- 2.34.1 From 7f28d161bfc2a47a1036746e6fc5e974f06fa019 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:27:06 +0800 Subject: [PATCH 126/274] vault backup: 2025-12-31 19:27:06 --- .../试卷/1231线性代数考试卷（解析版）.md | 7 +++++++ 1 file changed, 7 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 26a0a51..4a15329 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -203,12 +203,19 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ --- + 解析：观察得每乘一次第一行少 $m$ 个0，故最少进行 $n$ 次即可 +--- 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ +--- +又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 +$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ + +--- ## 三、解答题，共五道，共64分 --- -- 2.34.1 From 0c1ffd9a966f08adc669818d79ab10cd595eb1f9 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:27:17 +0800 Subject: [PATCH 127/274] vault backup: 2025-12-31 19:27:17 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 4a15329..f08d752 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -212,7 +212,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -又方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 +由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- -- 2.34.1 From 60a8ba6da0b91b27128986234deff64acccd7238 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:28:05 +0800 Subject: [PATCH 128/274] vault backup: 2025-12-31 19:28:05 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index f08d752..34a0edf 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -212,7 +212,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$,由初等变换不改变秩得 +由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- -- 2.34.1 From 10b6879ece9ff901d88e116c3544b155c027c50a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:39:44 +0800 Subject: [PATCH 129/274] vault backup: 2025-12-31 19:39:44 --- .../1231线性代数考试卷（解析版）.md | 15 ++++++++++++++- 1 file changed, 14 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 34a0edf..e875ebf 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -162,6 +162,19 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ $$ a = \underline{\qquad\qquad}. $$ +--- + +【答】5. + +【解析】（方法一）依题意知 $\beta_1, \beta_2, \beta_3$ 线性相关，否则，若 $\beta_1, \beta_2, \beta_3$ 线性无关，则 $\beta_1, \beta_2, \beta_3$ 为向量空间 $R^3$ 的一组基，$\alpha_1, \alpha_2, \alpha_3$ 能由 $\beta_1, \beta_2, \beta_3$ 线性表示，矛盾。记 $B = [\beta_1, \beta_2, \beta_3]$，则 $|B| = 0$，解得 $a = 5$。 + +（方法二）令 +$$A = [\beta_1, \beta_2, \beta_3, \alpha_1, \alpha_2, \alpha_3] = \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 1 & 2 & 4 & 0 & 1 & 3 \\ 1 & 3 & a & 1 & 1 & 5 \end{bmatrix}$$ +对其进行初等行变换 + +$$A \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 2 & a-3 & 0 & 1 & 4 \end{bmatrix} \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 0 & a-5 & 2 & -1 & 0 \end{bmatrix} = \begin{bmatrix} \gamma_1 & \gamma_2 & \gamma_3 & \gamma_4 & \gamma_5 & \gamma_6 \end{bmatrix}$$ + +由 $\alpha_1, \alpha_2, \alpha_3$ 不能由 $\beta_1, \beta_2, \beta_3$ 线性表示可知 $\gamma_4, \gamma_5, \gamma_6$ 不能由 $\gamma_1, \gamma_2, \gamma_3$ 线性表示，从而 $a = 5$。 --- @@ -212,7 +225,7 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ --- -由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得 +解析：由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得 $\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ --- -- 2.34.1 From 4f70171fc76b0b1792a0433f1d8250e157a24dcc Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 19:49:32 +0800 Subject: [PATCH 130/274] vault backup: 2025-12-31 19:49:32 --- ...231线性代数考试卷（解析版）.md | 38 +++++++++++++++++-- 1 file changed, 35 insertions(+), 3 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index e875ebf..2c4fd4a 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -194,10 +194,42 @@ $$A \to \begin{bmatrix} 1 & 1 & 3 & 1 & 0 & 1 \\ 0 & 1 & 1 & -1 & 1 & 2 \\ 0 & 2 \underline{\qquad\qquad\qquad\qquad}. $$ --- -解析： -一眼顶针，鉴定为： $$ - x = (1,0,0,0)^T$$ +**答**：$(1,0,0,0)^T$。 + +**解析**：由范德蒙行列式的性质可知 $|A| \neq 0$，从而线性方程组 $Ax = b$ 有唯一解。 + +又由 +$$ +\begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix} +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} += +\begin{bmatrix} +1 \\ +1 \\ +1 \\ +1 +\end{bmatrix} +$$ +可知 $Ax = b$ 的解为 +$$ +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} +$$ --- -- 2.34.1 From eae2299e451deb33224d6fc1594302829c94d7d6 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:02:39 +0800 Subject: [PATCH 131/274] vault backup: 2025-12-31 20:02:39 --- .../试卷/1231线性代数考试卷（解析版）.md | 8 ++++++++ 1 file changed, 8 insertions(+) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 2c4fd4a..bac74eb 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -125,12 +125,20 @@ $\quad T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}$是坐标变换矩 $$ \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}. $$ +--- + +【答】4 + +【解析】由内积的性质可知 $$ \begin{aligned} \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle &= \langle \alpha_1, 2\alpha_1 + \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle \\[1em] &= \langle \alpha_1, 2\alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + \langle 2\alpha_2, 2\alpha_1 \rangle + \langle 2\alpha_2, \alpha_3 \rangle \\[1em] &= 2\langle \alpha_1, \alpha_1 \rangle + \langle \alpha_1, \alpha_3 \rangle + 4\langle \alpha_2, \alpha_1 \rangle + 2\langle \alpha_2, \alpha_3 \rangle, \end{aligned} $$ 再由题意，可知 $$ \langle \alpha_1, \alpha_1 \rangle = 2,\quad \langle \alpha_1, \alpha_3 \rangle = -2,\quad \langle \alpha_2, \alpha_1 \rangle = 2,\quad \langle \alpha_2, \alpha_3 \rangle = -3. $$ 从而 $$ \langle \alpha_1 + 2\alpha_2, 2\alpha_1 + \alpha_3 \rangle = 4. $$ + +--- 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 --- 解析： + 先计算$A^2$： $$A^2 -- 2.34.1 From db55391fd97df9be6fe8316f4036246b2f63ae8f Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 31 Dec 2025 20:15:13 +0800 Subject: [PATCH 132/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E7=9A=84=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A1=E5=88=B06=E9=A2=98=E8=A7=A3=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...231线性代数考试卷（解析版）.md | 601 ++---------------- 1 file changed, 57 insertions(+), 544 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 645fe8b..a874a1f 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -2,191 +2,91 @@ tags: - 编写小组 --- +# 线性代数适应性调研 ## 一、选择题，共六道，每题3分，共18分 -1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【 】 - +1. 设 $A$ 为 $n$ 阶对称矩阵，$B$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是 (A) $AB - BA$; (B) $AB + BA$; (C) $BAB$; (D) $(AB)^2$. -2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 -$$ -\varepsilon_1 = e_1 + 5e_2,\quad \varepsilon_2 = e_2, -$$ -则由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵是 - -$$ -(A) \begin{bmatrix} --1 & 0 \\ -5 & -1 -\end{bmatrix} \quad -(B) \begin{bmatrix} -0 & -1 \\ --6 & 0 -\end{bmatrix} \quad -(C) \begin{bmatrix} -1 & 0 \\ --5 & -1 -\end{bmatrix} \quad -(D) \begin{bmatrix} -1 & 0 \\ --5 & 1 -\end{bmatrix}. -$$ - ---- - -解析： - -$$ -\varepsilon_1 = e_1 + 5e_2,\quad \varepsilon_2 = 0e_1 + 1e_2. -$$ - -把它们按列排成矩阵形式： - -$$ -[\varepsilon_1, \varepsilon_2] = [e_1, e_2] -\begin{bmatrix} -1 & 0 \\ -5 & 1 -\end{bmatrix}. -$$ - -基变换矩阵为： - -$$ -T = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}. -$$ -$$ -\quad -T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}. -$$ -$\quad T^{-1} = \begin{bmatrix} 1 & 0 \\ -5 & 1 \end{bmatrix}$是坐标变换矩阵，即为过渡矩阵，选D - ---- - -3. 设向量组 - $$ - \alpha_1 = (0, 0, c_1)^T,\quad - \alpha_2 = (0, 1, c_2)^T,\quad - \alpha_3 = (1, -1, c_3)^T,\quad - \alpha_4 = (-1, 1, c_4)^T, - $$ - 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是【 】 - +>答案：**B** +>解析：$A$ 为 $n$ 阶对称矩阵 $\Rightarrow$ $A^T=A$ ， $B$ 为 $n$ 阶反称矩阵 $\Rightarrow$ $B^T=-B$ ； +>逐个选项分析： +>(A) $(AB-BA)^T=(AB)^T-(BA)^T=B^TA^T-A^TB^T=-BA+AB$，故 $AB-BA$ 是对称矩阵 +>(B) $(AB+BA)^T=(AB)^T+(BA)^T=B^TA^T+A^TB^T=-BA-AB$，故 $(AB+BA)$ 是反称矩阵 +>(C) $(BAB)^T=B^TA^TB^T=BAB$，故 $BAB$ 是对称矩阵 +>(D) $((AB)^2)^T=(ABAB)^T=B^TA^TB^TA^T=(BA)^2$，故 $(AB)^2$ 不一定有对称性 + +2.  设 $e_1, e_2$ 和 $\varepsilon_1, \varepsilon_2$ 是线性空间 $\mathbb{R}^2$ 的两组基，并且已知关系式 $\varepsilon_1 = e_1 + 5e_2,\ \varepsilon_2 = e_2,$ 则由基 $\varepsilon_1, \varepsilon_2$ 到基 $e_1, e_2$ 到基的过渡矩阵是 + (A) $\begin{bmatrix}0 & -1 \\ -6 & 0\end{bmatrix}$ + (B) $\begin{bmatrix}-1 & 0 \\5 & -1\end{bmatrix}$ + (C) $\begin{bmatrix}1 & 0 \\-5 & -1\end{bmatrix}$ + (D) $\begin{bmatrix}1 & 0 \\-5 & 1\end{bmatrix}$ + +>答案：**D** +>解析：$\varepsilon_1 = e_1 + 5e_2,\quad \varepsilon_2 = 0e_1 + 1e_2.$ +>把它们按列排成矩阵形式：$[\varepsilon_1, \varepsilon_2] = [e_1, e_2] \begin{bmatrix}1 & 0 \\5 & 1\end{bmatrix}.$ +>而由基 $e_1, e_2$ 到基 $\varepsilon_1, \varepsilon_2$ 的过渡矩阵 $T$ 应该满足$\begin{bmatrix}e_1&e_2\end{bmatrix}=\begin{bmatrix}\varepsilon_1&\varepsilon_2\end{bmatrix}T$，即： +>$\begin{bmatrix}e_1&e_2\end{bmatrix}=\begin{bmatrix}e_1&e_2\end{bmatrix}\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}T$，所以 $T=\begin{bmatrix}1 & 0 \\ 5 & 1\end{bmatrix}^{-1}=\begin{bmatrix}1 & 0 \\ -5 & 1\end{bmatrix}$ + +3. 设向量组 $\alpha_1 = (0, 0, c_1)^T,\quad \alpha_2 = (0, 1, c_2)^T,\quad \alpha_3 = (1, -1, c_3)^T,\quad \alpha_4 = (-1, 1, c_4)^T,$ + 其中 $c_1, c_2, c_3, c_4$ 为任意常数，则下列向量组线性相关的是 (A) $\alpha_1, \alpha_2, \alpha_3$; (B) $\alpha_1, \alpha_2, \alpha_4$; (C) $\alpha_1, \alpha_3, \alpha_4$; (D) $\alpha_2, \alpha_3, \alpha_4$. -4. 设 $A, B$ 为 $n$ 阶矩阵，则【 】 +>答案：**C** +>解析：逐选项分析： +>A. $\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}=\begin{bmatrix}0&0&1\\0&1&-1\\c_1&c_2&c_3\end{bmatrix}$，当$c_1\ne 0$时，作初等列变换，$\text{rank}\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}=\text{rank}\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}=3$，故线性无关； +>B. $\begin{bmatrix}\alpha_1&\alpha_2&\alpha_4\end{bmatrix}=\begin{bmatrix}0&0&-1\\0&1&1\\c_1&c_2&c_4\end{bmatrix}$，当$c_1\ne 0$时，作初等列变换，$\text{rank}\begin{bmatrix}0&0&-1\\0&1&1\\c_1&c_2&c_4\end{bmatrix}=\text{rank}\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}=3$，故线性无关； +>C. $\begin{bmatrix}\alpha_1&\alpha_3&\alpha_4\end{bmatrix}=\begin{bmatrix}0&1&-1\\0&-1&1\\c_1&c_3&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&-1\\0&0&1\\c_1&c_3+c_4&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&0\\0&0&1\\c_1&c_3+c_4&c_4\end{bmatrix}$，故其必定线性相关； +>D. $\begin{bmatrix}\alpha_1&\alpha_2&\alpha_4\end{bmatrix}=\begin{bmatrix}0&1&-1\\1&-1&1\\c_2&c_3&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&-1\\1&0&1\\c_2&c_3+c_4&c_4\end{bmatrix}$，当$c_3+c_4\ne 0$时，作初等列变换，$\text{rank}\begin{bmatrix}0&0&-1\\1&0&1\\c_2&c_3+c_4&c_4\end{bmatrix}=\text{rank}\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}=3$，故线性无关； +4. 设 $A, B$ 为 $n$ 阶矩阵，则 (A) $\text{rank}[A \ AB] = \text{rank} A$; (B) $\text{rank}[A \ BA] = \text{rank} A$; (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. -5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足【 】 +>答案：**A** +>重点：$AB$的每一列都是 $A$ 的列向量的线性组合，因此，$\text{rank}[A \quad AB]=\text{rank}A$ ，因为 $AB$ 的列都在 $A$ 的列空间中 +5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; (B) 将 $A^*$ 的第一行加上第二行的 2 倍得到 $B^*$; (C) 将 $A^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $B^*$; (D) 将 $A^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $B^*$. -6. 已知方程组 - $$ - \text{(I)} \quad - \begin{cases} - x_1 + 2x_2 + 3x_3 = 0, \\ - 2x_1 + 3x_2 + 5x_3 = 0, \\ - x_1 + x_2 + ax_3 = 0, - \end{cases} - $$ - 与 - $$ - \text{(II)} \quad - \begin{cases} - x_1 + bx_2 + cx_3 = 0, \\ - 2x_1 + b^2x_2 + (c+1)x_3 = 0 - \end{cases} - $$ - 同解，则【 】 +>答案：**D** +>解析：$B=AP(2,1(2))$，故$B^{-1}=P(2,1(-2))A^{-1}$， +>又$|A|=|B|$，$A^*=|A|A^{-1}$， +>可得$B^*=|B|B^{-1}=|A|P(2,1(-2))A^{-1}=P(2,1(-2))A^*$，根据初等矩阵的性质，$B^*$应该是$A^*$的第一行乘以$(-2)$加到第二行的结果。 +6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 (A) $a = 1, b = 0, c = 1$; (B) $a = 1, b = 1, c = 2$; (C) $a = 2, b = 0, c = 1$; (D) $a = 2, b = 1, c = 2$. -## 二、填空题，共六道，每题3分，共18分 - -7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积 - $$ - \langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}. - $$ - -8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 - ---- - -解析： -先计算$A^2$： - -$$A^2 -= \begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ -$$= \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix} -$$$$= \begin{bmatrix}18&-6\\-54&18\end{bmatrix} $$ -$$= 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ - -由此递推： -- $$A^3 = A^2 \cdot A = 6A \cdot A = 6A^2 = 6\times6A = 6^2A$$ -- 归纳可得当$n \geq 1$时，$A^n = 6^{n-1}A$ - - - -将A代入得： -$$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ - --- +>答案：**D** +>解析：$\text{(I)}:\begin{bmatrix}1&2&3\\2&3&5\\1&1&a\end{bmatrix}x=0$，$\text{(II)}: \begin{bmatrix}1&b&c\\2&b^2&c+1\end{bmatrix}x=0$，对方程做初等行变换： +>$\text{(I)}:\begin{bmatrix}1&0&1\\0&1&1\\0&0&a-2\end{bmatrix}x=0$，$\text{(II)}: \begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}x=0$，记系数矩阵分别为$A,B$ +>因为方程(I),(II)同解，所以$\text{rank}A=\text{rank}B$，而$\text{rank}A\ge 2,\text{rank}B\le 2$，故$\text{rank}A=\text{rank}B=2$，故$a-2=0 \to a=2$；所以方程组(I)的解为$x=k(1,1,-1)^T$； +>令$k=1,x=(1,1,-1)^T$代入方程组(II)得$\begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix}=\begin{bmatrix}1+b-c\\b^2-2b+c-1\end{bmatrix}=0$，解得$\begin{cases}b=0\\c=1\end{cases}$或$\begin{cases}b=1\\c=2\end{cases}$；然而，当$\begin{cases}b=0\\c=1\end{cases}$时，$\begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}=\begin{bmatrix}1&0&1\\0&0&0\end{bmatrix}$，不符合$\text{rank}B=2$的约束，故舍去； +>综上，$\begin{cases}a=2\\b=1\\c=2\end{cases}$ +## 二、填空题，共六道，每题3分，共18分 -9. 若向量组 - $$ - \alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T - $$ - 不能由向量组 - $$ - \beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 = (3,4,a)^T - $$ - 线性表示，则 - $$ - a = \underline{\qquad\qquad}. - $$ - ---- +7. 已知向量 $\alpha_1 = (1,0,-1,0)^T$，$\alpha_2 = (1,1,-1,-1)^T$，$\alpha_3 = (-1,0,1,1)^T$，则向量 $\alpha_1 + 2\alpha_2$ 与 $2\alpha_1 + \alpha_3$ 的内积$\langle \alpha_1 + 2\alpha_2,\, 2\alpha_1 + \alpha_3 \rangle = \underline{\qquad\qquad}.$ -10. 设矩阵 - $$ - A = \begin{bmatrix} - 1 & a_1 & a_1^2 & a_1^3 \\ - 1 & a_2 & a_2^2 & a_2^3 \\ - 1 & a_3 & a_3^2 & a_3^3 \\ - 1 & a_4 & a_4^2 & a_4^3 - \end{bmatrix},\quad - x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},\quad - b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, - $$ - 其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为 - $$ - \underline{\qquad\qquad\qquad\qquad}. - $$ ---- -解析： +8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\qquad\qquad\quad\quad}$。 -一眼顶针，鉴定为： $$ - x = (1,0,0,0)^T$$ ---- +9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad\beta_2 = (1,2,3)^T,\quad\beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ +10. 设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ 11. 矩阵(陈峰华原创题)$$A=\begin{bmatrix} 0 & 0 & 0 & \cdots & 0 & 1 & 1 & \cdots & 1 & 1 & 1 \\ @@ -202,19 +102,12 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ 0 & 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & 0 & 0 & 0 \end{bmatrix}_{(mn) \times (mn)}$$其中第一行有$m$个$0$.若$A^k=O$，则 $k$ 的最小值为$\underline{\qquad\qquad\qquad\qquad}.$ ---- -解析： - - 12. 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ - ## 三、解答题，共五道，共64分 ---- 13. （20 分）计算 下面的两个$n$阶行列式 - - $$ +$$ K_n = \begin{vmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 1 & 2 & \cdots & n-2 & n-1 \\ @@ -224,8 +117,6 @@ $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ n & n-1 & n-2 & \cdots & 2 & 1 \end{vmatrix}. $$ - - $$ M_n =\begin{vmatrix} 1+x_1 & 1+x_1^2 & \cdots & 1+x_1^n \\ @@ -233,392 +124,14 @@ $$ \vdots & \vdots & \ddots & \vdots \\ 1+x_n & 1+x_n^2 & \cdots & 1+x_n^n \end{vmatrix} - -$$ - ---- - -解析 -（1）$K_n$： -从第 $n-1$ 行开始，依次乘以 $(-1)$ 加到下一行，再把第 $n$ 列加到前面各列，得 - -$$ -\begin{aligned} -K_n &= -\begin{vmatrix} -1 & 2 & 3 & \cdots & n-1 & n \\ -2 & 1 & 2 & \cdots & n-2 & n-1 \\ -3 & 2 & 1 & \cdots & n-3 & n-2 \\ -\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -n-1 & n-2 & n-3 & \cdots & 1 & 2 \\ -n & n-1 & n-2 & \cdots & 2 & 1 -\end{vmatrix} \\[4pt] -&= -\begin{vmatrix} -1 & 2 & 3 & \cdots & n-1 & n \\ -1 & -1 & -1 & \cdots & -1 & -1 \\ -1 & 1 & -1 & \cdots & -1 & -1 \\ -\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -1 & 1 & 1 & \cdots & -1 & -1 \\ -1 & 1 & 1 & \cdots & 1 & -1 -\end{vmatrix} - -\end{aligned} -$$ - -继续化简： - -$$ -\begin{aligned} -&= -\begin{vmatrix} -n+1 & n+2 & n+3 & \cdots & 2n-1 & n \\ -0 & -2 & -2 & \cdots & -2 & -1 \\ -0 & 0 & -2 & \cdots & -2 & -1 \\ -\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ -0 & 0 & 0 & \cdots & -2 & -1 \\ -0 & 0 & 0 & \cdots & 0 & -1 -\end{vmatrix} - -\end{aligned} -$$ - -这是一个上三角行列式，因此 - -$$ -D_n = (-1)^{n-1} \cdot 2^{n-2} \cdot (n+1) - -$$ ---- -（2）加边 - - -$$ -\tilde{D} = -\begin{vmatrix} -1 & 0 & 0 & \cdots & 0 \\ -1 & 1 + x_1 & 1 + x_1^2 & \cdots & 1 + x_1^n \\ -1 & 1 + x_2 & 1 + x_2^2 & \cdots & 1 + x_2^n \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & 1 + x_n & 1 + x_n^2 & \cdots & 1 + x_n^n -\end{vmatrix} - -$$ - -$$ -\tilde{D} = -\begin{vmatrix} -1 & -1 & -1 & \cdots & -1 \\ -1 & x_1 & x_1^2 & \cdots & x_1^n \\ -1 & x_2 & x_2^2 & \cdots & x_2^n \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & x_n & x_n^2 & \cdots & x_n^n -\end{vmatrix} -$$ - -将第一行拆为 $(2,0,0,\dots,0)$ 与 $(-1,-1,\dots,-1)$ 之和： - -$$ -\tilde{D} = -\begin{vmatrix} -2 & 0 & 0 & \cdots & 0 \\ -1 & x_1 & x_1^2 & \cdots & x_1^n \\ -1 & x_2 & x_2^2 & \cdots & x_2^n \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & x_n & x_n^2 & \cdots & x_n^n -\end{vmatrix} -+ -\begin{vmatrix} --1 & -1 & -1 & \cdots & -1 \\ -1 & x_1 & x_1^2 & \cdots & x_1^n \\ -1 & x_2 & x_2^2 & \cdots & x_2^n \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & x_n & x_n^2 & \cdots & x_n^n -\end{vmatrix} -$$ - -令左边为 $A$，右边为 $B$。 - -计算 $A$，按第一行展开： - -$$ -A = 2 \cdot -\begin{vmatrix} -x_1 & x_1^2 & \cdots & x_1^n \\ -x_2 & x_2^2 & \cdots & x_2^n \\ -\vdots & \vdots & \ddots & \vdots \\ -x_n & x_n^2 & \cdots & x_n^n -\end{vmatrix} -= 2 \cdot \left( \prod_{i=1}^{n} x_i \right) \cdot -\begin{vmatrix} -1 & x_1 & \cdots & x_1^{n-1} \\ -1 & x_2 & \cdots & x_2^{n-1} \\ -\vdots & \vdots & \ddots & \vdots \\ -1 & x_n & \cdots & x_n^{n-1} -\end{vmatrix} -$$ - -右边为范德蒙德行列式： - -$$ -A = 2 \prod_{i=1}^{n} x_i \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) -$$ - -计算 $B$，提出第一行的因子 $-1$： - -$$ -B = (-1) \cdot -\begin{vmatrix} -1 & 1 & 1 & \cdots & 1 \\ -1 & x_1 & x_1^2 & \cdots & x_1^n \\ -1 & x_2 & x_2^2 & \cdots & x_2^n \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -1 & x_n & x_n^2 & \cdots & x_n^n -\end{vmatrix} -$$ - -该行列式为 $n+1$ 阶范德蒙德行列式，变量为 $1, x_1, x_2, \dots, x_n$： - -$$ -B = (-1) \cdot \prod_{i=1}^{n} (x_i - 1) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) -$$ - -因此： - -$$ -\tilde{D} = A + B = \left( 2 \prod_{i=1}^{n} x_i - \prod_{i=1}^{n} (x_i - 1) \right) \cdot \prod_{1 \leq i < j \leq n} (x_j - x_i) -$$ - -$$ -\boxed{\tilde{D} = \left(2\prod\limits_{i=1}^{n}x_i - \prod\limits_{i=1}^{n}(x_i-1)\right) \prod\limits_{1\leq i<j\leq n}(x_j-x_i)} -$$ - - ---- -14. 设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. - - (1)证明：方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解； - (2)若方程组$Ax=\alpha$与方程组$Bx=\beta$不同解，求$a$的值. - ---- - -解析： -(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. - -(2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. - ---- - -15. （10 分）设 - $$ - \alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T - $$ - 和 - $$ - \beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T - $$ - 是 $\mathbb{R}^3$ 的两组基，求向量 - $$ - u = \alpha_1 + 2\alpha_2 - 3\alpha_3 - $$ - 在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 - ---- - -解析： - -已知： - -$$ -A = (\alpha_1, \alpha_2, \alpha_3) = -\begin{bmatrix} -1 & 2 & 1 \\ -0 & 1 & 1 \\ --1 & 1 & 1 -\end{bmatrix}, -$$ - -$$ -B = (\beta_1, \beta_2, \beta_3) = -\begin{bmatrix} -0 & -1 & 0 \\ -1 & 1 & 2 \\ -1 & 0 & 1 -\end{bmatrix}. -$$ - -设 $u$ 在基 $\alpha_1, \alpha_2, \alpha_3$ 下的坐标为 $x = (1, 2, -3)^T$，在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 $y$，则 - -$$ -u = (\alpha_1, \alpha_2, \alpha_3) x = (\beta_1, \beta_2, \beta_3) y, -$$ - -即 - -$$ -Ax = By. -$$ - -因为 $B$ 可逆，所以 - -$$ -y = B^{-1} A x. $$ - -用增广矩阵求解 $y$： - -$$ -(B, Ax) = -\begin{bmatrix} -0 & -1 & 0 & \vert & 2 \\ -1 & 1 & 2 & \vert & -1 \\ -1 & 0 & 1 & \vert & -2 -\end{bmatrix} -$$ - -作行初等变换： - -$$ -\begin{aligned} -&\rightarrow -\begin{bmatrix} -1 & 0 & 1 & \vert & -2 \\ -0 & 1 & 1 & \vert & 1 \\ -0 & -1 & 0 & \vert & 2 -\end{bmatrix} \\[1em] -&\rightarrow -\begin{bmatrix} -1 & 0 & 1 & \vert & -2 \\ -0 & 1 & 1 & \vert & 1 \\ -0 & 0 & 1 & \vert & 3 -\end{bmatrix} \\[1em] -&\rightarrow -\begin{bmatrix} -1 & 0 & 0 & \vert & -5 \\ -0 & 1 & 0 & \vert & -2 \\ -0 & 0 & 1 & \vert & 3 -\end{bmatrix}. -\end{aligned} -$$ - -因此向量 - -$$ -u = \alpha_1 + 2\alpha_2 - 3\alpha_3 -$$ - -在基 $\beta_1, \beta_2, \beta_3$ 下的坐标为 - -$$ -y = (-5, -2, 3)^T. -$$ - ---- - - +14. （10 分）设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. + (1)证明：方程组 $Ax=\alpha$ 的解均为方程组 $Bx=\beta$ 的解； + (2)若方程组 $Ax=\alpha$ 与方程组 $Bx=\beta$ 不同解，求 $a$ 的值. +15. （10 分）设 $\alpha_1 = (1,0,-1)^T,\quad \alpha_2 = (2,1,1)^T,\quad \alpha_3 = (1,1,1)^T$和$\beta_1 = (0,1,1)^T,\quad \beta_2 = (-1,1,0)^T,\quad \beta_3 = (0,2,1)^T$是 $\mathbb{R}^3$ 的两组基，求向量 $u = \alpha_1 + 2\alpha_2 - 3\alpha_3$在基 $\beta_1, \beta_2, \beta_3$ 下的坐标。 16. （12 分）设 $n$ 阶方阵 $A, B$ 满足 $AB = A + B$。 - （1）证明 $A - E$ 可逆； - （2）证明 $AB = BA$； - （3）证明 $\mathrm{rank}(A) = \mathrm{rank}(B)$； - - （4）若矩阵 - $$ - B = \begin{bmatrix} - 1 & -3 & 0 \\ - 2 & 1 & 0 \\ - 0 & 0 & 2 - \end{bmatrix}, - $$ - 求矩阵 $A$。 - - ---- - - -**【解】** - -**(1)** -由 $AB = A + B$ 得 $(A - E)(B - E) = E$，因此 $A - E$ 可逆。 -$$\text{……3 分}$$ - -**(2)** -由 $(A - E)(B - E) = E$ 得 $(B - E)(A - E) = E$，因此 $AB = BA$。 -$$\text{……6 分}$$ - -**(3)** -由 $AB = A + B$ 得 $A = (A - E)B$，而 $A - E$ 可逆，故 -$$ -\mathrm{rank}(A) = \mathrm{rank}(B). -$$ -$$\text{……9 分}$$ - -**(4)** -由 $AB = A + B$ 得 $A(B - E) = B$，而 $B - E$ 可逆，故 -$$ -A = B(B - E)^{-1}. -$$ -已知 -$$ -B = \begin{bmatrix} -1 & -3 & 0 \\ -2 & 1 & 0 \\ -0 & 0 & 2 -\end{bmatrix}, -$$ -则 -$$ -B - E = \begin{bmatrix} -0 & -3 & 0 \\ -2 & 0 & 0 \\ -0 & 0 & 1 -\end{bmatrix}. -$$ -求逆得 -$$ -(B - E)^{-1} = \begin{bmatrix} -0 & \frac12 & 0 \\[2pt] --\frac13 & 0 & 0 \\[2pt] -0 & 0 & 1 -\end{bmatrix}. -$$ -于是 -$$ -A = B(B - E)^{-1} = \begin{bmatrix} -1 & -3 & 0 \\ -2 & 1 & 0 \\ -0 & 0 & 2 -\end{bmatrix} -\begin{bmatrix} -0 & \frac12 & 0 \\[2pt] --\frac13 & 0 & 0 \\[2pt] -0 & 0 & 1 -\end{bmatrix} -= \begin{bmatrix} -1 & \frac12 & 0 \\[2pt] --\frac13 & 1 & 0 \\[2pt] -0 & 0 & 2 -\end{bmatrix}. -$$ -$$\text{……12 分}$$ - ---- - -17. 设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组Ax=0的基础解系中含有两个解向量，求Ax=0的通解。 - ---- - -解析： -因为n=4，$n-\text{rank}A=2$，所以$\text{rank}A=2$。 -对A施行初等行变换，得 -$$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&t-2&-1&-1\end{bmatrix}$$ - -$$\to\begin{bmatrix}1&2&1&2\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}\to\begin{bmatrix}1&0&1-2t&2-2t\\0&1&t&t\\0&0&-(1-t)^2&-(1-t)^2\end{bmatrix}$$ - - -要使$\text{rank}A=2$，则必有t=1。 -此时，与Ax=0同解的方程组为$\begin{cases}x_1=x_3\\x_2=-x_3-x_4\end{cases}$，得基础解系为 -$$\boldsymbol{\xi}_1=\begin{bmatrix}1\\-1\\1\\0\end{bmatrix},\ \boldsymbol{\xi}_2=\begin{bmatrix}0\\-1\\0\\1\end{bmatrix}$$ -方程组的通解为$$\boldsymbol{x}=k_1\boldsymbol{\xi}_1+k_2\boldsymbol{\xi}_2，（k_1,k_2为任意常数）$$ \ No newline at end of file + （4）若矩阵$B = \begin{bmatrix}1 & -3 & 0 \\2 & 1 & 0 \\0 & 0 & 2\end{bmatrix}$，求矩阵 $A$。 +17. （12 分）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组 $Ax=0$ 的基础解系中含有两个解向量，求 $Ax=0$ 的通解。 \ No newline at end of file -- 2.34.1 From d1e0d0e235b63490232f89c63c5ff2aef70d2981 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:26:37 +0800 Subject: [PATCH 133/274] vault backup: 2025-12-31 20:26:37 --- .../试卷/1231线性代数考试卷（解析版）.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index ee3669b..19664aa 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -220,8 +220,7 @@ $$ --- -解析：由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得 -$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$ +解析：类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ --- ## 三、解答题，共五道，共64分 -- 2.34.1 From c567ab0917f21546bf684bd872a85cb6243c0a7b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 31 Dec 2025 20:35:13 +0800 Subject: [PATCH 134/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 未命名 1.canvas | 1 + 未命名.canvas | 1 + 编写小组/试卷/0103高数模拟试卷.md | 567 +++++++++++++++++- 3 files changed, 568 insertions(+), 1 deletion(-) create mode 100644 未命名 1.canvas create mode 100644 未命名.canvas diff --git a/未命名 1.canvas b/未命名 1.canvas new file mode 100644 index 0000000..9e26dfe --- /dev/null +++ b/未命名 1.canvas @@ -0,0 +1 @@ +{} \ No newline at end of file diff --git a/未命名.canvas b/未命名.canvas new file mode 100644 index 0000000..9e26dfe --- /dev/null +++ b/未命名.canvas @@ -0,0 +1 @@ +{} \ No newline at end of file diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index ed86bba..2e28b62 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -2,11 +2,576 @@ tags: - 编写小组 --- -## 一、单选题（共5小题，每小题2分，共10分） +## 一、单选题（共5小题，每小题2分，共10分） +1.设函数 $f(x)$ 在 $x = 0$ 处可导，且 $f(0) = 0$。若 +$$ +\lim_{x \to 0} \frac{f(2x) - 3f(x) + f(-x)}{x} = 2, +$$ +则 $f'(0)$ 的值为（ ）。 +A. $-2$ +B. $-1$ +C. $1$ +D. $2$ + +## 解析 +因为 $f(x)$ 在 $x=0$ 处可导且 $f(0)=0$，所以 $f'(0) = \lim\limits_{x \to 0} \frac{f(x)}{x}$ 存在。 + +将极限式分解： +$$ +\frac{f(2x) - 3f(x) + f(-x)}{x} = \frac{f(2x)}{x} - 3\frac{f(x)}{x} + \frac{f(-x)}{x}. +$$ + +分别计算各项极限： +$$ +\begin{aligned} +\lim_{x \to 0} \frac{f(2x)}{x} &= 2 \lim_{x \to 0} \frac{f(2x)}{2x} = 2f'(0), \\ +\lim_{x \to 0} \frac{f(x)}{x} &= f'(0), \\ +\lim_{x \to 0} \frac{f(-x)}{x} &= -\lim_{x \to 0} \frac{f(-x)}{-x} = -f'(0). +\end{aligned} +$$ + +因此， +$$ +\lim_{x \to 0} \frac{f(2x) - 3f(x) + f(-x)}{x} = 2f'(0) - 3f'(0) + (-f'(0)) = -2f'(0). +$$ + +由已知条件 $-2f'(0) = 2$，解得 $f'(0) = -1$。 + +**答案：B** + + +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2}$ +A.$-\frac{1}{12}$ +B.$\frac{1}{12}$ +C.$-\frac{1}{6}$ +D$\frac{1}{6}$ + +**解：** + +分别对分子和分母进行泰勒展开： + +分子： +$$ +e^{x^2} \cos x = \left(1 + x^2 + \frac{x^4}{2} + o(x^4)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)\right) = 1 + \frac{1}{2}x^2 + \left(-\frac{1}{2} + \frac{1}{2} + \frac{1}{24}\right)x^4 + o(x^4) = 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) +$$ +所以 $e^{x^2} \cos x - 1 - \frac{1}{2}x^2 = \frac{1}{24}x^4 + o(x^4)$ + +分母： +$$ +\ln(1+x^2) = x^2 - \frac{1}{2}x^4 + o(x^4) +$$ +所以 $\ln(1+x^2) - x^2 = -\frac{1}{2}x^4 + o(x^4)$ + +因此 +$$ +\lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2} = \lim_{x \to 0} \frac{\frac{1}{24}x^4 + o(x^4)}{-\frac{1}{2}x^4 + o(x^4)} = \frac{1/24}{-1/2} = -\frac{1}{12} +$$ + + +3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) dx$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 + +(A) $a = 1, \, b = 1$ +(B) $a = 1, \, b = 2$ +(C) $a = 2, \, b = 1$ +(D) $a = 2, \, b = 2$ + +**解：** +为使 $f(x)$ 与 $g(x)$ 在 $x \to 0$ 时为同阶无穷小，需使二者最低阶非零项的阶数相同。下面分别展开 $f(x)$ 和 $g(x)$ 的麦克劳林公式。 + +对于 $f(x)$，利用已知展开式： +$$ +e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), +$$ +则 +$$ +e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). +$$ +因此 +$$ +2e^x \sin x = 2x + 2x^2 + \frac{2}{3}x^3 + O(x^5). +$$ +代入 $f(x)$ 得 +$$ +f(x) = 2x + 2x^2 + \frac{2}{3}x^3 - 2ax - bx^2 + O(x^5) = (2 - 2a)x + (2 - b)x^2 + \frac{2}{3}x^3 + O(x^5). +$$ + +对于 $g(x)$，由 $g(0)=0$ 及 $\arctan(x^2)$ 的展开： +$$ +\arctan(x^2) = x^2 - \frac{x^6}{3} + O(x^{10}), +$$ +积分得 +$$ +g(x) = \int_0^x \arctan(t^2) dt = \frac{x^3}{3} - \frac{x^7}{21} + O(x^{11}) = \frac{1}{3}x^3 + O(x^7). +$$ +可见 $g(x)$ 是 $x \to 0$ 时的三阶无穷小，其主项为 $\dfrac{1}{3}x^3$。 + +为使 $f(x)$ 也是三阶无穷小（即与 $g(x)$ 同阶），$f(x)$ 中 $x$ 和 $x^2$ 的系数必须为零： +$$ +2 - 2a = 0, \quad 2 - b = 0, +$$ +解得 $a = 1$, $b = 2$。此时 +$$ +f(x) = \frac{2}{3}x^3 + O(x^5), +$$ +与 $g(x)$ 同阶（但不等价，因为系数比值为 $2$）。 + +因此正确选项为 (B)。 + +**答案：** (B) ## 二、填空题（共5小题，每小题2分，共10分） +1.设 $y = \arctan x$，求 $y^{(n)}(0)$ + +**分析** +逐次求导以找到 $n$ 阶导数的规律。由于 $y' = \frac{1}{1 + x^2}$，即 $(1 + x^2)y' = 1$，故想到用莱布尼茨公式。 + +**解** + +**方法1** +因为 $y' = \frac{1}{1 + x^2}$，所以 $(1 + x^2)y' = 1$。 +上述两端对 $x$ 求 $n$ 阶导数，并利用莱布尼茨公式，得 +$$\sum_{k=0}^{n} C_n^k (1 + x^2)^{(k)} (y')^{(n-k)} = (1 + x^2)y^{(n+1)} + 2nxy^{(n)} + n(n-1)y^{(n-1)} = 0.$$ +在上式中令 $x = 0$，得 +$$y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0).$$ +由此递推公式，再加上 $y'(0) = 1, y''(0) = 0$，可得： +$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ + +**方法2** +因为 $y' = \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， +所以 +$$y = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, (|x| < 1).$$ +又知 $y(x)$ 在点 $x = 0$ 处的麦克劳林展开式为 +$$y(x) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!} x^n.$$ +比较系数可得 +$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ + +**方法3** +由 $y = \arctan x$，得 $x = \tan y$，则 +$$y' = \frac{1}{1 + x^2} = \frac{1}{1 + \tan^2 y} = \cos^2 y.$$ +利用复合函数求导法则： +$$y'' = -2 \cos y \sin y \cdot y' = -\sin(2y) \cos^2 y = \cos^2 y \sin 2\left( y + \frac{\pi}{2} \right),$$ +$$y''' = \left[ -2 \cos y \sin y \sin 2\left( y + \frac{\pi}{2} \right) + 2 \cos^2 y \cos 2\left( y + \frac{\pi}{2} \right) \right] y' = 2 \cos^3 y \cos\left( y + \frac{\pi}{2} \right) y' + 2 \cos^3 y \sin 3\left( y + \frac{\pi}{2} \right),$$ +$$y^{(4)} = 6 \cos^4 y \cos\left( y + \frac{\pi}{2} \right) y' + 3 \cos^4 y \sin 4\left( y + \frac{\pi}{2} \right).$$ +由此归纳出 +$$y^{(n)} = (n-1)! \cos^n y \sin n\left( y + \frac{\pi}{2} \right).$$ +下面用归纳法证明以上结论： +当 $n = 1$ 时结论成立； +假设 $n = k$ 时结论成立，则当 $n = k + 1$ 时， +$$\begin{aligned} y^{(k+1)} &= (y^{(k)})' \\ &= (k-1)! \left[ -k \cos^{k+1} y \sin y \sin k\left( y + \frac{\pi}{2} \right) + k \cos^k y \cos k\left( y + \frac{\pi}{2} \right) \right] y' \\ &= k \cos^{k+1} y \cos\left( y + \frac{\pi}{2} \right) y' + k \cos^{k+1} y \sin (k+1)\left( y + \frac{\pi}{2} \right). \end{aligned}$$ +由于 $y(0) = \arctan 0 = 0, \cos 0 = 1$，因此 +$$y^{(n)}(0) = (n-1)! \sin \frac{n\pi}{2}.$$ +根据 $\sin \frac{n\pi}{2}$ 的取值（$n$ 为偶数时为 $0$，$n=4k+1$ 时为 $1$，$n=4k+3$ 时为 $-1$），可得到与方法1、2相同的结果。 + + + +2.已知 $f(x)$ 是三次多项式，且有 $\lim_{x \to 2a} \frac{f(x)}{x-2a} = \lim_{x \to 4a} \frac{f(x)}{x-4a} = 1$，求 $\lim_{x \to 3a} \frac{f(x)}{x-3a}$ + +**分析** +由已知的两个极限式可确定 $f(x)$ 的两个一次因子以及两个待定系数，从而完全确定 $f(x)$。 + +**解** +由已知有 $\lim_{x \to 2a} f(x) = \lim_{x \to 4a} f(x) = 0$。 +由于 $f(x)$ 处处连续，故 $f(2a) = f(4a) = 0$。 +因此 $f(x)$ 含有因式 $(x-2a)(x-4a)$，可设 +$$f(x) = (Ax + B)(x - 2a)(x - 4a).$$ +由条件： +$$\lim_{x \to 2a} \frac{f(x)}{x-2a} = \lim_{x \to 2a} (Ax + B)(x - 4a) = (2aA + B)(-2a) = 1,$$ +$$\lim_{x \to 4a} \frac{f(x)}{x-4a} = \lim_{x \to 4a} (Ax + B)(x - 2a) = (4aA + B)(2a) = 1.$$ +解方程组： +$$\begin{cases} (2aA + B)(-2a) = 1 \\ (4aA + B)(2a) = 1 \end{cases} \Rightarrow \begin{cases} -4a^2A - 2aB = 1 \\ 8a^2A + 2aB = 1 \end{cases}.$$ +相加得 $4a^2A = 2$，故 $A = \frac{1}{2a^2}$；代入第一个方程得 $-4a^2 \cdot \frac{1}{2a^2} - 2aB = 1$，即 $-2 - 2aB = 1$，解得 $B = -\frac{3}{2a}$。 +于是 +$$f(x) = \left( \frac{1}{2a^2}x - \frac{3}{2a} \right)(x - 2a)(x - 4a) = \frac{1}{2a^2}(x - 3a)(x - 2a)(x - 4a).$$ +最后， +$$\lim_{x \to 3a} \frac{f(x)}{x-3a} = \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} = \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) = \frac{1}{2a^2} \cdot a \cdot (-a) = -\frac{1}{2}.$$ + +**答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. +# 高等数学题解集 + +## 1. 级数收敛性选择题 + +### 题目 +下列级数中收敛的是（ ）。 + +(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ + +(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ + +(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ + +(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ + +### 解析 +- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 +- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 +- **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 +- **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 + +**答案：(B)** + +--- + + +--- + + +--- + ## 三、解答题（共11小题，共80分） +1.求下列不定积分(提示，换元），其中 $a > 0$ + +(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} dx$ + +**解：** +令 $x = a \sin t$，则： + +$$ +\begin{aligned} +\int \frac{x^2}{\sqrt{a^2 - x^2}} dx &= \int \frac{a^2 \sin^2 t}{a \cos t} \cdot a \cos t \, dt \\ +&= a^2 \int \sin^2 t \, dt \\ +&= \frac{a^2}{2} \int (1 - \cos 2t) \, dt \\ +&= \frac{a^2}{2} \left( t - \frac{1}{2} \sin 2t \right) + C \\ +&= \frac{a^2}{2} \arcsin \frac{x}{a} - \frac{x}{2} \sqrt{a^2 - x^2} + C +\end{aligned} +$$ + +--- + + (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} dx$ + +**解：** +**方法一：** 令 $x = a \tan t$，则： + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 + a^2}}{x^2} dx &= \int \frac{a \sec t}{a^2 \tan^2 t} \cdot a \sec^2 t \, dt \\ +&= \int \frac{\sec^3 t}{\tan^2 t} \, dt \\ +&= \int \frac{1}{\sin^2 t \cos t} \, dt \\ +&= \int \frac{\sin^2 t + \cos^2 t}{\sin^2 t \cos t} \, dt \\ +&= \int \sec t \, dt + \int \frac{\cos t}{\sin^2 t} \, dt \\ +&= \ln |\sec t + \tan t| - \frac{1}{\sin t} + C \\ +&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C +\end{aligned} +$$ + +**方法二：** + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 + a^2}}{x^2} dx &= \int \frac{x^2 + a^2}{x^2 \sqrt{x^2 + a^2}} dx \\ +&= \int \frac{1}{\sqrt{x^2 + a^2}} dx + a^2 \int \frac{1}{x^2 \sqrt{x^2 + a^2}} dx \\ +&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C +\end{aligned} +$$ + +--- + +(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} dx$ + +**解：** +令 $x = a \sec t$，则： + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 - a^2}}{x} dx &= \int \frac{a \tan t}{a \sec t} \cdot a \sec t \tan t \, dt \\ +&= a \int \tan^2 t \, dt \\ +&= a \int (\sec^2 t - 1) \, dt \\ +&= a (\tan t - t) + C \\ +&= \sqrt{x^2 - a^2} - a \arccos \frac{a}{x} + C +\end{aligned} +$$ + +--- + +(4) $\displaystyle \int \sqrt{1 + e^x} dx$ + +**解：** +令 $t = \sqrt{1 + e^x}$，则 $e^x = t^2 - 1$，$x = \ln(t^2 - 1)$，$dx = \frac{2t}{t^2 - 1} dt$： + +$$ +\begin{aligned} +\int \sqrt{1 + e^x} dx &= \int t \cdot \frac{2t}{t^2 - 1} dt \\ +&= \int \frac{2t^2}{t^2 - 1} dt \\ +&= 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) dt \\ +&= 2 \int 1 \, dt + \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) dt \\ +&= 2t + \ln \left| \frac{t-1}{t+1} \right| + C \\ +&= 2 \sqrt{1 + e^x} + \ln \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} + C +\end{aligned} +$$ + + + +$$ + +$$ + +--- + + +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x}$。 + +**解：** + +当 $x \to 0$ 时，利用等价无穷小替换和泰勒展开： + +分母：$x^3 \arcsin x \sim x^3 \cdot x = x^4$ + +分子：$e^{x^2} - 1 - \ln(1+x^2) = \left(1+x^2+\frac{x^4}{2}+o(x^4)\right) - 1 - \left(x^2-\frac{x^4}{2}+o(x^4)\right) = x^4 + o(x^4)$ + +所以 +$$ +\lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x} = \lim_{x \to 0} \frac{x^4 + o(x^4)}{x^4} = 1 +$$ + + + +设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 +证明：存在 $\xi \in (0, \frac{1}{2})$，使得 $f''(\xi) = \frac{3f'(\xi)}{1-2\xi}$。 + +**证明：** +构造辅助函数 $g(x) = (1-2x)^{3/2} f'(x)$，$x \in [0, \frac{1}{2}]$。 +由于 $f$ 二阶可导，故 $g$ 在 $[0, \frac{1}{2}]$ 上连续，在 $(0, \frac{1}{2})$ 内可导。 +计算 $g$ 的导数： + +$$ +\begin{aligned} +g'(x) &= (1-2x)^{3/2} f''(x) + \frac{3}{2}(1-2x)^{1/2} \cdot (-2) \cdot f'(x) \\ +&= (1-2x)^{1/2} \left[ (1-2x) f''(x) - 3f'(x) \right] +\end{aligned} +$$ + +当 $x \in (0, \frac{1}{2})$ 时，$(1-2x)^{1/2} > 0$，因此 $g'(x) = 0$ 当且仅当： + +$$ +(1-2x) f''(x) - 3f'(x) = 0 \quad \text{即} \quad f''(x) = \frac{3f'(x)}{1-2x} +$$ + +所以只需证明存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +由已知条件： + +$$ +g(0) = (1-0)^{3/2} f'(0) = f'(0) = f(0), \quad g\left(\frac{1}{2}\right) = (1-1)^{3/2} f'\left(\frac{1}{2}\right) = 0 +$$ + +分两种情况讨论： + +1. **若 $f'(0) = 0$** + 此时 $g(0) = 0$，所以 $g(0) = g\left(\frac{1}{2}\right) = 0$。由罗尔定理，存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +2. **若 $f'(0) \neq 0$** + 对 $f$ 在 $[0, \frac{1}{2}]$ 上应用拉格朗日中值定理，存在 $c \in (0, \frac{1}{2})$ 使得： + + $$ + f'(c) = \frac{f\left(\frac{1}{2}\right) - f(0)}{\frac{1}{2} - 0} = \frac{0 - f(0)}{\frac{1}{2}} = -2f(0) = -2f'(0) + $$ + + 由于 $f'(0) \neq 0$，故 $f'(c) \neq 0$ 且与 $f'(0)$ 异号。 + 计算 $g(c) = (1-2c)^{3/2} f'(c)$，因为 $(1-2c)^{3/2} > 0$，所以 $g(c)$ 与 $f'(c)$ 同号，从而 $g(c)$ 与 $g(0) = f'(0)$ 异号。 + 由连续函数的介值定理，存在 $d \in (0, c)$ 使得 $g(d) = 0$。 + 在区间 $[d, \frac{1}{2}]$ 上，$g(d) = g\left(\frac{1}{2}\right) = 0$，由罗尔定理，存在 $\xi \in (d, \frac{1}{2}) \subset (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +综上所述，无论何种情况，均存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$，从而有： + +$$ +f''(\xi) = \frac{3f'(\xi)}{1-2\xi} +$$ + +原命题得证。 + + + +3.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 + + +### 方法一：逐阶求导归纳法 + +由 +$$ +y' = a e^{ax} \sin bx + b e^{ax} \cos bx += e^{ax} \bigl( a \sin bx + b \cos bx \bigr), +$$ +令 +$$ +\varphi = \arctan \frac{b}{a}, +$$ +则 +$$ +a \sin bx + b \cos bx = \sqrt{a^2 + b^2} \, \sin(bx + \varphi). +$$ +于是 +$$ +y' = e^{ax} \cdot \sqrt{a^2 + b^2} \, \sin(bx + \varphi). +$$ +同理， +$$ +y'' = \sqrt{a^2 + b^2} \cdot e^{ax} \bigl[ a \sin(bx + \varphi) + b \cos(bx + \varphi) \bigr] += (a^2 + b^2) e^{ax} \sin(bx + 2\varphi). +$$ +依此类推，由归纳法可得 +$$ +y^{(n)} = (a^2 + b^2)^{n/2} \, e^{ax} \sin(bx + n\varphi), +$$ +其中 +$$ +\varphi = \arctan \frac{b}{a}. +$$ + +--- + +### 方法二：欧拉公式法 + +设 +$$ +u = e^{ax} \cos bx, \quad v = e^{ax} \sin bx, +$$ +则 +$$ +u + iv = e^{ax} (\cos bx + i \sin bx) = e^{(a + bi)x}. +$$ +求 $n$ 阶导数： +$$ +u^{(n)} + i v^{(n)} = \bigl[ e^{(a + bi)x} \bigr]^{(n)} += (a + bi)^n e^{(a + bi)x}. +$$ +记 +$$ +a + bi = \sqrt{a^2 + b^2} \, e^{i\varphi}, +\quad \varphi = \arctan \frac{b}{a}, +$$ +则 +$$ +(a + bi)^n = (a^2 + b^2)^{n/2} e^{in\varphi}. +$$ +于是 +$$ +u^{(n)} + i v^{(n)} += (a^2 + b^2)^{n/2} e^{in\varphi} \cdot e^{ax} e^{ibx} += (a^2 + b^2)^{n/2} e^{ax} e^{i(bx + n\varphi)}. +$$ +取虚部，即得 +$$ +\bigl( e^{ax} \sin bx \bigr)^{(n)} += (a^2 + b^2)^{n/2} e^{ax} \sin(bx + n\varphi). +$$ +类似可得 +$$ +\bigl( e^{ax} \cos bx \bigr)^{(n)} += (a^2 + b^2)^{n/2} e^{ax} \cos(bx + n\varphi). +$$ + +4.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 + +**解：** +已知 $e^x$ 的麦克劳林展开为： +$$ +e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6). +$$ +则 +$$ +x^2 e^x = x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6)\right) = x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} + \frac{x^7}{5!} + \frac{x^8}{6!} + o(x^8). +$$ +由于我们只关心六阶展开，保留到 $x^6$ 项，$x^7$ 及更高次项可并入 $o(x^6)$，故 +$$ +x^2 e^x = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + o(x^6). +$$ +于是 +$$ +f(x) = x^2 e^x + x^6 = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + x^6 + o(x^6) = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{25}{24}x^6 + o(x^6). +$$ +这就是 $f(x)$ 的六阶带佩亚诺余项的麦克劳林公式。 + +由泰勒展开的唯一性，$x^6$ 项的系数为 $\frac{f^{(6)}(0)}{6!}$，即 +$$ +\frac{f^{(6)}(0)}{6!} = \frac{25}{24}. +$$ +所以 +$$ +f^{(6)}(0) = 6! \times \frac{25}{24} = 720 \times \frac{25}{24} = 30 \times 25 = 750. +$$ + +**答案：** +展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$。 + + + +**题目：** + + +5.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 +$$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ +证明： + +（1）数列 $\{x_n\}$ 收敛（记 $\lim_{n \to \infty} x_n = a$）； +（2）方程 $f(x) = x$ 有唯一实根 $x = a$。 + +**证明：** + +（1）首先证明 $\{x_n\}$ 是柯西数列。对任意 $n \geq 1$，由拉格朗日中值定理，存在 $\xi_n$ 介于 $x_n$ 和 $x_{n-1}$ 之间，使得 +$$|x_{n+1} - x_n| = |f(x_n) - f(x_{n-1})| = |f'(\xi_n)| \cdot |x_n - x_{n-1}| \leq r |x_n - x_{n-1}|.$$ +反复应用此不等式，得 +$$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdots \leq r^{n-1} |x_2 - x_1|.$$ +于是对任意正整数 $m > n$，有 +$$ +\begin{aligned} +|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\ +&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\ +&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\ +&\leq \frac{r^{n-1}}{1 - r} |x_2 - x_1|. +\end{aligned} +$$ +因为 $0 < r < 1$，所以当 $n \to \infty$ 时，$r^{n-1} \to 0$，从而对任意 $\varepsilon > 0$，存在 $N$，当 $m > n > N$ 时 $|x_m - x_n| < \varepsilon$。故 $\{x_n\}$ 是柯西数列，因此在实数域中收敛，记 $\lim_{n \to \infty} x_n = a$。 + +（2）由于 $f$ 可导，故连续。在递推式 $x_{n+1} = f(x_n)$ 两边取极限 $n \to \infty$，得 +$$a = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n) = f(a),$$ +即 $a$ 是方程 $f(x) = x$ 的一个实根。 + +下证唯一性。假设另有 $b \neq a$ 满足 $f(b) = b$，则由拉格朗日中值定理，存在 $\eta$ 介于 $a$ 和 $b$ 之间，使得 +$$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ +由于 $|a - b| > 0$，两边除以 $|a - b|$ 得 $1 \leq r$，与 $0 < r < 1$ 矛盾。故方程 $f(x) = x$ 有唯一实根 $x = a$。 + +综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。$\square$ + + + +6.设 $f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 + +**解** +令 $g(x) = \sin f(x)$，则 +$$ +g(x) - \frac{1}{3} g\left(\frac{1}{3}x\right) = x. +$$ + +依次将 $x$ 替换为 $\frac{x}{3}, \frac{x}{3^2}, \cdots, \frac{x}{3^{n-1}}$，并乘以相应系数，得 +$$ +\begin{aligned} +g(x) - \frac{1}{3} g\left(\frac{x}{3}\right) &= x, \\ +\frac{1}{3} g\left(\frac{x}{3}\right) - \frac{1}{3^2} g\left(\frac{x}{3^2}\right) &= \frac{1}{3^2} x, \\ +\frac{1}{3^2} g\left(\frac{x}{3^2}\right) - \frac{1}{3^3} g\left(\frac{x}{3^3}\right) &= \frac{1}{3^4} x, \\ +&\vdots \\ +\frac{1}{3^{n-1}} g\left(\frac{x}{3^{n-1}}\right) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) &= \frac{1}{3^{2(n-1)}} x. +\end{aligned} +$$ + +以上各式相加，得 +$$ +g(x) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = x \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right). +$$ + +因为 $|g(x)| \leq 1$，所以 $\lim\limits_{n \to \infty} \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = 0$。 +而 $\lim\limits_{n \to \infty} \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right) = \frac{1}{1 - \frac{1}{9}} = \frac{9}{8}$， +因此 +$$ +g(x) = \frac{9}{8} x. +$$ + +于是 $\sin f(x) = \frac{9}{8} x$，解得 +$$ +f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). +$$ + + -- 2.34.1 From 8563ec763978665d611cb9960c17588758d29c28 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:38:07 +0800 Subject: [PATCH 135/274] vault backup: 2025-12-31 20:38:06 --- 未命名 1.canvas | 1 - 未命名.canvas | 1 - 2 files changed, 2 deletions(-) delete mode 100644 未命名 1.canvas delete mode 100644 未命名.canvas diff --git a/未命名 1.canvas b/未命名 1.canvas deleted file mode 100644 index 9e26dfe..0000000 --- a/未命名 1.canvas +++ /dev/null @@ -1 +0,0 @@ -{} \ No newline at end of file diff --git a/未命名.canvas b/未命名.canvas deleted file mode 100644 index 9e26dfe..0000000 --- a/未命名.canvas +++ /dev/null @@ -1 +0,0 @@ -{} \ No newline at end of file -- 2.34.1 From 855f960b90db34b04683866fd78dd06ab4b5b2a7 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:49:39 +0800 Subject: [PATCH 136/274] vault backup: 2025-12-31 20:49:39 --- 编写小组/试卷/0103高数模拟试卷.md | 6 ------ 1 file changed, 6 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 2e28b62..62dc598 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -302,12 +302,6 @@ $$ \end{aligned} $$ - - -$$ - -$$ - --- -- 2.34.1 From 44eff74db1a2cfc0b5397745adbeef23e48f1dc2 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:52:48 +0800 Subject: [PATCH 137/274] vault backup: 2025-12-31 20:52:48 --- 编写小组/试卷/0103高数模拟试卷.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 62dc598..68aea5a 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -466,7 +466,9 @@ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} $$ 则 $$ -x^2 e^x = x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6)\right) = x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} + \frac{x^7}{5!} + \frac{x^8}{6!} + o(x^8). +\begin{aligned} +x^2 e^x &= x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6)\right)\\[1em] &= x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} + \frac{x^7}{5!} + \frac{x^8}{6!} + o(x^8). +\end{aligned} $$ 由于我们只关心六阶展开，保留到 $x^6$ 项，$x^7$ 及更高次项可并入 $o(x^6)$，故 $$ @@ -474,7 +476,9 @@ x^2 e^x = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + o(x^6). $$ 于是 $$ -f(x) = x^2 e^x + x^6 = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + x^6 + o(x^6) = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{25}{24}x^6 + o(x^6). +\begin{aligned} +f(x) &= x^2 e^x + x^6\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + x^6 + o(x^6)\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{25}{24}x^6 + o(x^6). +\end{aligned} $$ 这就是 $f(x)$ 的六阶带佩亚诺余项的麦克劳林公式。 -- 2.34.1 From 910eacbab9f6a521981c993e4ff073e630932df2 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:53:23 +0800 Subject: [PATCH 138/274] vault backup: 2025-12-31 20:53:23 --- 编写小组/试卷/0103高数模拟试卷.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 68aea5a..824a21b 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -515,9 +515,9 @@ $$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdot 于是对任意正整数 $m > n$，有 $$ \begin{aligned} -|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\ -&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\ -&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\ +|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\[1em] +&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\[1em] +&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\[1em] &\leq \frac{r^{n-1}}{1 - r} |x_2 - x_1|. \end{aligned} $$ -- 2.34.1 From 14db05f2e07f1fc35df566a7858c261642ae3e0c Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:53:40 +0800 Subject: [PATCH 139/274] vault backup: 2025-12-31 20:53:40 --- 编写小组/试卷/0103高数模拟试卷.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 824a21b..9c9cb10 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -529,9 +529,9 @@ $$a = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(\lim_{n \to \ 下证唯一性。假设另有 $b \neq a$ 满足 $f(b) = b$，则由拉格朗日中值定理，存在 $\eta$ 介于 $a$ 和 $b$ 之间，使得 $$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ -由于 $|a - b| > 0$，两边除以 $|a - b|$ 得 $1 \leq r$，与 $0 < r < 1$ 矛盾。故方程 $f(x) = x$ 有唯一实根 $x = a$。 +由于 $|a - b| > 0$，两边除以 $|a - b|$ 得 $1 \leq r$，与 $0 < r < 1$ 矛盾。故方程 $f(x) = x$ 有唯一实根 $x = a$ -综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。$\square$ +综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。 -- 2.34.1 From f6e27458221a52693029fa888e2e8209ad467ad0 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 20:54:00 +0800 Subject: [PATCH 140/274] vault backup: 2025-12-31 20:54:00 --- 编写小组/试卷/0103高数模拟试卷.md | 1 - 1 file changed, 1 deletion(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 9c9cb10..f0a55ee 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -534,7 +534,6 @@ $$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ 综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。 - 6.设 $f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 **解** -- 2.34.1 From b04f79a2126593a573845021d5de20aa7f043d6b Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 31 Dec 2025 21:24:58 +0800 Subject: [PATCH 141/274] vault backup: 2025-12-31 21:24:58 --- 编写小组/试卷/0103高数模拟试卷.md | 61 +++++++++++++++++++ 1 file changed, 61 insertions(+) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index f0a55ee..634bcb0 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -570,5 +570,66 @@ $$ f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). $$ +7.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 +**补充整理：** + +1. 若两级数均收敛，则其和也收敛；若两个都绝对收敛，则和也绝对收敛；若一个绝对收敛一个条件收敛，则和条件收敛。 + +2. 若两级数仅一个收敛，则其和是发散的。 + +**级数添加与去掉括号的敛散性有以下结论：** + +1. 收敛级数任意添加括号也收敛。 +2. 若收敛级数去掉括号后的通项仍以0为极限，则去掉括号后的级数也收敛，且和不变。 + +**去括号情况的证明** + +设 $(a_1 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + \cdots + (a_{n_{k-1}+1} + \cdots + a_{n_k}) + \cdots$ 收敛于 $S$，且 $\lim_{n \to \infty} a_n = 0$。记该级数的部分和为 $T_k$，$\sum_{n=1}^{\infty} a_n$ 的部分和为 $S_n$，则 $T_k = S_{n_k}$，$\lim_{k \to \infty} S_{n_k} = \lim_{k \to \infty} T_k = S$。 + +由于 $\lim_{n \to \infty} a_n = 0$，对 $\forall i \in \{1, 2, \cdots, n_{k+1} - n_k - 1\}$，有 + +$lim_{k \to \infty} S_{n_k+i} = \lim_{k \to \infty} \left( T_k + a_{n_k+1} + \cdots + a_{n_k+i} \right) = S$ + +所以 $\lim_{n \to \infty} S_n = S$，即 $\sum_{n=1}^{\infty} a_n$ 也收敛于 $S$。 + +**分析** 这是交错级数，且通项趋于0，但通项不单调，不适用莱布尼茨准则。可考虑用添加括号的方式来证明。也可采用交换相邻两项顺序的方式使通项满足单调性。 + +**解** +$$|a_n| = \left| \frac{(-1)^n}{[n+(-1)^n]^p} \right| = \frac{1}{n^p} \cdot \frac{1}{\left[ 1 + \frac{(-1)^n}{n} \right]^p} \sim \frac{1}{n^p}$$ ($n \to \infty$) + +当 $p>1$ 时，级数绝对收敛；当 $0<p<1$ 时，级数不绝对收敛。 +下面讨论 $0<p\leq 1$ 时，级数的收敛性。 +首先，级数的通项 $a_n \to 0$。 + +**方法1** 将原级数按如下方式添加括号 +$$\left( \frac{1}{3^p} - \frac{1}{2^p} \right) + \left( \frac{1}{5^p} - \frac{1}{4^p} \right) + \cdots + \left( \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} \right) + \cdots$$ + +记 $b_n = \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p}$，则 $b_n < 0$，$\sum_{n=2}^{\infty} (-b_n)$是正项级数。由于 +$$-b_n = \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} = \frac{1}{(2n+1)^p} \left[ \left( 1 + \frac{1}{2n} \right)^p - 1 \right] \sim \frac{1}{(2n+1)^p} \cdot \frac{p}{2n} \sim \frac{p}{(2n)^{p+1}}$$ + +而 $p+1>1$，所以 $\sum_{n=2}^{\infty} (-b_n)$ 收敛，由上面补充中去括号的讨论知，原级数收敛。 + +**方法2** 同样考虑方法1中的级数 $\sum_{n=2}^{\infty} (-b_n)$，其部分和为 +$$ +\begin{aligned} +S_n &= \left( \frac{1}{2^p} - \frac{1}{3^p} \right) + \left( \frac{1}{4^p} - \frac{1}{5^p} \right) + \cdots + \left( \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} \right) \\ +&= \frac{1}{2^p} \left( \frac{1}{3^p} - \frac{1}{4^p} \right) - \left( \frac{1}{5^p} - \frac{1}{6^p} \right) - \cdots - \left( \frac{1}{(2n-1)^p} - \frac{1}{(2n)^p} \right) - \frac{1}{(2n+1)^p} < \frac{1}{2^p} +\end{aligned} +$$ +正项级数部分和数列有界，级数收敛，从而原级数收敛。 + +**方法3** 原级数是 +$$\frac{1}{3^p} - \frac{1}{2^p} + \frac{1}{5^p} - \frac{1}{4^p} + \cdots + \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} + \cdots$$ +奇偶项互换后的新级数为 +$$\frac{1}{2^p} - \frac{1}{3^p} + \frac{1}{4^p} - \frac{1}{5^p} + \cdots + \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} + \cdots$$ + +记 $c_n = \frac{1}{n^p}$，该级数为 $\sum_{n=2}^{\infty}(-1)^{n-1}c_n$，由于 $c_n$ 单减趋于0，由莱布尼茨判别法知，该交错级数收敛，从而原级数收敛。 + +**评注** “方法3” 用到了收敛级数的性质：收敛级数交换相邻两项的位置后的级数仍收敛，且和不变。 + +证明如下： +设 $a_1 + a_2 + a_3 + a_4 + \cdots + a_{2n-1} + a_{2n} + \cdots$ 收敛于 $S$，其部分和为 $S_n$。交换相邻两项的位置后的级数为 $a_2 + a_1 + a_4 + a_3 + \cdots + a_{2n} + a_{2n-1} + \cdots$，其部分和为 $T_n$，则 +$$T_{2n} = S_{2n} \Rightarrow \lim_{n \to \infty} T_{2n} = \lim_{n \to \infty} S_{2n} = S, \quad \lim_{n \to \infty} T_{2n+1} = \lim_{n \to \infty} T_{2n} + \lim_{n \to \infty} a_{2n+2} = S,$$ +所以 $\lim_{n \to \infty} T_n = S$。 -- 2.34.1 From 445f632860fa8b3e1584fdccc42a9abd903241ff Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 31 Dec 2025 21:43:47 +0800 Subject: [PATCH 142/274] =?UTF-8?q?=E8=B0=83=E6=95=B4=E8=A7=A3=E6=9E=90?= =?UTF-8?q?=E6=A0=BC=E5=BC=8F+=E9=AB=98=E6=95=B0=E6=A8=A1=E6=8B=9F?= =?UTF-8?q?=E8=AF=95=E5=8D=B7=E5=A1=AB=E7=A9=BA=E9=A2=98?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 50 +++++++++++++++++++ ...231线性代数考试卷（解析版）.md | 36 ++++--------- 2 files changed, 61 insertions(+), 25 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 634bcb0..b6902eb 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -188,6 +188,56 @@ $$f(x) = \left( \frac{1}{2a^2}x - \frac{3}{2a} \right)(x - 2a)(x - 4a) = \frac{1 $$\lim_{x \to 3a} \frac{f(x)}{x-3a} = \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} = \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) = \frac{1}{2a^2} \cdot a \cdot (-a) = -\frac{1}{2}.$$ **答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. + +若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ + 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}|\sin(\frac{1}{\sqrt{n}})|$ + 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ + 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ + 根据**p级数**的收敛性结论，级数$\sum\limits_{n=1}^{\infty}\frac{1}{n^{p+\frac{1}{2}}}$当且仅当$p+\frac{1}{2}>1$时收敛，即$p>\frac{1}{2}$， +所以，$p\in(\frac{1}{2},+\infty)$ +$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ + 方法1： + 令$x=2\sin t$，则$\mathrm{d}x=2\cos t \mathrm{d}t$， +$$\begin{align}\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=\int{(2\sin t)^3\sqrt{4-4\sin^2 t} \cdot 2\cos t\mathrm{d}t}\\ +&=32\int{\sin^3t\cos^2t\mathrm{d}t}\\ +&=32\int{\sin t(1-\cos^2t)\cos^2t\mathrm{d}t}\\ +&=-32\int{(\cos^2t-cos^4t)\mathrm{d}\cos t}\\ +&=-32(\frac{\cos^3t}{3}-\frac{cos^5t}{5})+C\\ +&=-\frac{4}{3}(\sqrt{4-x^2})^3+\frac{1}{5}(\sqrt{4-x^2})^5+C +\end{align} +$$ + 方法2 + 令$\sqrt{4-x^2}=t$，$x^2=4-t^2$，$x\mathrm{d}x=-t\mathrm{d}t$， +$$ +\begin{align} +\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=-\int{(4-t^2)t^2\mathrm{d}t}\\ +&=\frac{t^5}{5}-\frac{4t^3}{3}+C\\ +&=\frac{(\sqrt{4-x^2})^5}{5}-\frac{4(\sqrt{4-x^2})^3}{3}+C +\end{align} +$$ + +- 设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ +解：方程两边对$t$求导，得 +$$ +\begin{cases} +\frac{\mathrm{d}x}{\mathrm{d}t}=2t+2\\ +2t-\frac{\mathrm{d}y}{\mathrm{d}t}+a\frac{\mathrm{d}y}{\mathrm{d}t}\cos y=0 +\end{cases} +\Rightarrow +\begin{cases} +\frac{\mathrm{d}x}{\mathrm{d}t}=2(t+1)\\ +\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{2t}{1-a\cos y} +\end{cases} +\Rightarrow +\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t}{(t+1)(1-a\cos y)} +$$ +$$ +\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}x}=\frac{\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{\frac{(1-a\cos y)-at(t+1)\frac{\mathrm{d}y}{\mathrm{d}t}\sin y}{(t+1)^2(1-a\cos y)^2}}{2(t+1)} +$$ +注意到$y|_{t=0}=b,\frac{\mathrm{d}y}{\mathrm{d}t}|_{t=0}=0$，得 +$$ +\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\frac{1}{2(1-a\cos b)} +$$ # 高等数学题解集 ## 1. 级数收敛性选择题 diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index 19664aa..aa594eb 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -52,7 +52,7 @@ tags: (D) $\text{rank}[A \ B] = \text{rank}[A^T \ B^T]$. >答案：**A** ->重点：$AB$的每一列都是 $A$ 的列向量的线性组合，因此，$\text{rank}[A \quad AB]=\text{rank}A$ ，因为 $AB$ 的列都在 $A$ 的列空间中 +>重点：$AB$ 的每一列都是 $A$ 的列向量的线性组合，因此，$\text{rank}[A\quad AB]=\text{rank}A$ ，因为 $AB$ 的列都在 $A$ 的列空间中 5. 设 $A$ 可逆，将 $A$ 的第一列加上第二列的 2 倍得到 $B$，则 $A^*$ 与 $B^*$ 满足 (A) 将 $A^*$ 的第一列加上第二列的 2 倍得到 $B^*$; @@ -93,38 +93,24 @@ tags: 9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad\beta_2 = (1,2,3)^T,\quad\beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ 解析： - 先计算$A^2$： -$$A^2 -= \begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ -$$= \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix} -$$$$= \begin{bmatrix}18&-6\\-54&18\end{bmatrix} $$ -$$= 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} = 6A$$ +$$\begin{align}A^2&=\begin{bmatrix}3&-1\\-9&3\end{bmatrix}\begin{bmatrix}3&-1\\-9&3\end{bmatrix}\\[1em] +&= \begin{bmatrix}3\times3 + (-1)\times(-9)&3\times(-1) + (-1)\times3\\-9\times3 + 3\times(-9)&-9\times(-1) + 3\times3\end{bmatrix}\\[1em] +&=\begin{bmatrix}18&-6\\-54&18\end{bmatrix}\\[1em] +&= 6\begin{bmatrix}3&-1\\-9&3\end{bmatrix} \\[1em] &= 6A +\end{align} +$$ -由此递推： -- $$A^3 = A^2 \cdot A = 6A \cdot A = 6A^2 = 6\times6A = 6^2A$$ -- 归纳可得当$n \geq 1$时，$A^n = 6^{n-1}A$ +由此递推：$A^3 = A^2 \cdot A = 6A \cdot A = 6A^2 = 6\times6A = 6^2A$，归纳可得当$n \geq 1$时，$A^n = 6^{n-1}A$ 将A代入得： $$A^n = 6^{n-1}\begin{bmatrix}3&-1\\-9&3\end{bmatrix} $$ - --- - - -9. 若向量组 - $$ - \alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T - $$ - 不能由向量组 - $$ - \beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 = (3,4,a)^T - $$ - 线性表示，则 - $$ - a = \underline{\qquad\qquad}. - $$ + +9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 =(1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad \beta_2 = (1,2,3)^T,\quad \beta_3 =(3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ + --- 【答】5. -- 2.34.1 From a547b90b10c3928323e79ae168f826758134d481 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 31 Dec 2025 21:44:56 +0800 Subject: [PATCH 143/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 83 ++++++++++++++++++- 1 file changed, 82 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index b6902eb..39ab452 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -620,7 +620,88 @@ $$ f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). $$ -7.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 +7.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 + +(1) 求参数$a, b$的值；（5分） +(2) 计算极限$$ +\lim_{x \to 0} \frac{f(x) - x^2}{x^4} +$$的值。（5分） + +**解答** + +**(1)** +由于 +$$ +\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( a + bx^2 - \cos x \right) = a - 1 = 0, +$$ +故$a = 1$。 +又因为$f(x)$与$x^2$等价无穷小，所以 +$$ +\lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{1 + bx^2 - \cos x}{x^2} = b + \lim_{x \to 0} \frac{1 - \cos x}{x^2} = b + \frac{1}{2} = 1, +$$ +因此$b = \frac{1}{2}$。 + +**(2)** +由$\cos x$的麦克劳林展开： +$$ +\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4). +$$ +代入$f(x)$： +$$ +f(x) = 1 + \frac{1}{2}x^2 - \left( 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) \right) = x^2 - \frac{1}{24}x^4 + o(x^4). +$$ +于是 +$$ +\lim_{x \to 0} \frac{f(x) - x^2}{x^4} = \lim_{x \to 0} \frac{-\frac{1}{24}x^4 + o(x^4)}{x^4} = -\frac{1}{24}. +$$ + +--- + + +8.设$f(x) \in C[0,1] \cap D(0,1), f(0)=0, f(1)=1$。试证： +（1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； +（2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 + +**分析** +（1）只需将$[0,1]$分成两个区间，使$f(x)$在两个区间各用一次微分中值定理。设分点为$x_0 \in (0,1)$，由 +$$ +f(x_0)-f(0)=f'(\xi)(x_0-0),\quad f(1)-f(x_0)=f'(\eta)(1-x_0) \quad (0<\xi<x_0<\eta<1), +$$ +得$f'(\xi)=\frac{f(x_0)}{x_0}$，$f'(\eta)=\frac{1-f(x_0)}{1-x_0}$，则 +$$ +f'(\xi)f'(\eta)=1 \Leftrightarrow \frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = 1. +$$ +等价于$x_0$是方程$f(x)[1-f(x)] = x(1-x)$的根。取$x_0$满足$f(x_0)=1-x_0$即可。 + +**证明** +（1）令$F(x)=f(x)-1+x$，则$F(x)$在$[0,1]$上连续，且$F(0)=-1<0$，$F(1)=1>0$。由介值定理知，存在$x_0 \in (0,1)$使$F(x_0)=0$，即$f(x_0)=1-x_0$。 +在$[0,x_0]$和$[x_0,1]$上分别应用拉格朗日中值定理，存在$\xi \in (0,x_0)$，$\eta \in (x_0,1)$，使得 +$$ +f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}. +$$ +于是 +$$ +f'(\xi)f'(\eta)=\frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = \frac{1-x_0}{x_0} \cdot \frac{x_0}{1-x_0} = 1. +$$ **（2）** +给定正数$a, b$，令$c = \frac{a}{a+b}$，则$0<c<1$。由于$f(x)$在$[0,1]$上连续，且$f(0)=0$, $f(1)=1$，由介值定理，存在$x_1 \in (0,1)$使得$f(x_1) = c = \frac{a}{a+b}$。 +在区间$[0, x_1]$和$[x_1, 1]$上分别应用拉格朗日中值定理，存在$\xi \in (0, x_1)$, $\eta \in (x_1, 1)$，使得 +$$ +f'(\xi) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{f(x_1)}{x_1}, \quad +f'(\eta) = \frac{f(1) - f(x_1)}{1 - x_1} = \frac{1 - f(x_1)}{1 - x_1}. +$$ +于是 +$$ +\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{f(x_1)} + b \cdot \frac{1 - x_1}{1 - f(x_1)}. +$$ +代入$f(x_1) = \frac{a}{a+b}$, $1 - f(x_1) = \frac{b}{a+b}$，得 +$$ +\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot \frac{1 - x_1}{b/(a+b)} = (a+b)x_1 + (a+b)(1 - x_1) = a+b. +$$ +故命题得证。 + + + +9.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 **补充整理：** -- 2.34.1 From d9e242ee38d97a47105f743bd5c949a290543319 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Thu, 1 Jan 2026 10:34:23 +0800 Subject: [PATCH 144/274] vault backup: 2026-01-01 10:34:23 --- 笔记分享/线性映射与矩阵.md | 10 ++++++++++ 1 file changed, 10 insertions(+) create mode 100644 笔记分享/线性映射与矩阵.md diff --git a/笔记分享/线性映射与矩阵.md b/笔记分享/线性映射与矩阵.md new file mode 100644 index 0000000..0d82fe9 --- /dev/null +++ b/笔记分享/线性映射与矩阵.md @@ -0,0 +1,10 @@ +课上我们已经学了线性变换和方阵的关系，即任意一个线性变换都与一个方阵相对应.具体的描述可以写成如下的形式： +>[!note] 定理1： +>设$V$是在数域$\mathbb{P}$上的线性空间，$T$是线性空间$V$上的一个线性变换，若$V$上的一组基为$$\alpha_1,\alpha_2,\cdots,\alpha_r,$$则$\forall\alpha\in V$,有$$T(\alpha)=[T(\alpha_1),T(\alpha_2),\cdots,T(\alpha_r)]\begin{bmatrix}x_1\\x_2\\\cdots\\x_r\end{bmatrix}=[\alpha_1,\alpha_2,\cdots,\alpha_r]A\begin{bmatrix}x_1\\x_2\\\cdots\\x_r\end{bmatrix}$$其中$\begin{bmatrix}x_1\\x_2\\\cdots\\x_r\end{bmatrix}$是$\alpha$在这组基下的坐标，方阵$A$称为$T$在这组基下的方阵. + +我们考虑较为熟悉的情形，不妨令$\mathbb{P}=\mathbb{R},V=\mathbb{R}^n$，那么我们就可以认为方阵$A$与一个从$\mathbb{R}^n$到$\mathbb{R}^n$的线性映射相对应.自然可以想到：如果考虑线性映射$$\sigma:\mathbb{R}^m\rightarrow\mathbb{R}^n$$我们是否也可以得到一个矩阵呢？而且我们会设想它是一个$m\times n$或者$n\times m$的矩阵.为使表述简单，我们取标准正交基$$\varepsilon_i=\begin{bmatrix}0\\\vdots\\1\\\vdots\\0\end{bmatrix}\in\mathbb{R}^m,其中1在第i行.$$设$$\sigma(\varepsilon_i)=\mathcal{A}_i\in\mathbb{R}^n.$$取$\alpha\in\mathbb{R}^m$,设$\alpha=\begin{bmatrix}a_1 \\a_2\\ \vdots \\a_m\end{bmatrix}=a_1\varepsilon_1+a_2+\varepsilon_2+\cdots+a_m\epsilon_m$,则$$\sigma(\alpha)=\sigma(a_1\varepsilon_1+a_2\varepsilon_2+\cdots+a_m\varepsilon_m)=a_1\sigma(\varepsilon_1)+a_2\sigma(\varepsilon_2)+\cdots+a_m\sigma(\varepsilon_m)$$带入得$$\sigma(\alpha)=a_1\mathcal{A}_1+a_2\mathcal{A}_2+\cdots+a_m\mathcal{A}_m=\begin{bmatrix}\mathcal{A}_1\ \mathcal{A}_2\ \cdots\ \mathcal{A}_m\end{bmatrix}\begin{bmatrix}a_1\\a_2\\\vdots\\a_m\end{bmatrix}$$也就是说，对$\alpha$作用$\sigma$等价于对$\alpha$的坐标左乘一个$n$行$m$列的矩阵$$A=\begin{bmatrix}\mathcal{A}_1\ \mathcal{A}_2\ \cdots\ \mathcal{A}_m\end{bmatrix},$$也就是说，一个线性映射对应一个矩阵.反过来，一个矩阵能否对应一个从$\mathbb{R}^m$到$\mathbb{R}^n$的线性映射呢？ +取$A\in\mathbb{R}^{n\times m}$,设$A=[\mathcal{A}_1\ \mathcal{A}_2\ \cdots\ \mathcal{A}_m\ ]$,取线性映射$\sigma:\mathbb{R}^m\to\mathbb{R}^n$，满足$$\sigma(\varepsilon_i)=\mathcal{A_i}$$于是若$\alpha=\begin{bmatrix}a_1\\a_2\\\vdots\\a_m\end{bmatrix}\in\mathbb{R}^m$,有$$\begin{aligned}\sigma(\alpha)&=\sigma(a_1\varepsilon_1+a_2\varepsilon_2+\cdots+a_m\varepsilon_m)\\&=a_1\sigma(\varepsilon_1)+a_2\sigma(\varepsilon_2)+\cdots+a_m\sigma(\varepsilon_m)\\&=a_1\mathcal{A}_1+a_2\mathcal{A}_2+\cdots+a_m\mathcal{A}_m\\&=A\begin{bmatrix}a_1\\a_2\\\vdots\\a_m\end{bmatrix}\\&=A\alpha\end{aligned}$$于是一个矩阵也对应一个线性映射.从而有如下定理： +>[!note] 定理2： +>从$\mathbb{R}^m$到$\mathbb{R}^n$的线性映射全体与$\mathbb{R}^{n\times m}$中的矩阵一一对应. + +由于每个线性空间都能由其基与一个向量空间构成同构，我们就能凭借上面这个定理把任意线性空间中的一个线性映射与一个矩阵一一对应起来. \ No newline at end of file -- 2.34.1 From bd8ad847f2a5b7677cbb7256d2655a3de29d1dc6 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Thu, 1 Jan 2026 10:42:32 +0800 Subject: [PATCH 145/274] =?UTF-8?q?=E6=94=B9=E4=BA=86=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E5=B0=8F=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/洛必达法则-注意事项.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/洛必达法则-注意事项.md b/素材/洛必达法则-注意事项.md index 9dbabd4..fee34d8 100644 --- a/素材/洛必达法则-注意事项.md +++ b/素材/洛必达法则-注意事项.md @@ -46,5 +46,5 @@ $$\boxed{\infty-\infty}\overset{通分}{\Longrightarrow}\boxed{\frac{0}{0}或\fr >(2)若$\forall x\in(0,1)$,有$f''(x)\neq2$,则当$x\in(0,1)$时，恒有$f(x)>x^2$. 证明： -（1）令$F(x)=f(x)-x^2$,则$F(0)=F(1)=0,F(\frac{1}{2})>0$.显然$F(x)$在$[0,\frac{1}{2}]$和$[\frac{1}{2},1]$上都满足拉格朗日中值定理得条件，所以由拉格朗日中值定理得：$$\exists\xi_1\in(0,\frac{1}{2}),\xi_2\in(\frac{1}{2},1),有\frac{F(\frac{1}{2})-F(0)}{\frac{1}{2}-0}=F'(\xi_1)>0,\frac{F(1)-F(\frac{1}{2})}{1-\frac{1}{2}}=F'(\xi_2)<0,$$于是再对$F(x)$在$[\xi_1,\xi_2]$上用拉格朗日中值定理得：$$\exists\xi\in(\xi_1,\xi_2),有F''(\xi)=\frac{F(\xi_2)-F(\xi_1)}{\xi_2-\xi_1}<0,即f'(\xi)<2.$$ +（1）令$F(x)=f(x)-x^2$,则$F(0)=F(1)=0,F(\frac{1}{2})>0$.显然$F(x)$在$[0,\frac{1}{2}]$和$[\frac{1}{2},1]$上都满足拉格朗日中值定理得条件，所以由拉格朗日中值定理得：$$\exists\xi_1\in(0,\frac{1}{2}),\xi_2\in(\frac{1}{2},1),有\frac{F(\frac{1}{2})-F(0)}{\frac{1}{2}-0}=F'(\xi_1)>0,\frac{F(1)-F(\frac{1}{2})}{1-\frac{1}{2}}=F'(\xi_2)<0,$$于是再对$F(x)$在$[\xi_1,\xi_2]$上用拉格朗日中值定理得：$$\exists\xi\in(\xi_1,\xi_2),有F''(\xi)=\frac{F(\xi_2)-F(\xi_1)}{\xi_2-\xi_1}<0,即f''(\xi)<2.$$ （2）用反证法，假设$\exists \eta\in(0,1),f(\eta)\le \eta^2,则F(\eta)\le 0$, \ No newline at end of file -- 2.34.1 From 20ca449ab992b896daba7fbaf5aad9d6930e93ff Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Thu, 1 Jan 2026 17:49:42 +0800 Subject: [PATCH 146/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E9=83=A8?= =?UTF-8?q?=E5=88=86=E6=A0=BC=E5=BC=8F=EF=BC=8C=E5=85=84=E5=BC=9F=E5=A7=90?= =?UTF-8?q?=E5=A6=B9=E4=BB=AC=E6=B3=A8=E6=84=8F=E8=BE=93=E5=85=A5=E8=A7=84?= =?UTF-8?q?=E8=8C=83=E5=95=8A=EF=BC=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 76 +++++++++---------- 1 file changed, 38 insertions(+), 38 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 39ab452..26ccfb7 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -53,9 +53,9 @@ D$\frac{1}{6}$ 分别对分子和分母进行泰勒展开： 分子： -$$ -e^{x^2} \cos x = \left(1 + x^2 + \frac{x^4}{2} + o(x^4)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)\right) = 1 + \frac{1}{2}x^2 + \left(-\frac{1}{2} + \frac{1}{2} + \frac{1}{24}\right)x^4 + o(x^4) = 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) -$$ +$$\begin{aligned} +e^{x^2} \cos x &= \left(1 + x^2 + \frac{x^4}{2} + o(x^4)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)\right) \\&= 1 + \frac{1}{2}x^2 + \left(-\frac{1}{2} + \frac{1}{2} + \frac{1}{24}\right)x^4 + o(x^4) \\&= 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) +\end{aligned}$$ 所以 $e^{x^2} \cos x - 1 - \frac{1}{2}x^2 = \frac{1}{24}x^4 + o(x^4)$ 分母： @@ -70,7 +70,7 @@ $$ $$ -3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) dx$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 +3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) \mathrm{d}x$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 (A) $a = 1, \, b = 1$ (B) $a = 1, \, b = 2$ @@ -84,7 +84,7 @@ $$ $$ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), $$ -则 +则（把大于等于$5$次的项全都放在高阶无穷小里边） $$ e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). $$ @@ -103,7 +103,7 @@ $$ $$ 积分得 $$ -g(x) = \int_0^x \arctan(t^2) dt = \frac{x^3}{3} - \frac{x^7}{21} + O(x^{11}) = \frac{1}{3}x^3 + O(x^7). +g(x) = \int_0^x \arctan(t^2) \mathrm{d}t = \frac{x^3}{3} - \frac{x^7}{21} + O(x^{11}) = \frac{1}{3}x^3 + O(x^7). $$ 可见 $g(x)$ 是 $x \to 0$ 时的三阶无穷小，其主项为 $\dfrac{1}{3}x^3$。 @@ -140,7 +140,7 @@ $$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{ **方法2** 因为 $y' = \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， -所以 +所以 （积分得） $$y = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, (|x| < 1).$$ 又知 $y(x)$ 在点 $x = 0$ 处的麦克劳林展开式为 $$y(x) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!} x^n.$$ @@ -152,7 +152,7 @@ $$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{ $$y' = \frac{1}{1 + x^2} = \frac{1}{1 + \tan^2 y} = \cos^2 y.$$ 利用复合函数求导法则： $$y'' = -2 \cos y \sin y \cdot y' = -\sin(2y) \cos^2 y = \cos^2 y \sin 2\left( y + \frac{\pi}{2} \right),$$ -$$y''' = \left[ -2 \cos y \sin y \sin 2\left( y + \frac{\pi}{2} \right) + 2 \cos^2 y \cos 2\left( y + \frac{\pi}{2} \right) \right] y' = 2 \cos^3 y \cos\left( y + \frac{\pi}{2} \right) y' + 2 \cos^3 y \sin 3\left( y + \frac{\pi}{2} \right),$$ +$$\begin{aligned}y''' &= \left[ -2 \cos y \sin y \sin 2\left( y + \frac{\pi}{2} \right) + 2 \cos^2 y \cos 2\left( y + \frac{\pi}{2} \right) \right] y'\\& = 2 \cos^3 y \cos\left( y + \frac{\pi}{2} \right) y' + 2 \cos^3 y \sin 3\left( y + \frac{\pi}{2} \right),\end{aligned}$$ $$y^{(4)} = 6 \cos^4 y \cos\left( y + \frac{\pi}{2} \right) y' + 3 \cos^4 y \sin 4\left( y + \frac{\pi}{2} \right).$$ 由此归纳出 $$y^{(n)} = (n-1)! \cos^n y \sin n\left( y + \frac{\pi}{2} \right).$$ @@ -166,13 +166,13 @@ $$y^{(n)}(0) = (n-1)! \sin \frac{n\pi}{2}.$$ -2.已知 $f(x)$ 是三次多项式，且有 $\lim_{x \to 2a} \frac{f(x)}{x-2a} = \lim_{x \to 4a} \frac{f(x)}{x-4a} = 1$，求 $\lim_{x \to 3a} \frac{f(x)}{x-3a}$ +2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，求 $\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}.$ **分析** 由已知的两个极限式可确定 $f(x)$ 的两个一次因子以及两个待定系数，从而完全确定 $f(x)$。 **解** -由已知有 $\lim_{x \to 2a} f(x) = \lim_{x \to 4a} f(x) = 0$。 +由已知有 $\lim\limits_{x \to 2a} f(x) = \lim\limits_{x \to 4a} f(x) = 0$。 由于 $f(x)$ 处处连续，故 $f(2a) = f(4a) = 0$。 因此 $f(x)$ 含有因式 $(x-2a)(x-4a)$，可设 $$f(x) = (Ax + B)(x - 2a)(x - 4a).$$ @@ -185,7 +185,7 @@ $$\begin{cases} (2aA + B)(-2a) = 1 \\ (4aA + B)(2a) = 1 \end{cases} \Rightarrow 于是 $$f(x) = \left( \frac{1}{2a^2}x - \frac{3}{2a} \right)(x - 2a)(x - 4a) = \frac{1}{2a^2}(x - 3a)(x - 2a)(x - 4a).$$ 最后， -$$\lim_{x \to 3a} \frac{f(x)}{x-3a} = \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} = \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) = \frac{1}{2a^2} \cdot a \cdot (-a) = -\frac{1}{2}.$$ +$$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} \\&= \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) \\&= \frac{1}{2a^2} \cdot a \cdot (-a) \\&= -\frac{1}{2}.\end{aligned}$$ **答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. @@ -273,16 +273,16 @@ $$ ## 三、解答题（共11小题，共80分） 1.求下列不定积分(提示，换元），其中 $a > 0$ -(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} dx$ +(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x$ **解：** 令 $x = a \sin t$，则： $$ \begin{aligned} -\int \frac{x^2}{\sqrt{a^2 - x^2}} dx &= \int \frac{a^2 \sin^2 t}{a \cos t} \cdot a \cos t \, dt \\ -&= a^2 \int \sin^2 t \, dt \\ -&= \frac{a^2}{2} \int (1 - \cos 2t) \, dt \\ +\int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x &= \int \frac{a^2 \sin^2 t}{a \cos t} \cdot a \cos t \, \mathrm{d}t \\ +&= a^2 \int \sin^2 t \, \mathrm{d}t \\ +&= \frac{a^2}{2} \int (1 - \cos 2t) \, \mathrm{d}t \\ &= \frac{a^2}{2} \left( t - \frac{1}{2} \sin 2t \right) + C \\ &= \frac{a^2}{2} \arcsin \frac{x}{a} - \frac{x}{2} \sqrt{a^2 - x^2} + C \end{aligned} @@ -290,18 +290,18 @@ $$ --- - (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} dx$ + (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x$ **解：** **方法一：** 令 $x = a \tan t$，则： $$ \begin{aligned} -\int \frac{\sqrt{x^2 + a^2}}{x^2} dx &= \int \frac{a \sec t}{a^2 \tan^2 t} \cdot a \sec^2 t \, dt \\ -&= \int \frac{\sec^3 t}{\tan^2 t} \, dt \\ -&= \int \frac{1}{\sin^2 t \cos t} \, dt \\ -&= \int \frac{\sin^2 t + \cos^2 t}{\sin^2 t \cos t} \, dt \\ -&= \int \sec t \, dt + \int \frac{\cos t}{\sin^2 t} \, dt \\ +\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{a \sec t}{a^2 \tan^2 t} \cdot a \sec^2 t \, \mathrm{d}t \\ +&= \int \frac{\sec^3 t}{\tan^2 t} \, \mathrm{d}t \\ +&= \int \frac{1}{\sin^2 t \cos t} \, \mathrm{d}t \\ +&= \int \frac{\sin^2 t + \cos^2 t}{\sin^2 t \cos t} \, \mathrm{d}t \\ +&= \int \sec t \, \mathrm{d}t + \int \frac{\cos t}{\sin^2 t} \, \mathrm{d}t \\ &= \ln |\sec t + \tan t| - \frac{1}{\sin t} + C \\ &= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C \end{aligned} @@ -311,24 +311,24 @@ $$ $$ \begin{aligned} -\int \frac{\sqrt{x^2 + a^2}}{x^2} dx &= \int \frac{x^2 + a^2}{x^2 \sqrt{x^2 + a^2}} dx \\ -&= \int \frac{1}{\sqrt{x^2 + a^2}} dx + a^2 \int \frac{1}{x^2 \sqrt{x^2 + a^2}} dx \\ +\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{x^2 + a^2}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ +&= \int \frac{1}{\sqrt{x^2 + a^2}} \mathrm{d}x + a^2 \int \frac{1}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ &= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C \end{aligned} $$ --- -(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} dx$ +(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x$ **解：** 令 $x = a \sec t$，则： $$ \begin{aligned} -\int \frac{\sqrt{x^2 - a^2}}{x} dx &= \int \frac{a \tan t}{a \sec t} \cdot a \sec t \tan t \, dt \\ -&= a \int \tan^2 t \, dt \\ -&= a \int (\sec^2 t - 1) \, dt \\ +\int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x &= \int \frac{a \tan t}{a \sec t} \cdot a \sec t \tan t \, \mathrm{d}t \\ +&= a \int \tan^2 t \, \mathrm{d}t \\ +&= a \int (\sec^2 t - 1) \, \mathrm{d}t \\ &= a (\tan t - t) + C \\ &= \sqrt{x^2 - a^2} - a \arccos \frac{a}{x} + C \end{aligned} @@ -336,17 +336,17 @@ $$ --- -(4) $\displaystyle \int \sqrt{1 + e^x} dx$ +(4) $\displaystyle \int \sqrt{1 + e^x} \mathrm{d}x$ **解：** -令 $t = \sqrt{1 + e^x}$，则 $e^x = t^2 - 1$，$x = \ln(t^2 - 1)$，$dx = \frac{2t}{t^2 - 1} dt$： +令 $t = \sqrt{1 + e^x}$，则 $e^x = t^2 - 1$，$x = \ln(t^2 - 1)$，$\mathrm{d}x = \frac{2t}{t^2 - 1} \mathrm{d}t$： $$ \begin{aligned} -\int \sqrt{1 + e^x} dx &= \int t \cdot \frac{2t}{t^2 - 1} dt \\ -&= \int \frac{2t^2}{t^2 - 1} dt \\ -&= 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) dt \\ -&= 2 \int 1 \, dt + \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) dt \\ +\int \sqrt{1 + e^x} \mathrm{d}x &= \int t \cdot \frac{2t}{t^2 - 1} \mathrm{d}t \\ +&= \int \frac{2t^2}{t^2 - 1} \mathrm{d}t \\ +&= 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) \mathrm{d}t \\ +&= 2 \int 1 \, \mathrm{d}t + \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) \mathrm{d}t \\ &= 2t + \ln \left| \frac{t-1}{t+1} \right| + C \\ &= 2 \sqrt{1 + e^x} + \ln \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} + C \end{aligned} @@ -553,7 +553,7 @@ $$ $$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ 证明： -（1）数列 $\{x_n\}$ 收敛（记 $\lim_{n \to \infty} x_n = a$）； +（1）数列 $\{x_n\}$ 收敛（记 $\lim\limits_{n \to \infty} x_n = a$）； （2）方程 $f(x) = x$ 有唯一实根 $x = a$。 **证明：** @@ -716,13 +716,13 @@ $$ **去括号情况的证明** -设 $(a_1 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + \cdots + (a_{n_{k-1}+1} + \cdots + a_{n_k}) + \cdots$ 收敛于 $S$，且 $\lim_{n \to \infty} a_n = 0$。记该级数的部分和为 $T_k$，$\sum_{n=1}^{\infty} a_n$ 的部分和为 $S_n$，则 $T_k = S_{n_k}$，$\lim_{k \to \infty} S_{n_k} = \lim_{k \to \infty} T_k = S$。 +设 $(a_1 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + \cdots + (a_{n_{k-1}+1} + \cdots + a_{n_k}) + \cdots$ 收敛于 $S$，且 $\lim\limits_{n \to \infty} a_n = 0$。记该级数的部分和为 $T_k$，$\sum_{n=1}^{\infty} a_n$ 的部分和为 $S_n$，则 $T_k = S_{n_k}$，$\lim\limits_{k \to \infty} S_{n_k} = \lim\limits_{k \to \infty} T_k = S$。 -由于 $\lim_{n \to \infty} a_n = 0$，对 $\forall i \in \{1, 2, \cdots, n_{k+1} - n_k - 1\}$，有 +由于 $\lim\limits_{n \to \infty} a_n = 0$，对 $\forall i \in \{1, 2, \cdots, n_{k+1} - n_k - 1\}$，有 -$lim_{k \to \infty} S_{n_k+i} = \lim_{k \to \infty} \left( T_k + a_{n_k+1} + \cdots + a_{n_k+i} \right) = S$ +$\lim\limits_{k \to \infty} S_{n_k+i} = \lim\limits_{k \to \infty} ( T_k + a_{n_k+1} + \cdots + a_{n_k+i} ) = S$ -所以 $\lim_{n \to \infty} S_n = S$，即 $\sum_{n=1}^{\infty} a_n$ 也收敛于 $S$。 +所以 $\lim\limits_{n \to \infty} S_n = S$，即 $\sum\limits_{n=1}^{\infty} a_n$ 也收敛于 $S$。 **分析** 这是交错级数，且通项趋于0，但通项不单调，不适用莱布尼茨准则。可考虑用添加括号的方式来证明。也可采用交换相邻两项顺序的方式使通项满足单调性。 -- 2.34.1 From fe148fb10a24309068b9214e24cfcc864dc3ef1e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Thu, 1 Jan 2026 17:51:51 +0800 Subject: [PATCH 147/274] =?UTF-8?q?=E6=94=B9=E4=BA=86=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E7=AD=94=E6=A1=88=E9=94=99=E8=AF=AF?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 26ccfb7..f4cca66 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -190,7 +190,7 @@ $$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\fra **答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. 若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ - 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}|\sin(\frac{1}{\sqrt{n}})|$ + 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ 根据**p级数**的收敛性结论，级数$\sum\limits_{n=1}^{\infty}\frac{1}{n^{p+\frac{1}{2}}}$当且仅当$p+\frac{1}{2}>1$时收敛，即$p>\frac{1}{2}$， -- 2.34.1 From 18da8b2f677bfe00525d54dca47bd1dbaf82650c Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Fri, 2 Jan 2026 01:10:35 +0800 Subject: [PATCH 148/274] vault backup: 2026-01-02 01:10:35 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index aa594eb..f9048b1 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -45,7 +45,7 @@ tags: >C. $\begin{bmatrix}\alpha_1&\alpha_3&\alpha_4\end{bmatrix}=\begin{bmatrix}0&1&-1\\0&-1&1\\c_1&c_3&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&-1\\0&0&1\\c_1&c_3+c_4&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&0\\0&0&1\\c_1&c_3+c_4&c_4\end{bmatrix}$，故其必定线性相关； >D. $\begin{bmatrix}\alpha_1&\alpha_2&\alpha_4\end{bmatrix}=\begin{bmatrix}0&1&-1\\1&-1&1\\c_2&c_3&c_4\end{bmatrix}\cong\begin{bmatrix}0&0&-1\\1&0&1\\c_2&c_3+c_4&c_4\end{bmatrix}$，当$c_3+c_4\ne 0$时，作初等列变换，$\text{rank}\begin{bmatrix}0&0&-1\\1&0&1\\c_2&c_3+c_4&c_4\end{bmatrix}=\text{rank}\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}=3$，故线性无关； -4. 设 $A, B$ 为 $n$ 阶矩阵，则 +4. 设 $A, B$ 为 $n$ 阶矩阵，则 (A) $\text{rank}[A \ AB] = \text{rank} A$; (B) $\text{rank}[A \ BA] = \text{rank} A$; (C) $\text{rank}[A \ B] = \max\{\text{rank} A, \text{rank} B\}$; -- 2.34.1 From 2ab3fecec6a60c7ab3f3fbe6b2148b2f7f0b8e75 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 12:14:56 +0800 Subject: [PATCH 149/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=A0=BC?= =?UTF-8?q?=E5=BC=8F=EF=BC=8C=E7=BB=99=E9=83=A8=E5=88=86=E9=A2=98=E7=9B=AE?= =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E6=96=B0=E6=96=B9=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 笔记分享/LaTeX（KaTeX）输入规范.md | 3 +- 编写小组/试卷/0103高数模拟试卷.md | 105 ++++++++++++------ 2 files changed, 76 insertions(+), 32 deletions(-) diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md index e9538a1..0b40a08 100644 --- a/笔记分享/LaTeX（KaTeX）输入规范.md +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -1,4 +1,5 @@ 通常的，记号严格按照教材中的规范。 1. 矩阵使用bmatrix 2. 自然常数或电荷量e、虚数单位i应当为**正体**，需要用mathrm记号包裹，例：$\mathrm{e}^{\mathrm{i}\pi}+1=0$；然而，当e,i作为变量时，应当用正常的斜体。例：$\sum\limits_{i=1}^{n}a_i$ -3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ \ No newline at end of file +3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ +4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ \ No newline at end of file diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index f4cca66..88b7e3b 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -78,15 +78,15 @@ $$ (D) $a = 2, \, b = 2$ **解：** -为使 $f(x)$ 与 $g(x)$ 在 $x \to 0$ 时为同阶无穷小，需使二者最低阶非零项的阶数相同。下面分别展开 $f(x)$ 和 $g(x)$ 的麦克劳林公式。 +**法一**：为使 $f(x)$ 与 $g(x)$ 在 $x \to 0$ 时为同阶无穷小，需使二者最低阶非零项的阶数相同。下面分别展开 $f(x)$ 和 $g(x)$ 的麦克劳林公式。 对于 $f(x)$，利用已知展开式： $$ -e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^4), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), +e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^3), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), $$ 则（把大于等于$5$次的项全都放在高阶无穷小里边） $$ -e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). +e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^3)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). $$ 因此 $$ @@ -119,10 +119,75 @@ $$ 因此正确选项为 (B)。 +**法二**：洛必达＋泰勒 +先考虑两者之比的极限$$\lim\limits_{x\to0}\frac{f(x)}{g(x)}=\lim\limits_{x\to0}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to0}\frac{2\mathrm{e}^x(\sin x+\cos x)-2a-2bx}{\arctan x^2}$$由麦克劳林公式得$$\begin{aligned}\mathrm{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+O(x^3)\\\sin x&=x-\frac{x^3}{6}+O(x^3)\\\cos x&=1-\frac{x^2}{2}+O(x^3)\end{aligned}$$于是（把所有大于3次的项都放在高阶无穷小里面，这样可以简化计算）$$\mathrm{e}^x(\sin x+\cos x)=1+2x+x^2+O(x^3)$$从而极限式的分子等于$$2-2a+(4-2b)+2x^2+O(x^3).$$又由麦克劳林公式得$$\arctan x=x-\frac{x^3}{6}+O(x^3)$$故分母为$$x^2-\frac{x^6}{6}+O(x^6)=x^2+O(x^3).$$上面的分子分母带入极限式中得$$\lim\limits_{x\to0}\frac{2-2a+(4-2b)+2x^2+O(x^3)}{x^2+O(x^3)}$$要让上式为有限值且不为$0$,只有$$\begin{cases}2-2a&=0\\4-2b&=0\end{cases}\implies\begin{cases}a&=1\\b&=2\end{cases}$$故选(B) **答案：** (B) + +4.下列级数中收敛的是（ ）。 + +(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ + +(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ + +(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ + +(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ + +### 解析 +- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 +- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 +- **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 +- **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 + +**答案：(B)** + +5.一个倒置的圆锥形容器（顶点在下，底面在上），高度为 10 米，底面半径为 5 米。容器内装有水，水从底部的一个小孔（面积为 $0.1\pi$ 平方米）流出，流速为 $v = 0.6\sqrt{2gh}$ 米/秒，其中 $g = 10$ 米/秒$^2$。同时，以恒定速率 $Q = 0.3\pi$ 立方米/秒从顶部注入水。当水面高度为 5 米时，水面高度的瞬时变化率是多少？ + +选项: +A. $-0.048$ 米/秒 +B. $-0.024$ 米/秒 +C. 0 +D. 0.048 米/秒 +**解析：** +设水面高度为 $h$（从圆锥顶点算起）。由相似关系，水面半径 +$$ +r = \frac{R}{H}h = \frac{5}{10}h = 0.5h, +$$ +水的体积为 +$$ +V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3. +$$ +对时间 $t$ 求导，得 +$$ +\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}. +$$ +另一方面，体积变化率由注入速率 $Q$ 和流出速率 $A \cdot v$ 决定： +$$ +\frac{dV}{dt} = Q - A \cdot v = 0.3\pi - 0.1\pi \cdot 0.6\sqrt{2gh}. +$$ +代入 $h=5$，$g=10$，则 +$$ +\sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10, +$$ +于是 +$$ +A \cdot v = 0.1\pi \cdot 0.6 \cdot 10 = 0.6\pi, +$$ +$$ +\frac{dV}{dt} = 0.3\pi - 0.6\pi = -0.3\pi. +$$ +代入微分式： +$$ +\frac{1}{4}\pi \cdot 5^2 \cdot \frac{dh}{dt} = -0.3\pi \quad \Rightarrow \quad \frac{25}{4}\pi \frac{dh}{dt} = -0.3\pi. +$$ +解得 +$$ +\frac{dh}{dt} = -0.3 \times \frac{4}{25} = -0.048 \ \text{米/秒}. +$$ +因此水面高度以每秒 $0.048$ 米的速度下降，故选 A。 ## 二、填空题（共5小题，每小题2分，共10分） -1.设 $y = \arctan x$，求 $y^{(n)}(0)$ +1.设 $y = \arctan x$，则 $y^{(n)}(0)=\_\_\_\_.$ **分析** 逐次求导以找到 $n$ 阶导数的规律。由于 $y' = \frac{1}{1 + x^2}$，即 $(1 + x^2)y' = 1$，故想到用莱布尼茨公式。 @@ -139,7 +204,7 @@ $$y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0).$$ $$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ **方法2** -因为 $y' = \frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， +因为 $y' = \frac{1}{1 + x^2} = \sum\limits_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， 所以 （积分得） $$y = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, (|x| < 1).$$ 又知 $y(x)$ 在点 $x = 0$ 处的麦克劳林展开式为 @@ -166,7 +231,7 @@ $$y^{(n)}(0) = (n-1)! \sin \frac{n\pi}{2}.$$ -2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，求 $\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}.$ +2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，则$\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}=\_\_\_.$ **分析** 由已知的两个极限式可确定 $f(x)$ 的两个一次因子以及两个待定系数，从而完全确定 $f(x)$。 @@ -189,7 +254,7 @@ $$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\fra **答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. -若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ +3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ @@ -216,7 +281,7 @@ $$ \end{align} $$ -- 设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ +4..设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ 解：方程两边对$t$求导，得 $$ \begin{cases} @@ -238,28 +303,6 @@ $$ $$ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\frac{1}{2(1-a\cos b)} $$ -# 高等数学题解集 - -## 1. 级数收敛性选择题 - -### 题目 -下列级数中收敛的是（ ）。 - -(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ - -(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ - -(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ - -(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ - -### 解析 -- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 -- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 -- **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 -- **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 - -**答案：(B)** --- @@ -276,7 +319,7 @@ $$ (1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x$ **解：** -令 $x = a \sin t$，则： +令 $x = a \sin t$，有$\mathrm{d}x=a\cos t\mathrm{d}t$,则： $$ \begin{aligned} -- 2.34.1 From 38178f6560a0da3dfab7b04a5df0d5247b983981 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 12:41:42 +0800 Subject: [PATCH 150/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=A0=BC?= =?UTF-8?q?=E5=BC=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 52 ++++++++++++++++--- 1 file changed, 44 insertions(+), 8 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 88b7e3b..6e2d32a 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -585,7 +585,7 @@ f^{(6)}(0) = 6! \times \frac{25}{24} = 720 \times \frac{25}{24} = 30 \times 25 = $$ **答案：** -展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$。 +展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$. @@ -597,11 +597,11 @@ $$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ 证明： （1）数列 $\{x_n\}$ 收敛（记 $\lim\limits_{n \to \infty} x_n = a$）； -（2）方程 $f(x) = x$ 有唯一实根 $x = a$。 +（2）方程 $f(x) = x$ 有唯一实根 $x = a$. **证明：** -（1）首先证明 $\{x_n\}$ 是柯西数列。对任意 $n \geq 1$，由拉格朗日中值定理，存在 $\xi_n$ 介于 $x_n$ 和 $x_{n-1}$ 之间，使得 +（1）首先证明 $\{x_n\}$ 是柯西数列.对任意 $n \geq 1$，由拉格朗日中值定理，存在 $\xi_n$ 介于 $x_n$ 和 $x_{n-1}$ 之间，使得 $$|x_{n+1} - x_n| = |f(x_n) - f(x_{n-1})| = |f'(\xi_n)| \cdot |x_n - x_{n-1}| \leq r |x_n - x_{n-1}|.$$ 反复应用此不等式，得 $$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdots \leq r^{n-1} |x_2 - x_1|.$$ @@ -701,7 +701,7 @@ $$ --- -8.设$f(x) \in C[0,1] \cap D(0,1), f(0)=0, f(1)=1$。试证： +8.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： （1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； （2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 @@ -769,7 +769,7 @@ $\lim\limits_{k \to \infty} S_{n_k+i} = \lim\limits_{k \to \infty} ( T_k + a_{n_ **分析** 这是交错级数，且通项趋于0，但通项不单调，不适用莱布尼茨准则。可考虑用添加括号的方式来证明。也可采用交换相邻两项顺序的方式使通项满足单调性。 -**解** +**解**： $$|a_n| = \left| \frac{(-1)^n}{[n+(-1)^n]^p} \right| = \frac{1}{n^p} \cdot \frac{1}{\left[ 1 + \frac{(-1)^n}{n} \right]^p} \sim \frac{1}{n^p}$$ ($n \to \infty$) 当 $p>1$ 时，级数绝对收敛；当 $0<p<1$ 时，级数不绝对收敛。 @@ -784,7 +784,7 @@ $$-b_n = \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} = \frac{1}{(2n+1)^p} \left[ \left 而 $p+1>1$，所以 $\sum_{n=2}^{\infty} (-b_n)$ 收敛，由上面补充中去括号的讨论知，原级数收敛。 -**方法2** 同样考虑方法1中的级数 $\sum_{n=2}^{\infty} (-b_n)$，其部分和为 +**方法2** 同样考虑方法1中的级数 $\sum\limits_{n=2}^{\infty} (-b_n)$，其部分和为 $$ \begin{aligned} S_n &= \left( \frac{1}{2^p} - \frac{1}{3^p} \right) + \left( \frac{1}{4^p} - \frac{1}{5^p} \right) + \cdots + \left( \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} \right) \\ @@ -798,12 +798,48 @@ $$\frac{1}{3^p} - \frac{1}{2^p} + \frac{1}{5^p} - \frac{1}{4^p} + \cdots + \frac 奇偶项互换后的新级数为 $$\frac{1}{2^p} - \frac{1}{3^p} + \frac{1}{4^p} - \frac{1}{5^p} + \cdots + \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} + \cdots$$ -记 $c_n = \frac{1}{n^p}$，该级数为 $\sum_{n=2}^{\infty}(-1)^{n-1}c_n$，由于 $c_n$ 单减趋于0，由莱布尼茨判别法知，该交错级数收敛，从而原级数收敛。 +记 $c_n = \frac{1}{n^p}$，该级数为 $\sum\limits_{n=2}^{\infty}(-1)^{n-1}c_n$，由于 $c_n$ 单减趋于0，由莱布尼茨判别法知，该交错级数收敛，从而原级数收敛。 **评注** “方法3” 用到了收敛级数的性质：收敛级数交换相邻两项的位置后的级数仍收敛，且和不变。 证明如下： 设 $a_1 + a_2 + a_3 + a_4 + \cdots + a_{2n-1} + a_{2n} + \cdots$ 收敛于 $S$，其部分和为 $S_n$。交换相邻两项的位置后的级数为 $a_2 + a_1 + a_4 + a_3 + \cdots + a_{2n} + a_{2n-1} + \cdots$，其部分和为 $T_n$，则 $$T_{2n} = S_{2n} \Rightarrow \lim_{n \to \infty} T_{2n} = \lim_{n \to \infty} S_{2n} = S, \quad \lim_{n \to \infty} T_{2n+1} = \lim_{n \to \infty} T_{2n} + \lim_{n \to \infty} a_{2n+2} = S,$$ -所以 $\lim_{n \to \infty} T_n = S$。 +所以 $\lim\limits_{n \to \infty} T_n = S$。 +10.(10分) + +(1)证明： 对任意的正整数 $n$，方程 +$$ x^n + n^2 x - 1 = 0 $$ +有唯一正实根（记为 $x_n$）。 + +**证明：** +令 $f_n(x) = x^n + n^2x - 1$ 。显然 $f_n(x)$ 在 $[0,1]$ 上连续，且 +$$ f_n(0) = -1 < 0, \quad f_n(1) = n^2 > 0, $$ +由闭区间上连续函数的零值定理可知，至少存在一点 $\xi \in (0,1)$，使得 $f_n(\xi) = 0$，即 +$$ x^n + n^2x - 1 = 0 $$ +至少有一个正实根。 + +又 +$$ f'_n(x) = nx^{n-1} + n^2, $$ +易知当 $x > 0$ 时，$f'_n(x) > 0$，故函数 $f_n(x)$ 在 $(0,+\infty)$ 内严格单调增加，因此 $f_n(x)$ 在 $(0,+\infty)$ 内至多只有一个零点。综上，函数 $f_n(x)$ 在$(0,+\infty)$ 内有唯一零点，即方程 $x^n + n^2x - 1 = 0$ 有唯一正实根。 + +--- + + (2) 证明： 级数 +$$ \sum_{n=1}^\infty x_n $$ +收敛，且其和不超过 2。 + +**证明：** +记方程$x^n + n^2 x - 1 = 0$ 的唯一正实根为 $x_n$，则 $x_n^n + n^2 x_n - 1 = 0$，故 +$$ 0 < x_n = \frac{1}{n^2} - \frac{x_n^n}{n^2} < \frac{1}{n^2}. $$ +根据比较判别法，由于级数 $\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$ 收敛，故级数$\sum\limits_{n=1}^{\infty} x_n$收敛。 + +记$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$的前$n$ 项部分和为 $S_n$,$\sum\limits_{n=1}^{\infty} x_n$的前$n$ 项部分和为$T_n$，显然$T_n < S_n$，且 +$$\begin{aligned} +S_n &= \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < \frac{1}{1} + \frac{1}{1 \cdot 2} + \cdots + \frac{1}{(n-1)n} + \\&= 1 + 1 - \frac{1}{2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n-1} - \frac{1}{n} + \\&= 2 - \frac{1}{n}<2, + \end{aligned}$$ +根据数列极限的保号性（更准确地说是保序性）， +$$ \sum_{n=1}^{\infty} x_n = \lim_{n \to \infty} T_n \leq \lim_{n \to \infty} S_n \leq 2. $$ \ No newline at end of file -- 2.34.1 From f68c0c2afc6b2760f061cfe63f088958a9073d65 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 13:31:03 +0800 Subject: [PATCH 151/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E6=A0=BC=E5=BC=8F?= =?UTF-8?q?=E5=92=8C=E7=AD=94=E6=A1=88=EF=BC=8C=E5=B9=B6=E5=87=BA=E4=BA=86?= =?UTF-8?q?=E7=BA=AF=E8=AF=95=E9=A2=98=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 871 +++++------------- .../0103高数模拟试卷（解析版）.md | 842 +++++++++++++++++ 2 files changed, 1050 insertions(+), 663 deletions(-) create mode 100644 编写小组/试卷/0103高数模拟试卷（解析版）.md diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 6e2d32a..35a8be1 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -1,7 +1,3 @@ ---- -tags: - - 编写小组 ---- ## 一、单选题（共5小题，每小题2分，共10分） 1.设函数 $f(x)$ 在 $x = 0$ 处可导，且 $f(0) = 0$。若 @@ -15,831 +11,380 @@ B. $-1$ C. $1$ D. $2$ -## 解析 -因为 $f(x)$ 在 $x=0$ 处可导且 $f(0)=0$，所以 $f'(0) = \lim\limits_{x \to 0} \frac{f(x)}{x}$ 存在。 +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2}$ +A.$-\frac{1}{12}$ +B.$\frac{1}{12}$ +C.$-\frac{1}{6}$ +D$\frac{1}{6}$ -将极限式分解： -$$ -\frac{f(2x) - 3f(x) + f(-x)}{x} = \frac{f(2x)}{x} - 3\frac{f(x)}{x} + \frac{f(-x)}{x}. -$$ +3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) \mathrm{d}x$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 -分别计算各项极限： -$$ -\begin{aligned} -\lim_{x \to 0} \frac{f(2x)}{x} &= 2 \lim_{x \to 0} \frac{f(2x)}{2x} = 2f'(0), \\ -\lim_{x \to 0} \frac{f(x)}{x} &= f'(0), \\ -\lim_{x \to 0} \frac{f(-x)}{x} &= -\lim_{x \to 0} \frac{f(-x)}{-x} = -f'(0). -\end{aligned} -$$ +(A) $a = 1, \, b = 1$ +(B) $a = 1, \, b = 2$ +(C) $a = 2, \, b = 1$ +(D) $a = 2, \, b = 2$ -因此， -$$ -\lim_{x \to 0} \frac{f(2x) - 3f(x) + f(-x)}{x} = 2f'(0) - 3f'(0) + (-f'(0)) = -2f'(0). -$$ +4.下列级数中收敛的是（ ）。 -由已知条件 $-2f'(0) = 2$，解得 $f'(0) = -1$。 +(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ -**答案：B** +(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ +(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ -2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2}$ -A.$-\frac{1}{12}$ -B.$\frac{1}{12}$ -C.$-\frac{1}{6}$ -D$\frac{1}{6}$ +(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ -**解：** +## 二、填空题（共5小题，每小题2分，共10分） -分别对分子和分母进行泰勒展开： +1.设 $y = \arctan x$，则 $y^{(n)}(0)=\_\_\_\_.$ -分子： -$$\begin{aligned} -e^{x^2} \cos x &= \left(1 + x^2 + \frac{x^4}{2} + o(x^4)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)\right) \\&= 1 + \frac{1}{2}x^2 + \left(-\frac{1}{2} + \frac{1}{2} + \frac{1}{24}\right)x^4 + o(x^4) \\&= 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) -\end{aligned}$$ -所以 $e^{x^2} \cos x - 1 - \frac{1}{2}x^2 = \frac{1}{24}x^4 + o(x^4)$ +2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，则$\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}=\_\_\_.$ -分母： -$$ -\ln(1+x^2) = x^2 - \frac{1}{2}x^4 + o(x^4) -$$ -所以 $\ln(1+x^2) - x^2 = -\frac{1}{2}x^4 + o(x^4)$ +3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ -因此 -$$ -\lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2} = \lim_{x \to 0} \frac{\frac{1}{24}x^4 + o(x^4)}{-\frac{1}{2}x^4 + o(x^4)} = \frac{1/24}{-1/2} = -\frac{1}{12} -$$ +4.求不定积分$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ +5.设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ -3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) \mathrm{d}x$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 +## 三、解答题（共11小题，共80分） +1.求下列不定积分(提示，换元），其中 $a > 0$ -(A) $a = 1, \, b = 1$ -(B) $a = 1, \, b = 2$ -(C) $a = 2, \, b = 1$ -(D) $a = 2, \, b = 2$ +(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x$; (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x$; +(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x$; (4) $\displaystyle \int \sqrt{1 + e^x} \mathrm{d}x$. +```text -**解：** -**法一**：为使 $f(x)$ 与 $g(x)$ 在 $x \to 0$ 时为同阶无穷小，需使二者最低阶非零项的阶数相同。下面分别展开 $f(x)$ 和 $g(x)$ 的麦克劳林公式。 -对于 $f(x)$，利用已知展开式： -$$ -e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^3), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), -$$ -则（把大于等于$5$次的项全都放在高阶无穷小里边） -$$ -e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^3)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). -$$ -因此 -$$ -2e^x \sin x = 2x + 2x^2 + \frac{2}{3}x^3 + O(x^5). -$$ -代入 $f(x)$ 得 -$$ -f(x) = 2x + 2x^2 + \frac{2}{3}x^3 - 2ax - bx^2 + O(x^5) = (2 - 2a)x + (2 - b)x^2 + \frac{2}{3}x^3 + O(x^5). -$$ -对于 $g(x)$，由 $g(0)=0$ 及 $\arctan(x^2)$ 的展开： -$$ -\arctan(x^2) = x^2 - \frac{x^6}{3} + O(x^{10}), -$$ -积分得 -$$ -g(x) = \int_0^x \arctan(t^2) \mathrm{d}t = \frac{x^3}{3} - \frac{x^7}{21} + O(x^{11}) = \frac{1}{3}x^3 + O(x^7). -$$ -可见 $g(x)$ 是 $x \to 0$ 时的三阶无穷小，其主项为 $\dfrac{1}{3}x^3$。 -为使 $f(x)$ 也是三阶无穷小（即与 $g(x)$ 同阶），$f(x)$ 中 $x$ 和 $x^2$ 的系数必须为零： -$$ -2 - 2a = 0, \quad 2 - b = 0, -$$ -解得 $a = 1$, $b = 2$。此时 -$$ -f(x) = \frac{2}{3}x^3 + O(x^5), -$$ -与 $g(x)$ 同阶（但不等价，因为系数比值为 $2$）。 -因此正确选项为 (B)。 -**法二**：洛必达＋泰勒 -先考虑两者之比的极限$$\lim\limits_{x\to0}\frac{f(x)}{g(x)}=\lim\limits_{x\to0}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to0}\frac{2\mathrm{e}^x(\sin x+\cos x)-2a-2bx}{\arctan x^2}$$由麦克劳林公式得$$\begin{aligned}\mathrm{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+O(x^3)\\\sin x&=x-\frac{x^3}{6}+O(x^3)\\\cos x&=1-\frac{x^2}{2}+O(x^3)\end{aligned}$$于是（把所有大于3次的项都放在高阶无穷小里面，这样可以简化计算）$$\mathrm{e}^x(\sin x+\cos x)=1+2x+x^2+O(x^3)$$从而极限式的分子等于$$2-2a+(4-2b)+2x^2+O(x^3).$$又由麦克劳林公式得$$\arctan x=x-\frac{x^3}{6}+O(x^3)$$故分母为$$x^2-\frac{x^6}{6}+O(x^6)=x^2+O(x^3).$$上面的分子分母带入极限式中得$$\lim\limits_{x\to0}\frac{2-2a+(4-2b)+2x^2+O(x^3)}{x^2+O(x^3)}$$要让上式为有限值且不为$0$,只有$$\begin{cases}2-2a&=0\\4-2b&=0\end{cases}\implies\begin{cases}a&=1\\b&=2\end{cases}$$故选(B) -**答案：** (B) -4.下列级数中收敛的是（ ）。 -(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ -(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ -(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ -(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ -### 解析 -- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 -- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 -- **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 -- **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 -**答案：(B)** -5.一个倒置的圆锥形容器（顶点在下，底面在上），高度为 10 米，底面半径为 5 米。容器内装有水，水从底部的一个小孔（面积为 $0.1\pi$ 平方米）流出，流速为 $v = 0.6\sqrt{2gh}$ 米/秒，其中 $g = 10$ 米/秒$^2$。同时，以恒定速率 $Q = 0.3\pi$ 立方米/秒从顶部注入水。当水面高度为 5 米时，水面高度的瞬时变化率是多少？ -选项: -A. $-0.048$ 米/秒 -B. $-0.024$ 米/秒 -C. 0 -D. 0.048 米/秒 -**解析：** -设水面高度为 $h$（从圆锥顶点算起）。由相似关系，水面半径 -$$ -r = \frac{R}{H}h = \frac{5}{10}h = 0.5h, -$$ -水的体积为 -$$ -V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3. -$$ -对时间 $t$ 求导，得 -$$ -\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}. -$$ -另一方面，体积变化率由注入速率 $Q$ 和流出速率 $A \cdot v$ 决定： -$$ -\frac{dV}{dt} = Q - A \cdot v = 0.3\pi - 0.1\pi \cdot 0.6\sqrt{2gh}. -$$ -代入 $h=5$，$g=10$，则 -$$ -\sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10, -$$ -于是 -$$ -A \cdot v = 0.1\pi \cdot 0.6 \cdot 10 = 0.6\pi, -$$ -$$ -\frac{dV}{dt} = 0.3\pi - 0.6\pi = -0.3\pi. -$$ -代入微分式： -$$ -\frac{1}{4}\pi \cdot 5^2 \cdot \frac{dh}{dt} = -0.3\pi \quad \Rightarrow \quad \frac{25}{4}\pi \frac{dh}{dt} = -0.3\pi. -$$ -解得 -$$ -\frac{dh}{dt} = -0.3 \times \frac{4}{25} = -0.048 \ \text{米/秒}. -$$ -因此水面高度以每秒 $0.048$ 米的速度下降，故选 A。 -## 二、填空题（共5小题，每小题2分，共10分） +``` +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x}$。 +```text -1.设 $y = \arctan x$，则 $y^{(n)}(0)=\_\_\_\_.$ -**分析** -逐次求导以找到 $n$ 阶导数的规律。由于 $y' = \frac{1}{1 + x^2}$，即 $(1 + x^2)y' = 1$，故想到用莱布尼茨公式。 - -**解** - -**方法1** -因为 $y' = \frac{1}{1 + x^2}$，所以 $(1 + x^2)y' = 1$。 -上述两端对 $x$ 求 $n$ 阶导数，并利用莱布尼茨公式，得 -$$\sum_{k=0}^{n} C_n^k (1 + x^2)^{(k)} (y')^{(n-k)} = (1 + x^2)y^{(n+1)} + 2nxy^{(n)} + n(n-1)y^{(n-1)} = 0.$$ -在上式中令 $x = 0$，得 -$$y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0).$$ -由此递推公式，再加上 $y'(0) = 1, y''(0) = 0$，可得： -$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ - -**方法2** -因为 $y' = \frac{1}{1 + x^2} = \sum\limits_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， -所以 （积分得） -$$y = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, (|x| < 1).$$ -又知 $y(x)$ 在点 $x = 0$ 处的麦克劳林展开式为 -$$y(x) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!} x^n.$$ -比较系数可得 -$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ - -**方法3** -由 $y = \arctan x$，得 $x = \tan y$，则 -$$y' = \frac{1}{1 + x^2} = \frac{1}{1 + \tan^2 y} = \cos^2 y.$$ -利用复合函数求导法则： -$$y'' = -2 \cos y \sin y \cdot y' = -\sin(2y) \cos^2 y = \cos^2 y \sin 2\left( y + \frac{\pi}{2} \right),$$ -$$\begin{aligned}y''' &= \left[ -2 \cos y \sin y \sin 2\left( y + \frac{\pi}{2} \right) + 2 \cos^2 y \cos 2\left( y + \frac{\pi}{2} \right) \right] y'\\& = 2 \cos^3 y \cos\left( y + \frac{\pi}{2} \right) y' + 2 \cos^3 y \sin 3\left( y + \frac{\pi}{2} \right),\end{aligned}$$ -$$y^{(4)} = 6 \cos^4 y \cos\left( y + \frac{\pi}{2} \right) y' + 3 \cos^4 y \sin 4\left( y + \frac{\pi}{2} \right).$$ -由此归纳出 -$$y^{(n)} = (n-1)! \cos^n y \sin n\left( y + \frac{\pi}{2} \right).$$ -下面用归纳法证明以上结论： -当 $n = 1$ 时结论成立； -假设 $n = k$ 时结论成立，则当 $n = k + 1$ 时， -$$\begin{aligned} y^{(k+1)} &= (y^{(k)})' \\ &= (k-1)! \left[ -k \cos^{k+1} y \sin y \sin k\left( y + \frac{\pi}{2} \right) + k \cos^k y \cos k\left( y + \frac{\pi}{2} \right) \right] y' \\ &= k \cos^{k+1} y \cos\left( y + \frac{\pi}{2} \right) y' + k \cos^{k+1} y \sin (k+1)\left( y + \frac{\pi}{2} \right). \end{aligned}$$ -由于 $y(0) = \arctan 0 = 0, \cos 0 = 1$，因此 -$$y^{(n)}(0) = (n-1)! \sin \frac{n\pi}{2}.$$ -根据 $\sin \frac{n\pi}{2}$ 的取值（$n$ 为偶数时为 $0$，$n=4k+1$ 时为 $1$，$n=4k+3$ 时为 $-1$），可得到与方法1、2相同的结果。 -2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，则$\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}=\_\_\_.$ -**分析** -由已知的两个极限式可确定 $f(x)$ 的两个一次因子以及两个待定系数，从而完全确定 $f(x)$。 - -**解** -由已知有 $\lim\limits_{x \to 2a} f(x) = \lim\limits_{x \to 4a} f(x) = 0$。 -由于 $f(x)$ 处处连续，故 $f(2a) = f(4a) = 0$。 -因此 $f(x)$ 含有因式 $(x-2a)(x-4a)$，可设 -$$f(x) = (Ax + B)(x - 2a)(x - 4a).$$ -由条件： -$$\lim_{x \to 2a} \frac{f(x)}{x-2a} = \lim_{x \to 2a} (Ax + B)(x - 4a) = (2aA + B)(-2a) = 1,$$ -$$\lim_{x \to 4a} \frac{f(x)}{x-4a} = \lim_{x \to 4a} (Ax + B)(x - 2a) = (4aA + B)(2a) = 1.$$ -解方程组： -$$\begin{cases} (2aA + B)(-2a) = 1 \\ (4aA + B)(2a) = 1 \end{cases} \Rightarrow \begin{cases} -4a^2A - 2aB = 1 \\ 8a^2A + 2aB = 1 \end{cases}.$$ -相加得 $4a^2A = 2$，故 $A = \frac{1}{2a^2}$；代入第一个方程得 $-4a^2 \cdot \frac{1}{2a^2} - 2aB = 1$，即 $-2 - 2aB = 1$，解得 $B = -\frac{3}{2a}$。 -于是 -$$f(x) = \left( \frac{1}{2a^2}x - \frac{3}{2a} \right)(x - 2a)(x - 4a) = \frac{1}{2a^2}(x - 3a)(x - 2a)(x - 4a).$$ -最后， -$$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} \\&= \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) \\&= \frac{1}{2a^2} \cdot a \cdot (-a) \\&= -\frac{1}{2}.\end{aligned}$$ - -**答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. -3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ - 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ - 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ - 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ - 根据**p级数**的收敛性结论，级数$\sum\limits_{n=1}^{\infty}\frac{1}{n^{p+\frac{1}{2}}}$当且仅当$p+\frac{1}{2}>1$时收敛，即$p>\frac{1}{2}$， -所以，$p\in(\frac{1}{2},+\infty)$ -$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ - 方法1： - 令$x=2\sin t$，则$\mathrm{d}x=2\cos t \mathrm{d}t$， -$$\begin{align}\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=\int{(2\sin t)^3\sqrt{4-4\sin^2 t} \cdot 2\cos t\mathrm{d}t}\\ -&=32\int{\sin^3t\cos^2t\mathrm{d}t}\\ -&=32\int{\sin t(1-\cos^2t)\cos^2t\mathrm{d}t}\\ -&=-32\int{(\cos^2t-cos^4t)\mathrm{d}\cos t}\\ -&=-32(\frac{\cos^3t}{3}-\frac{cos^5t}{5})+C\\ -&=-\frac{4}{3}(\sqrt{4-x^2})^3+\frac{1}{5}(\sqrt{4-x^2})^5+C -\end{align} -$$ - 方法2 - 令$\sqrt{4-x^2}=t$，$x^2=4-t^2$，$x\mathrm{d}x=-t\mathrm{d}t$， -$$ -\begin{align} -\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=-\int{(4-t^2)t^2\mathrm{d}t}\\ -&=\frac{t^5}{5}-\frac{4t^3}{3}+C\\ -&=\frac{(\sqrt{4-x^2})^5}{5}-\frac{4(\sqrt{4-x^2})^3}{3}+C -\end{align} -$$ - -4..设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ -解：方程两边对$t$求导，得 -$$ -\begin{cases} -\frac{\mathrm{d}x}{\mathrm{d}t}=2t+2\\ -2t-\frac{\mathrm{d}y}{\mathrm{d}t}+a\frac{\mathrm{d}y}{\mathrm{d}t}\cos y=0 -\end{cases} -\Rightarrow -\begin{cases} -\frac{\mathrm{d}x}{\mathrm{d}t}=2(t+1)\\ -\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{2t}{1-a\cos y} -\end{cases} -\Rightarrow -\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t}{(t+1)(1-a\cos y)} -$$ -$$ -\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}x}=\frac{\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{\frac{(1-a\cos y)-at(t+1)\frac{\mathrm{d}y}{\mathrm{d}t}\sin y}{(t+1)^2(1-a\cos y)^2}}{2(t+1)} -$$ -注意到$y|_{t=0}=b,\frac{\mathrm{d}y}{\mathrm{d}t}|_{t=0}=0$，得 -$$ -\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\frac{1}{2(1-a\cos b)} -$$ ---- ---- ---- -## 三、解答题（共11小题，共80分） -1.求下列不定积分(提示，换元），其中 $a > 0$ -(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x$ -**解：** -令 $x = a \sin t$，有$\mathrm{d}x=a\cos t\mathrm{d}t$,则： -$$ -\begin{aligned} -\int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x &= \int \frac{a^2 \sin^2 t}{a \cos t} \cdot a \cos t \, \mathrm{d}t \\ -&= a^2 \int \sin^2 t \, \mathrm{d}t \\ -&= \frac{a^2}{2} \int (1 - \cos 2t) \, \mathrm{d}t \\ -&= \frac{a^2}{2} \left( t - \frac{1}{2} \sin 2t \right) + C \\ -&= \frac{a^2}{2} \arcsin \frac{x}{a} - \frac{x}{2} \sqrt{a^2 - x^2} + C -\end{aligned} -$$ ---- +``` - (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x$ +3.设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 +证明：存在 $\xi \in (0, \frac{1}{2})$，使得 $f''(\xi) = \frac{3f'(\xi)}{1-2\xi}$。 +```text -**解：** -**方法一：** 令 $x = a \tan t$，则： -$$ -\begin{aligned} -\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{a \sec t}{a^2 \tan^2 t} \cdot a \sec^2 t \, \mathrm{d}t \\ -&= \int \frac{\sec^3 t}{\tan^2 t} \, \mathrm{d}t \\ -&= \int \frac{1}{\sin^2 t \cos t} \, \mathrm{d}t \\ -&= \int \frac{\sin^2 t + \cos^2 t}{\sin^2 t \cos t} \, \mathrm{d}t \\ -&= \int \sec t \, \mathrm{d}t + \int \frac{\cos t}{\sin^2 t} \, \mathrm{d}t \\ -&= \ln |\sec t + \tan t| - \frac{1}{\sin t} + C \\ -&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C -\end{aligned} -$$ -**方法二：** -$$ -\begin{aligned} -\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{x^2 + a^2}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ -&= \int \frac{1}{\sqrt{x^2 + a^2}} \mathrm{d}x + a^2 \int \frac{1}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ -&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C -\end{aligned} -$$ ---- -(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x$ -**解：** -令 $x = a \sec t$，则： -$$ -\begin{aligned} -\int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x &= \int \frac{a \tan t}{a \sec t} \cdot a \sec t \tan t \, \mathrm{d}t \\ -&= a \int \tan^2 t \, \mathrm{d}t \\ -&= a \int (\sec^2 t - 1) \, \mathrm{d}t \\ -&= a (\tan t - t) + C \\ -&= \sqrt{x^2 - a^2} - a \arccos \frac{a}{x} + C -\end{aligned} -$$ ---- -(4) $\displaystyle \int \sqrt{1 + e^x} \mathrm{d}x$ -**解：** -令 $t = \sqrt{1 + e^x}$，则 $e^x = t^2 - 1$，$x = \ln(t^2 - 1)$，$\mathrm{d}x = \frac{2t}{t^2 - 1} \mathrm{d}t$： -$$ -\begin{aligned} -\int \sqrt{1 + e^x} \mathrm{d}x &= \int t \cdot \frac{2t}{t^2 - 1} \mathrm{d}t \\ -&= \int \frac{2t^2}{t^2 - 1} \mathrm{d}t \\ -&= 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) \mathrm{d}t \\ -&= 2 \int 1 \, \mathrm{d}t + \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) \mathrm{d}t \\ -&= 2t + \ln \left| \frac{t-1}{t+1} \right| + C \\ -&= 2 \sqrt{1 + e^x} + \ln \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} + C -\end{aligned} -$$ ---- -2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x}$。 -**解：** -当 $x \to 0$ 时，利用等价无穷小替换和泰勒展开： -分母：$x^3 \arcsin x \sim x^3 \cdot x = x^4$ -分子：$e^{x^2} - 1 - \ln(1+x^2) = \left(1+x^2+\frac{x^4}{2}+o(x^4)\right) - 1 - \left(x^2-\frac{x^4}{2}+o(x^4)\right) = x^4 + o(x^4)$ -所以 -$$ -\lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x} = \lim_{x \to 0} \frac{x^4 + o(x^4)}{x^4} = 1 -$$ - -设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 -证明：存在 $\xi \in (0, \frac{1}{2})$，使得 $f''(\xi) = \frac{3f'(\xi)}{1-2\xi}$。 -**证明：** -构造辅助函数 $g(x) = (1-2x)^{3/2} f'(x)$，$x \in [0, \frac{1}{2}]$。 -由于 $f$ 二阶可导，故 $g$ 在 $[0, \frac{1}{2}]$ 上连续，在 $(0, \frac{1}{2})$ 内可导。 -计算 $g$ 的导数： -$$ -\begin{aligned} -g'(x) &= (1-2x)^{3/2} f''(x) + \frac{3}{2}(1-2x)^{1/2} \cdot (-2) \cdot f'(x) \\ -&= (1-2x)^{1/2} \left[ (1-2x) f''(x) - 3f'(x) \right] -\end{aligned} -$$ -当 $x \in (0, \frac{1}{2})$ 时，$(1-2x)^{1/2} > 0$，因此 $g'(x) = 0$ 当且仅当： -$$ -(1-2x) f''(x) - 3f'(x) = 0 \quad \text{即} \quad f''(x) = \frac{3f'(x)}{1-2x} -$$ -所以只需证明存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 -由已知条件： -$$ -g(0) = (1-0)^{3/2} f'(0) = f'(0) = f(0), \quad g\left(\frac{1}{2}\right) = (1-1)^{3/2} f'\left(\frac{1}{2}\right) = 0 -$$ -分两种情况讨论： +``` -1. **若 $f'(0) = 0$** - 此时 $g(0) = 0$，所以 $g(0) = g\left(\frac{1}{2}\right) = 0$。由罗尔定理，存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 +4.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 +```text -2. **若 $f'(0) \neq 0$** - 对 $f$ 在 $[0, \frac{1}{2}]$ 上应用拉格朗日中值定理，存在 $c \in (0, \frac{1}{2})$ 使得： - $$ - f'(c) = \frac{f\left(\frac{1}{2}\right) - f(0)}{\frac{1}{2} - 0} = \frac{0 - f(0)}{\frac{1}{2}} = -2f(0) = -2f'(0) - $$ - 由于 $f'(0) \neq 0$，故 $f'(c) \neq 0$ 且与 $f'(0)$ 异号。 - 计算 $g(c) = (1-2c)^{3/2} f'(c)$，因为 $(1-2c)^{3/2} > 0$，所以 $g(c)$ 与 $f'(c)$ 同号，从而 $g(c)$ 与 $g(0) = f'(0)$ 异号。 - 由连续函数的介值定理，存在 $d \in (0, c)$ 使得 $g(d) = 0$。 - 在区间 $[d, \frac{1}{2}]$ 上，$g(d) = g\left(\frac{1}{2}\right) = 0$，由罗尔定理，存在 $\xi \in (d, \frac{1}{2}) \subset (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 -综上所述，无论何种情况，均存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$，从而有： -$$ -f''(\xi) = \frac{3f'(\xi)}{1-2\xi} -$$ -原命题得证。 -3.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 -### 方法一：逐阶求导归纳法 -由 -$$ -y' = a e^{ax} \sin bx + b e^{ax} \cos bx -= e^{ax} \bigl( a \sin bx + b \cos bx \bigr), -$$ -令 -$$ -\varphi = \arctan \frac{b}{a}, -$$ -则 -$$ -a \sin bx + b \cos bx = \sqrt{a^2 + b^2} \, \sin(bx + \varphi). -$$ -于是 -$$ -y' = e^{ax} \cdot \sqrt{a^2 + b^2} \, \sin(bx + \varphi). -$$ -同理， -$$ -y'' = \sqrt{a^2 + b^2} \cdot e^{ax} \bigl[ a \sin(bx + \varphi) + b \cos(bx + \varphi) \bigr] -= (a^2 + b^2) e^{ax} \sin(bx + 2\varphi). -$$ -依此类推，由归纳法可得 -$$ -y^{(n)} = (a^2 + b^2)^{n/2} \, e^{ax} \sin(bx + n\varphi), -$$ -其中 -$$ -\varphi = \arctan \frac{b}{a}. -$$ ---- -### 方法二：欧拉公式法 -设 -$$ -u = e^{ax} \cos bx, \quad v = e^{ax} \sin bx, -$$ -则 -$$ -u + iv = e^{ax} (\cos bx + i \sin bx) = e^{(a + bi)x}. -$$ -求 $n$ 阶导数： -$$ -u^{(n)} + i v^{(n)} = \bigl[ e^{(a + bi)x} \bigr]^{(n)} -= (a + bi)^n e^{(a + bi)x}. -$$ -记 -$$ -a + bi = \sqrt{a^2 + b^2} \, e^{i\varphi}, -\quad \varphi = \arctan \frac{b}{a}, -$$ -则 -$$ -(a + bi)^n = (a^2 + b^2)^{n/2} e^{in\varphi}. -$$ -于是 -$$ -u^{(n)} + i v^{(n)} -= (a^2 + b^2)^{n/2} e^{in\varphi} \cdot e^{ax} e^{ibx} -= (a^2 + b^2)^{n/2} e^{ax} e^{i(bx + n\varphi)}. -$$ -取虚部，即得 -$$ -\bigl( e^{ax} \sin bx \bigr)^{(n)} -= (a^2 + b^2)^{n/2} e^{ax} \sin(bx + n\varphi). -$$ -类似可得 -$$ -\bigl( e^{ax} \cos bx \bigr)^{(n)} -= (a^2 + b^2)^{n/2} e^{ax} \cos(bx + n\varphi). -$$ -4.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 -**解：** -已知 $e^x$ 的麦克劳林展开为： -$$ -e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6). -$$ -则 -$$ -\begin{aligned} -x^2 e^x &= x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6)\right)\\[1em] &= x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} + \frac{x^7}{5!} + \frac{x^8}{6!} + o(x^8). -\end{aligned} -$$ -由于我们只关心六阶展开，保留到 $x^6$ 项，$x^7$ 及更高次项可并入 $o(x^6)$，故 -$$ -x^2 e^x = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + o(x^6). -$$ -于是 -$$ -\begin{aligned} -f(x) &= x^2 e^x + x^6\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + x^6 + o(x^6)\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{25}{24}x^6 + o(x^6). -\end{aligned} -$$ -这就是 $f(x)$ 的六阶带佩亚诺余项的麦克劳林公式。 -由泰勒展开的唯一性，$x^6$ 项的系数为 $\frac{f^{(6)}(0)}{6!}$，即 -$$ -\frac{f^{(6)}(0)}{6!} = \frac{25}{24}. -$$ -所以 -$$ -f^{(6)}(0) = 6! \times \frac{25}{24} = 720 \times \frac{25}{24} = 30 \times 25 = 750. -$$ -**答案：** -展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$. -**题目：** -5.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 -$$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ -证明： + + + +``` + +5.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 +```text + + + + + + + + + + + + + + + + + + + + + + + + + +``` + +6.已知对于一数列$\{\alpha_n\}$，若满足$$\forall N\in\mathbb{N}_+,\exists\delta>0,当n>N时 ，有,\lvert \alpha_{n+1}-\alpha_n\rvert<\delta,$$则数列$\{\alpha_n\}$收敛.满足以上条件的数列称为柯西数列.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 +$$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ 试证明： （1）数列 $\{x_n\}$ 收敛（记 $\lim\limits_{n \to \infty} x_n = a$）； （2）方程 $f(x) = x$ 有唯一实根 $x = a$. +```text -**证明：** -（1）首先证明 $\{x_n\}$ 是柯西数列.对任意 $n \geq 1$，由拉格朗日中值定理，存在 $\xi_n$ 介于 $x_n$ 和 $x_{n-1}$ 之间，使得 -$$|x_{n+1} - x_n| = |f(x_n) - f(x_{n-1})| = |f'(\xi_n)| \cdot |x_n - x_{n-1}| \leq r |x_n - x_{n-1}|.$$ -反复应用此不等式，得 -$$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdots \leq r^{n-1} |x_2 - x_1|.$$ -于是对任意正整数 $m > n$，有 -$$ -\begin{aligned} -|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\[1em] -&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\[1em] -&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\[1em] -&\leq \frac{r^{n-1}}{1 - r} |x_2 - x_1|. -\end{aligned} -$$ -因为 $0 < r < 1$，所以当 $n \to \infty$ 时，$r^{n-1} \to 0$，从而对任意 $\varepsilon > 0$，存在 $N$，当 $m > n > N$ 时 $|x_m - x_n| < \varepsilon$。故 $\{x_n\}$ 是柯西数列，因此在实数域中收敛，记 $\lim_{n \to \infty} x_n = a$。 -（2）由于 $f$ 可导，故连续。在递推式 $x_{n+1} = f(x_n)$ 两边取极限 $n \to \infty$，得 -$$a = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n) = f(a),$$ -即 $a$ 是方程 $f(x) = x$ 的一个实根。 -下证唯一性。假设另有 $b \neq a$ 满足 $f(b) = b$，则由拉格朗日中值定理，存在 $\eta$ 介于 $a$ 和 $b$ 之间，使得 -$$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ -由于 $|a - b| > 0$，两边除以 $|a - b|$ 得 $1 \leq r$，与 $0 < r < 1$ 矛盾。故方程 $f(x) = x$ 有唯一实根 $x = a$ -综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。 -6.设 $f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 -**解** -令 $g(x) = \sin f(x)$，则 -$$ -g(x) - \frac{1}{3} g\left(\frac{1}{3}x\right) = x. -$$ -依次将 $x$ 替换为 $\frac{x}{3}, \frac{x}{3^2}, \cdots, \frac{x}{3^{n-1}}$，并乘以相应系数，得 -$$ -\begin{aligned} -g(x) - \frac{1}{3} g\left(\frac{x}{3}\right) &= x, \\ -\frac{1}{3} g\left(\frac{x}{3}\right) - \frac{1}{3^2} g\left(\frac{x}{3^2}\right) &= \frac{1}{3^2} x, \\ -\frac{1}{3^2} g\left(\frac{x}{3^2}\right) - \frac{1}{3^3} g\left(\frac{x}{3^3}\right) &= \frac{1}{3^4} x, \\ -&\vdots \\ -\frac{1}{3^{n-1}} g\left(\frac{x}{3^{n-1}}\right) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) &= \frac{1}{3^{2(n-1)}} x. -\end{aligned} -$$ -以上各式相加，得 -$$ -g(x) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = x \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right). -$$ -因为 $|g(x)| \leq 1$，所以 $\lim\limits_{n \to \infty} \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = 0$。 -而 $\lim\limits_{n \to \infty} \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right) = \frac{1}{1 - \frac{1}{9}} = \frac{9}{8}$， -因此 -$$ -g(x) = \frac{9}{8} x. -$$ -于是 $\sin f(x) = \frac{9}{8} x$，解得 -$$ -f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). -$$ -7.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 + + + + + + + + + + + + + +``` + +7.设在$\mathbb{R}$上的连续函数$f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 + +```text + + + + + + + + + + + + + + + + + + + + + + + + + +``` + +8.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 (1) 求参数$a, b$的值；（5分） (2) 计算极限$$ \lim_{x \to 0} \frac{f(x) - x^2}{x^4} $$的值。（5分） +```text + + + + + + + + + -**解答** -**(1)** -由于 -$$ -\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( a + bx^2 - \cos x \right) = a - 1 = 0, -$$ -故$a = 1$。 -又因为$f(x)$与$x^2$等价无穷小，所以 -$$ -\lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{1 + bx^2 - \cos x}{x^2} = b + \lim_{x \to 0} \frac{1 - \cos x}{x^2} = b + \frac{1}{2} = 1, -$$ -因此$b = \frac{1}{2}$。 -**(2)** -由$\cos x$的麦克劳林展开： -$$ -\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4). -$$ -代入$f(x)$： -$$ -f(x) = 1 + \frac{1}{2}x^2 - \left( 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) \right) = x^2 - \frac{1}{24}x^4 + o(x^4). -$$ -于是 -$$ -\lim_{x \to 0} \frac{f(x) - x^2}{x^4} = \lim_{x \to 0} \frac{-\frac{1}{24}x^4 + o(x^4)}{x^4} = -\frac{1}{24}. -$$ ---- -8.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： + + + + + + +``` + +9.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： （1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； （2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 +```text -**分析** -（1）只需将$[0,1]$分成两个区间，使$f(x)$在两个区间各用一次微分中值定理。设分点为$x_0 \in (0,1)$，由 -$$ -f(x_0)-f(0)=f'(\xi)(x_0-0),\quad f(1)-f(x_0)=f'(\eta)(1-x_0) \quad (0<\xi<x_0<\eta<1), -$$ -得$f'(\xi)=\frac{f(x_0)}{x_0}$，$f'(\eta)=\frac{1-f(x_0)}{1-x_0}$，则 -$$ -f'(\xi)f'(\eta)=1 \Leftrightarrow \frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = 1. -$$ -等价于$x_0$是方程$f(x)[1-f(x)] = x(1-x)$的根。取$x_0$满足$f(x_0)=1-x_0$即可。 -**证明** -（1）令$F(x)=f(x)-1+x$，则$F(x)$在$[0,1]$上连续，且$F(0)=-1<0$，$F(1)=1>0$。由介值定理知，存在$x_0 \in (0,1)$使$F(x_0)=0$，即$f(x_0)=1-x_0$。 -在$[0,x_0]$和$[x_0,1]$上分别应用拉格朗日中值定理，存在$\xi \in (0,x_0)$，$\eta \in (x_0,1)$，使得 -$$ -f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}. -$$ -于是 -$$ -f'(\xi)f'(\eta)=\frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = \frac{1-x_0}{x_0} \cdot \frac{x_0}{1-x_0} = 1. -$$ **（2）** -给定正数$a, b$，令$c = \frac{a}{a+b}$，则$0<c<1$。由于$f(x)$在$[0,1]$上连续，且$f(0)=0$, $f(1)=1$，由介值定理，存在$x_1 \in (0,1)$使得$f(x_1) = c = \frac{a}{a+b}$。 -在区间$[0, x_1]$和$[x_1, 1]$上分别应用拉格朗日中值定理，存在$\xi \in (0, x_1)$, $\eta \in (x_1, 1)$，使得 -$$ -f'(\xi) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{f(x_1)}{x_1}, \quad -f'(\eta) = \frac{f(1) - f(x_1)}{1 - x_1} = \frac{1 - f(x_1)}{1 - x_1}. -$$ -于是 -$$ -\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{f(x_1)} + b \cdot \frac{1 - x_1}{1 - f(x_1)}. -$$ -代入$f(x_1) = \frac{a}{a+b}$, $1 - f(x_1) = \frac{b}{a+b}$，得 -$$ -\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot \frac{1 - x_1}{b/(a+b)} = (a+b)x_1 + (a+b)(1 - x_1) = a+b. -$$ -故命题得证。 -9.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 -**补充整理：** -1. 若两级数均收敛，则其和也收敛；若两个都绝对收敛，则和也绝对收敛；若一个绝对收敛一个条件收敛，则和条件收敛。 -2. 若两级数仅一个收敛，则其和是发散的。 -**级数添加与去掉括号的敛散性有以下结论：** -1. 收敛级数任意添加括号也收敛。 -2. 若收敛级数去掉括号后的通项仍以0为极限，则去掉括号后的级数也收敛，且和不变。 -**去括号情况的证明** -设 $(a_1 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + \cdots + (a_{n_{k-1}+1} + \cdots + a_{n_k}) + \cdots$ 收敛于 $S$，且 $\lim\limits_{n \to \infty} a_n = 0$。记该级数的部分和为 $T_k$，$\sum_{n=1}^{\infty} a_n$ 的部分和为 $S_n$，则 $T_k = S_{n_k}$，$\lim\limits_{k \to \infty} S_{n_k} = \lim\limits_{k \to \infty} T_k = S$。 -由于 $\lim\limits_{n \to \infty} a_n = 0$，对 $\forall i \in \{1, 2, \cdots, n_{k+1} - n_k - 1\}$，有 -$\lim\limits_{k \to \infty} S_{n_k+i} = \lim\limits_{k \to \infty} ( T_k + a_{n_k+1} + \cdots + a_{n_k+i} ) = S$ -所以 $\lim\limits_{n \to \infty} S_n = S$，即 $\sum\limits_{n=1}^{\infty} a_n$ 也收敛于 $S$。 -**分析** 这是交错级数，且通项趋于0，但通项不单调，不适用莱布尼茨准则。可考虑用添加括号的方式来证明。也可采用交换相邻两项顺序的方式使通项满足单调性。 -**解**： -$$|a_n| = \left| \frac{(-1)^n}{[n+(-1)^n]^p} \right| = \frac{1}{n^p} \cdot \frac{1}{\left[ 1 + \frac{(-1)^n}{n} \right]^p} \sim \frac{1}{n^p}$$ ($n \to \infty$) -当 $p>1$ 时，级数绝对收敛；当 $0<p<1$ 时，级数不绝对收敛。 -下面讨论 $0<p\leq 1$ 时，级数的收敛性。 -首先，级数的通项 $a_n \to 0$。 -**方法1** 将原级数按如下方式添加括号 -$$\left( \frac{1}{3^p} - \frac{1}{2^p} \right) + \left( \frac{1}{5^p} - \frac{1}{4^p} \right) + \cdots + \left( \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} \right) + \cdots$$ -记 $b_n = \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p}$，则 $b_n < 0$，$\sum_{n=2}^{\infty} (-b_n)$是正项级数。由于 -$$-b_n = \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} = \frac{1}{(2n+1)^p} \left[ \left( 1 + \frac{1}{2n} \right)^p - 1 \right] \sim \frac{1}{(2n+1)^p} \cdot \frac{p}{2n} \sim \frac{p}{(2n)^{p+1}}$$ -而 $p+1>1$，所以 $\sum_{n=2}^{\infty} (-b_n)$ 收敛，由上面补充中去括号的讨论知，原级数收敛。 -**方法2** 同样考虑方法1中的级数 $\sum\limits_{n=2}^{\infty} (-b_n)$，其部分和为 -$$ -\begin{aligned} -S_n &= \left( \frac{1}{2^p} - \frac{1}{3^p} \right) + \left( \frac{1}{4^p} - \frac{1}{5^p} \right) + \cdots + \left( \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} \right) \\ -&= \frac{1}{2^p} \left( \frac{1}{3^p} - \frac{1}{4^p} \right) - \left( \frac{1}{5^p} - \frac{1}{6^p} \right) - \cdots - \left( \frac{1}{(2n-1)^p} - \frac{1}{(2n)^p} \right) - \frac{1}{(2n+1)^p} < \frac{1}{2^p} -\end{aligned} -$$ -正项级数部分和数列有界，级数收敛，从而原级数收敛。 -**方法3** 原级数是 -$$\frac{1}{3^p} - \frac{1}{2^p} + \frac{1}{5^p} - \frac{1}{4^p} + \cdots + \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} + \cdots$$ -奇偶项互换后的新级数为 -$$\frac{1}{2^p} - \frac{1}{3^p} + \frac{1}{4^p} - \frac{1}{5^p} + \cdots + \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} + \cdots$$ -记 $c_n = \frac{1}{n^p}$，该级数为 $\sum\limits_{n=2}^{\infty}(-1)^{n-1}c_n$，由于 $c_n$ 单减趋于0，由莱布尼茨判别法知，该交错级数收敛，从而原级数收敛。 -**评注** “方法3” 用到了收敛级数的性质：收敛级数交换相邻两项的位置后的级数仍收敛，且和不变。 -证明如下： -设 $a_1 + a_2 + a_3 + a_4 + \cdots + a_{2n-1} + a_{2n} + \cdots$ 收敛于 $S$，其部分和为 $S_n$。交换相邻两项的位置后的级数为 $a_2 + a_1 + a_4 + a_3 + \cdots + a_{2n} + a_{2n-1} + \cdots$，其部分和为 $T_n$，则 -$$T_{2n} = S_{2n} \Rightarrow \lim_{n \to \infty} T_{2n} = \lim_{n \to \infty} S_{2n} = S, \quad \lim_{n \to \infty} T_{2n+1} = \lim_{n \to \infty} T_{2n} + \lim_{n \to \infty} a_{2n+2} = S,$$ -所以 $\lim\limits_{n \to \infty} T_n = S$。 -10.(10分) +``` + + +10.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 +```text + + + + + + + + + + + + + + + + + -(1)证明： 对任意的正整数 $n$，方程 -$$ x^n + n^2 x - 1 = 0 $$ -有唯一正实根（记为 $x_n$）。 -**证明：** -令 $f_n(x) = x^n + n^2x - 1$ 。显然 $f_n(x)$ 在 $[0,1]$ 上连续，且 -$$ f_n(0) = -1 < 0, \quad f_n(1) = n^2 > 0, $$ -由闭区间上连续函数的零值定理可知，至少存在一点 $\xi \in (0,1)$，使得 $f_n(\xi) = 0$，即 -$$ x^n + n^2x - 1 = 0 $$ -至少有一个正实根。 -又 -$$ f'_n(x) = nx^{n-1} + n^2, $$ -易知当 $x > 0$ 时，$f'_n(x) > 0$，故函数 $f_n(x)$ 在 $(0,+\infty)$ 内严格单调增加，因此 $f_n(x)$ 在 $(0,+\infty)$ 内至多只有一个零点。综上，函数 $f_n(x)$ 在$(0,+\infty)$ 内有唯一零点，即方程 $x^n + n^2x - 1 = 0$ 有唯一正实根。 ---- + + + + + + + + +``` + +11.(10分) + +(1)证明： 对任意的正整数 $n$，方程 +$$ x^n + n^2 x - 1 = 0 $$ +有唯一正实根（记为 $x_n$）。 (2) 证明： 级数 $$ \sum_{n=1}^\infty x_n $$ 收敛，且其和不超过 2。 +```text + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + -**证明：** -记方程$x^n + n^2 x - 1 = 0$ 的唯一正实根为 $x_n$，则 $x_n^n + n^2 x_n - 1 = 0$，故 -$$ 0 < x_n = \frac{1}{n^2} - \frac{x_n^n}{n^2} < \frac{1}{n^2}. $$ -根据比较判别法，由于级数 $\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$ 收敛，故级数$\sum\limits_{n=1}^{\infty} x_n$收敛。 - -记$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$的前$n$ 项部分和为 $S_n$,$\sum\limits_{n=1}^{\infty} x_n$的前$n$ 项部分和为$T_n$，显然$T_n < S_n$，且 -$$\begin{aligned} -S_n &= \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < \frac{1}{1} + \frac{1}{1 \cdot 2} + \cdots + \frac{1}{(n-1)n} - \\&= 1 + 1 - \frac{1}{2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n-1} - \frac{1}{n} - \\&= 2 - \frac{1}{n}<2, - \end{aligned}$$ -根据数列极限的保号性（更准确地说是保序性）， -$$ \sum_{n=1}^{\infty} x_n = \lim_{n \to \infty} T_n \leq \lim_{n \to \infty} S_n \leq 2. $$ \ No newline at end of file +``` diff --git a/编写小组/试卷/0103高数模拟试卷（解析版）.md b/编写小组/试卷/0103高数模拟试卷（解析版）.md new file mode 100644 index 0000000..b8deec1 --- /dev/null +++ b/编写小组/试卷/0103高数模拟试卷（解析版）.md @@ -0,0 +1,842 @@ +--- +tags: + - 编写小组 +--- +## 一、单选题（共5小题，每小题2分，共10分） + +1.设函数 $f(x)$ 在 $x = 0$ 处可导，且 $f(0) = 0$。若 +$$ +\lim_{x \to 0} \frac{f(2x) - 3f(x) + f(-x)}{x} = 2, +$$ +则 $f'(0)$ 的值为（ ）。 + +A. $-2$ +B. $-1$ +C. $1$ +D. $2$ + +## 解析 +因为 $f(x)$ 在 $x=0$ 处可导且 $f(0)=0$，所以 $f'(0) = \lim\limits_{x \to 0} \frac{f(x)}{x}$ 存在。 + +将极限式分解： +$$ +\frac{f(2x) - 3f(x) + f(-x)}{x} = \frac{f(2x)}{x} - 3\frac{f(x)}{x} + \frac{f(-x)}{x}. +$$ + +分别计算各项极限： +$$ +\begin{aligned} +\lim_{x \to 0} \frac{f(2x)}{x} &= 2 \lim_{x \to 0} \frac{f(2x)}{2x} = 2f'(0), \\ +\lim_{x \to 0} \frac{f(x)}{x} &= f'(0), \\ +\lim_{x \to 0} \frac{f(-x)}{x} &= -\lim_{x \to 0} \frac{f(-x)}{-x} = -f'(0). +\end{aligned} +$$ + +因此， +$$ +\lim_{x \to 0} \frac{f(2x) - 3f(x) + f(-x)}{x} = 2f'(0) - 3f'(0) + (-f'(0)) = -2f'(0). +$$ + +由已知条件 $-2f'(0) = 2$，解得 $f'(0) = -1$。 + +**答案：B** + + +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2}$ +A.$-\frac{1}{12}$ +B.$\frac{1}{12}$ +C.$-\frac{1}{6}$ +D$\frac{1}{6}$ + +**解：** + +分别对分子和分母进行泰勒展开： + +分子： +$$\begin{aligned} +e^{x^2} \cos x &= \left(1 + x^2 + \frac{x^4}{2} + o(x^4)\right) \left(1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4)\right) \\&= 1 + \frac{1}{2}x^2 + \left(-\frac{1}{2} + \frac{1}{2} + \frac{1}{24}\right)x^4 + o(x^4) \\&= 1 + \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) +\end{aligned}$$ +所以 $e^{x^2} \cos x - 1 - \frac{1}{2}x^2 = \frac{1}{24}x^4 + o(x^4)$ + +分母： +$$ +\ln(1+x^2) = x^2 - \frac{1}{2}x^4 + o(x^4) +$$ +所以 $\ln(1+x^2) - x^2 = -\frac{1}{2}x^4 + o(x^4)$ + +因此 +$$ +\lim_{x \to 0} \frac{e^{x^2} \cos x - 1 - \frac{1}{2}x^2}{\ln(1 + x^2) - x^2} = \lim_{x \to 0} \frac{\frac{1}{24}x^4 + o(x^4)}{-\frac{1}{2}x^4 + o(x^4)} = \frac{1/24}{-1/2} = -\frac{1}{12} +$$ + + +3.已知函数 $f(x) = 2e^x \sin x - 2ax - bx^2$ 与 $g(x) = \int \arctan(x^2) \mathrm{d}x$（取满足 $g(0) = 0$ 的那个原函数）是 $x \to 0$ 过程的同阶无穷小量，则（ ）。 + +(A) $a = 1, \, b = 1$ +(B) $a = 1, \, b = 2$ +(C) $a = 2, \, b = 1$ +(D) $a = 2, \, b = 2$ + +**解：** +**法一**：为使 $f(x)$ 与 $g(x)$ 在 $x \to 0$ 时为同阶无穷小，需使二者最低阶非零项的阶数相同。下面分别展开 $f(x)$ 和 $g(x)$ 的麦克劳林公式。 + +对于 $f(x)$，利用已知展开式： +$$ +e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + O(x^3), \quad \sin x = x - \frac{x^3}{3!} + O(x^5), +$$ +则（把大于等于$5$次的项全都放在高阶无穷小里边） +$$ +e^x \sin x = \left(1 + x + \frac{x^2}{2} + \frac{x^3}{6} + O(x^3)\right)\left(x - \frac{x^3}{6} + O(x^5)\right) = x + x^2 + \frac{x^3}{3} + O(x^5). +$$ +因此 +$$ +2e^x \sin x = 2x + 2x^2 + \frac{2}{3}x^3 + O(x^5). +$$ +代入 $f(x)$ 得 +$$ +f(x) = 2x + 2x^2 + \frac{2}{3}x^3 - 2ax - bx^2 + O(x^5) = (2 - 2a)x + (2 - b)x^2 + \frac{2}{3}x^3 + O(x^5). +$$ + +对于 $g(x)$，由 $g(0)=0$ 及 $\arctan(x^2)$ 的展开： +$$ +\arctan(x^2) = x^2 - \frac{x^6}{3} + O(x^{10}), +$$ +积分得 +$$ +g(x) = \int_0^x \arctan(t^2) \mathrm{d}t = \frac{x^3}{3} - \frac{x^7}{21} + O(x^{11}) = \frac{1}{3}x^3 + O(x^7). +$$ +可见 $g(x)$ 是 $x \to 0$ 时的三阶无穷小，其主项为 $\dfrac{1}{3}x^3$。 + +为使 $f(x)$ 也是三阶无穷小（即与 $g(x)$ 同阶），$f(x)$ 中 $x$ 和 $x^2$ 的系数必须为零： +$$ +2 - 2a = 0, \quad 2 - b = 0, +$$ +解得 $a = 1$, $b = 2$。此时 +$$ +f(x) = \frac{2}{3}x^3 + O(x^5), +$$ +与 $g(x)$ 同阶（但不等价，因为系数比值为 $2$）。 + +因此正确选项为 (B)。 + +**法二**：洛必达＋泰勒 +先考虑两者之比的极限$$\lim\limits_{x\to0}\frac{f(x)}{g(x)}=\lim\limits_{x\to0}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to0}\frac{2\mathrm{e}^x(\sin x+\cos x)-2a-2bx}{\arctan x^2}$$由麦克劳林公式得$$\begin{aligned}\mathrm{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+O(x^3)\\\sin x&=x-\frac{x^3}{6}+O(x^3)\\\cos x&=1-\frac{x^2}{2}+O(x^3)\end{aligned}$$于是（把所有大于3次的项都放在高阶无穷小里面，这样可以简化计算）$$\mathrm{e}^x(\sin x+\cos x)=1+2x+x^2+O(x^3)$$从而极限式的分子等于$$2-2a+(4-2b)+2x^2+O(x^3).$$又由麦克劳林公式得$$\arctan x=x-\frac{x^3}{6}+O(x^3)$$故分母为$$x^2-\frac{x^6}{6}+O(x^6)=x^2+O(x^3).$$上面的分子分母带入极限式中得$$\lim\limits_{x\to0}\frac{2-2a+(4-2b)+2x^2+O(x^3)}{x^2+O(x^3)}$$要让上式为有限值且不为$0$,只有$$\begin{cases}2-2a&=0\\4-2b&=0\end{cases}\implies\begin{cases}a&=1\\b&=2\end{cases}$$故选(B) +**答案：** (B) + +4.下列级数中收敛的是（ ）。 + +(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ + +(B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ + +(C) $\displaystyle \sum_{n=1}^{\infty} \frac{n! \cdot 3^n}{n^n}$ + +(D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ + +### 解析 +- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 +- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 +- **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 +- **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 + +**答案：(B)** + +5.一个倒置的圆锥形容器（顶点在下，底面在上），高度为 10 米，底面半径为 5 米。容器内装有水，水从底部的一个小孔（面积为 $0.1\pi$ 平方米）流出，流速为 $v = 0.6\sqrt{2gh}$ 米/秒，其中 $g = 10$ 米/秒$^2$。同时，以恒定速率 $Q = 0.3\pi$ 立方米/秒从顶部注入水。当水面高度为 5 米时，水面高度的瞬时变化率是多少？ + +选项: +A. $-0.048$ 米/秒 +B. $-0.024$ 米/秒 +C. 0 +D. 0.048 米/秒 +**解析：** +设水面高度为 $h$（从圆锥顶点算起）。由相似关系，水面半径 +$$ +r = \frac{R}{H}h = \frac{5}{10}h = 0.5h, +$$ +水的体积为 +$$ +V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3. +$$ +对时间 $t$ 求导，得 +$$ +\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}. +$$ +另一方面，体积变化率由注入速率 $Q$ 和流出速率 $A \cdot v$ 决定： +$$ +\frac{dV}{dt} = Q - A \cdot v = 0.3\pi - 0.1\pi \cdot 0.6\sqrt{2gh}. +$$ +代入 $h=5$，$g=10$，则 +$$ +\sqrt{2gh} = \sqrt{2 \cdot 10 \cdot 5} = \sqrt{100} = 10, +$$ +于是 +$$ +A \cdot v = 0.1\pi \cdot 0.6 \cdot 10 = 0.6\pi, +$$ +$$ +\frac{dV}{dt} = 0.3\pi - 0.6\pi = -0.3\pi. +$$ +代入微分式： +$$ +\frac{1}{4}\pi \cdot 5^2 \cdot \frac{dh}{dt} = -0.3\pi \quad \Rightarrow \quad \frac{25}{4}\pi \frac{dh}{dt} = -0.3\pi. +$$ +解得 +$$ +\frac{dh}{dt} = -0.3 \times \frac{4}{25} = -0.048 \ \text{米/秒}. +$$ +因此水面高度以每秒 $0.048$ 米的速度下降，故选 A。 +## 二、填空题（共5小题，每小题2分，共10分） + +1.设 $y = \arctan x$，则 $y^{(n)}(0)=\_\_\_\_.$ + +**分析** +逐次求导以找到 $n$ 阶导数的规律。由于 $y' = \frac{1}{1 + x^2}$，即 $(1 + x^2)y' = 1$，故想到用莱布尼茨公式。 + +**解** + +**方法1** +因为 $y' = \frac{1}{1 + x^2}$，所以 $(1 + x^2)y' = 1$。 +上述两端对 $x$ 求 $n$ 阶导数，并利用莱布尼茨公式，得 +$$\sum_{k=0}^{n} C_n^k (1 + x^2)^{(k)} (y')^{(n-k)} = (1 + x^2)y^{(n+1)} + 2nxy^{(n)} + n(n-1)y^{(n-1)} = 0.$$ +在上式中令 $x = 0$，得 +$$y^{(n+1)}(0) = -n(n-1)y^{(n-1)}(0).$$ +由此递推公式，再加上 $y'(0) = 1, y''(0) = 0$，可得： +$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ + +**方法2** +因为 $y' = \frac{1}{1 + x^2} = \sum\limits_{n=0}^{\infty} (-1)^n x^{2n}, (|x| < 1)$， +所以 （积分得） +$$y = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1}, (|x| < 1).$$ +又知 $y(x)$ 在点 $x = 0$ 处的麦克劳林展开式为 +$$y(x) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!} x^n.$$ +比较系数可得 +$$y^{(n)}(0) = \begin{cases} 0, & n\text{为偶数} \\ (-1)^n (n-1)!, & n\text{为奇数} \end{cases}.$$ + +**方法3** +由 $y = \arctan x$，得 $x = \tan y$，则 +$$y' = \frac{1}{1 + x^2} = \frac{1}{1 + \tan^2 y} = \cos^2 y.$$ +利用复合函数求导法则： +$$y'' = -2 \cos y \sin y \cdot y' = -\sin(2y) \cos^2 y = \cos^2 y \sin 2\left( y + \frac{\pi}{2} \right),$$ +$$\begin{aligned}y''' &= \left[ -2 \cos y \sin y \sin 2\left( y + \frac{\pi}{2} \right) + 2 \cos^2 y \cos 2\left( y + \frac{\pi}{2} \right) \right] y'\\& = 2 \cos^3 y \cos\left( y + \frac{\pi}{2} \right) y' + 2 \cos^3 y \sin 3\left( y + \frac{\pi}{2} \right),\end{aligned}$$ +$$y^{(4)} = 6 \cos^4 y \cos\left( y + \frac{\pi}{2} \right) y' + 3 \cos^4 y \sin 4\left( y + \frac{\pi}{2} \right).$$ +由此归纳出 +$$y^{(n)} = (n-1)! \cos^n y \sin n\left( y + \frac{\pi}{2} \right).$$ +下面用归纳法证明以上结论： +当 $n = 1$ 时结论成立； +假设 $n = k$ 时结论成立，则当 $n = k + 1$ 时， +$$\begin{aligned} y^{(k+1)} &= (y^{(k)})' \\ &= (k-1)! \left[ -k \cos^{k+1} y \sin y \sin k\left( y + \frac{\pi}{2} \right) + k \cos^k y \cos k\left( y + \frac{\pi}{2} \right) \right] y' \\ &= k \cos^{k+1} y \cos\left( y + \frac{\pi}{2} \right) y' + k \cos^{k+1} y \sin (k+1)\left( y + \frac{\pi}{2} \right). \end{aligned}$$ +由于 $y(0) = \arctan 0 = 0, \cos 0 = 1$，因此 +$$y^{(n)}(0) = (n-1)! \sin \frac{n\pi}{2}.$$ +根据 $\sin \frac{n\pi}{2}$ 的取值（$n$ 为偶数时为 $0$，$n=4k+1$ 时为 $1$，$n=4k+3$ 时为 $-1$），可得到与方法1、2相同的结果。 + + + +2.已知 $f(x)$ 是三次多项式，且有 $\lim\limits_{x \to 2a} \frac{f(x)}{x-2a} = \lim\limits_{x \to 4a} \frac{f(x)}{x-4a} = 1$，则$\lim\limits_{x \to 3a} \frac{f(x)}{x-3a}=\_\_\_.$ + +**分析** +由已知的两个极限式可确定 $f(x)$ 的两个一次因子以及两个待定系数，从而完全确定 $f(x)$。 + +**解** +由已知有 $\lim\limits_{x \to 2a} f(x) = \lim\limits_{x \to 4a} f(x) = 0$。 +由于 $f(x)$ 处处连续，故 $f(2a) = f(4a) = 0$。 +因此 $f(x)$ 含有因式 $(x-2a)(x-4a)$，可设 +$$f(x) = (Ax + B)(x - 2a)(x - 4a).$$ +由条件： +$$\lim_{x \to 2a} \frac{f(x)}{x-2a} = \lim_{x \to 2a} (Ax + B)(x - 4a) = (2aA + B)(-2a) = 1,$$ +$$\lim_{x \to 4a} \frac{f(x)}{x-4a} = \lim_{x \to 4a} (Ax + B)(x - 2a) = (4aA + B)(2a) = 1.$$ +解方程组： +$$\begin{cases} (2aA + B)(-2a) = 1 \\ (4aA + B)(2a) = 1 \end{cases} \Rightarrow \begin{cases} -4a^2A - 2aB = 1 \\ 8a^2A + 2aB = 1 \end{cases}.$$ +相加得 $4a^2A = 2$，故 $A = \frac{1}{2a^2}$；代入第一个方程得 $-4a^2 \cdot \frac{1}{2a^2} - 2aB = 1$，即 $-2 - 2aB = 1$，解得 $B = -\frac{3}{2a}$。 +于是 +$$f(x) = \left( \frac{1}{2a^2}x - \frac{3}{2a} \right)(x - 2a)(x - 4a) = \frac{1}{2a^2}(x - 3a)(x - 2a)(x - 4a).$$ +最后， +$$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\frac{1}{2a^2}(x-3a)(x-2a)(x-4a)}{x-3a} \\&= \frac{1}{2a^2} \cdot (3a-2a)(3a-4a) \\&= \frac{1}{2a^2} \cdot a \cdot (-a) \\&= -\frac{1}{2}.\end{aligned}$$ + +**答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. + +3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ + 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ + 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ + 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ + 根据**p级数**的收敛性结论，级数$\sum\limits_{n=1}^{\infty}\frac{1}{n^{p+\frac{1}{2}}}$当且仅当$p+\frac{1}{2}>1$时收敛，即$p>\frac{1}{2}$， +所以，$p\in(\frac{1}{2},+\infty)$ +4.$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ + 方法1： + 令$x=2\sin t$，则$\mathrm{d}x=2\cos t \mathrm{d}t$， +$$\begin{align}\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=\int{(2\sin t)^3\sqrt{4-4\sin^2 t} \cdot 2\cos t\mathrm{d}t}\\ +&=32\int{\sin^3t\cos^2t\mathrm{d}t}\\ +&=32\int{\sin t(1-\cos^2t)\cos^2t\mathrm{d}t}\\ +&=-32\int{(\cos^2t-cos^4t)\mathrm{d}\cos t}\\ +&=-32(\frac{\cos^3t}{3}-\frac{cos^5t}{5})+C\\ +&=-\frac{4}{3}(\sqrt{4-x^2})^3+\frac{1}{5}(\sqrt{4-x^2})^5+C +\end{align} +$$ + 方法2 + 令$\sqrt{4-x^2}=t$，$x^2=4-t^2$，$x\mathrm{d}x=-t\mathrm{d}t$， +$$ +\begin{align} +\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=-\int{(4-t^2)t^2\mathrm{d}t}\\ +&=\frac{t^5}{5}-\frac{4t^3}{3}+C\\ +&=\frac{(\sqrt{4-x^2})^5}{5}-\frac{4(\sqrt{4-x^2})^3}{3}+C +\end{align} +$$ + +5.设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ +解：方程两边对$t$求导，得 +$$ +\begin{cases} +\frac{\mathrm{d}x}{\mathrm{d}t}=2t+2\\ +2t-\frac{\mathrm{d}y}{\mathrm{d}t}+a\frac{\mathrm{d}y}{\mathrm{d}t}\cos y=0 +\end{cases} +\Rightarrow +\begin{cases} +\frac{\mathrm{d}x}{\mathrm{d}t}=2(t+1)\\ +\frac{\mathrm{d}y}{\mathrm{d}t}=\frac{2t}{1-a\cos y} +\end{cases} +\Rightarrow +\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{t}{(t+1)(1-a\cos y)} +$$ +$$ +\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}x}=\frac{\frac{\mathrm{d}\frac{\mathrm{d}y}{\mathrm{d}x}}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{\frac{(1-a\cos y)-at(t+1)\frac{\mathrm{d}y}{\mathrm{d}t}\sin y}{(t+1)^2(1-a\cos y)^2}}{2(t+1)} +$$ +注意到$y|_{t=0}=b,\frac{\mathrm{d}y}{\mathrm{d}t}|_{t=0}=0$，得 +$$ +\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\frac{1}{2(1-a\cos b)} +$$ + +--- + + +--- + + +--- + + +## 三、解答题（共11小题，共80分） +1.求下列不定积分(提示，换元），其中 $a > 0$ + +(1) $\displaystyle \int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x$ + +**解：** +令 $x = a \sin t$，有$\mathrm{d}x=a\cos t\mathrm{d}t$,则： + +$$ +\begin{aligned} +\int \frac{x^2}{\sqrt{a^2 - x^2}} \mathrm{d}x &= \int \frac{a^2 \sin^2 t}{a \cos t} \cdot a \cos t \, \mathrm{d}t \\ +&= a^2 \int \sin^2 t \, \mathrm{d}t \\ +&= \frac{a^2}{2} \int (1 - \cos 2t) \, \mathrm{d}t \\ +&= \frac{a^2}{2} \left( t - \frac{1}{2} \sin 2t \right) + C \\ +&= \frac{a^2}{2} \arcsin \frac{x}{a} - \frac{x}{2} \sqrt{a^2 - x^2} + C +\end{aligned} +$$ + +--- + + (2) $\displaystyle \int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x$ + +**解：** +**方法一：** 令 $x = a \tan t$，则： + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{a \sec t}{a^2 \tan^2 t} \cdot a \sec^2 t \, \mathrm{d}t \\ +&= \int \frac{\sec^3 t}{\tan^2 t} \, \mathrm{d}t \\ +&= \int \frac{1}{\sin^2 t \cos t} \, \mathrm{d}t \\ +&= \int \frac{\sin^2 t + \cos^2 t}{\sin^2 t \cos t} \, \mathrm{d}t \\ +&= \int \sec t \, \mathrm{d}t + \int \frac{\cos t}{\sin^2 t} \, \mathrm{d}t \\ +&= \ln |\sec t + \tan t| - \frac{1}{\sin t} + C \\ +&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C +\end{aligned} +$$ + +**方法二：** + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 + a^2}}{x^2} \mathrm{d}x &= \int \frac{x^2 + a^2}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ +&= \int \frac{1}{\sqrt{x^2 + a^2}} \mathrm{d}x + a^2 \int \frac{1}{x^2 \sqrt{x^2 + a^2}} \mathrm{d}x \\ +&= \ln (x + \sqrt{x^2 + a^2}) - \frac{\sqrt{x^2 + a^2}}{x} + C +\end{aligned} +$$ + +--- + +(3) $\displaystyle \int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x$ + +**解：** +令 $x = a \sec t$，则： + +$$ +\begin{aligned} +\int \frac{\sqrt{x^2 - a^2}}{x} \mathrm{d}x &= \int \frac{a \tan t}{a \sec t} \cdot a \sec t \tan t \, \mathrm{d}t \\ +&= a \int \tan^2 t \, \mathrm{d}t \\ +&= a \int (\sec^2 t - 1) \, \mathrm{d}t \\ +&= a (\tan t - t) + C \\ +&= \sqrt{x^2 - a^2} - a \arccos \frac{a}{x} + C +\end{aligned} +$$ + +--- + +(4) $\displaystyle \int \sqrt{1 + e^x} \mathrm{d}x$ + +**解：** +令 $t = \sqrt{1 + e^x}$，则 $e^x = t^2 - 1$，$x = \ln(t^2 - 1)$，$\mathrm{d}x = \frac{2t}{t^2 - 1} \mathrm{d}t$： + +$$ +\begin{aligned} +\int \sqrt{1 + e^x} \mathrm{d}x &= \int t \cdot \frac{2t}{t^2 - 1} \mathrm{d}t \\ +&= \int \frac{2t^2}{t^2 - 1} \mathrm{d}t \\ +&= 2 \int \left( 1 + \frac{1}{t^2 - 1} \right) \mathrm{d}t \\ +&= 2 \int 1 \, \mathrm{d}t + \int \left( \frac{1}{t-1} - \frac{1}{t+1} \right) \mathrm{d}t \\ +&= 2t + \ln \left| \frac{t-1}{t+1} \right| + C \\ +&= 2 \sqrt{1 + e^x} + \ln \frac{\sqrt{1 + e^x} - 1}{\sqrt{1 + e^x} + 1} + C +\end{aligned} +$$ + +--- + + +2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x}$。 + +**解：** + +当 $x \to 0$ 时，利用等价无穷小替换和泰勒展开： + +分母：$x^3 \arcsin x \sim x^3 \cdot x = x^4$ + +分子：$e^{x^2} - 1 - \ln(1+x^2) = \left(1+x^2+\frac{x^4}{2}+o(x^4)\right) - 1 - \left(x^2-\frac{x^4}{2}+o(x^4)\right) = x^4 + o(x^4)$ + +所以 +$$ +\lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x} = \lim_{x \to 0} \frac{x^4 + o(x^4)}{x^4} = 1 +$$ + + + +设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 +证明：存在 $\xi \in (0, \frac{1}{2})$，使得 $f''(\xi) = \frac{3f'(\xi)}{1-2\xi}$。 + +**证明：** +构造辅助函数 $g(x) = (1-2x)^{3/2} f'(x)$，$x \in [0, \frac{1}{2}]$。 +由于 $f$ 二阶可导，故 $g$ 在 $[0, \frac{1}{2}]$ 上连续，在 $(0, \frac{1}{2})$ 内可导。 +计算 $g$ 的导数： + +$$ +\begin{aligned} +g'(x) &= (1-2x)^{3/2} f''(x) + \frac{3}{2}(1-2x)^{1/2} \cdot (-2) \cdot f'(x) \\ +&= (1-2x)^{1/2} \left[ (1-2x) f''(x) - 3f'(x) \right] +\end{aligned} +$$ + +当 $x \in (0, \frac{1}{2})$ 时，$(1-2x)^{1/2} > 0$，因此 $g'(x) = 0$ 当且仅当： + +$$ +(1-2x) f''(x) - 3f'(x) = 0 \quad \text{即} \quad f''(x) = \frac{3f'(x)}{1-2x} +$$ + +所以只需证明存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +由已知条件： + +$$ +g(0) = (1-0)^{3/2} f'(0) = f'(0) = f(0), \quad g\left(\frac{1}{2}\right) = (1-1)^{3/2} f'\left(\frac{1}{2}\right) = 0 +$$ + +分两种情况讨论： + +1. **若 $f'(0) = 0$** + 此时 $g(0) = 0$，所以 $g(0) = g\left(\frac{1}{2}\right) = 0$。由罗尔定理，存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +2. **若 $f'(0) \neq 0$** + 对 $f$ 在 $[0, \frac{1}{2}]$ 上应用拉格朗日中值定理，存在 $c \in (0, \frac{1}{2})$ 使得： + + $$ + f'(c) = \frac{f\left(\frac{1}{2}\right) - f(0)}{\frac{1}{2} - 0} = \frac{0 - f(0)}{\frac{1}{2}} = -2f(0) = -2f'(0) + $$ + + 由于 $f'(0) \neq 0$，故 $f'(c) \neq 0$ 且与 $f'(0)$ 异号。 + 计算 $g(c) = (1-2c)^{3/2} f'(c)$，因为 $(1-2c)^{3/2} > 0$，所以 $g(c)$ 与 $f'(c)$ 同号，从而 $g(c)$ 与 $g(0) = f'(0)$ 异号。 + 由连续函数的介值定理，存在 $d \in (0, c)$ 使得 $g(d) = 0$。 + 在区间 $[d, \frac{1}{2}]$ 上，$g(d) = g\left(\frac{1}{2}\right) = 0$，由罗尔定理，存在 $\xi \in (d, \frac{1}{2}) \subset (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$。 + +综上所述，无论何种情况，均存在 $\xi \in (0, \frac{1}{2})$ 使得 $g'(\xi) = 0$，从而有： + +$$ +f''(\xi) = \frac{3f'(\xi)}{1-2\xi} +$$ + +原命题得证。 + + + +3.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 + + +### 方法一：逐阶求导归纳法 + +由 +$$ +y' = a e^{ax} \sin bx + b e^{ax} \cos bx += e^{ax} \bigl( a \sin bx + b \cos bx \bigr), +$$ +令 +$$ +\varphi = \arctan \frac{b}{a}, +$$ +则 +$$ +a \sin bx + b \cos bx = \sqrt{a^2 + b^2} \, \sin(bx + \varphi). +$$ +于是 +$$ +y' = e^{ax} \cdot \sqrt{a^2 + b^2} \, \sin(bx + \varphi). +$$ +同理， +$$ +y'' = \sqrt{a^2 + b^2} \cdot e^{ax} \bigl[ a \sin(bx + \varphi) + b \cos(bx + \varphi) \bigr] += (a^2 + b^2) e^{ax} \sin(bx + 2\varphi). +$$ +依此类推，由归纳法可得 +$$ +y^{(n)} = (a^2 + b^2)^{n/2} \, e^{ax} \sin(bx + n\varphi), +$$ +其中 +$$ +\varphi = \arctan \frac{b}{a}. +$$ +下用数学归纳法证明.设$y^{(n)}=(a^2+b^2)^{n/2}\mathrm{e}^{ax}\sin(bx+n\varphi)$,则$$\begin{aligned}y^{(n+1)}&=(a^2+b^2)^{n/2}(a\mathrm{e}^{ax}\sin(bx+n\varphi)+b\mathrm{e}^{ax}\cos(bx+n\varphi))\\&=(a^2+b^2)^{(n+1)/2}\mathrm{e}^{ax}\sin(bx+(n+1)\varphi)\end{aligned}$$得证. + +--- + +### 方法二：欧拉公式法 + +设 +$$ +u = e^{ax} \cos bx, \quad v = e^{ax} \sin bx, +$$ +则 +$$ +u + \mathrm{i}v = e^{ax} (\cos bx + \mathrm{i} \sin bx) = e^{(a + b\mathrm{i})x}. +$$ +求 $n$ 阶导数： +$$ +u^{(n)} + \mathrm{i} v^{(n)} = \bigl[ e^{(a + b\mathrm{i})x} \bigr]^{(n)} += (a + b\mathrm{i})^n e^{(a + b\mathrm{i})x}. +$$ +记 +$$ +a + bi = \sqrt{a^2 + b^2} \, e^{\mathrm{i}\varphi}, +\quad \varphi = \arctan \frac{b}{a}, +$$ +则 +$$ +(a + b\mathrm{i})^n = (a^2 + b^2)^{n/2} e^{n\varphi\mathrm{i}}. +$$ +于是 +$$ +u^{(n)} + \mathrm{i} v^{(n)} += (a^2 + b^2)^{n/2} e^{ n\varphi\mathrm{i}} \cdot e^{ax} e^{\mathrm{i}bx} += (a^2 + b^2)^{n/2} e^{ax} e^{\mathrm{i}(bx + n\varphi)}. +$$ +取虚部，即得 +$$ +\bigl( e^{ax} \sin bx \bigr)^{(n)} += (a^2 + b^2)^{n/2} e^{ax} \sin(bx + n\varphi). +$$ +类似可得 +$$ +\bigl( e^{ax} \cos bx \bigr)^{(n)} += (a^2 + b^2)^{n/2} e^{ax} \cos(bx + n\varphi). +$$ + +4.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 + +**解：** +已知 $e^x$ 的麦克劳林展开为： +$$ +e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6). +$$ +则 +$$ +\begin{aligned} +x^2 e^x &= x^2 \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!} + o(x^6)\right)\\[1em] &= x^2 + x^3 + \frac{x^4}{2!} + \frac{x^5}{3!} + \frac{x^6}{4!} + \frac{x^7}{5!} + \frac{x^8}{6!} + o(x^8). +\end{aligned} +$$ +由于我们只关心六阶展开，保留到 $x^6$ 项，$x^7$ 及更高次项可并入 $o(x^6)$，故 +$$ +x^2 e^x = x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + o(x^6). +$$ +于是 +$$ +\begin{aligned} +f(x) &= x^2 e^x + x^6\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{x^6}{24} + x^6 + o(x^6)\\[1em] &= x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6} + \frac{25}{24}x^6 + o(x^6). +\end{aligned} +$$ +这就是 $f(x)$ 的六阶带佩亚诺余项的麦克劳林公式。 + +由泰勒展开的唯一性，$x^6$ 项的系数为 $\frac{f^{(6)}(0)}{6!}$，即 +$$ +\frac{f^{(6)}(0)}{6!} = \frac{25}{24}. +$$ +所以 +$$ +f^{(6)}(0) = 6! \times \frac{25}{24} = 720 \times \frac{25}{24} = 30 \times 25 = 750. +$$ + +**答案：** +展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$. + + +5.已知对于一数列$\{\alpha_n\}$，若满足$$\forall N\in\mathbb{N}_+,\exists\delta>0,当n>N时 ，有,\lvert \alpha_{n+1}-\alpha_n\rvert<\delta,$$则数列$\{\alpha_n\}$收敛.满足以上条件的数列称为柯西数列.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 +$$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ +证明： + +（1）数列 $\{x_n\}$ 收敛（记 $\lim\limits_{n \to \infty} x_n = a$）； +（2）方程 $f(x) = x$ 有唯一实根 $x = a$. + +**证明：** + +（1）首先证明 $\{x_n\}$ 是柯西数列.对任意 $n \geq 1$，由拉格朗日中值定理，存在 $\xi_n$ 介于 $x_n$ 和 $x_{n-1}$ 之间，使得 +$$|x_{n+1} - x_n| = |f(x_n) - f(x_{n-1})| = |f'(\xi_n)| \cdot |x_n - x_{n-1}| \leq r |x_n - x_{n-1}|.$$ +反复应用此不等式，得 +$$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdots \leq r^{n-1} |x_2 - x_1|.$$ +于是对任意正整数 $m > n$，有 +$$ +\begin{aligned} +|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\[1em] +&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\[1em] +&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\[1em] +&\leq \frac{r^{n-1}}{1 - r} |x_2 - x_1|. +\end{aligned} +$$ +因为 $0 < r < 1$，所以当 $n \to \infty$ 时，$r^{n-1} \to 0$，从而对任意 $\varepsilon > 0$，存在 $N$，当 $m > n > N$ 时 $|x_m - x_n| < \varepsilon$。故 $\{x_n\}$ 是柯西数列，因此在实数域中收敛，记 $\lim_{n \to \infty} x_n = a$。 + +（2）由于 $f$ 可导，故连续。在递推式 $x_{n+1} = f(x_n)$ 两边取极限 $n \to \infty$，得 +$$a = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} f(x_n) = f(\lim_{n \to \infty} x_n) = f(a),$$ +即 $a$ 是方程 $f(x) = x$ 的一个实根。 + +下证唯一性。假设另有 $b \neq a$ 满足 $f(b) = b$，则由拉格朗日中值定理，存在 $\eta$ 介于 $a$ 和 $b$ 之间，使得 +$$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ +由于 $|a - b| > 0$，两边除以 $|a - b|$ 得 $1 \leq r$，与 $0 < r < 1$ 矛盾。故方程 $f(x) = x$ 有唯一实根 $x = a$ + +综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。 + + +6.设在$\mathbb{R}$上的连续函数$f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 + +**解** +令 $g(x) = \sin f(x)$，则 +$$ +g(x) - \frac{1}{3} g\left(\frac{1}{3}x\right) = x. +$$ + +依次将 $x$ 替换为 $\frac{x}{3}, \frac{x}{3^2}, \cdots, \frac{x}{3^{n-1}}$，并乘以相应系数，得 +$$ +\begin{aligned} +g(x) - \frac{1}{3} g\left(\frac{x}{3}\right) &= x, \\ +\frac{1}{3} g\left(\frac{x}{3}\right) - \frac{1}{3^2} g\left(\frac{x}{3^2}\right) &= \frac{1}{3^2} x, \\ +\frac{1}{3^2} g\left(\frac{x}{3^2}\right) - \frac{1}{3^3} g\left(\frac{x}{3^3}\right) &= \frac{1}{3^4} x, \\ +&\vdots \\ +\frac{1}{3^{n-1}} g\left(\frac{x}{3^{n-1}}\right) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) &= \frac{1}{3^{2(n-1)}} x. +\end{aligned} +$$ + +以上各式相加，得 +$$ +g(x) - \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = x \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right). +$$ + +因为 $|g(x)| \leq 1$，所以 $\lim\limits_{n \to \infty} \frac{1}{3^n} g\left(\frac{x}{3^n}\right) = 0$。 +而 $\lim\limits_{n \to \infty} \left(1 + \frac{1}{9} + \frac{1}{9^2} + \cdots + \frac{1}{9^{n-1}}\right) = \frac{1}{1 - \frac{1}{9}} = \frac{9}{8}$， +因此 由函数的连续性， +$$ +g(x) = \frac{9}{8} x. +$$ + +于是 $\sin f(x) = \frac{9}{8} x$，解得 +$$ +f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). +$$ + +7.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 + +(1) 求参数$a, b$的值；（5分） +(2) 计算极限$$ +\lim_{x \to 0} \frac{f(x) - x^2}{x^4} +$$的值。（5分） + +**解答** + +**(1)** +由于 +$$ +\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( a + bx^2 - \cos x \right) = a - 1 = 0, +$$ +故$a = 1$。 +又因为$f(x)$与$x^2$等价无穷小，所以 +$$ +\lim_{x \to 0} \frac{f(x)}{x^2} = \lim_{x \to 0} \frac{1 + bx^2 - \cos x}{x^2} = b + \lim_{x \to 0} \frac{1 - \cos x}{x^2} = b + \frac{1}{2} = 1, +$$ +因此$b = \frac{1}{2}$。 + +**(2)** +由$\cos x$的麦克劳林展开： +$$ +\cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4). +$$ +代入$f(x)$： +$$ +f(x) = 1 + \frac{1}{2}x^2 - \left( 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4) \right) = x^2 - \frac{1}{24}x^4 + o(x^4). +$$ +于是 +$$ +\lim_{x \to 0} \frac{f(x) - x^2}{x^4} = \lim_{x \to 0} \frac{-\frac{1}{24}x^4 + o(x^4)}{x^4} = -\frac{1}{24}. +$$ + +--- + + +8.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： +（1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； +（2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 + +**分析** +（1）只需将$[0,1]$分成两个区间，使$f(x)$在两个区间各用一次微分中值定理。设分点为$x_0 \in (0,1)$，由 +$$ +f(x_0)-f(0)=f'(\xi)(x_0-0),\quad f(1)-f(x_0)=f'(\eta)(1-x_0) \quad (0<\xi<x_0<\eta<1), +$$ +得$f'(\xi)=\frac{f(x_0)}{x_0}$，$f'(\eta)=\frac{1-f(x_0)}{1-x_0}$，则 +$$ +f'(\xi)f'(\eta)=1 \Leftrightarrow \frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = 1. +$$ +等价于$x_0$是方程$f(x)[1-f(x)] = x(1-x)$的根。取$x_0$满足$f(x_0)=1-x_0$即可。 + +**证明** +（1）令$F(x)=f(x)-1+x$，则$F(x)$在$[0,1]$上连续，且$F(0)=-1<0$，$F(1)=1>0$。由介值定理知，存在$x_0 \in (0,1)$使$F(x_0)=0$，即$f(x_0)=1-x_0$。 +在$[0,x_0]$和$[x_0,1]$上分别应用拉格朗日中值定理，存在$\xi \in (0,x_0)$，$\eta \in (x_0,1)$，使得 +$$ +f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}. +$$ +于是 +$$ +f'(\xi)f'(\eta)=\frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = \frac{1-x_0}{x_0} \cdot \frac{x_0}{1-x_0} = 1. +$$ （2） +给定正数$a, b$，令$c = \frac{a}{a+b}$，则$0<c<1$。由于$f(x)$在$[0,1]$上连续，且$f(0)=0$, $f(1)=1$，由介值定理，存在$x_1 \in (0,1)$使得$f(x_1) = c = \frac{a}{a+b}$。 +在区间$[0, x_1]$和$[x_1, 1]$上分别应用拉格朗日中值定理，存在$\xi \in (0, x_1)$, $\eta \in (x_1, 1)$，使得 +$$ +f'(\xi) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{f(x_1)}{x_1}, \quad +f'(\eta) = \frac{f(1) - f(x_1)}{1 - x_1} = \frac{1 - f(x_1)}{1 - x_1}. +$$ +于是 +$$ +\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{f(x_1)} + b \cdot \frac{1 - x_1}{1 - f(x_1)}. +$$ +代入$f(x_1) = \frac{a}{a+b}$, $1 - f(x_1) = \frac{b}{a+b}$，得 +$$ +\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot \frac{1 - x_1}{b/(a+b)} = (a+b)x_1 + (a+b)(1 - x_1) = a+b. +$$ +故命题得证。 + + + +9.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 + +**补充整理：** + +1. 若两级数均收敛，则其和也收敛；若两个都绝对收敛，则和也绝对收敛；若一个绝对收敛一个条件收敛，则和条件收敛。 + +2. 若两级数仅一个收敛，则其和是发散的。 + +**级数添加与去掉括号的敛散性有以下结论：** + +1. 收敛级数任意添加括号也收敛。 +2. 若收敛级数去掉括号后的通项仍以0为极限，则去掉括号后的级数也收敛，且和不变。 + +**去括号情况的证明** + +设 $(a_1 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + \cdots + (a_{n_{k-1}+1} + \cdots + a_{n_k}) + \cdots$ 收敛于 $S$，且 $\lim\limits_{n \to \infty} a_n = 0$。记该级数的部分和为 $T_k$，$\sum_{n=1}^{\infty} a_n$ 的部分和为 $S_n$，则 $T_k = S_{n_k}$，$\lim\limits_{k \to \infty} S_{n_k} = \lim\limits_{k \to \infty} T_k = S$。 + +由于 $\lim\limits_{n \to \infty} a_n = 0$，对 $\forall i \in \{1, 2, \cdots, n_{k+1} - n_k - 1\}$，有 + +$\lim\limits_{k \to \infty} S_{n_k+i} = \lim\limits_{k \to \infty} ( T_k + a_{n_k+1} + \cdots + a_{n_k+i} ) = S$ + +所以 $\lim\limits_{n \to \infty} S_n = S$，即 $\sum\limits_{n=1}^{\infty} a_n$ 也收敛于 $S$。 + +**分析** 这是交错级数，且通项趋于0，但通项不单调，不适用莱布尼茨准则。可考虑用添加括号的方式来证明。也可采用交换相邻两项顺序的方式使通项满足单调性。 + +**解**： +$$|a_n| = \left| \frac{(-1)^n}{[n+(-1)^n]^p} \right| = \frac{1}{n^p} \cdot \frac{1}{\left[ 1 + \frac{(-1)^n}{n} \right]^p} \sim \frac{1}{n^p}$$ ($n \to \infty$) + +当 $p>1$ 时，级数绝对收敛；当 $0<p<1$ 时，级数不绝对收敛。 +下面讨论 $0<p\leq 1$ 时，级数的收敛性。 +首先，级数的通项 $a_n \to 0$。 + +**方法1** 将原级数按如下方式添加括号 +$$\left( \frac{1}{3^p} - \frac{1}{2^p} \right) + \left( \frac{1}{5^p} - \frac{1}{4^p} \right) + \cdots + \left( \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} \right) + \cdots$$ + +记 $b_n = \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p}$，则 $b_n < 0$，$\sum_{n=2}^{\infty} (-b_n)$是正项级数。由于 +$$-b_n = \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} = \frac{1}{(2n+1)^p} \left[ \left( 1 + \frac{1}{2n} \right)^p - 1 \right] \sim \frac{1}{(2n+1)^p} \cdot \frac{p}{2n} \sim \frac{p}{(2n)^{p+1}}$$ + +而 $p+1>1$，所以 $\sum_{n=2}^{\infty} (-b_n)$ 收敛，由上面补充中去括号的讨论知，原级数收敛。 + +**方法2** 同样考虑方法1中的级数 $\sum\limits_{n=2}^{\infty} (-b_n)$，其部分和为 +$$ +\begin{aligned} +S_n &= \left( \frac{1}{2^p} - \frac{1}{3^p} \right) + \left( \frac{1}{4^p} - \frac{1}{5^p} \right) + \cdots + \left( \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} \right) \\ +&= \frac{1}{2^p} \left( \frac{1}{3^p} - \frac{1}{4^p} \right) - \left( \frac{1}{5^p} - \frac{1}{6^p} \right) - \cdots - \left( \frac{1}{(2n-1)^p} - \frac{1}{(2n)^p} \right) - \frac{1}{(2n+1)^p} < \frac{1}{2^p} +\end{aligned} +$$ +正项级数部分和数列有界，级数收敛，从而原级数收敛。 + +**方法3** 原级数是 +$$\frac{1}{3^p} - \frac{1}{2^p} + \frac{1}{5^p} - \frac{1}{4^p} + \cdots + \frac{1}{(2n+1)^p} - \frac{1}{(2n)^p} + \cdots$$ +奇偶项互换后的新级数为 +$$\frac{1}{2^p} - \frac{1}{3^p} + \frac{1}{4^p} - \frac{1}{5^p} + \cdots + \frac{1}{(2n)^p} - \frac{1}{(2n+1)^p} + \cdots$$ + +记 $c_n = \frac{1}{n^p}$，该级数为 $\sum\limits_{n=2}^{\infty}(-1)^{n-1}c_n$，由于 $c_n$ 单减趋于0，由莱布尼茨判别法知，该交错级数收敛，从而原级数收敛。 + +**评注** “方法3” 用到了收敛级数的性质：收敛级数交换相邻两项的位置后的级数仍收敛，且和不变。 + +证明如下： +设 $a_1 + a_2 + a_3 + a_4 + \cdots + a_{2n-1} + a_{2n} + \cdots$ 收敛于 $S$，其部分和为 $S_n$。交换相邻两项的位置后的级数为 $a_2 + a_1 + a_4 + a_3 + \cdots + a_{2n} + a_{2n-1} + \cdots$，其部分和为 $T_n$，则 +$$T_{2n} = S_{2n} \Rightarrow \lim_{n \to \infty} T_{2n} = \lim_{n \to \infty} S_{2n} = S, \quad \lim_{n \to \infty} T_{2n+1} = \lim_{n \to \infty} T_{2n} + \lim_{n \to \infty} a_{2n+2} = S,$$ +所以 $\lim\limits_{n \to \infty} T_n = S$。 + +10.(10分) + +(1)证明： 对任意的正整数 $n$，方程 +$$ x^n + n^2 x - 1 = 0 $$ +有唯一正实根（记为 $x_n$）。 + +**证明：** +令 $f_n(x) = x^n + n^2x - 1$ 。显然 $f_n(x)$ 在 $[0,1]$ 上连续，且 +$$ f_n(0) = -1 < 0, \quad f_n(1) = n^2 > 0, $$ +由闭区间上连续函数的零值定理可知，至少存在一点 $\xi \in (0,1)$，使得 $f_n(\xi) = 0$，即 +$$ x^n + n^2x - 1 = 0 $$ +至少有一个正实根。 + +又 +$$ f'_n(x) = nx^{n-1} + n^2, $$ +易知当 $x > 0$ 时，$f'_n(x) > 0$，故函数 $f_n(x)$ 在 $(0,+\infty)$ 内严格单调增加，因此 $f_n(x)$ 在 $(0,+\infty)$ 内至多只有一个零点。综上，函数 $f_n(x)$ 在$(0,+\infty)$ 内有唯一零点，即方程 $x^n + n^2x - 1 = 0$ 有唯一正实根。 + +--- + + (2) 证明： 级数 +$$ \sum_{n=1}^\infty x_n $$ +收敛，且其和不超过 2。 + +**证明：** +记方程$x^n + n^2 x - 1 = 0$ 的唯一正实根为 $x_n$，则 $x_n^n + n^2 x_n - 1 = 0$，故 +$$ 0 < x_n = \frac{1}{n^2} - \frac{x_n^n}{n^2} < \frac{1}{n^2}. $$ +根据比较判别法，由于级数 $\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$ 收敛，故级数$\sum\limits_{n=1}^{\infty} x_n$收敛。 + +记$\sum\limits_{n=1}^{\infty} \frac{1}{n^2}$的前$n$ 项部分和为 $S_n$,$\sum\limits_{n=1}^{\infty} x_n$的前$n$ 项部分和为$T_n$，显然$T_n < S_n$，且 +$$\begin{aligned} +S_n &= \frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < \frac{1}{1} + \frac{1}{1 \cdot 2} + \cdots + \frac{1}{(n-1)n} + \\&= 1 + 1 - \frac{1}{2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n-1} - \frac{1}{n} + \\&= 2 - \frac{1}{n}<2, + \end{aligned}$$ +根据数列极限的保号性（更准确地说是保序性）， +$$ \sum_{n=1}^{\infty} x_n = \lim_{n \to \infty} T_n \leq \lim_{n \to \infty} S_n \leq 2. $$ \ No newline at end of file -- 2.34.1 From 7dc91dbfd10e1ab53fa1aa14a14668d3181893f4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 13:37:45 +0800 Subject: [PATCH 152/274] =?UTF-8?q?=E5=8F=88=E5=8F=8C=E5=8F=92=E4=BF=AE?= =?UTF-8?q?=E6=94=B9=E4=BA=86=E9=83=A8=E5=88=86=E6=A0=BC=E5=BC=8F=EF=BC=8C?= =?UTF-8?q?=E4=BD=BF=E5=BE=97pdf=E7=89=88=E7=9C=8B=E8=B5=B7=E6=9D=A5?= =?UTF-8?q?=E6=AD=A3=E5=B8=B8?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 20 +++++++------------ 1 file changed, 7 insertions(+), 13 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index 35a8be1..3f3544f 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -250,11 +250,8 @@ $$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ 试证明： ``` 8.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 - -(1) 求参数$a, b$的值；（5分） -(2) 计算极限$$ -\lim_{x \to 0} \frac{f(x) - x^2}{x^4} -$$的值。（5分） +(1) 求参数$a, b$的值;（5分） +(2) 计算极限$$\lim\limits_{x \to 0} \frac{f(x) - x^2}{x^4}$$的值.（5分） ```text @@ -279,9 +276,9 @@ $$的值。（5分） ``` -9.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： -（1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； -（2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 +9.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： +（1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$; +（2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$. ```text @@ -349,12 +346,9 @@ $$的值。（5分） 11.(10分) -(1)证明： 对任意的正整数 $n$，方程 -$$ x^n + n^2 x - 1 = 0 $$ +(1)证明： 对任意的正整数 $n$，方程 $$ x^n + n^2 x - 1 = 0 $$ 有唯一正实根（记为 $x_n$）。 - (2) 证明： 级数 -$$ \sum_{n=1}^\infty x_n $$ -收敛，且其和不超过 2。 + (2) 证明： 级数$\sum\limits_{n=1}^{\infty} x_n$收敛，且其和不超过 2。 ```text -- 2.34.1 From 882456d9a04df4c2b3e7ea41b9a405a6c744c82f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 13:37:57 +0800 Subject: [PATCH 153/274] vault backup: 2026-01-02 13:37:57 --- .../0103高数模拟试卷（解析版）.md | 26 +++++-------------- 1 file changed, 7 insertions(+), 19 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷（解析版）.md b/编写小组/试卷/0103高数模拟试卷（解析版）.md index b8deec1..2a5937a 100644 --- a/编写小组/试卷/0103高数模拟试卷（解析版）.md +++ b/编写小组/试卷/0103高数模拟试卷（解析版）.md @@ -663,9 +663,7 @@ $$ 7.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 (1) 求参数$a, b$的值；（5分） -(2) 计算极限$$ -\lim_{x \to 0} \frac{f(x) - x^2}{x^4} -$$的值。（5分） +(2) 计算极限$$\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$$的值。（5分） **解答** @@ -716,27 +714,17 @@ $$ **证明** （1）令$F(x)=f(x)-1+x$，则$F(x)$在$[0,1]$上连续，且$F(0)=-1<0$，$F(1)=1>0$。由介值定理知，存在$x_0 \in (0,1)$使$F(x_0)=0$，即$f(x_0)=1-x_0$。 在$[0,x_0]$和$[x_0,1]$上分别应用拉格朗日中值定理，存在$\xi \in (0,x_0)$，$\eta \in (x_0,1)$，使得 -$$ -f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}. -$$ +$$f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}.$$ 于是 -$$ -f'(\xi)f'(\eta)=\frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = \frac{1-x_0}{x_0} \cdot \frac{x_0}{1-x_0} = 1. -$$ （2） +$$f'(\xi)f'(\eta)=\frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0} = \frac{1-x_0}{x_0} \cdot \frac{x_0}{1-x_0} = 1.$$ （2） 给定正数$a, b$，令$c = \frac{a}{a+b}$，则$0<c<1$。由于$f(x)$在$[0,1]$上连续，且$f(0)=0$, $f(1)=1$，由介值定理，存在$x_1 \in (0,1)$使得$f(x_1) = c = \frac{a}{a+b}$。 在区间$[0, x_1]$和$[x_1, 1]$上分别应用拉格朗日中值定理，存在$\xi \in (0, x_1)$, $\eta \in (x_1, 1)$，使得 -$$ -f'(\xi) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{f(x_1)}{x_1}, \quad -f'(\eta) = \frac{f(1) - f(x_1)}{1 - x_1} = \frac{1 - f(x_1)}{1 - x_1}. -$$ +$$f'(\xi) = \frac{f(x_1) - f(0)}{x_1 - 0} = \frac{f(x_1)}{x_1}, \quad +f'(\eta) = \frac{f(1) - f(x_1)}{1 - x_1} = \frac{1 - f(x_1)}{1 - x_1}.$$ 于是 -$$ -\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{f(x_1)} + b \cdot \frac{1 - x_1}{1 - f(x_1)}. -$$ +$$\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{f(x_1)} + b \cdot \frac{1 - x_1}{1 - f(x_1)}.$$ 代入$f(x_1) = \frac{a}{a+b}$, $1 - f(x_1) = \frac{b}{a+b}$，得 -$$ -\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot \frac{1 - x_1}{b/(a+b)} = (a+b)x_1 + (a+b)(1 - x_1) = a+b. -$$ +$$\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot \frac{1 - x_1}{b/(a+b)} = (a+b)x_1 + (a+b)(1 - x_1) = a+b.$$ 故命题得证。 -- 2.34.1 From 5c84acd16647e341ecc8927edbbcea1cc8c74368 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 2 Jan 2026 13:51:53 +0800 Subject: [PATCH 154/274] vault backup: 2026-01-02 13:51:52 --- .../0103高数模拟试卷（解析版）.md | 70 +++++++++---------- 1 file changed, 33 insertions(+), 37 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷（解析版）.md b/编写小组/试卷/0103高数模拟试卷（解析版）.md index 2a5937a..5a3ea54 100644 --- a/编写小组/试卷/0103高数模拟试卷（解析版）.md +++ b/编写小组/试卷/0103高数模拟试卷（解析版）.md @@ -15,7 +15,7 @@ B. $-1$ C. $1$ D. $2$ -## 解析 + **解析** 因为 $f(x)$ 在 $x=0$ 处可导且 $f(0)=0$，所以 $f'(0) = \lim\limits_{x \to 0} \frac{f(x)}{x}$ 存在。 将极限式分解： @@ -133,7 +133,7 @@ $$ (D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ -### 解析 + **解析** - **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 - **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 - **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 @@ -159,11 +159,11 @@ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (0.5h)^2 h = \frac{1}{12}\pi h^3. $$ 对时间 $t$ 求导，得 $$ -\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}. +\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1}{4}\pi h^2 \frac{\mathrm{d}h}{\mathrm{d}t}. $$ 另一方面，体积变化率由注入速率 $Q$ 和流出速率 $A \cdot v$ 决定： $$ -\frac{dV}{dt} = Q - A \cdot v = 0.3\pi - 0.1\pi \cdot 0.6\sqrt{2gh}. +\frac{\mathrm{d}V}{\mathrm{d}t} = Q - A \cdot v = 0.3\pi - 0.1\pi \cdot 0.6\sqrt{2gh}. $$ 代入 $h=5$，$g=10$，则 $$ @@ -178,11 +178,11 @@ $$ $$ 代入微分式： $$ -\frac{1}{4}\pi \cdot 5^2 \cdot \frac{dh}{dt} = -0.3\pi \quad \Rightarrow \quad \frac{25}{4}\pi \frac{dh}{dt} = -0.3\pi. +\frac{1}{4}\pi \cdot 5^2 \cdot \frac{\mathrm{d}h}{\mathrm{d}t} = -0.3\pi \quad \Rightarrow \quad \frac{25}{4}\pi \frac{\mathrm{d}h}{\mathrm{d}t} = -0.3\pi. $$ 解得 $$ -\frac{dh}{dt} = -0.3 \times \frac{4}{25} = -0.048 \ \text{米/秒}. +\frac{\mathrm{d}h}{\mathrm{d}t} = -0.3 \times \frac{4}{25} = -0.048 \ \text{米/秒}. $$ 因此水面高度以每秒 $0.048$ 米的速度下降，故选 A。 ## 二、填空题（共5小题，每小题2分，共10分） @@ -255,13 +255,16 @@ $$\begin{aligned}\lim_{x \to 3a} \frac{f(x)}{x-3a} &= \lim_{x \to 3a} \frac{\fra **答案**：$\displaystyle \lim_{x \to 3a} \frac{f(x)}{x-3a} = -\frac{1}{2}$. 3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ - 首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ + +解：首先，考虑级数$\sum\limits_{n=1}^{\infty}|\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})|=\sum\limits_{n=1}^{\infty}{n^p}|\sin(\frac{1}{\sqrt{n}})|$ 当$n\to\infty$时，$\frac{1}{\sqrt{n}}\to 0$，此时有等价无穷小关系：$\sin(\frac{1}{\sqrt{n}}) \sim \frac{1}{\sqrt{n}}.$ 因此，级数的通项可以近似为$\frac{1}{n^p}\frac{1}{\sqrt{n}}=\frac{1}{n^{p+\frac{1}{2}}}$ 根据**p级数**的收敛性结论，级数$\sum\limits_{n=1}^{\infty}\frac{1}{n^{p+\frac{1}{2}}}$当且仅当$p+\frac{1}{2}>1$时收敛，即$p>\frac{1}{2}$， 所以，$p\in(\frac{1}{2},+\infty)$ -4.$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ - 方法1： + +4.求不定积分$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ + +方法1： 令$x=2\sin t$，则$\mathrm{d}x=2\cos t \mathrm{d}t$， $$\begin{align}\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=\int{(2\sin t)^3\sqrt{4-4\sin^2 t} \cdot 2\cos t\mathrm{d}t}\\ &=32\int{\sin^3t\cos^2t\mathrm{d}t}\\ @@ -271,7 +274,8 @@ $$\begin{align}\int{x^3\sqrt{4-x^2}\mathrm{d}x}&=\int{(2\sin t)^3\sqrt{4-4\sin^2 &=-\frac{4}{3}(\sqrt{4-x^2})^3+\frac{1}{5}(\sqrt{4-x^2})^5+C \end{align} $$ - 方法2 + +方法2： 令$\sqrt{4-x^2}=t$，$x^2=4-t^2$，$x\mathrm{d}x=-t\mathrm{d}t$， $$ \begin{align} @@ -280,9 +284,11 @@ $$ &=\frac{(\sqrt{4-x^2})^5}{5}-\frac{4(\sqrt{4-x^2})^3}{3}+C \end{align} $$ - + 5.设$y=f(x)$由$\begin{cases}x=t^2+2t\\t^2-y+a\sin y=1\end{cases}$确定，若$y(0)=b$，$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\underline{\quad\quad\quad}.$ -解：方程两边对$t$求导，得 + +**解**： +方程两边对$t$求导，得 $$ \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t}=2t+2\\ @@ -304,14 +310,6 @@ $$ \frac{\mathrm{d}^2y}{\mathrm{d}x^2}|_{t=0}=\frac{1}{2(1-a\cos b)} $$ ---- - - ---- - - ---- - ## 三、解答题（共11小题，共80分） 1.求下列不定积分(提示，换元），其中 $a > 0$ @@ -395,8 +393,6 @@ $$ \end{aligned} $$ ---- - 2.求极限 $\displaystyle \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln\left(x^2 + 1\right)}{x^3 \arcsin x}$。 @@ -415,7 +411,7 @@ $$ -设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 +3.设 $f(x)$ 在 $[0, \frac{1}{2}]$ 上二阶可导，$f(0) = f'(0)$，$f\left(\frac{1}{2}\right) = 0$。 证明：存在 $\xi \in (0, \frac{1}{2})$，使得 $f''(\xi) = \frac{3f'(\xi)}{1-2\xi}$。 **证明：** @@ -471,10 +467,10 @@ $$ -3.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 +4.求 $y = e^{ax} \sin bx$ 的 $n$ 阶导数 $y^{(n)}\text{,其中a，b为非零常数}$。 -### 方法一：逐阶求导归纳法 + **方法一：逐阶求导归纳法** 由 $$ @@ -510,7 +506,7 @@ $$ --- -### 方法二：欧拉公式法 + **方法二：欧拉公式法** 设 $$ @@ -551,7 +547,7 @@ $$ = (a^2 + b^2)^{n/2} e^{ax} \cos(bx + n\varphi). $$ -4.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 +5.将函数 $f(x) = x^2 e^x + x^6$ 展开成六阶带佩亚诺余项的麦克劳林公式，并求 $f^{(6)}(0)$ 的值。 **解：** 已知 $e^x$ 的麦克劳林展开为： @@ -589,7 +585,7 @@ $$ 展开式为 $f(x) = x^2 + x^3 + \frac{1}{2}x^4 + \frac{1}{6}x^5 + \frac{25}{24}x^6 + o(x^6)$，$f^{(6)}(0) = 750$. -5.已知对于一数列$\{\alpha_n\}$，若满足$$\forall N\in\mathbb{N}_+,\exists\delta>0,当n>N时 ，有,\lvert \alpha_{n+1}-\alpha_n\rvert<\delta,$$则数列$\{\alpha_n\}$收敛.满足以上条件的数列称为柯西数列.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 +6.已知对于一数列$\{\alpha_n\}$，若满足$$\forall N\in\mathbb{N}_+,\exists\delta>0,当n>N时 ，有,\lvert \alpha_{n+1}-\alpha_n\rvert<\delta,$$则数列$\{\alpha_n\}$收敛.满足以上条件的数列称为柯西数列.设函数 $f(x)$ 在 $(-\infty, +\infty)$ 内可导，且 $|f'(x)| \leq r$ ($0 < r < 1$)。取实数 $x_1$，记 $$x_{n+1} = f(x_n), \quad n = 1, 2, \cdots.$$ 证明： @@ -624,7 +620,7 @@ $$|a - b| = |f(a) - f(b)| = |f'(\eta)| \cdot |a - b| \leq r |a - b|.$$ 综上所述，数列 $\{x_n\}$ 收敛于 $a$，且 $a$ 是方程 $f(x) = x$ 的唯一实根。 -6.设在$\mathbb{R}$上的连续函数$f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 +7.设在$\mathbb{R}$上的连续函数$f(x)$ 满足 $\sin f(x) - \frac{1}{3} \sin f\left(\frac{1}{3}x\right) = x$，求 $f(x)$。 **解** 令 $g(x) = \sin f(x)$，则 @@ -660,14 +656,14 @@ $$ f(x) = 2k\pi + \arcsin \frac{9}{8} x \quad \text{或} \quad f(x) = (2k-1)\pi - \arcsin \frac{9}{8} x \quad (k \in \mathbb{Z}). $$ -7.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 +8.已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 (1) 求参数$a, b$的值；（5分） (2) 计算极限$$\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$$的值。（5分） -**解答** +**解** -**(1)** +(1) 由于 $$ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( a + bx^2 - \cos x \right) = a - 1 = 0, @@ -679,7 +675,7 @@ $$ $$ 因此$b = \frac{1}{2}$。 -**(2)** +(2) 由$\cos x$的麦克劳林展开： $$ \cos x = 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 + o(x^4). @@ -696,7 +692,7 @@ $$ --- -8.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： +9.设$f(x)$在$[0,1]$上连续，在$(0,1)$上可导,且 $f(0)=0, f(1)=1$。试证： （1）在$(0,1)$内存在不同的$\xi, \eta$使$f'(\xi)f'(\eta)=1$； （2）对任意给定的正数$a, b$，在$(0,1)$内存在不同的$\xi, \eta$使$\frac{a}{f'(\xi)}+\frac{b}{f'(\eta)}=a+b$。 @@ -711,7 +707,7 @@ f'(\xi)f'(\eta)=1 \Leftrightarrow \frac{f(x_0)}{x_0} \cdot \frac{1-f(x_0)}{1-x_0 $$ 等价于$x_0$是方程$f(x)[1-f(x)] = x(1-x)$的根。取$x_0$满足$f(x_0)=1-x_0$即可。 -**证明** +**证明**： （1）令$F(x)=f(x)-1+x$，则$F(x)$在$[0,1]$上连续，且$F(0)=-1<0$，$F(1)=1>0$。由介值定理知，存在$x_0 \in (0,1)$使$F(x_0)=0$，即$f(x_0)=1-x_0$。 在$[0,x_0]$和$[x_0,1]$上分别应用拉格朗日中值定理，存在$\xi \in (0,x_0)$，$\eta \in (x_0,1)$，使得 $$f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0},\quad f'(\eta)=\frac{f(1)-f(x_0)}{1-x_0}.$$ @@ -729,7 +725,7 @@ $$\frac{a}{f'(\xi)} + \frac{b}{f'(\eta)} = a \cdot \frac{x_1}{a/(a+b)} + b \cdot -9.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 +10.讨论级数 $$\sum_{n=2}^{\infty} \frac{(-1)^n}{[n+(-1)^n]^p}$$ ( $p>0$ )的敛散性。 **补充整理：** @@ -792,7 +788,7 @@ $$\frac{1}{2^p} - \frac{1}{3^p} + \frac{1}{4^p} - \frac{1}{5^p} + \cdots + \frac $$T_{2n} = S_{2n} \Rightarrow \lim_{n \to \infty} T_{2n} = \lim_{n \to \infty} S_{2n} = S, \quad \lim_{n \to \infty} T_{2n+1} = \lim_{n \to \infty} T_{2n} + \lim_{n \to \infty} a_{2n+2} = S,$$ 所以 $\lim\limits_{n \to \infty} T_n = S$。 -10.(10分) +11.(10分) (1)证明： 对任意的正整数 $n$，方程 $$ x^n + n^2 x - 1 = 0 $$ -- 2.34.1 From beb79bb015bc718d46553775332b4b31b4444b4b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 2 Jan 2026 18:47:04 +0800 Subject: [PATCH 155/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 编写小组/试卷/0103高数模拟试卷.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index f4cca66..9637675 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -261,13 +261,13 @@ $$ **答案：(B)** ---- ---- ---- + + + ## 三、解答题（共11小题，共80分） -- 2.34.1 From 4f371a405efcf1484c050016fced0f5ea65a1801 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Fri, 2 Jan 2026 18:49:56 +0800 Subject: [PATCH 156/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- conflict-files-obsidian-git.md | 17 ----------------- 1 file changed, 17 deletions(-) delete mode 100644 conflict-files-obsidian-git.md diff --git a/conflict-files-obsidian-git.md b/conflict-files-obsidian-git.md deleted file mode 100644 index 8b99edd..0000000 --- a/conflict-files-obsidian-git.md +++ /dev/null @@ -1,17 +0,0 @@ -# Conflicts -Please resolve them and commit them using the commands `Git: Commit all changes` followed by `Git: Push` -(This file will automatically be deleted before commit) -[[#Additional Instructions]] available below file list - -- [[0103高数模拟试卷]] - -# Additional Instructions -I strongly recommend to use "Source mode" for viewing the conflicted files. For simple conflicts, in each file listed above replace every occurrence of the following text blocks with the desired text. - -```diff -<<<<<<< HEAD - File changes in local repository -======= - File changes in remote repository ->>>>>>> origin/main -``` \ No newline at end of file -- 2.34.1 From e640a5a5e5f3e8e5aafd800a5cc7634eb0700478 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 2 Jan 2026 21:29:28 +0800 Subject: [PATCH 157/274] =?UTF-8?q?=E8=A1=A5=E5=85=85=E8=BE=93=E5=85=A5?= =?UTF-8?q?=E8=A7=84=E8=8C=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 笔记分享/LaTeX（KaTeX）输入规范.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md index 0b40a08..84987d5 100644 --- a/笔记分享/LaTeX（KaTeX）输入规范.md +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -2,4 +2,5 @@ 1. 矩阵使用bmatrix 2. 自然常数或电荷量e、虚数单位i应当为**正体**，需要用mathrm记号包裹，例：$\mathrm{e}^{\mathrm{i}\pi}+1=0$；然而，当e,i作为变量时，应当用正常的斜体。例：$\sum\limits_{i=1}^{n}a_i$ 3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ -4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ \ No newline at end of file +4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ +5. \$\$双美元符号之间不要打回车！除非你有\begin{...}\end{...}\$\$ \ No newline at end of file -- 2.34.1 From 21185704ce19d0a56608945ad6cb8e039b197f1d Mon Sep 17 00:00:00 2001 From: Elwood <3286545699@qq.com> Date: Sat, 3 Jan 2026 16:17:09 +0800 Subject: [PATCH 158/274] vault backup: 2026-01-03 16:17:09 --- 编写小组/试卷/0103高数模拟试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/0103高数模拟试卷（解析版）.md b/编写小组/试卷/0103高数模拟试卷（解析版）.md index 5a3ea54..2616d35 100644 --- a/编写小组/试卷/0103高数模拟试卷（解析版）.md +++ b/编写小组/试卷/0103高数模拟试卷（解析版）.md @@ -398,7 +398,7 @@ $$ **解：** -当 $x \to 0$ 时，利用等价无穷小替换和泰勒展开： +当 $x \to 0$ 时，利用等价无穷小替换和泰勒展开：9 分母：$x^3 \arcsin x \sim x^3 \cdot x = x^4$ -- 2.34.1 From a53297d5345deaf30c57cf1a25413f8d2aeafff0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sat, 3 Jan 2026 20:12:18 +0800 Subject: [PATCH 159/274] vault backup: 2026-01-03 20:12:18 --- 编写小组/试卷/0103高数模拟试卷.md | 4 ++-- .../试卷/0103高数模拟试卷（解析版）.md | 8 ++++---- 2 files changed, 6 insertions(+), 6 deletions(-) diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index a2fb50a..bd29779 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -24,9 +24,9 @@ D$\frac{1}{6}$ (C) $a = 2, \, b = 1$ (D) $a = 2, \, b = 2$ -4.下列级数中收敛的是（ ）。 +4.下列级数中绝对收敛的是（ ）。 -(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ +(A) $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ (B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ diff --git a/编写小组/试卷/0103高数模拟试卷（解析版）.md b/编写小组/试卷/0103高数模拟试卷（解析版）.md index 5a3ea54..d4437d3 100644 --- a/编写小组/试卷/0103高数模拟试卷（解析版）.md +++ b/编写小组/试卷/0103高数模拟试卷（解析版）.md @@ -123,9 +123,9 @@ $$ 先考虑两者之比的极限$$\lim\limits_{x\to0}\frac{f(x)}{g(x)}=\lim\limits_{x\to0}\frac{f'(x)}{g'(x)}=\lim\limits_{x\to0}\frac{2\mathrm{e}^x(\sin x+\cos x)-2a-2bx}{\arctan x^2}$$由麦克劳林公式得$$\begin{aligned}\mathrm{e}^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+O(x^3)\\\sin x&=x-\frac{x^3}{6}+O(x^3)\\\cos x&=1-\frac{x^2}{2}+O(x^3)\end{aligned}$$于是（把所有大于3次的项都放在高阶无穷小里面，这样可以简化计算）$$\mathrm{e}^x(\sin x+\cos x)=1+2x+x^2+O(x^3)$$从而极限式的分子等于$$2-2a+(4-2b)+2x^2+O(x^3).$$又由麦克劳林公式得$$\arctan x=x-\frac{x^3}{6}+O(x^3)$$故分母为$$x^2-\frac{x^6}{6}+O(x^6)=x^2+O(x^3).$$上面的分子分母带入极限式中得$$\lim\limits_{x\to0}\frac{2-2a+(4-2b)+2x^2+O(x^3)}{x^2+O(x^3)}$$要让上式为有限值且不为$0$,只有$$\begin{cases}2-2a&=0\\4-2b&=0\end{cases}\implies\begin{cases}a&=1\\b&=2\end{cases}$$故选(B) **答案：** (B) -4.下列级数中收敛的是（ ）。 +4.下列级数中绝对收敛的是（ ）。 -(A) $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ +(A) $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^n}{\sqrt{n+(-1)^n}}$ (B) $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln n)^{\ln n}}$ @@ -134,8 +134,8 @@ $$ (D) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1+1/n}}$ **解析** -- **(A)** 当 $n=1$ 时，分母 $\sqrt{1+(-1)^1}=0$，项无定义，即便忽略此项，级数条件收敛，但整体不收敛。 -- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数收敛。 +- **(A)** g根据莱布尼兹判别法，级数条件收敛，但其绝对值接近于$p-$级数（$p=\frac{1}{2}$）的情形，故不绝对收敛。 +- **(B)** 由于 $(\ln n)^{\ln n} = n^{\ln \ln n}$，当 $n$ 足够大时，$\ln \ln n > 2$，故 $\frac{1}{(\ln n)^{\ln n}} < \frac{1}{n^2}$，由 $p$-级数收敛知原级数绝对收敛。 - **(C)** 用比值判别法：$\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim\limits_{n\to\infty} \frac{3}{(1+1/n)^n} = \frac{3}{e} > 1$，发散。 - **(D)** 由于 $n^{1/n} \to 1$，故 $\frac{1}{n^{1+1/n}} \sim \frac{1}{n}$，与调和级数比较，发散。 -- 2.34.1 From e491870c62fbfeb172e9cda6df41379976fa9e59 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 5 Jan 2026 00:57:43 +0800 Subject: [PATCH 160/274] vault backup: 2026-01-05 00:57:43 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 -- 1 file changed, 2 deletions(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index aa594eb..e1d2cc6 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -90,8 +90,6 @@ tags: 8. 设2阶矩阵A=$\begin{bmatrix}3&-1\\-9&3\end{bmatrix}$，n为正整数，则$A^n=\underline{\quad\quad}$。 -9. 若向量组$\alpha_1 = (1,0,1)^T,\quad \alpha_2 = (0,1,1)^T,\quad \alpha_3 = (1,3,5)^T$不能由向量组$\beta_1 = (1,1,1)^T,\quad\beta_2 = (1,2,3)^T,\quad\beta_3 = (3,4,a)^T$线性表示，则$a = \underline{\qquad\qquad}.$ - 解析： 先计算$A^2$： -- 2.34.1 From 8058bbd253134537b97bbd3e4262474839262fbe Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 5 Jan 2026 11:24:34 +0800 Subject: [PATCH 161/274] vault backup: 2026-01-05 11:24:33 --- 编写小组/试卷/1231线性代数考试卷（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/1231线性代数考试卷（解析版）.md b/编写小组/试卷/1231线性代数考试卷（解析版）.md index e1d2cc6..ea6f19f 100644 --- a/编写小组/试卷/1231线性代数考试卷（解析版）.md +++ b/编写小组/试卷/1231线性代数考试卷（解析版）.md @@ -399,7 +399,7 @@ $$ --- 解析： -(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. +(1)证明：$[A\ \ \alpha] \rightarrow \begin{bmatrix}1 & 0 & 0 & 1 & 1\\0 & 1 & 0 & 2 & 1\\0 & 0 & 1 & 1 & 0\end{bmatrix}$,于是$Ax=\alpha$的通解为$$x=k\begin{bmatrix}-1\\-2\\-1\\1\end{bmatrix}+\begin{bmatrix}1\\1\\0\\0\end{bmatrix},$$把方程$Bx=\beta$还原成方程组得$$\begin{cases}x_1&+0x_2&+x_3&+2x_4&=1\\x_1&-x_2&+ax_3&+(a-1)x_4&=1\\2x_1&-3x_2&+2x_3&-2x_4&=-1\end{cases}$$把$Ax=\alpha$的解带入上方程组，显然符合，故方程组$Ax=\alpha$的解均为方程组$Bx=\beta$的解. (2)方程组$Bx=\beta$与方程组$Ax=\alpha$不同解，而由上一题，方程组$Ax=\alpha$的解是$Bx=\beta$的解的真子集，于是$\dim N(A)<\dim N(B),r(A)=3>r(B),r(B)\le2$.对$B$进行初等行变换得$$B\rightarrow\begin{bmatrix}1&0&1&2\\0&1&0&2\\0&0&a-1&a-1\end{bmatrix},$$于是$a=1$. -- 2.34.1 From 3e6819c229c55ac5d83c79e21b7125cdba2ba1e3 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 7 Jan 2026 12:01:21 +0800 Subject: [PATCH 162/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E7=9A=84=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/拐点是一个点.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) create mode 100644 素材/拐点是一个点.md diff --git a/素材/拐点是一个点.md b/素材/拐点是一个点.md new file mode 100644 index 0000000..e69de29 -- 2.34.1 From f98e7550d7a09679f3eb3d4dab6bc52fd0b3fe04 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 7 Jan 2026 16:58:39 +0800 Subject: [PATCH 163/274] vault backup: 2026-01-07 16:58:39 --- .../1.10高数限时练.md | 4 + .../2020高数期末考试卷.md | 147 ++++++++++++++++++ 2 files changed, 151 insertions(+) create mode 100644 编写小组/试卷/高数期末复习模拟/1.10高数限时练.md create mode 100644 编写小组/试卷/高数期末真题/2020高数期末考试卷.md diff --git a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md new file mode 100644 index 0000000..41ba6e5 --- /dev/null +++ b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md @@ -0,0 +1,4 @@ +--- +tags: + - 高数复习模拟 +--- diff --git a/编写小组/试卷/高数期末真题/2020高数期末考试卷.md b/编写小组/试卷/高数期末真题/2020高数期末考试卷.md new file mode 100644 index 0000000..869962d --- /dev/null +++ b/编写小组/试卷/高数期末真题/2020高数期末考试卷.md @@ -0,0 +1,147 @@ +--- +tags: + - 官方试卷 +--- +# 2020—2021 学年秋季学期 +## 《高等数学》（I）考试试卷（A）卷 + +**考试形式：闭卷** +**考试时间：150 分钟** +**满分：100 分** + +| 题号 | 一 | 二 | 三 | 总分 | +|------|----|----|----|------| +| 得分 | | | | | +| 评阅人 | | | | | + +**注意：** +1. 所有答题都须写在答题卡对应位置，写在其它纸上一律无效。 +2. 答题卡密封线外不得有姓名及相关标记。 + +--- + +### 一、单选题（共5小题，每小题2分，共10分） + +1. 函数$f(x) = \frac{|x+1|}{x(x-1)(x+1)} \arctan x + \sin \frac{1}{x-2}$的可去间断点为（ ）。 + (A)$x = 0$ + (B)$x = 1$ + (C)$x = 2$ + (D)$x = -1$ + +2. 已知函数$f(x)$可导，$y = \int_0^x f(t^2) dt$，则$y''(x)$等于（ ）。 + (A)$f(x^2)$ + (B)$2xf'(x^2)$ + (C)$2xf(x^2)$ + (D)$2f(x^2) + 4x^2 f'(x^2)$ + +3. “级数$\sum_{n=1}^\infty a_n$收敛”是“级数$a_1 - a_1 + a_2 - a_2 + \cdots + a_n - a_n + \cdots$收敛”的（ ）。 + (A) 充分但非必要条件 + (B) 必要但非充分条件 + (C) 充要条件 + (D) 既非充分也非必要条件 + +4. 已知函数$f(x), g(x)$在$(-\infty, +\infty)$内可导，且$f'(x) > 0, g'(x) < 0$，则（ ）。 + (A)$\int_0^1 f(x) dx > \int_1^2 f(x) dx$ + (B)$\int_0^1 [f(x)] dx > \int_1^2 [f(x)] dx$ + (C)$\int_0^1 f(x) g(x) dx > \int_1^2 f(x) g(x) dx$ + (D)$\int_0^1 [g(x)] dx > \int_1^2 [g(x)] dx$ + +5. 已知函数$f(x) = x^2 \sin \frac{1}{x}$，则下列四个判断： + ①$f(x)$在$(0, +\infty)$上有界 + ②$f(x)$是$x \to +\infty$过程的无穷大量 + ③$f'(x)$在$(0, +\infty)$上有界 + ④$f'(x)$是$x \to +\infty$过程的无穷大量 + 正确的是（ ）。 + (A) ①③ + (B) ①④ + (C) ②③ + (D) ②④ + +--- + +### 二、填空题（共5小题，每小题2分，共10分） + +6. 设$y = y(x)$是由方程$xy + e^{2y} = \sin 3x + 1$确定的隐函数，则$dy|_{x=0} = \underline{\qquad}$。 + +7. 极限 + $$ + \lim_{n \to \infty} \left( \frac{1}{\sqrt{3n^4 + 1}} + \frac{2}{\sqrt{3n^4 + 2}} + \cdots + \frac{n}{\sqrt{3n^4 + n}} \right) + $$ + 的值为 $\underline{\qquad}$。 + +8. 定积分 + $$ + \int \frac{\pi}{2} \left( |\sin x| + x \right) \cos^2 x \, dx + $$ + 的值为 $\underline{\qquad}$。 + +9. 不定积分 + $$ + \int x \tan^2 x \, dx = \underline{\qquad}。 + $$ + +10. 已知 + $$ + \int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C, + $$ + 则曲线$y = f(x) (x > 0)$的拐点为 \underline{\qquad}。 + +--- + +### 三、解答题（共11小题，共80分） + +11. （6分）求极限 + $$ + \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln \left( x^2 + 1 \right)}{x^3 \arcsin x}. + $$ + +12. （6分）设函数 + $$ + f(x) = + \begin{cases} + |x|^x, & x \neq 0, \\ + a, & x = 0, + \end{cases} + $$ + 试求常数$a$的值，使得$f(x)$在$x = 0$处连续，并讨论此时$f(x)$在$x = 0$处的可导性。 + +13. （6分）将函数$f(x) = x^2 e^x + x^6$展开成六阶带佩亚诺余项的麦克劳林公式，并求$f^{(6)} (0)$的值。 + +14. （6分）已知$f(x)$是周期为 2 的可导函数，且曲线$y = f(x)$与曲线$y = \int_0^x e^{-t^2} dt$在点$(0,0)$处相切。求曲线$y = f(x)$在$x = 4$处的切线方程。 + +15. （6分）已知函数$f(x)$在$[0,1]$上可导，$f(0)=0$。证明：至少存在一点$\xi \in (0,1)$，使得 + $$ + f'(1-\xi)=\frac{2}{\xi}f(1-\xi). + $$ + +16. （6分）在某直线公路上有 A, B, C, D 四个加油站，依次相距 20 km。现计划在该公路上建一个加油总站 M，并配备一台供油车给各加油站供油。已知 A, B, C, D 四个加油站每天所需的油量依次为 2 车、3 车、5 车及 4 车。问：加油总站 M 建在何处可使供油车每天行驶的总路程最少？并说明理由。 + + （图略：A、B、C、D 依次排列，间距均为 20 km） + +17. （8分）已知函数$f(x) = \frac{\ln(1+x)}{1+x}$。 + (1) 求函数$f(x)$的单调区间与极值；（4分） + (2) 求曲线$y = f(x)$的渐近线。（4分） + +18. （8分）求极坐标曲线$C: \rho = 1 + \cos \theta$在$\theta = \frac{\pi}{2}$对应点处的曲率与曲率半径。 + +19. （8分）已知函数$f(x)$在$(-\infty, +\infty)$内连续，且 + $$ + \int_{0}^{x} tf(x-t) dt = x^3 - \int_{0}^{x} f(t) dt. + $$ + (1) 验证：$f'(x) + f(x) = 6x$且$f(0) = 0$；（5分） + (2) 计算定积分$\int_{0}^{1} e^{x} f(x) dx$。（3分） + +20. （10分）设函数$f(x)$在$(-\infty, +\infty)$内可导，$f'(0) = 2$，且对任意$x, y$，恒有 + $$ + f(x+y) = f(x) + f(y) + 2xy. + $$ + (1) 求函数$f(x)$的表达式；（6分） + (2) 求曲线$y = f(x)$与$x$轴所围成的平面图形的面积。（4分） + +21. （10分） + (1) 证明：对任意的正整数$n$，方程 + $$ + x^n + n^2 x - 1 = 0 + $$ + 有唯一正实根（记为$x_n$）；（6分） + (2) 证明级数$\sum_{n=1}^{\infty} x_n$收敛，且其和不超过 2。（4分） \ No newline at end of file -- 2.34.1 From c422f056c05c79875fa5563fe6203f36dd5ceb0d Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 7 Jan 2026 17:18:43 +0800 Subject: [PATCH 164/274] =?UTF-8?q?1.10=E7=BA=BF=E4=BB=A3=E9=99=90?= =?UTF-8?q?=E6=97=B6=E7=BB=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../线代期末复习模拟/1.10线代限时练.md | 8 ++++++++ 1 file changed, 8 insertions(+) create mode 100644 编写小组/试卷/线代期末复习模拟/1.10线代限时练.md diff --git a/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md new file mode 100644 index 0000000..cb361f3 --- /dev/null +++ b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md @@ -0,0 +1,8 @@ +一、填空题 + +二、解答题 +7. 计算$n$阶行列式 $D_n=\begin{vmatrix}2+a_1^2 & a_1a_2 & \cdots&a_1a_n \\ a_2a_1& 2+a_2^2 & \cdots & a_2a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_na_1&a_na_2&\cdots& 2+a_n^2\end{vmatrix}$. +8. 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. +9. 设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： + (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . + (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解. \ No newline at end of file -- 2.34.1 From 071179dd43bf5e0dfe0035d517d1a3c5e30065a9 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 7 Jan 2026 17:33:44 +0800 Subject: [PATCH 165/274] vault backup: 2026-01-07 17:33:44 --- 编写小组/试卷/0103高数模拟试卷.md | 6 - .../1.10高数限时练.md | 131 +++++++++++++++ .../2017高数期末考试卷.md | 146 +++++++++++++++++ .../2018高数期末考试卷.md | 149 ++++++++++++++++++ .../2019高数期末考试卷.md | 144 +++++++++++++++++ 5 files changed, 570 insertions(+), 6 deletions(-) create mode 100644 编写小组/试卷/高数期末真题/2017高数期末考试卷.md create mode 100644 编写小组/试卷/高数期末真题/2018高数期末考试卷.md create mode 100644 编写小组/试卷/高数期末真题/2019高数期末考试卷.md diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md index bd29779..33fabc5 100644 --- a/编写小组/试卷/0103高数模拟试卷.md +++ b/编写小组/试卷/0103高数模拟试卷.md @@ -45,12 +45,6 @@ D$\frac{1}{6}$ 3.若级数$\sum\limits_{n=1}^{\infty}\frac{n^p}{(-1)^n}\sin(\frac{1}{\sqrt{n}})$绝对收敛，则常数$p$的取值范围是$\underline{\quad\quad\quad}.$ - - - - - -======= 4.求不定积分$\int{x^3\sqrt{4-x^2}\mathrm{d}x}=\underline{\quad\quad\quad}.$ diff --git a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md index 41ba6e5..2ae0acd 100644 --- a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md +++ b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md @@ -2,3 +2,134 @@ tags: - 高数复习模拟 --- +# 1.10高数限时练 + +### 一、单选题（共5小题，每小题2分，共10分） + +1. 函数$f(x) = \frac{|x+1|}{x(x-1)(x+1)} \arctan x + \sin \frac{1}{x-2}$的可去间断点为（ ）。 + (A)$x = 0$ + (B)$x = 1$ + (C)$x = 2$ + (D)$x = -1$ + +2. “级数$\sum_{n=1}^\infty a_n$收敛”是“级数$a_1 - a_1 + a_2 - a_2 + \cdots + a_n - a_n + \cdots$收敛”的（ ）。 + (A) 充分但非必要条件 + (B) 必要但非充分条件 + (C) 充要条件 + (D) 既非充分也非必要条件 + +3. 已知函数$f(x) = x^2 \sin \frac{1}{x}$，则下列四个判断： + ①$f(x)$在$(0, +\infty)$上有界 + ②$f(x)$是$x \to +\infty$过程的无穷大量 + ③$f'(x)$在$(0, +\infty)$上有界 + ④$f'(x)$是$x \to +\infty$过程的无穷大量 + 正确的是（ ）。 + (A) ①③ + (B) ①④ + (C) ②③ + (D) ②④ + +--- + +### 二、填空题（共5小题，每小题2分，共10分） + +4. 设$y = y(x)$是由方程$xy + e^{2y} = \sin 3x + 1$确定的隐函数，则$dy|_{x=0} = \underline{\qquad}$。 + +5. 极限 + $$ + \lim_{n \to \infty} \left( \frac{1}{\sqrt{3n^4 + 1}} + \frac{2}{\sqrt{3n^4 + 2}} + \cdots + \frac{n}{\sqrt{3n^4 + n}} \right) + $$ + 的值为 $\underline{\qquad}$。 + +6. 已知 +$$\int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C$$ + 则曲线$y = f(x) (x > 0)$的拐点为 $\underline{\qquad}$。 + +--- + +### 三、解答题（共11小题，共80分） + +7. （6分）求极限 + $$ + \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln \left( x^2 + 1 \right)}{x^3 \arcsin x}. + $$ +```text + + + + + + + + + + + + + + + +``` +8. （6分）设函数 + $$ + f(x) = + \begin{cases} + |x|^x, & x \neq 0, \\ + a, & x = 0, + \end{cases} + $$ + 试求常数$a$的值，使得$f(x)$在$x = 0$处连续，并讨论此时$f(x)$在$x = 0$处的可导性。 +```text + + + + + + + + + + + + + + + +``` +9. （6分）将函数$f(x) = x^2 e^x + x^6$展开成六阶带佩亚诺余项的麦克劳林公式，并求$f^{(6)} (0)$的值。 +```text + + + + + + + + + + + + + + + +``` +10. （6分）已知函数$f(x)$在$[0,1]$上可导，$f(0)=0$。证明：至少存在一点$\xi \in (0,1)$，使得 + $$f'(1-\xi)=\frac{2}{\xi}f(1-\xi).$$ +```text + + + + + + + + + + + + + + + +``` \ No newline at end of file diff --git a/编写小组/试卷/高数期末真题/2017高数期末考试卷.md b/编写小组/试卷/高数期末真题/2017高数期末考试卷.md new file mode 100644 index 0000000..471e755 --- /dev/null +++ b/编写小组/试卷/高数期末真题/2017高数期末考试卷.md @@ -0,0 +1,146 @@ +--- +tags: + - 官方试卷 +--- +# 国防科技大学 2017—2018 学年秋季学期 +## 《高等数学》考试试卷（A）卷 + +**(2018 年 1 月 26 日)** + +**考试形式：闭卷** +**考试时间：150 分钟** +**满分：100 分** + +| 题号 | 一 | 二 | 三 | 四 | 五 | 六 | 七 | 八 | 九 | 十 | 十一 | 十二 | 总分 | 核分 | +|------|----|----|----|----|----|----|----|----|----|----|-----|-----|------|------| +| 满分 | 15 | 15 | 6 | 6 | 6 | 6 | 6 | 8 | 8 | 8 | 8 | 8 | 100 | | +| 得分 | | | | | | | | | | | | | | | +| 评阅人 | | | | | | | | | | | | | | | + +**注意：** +1. 所有答题都须写在此试卷纸密封线右边，写在其它纸上一律无效。 +2. 密封线外不得有姓名及相关标记。 +3. 当题目留空不够时，可写在试卷反面，但密封线内请勿答题。 + +--- + +### 一、填空题（共 5 小题，每小题 3 分，共 15 分） + +1. 函数$y = \ln(2 - x^2)$在$x = 1$处的微分为 ______。 + +2. 曲线$y = 1 + xe^x$在点 (0,1) 处的曲率为 ______。 + +3. 函数$y = \frac{x^3 + 4}{x^2}$的单调递减区间为 ______。 + +4. 已知$\int f(x) dx = \arctan x + C$，则$f'(x) =$______。 + +5. 已知级数$\sum_{n=1}^{\infty} \frac{a^{2n}}{a^{2n} + 1}$收敛，则常数$a$的最大取值范围为 ______。 + +--- + +### 二、选择题（共 5 小题，每小题 3 分，共 15 分） + +1. 点$x = 0$为函数$f(x) = \frac{2e^x + 3}{3e^x + 2}$的（ ）。 + (A) 可去间断点 + (B) 跳跃间断点 + (C) 无穷间断点 + (D) 振荡间断点 + +2. 设函数$f(x), g(x)$在$[-a, a]$上均具有连续导数，且$f(x)$为奇函数，$g(x)$为偶函数，则定积分 + $$ + \int_{-a}^{a} [f'(x) + g'(x)] dx = \text{（ ）}。 + $$ + (A)$f(a) + g(a)$ + (B)$f(a) - g(a)$ + (C)$2g(a)$ + (D)$2f(a)$ + +3. 设$f(x)$是定义在$(-\infty, +\infty)$内的连续的奇函数，下列表格中给出了它的二阶导数$f''(x)$在$(0, +\infty)$内的符号信息，则曲线$y = f(x) (-\infty < x < +\infty)$的拐点个数为（ ）。 + + |$x$| (0,1) | 1 | (1,2) | 2 | (2,+\infty) | + |--------|--------|---|--------|---|-------------| + |$f''(x)$| + | 0 | - | 不存在 | - | + + (A) 1 + (B) 2 + (C) 3 + (D) 4 + +4. 下列级数中条件收敛的是（ ）。 + (A)$\sum_{n=1}^{\infty} (-1)^n \frac{n^2}{2^n + 1}$ + (B)$\sum_{n=1}^{\infty} (-1)^n \frac{2^n}{n! + 1}$ + (C)$\sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 1}$ + (D)$\sum_{n=1}^{\infty} (-1)^n \frac{2^n}{n^2 + 1}$ + +5. 极限 + $$ + \lim_{n \to \infty} \left( \frac{1}{n^2 + 1^2} + \frac{2}{n^2 + 2^2} + \cdots + \frac{n}{n^2 + n^2} \right) + $$ + 的值为（ ）。 + (A)$\frac{\ln 2}{2}$ + (B)$\frac{\pi}{4}$ + (C)$\frac{1}{2}$ + (D) 0 + +--- + +### 三、（6 分）求曲线$y = \frac{\ln(1-x)}{x}$的所有渐近线方程。 + +--- + +### 四、（6 分）已知函数$y = y(x)$由方程$x \cos y + e^y = 1$所确定，求曲线$y = y(x)$在点 (0,0) 处的切线方程。 + +--- + +### 五、（6 分）计算极限$\lim_{x \to +\infty} \left( 1 + 2^x + 3^x \right)^{\frac{1}{x}}$。 + +--- + +### 六、（6 分）已知 +$$ +\begin{cases} +x = \int_{0}^{t} \frac{\sin u}{u} du, \\ +y = \int_{0}^{t} \sin u^2 du, +\end{cases} +$$ +求$\frac{dy}{dx}$和$\frac{d^2 y}{dx^2}$。 + +--- + +### 七、（6 分）计算反常积分 +$$ +\int_{0}^{\infty} \frac{x^2}{\left(1 + x^2\right)^{\frac{5}{2}}} dx。 +$$ + +--- + +### 八、（8 分） +（I）写出函数$f(x) = \ln \sqrt{\frac{1 - x^2}{1 + x^2}}$带佩亚诺余项的 6 阶麦克劳林公式，并求$f''(0)$和$f'''(0)$的值； + +（II）已知当$x \to 0$时，$f(x)$与$g(x) = ae^{-x^2} + b\cos x$是等价无穷小，求常数$a, b$的值。 + +--- + +### 九、（8 分）求曲线$y = \sqrt{x}\sin x \ (0 \leq x \leq \pi)$与$x$轴所围的平面图形绕$x$轴旋转一周所得的旋转体体积。 + +--- + +### 十、（8 分）设$x_1 = 1, x_n = 1 + \frac{x_{n-1}}{1 + x_{n-1}}, n = 2, 3, \cdots$，试用单调有界原理证明极限$\lim_{n \to \infty} x_n$存在，并求出其值。 + +--- + +### 十一、（8 分）要造一个壁和底厚为$a(m)$、容积为$V(m^3)$、上端开口的圆柱形容器，问：当容器端口内半径尺寸$r(m)$为何值时，所用材料最省？ + +--- + +### 十二、（8 分）已知函数$f(x)$在$[0,+\infty)$内二阶可导，且 +$$ +\lim_{x \to 0^+} \frac{f(x)}{x} = \lambda < 0, \quad \lim_{x \to +\infty} f(x) = +\infty。 +$$ +证明： +（I）方程$f(x) = 0$在$(0,+\infty)$内至少有一个实根； +（II）方程$f(x)f''(x)+[f'(x)]^2=0$在$(0,+\infty)$内至少有两个实根。 + +--- + +**（试卷结束）** \ No newline at end of file diff --git a/编写小组/试卷/高数期末真题/2018高数期末考试卷.md b/编写小组/试卷/高数期末真题/2018高数期末考试卷.md new file mode 100644 index 0000000..a31ce35 --- /dev/null +++ b/编写小组/试卷/高数期末真题/2018高数期末考试卷.md @@ -0,0 +1,149 @@ +--- +tags: + - 官方试卷 +--- +# 国防科技大学 2018—2019 学年秋季学期 +## 《高等数学》(I) 考试试卷 (A) 卷 + +**(2019 年 1 月 25 日 8:00—10:30)** + +**考试形式：闭卷** +**考试时间：150 分钟** +**满分：100 分** + +**注意：** +1. 所有答题都须写在答题卡指定区域内，写在其它纸上一律无效。 +2. 答题区域内不得有姓名及相关标记。 + +--- + +### 一、单选题（共 5 小题，每小题 3 分，共 15 分） + +1. 点$x = 0$是函数$f(x) = \frac{e^x - 1}{|x|}$的（ ）。 + (A) 可去间断点 + (B) 跳跃间断点 + (C) 无穷间断点 + (D) 振荡间断点 + +2. 下列级数发散的是（ ）。 + (A)$\sum_{n=1}^{\infty} \frac{4^n + 1}{n^4 + 1}$ + (B)$\sum_{n=1}^{\infty} \frac{n^4 + 1}{4^n + 1}$ + (C)$\sum_{n=1}^{\infty} (-1)^n \frac{1}{\sqrt{n+1}}$ + (D)$\sum_{n=1}^{\infty} (-1)^n \frac{\sqrt{n}}{n+1}$ + +3. 当$x \to 0$时，与$x - \sin x$同阶的无穷小是（ ）。 + (A)$x + \tan x$ + (B)$x \tan x$ + (C)$x^2 + \tan x$ + (D)$x^2 \tan x$ + +4. 已知函数$f(x)$在$x = 1$的某邻域内三阶可导，且$\lim_{x \to 1} \frac{f(x) - 2}{(x-1)^2 \ln x} = \frac{1}{3}$，则（ ）。 + (A)$x = 1$为函数$f(x)$的极大值点 + (B) 点 (1, 2) 为曲线$y = f(x)$的拐点 + (C)$x = 1$为函数$f(x)$的极小值点 + (D) 点 (1, 0) 为曲线$y = f(x)$的拐点 + +5. 设$f(x)$与$g(x)$互为反函数，已知$f(x)$二阶可导，则依据下表知$g'(2)$与$g''(2)$的值分别为（ ）。 + + |$x$| 1 | 2 | + |--------|---|---| + |$f(x)$| 2 | 1 | + |$f'(x)$| -3 | -1 | + |$f''(x)$| 3 | 2 | + + (A) -1, 2 + (B) -1, -2 + (C)$\frac{1}{3}, \frac{1}{9}$ + (D)$\frac{1}{3}, -\frac{1}{3}$ + +--- + +### 二、填空题（共 5 小题，每小题 3 分，共 15 分） + +6. 函数$y = \frac{1 - x + x^2}{1 + x + x^2}$在$x = 0$处的微分是 ______。 + +7. 极限 + $$ + \lim_{n \to \infty} \frac{1}{n} \left( \tan \frac{\pi}{4n} + \tan \frac{2\pi}{4n} + \cdots + \tan \frac{n\pi}{4n} \right) + $$ + 的值为 ______。 + +8. 定积分 + $$ + \int_{-1}^{1} \frac{x(\cos x + x)}{1 + x^2} dx + $$ + 的值为 ______。 + +9. 已知$\int f(x^2) dx = x \ln x + C$，则$f'(2) =$______。 + +10. 极坐标曲线$\rho = \theta$在点$(\rho, \theta) = (\pi, \pi)$处的切线的直角坐标方程为 ______。 + +--- + +### 三、解答题（共 8 小题，共 54 分） + +11. （6分）计算极限 + $$ + \lim_{x \to 0} \frac{x \int_{0}^{x} \sqrt{1 + t^4} dt}{x - \ln (1 + x)}。 + $$ + +12. （6分）试确定常数$a, b$的值，使函数 + $$ + f(x) = + \begin{cases} + ae^x + b, & x > 0, \\ + \cos 3x, & x \leq 0 + \end{cases} + $$ + 在$x = 0$处可导。 + +13. （6分）计算不定积分 + $$ + \int \frac{2 - \sqrt{2x + 1}}{2 + \sqrt{2x + 1}} dx。 + $$ + +14. （6分）计算定积分 + $$ + \int_{0}^{\pi/2} x \sin x \cos^3 x dx。 + $$ + +15. （6分）试写出函数$f(x) = x \ln (2 + x) + \cos x$的 4 阶带佩亚诺余项的麦克劳林公式，并求函数图形在点$(0, 1)$处的曲率。 + +16. （8分）已知函数$f(x) = (2x + 3) e^{\frac{2}{x}}$。 + （1）求函数$f(x)$的单调区间与极值。 + （2）求曲线$y = f(x)$的渐近线方程。 + +17. （8分） + （1）证明：存在$\theta \in (0, 1)$使得 + $$ + \ln (1 + x) - \ln \left( 1 + \frac{x}{2} \right) = \frac{x}{2 + (1 + \theta)x}, \quad x > 0； + $$ + （2）证明不等式 + $$ + \left( 1 + \frac{1}{n} \right)^{n+1} < e \left( 1 + \frac{1}{2n} \right)，\quad n \text{为正整数}。 + $$ + +18. （8分）已知$y = y(x)$是由方程$x \cos y + \sin x + e^y = 1$所确定的隐函数。 + （1）求$\frac{dy}{dx}$； + （2）计算极限 + $$ + \lim_{x \to 0} \left( \frac{1 - y(x)}{1 + y(x)} \right)^{\frac{1}{x}} 的值。 + $$ + +--- + +### 四、应用题（8 分） + +19. 如图所示，某人在离水面 3m 高的岸上，用缆绳拉船靠岸。已知初始时刻船距离岸壁 20m，假设船的运动方向始终与岸壁垂直。问：当船离岸壁 4m，收缆速度为 2m/s 时，船靠岸的速度大小为多少？此时，缆绳倾斜角$\theta$关于时间的变化率是多少？ + +--- + +### 五、证明题（8 分） + +20. 设函数$f(x)$在$(-\infty, +\infty)$内可导，且$|f'(x)| \leq \lambda$，其中$\lambda (0 < \lambda < 1)$为常数。给定实数$a_1$，令$a_{n+1} = f(a_n) \quad (n=1,2,\cdots)$，证明： + （1）级数$\sum_{n=1}^{\infty} (a_{n+1} - a_n)$是绝对收敛的； + （2）数列$\{a_n\}$收敛，且其极限值为方程$f(x) = x$的唯一实数根。 + +--- + +**（试卷结束）** \ No newline at end of file diff --git a/编写小组/试卷/高数期末真题/2019高数期末考试卷.md b/编写小组/试卷/高数期末真题/2019高数期末考试卷.md new file mode 100644 index 0000000..2db1129 --- /dev/null +++ b/编写小组/试卷/高数期末真题/2019高数期末考试卷.md @@ -0,0 +1,144 @@ +--- +tags: + - 官方试卷 +--- +# 2019—2020学年秋季学期 +## 《高等数学》（I）考试试卷（A）卷 + +**考试形式：闭卷** +**考试时间：150分钟** +**满分：100分** + +| 题号 | 一 | 二 | 三 | 总分 | +|------|----|----|----|------| +| 得分 | | | | | +| 评阅人 | | | | | + +**注意：** +1. 所有答题都须写在答题卡对应位置，写在其它纸上一律无效。 +2. 答题卡密封线外不得有姓名及相关标记。 +3. 所有教学班做1-19题，空医、陆医三个教学班做20*题和21*题，其他教学班做20题和21题。 + +--- + +### 一、选择题（共5小题，每小题2分，共10分） + +1. 函数 + $$ + f(x) = \begin{cases} + \sqrt{x} \sin \frac{1}{x}, & x > 0, \\ + 0, & x \leq 0 + \end{cases} + $$ + 在点$x = 0$处（ ）。 + (A) 不连续 + (B) 连续但不可导 + (C) 可导且$f'(0) = 0$ + (D) 可导且$f'(0) \neq 0$ + +2. 数列极限$\lim_{n \to \infty} (e^{-n} + \pi^{-n})^{\frac{1}{n}}$的值为（ ）。 + (A)$e$ + (B)$\pi$ + (C)$\frac{1}{e}$ + (D)$\frac{1}{\pi}$ + +3. 曲线$y = \frac{x^3 - x^2}{2 + x^2}$的渐近线为（ ）。 + (A)$y = x - 1$ + (B)$y = x + 1$ + (C)$y = x$ + (D)$y = \frac{1}{2}x$ + +4. 图1中曲线分别为函数$y = f(x)$及其一、二阶导函数的图形，则编号为①、②、③的曲线依次对应的函数是（ ）。 + (A)$y = f(x)$、$y = f'(x)$、$y = f''(x)$ + (B)$y = f'(x)$、$y = f(x)$、$y = f''(x)$ + (C)$y = f'(x)$、$y = f''(x)$、$y = f(x)$ + (D)$y = f(x)$、$y = f''(x)$、$y = f'(x)$ + +5. 设$y = f(x)$为区间$[0,1]$上单调增加的连续函数，且$f(0) = 0$，$f(1) = 2$，$x = g(y)$为$y = f(x)$的反函数。若$\int_{0}^{1} f(x) dx = \frac{1}{3}$，则$\int_{0}^{2} g(y) dy$的值为（ ）。 + (A)$\frac{1}{3}$ + (B)$\frac{2}{3}$ + (C)$\frac{4}{3}$ + (D)$\frac{5}{3}$ + +--- + +### 二、填空题（共5小题，每小题2分，共10分） + +6. 已知函数$y = e^{\sin(2x)}$，则在$x = 0$处的微分$dy|_{x=0} = \underline{\qquad}$。 + +7. 函数$f(x) = xe^{-x^2}$在$(-\infty,+\infty)$上的最大值为 \underline{\qquad}。 + +8. 曲线$C: x = \frac{1}{2} \cos t, y = \sin t, t \in [0,2\pi]$在点$(0,-1)$处的曲率为 \underline{\qquad}。 + +9. 已知函数$f(x) = x^2 \sin x$，则$f^{(7)}(0) = \underline{\qquad}$。 + +10. 不定积分$\int \frac{1}{x(1+2\ln x)} dx = \underline{\qquad}$。 + +--- + +### 三、解答题（共11小题，共80分） + +11. 计算数列极限 + $$ + \lim_{n \to \infty} \left( \frac{1}{2n+1} + \frac{1}{2n+2} + \cdots + \frac{1}{2n+n} \right)。 + $$ + （6分） + +12. 设$y(x)$是由曲线方程$\sin x + y + e^x = 2$确定的隐函数，试计算$\frac{dy}{dx} \bigg|_{x=0}$的值，并求该曲线在点$P(0,1)$处的切线方程。（6分） + +13. 计算不定积分 + $$ + \int \frac{x}{1+\sqrt{1-x^2}} dx。 + $$ + （6分） + +14. 设曲线$f(x) = x^3 + ax^2 + 18x$（$a$为大于零的常数）的拐点正好位于$x$轴上，试求$a$的值及曲线$y = f(x)$的拐点坐标。（6分） + +15. 计算极限 + $$ + \lim_{x \to +\infty} \left[ x + x^2 \ln \left( 1 - \frac{1}{x} \right) \right]。 + $$ + （6分） + +16. 设函数$f(x)$在$[0,1]$上连续，且$f(x) = \arcsin x + x \int_0^1 f(x) \, dx$，试求$f(x)$的表达式。（6分） + +17. 无人机在海面执勤时，发现可疑目标后将实时拍照传回指挥部。已知无人机按直线飞行，高度为150米，飞行过程中摄像头始终对准可疑目标（如图2所示）。当无人机飞临目标正上方时，速度为3米/秒。问：此时无人机摄像头转动角速度为多少？（8分） + +18. 证明不等式 + $$ + 1 + x \int_0^x \frac{dt}{\sqrt{1 + t^2}} > \sqrt{1 + x^2} \quad (x > 0)。 + $$ + （8分） + +19. 记$I_n = \int_{0}^{\frac{\pi}{4}} \sec^n x \, dx, (n=0,1,2,\cdots)$。 + （1）证明：当$n \geq 2$时， + $$ + I_n = \frac{2^{\frac{n-2}{2}}}{n-1} + \frac{n-2}{n-1} I_{n-2}； + $$ + （2）计算$I_3$的值。（8分） + +--- + +**（以下为选做题，普通班做20、21题；空医、陆医班做20*、21*题）** + +20. （10分）已知当$x \to 0$时，函数$f(x) = \sqrt{a + bx^2} - \cos x$与$x^2$是等价无穷小。 + （1）求参数$a, b$的值；（5分） + （2）计算极限$\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（5分） + +20*. （10分）已知当$x \to 0$时，函数$f(x) = a + bx^2 - \cos x$与$x^2$是等价无穷小。 + （1）求参数$a, b$的值；（5分） + （2）计算极限$\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（5分） + +21. （10分）设函数$f(x)$在$[0,1]$上可导，$f(1) = 1$且$2 \int_{0}^{1} x^2 f(x) dx = 1$。证明存在$\xi \in (0,1)$，使得 + $$ + f(\xi) = -\frac{\xi}{2} f'(\xi)。 + $$ + +21*. （10分）设函数$f(x)$在$[0,1]$上可导，$\int_{0}^{\frac{1}{2}} f(x) dx = 0$。证明存在$\xi \in (0,1)$，使得 + $$ + f(\xi) = (1 - \xi) f'(\xi)。 + $$ + +--- + +**（试卷结束）** \ No newline at end of file -- 2.34.1 From 56738ef548a27358b662a3458b32ec710c829537 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Wed, 7 Jan 2026 17:35:39 +0800 Subject: [PATCH 166/274] =?UTF-8?q?1.10=E7=BA=BF=E4=BB=A3=E9=99=90?= =?UTF-8?q?=E6=97=B6=E7=BB=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../试卷/线代期末复习模拟/1.10线代限时练.md | 7 ++++++- 1 file changed, 6 insertions(+), 1 deletion(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md index cb361f3..a837904 100644 --- a/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md +++ b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md @@ -1,5 +1,10 @@ 一、填空题 - +1. 设 $A=\begin{bmatrix}1&2&0\\0&2&0\\-2&-1&-1\end{bmatrix}$，则$A^{100}=$\_\_\_\_\_\_\_\_\_\_\_ +2. 线性空间 $V = \{A\in\mathbb{R}^{n\times n}|A = -A^\mathrm{T}\}$ 的维数是\_\_\_\_\_\_\_\_\_\_\_ +3. 设向量组 $\alpha_1,\alpha_2,\cdots,\alpha_s$ 线性无关，$\beta_1=\alpha_1+\alpha_2,\beta_2=\alpha_2+\alpha_3,\cdots,\beta_{s-1}=\alpha_{s-1}+\alpha_s$，则向量组 $\beta_1,\beta_2,\cdots,\beta_s$ 线性无关的充要条件是\_\_\_\_\_\_\_\_\_\_\_ +4. 已知 $A,B$ 均为 $n$ 阶正交矩阵，且 $|A|=-|B|$，则 $|A+B|$ 的值为\_\_\_\_\_\_\_\_\_\_\_ +5. 已知 $4$ 阶矩阵 $A$ 与 $B$ 相似，$A$ 的全部特征值为 $1,2,3,4$，则行列式 $|B^{-1}-E|$ 为\_\_\_\_\_\_\_\_\_\_\_ +6. 设 $\alpha,\beta,\gamma$ 为 $x^3+px+q=0$ 的三个根，则行列式 $\begin{vmatrix}\alpha&\beta&\gamma\\\gamma&\alpha&\beta\\\beta&\gamma&\alpha\end{vmatrix}$ 的值为\_\_\_\_\_\_\_\_\_\_\_ 二、解答题 7. 计算$n$阶行列式 $D_n=\begin{vmatrix}2+a_1^2 & a_1a_2 & \cdots&a_1a_n \\ a_2a_1& 2+a_2^2 & \cdots & a_2a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_na_1&a_na_2&\cdots& 2+a_n^2\end{vmatrix}$. 8. 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. -- 2.34.1 From a66ce2d2fa7b91611a6bf81a720c1a5d05a0128e Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 7 Jan 2026 17:37:23 +0800 Subject: [PATCH 167/274] vault backup: 2026-01-07 17:37:23 --- .../1.10高数限时练.md | 14 +++++++------- 1 file changed, 7 insertions(+), 7 deletions(-) diff --git a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md index 2ae0acd..0c34b3f 100644 --- a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md +++ b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md @@ -4,7 +4,7 @@ tags: --- # 1.10高数限时练 -### 一、单选题（共5小题，每小题2分，共10分） +### 一、单选题（共3小题，每小题6分，共18分） 1. 函数$f(x) = \frac{|x+1|}{x(x-1)(x+1)} \arctan x + \sin \frac{1}{x-2}$的可去间断点为（ ）。 (A)$x = 0$ @@ -31,7 +31,7 @@ tags: --- -### 二、填空题（共5小题，每小题2分，共10分） +### 二、填空题（共3小题，每小题6分，共18分） 4. 设$y = y(x)$是由方程$xy + e^{2y} = \sin 3x + 1$确定的隐函数，则$dy|_{x=0} = \underline{\qquad}$。 @@ -47,9 +47,9 @@ $$\int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C$$ --- -### 三、解答题（共11小题，共80分） +### 三、解答题（共4小题，共64分） -7. （6分）求极限 +7. （16分）求极限 $$ \lim_{x \to 0} \frac{e^{x^2} - 1 - \ln \left( x^2 + 1 \right)}{x^3 \arcsin x}. $$ @@ -70,7 +70,7 @@ $$\int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C$$ ``` -8. （6分）设函数 +8. （16分）设函数 $$ f(x) = \begin{cases} @@ -96,7 +96,7 @@ $$\int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C$$ ``` -9. （6分）将函数$f(x) = x^2 e^x + x^6$展开成六阶带佩亚诺余项的麦克劳林公式，并求$f^{(6)} (0)$的值。 +9. （16分）将函数$f(x) = x^2 e^x + x^6$展开成六阶带佩亚诺余项的麦克劳林公式，并求$f^{(6)} (0)$的值。 ```text @@ -114,7 +114,7 @@ $$\int x f(x) \, dx = \ln \left( 1 + x^2 \right) + C$$ ``` -10. （6分）已知函数$f(x)$在$[0,1]$上可导，$f(0)=0$。证明：至少存在一点$\xi \in (0,1)$，使得 +10. （16分）已知函数$f(x)$在$[0,1]$上可导，$f(0)=0$。证明：至少存在一点$\xi \in (0,1)$，使得 $$f'(1-\xi)=\frac{2}{\xi}f(1-\xi).$$ ```text -- 2.34.1 From 214503fbc649d7c49dd1bde7092fc0d6fd17bb6a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 7 Jan 2026 17:38:56 +0800 Subject: [PATCH 168/274] vault backup: 2026-01-07 17:38:56 --- .../试卷/高数期末复习模拟/1.10高数限时练.md | 2 -- 1 file changed, 2 deletions(-) diff --git a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md index 0c34b3f..536e7fe 100644 --- a/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md +++ b/编写小组/试卷/高数期末复习模拟/1.10高数限时练.md @@ -2,8 +2,6 @@ tags: - 高数复习模拟 --- -# 1.10高数限时练 - ### 一、单选题（共3小题，每小题6分，共18分） 1. 函数$f(x) = \frac{|x+1|}{x(x-1)(x+1)} \arctan x + \sin \frac{1}{x-2}$的可去间断点为（ ）。 -- 2.34.1 From 561729887834b4530ce9e35402e57325648ba7ed Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Thu, 8 Jan 2026 23:47:36 +0800 Subject: [PATCH 169/274] vault backup: 2026-01-08 23:47:36 --- 素材/拐点是一个点.md | 1 + 1 file changed, 1 insertion(+) diff --git a/素材/拐点是一个点.md b/素材/拐点是一个点.md index e69de29..65cf990 100644 --- a/素材/拐点是一个点.md +++ b/素材/拐点是一个点.md @@ -0,0 +1 @@ +拐点一定要是点 \ No newline at end of file -- 2.34.1 From 7d2d7d0662f7ef95da36525b885f03de2fb2c018 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 9 Jan 2026 00:03:23 +0800 Subject: [PATCH 170/274] =?UTF-8?q?=E7=A0=94=E8=AE=A8=E8=AE=B0=E5=BD=95202?= =?UTF-8?q?6.1.8=E6=99=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 研讨记录/2026.1.8.md | 23 +++++++++++++++++++++++ 1 file changed, 23 insertions(+) create mode 100644 研讨记录/2026.1.8.md diff --git a/研讨记录/2026.1.8.md b/研讨记录/2026.1.8.md new file mode 100644 index 0000000..7f8316a --- /dev/null +++ b/研讨记录/2026.1.8.md @@ -0,0 +1,23 @@ +平时进行雨课堂等上题目的交流 +打电话交流？no +之后自习会增多，我们可以争取一些自习课来，这个后面再说。体能。 +集中—>解答困惑，素材 +临时部署 +雨课堂和线代作业的整理，加班时间可能会增加 +**作业上找题目（平时小测）** 2~3道题目 +每两天进行一次研讨小会（一三五日，晚或体能）完成作业，分享思路——强迫比较好地完成作业。选择小测题目。复习。 +陈搜集资料（教员习题课），王嘉兴 +线代：真题优先，习题册（关键在于没有答案），期末考和习题册是有很大重叠的 +**提前确定会议主题**（如）周三：线代作业 +讨论主题：主要不是讨论答案是什么，而是使用的方法的适用条件，让一类题的做法形成一套’机械化‘的流程（过于理想化了） +**每周一套/两套试卷**，在什么量的情况下是可以接受而且必须要给自己逼一把的；灵活处理，不需要固定在哪天写哪些题。 +**确保作业全对。** 用py计算线性代数来检查。 +原则：**把大的目标落实到每一次研讨中** +**灵活处理，下次之前决定下次的议题** +确保出的题目得是自己做过的 +**讲义和学习资料** +安排几个平时笔记做得好的去跟线代习题课 + +分工问题。研讨中出试卷、讲义分工问题的解决。挑题、讨论、成文。 + +下次开会时间：1.11周日晚 \ No newline at end of file -- 2.34.1 From 829cc1260feeade648acf9c0f653a3c1d06f43d8 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 9 Jan 2026 00:03:52 +0800 Subject: [PATCH 171/274] Modify README --- README.md | 19 +++++++++++++++++-- 素材/图片/README/edit1.png | Bin 0 -> 397820 bytes 2 files changed, 17 insertions(+), 2 deletions(-) create mode 100644 素材/图片/README/edit1.png diff --git a/README.md b/README.md index 33c9ffe..390c1bd 100644 --- a/README.md +++ b/README.md @@ -1,2 +1,17 @@ -# 一起数学----高等数学笔记协作平台 -大家可以把Markdown高数笔记上传到此处 \ No newline at end of file +# 讲义编写研讨组欢迎你！ +### 开始工作前的准备： +##### 概述 +1. 下载Obsidian和Git For Windows（如果官网下不来，会给你镜像） +2. 头歌账号绑定邮箱，并且在本地准备Git +3. 在Obsidian里面安装插件Git +##### 详细步骤 +1. 下载Obsidian和Git For Windows（如果官网下不来，会给你镜像） +2. 打开头歌，打开个人主页，进入开发项目，新建一个项目 +![[edit1.png]] +新建项目的时候，如果没有绑定邮箱，会要求你绑定邮箱； +如果没有 +### 前置知识（需要在进入研讨组后尽快掌握） +1. KaTeX数学公式排版知识 +2. 基础Markdown知识 +3. 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Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1.10线代限时练-答案.md | 54 +++++++++++++++++++ .../1.10线代限时练.md | 18 +++---- 2 files changed, 63 insertions(+), 9 deletions(-) create mode 100644 编写小组/试卷/线代期末复习模拟/1.10线代限时练-答案.md diff --git a/编写小组/试卷/线代期末复习模拟/1.10线代限时练-答案.md b/编写小组/试卷/线代期末复习模拟/1.10线代限时练-答案.md new file mode 100644 index 0000000..ce24dbe --- /dev/null +++ b/编写小组/试卷/线代期末复习模拟/1.10线代限时练-答案.md @@ -0,0 +1,54 @@ +1. （7分） $\begin{bmatrix}1&2^{101}-2&0\\0&2^{100}&0\\0&\frac{5}{3}(1-2^{100})&1\end{bmatrix}$ +>解析： +>$A$ 的特征值为 $1,1,2$， +>$A=\begin{bmatrix}0&1&2\\0&0&1\\1&-1&-\frac{5}{3}\end{bmatrix}\begin{bmatrix}1&0&0\\0&1&0\\0&0&2\end{bmatrix}\begin{bmatrix}0&1&2\\0&0&1\\1&-1&-\frac{5}{3}\end{bmatrix}^{-1}$ +2. （6分） $\frac{n(n-1)}{2}$ +>解析：略 +3. $s$ 是奇数 +>解析：向量组 $\beta_1,\beta_2,\cdots,\beta_s$ 线性无关，即 $\alpha_1,\alpha_2,\cdots,\alpha_s$ 到 $\beta_1,\beta_2,\cdots,\beta_s$ 的过渡矩阵可逆（此时两个向量组等价），过渡矩阵为$\begin{bmatrix}1&&&\cdots&1\\1&1&&&\\&1&1&&\\&&\ddots&\ddots&\\&&&1&1\end{bmatrix}_{s\times s}$，过渡矩阵的行列式不为$0$ +>按照第一行展开： +>$\begin{vmatrix}1&&&\cdots&1\\1&1&&&\\&1&1&&\\&&\ddots&\ddots&\\&&&1&1\end{vmatrix}_{s\times s}=(-1)^2\begin{vmatrix}1&&&&\\1&1&&&\\&1&1&&\\&&\ddots&\ddots&\\&&&1&1\end{vmatrix}_{(s-1)\times (s-1)}+(-1)^{s+1}\begin{vmatrix}1&1&&&\\&1&1&&\\&&1&\ddots&\\&&&\ddots&1\\&&&&1\end{vmatrix}_{(s-1)\times (s-1)}$ +>即$\begin{vmatrix}1&&&\cdots&1\\1&1&&&\\&1&1&&\\&&\ddots&\ddots&\\&&&1&1\end{vmatrix}_{s\times s}=1+(-1)^{s+1}=\begin{cases}0,s\text{为偶数}\\2,s\text{为奇数}\end{cases}$ +>所以 $s$ 是奇数 +4. （6分） 0 +>解析：因为$A,B$是正交矩阵且$|A|=-|B|$，所以$A^\mathrm{T}A=E, B^\mathrm{T}B=E, |A|=\pm 1, |B|=\mp 1$， +>而$A^{-1}(A+B)B^{-1}=B^{-1}+A^{-1}=A^\mathrm{T}+B^\mathrm{T}=(A+B)^\mathrm{T}$， +>所以$|A^{-1}||A+B||B^{-1}|=|(A+B)^\mathrm{T}|=|A+B|$，$|A^{-1}||B^{-1}|=-1$ +>所以$-|A+B|=|A+B|$，即$|A+B|=0$ +5. （7分） 0 +>$A\sim B\sim\begin{bmatrix}1&&&\\&2&&\\&&3&\\&&&4\end{bmatrix}$， +>所以$B^{-1}-E\sim\begin{bmatrix}1&&&\\&2&&\\&&3&\\&&&4\end{bmatrix}^{-1}-E=\begin{bmatrix}0&&&\\&-\frac{1}{2}&&\\&&-\frac{2}{3}&\\&&&-\frac{3}{4}\end{bmatrix}$， +>所以$|B^{-1}-E|=\begin{vmatrix}0&&&\\&-\frac{1}{2}&&\\&&-\frac{2}{3}&\\&&&-\frac{3}{4}\end{vmatrix}=0$ +6. （7分） 0 +>解析：设 $\alpha,\beta,\gamma$ 为 $x^3+px+q=0$ 的三个根，则 $x^3+px+q=$$(x-\alpha)(x-\beta)(x-\gamma)=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x+\alpha\beta\gamma$ ，所以$\alpha+\beta+\gamma=0$，将行列式第二、三行加在第一行，第一行全为$\alpha+\beta+\gamma=0$，故行列式为0 +7. （20分） $D_n=2^{n-1}(2+\sum\limits_{j=1}^n a_j^2)$ +>解析 +>方法1： +>行列式加边法：$D_n=\begin{vmatrix}1&a_1&a_2&\cdots&a_n\\0&2+a_1^2&a_1a_2&\cdots&a_1a_n\\0&a_2a_1&2+a_2^2&\cdots&a_2a_n\\\vdots&\vdots&\vdots&\ddots&\vdots\\0&a_na_1&a_na_2&\cdots&2+a_n^2\end{vmatrix}$ +>将第一行乘以 $(-a_i)$ 加到第 $i+1$ 行： +>$D_n=\begin{vmatrix}1&a_1&a_2&\cdots&a_n\\-a_1&2&&&\\-a_2&&2&&\\\vdots&&&\ddots&\\-a_n&&&&2\end{vmatrix}$（模板：箭头行列式） +>再将第 $j+1$ 列乘以 $\frac{a_j}{2}$ 加到第一列： +>$D_n=\begin{vmatrix}1+\sum\limits_{j=1}^n\frac{a_j^2}{2}&a_1&a_2&\cdots&a_n\\&2&&&\\&&2&&\\&&&\ddots&\\&&&&2\end{vmatrix}$ +>所以$D_n=2^{n-1}(2+\sum\limits_{j=1}^n a_j^2)$ +>方法2： +>核心结论：$|E+AB|=|E+BA|$ +>$D_n=|2E_n+xx^\mathrm{T}|=2^n|E_n+\frac{1}{2}xx^\mathrm{T}|=2^n|E_1+\frac{1}{2}x^\mathrm{T}x|=2^n(1+x^\mathrm{T}x)=2^{n-1}(2+\sum\limits_{j=1}^{n}a_j^2)$ +8. （20分） $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ +>解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ +>故 $Ax=0$ 的解空间维数是 $1$ （5分） +>$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） +>$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） +>解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 +9. 证明如下： +>(1)（10分） +> 对于方程组$Ax=0$ (a)和 $A^\mathrm{T}Ax=0$ (b)，b的解空间一定包含a的解空间（5分）； +>而方程b两边同时乘以$x^\mathrm{T}$，得 $x^\mathrm{T}A^\mathrm{T}Ax=0$ ，即 $(x^\mathrm{T}A^\mathrm{T})(Ax)=0 \to Ax=0$， +>所以a的解空间包含b的解空间（5分）， +>所以a,b同解，所以$\mathrm{rank}A=\mathrm{rank}{A^\mathrm{T}A}$ +>(2) （10分） +>$A^\mathrm{T}Ax=A^\mathrm{T}\beta \iff \mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ +>而由(1)的结论得等式左边 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}A$； +>等式右边 $\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})\ge \mathrm{rank}A^\mathrm{T}A=\mathrm{rank}A$ （5分） +>又$\mathrm{rank}(A^\mathrm{T}\begin{bmatrix}A&\beta\end{bmatrix})\le \min{(\mathrm{rank}A^\mathrm{T},\ \mathrm{rank}\begin{bmatrix}A&\beta\end{bmatrix})}=\mathrm{rank}A$ （5分） +>所以$\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})=\mathrm{rank}A$ +>故 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ 得证. \ No newline at end of file diff --git a/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md index a837904..c76c6ad 100644 --- a/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md +++ b/编写小组/试卷/线代期末复习模拟/1.10线代限时练.md @@ -1,13 +1,13 @@ 一、填空题 -1. 设 $A=\begin{bmatrix}1&2&0\\0&2&0\\-2&-1&-1\end{bmatrix}$，则$A^{100}=$\_\_\_\_\_\_\_\_\_\_\_ -2. 线性空间 $V = \{A\in\mathbb{R}^{n\times n}|A = -A^\mathrm{T}\}$ 的维数是\_\_\_\_\_\_\_\_\_\_\_ -3. 设向量组 $\alpha_1,\alpha_2,\cdots,\alpha_s$ 线性无关，$\beta_1=\alpha_1+\alpha_2,\beta_2=\alpha_2+\alpha_3,\cdots,\beta_{s-1}=\alpha_{s-1}+\alpha_s$，则向量组 $\beta_1,\beta_2,\cdots,\beta_s$ 线性无关的充要条件是\_\_\_\_\_\_\_\_\_\_\_ -4. 已知 $A,B$ 均为 $n$ 阶正交矩阵，且 $|A|=-|B|$，则 $|A+B|$ 的值为\_\_\_\_\_\_\_\_\_\_\_ -5. 已知 $4$ 阶矩阵 $A$ 与 $B$ 相似，$A$ 的全部特征值为 $1,2,3,4$，则行列式 $|B^{-1}-E|$ 为\_\_\_\_\_\_\_\_\_\_\_ -6. 设 $\alpha,\beta,\gamma$ 为 $x^3+px+q=0$ 的三个根，则行列式 $\begin{vmatrix}\alpha&\beta&\gamma\\\gamma&\alpha&\beta\\\beta&\gamma&\alpha\end{vmatrix}$ 的值为\_\_\_\_\_\_\_\_\_\_\_ +1. （7分）设 $A=\begin{bmatrix}1&2&0\\0&2&0\\-2&-1&-1\end{bmatrix}$，则$A^{100}=$\_\_\_\_\_\_\_\_\_\_\_ +2. （6分）线性空间 $V = \{A\in\mathbb{R}^{n\times n}|A = -A^\mathrm{T}\}$ 的维数是\_\_\_\_\_\_\_\_\_\_\_ +3. （7分）设向量组 $\alpha_1,\alpha_2,\cdots,\alpha_s$ 线性无关，$\beta_1=\alpha_1+\alpha_2,\beta_2=\alpha_2+\alpha_3,\cdots,\beta_{s-1}=\alpha_{s-1}+\alpha_s, \beta_{s}=\alpha_{s}+\alpha_1$，则向量组 $\beta_1,\beta_2,\cdots,\beta_s$ 线性无关的充要条件是\_\_\_\_\_\_\_\_\_\_\_ +4. （6分）知 $A,B$ 均为 $n$ 阶正交矩阵，且 $|A|=-|B|$，则 $|A+B|$ 的值为\_\_\_\_\_\_\_\_\_\_\_ +5. （7分）已知 $4$ 阶矩阵 $A$ 与 $B$ 相似，$A$ 的全部特征值为 $1,2,3,4$，则行列式 $|B^{-1}-E|$ 为\_\_\_\_\_\_\_\_\_\_\_ +6. （7分）设 $\alpha,\beta,\gamma$ 为 $x^3+px+q=0$ 的三个根，则行列式 $\begin{vmatrix}\alpha&\beta&\gamma\\\gamma&\alpha&\beta\\\beta&\gamma&\alpha\end{vmatrix}$ 的值为\_\_\_\_\_\_\_\_\_\_\_ 二、解答题 -7. 计算$n$阶行列式 $D_n=\begin{vmatrix}2+a_1^2 & a_1a_2 & \cdots&a_1a_n \\ a_2a_1& 2+a_2^2 & \cdots & a_2a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_na_1&a_na_2&\cdots& 2+a_n^2\end{vmatrix}$. -8. 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. -9. 设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： +7. （20分）计算$n$阶行列式 $D_n=\begin{vmatrix}2+a_1^2 & a_1a_2 & \cdots&a_1a_n \\ a_2a_1& 2+a_2^2 & \cdots & a_2a_n \\ \vdots & \vdots & \ddots & \vdots \\ a_na_1&a_na_2&\cdots& 2+a_n^2\end{vmatrix}$. +8. （20分）已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. +9. （20分）设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解. \ No newline at end of file -- 2.34.1 From d1d2eaf38feb37c4aa2fcca1d2e1a5e2b6df6e77 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 16:56:36 +0800 Subject: [PATCH 173/274] vault backup: 2026-01-11 16:56:36 --- 未命名.md | 0 编写小组/讲义/线性方程组解的问题.md | 8 ++++++++ 2 files changed, 8 insertions(+) create mode 100644 未命名.md create mode 100644 编写小组/讲义/线性方程组解的问题.md diff --git a/未命名.md b/未命名.md new file mode 100644 index 0000000..e69de29 diff --git a/编写小组/讲义/线性方程组解的问题.md b/编写小组/讲义/线性方程组解的问题.md new file mode 100644 index 0000000..ce84b16 --- /dev/null +++ b/编写小组/讲义/线性方程组解的问题.md @@ -0,0 +1,8 @@ +对非齐次线性方程组 $A_{m\times n}x=b$, +1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； +2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； +3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 +注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 + +把以上结论应用到齐次线性方程组，可得 +推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 \ No newline at end of file -- 2.34.1 From e229b80da390c9758bab9e3529afb6a086c447d3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 16:57:43 +0800 Subject: [PATCH 174/274] vault backup: 2026-01-11 16:57:43 --- 编写小组/讲义/线性方程组解的问题.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组解的问题.md b/编写小组/讲义/线性方程组解的问题.md index ce84b16..9e9e91d 100644 --- a/编写小组/讲义/线性方程组解的问题.md +++ b/编写小组/讲义/线性方程组解的问题.md @@ -1,3 +1,4 @@ +定理 对非齐次线性方程组 $A_{m\times n}x=b$, 1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； 2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； @@ -5,4 +6,4 @@ 注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 把以上结论应用到齐次线性方程组，可得 -推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 \ No newline at end of file +推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 -- 2.34.1 From fbc483913818ac0f105da6ee356314be3cfc74e4 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 16:57:58 +0800 Subject: [PATCH 175/274] vault backup: 2026-01-11 16:57:58 --- {编写小组/讲义 => 素材}/线性方程组解的问题.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename {编写小组/讲义 => 素材}/线性方程组解的问题.md (100%) diff --git a/编写小组/讲义/线性方程组解的问题.md b/素材/线性方程组解的问题.md similarity index 100% rename from 编写小组/讲义/线性方程组解的问题.md rename to 素材/线性方程组解的问题.md -- 2.34.1 From 68508dbe19782c58753e45d6c93a055cfbe647b5 Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Sun, 11 Jan 2026 16:58:00 +0800 Subject: [PATCH 176/274] vault backup: 2026-01-11 16:57:59 --- 研讨记录/2026.1.11.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) create mode 100644 研讨记录/2026.1.11.md diff --git a/研讨记录/2026.1.11.md b/研讨记录/2026.1.11.md new file mode 100644 index 0000000..b39612b --- /dev/null +++ b/研讨记录/2026.1.11.md @@ -0,0 +1,16 @@ +研讨流程： +1.逐渐分配任务，后续边推进工作边提意见 +2.后期对讲义进行优化 +3.确定主题，目的性寻找真题，思考解题策略，总结方法 +4.把相似的题目直接发群里 +讲义建议： +1.提前发 +2.难题不要和例题混在一起，在讲义后专门列一个模块 +3.高数：第五章开始，知识点细化 +4.线代：模块化复习，部分二级结论的汇总与应用。 +5.编写目标：方法集，直接从集合中调用方法解题 +线代： +1.两个线性方程组同解的必要不充分条件是两个方程组的增广矩阵秩相等，充要条件是增广矩阵对应最简行阶梯行矩阵相等 +2.两个矩阵等价，其对应齐次线性方程组不一定同解 +3.期中考T13：多个线性方程组解的相同<----->与r(An)相等与否<------>列向量组的线性组合 +4.线性方程组解的结构<---->维数 \ No newline at end of file -- 2.34.1 From e65660aa7a78358ad86dcac3795b6ef710601c3d Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Sun, 11 Jan 2026 16:59:22 +0800 Subject: [PATCH 177/274] vault backup: 2026-01-11 16:59:22 --- 研讨记录/2026.1.11.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/研讨记录/2026.1.11.md b/研讨记录/2026.1.11.md index b39612b..efff04c 100644 --- a/研讨记录/2026.1.11.md +++ b/研讨记录/2026.1.11.md @@ -13,4 +13,5 @@ 1.两个线性方程组同解的必要不充分条件是两个方程组的增广矩阵秩相等，充要条件是增广矩阵对应最简行阶梯行矩阵相等 2.两个矩阵等价，其对应齐次线性方程组不一定同解 3.期中考T13：多个线性方程组解的相同<----->与r(An)相等与否<------>列向量组的线性组合 -4.线性方程组解的结构<---->维数 \ No newline at end of file +4.线性方程组解的结构<---->维数 +5.正交矩阵<----->施密特正交化法 \ No newline at end of file -- 2.34.1 From 52ac7df4afe9ce4cbe7df8d12c6289ead3c8df2d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 17:23:44 +0800 Subject: [PATCH 178/274] vault backup: 2026-01-11 17:23:44 --- 素材/线性方程组解的问题.md | 9 --------- 素材/解的问题.md | 29 +++++++++++++++++++++++++++ 2 files changed, 29 insertions(+), 9 deletions(-) delete mode 100644 素材/线性方程组解的问题.md create mode 100644 素材/解的问题.md diff --git a/素材/线性方程组解的问题.md b/素材/线性方程组解的问题.md deleted file mode 100644 index 9e9e91d..0000000 --- a/素材/线性方程组解的问题.md +++ /dev/null @@ -1,9 +0,0 @@ -定理 -对非齐次线性方程组 $A_{m\times n}x=b$, -1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； -2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； -3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 -注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 - -把以上结论应用到齐次线性方程组，可得 -推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 diff --git a/素材/解的问题.md b/素材/解的问题.md new file mode 100644 index 0000000..b3bfe9f --- /dev/null +++ b/素材/解的问题.md @@ -0,0 +1,29 @@ +### 1. 原理 +**线性方程组解的判定** +对非齐次线性方程组 $A_{m\times n}x=b$, +1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； +2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； +3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 +注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 + +把以上结论应用到齐次线性方程组，可得 +推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 + +**矩阵方程解的判定** +本质上和线性方程组是一脉相承的，只是形式上更一般化。 +最常见的矩阵方程是 $\boldsymbol{AX} = \boldsymbol{B}$，其中$\boldsymbol{A}$是 $m\times n$ 矩阵，$\boldsymbol{B}$ 是 $m\times p$ 矩阵，$\boldsymbol{X}$ 是待求的$n\times p$矩阵。 +1. 有解的充要条件： +矩阵方程有解的充要条件是系数矩阵 $\boldsymbol{A}$ 的秩等于增广矩阵 $[\boldsymbol{A} \ \boldsymbol{B}]$的秩，即： +$$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}])$$ +这个结论和非齐次线性方程组有解的条件完全一致。 +2. 解的结构： +- 唯一解：当$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) = n$ 时，方程有唯一解。 +- 无穷多解：当 $r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) < n$ 时，方程有无穷多解。 + +可逆矩阵 +- 当 $\boldsymbol{A}$ 是 n 阶可逆矩阵时，矩阵方程 $\boldsymbol{AX} = \boldsymbol{B}$有唯一解:$\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{B}$ + +其他形式的矩阵方程 +- 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法。 +- 对于 $\boldsymbol{AXB} = \boldsymbol{C}$ 形式的方程，当 $\boldsymbol{A} 和 \boldsymbol{B}$ 都可逆时，有唯一解 $\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{C}\boldsymbol{B}^{-1}$。 + \ No newline at end of file -- 2.34.1 From 45e33bf2143eb2eaa60a3cbaa0cfd6494273a511 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 17:28:18 +0800 Subject: [PATCH 179/274] vault backup: 2026-01-11 17:28:18 --- 素材/线性方程组同解.md | 13 +++++++++++++ 1 file changed, 13 insertions(+) create mode 100644 素材/线性方程组同解.md diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md new file mode 100644 index 0000000..dfc6600 --- /dev/null +++ b/素材/线性方程组同解.md @@ -0,0 +1,13 @@ +## **$Ax=0$与$Bx=0$同解问题**： +充要条件：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +$Ax=\alpha$ 与$Bx=\beta$同解问题： +充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. +如何理解（非严格证明，目的是便于理解）： +首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， +考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， +可以得到$N(A)\subset N(B)$,以及$N(B)\subset N(A)$. +$N(A)\subset N(B)$可以得到什么呢？ +说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格，换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$可以得到$Ax=0$的每个方程是由$Bx=0$的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. +需要注意的是,这个条件是充要的.非常的好用. +非齐次的时候同理. \ No newline at end of file -- 2.34.1 From c69084ad3f6a861fd8adf66b6638f11119a49951 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 17:38:38 +0800 Subject: [PATCH 180/274] vault backup: 2026-01-11 17:38:38 --- 素材/解的问题.md | 67 ++++++++++++++++++++++++++++++++++++++++-- 1 file changed, 64 insertions(+), 3 deletions(-) diff --git a/素材/解的问题.md b/素材/解的问题.md index b3bfe9f..af1d6f4 100644 --- a/素材/解的问题.md +++ b/素材/解的问题.md @@ -1,4 +1,4 @@ -### 1. 原理 +### 原理 **线性方程组解的判定** 对非齐次线性方程组 $A_{m\times n}x=b$, 1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； @@ -24,6 +24,67 @@ $$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}])$$ - 当 $\boldsymbol{A}$ 是 n 阶可逆矩阵时，矩阵方程 $\boldsymbol{AX} = \boldsymbol{B}$有唯一解:$\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{B}$ 其他形式的矩阵方程 -- 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法。 +- 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法，或类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} =\text{rank}A$ - 对于 $\boldsymbol{AXB} = \boldsymbol{C}$ 形式的方程，当 $\boldsymbol{A} 和 \boldsymbol{B}$ 都可逆时，有唯一解 $\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{C}\boldsymbol{B}^{-1}$。 - \ No newline at end of file + +>[!example] **例一** +>设矩阵 +>$$A = \begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix}, +\quad +x = \begin{bmatrix} +x_1 \\ x_2 \\ x_3 \\ x_4 +\end{bmatrix}, +\quad +b = \begin{bmatrix} +1 \\ 1 \\ 1 \\ 1 +\end{bmatrix}$$ +其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为 ______________。 + +**答**：$(1,0,0,0)^T$。 + +**解析**：由范德蒙行列式的性质可知 $|A| \neq 0$，从而线性方程组 $Ax = b$ 有唯一解。 + +又由 +$$ +\begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix} +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} += +\begin{bmatrix} +1 \\ +1 \\ +1 \\ +1 +\end{bmatrix} +$$ + 可知 $Ax = b$ 的解为 +$$ +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} +$$ + +>[!example] **例二** +> 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ + +--- + +解析： +类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ \ No newline at end of file -- 2.34.1 From 4d69ea2c7ed143714b8c6ea49f196d935be3c87e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 17:39:00 +0800 Subject: [PATCH 181/274] vault backup: 2026-01-11 17:39:00 --- 素材/解的问题.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/素材/解的问题.md b/素材/解的问题.md index af1d6f4..9486cda 100644 --- a/素材/解的问题.md +++ b/素材/解的问题.md @@ -84,7 +84,7 @@ $$ >[!example] **例二** > 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ ---- -解析： + +**解析**： 类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ \ No newline at end of file -- 2.34.1 From 245376ca0d0c0060258c358a0e2103149d589d9c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=88=98=E6=9F=AF=E5=A6=A4?= <2503393720@qq.com> Date: Sun, 11 Jan 2026 17:39:10 +0800 Subject: [PATCH 182/274] vault backup: 2026-01-11 17:39:10 --- 素材/解的问题.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/解的问题.md b/素材/解的问题.md index 9486cda..ff27b2e 100644 --- a/素材/解的问题.md +++ b/素材/解的问题.md @@ -1,4 +1,4 @@ -### 原理 +### 原理 **线性方程组解的判定** 对非齐次线性方程组 $A_{m\times n}x=b$, 1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； -- 2.34.1 From 513e099a3ad19062eedaf35aaa65dc40b1554bff Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 17:39:48 +0800 Subject: [PATCH 183/274] vault backup: 2026-01-11 17:39:48 --- 素材/线性方程组同解.md | 7 +++++-- 1 file changed, 5 insertions(+), 2 deletions(-) diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md index dfc6600..a589a92 100644 --- a/素材/线性方程组同解.md +++ b/素材/线性方程组同解.md @@ -1,13 +1,16 @@ ## **$Ax=0$与$Bx=0$同解问题**： 充要条件：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. -$Ax=\alpha$ 与$Bx=\beta$同解问题： +$Ax=\alpha$ 与 $Bx=\beta$ 同解问题： 充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. + 如何理解（非严格证明，目的是便于理解）： 首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， 考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， 可以得到$N(A)\subset N(B)$,以及$N(B)\subset N(A)$. $N(A)\subset N(B)$可以得到什么呢？ -说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格，换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$可以得到$Ax=0$的每个方程是由$Bx=0$的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格。换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$ 可以得到 $Ax=0$ 的每个方程是由 $Bx=0$ 的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. + 另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. + 需要注意的是,这个条件是充要的.非常的好用. 非齐次的时候同理. \ No newline at end of file -- 2.34.1 From efb560eed143b3bfb2a7a08e49ff9d436922e65a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 17:52:36 +0800 Subject: [PATCH 184/274] vault backup: 2026-01-11 17:52:36 --- 素材/线性方程组同解.md | 13 ++++++++++++- 1 file changed, 12 insertions(+), 1 deletion(-) diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md index a589a92..35ee541 100644 --- a/素材/线性方程组同解.md +++ b/素材/线性方程组同解.md @@ -13,4 +13,15 @@ $N(A)\subset N(B)$可以得到什么呢？ 另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. 需要注意的是,这个条件是充要的.非常的好用. -非齐次的时候同理. \ No newline at end of file +非齐次的时候同理. + +注意：由此，我们还能得到一些别的结论 +例如：$A$ 和 $B$ 等价，并不能得到两方程同解，因为等价的初等变换可能包括初等列变换，而列变换可能改变两方程的解 + +>[!example] 例1 +>6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$ 与$\quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + -- 2.34.1 From 1c1aa64321811a0e3f18648c83b0fd22cc5137c1 Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Sun, 11 Jan 2026 17:56:06 +0800 Subject: [PATCH 185/274] vault backup: 2026-01-11 17:56:06 --- 未命名.md | 0 研讨记录/2026.1.11.md | 2 ++ 线性方程组的系数矩阵与解关系.md | 5 +++++ 3 files changed, 7 insertions(+) delete mode 100644 未命名.md create mode 100644 线性方程组的系数矩阵与解关系.md diff --git a/未命名.md b/未命名.md deleted file mode 100644 index e69de29..0000000 diff --git a/研讨记录/2026.1.11.md b/研讨记录/2026.1.11.md index efff04c..fc64c20 100644 --- a/研讨记录/2026.1.11.md +++ b/研讨记录/2026.1.11.md @@ -3,12 +3,14 @@ 2.后期对讲义进行优化 3.确定主题，目的性寻找真题，思考解题策略，总结方法 4.把相似的题目直接发群里 + 讲义建议： 1.提前发 2.难题不要和例题混在一起，在讲义后专门列一个模块 3.高数：第五章开始，知识点细化 4.线代：模块化复习，部分二级结论的汇总与应用。 5.编写目标：方法集，直接从集合中调用方法解题 + 线代： 1.两个线性方程组同解的必要不充分条件是两个方程组的增广矩阵秩相等，充要条件是增广矩阵对应最简行阶梯行矩阵相等 2.两个矩阵等价，其对应齐次线性方程组不一定同解 diff --git a/线性方程组的系数矩阵与解关系.md b/线性方程组的系数矩阵与解关系.md new file mode 100644 index 0000000..1d721b7 --- /dev/null +++ b/线性方程组的系数矩阵与解关系.md @@ -0,0 +1,5 @@ +在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： +>[!note] 定理1： +>对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$rank\boldsymbol{A}=r$，则 +> $\qquad\qquad\qquad dimN(\boldsymbol{A})=n-r$ + -- 2.34.1 From bffc942ce75a29be410669c2b3b22be037ace62c Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 17:56:18 +0800 Subject: [PATCH 186/274] vault backup: 2026-01-11 17:56:18 --- 素材/线性方程组同解.md | 1 + 素材/解的问题.md | 6 ++---- 2 files changed, 3 insertions(+), 4 deletions(-) diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md index 35ee541..3211c95 100644 --- a/素材/线性方程组同解.md +++ b/素材/线性方程组同解.md @@ -25,3 +25,4 @@ $N(A)\subset N(B)$可以得到什么呢？ (C) $a = 2, b = 0, c = 1$; (D) $a = 2, b = 1, c = 2$. +解析：类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ \ No newline at end of file diff --git a/素材/解的问题.md b/素材/解的问题.md index ff27b2e..c21616c 100644 --- a/素材/解的问题.md +++ b/素材/解的问题.md @@ -27,7 +27,7 @@ $$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}])$$ - 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法，或类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} =\text{rank}A$ - 对于 $\boldsymbol{AXB} = \boldsymbol{C}$ 形式的方程，当 $\boldsymbol{A} 和 \boldsymbol{B}$ 都可逆时，有唯一解 $\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{C}\boldsymbol{B}^{-1}$。 ->[!example] **例一** +>[!example] **例1** >设矩阵 >$$A = \begin{bmatrix} 1 & a_1 & a_1^2 & a_1^3 \\ @@ -81,10 +81,8 @@ $$ \end{bmatrix} $$ ->[!example] **例二** +>[!example] **例2** > 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ - - **解析**： 类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ \ No newline at end of file -- 2.34.1 From b263a873cb4c10145a2f43155ed463154aa8c603 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sun, 11 Jan 2026 23:18:24 +0800 Subject: [PATCH 187/274] vault backup: 2026-01-11 23:18:24 --- 研讨记录/2026.1.11.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/研讨记录/2026.1.11.md b/研讨记录/2026.1.11.md index fc64c20..af30d2c 100644 --- a/研讨记录/2026.1.11.md +++ b/研讨记录/2026.1.11.md @@ -2,7 +2,8 @@ 1.逐渐分配任务，后续边推进工作边提意见 2.后期对讲义进行优化 3.确定主题，目的性寻找真题，思考解题策略，总结方法 -4.把相似的题目直接发群里 +4.先有初步的规划，避免效率太低 +5.把相似的题目直接发群里 讲义建议： 1.提前发 -- 2.34.1 From 5ad3e693d39034b1f186a596d7c51b8bf408d016 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 23:32:15 +0800 Subject: [PATCH 188/274] vault backup: 2026-01-11 23:32:15 --- 素材/秩的不等式.md | 63 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 63 insertions(+) create mode 100644 素材/秩的不等式.md diff --git a/素材/秩的不等式.md b/素材/秩的不等式.md new file mode 100644 index 0000000..fd57403 --- /dev/null +++ b/素材/秩的不等式.md @@ -0,0 +1,63 @@ +# 一般式 + +## 1. 和的秩不超过秩的和 + +设 $A, B$ 为同型矩阵，则 +$$ \operatorname{rank}(A+B) \leq \operatorname{rank} A + \operatorname{rank} B $$ + +## 2. 积的秩不超过任何因子的秩 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \leq \min\{\operatorname{rank} A, \operatorname{rank} B\} $$ + +## 3. 重要不等式 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \geq \operatorname{rank} A + \operatorname{rank} B - n $$ +特别地，当 $AB = 0$ 时，有 $\operatorname{rank} A + \operatorname{rank} B \leq n$。 + +# 分块式 + +设 $A_{n \times n}$， $B_{n \times n}$，则 + +$$(1)\ rank + +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } A, \quad \text{rank } +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } B +$$ + +$$(2)\ rank +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } A + \text{rank } B +$$ + +$$(3)\ rank +\begin{bmatrix} +A & E_n \\ +0 & B +\end{bmatrix} \geq \text{rank } A + \text{rank } B +$$ + +$$(4)\ rank +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A & B \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A + B & B \\ +B & B +\end{bmatrix} \geq \text{rank } (A + B) +$$ + -- 2.34.1 From 8e763b04a4f372af32ba354c0e4fa755861573a4 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Sun, 11 Jan 2026 23:34:45 +0800 Subject: [PATCH 189/274] vault backup: 2026-01-11 23:34:45 --- 素材/线性方程组同解.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md index 3211c95..1df3763 100644 --- a/素材/线性方程组同解.md +++ b/素材/线性方程组同解.md @@ -16,7 +16,7 @@ $N(A)\subset N(B)$可以得到什么呢？ 非齐次的时候同理. 注意：由此，我们还能得到一些别的结论 -例如：$A$ 和 $B$ 等价，并不能得到两方程同解，因为等价的初等变换可能包括初等列变换，而列变换可能改变两方程的解 +例如：$A$ 和 $B$ 等价（可以通过初等变换得到），并不能得到两方程同解，因为等价的初等变换可能包括初等列变换，而列变换可能改变两方程的解 >[!example] 例1 >6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$ 与$\quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 -- 2.34.1 From 243f301475ba04fbe971fd1e7388b0ca25e294a8 Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Sun, 11 Jan 2026 23:39:07 +0800 Subject: [PATCH 190/274] vault backup: 2026-01-11 23:39:07 --- 线性方程组的系数矩阵与解关系.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/线性方程组的系数矩阵与解关系.md b/线性方程组的系数矩阵与解关系.md index 1d721b7..91568da 100644 --- a/线性方程组的系数矩阵与解关系.md +++ b/线性方程组的系数矩阵与解关系.md @@ -1,5 +1,8 @@ 在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： >[!note] 定理1： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$rank\boldsymbol{A}=r$，则 -> $\qquad\qquad\qquad dimN(\boldsymbol{A})=n-r$ +> $$\dim N(\boldsymbol{A})=n-r$$ +>例一： +>设 $A=$ + -- 2.34.1 From 4cf0b6d9b9224e8a55134d8067fee533f2c28978 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sun, 11 Jan 2026 23:42:55 +0800 Subject: [PATCH 191/274] vault backup: 2026-01-11 23:42:55 --- 线性方程组的系数矩阵与解关系.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/线性方程组的系数矩阵与解关系.md b/线性方程组的系数矩阵与解关系.md index 91568da..3e40a9a 100644 --- a/线性方程组的系数矩阵与解关系.md +++ b/线性方程组的系数矩阵与解关系.md @@ -1,8 +1,9 @@ 在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： >[!note] 定理1： ->对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$rank\boldsymbol{A}=r$，则 +>对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ ->例一： ->设 $A=$ + +>[!example] 例一： +设 $A=$ -- 2.34.1 From a2d6bb97ff546f3443620a6dc1d94dadbd734c3c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Sun, 11 Jan 2026 23:46:17 +0800 Subject: [PATCH 192/274] vault backup: 2026-01-11 23:46:17 --- 素材/秩的不等式.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/素材/秩的不等式.md b/素材/秩的不等式.md index fd57403..3d28b8c 100644 --- a/素材/秩的不等式.md +++ b/素材/秩的不等式.md @@ -20,7 +20,7 @@ $$ \operatorname{rank}(AB) \geq \operatorname{rank} A + \operatorname{rank} B - 设 $A_{n \times n}$， $B_{n \times n}$，则 -$$(1)\ rank +$$(1)\ \mathrm{rank} \begin{bmatrix} A \\ @@ -32,21 +32,21 @@ B \end{bmatrix} \geq \text{rank } B $$ -$$(2)\ rank +$$(2)\ \mathrm{rank} \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix} = \text{rank } A + \text{rank } B $$ -$$(3)\ rank +$$(3)\ \mathrm{rank} \begin{bmatrix} A & E_n \\ 0 & B \end{bmatrix} \geq \text{rank } A + \text{rank } B $$ -$$(4)\ rank +$$(4)\ \mathrm{rank} \begin{bmatrix} A & 0 \\ 0 & B -- 2.34.1 From 93e2f0237b4c94ca12bf6b9ca72b6f6bb77b293a Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 12 Jan 2026 00:10:20 +0800 Subject: [PATCH 193/274] =?UTF-8?q?=E4=B8=8A=E4=BC=A0=E4=BA=861.11?= =?UTF-8?q?=E9=A2=98=E7=9B=AE=E7=9A=84=E7=B4=A0=E6=9D=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/1.11题目素材.md | 64 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 64 insertions(+) create mode 100644 素材/1.11题目素材.md diff --git a/素材/1.11题目素材.md b/素材/1.11题目素材.md new file mode 100644 index 0000000..35c8c84 --- /dev/null +++ b/素材/1.11题目素材.md @@ -0,0 +1,64 @@ +设 $A, B$ 为 $n$ 阶矩阵，则 +(A) $\mathrm{rank}\begin{bmatrix}A&AB\end{bmatrix}=\mathrm{rank}A$ +(B) $\mathrm{rank}\begin{bmatrix}A&BA\end{bmatrix}=\mathrm{rank}A$ +(C) $\mathrm{rank}\begin{bmatrix}A&B\end{bmatrix}=\max{\{\mathrm{rank}A,\mathrm{rank}B\}}$ +(D) $\mathrm{rank}\begin{bmatrix}A&B\end{bmatrix}=\mathrm{rank}\begin{bmatrix}A^T&B^T\end{bmatrix}$ + +（[[1.10线代限时练]]）设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： + (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . + (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解. + +设 $A$ 是 $m\times n$ 矩阵， $B$ 是 $n\times m$ 矩阵，$E$ 是 $m$ 阶单位矩阵，若 $AB=E$，则 +(A) $\mathrm{rank}A=m,\mathrm{rank}A=m$ +(B) $\mathrm{rank}A=m,\mathrm{rank}A=n$ +(C) $\mathrm{rank}A=n,\mathrm{rank}A=m$ +(D) $\mathrm{rank}A=n,\mathrm{rank}A=n$ + +已知 $A,B,C,D$ 都是 $4$ 阶非零矩阵，且 $ABCD=O$，如果 $|BC|\ne 0$，记 $\mathrm{rank}A+\mathrm{rank}B+\mathrm{rank}C+\mathrm{rank}A=r$，则 $r$ 的最大值是 +(A) $11$ +(B) $12$ +(C) $13$ +(D) $14$ + +已知 $A,B,C$ 都是 $n$ 阶非零矩阵，且 $ABC=O$，$E$ 是 $n$ 阶单位矩阵，记 $\begin{bmatrix}O&A\\BC&E\end{bmatrix}\begin{bmatrix}AB&C\\O&E\end{bmatrix},\begin{bmatrix}E&AB\\AB&O\end{bmatrix}$ 的秩分别是 $r_1,r_2,r_3$，则 +(A) $r_1 \le r_2 \le r_3$ +(B) $r_1 \le r_3 \le r_2$ +(C) $r_3 \le r_1 \le r_2$ +(D) $r_2 \le r_1 \le r_3$ + +设 $A,B$ 都是 $n$ 阶矩阵，求证：$\mathrm{rank}(AB-E) \le \mathrm{rank}(A-E)+\mathrm{rank}(B-E)$ + +设 $A$ 为 $n$ 阶矩阵，$1<r<n$，记 $s=\mathrm{rank}(\begin{bmatrix}E_r&O\\O&O\end{bmatrix}A)$，则 +(A) $s=r$ +(B) $s=\max{\{r,\mathrm{rank}A\}}$ +(C) $s\le\min{\{r,\mathrm{rank}A\}}$ +(D) $s=\mathrm{rank}A$ + +设 $A=\begin{bmatrix}3&1&2\\2&a&1\\1&-1&2\end{bmatrix}, B\in\mathbb{R}^{3\times 2}$ 是一个列满秩矩阵. +(1) 证明 $\mathrm{rank}(AB) \ge 1$ ; +(2) 若 $\mathrm{rank}(AB)=1$，求参数 $a$ 的值，并给出一个使此式成立的矩阵 $B$ ; +(3) 对于 (2) 给出的参数 $a$ 的值，举例说明存在这样的矩阵 $B$，使 $\mathrm{rank}(AB)=2$ . + +设 $A$ 是 $n$ 阶方阵，$A=A_1A_2A_3$，且 $A_i^2=A_i\ (i=1,2,3)$，证：$\mathrm{rank}(E-A)\le 3(n-\mathrm{rank}A)$ . + +已知 $A,B$ 均为 $m\times n$ 阶矩阵，$\beta_1,\beta_2$ 为 $m$ 维列向量，则下列选项中正确的有 +(A) 若 $\mathrm{rank}A=m$，则对于任意 $m$ 维列向量 $b$，$Ax=b$ 总有解. +(B) 若 $A$ 与 $B$ 等价，则齐次方程组 $Ax=0$ 与 $Bx=0$ 同解. +(C) 矩阵方程 $AX=B$ 有解，但 $BY=A$ 无界的充要条件是$\mathrm{rank}B<\mathrm{rank}A=\mathrm{rank}\begin{bmatrix}A&B\end{bmatrix}$ +(D) 线性方程组 $Ax=\beta_1$ 与 $Ax=\beta_2$ 同时有解当且仅当$\mathrm{rank}A=\mathrm{rank}\begin{bmatrix}A&\beta_1&\beta_2\end{bmatrix}$ + +（[[1231线性代数考试卷]]）已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$与$\text{(II)} \quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + +（[[1231线性代数考试卷]]）设矩阵$A = \begin{bmatrix}1 & a_1 & a_1^2 & a_1^3 \\1 & a_2 & a_2^2 & a_2^3 \\1 & a_3 & a_3^2 & a_3^3 \\1 & a_4 & a_4^2 & a_4^3\end{bmatrix},x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix},b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix},$其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为$\underline{\qquad\qquad\qquad\qquad}.$ + +（[[1231线性代数考试卷]]）设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ + +（[[1231线性代数考试卷]]）设$A=\begin{bmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6\end{bmatrix},B=\begin{bmatrix}1 & 0 & 1 & 2 \\ 1 & -1 & a & a-1 \\ 2 & -3 & 2 & -2\end{bmatrix}$,向量$\alpha=\begin{bmatrix}0\\2\\3\end{bmatrix},\beta=\begin{bmatrix}1\\0\\-1\end{bmatrix}$. + (1)证明：方程组 $Ax=\alpha$ 的解均为方程组 $Bx=\beta$ 的解； + (2)若方程组 $Ax=\alpha$ 与方程组 $Bx=\beta$ 不同解，求 $a$ 的值. + + （[[1231线性代数考试卷]]）设矩阵$A=\begin{bmatrix}1&2&1&2\\0&1&t&t\\1&t&0&1\end{bmatrix}$，齐次线性方程组 $Ax=0$ 的基础解系中含有两个解向量，求 $Ax=0$ 的通解。 \ No newline at end of file -- 2.34.1 From a33d99607614052afa0f9798c18e3f5980a56269 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E4=B8=8D=E8=B4=9F=E6=98=9F=E5=85=89?= <wjx070401@qq.com> Date: Mon, 12 Jan 2026 00:25:03 +0800 Subject: [PATCH 194/274] =?UTF-8?q?=E6=96=B0=E5=A2=9E=E7=BA=BF=E6=80=A7?= =?UTF-8?q?=E6=96=B9=E7=A8=8B=E7=BB=84=E5=90=8C=E8=A7=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/线性方程组同解.md | 33 +++++++++++++++++++++++++++++++++ 1 file changed, 33 insertions(+) create mode 100644 素材/线性方程组同解.md diff --git a/素材/线性方程组同解.md b/素材/线性方程组同解.md new file mode 100644 index 0000000..8b4f5ab --- /dev/null +++ b/素材/线性方程组同解.md @@ -0,0 +1,33 @@ +## **$Ax=0$与$Bx=0$同解问题**： +充要条件：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +$Ax=\alpha$ 与$Bx=\beta$同解问题： +充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. +如何理解（非严格证明，目的是便于理解）： +首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， +考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， +可以得到$N(A)\subset N(B)$,以及$N(B)\subset N(A)$. +$N(A)\subset N(B)$可以得到什么呢？ +说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格，换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$可以得到$Ax=0$的每个方程是由$Bx=0$的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. +需要注意的是,这个条件是充要的.非常的好用. +非齐次的时候同理. + + + + + + + + + + + + + + + + + + + + -- 2.34.1 From 2d12d1c923ea1f0270886e1b82725ee0848d0c1d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Mon, 12 Jan 2026 00:38:26 +0800 Subject: [PATCH 195/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 线性方程组的系数矩阵与解关系.md | 113 +++++++++++++++++- 1 file changed, 111 insertions(+), 2 deletions(-) diff --git a/线性方程组的系数矩阵与解关系.md b/线性方程组的系数矩阵与解关系.md index 3e40a9a..693d93a 100644 --- a/线性方程组的系数矩阵与解关系.md +++ b/线性方程组的系数矩阵与解关系.md @@ -3,7 +3,116 @@ >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ ->[!example] 例一： -设 $A=$ +已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. + $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ +>解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ +>故 $Ax=0$ 的解空间维数是 $1$ （5分） +>$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） +>$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） +>解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 + + + 设 +$$ +A = \begin{bmatrix} +1 & -1 & 0 & -1 \\ +1 & 1 & 0 & 3 \\ +2 & 1 & 2 & 6 +\end{bmatrix}, \quad +B = \begin{bmatrix} +1 & 0 & 1 & 2 \\ +1 & -1 & a & a-1 \\ +2 & -3 & 2 & -2 +\end{bmatrix}, +\quad +\alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad +\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} +$$ + +(1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； + +(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 +4. (10分) 设 +$$ +A = \begin{bmatrix} +1 & -1 & 0 & -1 \\ +1 & 1 & 0 & 3 \\ +2 & 1 & 2 & 6 +\end{bmatrix}, \quad +B = \begin{bmatrix} +1 & 0 & 1 & 2 \\ +1 & -1 & a & a-1 \\ +2 & -3 & 2 & -2 +\end{bmatrix}, +$$ +向量 +$$ +\alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad +\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}. +$$ + +(1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； + +(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 + +--- + +**解：** + +(1) 由于 +$$ +\left( \begin{array}{c} +A \quad \alpha \\ +B \quad \beta +\end{array} \right) = +\left( \begin{array}{ccccc} +1 & -1 & 0 & -1 & 0 \\ +1 & 1 & 0 & 3 & 2 \\ +2 & 1 & 2 & 6 & 3 \\ +1 & 0 & 1 & 2 & 1 \\ +1 & -1 & a & a-1 & 0 \\ +2 & -3 & 2 & -2 & -1 +\end{array} \right) +$$ +$$ +\rightarrow +\left( \begin{array}{ccccc} +1 & -1 & 0 & -1 & 0 \\ +0 & 1 & 0 & 2 & 1 \\ +0 & 0 & 2 & 2 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 +\end{array} \right), +$$ +故 +$$ +R \left( \begin{array}{c} +A \quad \alpha \\ +B \quad \beta +\end{array} \right) = R(A, \alpha), +$$ +从而方程组 +$$ +\begin{cases} +Ax = \alpha, \\ +Bx = \beta +\end{cases} +$$ +与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 + +(2) 由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 +$$ +4 - R(A) < 4 - R(B), +$$ +故 $R(A) > R(B)$。又因 $R(A) = 3$，故 $R(B) < 3$，则 +$$ +\left| \begin{array}{ccc} +1 & 0 & 1 \\ +1 & -1 & a \\ +2 & -3 & 2 +\end{array} \right| = 0, +$$ +解得 $a = 1$。 \ No newline at end of file -- 2.34.1 From bf8da1f7179057333e11c2b30be4818f5a8d74e4 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 12 Jan 2026 00:45:03 +0800 Subject: [PATCH 196/274] vault backup: 2026-01-12 00:45:03 --- .../线性方程组的系数矩阵与解关系.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) rename 线性方程组的系数矩阵与解关系.md => 素材/线性方程组的系数矩阵与解关系.md (99%) diff --git a/线性方程组的系数矩阵与解关系.md b/素材/线性方程组的系数矩阵与解关系.md similarity index 99% rename from 线性方程组的系数矩阵与解关系.md rename to 素材/线性方程组的系数矩阵与解关系.md index 693d93a..d2ae907 100644 --- a/线性方程组的系数矩阵与解关系.md +++ b/素材/线性方程组的系数矩阵与解关系.md @@ -1,10 +1,10 @@ 在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： + >[!note] 定理1： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ - 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ >解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ -- 2.34.1 From 99b5a6644c3521110b07a6fa846c8a7e15fa2d1a Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 12 Jan 2026 00:49:30 +0800 Subject: [PATCH 197/274] =?UTF-8?q?=E7=94=A8=E7=A7=A9=E7=9A=84=E4=B8=8D?= =?UTF-8?q?=E7=AD=89=E5=BC=8F=E2=80=9C=E5=A4=B9=E9=80=BC=E2=80=9D=E5=87=BA?= =?UTF-8?q?=E7=A1=AE=E5=88=87=E5=80=BC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/用秩的不等式“夹逼”出确切值.md | 9 +++++++++ 1 file changed, 9 insertions(+) create mode 100644 素材/用秩的不等式“夹逼”出确切值.md diff --git a/素材/用秩的不等式“夹逼”出确切值.md b/素材/用秩的不等式“夹逼”出确切值.md new file mode 100644 index 0000000..60acc1f --- /dev/null +++ b/素材/用秩的不等式“夹逼”出确切值.md @@ -0,0 +1,9 @@ + +>[!information] 做题思路 +>通过矩阵的秩的不等式，最大限度限制所求的表达式的取值范围，或者将其**限制到一个具体的值**. +>在希望求一个矩阵的秩的确切值时，也可以考虑用不等式关系来“夹逼”，常见的不等式： +>1. $\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB})\le\min{\{\mathrm{rank}\boldsymbol{A}, \mathrm{rank}\boldsymbol{B}\}}$ +>2. $\mathrm{rank}(\boldsymbol{A+B})<\mathrm{rank}\boldsymbol A+\mathrm{rank}\boldsymbol B$ +>3. 矩阵加边不会减小秩； +> +>特别的，在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ \ No newline at end of file -- 2.34.1 From c53e61bb518e7708f62a1d855a3282a0ac87c2a8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Mon, 12 Jan 2026 01:30:53 +0800 Subject: [PATCH 198/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/线性方程组的系数矩阵与解关系.md | 7 ++++--- 1 file changed, 4 insertions(+), 3 deletions(-) diff --git a/素材/线性方程组的系数矩阵与解关系.md b/素材/线性方程组的系数矩阵与解关系.md index d2ae907..ef43f52 100644 --- a/素材/线性方程组的系数矩阵与解关系.md +++ b/素材/线性方程组的系数矩阵与解关系.md @@ -1,12 +1,12 @@ -在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： ->[!note] 定理1： +>[!note] 解零度化定理： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ +>分析：在求解非齐次方程组通解的题目中，若是题目给出了特解与齐次方程组的解，那么大概率来说这个齐次方程组的解就可以拓展为齐次方程组通解（根据问题导向，不然写不出来了），那么如何由齐次方程组的解拓展为齐次方程组通解呢，那就要根据题目具体的条件进行分析了，这就要用到我们的解零度化定理来求齐次方程组解空间的维数 >解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ >故 $Ax=0$ 的解空间维数是 $1$ （5分） >$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） @@ -103,7 +103,8 @@ Bx = \beta $$ 与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 -(2) 由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 +(2) 分析： +由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 $$ 4 - R(A) < 4 - R(B), $$ -- 2.34.1 From 26f71bf895b7c7a9d3ba2365af6ff17115309522 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 12 Jan 2026 10:33:11 +0800 Subject: [PATCH 199/274] vault backup: 2026-01-12 10:33:11 --- .../讲义/线性方程组的解与秩的不等式.md | 6 ++++++ 1 file changed, 6 insertions(+) create mode 100644 编写小组/讲义/线性方程组的解与秩的不等式.md diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md new file mode 100644 index 0000000..31fdb0e --- /dev/null +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -0,0 +1,6 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 -- 2.34.1 From 15e01864220be3600ce263b59c8e745b1e7de125 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 12 Jan 2026 11:03:38 +0800 Subject: [PATCH 200/274] vault backup: 2026-01-12 11:03:38 --- ...线性方程组的解与秩的不等式.md | 298 ++++++++++++++++++ .../讲义/线性方程组解的问题.md | 8 - 2 files changed, 298 insertions(+), 8 deletions(-) delete mode 100644 编写小组/讲义/线性方程组解的问题.md diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 31fdb0e..0b8a51e 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -4,3 +4,301 @@ tags: --- **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 + +# 单方程组解的问题 + +### 线性方程组解的判定 + +对非齐次线性方程组 $A_{m\times n}x=b$, +1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； +2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； +3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 +注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 + +把以上结论应用到齐次线性方程组，可得 +推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 + +### 矩阵方程解的判定 + +本质上和线性方程组是一脉相承的，只是形式上更一般化。 +最常见的矩阵方程是 $\boldsymbol{AX} = \boldsymbol{B}$，其中$\boldsymbol{A}$是 $m\times n$ 矩阵，$\boldsymbol{B}$ 是 $m\times p$ 矩阵，$\boldsymbol{X}$ 是待求的$n\times p$矩阵。 +1. 有解的充要条件： +矩阵方程有解的充要条件是系数矩阵 $\boldsymbol{A}$ 的秩等于增广矩阵 $[\boldsymbol{A} \ \boldsymbol{B}]$的秩，即： +$$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}])$$ +这个结论和非齐次线性方程组有解的条件完全一致。 + +理解上可以将 $B$ 拆分成一列列 $b$ ，从而化归为上面的线性方程组问题 + +2. 解的结构： +- 唯一解：当$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) = n$ 时，方程有唯一解。 +- 无穷多解：当 $r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) < n$ 时，方程有无穷多解。 + +可逆矩阵 +- 当 $\boldsymbol{A}$ 是 n 阶可逆矩阵时，矩阵方程 $\boldsymbol{AX} = \boldsymbol{B}$有唯一解:$\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{B}$ + +其他形式的矩阵方程 +- 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法，或类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} =\text{rank}A$ +- 对于 $\boldsymbol{AXB} = \boldsymbol{C}$ 形式的方程，当 $\boldsymbol{A} 和 \boldsymbol{B}$ 都可逆时，有唯一解 $\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{C}\boldsymbol{B}^{-1}$。 + +>[!example] **例1** +>设矩阵 +>$$A = \begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix}, +\quad +x = \begin{bmatrix} +x_1 \\ x_2 \\ x_3 \\ x_4 +\end{bmatrix}, +\quad +b = \begin{bmatrix} +1 \\ 1 \\ 1 \\ 1 +\end{bmatrix}$$ +其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为 ______________。 + +**答**：$(1,0,0,0)^T$。 + +**解析**：由范德蒙行列式的性质可知 $|A| \neq 0$，从而线性方程组 $Ax = b$ 有唯一解。 + +又由 +$$ +\begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix} +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} += +\begin{bmatrix} +1 \\ +1 \\ +1 \\ +1 +\end{bmatrix} +$$ + 可知 $Ax = b$ 的解为 +$$ +\begin{bmatrix} +1 \\ +0 \\ +0 \\ +0 +\end{bmatrix} +$$ + +>[!example] **例2** +> 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ + +**解析**： +类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ +# 多方程组的问题（线性方程组同解） + +## **$Ax=0$与$Bx=0$同解问题**： + +充要条件：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +$Ax=\alpha$ 与$Bx=\beta$同解问题： +充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. + +如何理解（非严格证明，目的是便于理解）： +首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， +考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， +可以得到$N(A)\subset N(B)$,以及$N(B)\subset N(A)$. +$N(A)\subset N(B)$可以得到什么呢？ + +说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格。换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$ 可以得到 $Ax=0$ 的每个方程是由 $Bx=0$ 的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. + +另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. + +需要注意的是,这个条件是充要的.非常的好用. +非齐次的时候同理. + +注意：由此，我们还能得到一些别的结论 +例如：$A$ 和 $B$ 等价（可以通过初等变换得到），并不能得到两方程同解，因为等价的初等变换可能包括初等列变换，而列变换可能改变两方程的解 + +>[!example] 例1 +>6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$ 与$\quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + +解析：类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ +# 线性方程组的系数矩阵与解关系 + +在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： + +>[!note] 定理1： +>对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 +> $$\dim N(\boldsymbol{A})=n-r$$ + +> [!example] 例1 +> 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. + $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ + +>解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ +>故 $Ax=0$ 的解空间维数是 $1$ （5分） +>$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） +>$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） +>解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 + +> [!example] 例2 +> 设 $$ +A = \begin{bmatrix} +1 & -1 & 0 & -1 \\ +1 & 1 & 0 & 3 \\ +2 & 1 & 2 & 6 +\end{bmatrix}, \quad +B = \begin{bmatrix} +1 & 0 & 1 & 2 \\ +1 & -1 & a & a-1 \\ +2 & -3 & 2 & -2 +\end{bmatrix},$$ +向量 $$ +\alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad +\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.$$ + +(1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； + +(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 + +--- + +**解：** + +(1) 由于 +$$ +\left( \begin{array}{c} +A \quad \alpha \\ +B \quad \beta +\end{array} \right) = +\left( \begin{array}{ccccc} +1 & -1 & 0 & -1 & 0 \\ +1 & 1 & 0 & 3 & 2 \\ +2 & 1 & 2 & 6 & 3 \\ +1 & 0 & 1 & 2 & 1 \\ +1 & -1 & a & a-1 & 0 \\ +2 & -3 & 2 & -2 & -1 +\end{array} \right) +$$ +$$ +\rightarrow +\left( \begin{array}{ccccc} +1 & -1 & 0 & -1 & 0 \\ +0 & 1 & 0 & 2 & 1 \\ +0 & 0 & 2 & 2 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 & 0 +\end{array} \right), +$$ +故 +$$ +R \left( \begin{array}{c} +A \quad \alpha \\ +B \quad \beta +\end{array} \right) = R(A, \alpha), +$$ +从而方程组 +$$ +\begin{cases} +Ax = \alpha, \\ +Bx = \beta +\end{cases} +$$ +与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 + +(2) 由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 +$$ +4 - R(A) < 4 - R(B), +$$ +故 $R(A) > R(B)$。又因 $R(A) = 3$，故 $R(B) < 3$，则 +$$ +\left| \begin{array}{ccc} +1 & 0 & 1 \\ +1 & -1 & a \\ +2 & -3 & 2 +\end{array} \right| = 0, +$$ +解得 $a = 1$。 + +# 秩的不等式 + +### 1. 和的秩不超过秩的和 + +设 $A, B$ 为同型矩阵，则 +$$ \operatorname{rank}(A+B) \leq \operatorname{rank} A + \operatorname{rank} B $$ + +### 2. 积的秩不超过任何因子的秩 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \leq \min\{\operatorname{rank} A, \operatorname{rank} B\} $$ + +### 3. 重要不等式 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \geq \operatorname{rank} A + \operatorname{rank} B - n $$ +特别地，当 $AB = 0$ 时，有 $\operatorname{rank} A + \operatorname{rank} B \leq n$。 + +### 4. 分块式 + +设 $A_{n \times n}$， $B_{n \times n}$，则 + +$$(1)\ \mathrm{rank} + +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } A, \quad \text{rank } +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } B +$$ + +$$(2)\ \mathrm{rank} +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } A + \text{rank } B +$$ + +$$(3)\ \mathrm{rank} +\begin{bmatrix} +A & E_n \\ +0 & B +\end{bmatrix} \geq \text{rank } A + \text{rank } B +$$ + +$$(4)\ \mathrm{rank} +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A & B \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A + B & B \\ +B & B +\end{bmatrix} \geq \text{rank } (A + B) +$$ + +>[!information] 思路1 +>通过矩阵的秩的不等式，最大限度限制所求的表达式的取值范围，或者将其**限制到一个具体的值**. +>在希望求一个矩阵的秩的确切值时，也可以考虑用不等式关系来“夹逼”，常见的不等式： +>1. $\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB})\le\min{\{\mathrm{rank}\boldsymbol{A}, \mathrm{rank}\boldsymbol{B}\}}$ +>2. $\mathrm{rank}(\boldsymbol{A+B})<\mathrm{rank}\boldsymbol A+\mathrm{rank}\boldsymbol B$ +>3. 矩阵加边不会减小秩； +> + +> [!note] 思路2 +> 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ \ No newline at end of file diff --git a/编写小组/讲义/线性方程组解的问题.md b/编写小组/讲义/线性方程组解的问题.md deleted file mode 100644 index ce84b16..0000000 --- a/编写小组/讲义/线性方程组解的问题.md +++ /dev/null @@ -1,8 +0,0 @@ -对非齐次线性方程组 $A_{m\times n}x=b$, -1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； -2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； -3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 -注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 - -把以上结论应用到齐次线性方程组，可得 -推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 \ No newline at end of file -- 2.34.1 From 0e586aade933951d8f76f7d3203baeb22c774c96 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 12 Jan 2026 11:46:25 +0800 Subject: [PATCH 201/274] vault backup: 2026-01-12 11:46:25 --- ...线性方程组的解与秩的不等式.md | 27 ++++++++++++++++--- 1 file changed, 24 insertions(+), 3 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 0b8a51e..e68f0cc 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -164,9 +164,7 @@ B = \begin{bmatrix} 向量 $$ \alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad \beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.$$ - (1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； - (2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 --- @@ -301,4 +299,27 @@ $$ > > [!note] 思路2 -> 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ \ No newline at end of file +> 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ + + + +>[!example] 例3 +>已知$\boldsymbol{A},\boldsymbol{B}$均为$m\times n$矩阵，$\beta_1,\beta_2$为$m$维列向量，则下列选项正确的有[ ] +(A)若$\mathrm{rank}\boldsymbol{A}=m$，则对于任意$m$维列向量$\boldsymbol{b},\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解. +(B)若$\boldsymbol{A}$与$\boldsymbol{B}$等价，则齐次线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$与$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$同解. +(C)矩阵方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，但$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解的充要条件是$$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ B}].$$ +(D)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解当且仅当$$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}].$$ + +**解：** +(A)一方面$\mathrm{rank}[\boldsymbol{A\ b}]\ge \mathrm{rank}\boldsymbol{A}=m$,另一方面矩阵$[\boldsymbol{A\ b}]$只有$m$行，所以它的秩必然不大于$m$，所以$\mathrm{rank}[\boldsymbol{A\ b}]=m=\mathrm{rank}\boldsymbol{A}$，即方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解。 + +(B)等价的矩阵只需要是经过初等变换可以变成同一个矩阵就行了，但齐次线性方程组同解需要只经过初等行变换就能变成同一个矩阵才行，后一个条件明显更强，所以后一种更“难”达成，B就不对。 + +(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出C正确。 + +(D)我们同样有两种方法去解这道题，一种是形式化的、严谨的，另一种是理解性的、直观的。 +1)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解$\Leftrightarrow\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]=\mathrm{rank}[\boldsymbol{A\ \beta_2}]$,故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。 +2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 +3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 + +根据上面的题目，我们可不可以归纳出一种比较普遍的方式，去解决这种与秩和方程组解都有密切关系的题目呢？ \ No newline at end of file -- 2.34.1 From 48092f9a9c0429baaf08d27f9765d3a8d5a786c9 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Mon, 12 Jan 2026 11:51:26 +0800 Subject: [PATCH 202/274] vault backup: 2026-01-12 11:51:26 --- ... => 线性方程组的解与秩的不等式（解析版）.md} | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 编写小组/讲义/{线性方程组的解与秩的不等式.md => 线性方程组的解与秩的不等式（解析版）.md} (100%) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md similarity index 100% rename from 编写小组/讲义/线性方程组的解与秩的不等式.md rename to 编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md -- 2.34.1 From 0ed7438a95e6952ed92883be9b485ae54b9f90be Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Mon, 12 Jan 2026 11:57:25 +0800 Subject: [PATCH 203/274] vault backup: 2026-01-12 11:57:25 --- 笔记分享/LaTeX（KaTeX）输入规范.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md index 84987d5..9674991 100644 --- a/笔记分享/LaTeX（KaTeX）输入规范.md +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -3,4 +3,5 @@ 2. 自然常数或电荷量e、虚数单位i应当为**正体**，需要用mathrm记号包裹，例：$\mathrm{e}^{\mathrm{i}\pi}+1=0$；然而，当e,i作为变量时，应当用正常的斜体。例：$\sum\limits_{i=1}^{n}a_i$ 3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ 4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ -5. \$\$双美元符号之间不要打回车！除非你有\begin{...}\end{...}\$\$ \ No newline at end of file +5. \$\$双美元符号之间不要打回车！除非你有\begin{...}\end{...}\$\$ +6. 矩阵和向量要加粗，用\boldsymbol{}，比如$\boldsymbol{A},\boldsymbol{x}$。 \ No newline at end of file -- 2.34.1 From cc8f32434baa1a6f1cc2a527ef9ad0729ddeb8c6 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 12 Jan 2026 13:00:15 +0800 Subject: [PATCH 204/274] =?UTF-8?q?1.14=E7=BA=BF=E4=BB=A3=E9=99=90?= =?UTF-8?q?=E6=97=B6=E7=BB=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1.14线代限时练.md | 24 +++++++++++++++++++ 1 file changed, 24 insertions(+) create mode 100644 编写小组/试卷/线代期末复习模拟/1.14线代限时练.md diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md new file mode 100644 index 0000000..6c5de66 --- /dev/null +++ b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md @@ -0,0 +1,24 @@ +1. （2013秋A·三）设 + $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$证明：$D_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]$ . + +2. 已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. + +3. 设 $\boldsymbol{E}$ 为 $3$ 阶单位矩阵，$\boldsymbol\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $\boldsymbol E - \boldsymbol\alpha\boldsymbol\alpha^\mathrm{T}$ 的全部 $3$ 个特征值为\_\_\_\_\_\_\_. + +4. 设 $n$ 阶矩阵 $A=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^n(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2$ 的矩阵为\_\_\_\_\_\_\_. + +5. 若 $n$ 阶实对称矩阵 $\boldsymbol A$ 的特征值为 $\lambda_i=(-1)^i\ (i=1,2,\cdots,n)$，则 $\boldsymbol A^{100}=$ \_\_\_\_\_\_\_. + +6. 已知 $n(n\ge2)$ 维列向量 $\boldsymbol\alpha,\boldsymbol\beta$ 满足 $\boldsymbol\beta^\mathrm{T}\boldsymbol\alpha=-3$，则方阵 $(\boldsymbol{\beta\alpha}^\mathrm{T})^2$ 的非零特征值为\_\_\_\_\_\_\_\_\_\_. + +7. 已知向量组 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3$ 线性无关，其中 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3\in\mathbb{R}^3$，$\boldsymbol A$ 为 $3$ 阶方阵，且 +$$\boldsymbol{A\alpha_1}=2\boldsymbol\alpha_1-\boldsymbol\alpha_2-\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_2=\boldsymbol\alpha_1+2\boldsymbol\alpha_2+3\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_3=2\boldsymbol\alpha_1+4\boldsymbol\alpha_2+\boldsymbol\alpha_3.$$ + +(1) 证明 $A\boldsymbol\alpha_1, A\boldsymbol\alpha_2, A\boldsymbol\alpha_3$ 线性无关； +(2) 计算行列式 $\boldsymbol E-\boldsymbol A$，其中 $\boldsymbol E$ 是 $3$ 阶单位矩阵. + +8. 已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. + + + +9. 求 $n$ 阶方阵 $\boldsymbol{A} = \begin{bmatrix}1 & 1 & 1 & \dots & 1 \\1 & 0 & 1 & \dots & 1 \\1 & 1 & 0 & \dots & 1 \\\vdots & \vdots & \vdots & \ddots & \vdots \\1 & 1 & 1 & \dots & 0\end{bmatrix}$ 的逆。 -- 2.34.1 From 71f8aecb30a4f6c4aa224e328ae01ea2efbd5b9a Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 12 Jan 2026 13:53:41 +0800 Subject: [PATCH 205/274] =?UTF-8?q?=E5=88=9B=E5=BB=BA=E8=AF=95=E5=8D=B7?= =?UTF-8?q?=E5=BA=93?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1.14线代限时练.md | 64 ++++++++++++--- 试卷库/线性代数/线代2013秋A.md | 65 ++++++++++++++++ 试卷库/线性代数/线代2018秋A.md | 77 +++++++++++++++++++ 试卷库/线性代数/线代2019秋A.md | 65 ++++++++++++++++ 试卷库/线性代数/线代2021秋A.md | 3 + 试卷库/线性代数/线代2022秋A.md | 64 +++++++++++++++ 6 files changed, 327 insertions(+), 11 deletions(-) create mode 100644 试卷库/线性代数/线代2013秋A.md create mode 100644 试卷库/线性代数/线代2018秋A.md create mode 100644 试卷库/线性代数/线代2019秋A.md create mode 100644 试卷库/线性代数/线代2021秋A.md create mode 100644 试卷库/线性代数/线代2022秋A.md diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md index 6c5de66..df0efeb 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md @@ -1,24 +1,66 @@ -1. （2013秋A·三）设 - $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$证明：$D_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]$ . +1. 已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. -2. 已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. +2. （[[线代2013秋A]]·4）已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. -3. 设 $\boldsymbol{E}$ 为 $3$ 阶单位矩阵，$\boldsymbol\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $\boldsymbol E - \boldsymbol\alpha\boldsymbol\alpha^\mathrm{T}$ 的全部 $3$ 个特征值为\_\_\_\_\_\_\_. +3. （[[线代2019秋A]]·9）设 $\boldsymbol{E}$ 为 $3$ 阶单位矩阵，$\boldsymbol\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $\boldsymbol E - \boldsymbol\alpha\boldsymbol\alpha^\mathrm{T}$ 的全部 $3$ 个特征值为\_\_\_\_\_\_\_. -4. 设 $n$ 阶矩阵 $A=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^n(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2$ 的矩阵为\_\_\_\_\_\_\_. +4. （[[线代2018秋A]]·12）设 $n$ 阶矩阵 $A=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^n(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2$ 的矩阵为\_\_\_\_\_\_\_. -5. 若 $n$ 阶实对称矩阵 $\boldsymbol A$ 的特征值为 $\lambda_i=(-1)^i\ (i=1,2,\cdots,n)$，则 $\boldsymbol A^{100}=$ \_\_\_\_\_\_\_. +5. （[[线代2018秋A]]·11）若 $n$ 阶实对称矩阵 $\boldsymbol A$ 的特征值为 $\lambda_i=(-1)^i\ (i=1,2,\cdots,n)$，则 $\boldsymbol A^{100}=$ \_\_\_\_\_\_\_. -6. 已知 $n(n\ge2)$ 维列向量 $\boldsymbol\alpha,\boldsymbol\beta$ 满足 $\boldsymbol\beta^\mathrm{T}\boldsymbol\alpha=-3$，则方阵 $(\boldsymbol{\beta\alpha}^\mathrm{T})^2$ 的非零特征值为\_\_\_\_\_\_\_\_\_\_. +6. ([[线代2022秋A]]·5）已知 $n(n\ge2)$ 维列向量 $\boldsymbol\alpha,\boldsymbol\beta$ 满足 $\boldsymbol\beta^\mathrm{T}\boldsymbol\alpha=-3$，则方阵 $(\boldsymbol{\beta\alpha}^\mathrm{T})^2$ 的非零特征值为\_\_\_\_\_\_\_\_\_\_. -7. 已知向量组 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3$ 线性无关，其中 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3\in\mathbb{R}^3$，$\boldsymbol A$ 为 $3$ 阶方阵，且 +7. （[[线代2022秋A]]·七）已知向量组 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3$ 线性无关，其中 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3\in\mathbb{R}^3$，$\boldsymbol A$ 为 $3$ 阶方阵，且 $$\boldsymbol{A\alpha_1}=2\boldsymbol\alpha_1-\boldsymbol\alpha_2-\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_2=\boldsymbol\alpha_1+2\boldsymbol\alpha_2+3\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_3=2\boldsymbol\alpha_1+4\boldsymbol\alpha_2+\boldsymbol\alpha_3.$$ - (1) 证明 $A\boldsymbol\alpha_1, A\boldsymbol\alpha_2, A\boldsymbol\alpha_3$ 线性无关； (2) 计算行列式 $\boldsymbol E-\boldsymbol A$，其中 $\boldsymbol E$ 是 $3$ 阶单位矩阵. +```text + + + + + + + + + + + + + + + + + + + + + +``` + +8. （[[线代2013秋A]]）求 $n$ 阶方阵 $\boldsymbol{A} = \begin{bmatrix}1 & 1 & 1 & \dots & 1 \\1 & 0 & 1 & \dots & 1 \\1 & 1 & 0 & \dots & 1 \\\vdots & \vdots & \vdots & \ddots & \vdots \\1 & 1 & 1 & \dots & 0\end{bmatrix}$ 的逆。 +```text + + + + + + + + + + + + + + + + + -8. 已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. +``` -9. 求 $n$ 阶方阵 $\boldsymbol{A} = \begin{bmatrix}1 & 1 & 1 & \dots & 1 \\1 & 0 & 1 & \dots & 1 \\1 & 1 & 0 & \dots & 1 \\\vdots & \vdots & \vdots & \ddots & \vdots \\1 & 1 & 1 & \dots & 0\end{bmatrix}$ 的逆。 +9. （[[线代2013秋A]]·三）设 + $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$证明：$D_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]$ . \ No newline at end of file diff --git a/试卷库/线性代数/线代2013秋A.md b/试卷库/线性代数/线代2013秋A.md new file mode 100644 index 0000000..83209e6 --- /dev/null +++ b/试卷库/线性代数/线代2013秋A.md @@ -0,0 +1,65 @@ +## 一、填空题（共6小题，每小题3分，共18分） +1. 设行列式 $D=\begin{vmatrix}-1&2&-3\\1&2&0\\-1&3&2\end{vmatrix}$ ，则 $M_{12}+A_{21}-M_{32}=$ ______ +2. 设矩阵 $\boldsymbol{A}=\begin{bmatrix}\boldsymbol{B}&\boldsymbol{C}\\\boldsymbol{0}&\boldsymbol{D}\end{bmatrix}$ ，其中 $\boldsymbol{B}$ 、 $\boldsymbol{D}$ 皆为可逆矩阵，则 $\boldsymbol{A}^{-1}=$ ______ +3. 设矩阵 $\boldsymbol{A}=\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&-1&2\\0&0&-2&4\end{bmatrix}$ ， $n$ 为正整数，则 $\boldsymbol{A}^{n}=$ ______ +4. 已知向量空间 $V = \{(2a,2b,3b,3a)\mid a,b\in\mathbb{R}\}$ ，则 $V$ 的维数是______ +5. 已知矩阵 $\boldsymbol{A}=\begin{bmatrix}-2&1&1\\0&2&0\\-4&1&3\end{bmatrix}$ ，则 $\boldsymbol{A}$ 的特征值 $2$ 的几何重数是______ +6. 实二次型 $f(x_1,x_2,x_3)=2x_1x_2 - 2x_1x_3 + 2x_2x_3$ 的秩为______ + + +## 二、单选题（共6小题，每小题3分，共18分） +1. 在4阶行列式 $\det[a_{ij}]$ 的展开式中含有因子 $a_{31}$ 的项共有【】 + - (A) 4项 + - (B) 6项 + - (C) 8项 + - (D) 10项 +2. 设 $\boldsymbol{A}$ 、 $\boldsymbol{B}$ 是 $n$ 阶方阵，且 $\boldsymbol{B}$ 的第 $j$ 列元素全为零，则下列结论正确的是【】 + - (A) $\boldsymbol{AB}$ 的第 $j$ 列元素全等于零 + - (B) $\boldsymbol{AB}$ 的第 $j$ 行元素全等于零 + - (C) $\boldsymbol{BA}$ 的第 $j$ 列元素全等于零 + - (D) $\boldsymbol{BA}$ 的第 $j$ 行元素全等于零 +3. 设 $n$ 维向量组 $\boldsymbol{\alpha}_1$ 、 $\boldsymbol{\alpha}_2$ 、 $\boldsymbol{\alpha}_3$ 、 $\boldsymbol{\alpha}_4$ 、 $\boldsymbol{\alpha}_5$ 的秩为3，且满足 $\boldsymbol{\alpha}_1 + 2\boldsymbol{\alpha}_3 - 3\boldsymbol{\alpha}_5=\boldsymbol{0}$ ， $\boldsymbol{\alpha}_2 = 2\boldsymbol{\alpha}_4$ ，则该向量组的一个极大线性无关组为【】 + - (A) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_5$ + - (B) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_4$ + - (C) $\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3,\boldsymbol{\alpha}_5$ + - (D) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_3,\boldsymbol{\alpha}_5$ +4. 设 $\boldsymbol{A}$ 、 $\boldsymbol{B}$ 为 $n$ 阶方阵，给定以下命题：(1) $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 等价；(2) $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似；(3) $\boldsymbol{A}$ 、 $\boldsymbol{B}$ 的行向量组等价。下列命题正确的是【】 + - (A) (1) $\Rightarrow$ (2) $\Rightarrow$ (3) + - (B) (2) $\Rightarrow$ (1) $\Rightarrow$ (3) + - (C) (3) $\Rightarrow$ (2) $\Rightarrow$ (1) + - (D) 以上结论均不对 +1. 设矩阵 $\boldsymbol{A}\sim\boldsymbol{B}$ 、 $\boldsymbol{C}\sim\boldsymbol{D}$ ，则下列命题正确的是【】 + - (A) $\boldsymbol{A+B}\sim\boldsymbol{C+D}$ + - (B) $\boldsymbol{A-B}\sim\boldsymbol{C-D}$ + - (C) $\boldsymbol{A}^2\sim\boldsymbol{B}^2$ + - (D) $\boldsymbol{AB}\sim\boldsymbol{CD}$ +1. 如果 $\boldsymbol{A}$ 为反对称矩阵，那么 $\boldsymbol{B}=(\boldsymbol{E}-\boldsymbol{A})(\boldsymbol{E}+\boldsymbol{A})^{-1}$ 一定为【】 + - (A) 反对称矩阵 + - (B) 正交矩阵 + - (C) 对称矩阵 + - (D) 对角矩阵 + + +## 三、（10分） +设 $n$ 阶行列式 $D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$ ，证明： $D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$ + + +## 四、（10分） +求 $n$ 阶方阵 $\boldsymbol{A}$ （原题 $\boldsymbol{A}$ 具体元素排版混乱，推测可能为 $\boldsymbol{A}=\begin{bmatrix}1&1&0&\cdots&1&1\\1&1&1&\cdots&1&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\1&0&0&\cdots&1&1\end{bmatrix}$ ，具体以标准题型为准）的逆矩阵 $\boldsymbol{A}^{-1}$ + + +## 五、（10分） +求解非齐次线性方程组（原题方程排版混乱，整理后推测为）： + $\begin{cases}x_1 + 2x_2 + 3x_3 = 4\\2x_1 + 3x_2 + x_3 = -4\\3x_1 + 8x_2 - 2x_3 = 13\\4x_1 - x_2 + 9x_3 = -6\\x_1 - x_2 + 2x_3 = -5\end{cases}$ （具体方程以标准题型为准） + + +## 六、（10分） +设 $\boldsymbol{A}$ 、 $\boldsymbol{B}$ 为3阶矩阵， $\boldsymbol{A}$ 相似于 $\boldsymbol{B}$ ， $\lambda_1=-1$ 、 $\lambda_2=1$ 为 $\boldsymbol{A}$ 的两个特征值， $|\boldsymbol{B}^{-1}|=\frac{1}{3}$ ，求行列式 $\begin{vmatrix}-(\boldsymbol{A}-3\boldsymbol{E})^{-1}&\boldsymbol{0}\\\boldsymbol{0}&\boldsymbol{B}^*+\left(-\frac{1}{4}\boldsymbol{B}\right)^{-1}\end{vmatrix}$ （其中 $\boldsymbol{B}^*$ 为 $\boldsymbol{B}$ 的伴随矩阵） + + +## 七、（12分） +求一可逆线性变换 $\boldsymbol{x}=\boldsymbol{P}\boldsymbol{y}$ ，将二次型 $f=2x_1^2 + 9x_2^2 + 3x_3^2 + 8x_1x_2 - 4x_1x_3 - 10x_2x_3$ 化成二次型 $g=2y_1^2 + 3y_2^2 + 6y_3^2 - 4y_1y_2 - 4y_1y_3 + 8y_2y_3$ （或按“将 $f$ 化为标准形”的常规题型修正，具体以题目意图为准） + + +## 八、（12分） +设 $\boldsymbol{A}$ 是 $n$ 阶方阵，证明： $\boldsymbol{A}^2=\boldsymbol{E}$ 的充分必要条件是 $\mathrm{rank}(\boldsymbol{E}-\boldsymbol{A})+\mathrm{rank}(\boldsymbol{E}+\boldsymbol{A})=n$ （其中 $\mathrm{rank}$ 表示矩阵的秩） \ No newline at end of file diff --git a/试卷库/线性代数/线代2018秋A.md b/试卷库/线性代数/线代2018秋A.md new file mode 100644 index 0000000..19fbf41 --- /dev/null +++ b/试卷库/线性代数/线代2018秋A.md @@ -0,0 +1,77 @@ + +## 一、单选题(共6小题，每小题3分，共18分) +1. 设 $\boldsymbol{A}$ 为 $n$ 阶对称矩阵， $\boldsymbol{B}$ 为 $n$ 阶反对称矩阵，下列矩阵中为反对称矩阵的是【】 + (A) $\boldsymbol{AB}-\boldsymbol{BA}$ ； + (B) $\boldsymbol{AB}+\boldsymbol{BA}$ ； + (C) $\boldsymbol{BAB}$ ； + (D) $(\boldsymbol{AB})^2$ 。 + +2. 设 $\boldsymbol{A}$ ， $\boldsymbol{B}$ 是可逆矩阵，且 $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似，则下列结论错误的是【】 + (A) $\boldsymbol{A}^\mathrm{T}$ 与 $\boldsymbol{B}^\mathrm{T}$ 相似； + (B) $\boldsymbol{A}^{-1}$ 与 $\boldsymbol{B}^{-1}$ 相似； + (C) $\boldsymbol{A}+\boldsymbol{A}^\mathrm{T}$ 与 $\boldsymbol{B}+\boldsymbol{B}^\mathrm{T}$ 相似； + (D) $\boldsymbol{A}+\boldsymbol{A}^{-1}$ 与 $\boldsymbol{B}+\boldsymbol{B}^{-1}$ 相似。 + +3. 设向量组 $\boldsymbol{\alpha}_1=(0,0,c_1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_2=(0,1,c_2)^\mathrm{T}$ ， $\boldsymbol{\alpha}_3=(1,-1,c_3)^\mathrm{T}$ ， $\boldsymbol{\alpha}_4=(-1,1,c_4)^\mathrm{T}$ ，其中 $c_1$ ， $c_2$ ， $c_3$ ， $c_4$ 为任意常数，则下列向量组线性相关的是【】 + (A) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3$ ； + (B) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_4$ ； + (C) $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_3,\boldsymbol{\alpha}_4$ ； + (D) $\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3,\boldsymbol{\alpha}_4$ 。 + +4. 设 $\boldsymbol{A}$ ， $\boldsymbol{B}$ 为 $n$ 阶矩阵，则【】 + (A) $\mathrm{rank}\begin{bmatrix}\boldsymbol{A}&\boldsymbol{AB}\end{bmatrix}=\mathrm{rank}\boldsymbol{A}$ ； + (B) $\mathrm{rank}\begin{bmatrix}\boldsymbol{A}&\boldsymbol{BA}\end{bmatrix}=\mathrm{rank}\boldsymbol{A}$ ； + (C) $\mathrm{rank}\begin{bmatrix}\boldsymbol{A}&\boldsymbol{B}\end{bmatrix}=max\{\mathrm{rank}\boldsymbol{A},\mathrm{rank}\boldsymbol{B}\}$ ； + (D) $\mathrm{rank}\begin{bmatrix}\boldsymbol{A}&\boldsymbol{B}\end{bmatrix}=\mathrm{rank}\begin{bmatrix}\boldsymbol{A}^\mathrm{T}&\boldsymbol{B}^\mathrm{T}\end{bmatrix}$ 。 + +5. 设 $\boldsymbol{A}$ 可逆，将 $\boldsymbol{A}$ 的第一列加上第二列的2倍得到 $\boldsymbol{B}$ ，则 $\boldsymbol{A}^*$ 与 $\boldsymbol{B}^*$ 满足【】 + (A) 将 $\boldsymbol{A}^*$ 的第一列加上第二列的2倍得到 $\boldsymbol{B}^*$ ； + (B) 将 $\boldsymbol{A}^*$ 的第一行加上第二行的2倍得到 $\boldsymbol{B}^*$ ； + (C) 将 $\boldsymbol{A}^*$ 的第二列加上第一列的 $(-2)$ 倍得到 $\boldsymbol{B}^*$ ； + (D) 将 $\boldsymbol{A}^*$ 的第二行加上第一行的 $(-2)$ 倍得到 $\boldsymbol{B}^*$ 。 + +6. 设齐次线性方程组 + (I) $\begin{cases}x_1 + 2x_2 + 3x_3 = 0\\2x_1 + 3x_2 + 5x_3 = 0\\x_1 + x_2 + ax_3 = 0\end{cases}$ + (II) $\begin{cases}x_1 + bx_2 + cx_3 = 0\\2x_1 + b^2x_2 + (c + 1)x_3 = 0\end{cases}$ + 同解，则 $a,b,c$ 的值为【】 + (A) $a=1,b=0,c=1$ ； + (B) $a=1,b=1,c=2$ ； + (C) $a=2,b=0,c=1$ ； + (D) $a=2,b=1,c=2$ 。 + +## 二、填空题(共6小题，每小题3分，共18分) +7. 已知向量 $\boldsymbol{\alpha}_1=(1,0,-1,0)^\mathrm{T}$ ， $\boldsymbol{\alpha}_2=(1,1,-1,-1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_3=(-1,0,1,1)^\mathrm{T}$ ，则向量 $\boldsymbol{\alpha}_1 + 2\boldsymbol{\alpha}_2$ 与 $2\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_3$ 的内积 $<\boldsymbol{\alpha}_1 + 2\boldsymbol{\alpha}_2,2\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_3>=$ ________。 + +8. 设二阶矩阵 $\boldsymbol{A}$ 有两个相异特征值， $\boldsymbol{\alpha}_1$ ， $\boldsymbol{\alpha}_2$ 是 $\boldsymbol{A}$ 的线性无关的特征向量，且 $\boldsymbol{A}^2(\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_2)=\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_2$ ，则 $|\boldsymbol{A}|=$ ________。 + +9. 若向量组 $\boldsymbol{\alpha}_1=(1,0,1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_2=(0,1,1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_3=(1,3,5)^\mathrm{T}$ 不能由向量组 $\boldsymbol{\beta}_1=(1,1,1)^\mathrm{T}$ ， $\boldsymbol{\beta}_2=(1,2,3)^\mathrm{T}$ ， $\boldsymbol{\beta}_3=(3,4,a)^\mathrm{T}$ 线性表示，则 $a=$ ________。 + +10. 设矩阵 $\boldsymbol{A}=\begin{bmatrix}1&a_1&a_1^2&a_1^3\\1&a_2&a_2^2&a_2^3\\1&a_3&a_3^2&a_3^3\\1&a_4&a_4^2&a_4^3\end{bmatrix}$ ， $\boldsymbol{x}=\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$ ， $\boldsymbol{b}=\begin{bmatrix}1\\1\\1\\1\end{bmatrix}$ ，其中 $a_1,a_2,a_3,a_4$ 互不相同，则线性方程组 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的解为________。 + +11. 若 $n$ 阶实对称矩阵 $\boldsymbol{A}$ 的特征值为 $\lambda_i=(-1)^i(i=1,2,\cdots,n)$ ，则 $\boldsymbol{A}^{100}=$ ________。 + +12. 设 $n$ 阶矩阵 $\boldsymbol{A}=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum_{i=1}^n(a_{i1}x_1 + a_{i2}x_2 + \cdots + a_{in}x_n)^2$ 的矩阵为________。 + +## 三、计算与证明(共6小题，共64分) +13. (10分)计算 $n$ 阶行列式 $\begin{vmatrix}1&2&3&\cdots&n-1&n\\2&1&2&\cdots&n-2&n-1\\3&2&1&\cdots&n-3&n-2\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\n-1&n-2&n-3&\cdots&1&2\\n&n-1&n-2&\cdots&2&1\end{vmatrix}$ 。 + +14. (10分)设 $\boldsymbol{\alpha}_1=(1,0,-1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_2=(2,1,1)^\mathrm{T}$ ， $\boldsymbol{\alpha}_3=(1,1,1)^\mathrm{T}$ 和 $\boldsymbol{\beta}_1=(0,1,1)^\mathrm{T}$ ， $\boldsymbol{\beta}_2=(-1,1,0)^\mathrm{T}$ ， $\boldsymbol{\beta}_3=(0,2,1)^\mathrm{T}$ 是 $\mathbb{R}^3$ 的两组基，求向量 $\boldsymbol{u}=\boldsymbol{\alpha}_1 + 2\boldsymbol{\alpha}_2 - 3\boldsymbol{\alpha}_3$ 在基 $\boldsymbol{\beta}_1$ ， $\boldsymbol{\beta}_2$ ， $\boldsymbol{\beta}_3$ 下的坐标。 + +15. (10分)设实二次型 $f(x_1,x_2,x_3)=x_1^2 + x_2^2 + x_3^2 - 2x_1x_2 - 2x_1x_3 + 2ax_2x_3$ 通过正交变换可化为标准型 $f=2y_1^2 + 2y_2^2 + by_3^2$ 。 + (1)求 $a$ ， $b$ 及所用正交变换矩阵 $\boldsymbol{Q}$ ； + (2)证明 $\boldsymbol{A} + 2\boldsymbol{E}$ 为正定矩阵。 + +16. (10分)设三阶矩阵 $\boldsymbol{A}=[\boldsymbol{\alpha}_1\ \boldsymbol{\alpha}_2\ \boldsymbol{\alpha}_3]$ 有三个不同的特征值，且满足 $\boldsymbol{\alpha}_3=\boldsymbol{\alpha}_1 + 2\boldsymbol{\alpha}_2$ ， $\boldsymbol{\beta}=\boldsymbol{\alpha}_1 + \boldsymbol{\alpha}_2 + \boldsymbol{\alpha}_3$ 。 + (1)证明 $\mathrm{rank}\boldsymbol{A}=2$ ； + (2)求方程组 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta}$ 的通解。 + +17. (12分)设 $n$ 阶方阵 $\boldsymbol{A}$ ， $\boldsymbol{B}$ 满足 $\boldsymbol{AB}=\boldsymbol{A} + \boldsymbol{B}$ 。 + (1)证明 $\boldsymbol{A} - \boldsymbol{E}$ 可逆； + (2)证明 $\boldsymbol{AB}=\boldsymbol{BA}$ ； + (3)证明 $\mathrm{rank}\boldsymbol{A}=\mathrm{rank}\boldsymbol{B}$ ； + (4)若 $\boldsymbol{B}=\begin{bmatrix}1&-3&0\\2&1&0\\0&0&2\end{bmatrix}$ ，求矩阵 $\boldsymbol{A}$ 。 + +18. (12分)已知实矩阵 $\boldsymbol{A}=\begin{bmatrix}2&2\\2&a\end{bmatrix}$ ， $\boldsymbol{B}=\begin{bmatrix}4&b\\3&1\end{bmatrix}$ 。 + (1)证明矩阵方程 $\boldsymbol{AX}=\boldsymbol{B}$ 有解但 $\boldsymbol{BY}=\boldsymbol{A}$ 无解的充要条件是 $a\neq2$ ， $b=\frac{4}{3}$ ； + (2)证明矩阵 $\boldsymbol{A}$ 相似于 $\boldsymbol{B}$ 的充要条件是 $a=3$ ， $b=\frac{2}{3}$ ； + (3)证明矩阵 $\boldsymbol{A}$ 合同于 $\boldsymbol{B}$ 的充要条件是 $a<2$ ， $b=3$ 。 \ No newline at end of file diff --git a/试卷库/线性代数/线代2019秋A.md b/试卷库/线性代数/线代2019秋A.md new file mode 100644 index 0000000..ec00351 --- /dev/null +++ b/试卷库/线性代数/线代2019秋A.md @@ -0,0 +1,65 @@ + +## 一、单选题(共6小题,每小题3分,共18分) +1. n维向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, ..., \boldsymbol{\alpha}_{r}(3 \le r \le n)$ 线性无关的充要条件是【】 +A. 存在一组不全为零的数 $k_{1}, k_{2}, ..., k_{r}$ 使得 $\sum_{i=1}^{r} k_{i} \boldsymbol{\alpha}_{i} ≠0$ +B. $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, ..., \boldsymbol{\alpha}_{r}$ 中任意两个向量都线性无关 +C. $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, ..., \boldsymbol{\alpha}_{r}$ 中存在一个向量不能用其余向量线性表示 +D. $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, ..., \boldsymbol{\alpha}_{r}$ 中任意一个向量都不能用其余向量线性表示 + +2. 已知 $\boldsymbol{Q}$ 为n阶可逆矩阵， $\boldsymbol{P}$ 为n阶方阵，满足 $\boldsymbol{Q}\boldsymbol{P}=0$ ，则下列命题中正确的是【】 +A. $\mathrm{rank}(\boldsymbol{P})=0$ +B. $\boldsymbol{P}$ 可逆 +C. $\boldsymbol{P}$ 不可对角化 +D. 不存在这样的方阵 $\boldsymbol{P}$ + +3. 已知向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}$ 线性无关，则以下向量组中线性相关的是【】 +A. $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{4}$ +B. $\boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}-\boldsymbol{\alpha}_{1}$ +C. $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}-\boldsymbol{\alpha}_{1}$ +D. $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}, \boldsymbol{\alpha}_{3}-\boldsymbol{\alpha}_{4}, \boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{3}$ + +4. 设n阶非零方阵 $\boldsymbol{A}$ 满足 $\boldsymbol{A}^{2}=\boldsymbol{A}$ ，则下列命题中正确的是【】 +A. $\boldsymbol{A}$ 只有特征值1 +B. $\boldsymbol{A}$ 只有特征值0 +C. $\boldsymbol{A}$ 可对角化 +D. $\boldsymbol{A}$ 一定不可逆 + +5. 下列命题中正确的是【】 +A. 等价的矩阵必相似 +B. 合同的矩阵必相似 +C. 合同的矩阵必等价 +D. 等价的矩阵必合同 + +6. 设 $\boldsymbol{A}$ 是二次型 $f(x_{1}, x_{2}, ..., x_{n})$ 所对应的矩阵，则下列命题中正确的是【】 +A. 若 $f(x_{1}, x_{2}, ..., x_{n})$ 负定，则 $\boldsymbol{A}$ 的任意阶顺序主子式都大于零 +B. 若 $f(x_{1}, x_{2}, ..., x_{n})$ 正定，则 $\boldsymbol{A}$ 的任意阶顺序主子式都大于零 +C. 若 $f(x_{1}, x_{2}, ..., x_{n})$ 半负定，则 $\boldsymbol{A}$ 的任意阶顺序主子式都大于零 +D. 若 $f(x_{1}, x_{2}, ..., x_{n})$ 半正定，则 $\boldsymbol{A}$ 的任意阶顺序主子式都大于零 + +## 二、填空题(共6小题,每小题3分,共18分) +7. 如果 $\begin{vmatrix}1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 0 & 0 & 9 & x \\ 0 & 0 & 11 & 12\end{vmatrix}=0$ ，则 $x=$ _______ + +8. n阶方阵 $\begin{bmatrix}1 & a & \cdots & a \\ a & 1 & \cdots & a \\ \vdots & \vdots & \ddots & \vdots \\ a & a & \cdots & 1\end{bmatrix}$ 的秩为 $n - 1$ ，且 $n>2$ ，则 $a=$ _______ + +9. 设 $\boldsymbol{E}$ 为3阶单位矩阵， $\boldsymbol{\alpha}$ 为一个3维单位列向量，则矩阵 $\boldsymbol{E}-\boldsymbol{\alpha}\boldsymbol{\alpha}^{\mathrm{T}}$ 的全部3个特征值为_______ + +10. 设二次型 $f(x_{1}, x_{2}, x_{3})=x_{1}^{2}+4x_{2}^{2}+2x_{3}^{2}+2ax_{1}x_{2}+2x_{1}x_{3}$ 正定，则参数 $a$ 的取值范围是_______ + +11. $\begin{bmatrix}1 & a & 0 \\ 0 & 1 & a \\ 0 & 0 & 1\end{bmatrix}^{-1}=$ _______ + +12. 设 $\boldsymbol{A}=\begin{bmatrix}1 & 2 & 0 \\ 0 & 2 & 0 \\ -2 & -1 & -1\end{bmatrix}$ ，则 $\boldsymbol{A}^{100}=$ _______ + +## 三、计算与证明题(共6小题,共64分) +13. 计算行列式并解方程 $\begin{vmatrix}1 & 2 & 3 & 4 + x \\ 1 & 2 & 3 + x & 4 \\ 1 & 2 + x & 3 & 4 \\ 1 + x & 2 & 3 & 4\end{vmatrix}=0$ 。(10分) + +14. 设 $\boldsymbol{A}=\begin{bmatrix}1 & a & b \\ 0 & 1 & a \\ 0 & 0 & 1\end{bmatrix}$ ，其中 $a, b$ 都不等于0 +(1) 对任意自然数 $n \in \mathbb{N}$ ，计算 $\boldsymbol{A}^{n}$ 。(6分) +(2) 计算 $\boldsymbol{A}^{-1}$ 。(4分) + +15. 设 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 是线性方程组 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的解，其中 $\boldsymbol{x}=\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{bmatrix}$ ， $\boldsymbol{b}=\begin{bmatrix}b_{1} \\ b_{2} \\ b_{3} \\ b_{4}\end{bmatrix}$ ，现已知 $\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}=\begin{bmatrix}2 \\ 2 \\ 4 \\ 6\end{bmatrix}$ ， $\boldsymbol{\alpha}_{1}+2\boldsymbol{\alpha}_{3}=\begin{bmatrix}0 \\ 3 \\ 0 \\ 6\end{bmatrix}$ ，求该方程组的通解。(10分) + +16. 设方阵 $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似，其中 $\boldsymbol{A}=\begin{bmatrix}-2 & 0 & 0 \\ 2 & x & 2 \\ 3 & 1 & 1\end{bmatrix}$ ， $\boldsymbol{B}=\begin{bmatrix}-1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & y\end{bmatrix}$ ，求 $x, y$ 的值及可逆矩阵 $\boldsymbol{P}$ ，使得 $\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P}=\boldsymbol{B}$ 。(10分) + +17. 已知二次型 $f=x_{1}^{2}+ax_{2}^{2}+x_{3}^{2}+2bx_{1}x_{2}+2x_{1}x_{3}+2x_{2}x_{3}$ 可经过正交变换 $\begin{bmatrix}x_{1} \\ x_{2} \\ x_{3}\end{bmatrix}=\boldsymbol{P}\begin{bmatrix}y_{1} \\ y_{2} \\ y_{3}\end{bmatrix}$ 化为 $y_{2}^{2}+4y_{3}^{2}$ ，求 $a, b$ 的值和正交矩阵 $\boldsymbol{P}$ 。(12分) + +18. 设 $\boldsymbol{A}$ 为n阶可逆矩阵，证明 $\boldsymbol{A}$ 可以相似对角化当且仅当 $\boldsymbol{A}^{2}$ 可以相似对角化。(12分) \ No newline at end of file diff --git a/试卷库/线性代数/线代2021秋A.md b/试卷库/线性代数/线代2021秋A.md new file mode 100644 index 0000000..108577c --- /dev/null +++ b/试卷库/线性代数/线代2021秋A.md @@ -0,0 +1,3 @@ +1. 设三阶方阵 \(A=\begin{bmatrix}a&b&b\\b&a&b\\b&b&a\end{bmatrix}\)，其伴随矩阵 \(A^*\) 的秩等于1，则（ ） + A. \(a=b\) 且 \(a+2b\neq0\) + B. \(a=b\) 或 \(a+2b\neq0\) C. \(a\neq b\) 且 \(a+2b=0\) D. \(a\neq b\) 且 \(a+2b\neq0\) 2. 设 \(A, B, A+B, A^{-1}+B^{-1}\) 均为 \(n(n\geq2)\) 阶可逆矩阵，则 \((A^{-1}+B^{-1})^{-1}\) 等于（ ） A. \(A^{-1}+B^{-1}\) B. \(A+B\) C. \(A(A+B)^{-1}B\) D. \((A+B)^{-1}\) 3. 设 \(n\) 维向量 \(\alpha, \beta, \gamma\) 与数 \(k, l, m\) 满足 \(k\alpha+l\beta+m\gamma=0\)，且 \(km\neq0\)，则（ ） A. \(\alpha, \beta\) 与 \(\alpha, \gamma\) 等价 B. \(\alpha, \beta\) 与 \(\beta, \gamma\) 等价 C. \(\alpha, \gamma\) 与 \(\beta, \gamma\) 等价 D. \(\alpha\) 与 \(\gamma\) 等价 4. 下列矩阵中，与 \(\begin{bmatrix}4&2&0\\2&4&0\\0&0&-8\end{bmatrix}\) 合同的是（ ） A. \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&-1\end{bmatrix}\) B. \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}\) C. \(\begin{bmatrix}1&0&0\\0&-1&0\\0&0&0\end{bmatrix}\) D. \(\begin{bmatrix}1&0&0\\0&-1&0\\0&0&-1\end{bmatrix}\) 5. 设 \(\lambda_1, \lambda_2\) 是矩阵 \(A\) 的两个相异特征值，对应的特征向量分别为 \(\alpha_1, \alpha_2\)，则 \(A(\alpha_1+\alpha_2), \alpha_1\) 线性无关的充要条件为（ ） A. \(\lambda_1\neq0\) B. \(\lambda_2\neq0\) C. \(\lambda_1=0\) D. \(\lambda_2=0\) 6. 设 \(A, B\) 均为 \(n\) 阶正定矩阵，则下列矩阵中必为正定矩阵的是（ ） A. \(kAB\)，其中 \(k\) 为常数 B. \(kA^*+lB^*\)，其中 \(kl>0\) C. \(A^{-1}+B^{-1}\) D. \(A^{-1}-B^{-1}\) ## 二、填空题(共6小题,每小题3分,共18分) 7. 设矩阵 \(A=\begin{bmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{bmatrix}\)，则矩阵 \(A^3\) 的秩为________。 8. 已知 \(\alpha_1=\begin{bmatrix}1\\0\end{bmatrix}, \alpha_2=\begin{bmatrix}0\\1\end{bmatrix}\) 和 \(\beta_1=\begin{bmatrix}1\\2\end{bmatrix}, \beta_2=\begin{bmatrix}2\\3\end{bmatrix}\) 为向量空间的两组基，则从 \(\beta_1, \beta_2\) 到 \(\alpha_1, \alpha_2\) 的过渡矩阵为________。 9. 在实数域中，二次型 \(f(x_1, x_2, x_3)=2x_1x_2-2x_1x_3+2x_2x_3\) 的规范形为________。 10. 设5阶实对称矩阵 \(A\) 满足 \(A^2-2A=O\)，\(\text{rank}(A)=3\)，则 \(|A+E|\) 等于________。 11. 设 \(A=[a_{ij}]_{3\times3}\) 是正交矩阵，且 \(a_{33}=1\)，\(b=\begin{bmatrix}0\\0\\3\end{bmatrix}\)，则线性方程组 \(Ax=b\) 的解为________。 12. 设 \(A\) 为三阶方阵，将 \(A\) 的第2列加到第1列得到 \(B\)，再交换 \(B\) 的第2行与第3行得到单位矩阵，则 \(A\) 等于________。 ## 三、计算与证明题(共6小题,共64分) 13. 求多项式 \(f(x)=\begin{vmatrix}x&x&1&2x\\1&x&2&-1\\2&1&x&1\\2&-1&1&x\end{vmatrix}\) 中 \(x^3\) 的系数。(10分) 14. 设三阶方阵 \(A, B\) 满足 \(A^*BA=2BA-8E\)，其中 \(A=\begin{bmatrix}1&0&0\\0&-2&0\\0&0&1\end{bmatrix}\)，求 \(B\)。(10分) 15. 已知线性方程组 \(\begin{cases}x_1+x_2=0\\-2x_1+2x_2+(2-\lambda)x_3=1\\-4x_1+(5-\lambda)x_2+2x_3=2\end{cases}\)，讨论当 \(\lambda\) 取何值时，方程组有唯一解、无解、有无穷多解，并求出有无穷多解时的通解。(10分) 16. 设 \(\beta_1, \beta_2, \cdots, \beta_m\) 均为实数域上的 \(n\) 维列向量，其中 \(m<n\)，证明 \(\beta_1, \beta_2, \cdots, \beta_m\) 线性无关的充要条件为 \(B^TB\) 可逆（其中 \(B=[\beta_1, \beta_2, \cdots, \beta_m]\)）。(10分) 17. 已知 \(\alpha=\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}, \beta=\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}\) 为 \(n\) 维非零列向量，且 \(\beta^T\alpha\neq0\)，\(A=\alpha\beta^T\)，证明 \(A\) 可相似对角化。(12分) 18. 已知实二次型 \(f(x_1, x_2, x_3)=3x_1^2+2x_2^2+ax_3^2+bx_1x_3\)，经过正交变换 \(x=Py\) 得标准形 \(y_1^2+2y_2^2+5y_3^2\)，其中 \(a, b\) 都是非负实数，求正交变换矩阵 \(P\)。(12分) \ No newline at end of file diff --git a/试卷库/线性代数/线代2022秋A.md b/试卷库/线性代数/线代2022秋A.md new file mode 100644 index 0000000..e5f0236 --- /dev/null +++ b/试卷库/线性代数/线代2022秋A.md @@ -0,0 +1,64 @@ +## 一、单选题(共6小题,每小题3分,共18分) +1. 设 $\boldsymbol{A}$ 为 $n$ 阶方阵,且 $|\boldsymbol{A}| = 5$ ,则 $||\boldsymbol{A}|\boldsymbol{A}^{T}|$ =【】 +A. $5n$ +B. $5^{n + 1}$ +C. $5^{n - 1}$ +D. 25 + +2. 设 $\boldsymbol{A}$ , $\boldsymbol{B}$ 均为 $n$ 阶方阵,则下列命题中正确的是【】 +A. $|\boldsymbol{A}+\boldsymbol{B}|=|\boldsymbol{A}| + |\boldsymbol{B}|$ +B. $\boldsymbol{AB}=\boldsymbol{BA}$ +C. $|\boldsymbol{AB}|=|\boldsymbol{BA}|$ +D. $(\boldsymbol{A}+\boldsymbol{B})^{-1}=\boldsymbol{A}^{-1}+\boldsymbol{B}^{-1}$ + +3. 下列关于矩阵秩的结论中,错误的是【】 +A. 若矩阵 $\boldsymbol{A}$ , $\boldsymbol{B}$ , $\boldsymbol{C}$ 满足 $\boldsymbol{A}=\boldsymbol{BC}$ 且 $\boldsymbol{B}$ 是列满秩的,则 $\boldsymbol{A}$ 列满秩当且仅当 $\boldsymbol{C}$ 列满秩 +B. 行阶梯形矩阵中1的个数等于矩阵的秩 +C. 设 $\boldsymbol{A}$ , $\boldsymbol{B}$ 均为 $n$ 阶方阵,则 $R(\boldsymbol{AB})=R(\boldsymbol{B})$ 的充要条件是线性方程组 $(\boldsymbol{AB})\boldsymbol{x}=\boldsymbol{0}$ 与 $\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$ 同解 +D. 设 $\boldsymbol{A}$ 为 $n$ 阶方阵,则 $R(\boldsymbol{A}^{n})=R(\boldsymbol{A}^{n + 1})$ + +4. 设 $\boldsymbol{\eta}_{1}$ , $\boldsymbol{\eta}_{2}$ 是非齐次线性方程组 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的两个不同解, $\boldsymbol{\xi}_{1}$ , $\boldsymbol{\xi}_{2}$ 是导出方程组 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$ 的两个不同解,则:① $\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}$ 是 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$ 的解,② $3\boldsymbol{\xi}_{1}+\boldsymbol{\eta}_{1}$ 是 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的解,③ $2\boldsymbol{\xi}_{1}+2\boldsymbol{\xi}_{2}$ 是 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$ 的解,④ $2\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}$ 是 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的解,⑤ $\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}+\boldsymbol{\xi}_{1}$ 是 $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$ 的解。上述结论正确的有多少个?【】 +A. 2 +B. 3 +C. 4 +D. 5 + +5. 已知 $n$ 阶方阵 $\boldsymbol{A}$ 与 $\boldsymbol{B}$ 相似, $\boldsymbol{C}$ 与 $\boldsymbol{D}$ 相似,则下列命题中正确的是【】 +A. $\boldsymbol{A}+\boldsymbol{C}$ 与 $\boldsymbol{B}+\boldsymbol{D}$ 相似 +B. $\boldsymbol{AC}$ 与 $\boldsymbol{BD}$ 相似 +C. $\boldsymbol{A}^{2}+\boldsymbol{I}$ 与 $\boldsymbol{B}^{2}+\boldsymbol{I}$ 相似 +D. $\boldsymbol{A}^{T}+\boldsymbol{A}$ 与 $\boldsymbol{B}^{T}+\boldsymbol{B}$ 相似 + +6. 设 $f(x_{1},x_{2},x_{3})=2x_{1}^{2}+6x_{2}^{2}+x_{3}^{2}-4x_{1}x_{2}-2x_{1}x_{3}=C$ ,则此二次曲面是【】 +A. $C = 0$ 为锥面 +B. $C>0$ 为椭球面 +C. $C<0$ 为柱面 +D. $C = 1$ 为单叶双曲面 + +## 二、填空题(共6小题,每小题3分,共18分) +1. 设 $\alpha,\beta,\gamma$ 为 $x^{3}+px + q = 0$ 的三个根,则行列式 $\begin{vmatrix}\boldsymbol{\alpha}&\boldsymbol{\beta}&\boldsymbol{\gamma}\\\boldsymbol{\gamma}&\boldsymbol{\alpha}&\boldsymbol{\beta}\\\boldsymbol{\beta}&\boldsymbol{\gamma}&\boldsymbol{\alpha}\end{vmatrix}$ =________ +2. 设 $\boldsymbol{A}=\begin{bmatrix}1&1&1\\a_{1}&a_{2}&a_{3}\\a_{1}^{2}&a_{2}^{2}&a_{3}^{2}\end{bmatrix}$ , $\boldsymbol{b}=\begin{bmatrix}1\\1\\1\end{bmatrix}$ ,其中 $a_{1},a_{2},a_{3}$ 互不相同,线性方程组 $\boldsymbol{A}^{T}\boldsymbol{x}=\boldsymbol{b}$ 的解为________ +3. 设3阶方阵 $\boldsymbol{A}=\begin{bmatrix}2&-1&2\\1&0&1\\0&0&2\end{bmatrix}$ ,3维向量 $\boldsymbol{\alpha}=\begin{bmatrix}t\\2\\1\end{bmatrix}$ ,若 $\boldsymbol{A}\boldsymbol{\alpha}$ 与 $\boldsymbol{\alpha}$ 线性相关,则 $t=$ ________ +4. 已知 $\|\boldsymbol{a}\| = 2$ , $\|\boldsymbol{b}\| = 5$ ,且 $\boldsymbol{a}$ 和 $\boldsymbol{b}$ 的夹角为 $\frac{2}{3}\pi$ ,若向量 $\boldsymbol{A}=\lambda\boldsymbol{a}+17\boldsymbol{b}$ 与 $\boldsymbol{B}=3\boldsymbol{a}-\boldsymbol{b}$ 垂直,则 $\lambda=$ ________ +5. 已知 $n(n\geq2)$ 维列向量 $\boldsymbol{\alpha}$ , $\boldsymbol{\beta}$ 满足 $\boldsymbol{\beta}^{T}\boldsymbol{\alpha}=-3$ ,其中 $\boldsymbol{\beta}^{T}$ 为 $\boldsymbol{\beta}$ 的转置,则方阵 $(\boldsymbol{\beta}\boldsymbol{\alpha}^{T})^{2}$ 的非零特征值为________ +6. 已知 $\boldsymbol{A}$ 为 $m\times n$ 实矩阵(其中 $m < n$ ), $\boldsymbol{I}$ 为 $n$ 阶单位矩阵, $t$ 为实数,则 $\boldsymbol{A}^{T}\boldsymbol{A}+t\boldsymbol{I}$ 为正定矩阵的充分必要条件是实数 $t$ 满足关系式________ + +## 三、计算题(10分) +计算 $n$ 阶行列式 $\begin{vmatrix}a&a_{2}&a_{2}&\cdots&a_{2}\\a_{2}&a&a_{2}&\cdots&a_{2}\\a_{2}&a_{2}&a&\cdots&a_{2}\\\vdots&\vdots&\vdots&\ddots&\vdots\\a_{2}&a_{2}&a_{2}&\cdots&a\end{vmatrix}$ + +## 四、计算题(10分) +设直线 $L:\frac{x - 1}{2}=y=\frac{z - 2}{3}$ 与平面 $\pi:2x - y+z - 10 = 0$ 的交点为 $P$ ,求过点 $P$ 且与平面 $\pi$ 垂直的直线方程。 + +## 五、计算题(10分) +已知线性方程组 $\begin{cases}\lambda x_{1}-x_{2}-x_{3}=1\\-x_{1}+\lambda x_{2}-x_{3}=-\lambda\\-x_{1}-x_{2}+\lambda x_{3}=\lambda^{2}\end{cases}$ 至少存在两个不同的解,求该线性方程组的通解。 + +## 六、计算题(10分) +判定向量组 $\boldsymbol{\alpha}_{1}=\begin{bmatrix}1\\1\\1\\-1\end{bmatrix}$ , $\boldsymbol{\alpha}_{2}=\begin{bmatrix}1\\4\\1\\0\end{bmatrix}$ , $\boldsymbol{\alpha}_{3}=\begin{bmatrix}-1\\2\\-1\\2\end{bmatrix}$ , $\boldsymbol{\alpha}_{4}=\begin{bmatrix}1\\1\\2\\3\end{bmatrix}$ , $\boldsymbol{\alpha}_{5}=\begin{bmatrix}2\\-1\\4\\5\end{bmatrix}$ 的线性相关性,求其一个极大线性无关组,并将其余向量用该极大线性无关组线性表示。 + +## 七、证明与计算题(共2小题,第1小题4分,第2小题8分,共12分) +已知向量组 $\boldsymbol{\alpha}_{1}$ , $\boldsymbol{\alpha}_{2}$ , $\boldsymbol{\alpha}_{3}$ 线性无关,其中 $\boldsymbol{\alpha}_{1}$ , $\boldsymbol{\alpha}_{2}$ , $\boldsymbol{\alpha}_{3}\in\mathbb{R}^{3}$ , $\boldsymbol{A}$ 为3阶方阵,且 $\boldsymbol{A}\boldsymbol{\alpha}_{1}=2\boldsymbol{\alpha}_{1}-\boldsymbol{\alpha}_{2}-\boldsymbol{\alpha}_{3}$ , $\boldsymbol{A}\boldsymbol{\alpha}_{2}=\boldsymbol{\alpha}_{1}+2\boldsymbol{\alpha}_{2}+3\boldsymbol{\alpha}_{3}$ , $\boldsymbol{A}\boldsymbol{\alpha}_{3}=2\boldsymbol{\alpha}_{1}+4\boldsymbol{\alpha}_{2}+3\boldsymbol{\alpha}_{3}$ 。 +1. 证明 $\boldsymbol{A}\boldsymbol{\alpha}_{1}$ , $\boldsymbol{A}\boldsymbol{\alpha}_{2}$ , $\boldsymbol{A}\boldsymbol{\alpha}_{3}$ 线性无关； +2. 计算行列式 $|\boldsymbol{E}-\boldsymbol{A}|$ ,其中 $\boldsymbol{E}$ 是3阶单位矩阵。 + +## 八、计算题(12分) +已知实二次型 $f(x_{1},x_{2},x_{3})=\boldsymbol{x}^{T}\boldsymbol{A}\boldsymbol{x}$ 在正交变换 $\boldsymbol{x}=\boldsymbol{P}\boldsymbol{y}$ 下的标准形为 $3y_{1}^{2}+3y_{2}^{2}$ ,且 $\boldsymbol{P}$ 的第3列为 $\frac{1}{\sqrt{3}}\begin{bmatrix}1\\1\\1\end{bmatrix}$ ,求 $\boldsymbol{A}$ 。 \ No newline at end of file -- 2.34.1 From 2d8da7a660f690b5fba73a2fb1747c0d8ec7a393 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Mon, 12 Jan 2026 14:28:02 +0800 Subject: [PATCH 206/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/线性方程组的系数矩阵与解关系.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/素材/线性方程组的系数矩阵与解关系.md b/素材/线性方程组的系数矩阵与解关系.md index ef43f52..aaeb69a 100644 --- a/素材/线性方程组的系数矩阵与解关系.md +++ b/素材/线性方程组的系数矩阵与解关系.md @@ -1,4 +1,4 @@ - +这是一个链接了方程组解空间与方程组系数秩的公式 >[!note] 解零度化定理： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ @@ -103,7 +103,7 @@ Bx = \beta $$ 与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 -(2) 分析： +(2) 分析：不同解，却要可以求出a的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑解零度化定理 由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 $$ 4 - R(A) < 4 - R(B), -- 2.34.1 From c3036b20c9ed26ab67882ab3f045402036f20875 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Mon, 12 Jan 2026 14:34:55 +0800 Subject: [PATCH 207/274] minor edit --- .../试卷/线代期末复习模拟/1.14线代限时练.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md index df0efeb..111f58e 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md @@ -1,4 +1,4 @@ -1. 已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. +1. （[[线代2022秋B]]·1）已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. 2. （[[线代2013秋A]]·4）已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. -- 2.34.1 From 3847451d0a28e6554f9715bac581d9b0b8b2c610 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 12:52:00 +0800 Subject: [PATCH 208/274] vault backup: 2026-01-13 12:52:00 --- .../高数期末真题/2017高数期末考试卷.md | 10 +++++----- 1 file changed, 5 insertions(+), 5 deletions(-) diff --git a/编写小组/试卷/高数期末真题/2017高数期末考试卷.md b/编写小组/试卷/高数期末真题/2017高数期末考试卷.md index 471e755..082cd86 100644 --- a/编写小组/试卷/高数期末真题/2017高数期末考试卷.md +++ b/编写小组/试卷/高数期末真题/2017高数期末考试卷.md @@ -11,11 +11,11 @@ tags: **考试时间：150 分钟** **满分：100 分** -| 题号 | 一 | 二 | 三 | 四 | 五 | 六 | 七 | 八 | 九 | 十 | 十一 | 十二 | 总分 | 核分 | -|------|----|----|----|----|----|----|----|----|----|----|-----|-----|------|------| -| 满分 | 15 | 15 | 6 | 6 | 6 | 6 | 6 | 8 | 8 | 8 | 8 | 8 | 100 | | -| 得分 | | | | | | | | | | | | | | | -| 评阅人 | | | | | | | | | | | | | | | +| 题号 | 一 | 二 | 三 | 四 | 五 | 六 | 七 | 八 | 九 | 十 | 十一 | 十二 | 总分 | 核分 | +| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | +| 满分 | 15 | 15 | 6 | 6 | 6 | 6 | 6 | 8 | 8 | 8 | 8 | 8 | 100 | | +| 得分 | | | | | | | | | | | | | | | +| 评阅人 | | | | | | | | | | | | | | | **注意：** 1. 所有答题都须写在此试卷纸密封线右边，写在其它纸上一律无效。 -- 2.34.1 From 706bb1d989d09d123d4c87dca84337caf5f5b66f Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 13 Jan 2026 12:58:07 +0800 Subject: [PATCH 209/274] vault backup: 2026-01-13 12:58:07 --- 编写小组/讲义/微分中值定理（解析版）.md | 4 ++++ 1 file changed, 4 insertions(+) create mode 100644 编写小组/讲义/微分中值定理（解析版）.md diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md new file mode 100644 index 0000000..1fb86d3 --- /dev/null +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -0,0 +1,4 @@ +--- +tags: + - 编写小组 +--- -- 2.34.1 From 17c373cc914abdf1e4c8004b40c3fa6ea79b5eef Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 13 Jan 2026 13:00:58 +0800 Subject: [PATCH 210/274] vault backup: 2026-01-13 13:00:58 --- .../微分中值定理（解析版）.md | 3 ++ ...的解与秩的不等式（解析版）.md | 42 +++++++++---------- 2 files changed, 24 insertions(+), 21 deletions(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 1fb86d3..b52eaf9 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -2,3 +2,6 @@ tags: - 编写小组 --- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 + diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index e68f0cc..b858ee4 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -227,6 +227,27 @@ $$ $$ 解得 $a = 1$。 +# 通过秩反过来得方程是否有解 + +>[!example] 例3 +>已知$\boldsymbol{A},\boldsymbol{B}$均为$m\times n$矩阵，$\beta_1,\beta_2$为$m$维列向量，则下列选项正确的有[ ] +(A)若$\mathrm{rank}\boldsymbol{A}=m$，则对于任意$m$维列向量$\boldsymbol{b},\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解. +(B)若$\boldsymbol{A}$与$\boldsymbol{B}$等价，则齐次线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$与$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$同解. +(C)矩阵方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，但$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解的充要条件是$$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ B}].$$ +(D)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解当且仅当$$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}].$$ + +**解：** +(A)一方面$\mathrm{rank}[\boldsymbol{A\ b}]\ge \mathrm{rank}\boldsymbol{A}=m$,另一方面矩阵$[\boldsymbol{A\ b}]$只有$m$行，所以它的秩必然不大于$m$，所以$\mathrm{rank}[\boldsymbol{A\ b}]=m=\mathrm{rank}\boldsymbol{A}$，即方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解。 + +(B)等价的矩阵只需要是经过初等变换可以变成同一个矩阵就行了，但齐次线性方程组同解需要只经过初等行变换就能变成同一个矩阵才行，后一个条件明显更强，所以后一种更“难”达成，B就不对。 + +(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出C正确。 + +(D)我们同样有两种方法去解这道题，一种是形式化的、严谨的，另一种是理解性的、直观的。 +1)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解$\Leftrightarrow\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]=\mathrm{rank}[\boldsymbol{A\ \beta_2}]$,故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。 +2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 +3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 + # 秩的不等式 ### 1. 和的秩不超过秩的和 @@ -302,24 +323,3 @@ $$ > 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ - ->[!example] 例3 ->已知$\boldsymbol{A},\boldsymbol{B}$均为$m\times n$矩阵，$\beta_1,\beta_2$为$m$维列向量，则下列选项正确的有[ ] -(A)若$\mathrm{rank}\boldsymbol{A}=m$，则对于任意$m$维列向量$\boldsymbol{b},\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解. -(B)若$\boldsymbol{A}$与$\boldsymbol{B}$等价，则齐次线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$与$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$同解. -(C)矩阵方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，但$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解的充要条件是$$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ B}].$$ -(D)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解当且仅当$$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}].$$ - -**解：** -(A)一方面$\mathrm{rank}[\boldsymbol{A\ b}]\ge \mathrm{rank}\boldsymbol{A}=m$,另一方面矩阵$[\boldsymbol{A\ b}]$只有$m$行，所以它的秩必然不大于$m$，所以$\mathrm{rank}[\boldsymbol{A\ b}]=m=\mathrm{rank}\boldsymbol{A}$，即方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解。 - -(B)等价的矩阵只需要是经过初等变换可以变成同一个矩阵就行了，但齐次线性方程组同解需要只经过初等行变换就能变成同一个矩阵才行，后一个条件明显更强，所以后一种更“难”达成，B就不对。 - -(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出C正确。 - -(D)我们同样有两种方法去解这道题，一种是形式化的、严谨的，另一种是理解性的、直观的。 -1)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解$\Leftrightarrow\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]=\mathrm{rank}[\boldsymbol{A\ \beta_2}]$,故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。 -2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 -3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 - -根据上面的题目，我们可不可以归纳出一种比较普遍的方式，去解决这种与秩和方程组解都有密切关系的题目呢？ \ No newline at end of file -- 2.34.1 From 7b373c0b3ba1e9b3d63fc0f4d27ebab655c17f83 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Tue, 13 Jan 2026 13:05:17 +0800 Subject: [PATCH 211/274] vault backup: 2026-01-13 13:05:17 --- .../线性方程组的解与秩的不等式（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index b858ee4..d5e4d4f 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -229,7 +229,7 @@ $$ # 通过秩反过来得方程是否有解 ->[!example] 例3 +>[!example] 例1 >已知$\boldsymbol{A},\boldsymbol{B}$均为$m\times n$矩阵，$\beta_1,\beta_2$为$m$维列向量，则下列选项正确的有[ ] (A)若$\mathrm{rank}\boldsymbol{A}=m$，则对于任意$m$维列向量$\boldsymbol{b},\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解. (B)若$\boldsymbol{A}$与$\boldsymbol{B}$等价，则齐次线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$与$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$同解. -- 2.34.1 From 4eea080cd8ac66ea461fb643a904c6bba4eb30bd Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 13:39:30 +0800 Subject: [PATCH 212/274] =?UTF-8?q?1.14=E7=BA=BF=E4=BB=A3=E6=B5=8B?= =?UTF-8?q?=E8=AF=95=E7=AD=94=E6=A1=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 1.14线代测试答案.md | 218 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 218 insertions(+) create mode 100644 1.14线代测试答案.md diff --git a/1.14线代测试答案.md b/1.14线代测试答案.md new file mode 100644 index 0000000..c8a70a9 --- /dev/null +++ b/1.14线代测试答案.md @@ -0,0 +1,218 @@ +# **1.14 线性代数限时练（题目 + 答案与解析）** + +## **第一部分：题目** + +### **1.** + +已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum_{i=1}^{3}\sum_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ + +### **2.（2013 秋 A）** + +$已知向量空间 V=\{(2a,2b,3b,3a)\mid a,b\in\mathbb{R}\}，则 V 的维数是\underline{\qquad}。$ + +### **3.（2018 秋 A）** + +$设 E 为 3 阶单位矩阵，\alpha 为一个 3 维单位列向量，则矩阵 E-\alpha\alpha^T 的全部 3 个特征值为\underline{\qquad}。$ + +### **4.（2018 秋 A）** + +$设 n 阶矩阵 A=[a_{ij}]_{n\times n}，则二次型 f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^{n}(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2 的矩阵为\underline{\qquad}。$ +### **5.（2018 秋 A）** + +$若 n 阶实对称矩阵 A 的特征值为 \lambda_i=(-1)^i（i=1,2,\cdots,n），则 A^{100}=\underline{\qquad}。$ + +### **6.（2022 秋 A・5）** + +$已知 n（n\geq2）维列向量 \alpha,\beta 满足 \beta^T\alpha=-3，则方阵 (\beta\alpha^T)^2 的非零特征值为\underline{\qquad}。$ + +### **7.（2022 秋 A）** + +$已知向量组 \alpha_1,\alpha_2,\alpha_3 线性无关（\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}^3），A 为 3 阶方阵，且满足：$ + +$$\begin{aligned} +A\alpha_1&=2\alpha_1-\alpha_2-\alpha_3,\\A\alpha_2&=\alpha_1+2\alpha_2+3\alpha_3,\\A\alpha_3&=2\alpha_1+4\alpha_2+\alpha_3 +\end{aligned} +$$ +(1) $证明 A\alpha_1,A\alpha_2,A\alpha_3 线性无关$； + +(2) $计算行列式 |E-A|（E 是 3 阶单位矩阵）$。 + +### **8.（2013 秋 A）** + +$求 n 阶方阵 A=\begin{bmatrix}1&1&1&\cdots&1\\1&0&1&\cdots&1\\1&1&0&\cdots&1\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&1&1&\cdots&0\end{bmatrix} 的逆矩阵。$ + +### **9.（2013 秋 A・三）** + +设 n 阶行列式： + +$$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$ + +证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。 + +## **第二部分：答案与解析** + +### 1. 答案：$\boxed{4c}$ + +#### **解析：** + +$设原矩阵为 A（即 |A|=c），全 1 矩阵 J=\boldsymbol{e}\boldsymbol{e}^T（其中 \boldsymbol{e}=(1,1,1)^T），需求解的行列式为 |A+J|。$ +由代数余子式性质：$\sum_{i,j}A_{ij}=\boldsymbol{e}^T A^*\boldsymbol{e}=3c$，且可逆矩阵的伴随矩阵满足$A^*=|A|A^{-1}=cA^{-1}$（因 $|A|=c\neq0，A$ 可逆），代入得 $\boldsymbol{e}^T A^{-1}\boldsymbol{e}=3$。 + +利用**Sherman-Morrison 行列式公式**：对可逆矩阵$A$ 和向量 $\boldsymbol{u},\boldsymbol{v}$，有 $|A+\boldsymbol{u}\boldsymbol{v}^T|=|A|(1+\boldsymbol{v}^T A^{-1}\boldsymbol{u})$。 + +此处 $\boldsymbol{u}=\boldsymbol{v}=\boldsymbol{e}$，代入得 $|A+J|=c(1+\boldsymbol{e}^T A^{-1}\boldsymbol{e})=c(1+3)=4c$。 + +### **2. 答案：$\boxed{2}$ + +#### **解析：** + +将向量空间 V 中的元素拆分为线性组合形式： + +$$(2a,2b,3b,3a)=a(2,0,0,3)+b(0,2,3,0)$$ + +设 $\boldsymbol{\alpha}=(2,0,0,3)，\boldsymbol{\beta}=(0,2,3,0)$，需验证 $\boldsymbol{\alpha},\boldsymbol{\beta}$ 线性无关： + +若 $k_1\boldsymbol{\alpha}+k_2\boldsymbol{\beta}=\boldsymbol{0}$（零向量），则 $\begin{cases}2k_1=0\\2k_2=0\\3k_2=0\\3k_1=0\end{cases}$，解得 $k_1=k_2=0$，故 $\boldsymbol{\alpha},\boldsymbol{\beta}$是 V 的一组基。 + +向量空间的维数等于基的个数，因此 $V$ 的维数为 $2$。 + +### **3. 答案：$\boxed{1,1,0}$ + +#### **解析：** + +设 $B=\alpha\alpha^T$，该矩阵为**秩 $1$ 矩阵**（因 $\alpha$ 是单位列向量，$\alpha^T\alpha=1$）。 + +• 秩 $1$ 矩阵的特征值性质：非零特征值为矩阵的迹 $\text{tr}(B)=\alpha^T\alpha=1$，其余 $n-1=2$ 个特征值为 $0$（秩 $1$ 矩阵的非零特征值个数等于秩）。 + +• 若 $B\boldsymbol{x}=\lambda\boldsymbol{x}$（$\boldsymbol{x}$为特征向量），则 $(E-B)\boldsymbol{x}=(1-\lambda)\boldsymbol{x}$，即$E-B$ 的特征值为 $1-\lambda$。 + +$代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1-0=1，即 1,1,0$。 + +### **4. 答案：$\boxed{A^T A}$** + +#### **解析：** + +将二次型展开为矩阵形式： + +$$f(\boldsymbol{x})=\sum_{i=1}^{n}(A\boldsymbol{x})_i^2=(A\boldsymbol{x})^T(A\boldsymbol{x})=\boldsymbol{x}^T(A^T A)\boldsymbol{x}$$ + +• 二次型的矩阵需满足**对称性质**（即矩阵等于其转置），验证：$(A^T A)^T=A^T(A^T)^T=A^T A$，故 $A^T A$ 是对称矩阵，即为二次型的矩阵。 + +### **5. 答案：$\boxed{E}$（单位矩阵）** + +#### **解析：** + +实对称矩阵可对角化，即存在可逆矩阵 P，使得 $P^{-1}AP=\Lambda$（$\Lambda$ 为对角矩阵，对角元为 A 的特征值 $\lambda_i=(-1)^i$）。 + +• 矩阵幂运算性质：$A^{100}=P\Lambda^{100}P^{-1}$。 + +• 计算 $\Lambda^{100}$：对角元为 $\lambda_i^{100}=[(-1)^i]^{100}=1$，故 $\Lambda^{100}=E$（单位矩阵）。 + +• 因此 $A^{100}=P E P^{-1}=P P^{-1}=E$。 + +### **6. 答案：$\boxed{9}$** + +#### **解析：** + +设 $A=\beta\alpha^T$（秩 1 矩阵），计算 $A^2$： + +$A^2=(\beta\alpha^T)(\beta\alpha^T)=\beta(\alpha^T\beta)\alpha^T=(\alpha^T\beta)A$ + +• 注意：$\alpha^T\beta=(\beta^T\alpha)^T$（矩阵转置性质），而 $\beta^T\alpha=-3$（数，转置等于自身），故 $\alpha^T\beta=-3$，因此 $A^2=-3A$。 + +• 秩 $1$ 矩阵 $A$ 的非零特征值为 $\text{tr}(A)=\alpha^T\beta=-3$，设 $A\boldsymbol{x}=-3\boldsymbol{x}$，则 $A^2\boldsymbol{x}=(-3)A\boldsymbol{x}=(-3)^2\boldsymbol{x}=9\boldsymbol{x}$，即 $A^2$ 的非零特征值为 $9$。 + +### **7. 答案：(1) 证明见解析；(2) $\boxed{20}$** + +#### **(1) 证明 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关** + +因 $\alpha_1,\alpha_2,\alpha_3$ 线性无关，故矩阵 $P=(\alpha_1,\alpha_2,\alpha_3)$ 可逆（列向量线性无关的矩阵可逆）。 + +由题设条件，将 $A$ 对 $\alpha_1,\alpha_2,\alpha_3$ 的作用表示为矩阵乘法： + +$$A(\alpha_1,\alpha_2,\alpha_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}=P C$$ + +其中 $C=\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}$，计算 $|C|$： + +$$\begin{aligned} +|C|&=2\times(2\times1-4\times3)-1\times(-1\times1-4\times(-1))+2\times(-1\times3-2\times(-1))\\ +&=2\times(-10)-1\times3+2\times(-1)\\ +&=-25\neq0 +\end{aligned} +$$ + +• 因 $|C|\neq0$，故 $C$ 可逆，$\text{rank}(C)=3$。 + +• 又 $P$ 可逆，故 $\text{rank}(A\alpha_1,A\alpha_2,A\alpha_3)=\text{rank}(P C)=\text{rank}(C)=3$，即 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关。 + +#### **(2) 计算 $|E-A|$** + +由 (1) 知$P^{-1}AP=C（A 与 C 相似）$，则 $E-A$ 与 $E-C$ 相似（相似矩阵的 “单位矩阵减矩阵” 仍相似），而**相似矩阵的行列式相等**，故 $|E-A|=|E-C|$。 + +计算 $E-C$： + +$$E-C=\begin{bmatrix}1-2&0-1&0-2\\0-(-1)&1-2&0-4\\0-(-1)&0-3&1-1\end{bmatrix}=\begin{bmatrix}-1&-1&-2\\1&-1&-4\\1&-3&0\end{bmatrix}$$ + +按第三行展开计算行列式： + +$$ +\begin{aligned} +|E-C|&=1\times\begin{vmatrix}-1&-2\\-1&-4\end{vmatrix}-(-3)\times\begin{vmatrix}-1&-2\\1&-4\end{vmatrix}+0\times(\text{余子式})\\ +&=1\times(4-2)+3\times(4+2)\\ +&=2+18=20 +\end{aligned} +$$ + +故 $|E-A|=20$。 + +### **8. 答案：** + +$$A^{-1}=\begin{bmatrix}-(n-2)&1&1&\cdots&1\\1&-1&0&\cdots&0\\1&0&-1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&0&0&\cdots&-1\end{bmatrix} +$$ +#### **解析：** + +通过 “行变换法” 或 “规律归纳” 推导： + +• 观察矩阵 A 的结构：第一行全为 1，其余行的对角元为 0，非对角元为 1。可先计算 n=2,3 时的逆矩阵，归纳规律： + +◦ $当 n=2 时，A=\begin{bmatrix}1&1\\1&0\end{bmatrix}，逆矩阵为 \begin{bmatrix}0&1\\1&-1\end{bmatrix}（符合上述形式，-(2-2)=0）$； + +◦ 当 $n=3$ 时，$A=\begin{bmatrix}1&1&1\\1&0&1\\1&1&0\end{bmatrix}$，逆矩阵为 $\begin{bmatrix}-1&1&1\\1&-1&0\\1&0&-1\end{bmatrix}$（符合上述形式，$-(3-2)=-1$）。 + +• 验证规律：对 n 阶矩阵，逆矩阵的第一行第一列元素为 -(n-2)，第一行其余元素为 1，第一列其余元素为 1，对角元（除第一行第一列）为 -1，非对角元（除第一行、第一列）为 0，即为上述形式。 + +### **9. 证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$** + +#### **步骤 1：建立递推公式** + +对 $D_n$ 按**第一行展开**（第一行元素为 $a_{11}=1,a_{12}=1$，其余 $a_{1j}=0$）： + +$D_n=a_{11}\times(-1)^{1+1}M_{11}+a_{12}\times(-1)^{1+2}M_{12}$ + +• $M_{11}$：去掉第一行第一列后的子式，即 $n-1$ 阶行列式 $D_{n-1}$（结构与 $D_n$ 一致）； + +• $M_{12}$：去掉第一行第二列后的子式，按第一列展开（第一列仅首元素为 $-1$），得 $-D_{n-2}$（符号需结合 $(-1)^{1+2}=-1$）。 + +因此递推公式为： + +$$D_n=D_{n-1}+D_{n-2}$$ + +#### **步骤 2：确定初始条件** + +• 当 $n=1$ 时，$D_1=\begin{vmatrix}1\end{vmatrix}=1$； + +• 当 n=2 时，$D_2=\begin{vmatrix}1&1\\-1&1\end{vmatrix}=1\times1-1\times(-1)=2$。 + +#### **步骤 3：求解递推关系** + +递推公式 $D_n=D_{n-1}+D_{n-2}$ 是**斐波那契数列的变形**，斐波那契数列的通项公式（比内公式）为： + +$$F_k=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^k-\left(\frac{1-\sqrt{5}}{2}\right)^k\right]$$ + +对比初始条件：$D_1=1=F_2，D_2=2=F_3$，故 $D_n=F_{n+1}$（斐波那契数列的第 $n+1$ 项）。 + +将$ F_{n+1} 代入比内公式$，得： + +$$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$$ + +证毕。 \ No newline at end of file -- 2.34.1 From 1a75ff4da39606d3893922b18d2d3888837436b9 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 13:40:11 +0800 Subject: [PATCH 213/274] vault backup: 2026-01-13 13:40:11 --- 1.14线代测试答案.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/1.14线代测试答案.md b/1.14线代测试答案.md index c8a70a9..685aaee 100644 --- a/1.14线代测试答案.md +++ b/1.14线代测试答案.md @@ -4,7 +4,7 @@ ### **1.** -已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum_{i=1}^{3}\sum_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ +已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ ### **2.（2013 秋 A）** -- 2.34.1 From 92b0bb532fe300962e085d4ba7e8ad773a96a1c2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 13:41:32 +0800 Subject: [PATCH 214/274] vault backup: 2026-01-13 13:41:32 --- .../试卷/线代期末复习模拟/1.14线代测试答案.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 1.14线代测试答案.md => 编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md (100%) diff --git a/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md similarity index 100% rename from 1.14线代测试答案.md rename to 编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md -- 2.34.1 From 64e70d9c6ddb57179772a7c9290c4de0491554e1 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 13:54:02 +0800 Subject: [PATCH 215/274] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E4=B8=80?= =?UTF-8?q?=E9=81=93=E9=A2=98=E7=9A=84=E7=B4=A0=E6=9D=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/特征值与相似.md | 1 + 1 file changed, 1 insertion(+) create mode 100644 素材/特征值与相似.md diff --git a/素材/特征值与相似.md b/素材/特征值与相似.md new file mode 100644 index 0000000..2ce056b --- /dev/null +++ b/素材/特征值与相似.md @@ -0,0 +1 @@ +$设 E 为 3 阶单位矩阵，\alpha 为一个 3 维单位列向量，则矩阵 E-\alpha\alpha^T 的全部 3 个特征值为\underline{\qquad}。$ -- 2.34.1 From 38836d3e12a12e39f7775a31b74fba84a7d32cb8 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 14:49:04 +0800 Subject: [PATCH 216/274] vault backup: 2026-01-13 14:49:04 --- .../1.14线代测试答案.md | 21 ++++++++++--------- 1 file changed, 11 insertions(+), 10 deletions(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md index 685aaee..6afd7b5 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md @@ -54,14 +54,10 @@ $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vd ### 1. 答案：$\boxed{4c}$ #### **解析：** - -$设原矩阵为 A（即 |A|=c），全 1 矩阵 J=\boldsymbol{e}\boldsymbol{e}^T（其中 \boldsymbol{e}=(1,1,1)^T），需求解的行列式为 |A+J|。$ -由代数余子式性质：$\sum_{i,j}A_{ij}=\boldsymbol{e}^T A^*\boldsymbol{e}=3c$，且可逆矩阵的伴随矩阵满足$A^*=|A|A^{-1}=cA^{-1}$（因 $|A|=c\neq0，A$ 可逆），代入得 $\boldsymbol{e}^T A^{-1}\boldsymbol{e}=3$。 - -利用**Sherman-Morrison 行列式公式**：对可逆矩阵$A$ 和向量 $\boldsymbol{u},\boldsymbol{v}$，有 $|A+\boldsymbol{u}\boldsymbol{v}^T|=|A|(1+\boldsymbol{v}^T A^{-1}\boldsymbol{u})$。 - -此处 $\boldsymbol{u}=\boldsymbol{v}=\boldsymbol{e}$，代入得 $|A+J|=c(1+\boldsymbol{e}^T A^{-1}\boldsymbol{e})=c(1+3)=4c$。 - +没有全1的行列式：即$|A|=c$； +仅 1 列是全 1、其余为$A$中某一列的行列式：共 3 个，分别对应第 1、2、3 列拆分为全 1 列。此类行列式按全 1 列展开，值为该列的代数余子式之和，即$\sum\limits_{i=1}^3A_{i1},\sum\limits_{i=1}^3A_{i3},\sum\limits_{i=1}^3A_{i2}$ ，总和为$\sum\limits_{i,j=1}^3A_{ij}=3c$; +含两列及以上全1的行列式：等于0 +综上原式$=4c$. ### **2. 答案：$\boxed{2}$ #### **解析：** @@ -92,9 +88,14 @@ $代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1 #### **解析：** -将二次型展开为矩阵形式： +记$A_i$为$A$中把除了第$i$列之外全部都换成$0$的矩阵，则将二次型展开为矩阵形式： -$$f(\boldsymbol{x})=\sum_{i=1}^{n}(A\boldsymbol{x})_i^2=(A\boldsymbol{x})^T(A\boldsymbol{x})=\boldsymbol{x}^T(A^T A)\boldsymbol{x}$$ +$$\begin{aligned} +f(\boldsymbol{x})&=\sum_{i=1}^{n}(<A_i\boldsymbol{x},A_i\boldsymbol{x}>)^2\\ +&=\sum\limits_{i=1}^n((A_i\boldsymbol{x})^TA_i\boldsymbol{x})\\ +&=\boldsymbol{x}^T\sum\limits_{i=i}^{n}A_i^T\sum\limits_{i=1}^nA_i\boldsymbol{x}\\ +&=\boldsymbol{x}^T(A^T A)\boldsymbol{x} +\end{aligned}$$ • 二次型的矩阵需满足**对称性质**（即矩阵等于其转置），验证：$(A^T A)^T=A^T(A^T)^T=A^T A$，故 $A^T A$ 是对称矩阵，即为二次型的矩阵。 -- 2.34.1 From 79f732d0f54f5e803e494730d490244dccfb10ff Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Tue, 13 Jan 2026 15:25:23 +0800 Subject: [PATCH 217/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/微分中值定理.md | 218 +++++++++++++++++++++++++++++++++++ 1 file changed, 218 insertions(+) create mode 100644 素材/微分中值定理.md diff --git a/素材/微分中值定理.md b/素材/微分中值定理.md new file mode 100644 index 0000000..6304f45 --- /dev/null +++ b/素材/微分中值定理.md @@ -0,0 +1,218 @@ + +### 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $0 < a < b$，试证存在 $\xi, \eta \in (a, b)$，使得 +$$ +f'(\xi) = \frac{a + b}{2\eta} f'(\eta). +$$ + +**解析**： +本题结论中含有两个不同的中值 $\xi$ 和 $\eta$，且涉及两个不同的函数形式。可考虑分别对 $f(x)$ 和 $g(x)=x^2$ 在 $[a,b]$ 上应用柯西中值定理： +由柯西中值定理，存在 $\eta \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b^2-a^2} = \frac{f'(\eta)}{2\eta} +$$ +整理得 +$$ +\frac{f(b)-f(a)}{b-a} = \frac{a+b}{2\eta} f'(\eta) +$$ +再对 $f(x)$ 在 $[a,b]$ 上应用拉格朗日中值定理，存在 $\xi \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b-a} = f'(\xi) +$$ +比较两式即得结论。 + +--- + +### 例2 +设函数 $f(x)$ 在 $[0,3]$ 上连续，在 $(0,3)$ 内可导，且 $f(0) + f(1) + f(2) = 3$，$f(3) = 1$，试证必存在 $\xi \in (0, 3)$，使 $f'(\xi) = 0$。 + +**解析**： +由介值定理，$f(x)$ 在 $[0,2]$ 上的平均值为 $\frac{f(0)+f(1)+f(2)}{3} = 1$，又 $f(3)=1$，由连续函数介值定理，存在 $c \in [0,2]$，使得 $f(c)=1$，则在 $[c,3]$ 上，$f(c)=f(3)=1$，由罗尔定理存在 $\xi \in (c,3) \subset (0,3)$，使 $f'(\xi)=0$。 + +--- + +### 例3 +设 $f(x)$ 在区间 $[0,1]$ 上连续，在 $(0,1)$ 内可导，且 $f(0) = f(1) = 0$，$f(1/2) = 1$，试证： +1. 存在 $\eta \in (1/2, 1)$，使得 $f(\eta) = \eta$； +2. 对任意实数 $\lambda$，必存在 $\xi \in (0, \eta)$，使得 $f'(\xi) - \lambda [f(\xi) - \xi] = 1$。 + +**解析**： +(1) 令 $g(x)=f(x)-x$，则 $g(1/2)=1-1/2=1/2>0$，$g(1)=0-1=-1<0$，由零点定理，存在 $\eta \in (1/2,1)$，使 $g(\eta)=0$，即 $f(\eta)=\eta$。 +(2) 令 $h(x)=e^{-\lambda x}[f(x)-x]$，则 $h(0)=0$，$h(\eta)=0$，由罗尔定理，存在 $\xi \in (0,\eta)$，使 $h'(\xi)=0$，即 +$$ +e^{-\lambda \xi}[f'(\xi)-1] - \lambda e^{-\lambda \xi}[f(\xi)-\xi] = 0 +$$ +整理得 $f'(\xi) - \lambda [f(\xi) - \xi] = 1$。 + +--- + +## 5.2 微分中值定理及其应用 + +### 例1 +设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 +$$ +f(0) = 0, \quad |f'(x)| \leq 1, +$$ +证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 + +**解析**： +对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 +$$ +f(x) - f(0) = f'(\xi)(x-0) +$$ +即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 + +--- + +### 例2 +设 $a_i \in \mathbb{R} (i = 0,1,2,\cdots,n)$，且满足 +$$ +a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} = 0， +$$ +证明：方程 $a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0$ 在 $(0,1)$ 内至少有一个实根。 + +**解析**： +构造辅助函数 +$$ +F(x) = a_0x + \frac{a_1}{2}x^2 + \frac{a_2}{3}x^3 + \cdots + \frac{a_n}{n+1}x^{n+1} +$$ +则 $F(0)=0$，且由条件 $F(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使 $F'(\xi)=0$，即 +$$ +a_0 + a_1\xi + a_2\xi^2 + \cdots + a_n\xi^n = 0 +$$ + +--- + +### 例3 +设函数 $f(x)$ 在 $[a,b]$ 上可导，且 +$$ +f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0， +$$ +试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 + +**解析**： +由导数极限定理及 $f'_+(a)f'_-(b) > 0$，知在 $a$ 右侧和 $b$ 左侧，$f(x)$ 的符号相同，不妨设 $f'_+(a)>0$，$f'_-(b)>0$。则在 $a$ 右侧附近 $f(x)>0$，在 $b$ 左侧附近 $f(x)>0$。由于 $f(a)=f(b)=0$，由极值点的费马定理，$f(x)$ 在 $(a,b)$ 内至少有一个极大值点，该点处导数为零。又因为 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导，且 $f(a)=f(b)$，由罗尔定理至少存在一点 $c \in (a,b)$ 使 $f'(c)=0$。结合极大值点处的导数零点，可知至少有两个导数为零的点。 + +--- + +### 例4 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 + +**解析**： +弦 $AB$ 的方程为 +$$ +y = f(a) + \frac{f(b)-f(a)}{b-a}(x-a) +$$ +由条件，$f(c) = f(a) + \frac{f(b)-f(a)}{b-a}(c-a)$。分别对 $f(x)$ 在 $[a,c]$ 和 $[c,b]$ 上应用拉格朗日中值定理，存在 $\xi_1 \in (a,c)$，$\xi_2 \in (c,b)$，使得 +$$ +f'(\xi_1) = \frac{f(c)-f(a)}{c-a} = \frac{f(b)-f(a)}{b-a} +$$ +$$ +f'(\xi_2) = \frac{f(b)-f(c)}{b-c} = \frac{f(b)-f(a)}{b-a} +$$ +故 $f'(\xi_1)=f'(\xi_2)$。再对 $f'(x)$ 在 $[\xi_1,\xi_2]$ 上应用罗尔定理，存在 $\xi \in (\xi_1,\xi_2) \subset (a,b)$，使 $f''(\xi)=0$。 + +--- + +### 例5（柯西中值定理例） +试证至少存在一点 $\xi \in (1, e)$，使 $\sin 1 = \cos \ln \xi$。 + +**解析**： +考虑函数 $f(x)=\sin(\ln x)$，$g(x)=\ln x$，在 $[1,e]$ 上应用柯西中值定理： +存在 $\xi \in (1,e)$，使得 +$$ +\frac{f(e)-f(1)}{g(e)-g(1)} = \frac{f'(\xi)}{g'(\xi)} +$$ +计算得 $f(e)=\sin 1$，$f(1)=0$，$g(e)=1$，$g(1)=0$，$f'(x)=\frac{\cos(\ln x)}{x}$，$g'(x)=\frac{1}{x}$，代入得 +$$ +\frac{\sin 1 - 0}{1-0} = \frac{\cos(\ln \xi)/\xi}{1/\xi} = \cos(\ln \xi) +$$ +即 $\sin 1 = \cos(\ln \xi)$。 + +--- + +## 练习 + +### Ex3 +设 $f(x)$ 在 $(a, b)$ 内可导，且 $f'(x) \neq 1$。试证明 $f(x)$ 在 $(a, b)$ 内至多只有一个不动点，即方程 $f(x) = x$ 在 $(a, b)$ 内至多只有一个实根。 + +**解析**： +反证法。假设存在两个不动点 $x_1 < x_2$，即 $f(x_1)=x_1$，$f(x_2)=x_2$。由拉格朗日中值定理，存在 $\xi \in (x_1,x_2)$，使得 +$$ +f'(\xi) = \frac{f(x_2)-f(x_1)}{x_2-x_1} = \frac{x_2-x_1}{x_2-x_1} = 1 +$$ +与 $f'(x) \neq 1$ 矛盾。故至多只有一个不动点。 + +--- + +### Ex4 +设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 +$$ +f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4} +$$ +证明： +1. 至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； +2. 若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 + +**解析**： +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 +(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 + +--- + +### Ex5 +若 $f(x)$ 可导，试证在其两个零点间一定有 $f(x) + f'(x)$ 的零点。 + +**解析**： +设 $a<b$ 为 $f(x)$ 的两个零点，即 $f(a)=f(b)=0$。构造辅助函数 $F(x)=e^x f(x)$，则 $F(a)=F(b)=0$。由罗尔定理，存在 $\xi \in (a,b)$，使 $F'(\xi)=0$，即 +$$ +e^\xi f(\xi) + e^\xi f'(\xi) = 0 +$$ +因 $e^\xi \neq 0$，故 $f(\xi)+f'(\xi)=0$。 + +--- + +### Ex6 +设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 + +**解析**： +设辅助函数 $\varphi(x) = x^n f(x)$，则 $\varphi(0)=0$，$\varphi(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使得 $\varphi'(\xi)=0$，即 +$$ +n\xi^{n-1} f(\xi) + \xi^n f'(\xi) = 0 +$$ +两边除以 $\xi^{n-1}$ ($\xi>0$)，得 $nf(\xi) + \xi f'(\xi) = 0$。 + +--- + +### Ex7 +设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 +$$ +f(x_1 + x_2) < f(x_1) + f(x_2) +$$ + +**解析**： +不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： +$$ +f(x_1+x_2)-f(x_2) = f'(\xi_1)x_1, \quad \xi_1 \in (x_2, x_1+x_2) +$$ +$$ +f(x_1)-f(0) = f'(\xi_2)x_1, \quad \xi_2 \in (0, x_1) +$$ +于是 +$$ +f(x_1+x_2)-f(x_2)-f(x_1) = [f'(\xi_1)-f'(\xi_2)]x_1 +$$ +对 $f'(x)$ 在 $[\xi_2,\xi_1]$ 上应用中值定理，存在 $\xi \in (\xi_2,\xi_1)$，使 +$$ +f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 +$$ +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 + +--- + +## 解题方法总结 +1. **含一个中值的等式或根的存在**：多用罗尔定理，可用原函数法找辅助函数。 +2. **结论涉及含中值的两个不同函数**：可考虑用柯西中值定理。 +3. **结论中含两个或两个以上的中值**：必须多次应用中值定理。 +4. **已知条件中含高阶导数**：多考虑用泰勒公式，有时也可考虑对导数用中值定理。 +5. **结论为不等式**：要注意适当放大或缩小的技巧。 \ No newline at end of file -- 2.34.1 From 18c628a6f5a980b32a013c6c1a08e9ed0a9d644b Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Tue, 13 Jan 2026 15:32:55 +0800 Subject: [PATCH 218/274] vault backup: 2026-01-13 15:32:55 --- .../试卷/线代期末复习模拟/1.14线代测试答案.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md index 6afd7b5..d9519dd 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md @@ -212,7 +212,7 @@ $$F_k=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^k-\left(\frac{1- 对比初始条件：$D_1=1=F_2，D_2=2=F_3$，故 $D_n=F_{n+1}$（斐波那契数列的第 $n+1$ 项）。 -将$ F_{n+1} 代入比内公式$，得： +将$F_{n+1}$代入比内公式，得： $$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$$ -- 2.34.1 From 393b740fa0e694538356521cbc4d238058d2be19 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 15:42:38 +0800 Subject: [PATCH 219/274] vault backup: 2026-01-13 15:42:38 --- .../试卷/线代期末复习模拟/1.14线代测试答案.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md index 6afd7b5..ac44c30 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md @@ -212,7 +212,7 @@ $$F_k=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^k-\left(\frac{1- 对比初始条件：$D_1=1=F_2，D_2=2=F_3$，故 $D_n=F_{n+1}$（斐波那契数列的第 $n+1$ 项）。 -将$ F_{n+1} 代入比内公式$，得： +将$F_{n+1} 代入比内公式$，得： $$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$$ -- 2.34.1 From 95f6b4a8ba248266f33d88e2935e0be8a5fa1951 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 16:50:30 +0800 Subject: [PATCH 220/274] vault backup: 2026-01-13 16:50:30 --- .../1.14线代测试答案.md | 52 ++++++++++--------- 1 file changed, 28 insertions(+), 24 deletions(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md index d9519dd..189ee67 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md @@ -4,7 +4,7 @@ ### **1.** -已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ +已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum_{i=1}^{3}\sum_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ ### **2.（2013 秋 A）** @@ -54,10 +54,14 @@ $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vd ### 1. 答案：$\boxed{4c}$ #### **解析：** -没有全1的行列式：即$|A|=c$； -仅 1 列是全 1、其余为$A$中某一列的行列式：共 3 个，分别对应第 1、2、3 列拆分为全 1 列。此类行列式按全 1 列展开，值为该列的代数余子式之和，即$\sum\limits_{i=1}^3A_{i1},\sum\limits_{i=1}^3A_{i3},\sum\limits_{i=1}^3A_{i2}$ ，总和为$\sum\limits_{i,j=1}^3A_{ij}=3c$; -含两列及以上全1的行列式：等于0 -综上原式$=4c$. + +$设原矩阵为 A（即 |A|=c），全 1 矩阵 J=\boldsymbol{e}\boldsymbol{e}^T（其中 \boldsymbol{e}=(1,1,1)^T），需求解的行列式为 |A+J|。$ +由代数余子式性质：$\sum_{i,j}A_{ij}=\boldsymbol{e}^T A^*\boldsymbol{e}=3c$，且可逆矩阵的伴随矩阵满足$A^*=|A|A^{-1}=cA^{-1}$（因 $|A|=c\neq0，A$ 可逆），代入得 $\boldsymbol{e}^T A^{-1}\boldsymbol{e}=3$。 + +利用**Sherman-Morrison 行列式公式**：对可逆矩阵$A$ 和向量 $\boldsymbol{u},\boldsymbol{v}$，有 $|A+\boldsymbol{u}\boldsymbol{v}^T|=|A|(1+\boldsymbol{v}^T A^{-1}\boldsymbol{u})$。 + +此处 $\boldsymbol{u}=\boldsymbol{v}=\boldsymbol{e}$，代入得 $|A+J|=c(1+\boldsymbol{e}^T A^{-1}\boldsymbol{e})=c(1+3)=4c$。 + ### **2. 答案：$\boxed{2}$ #### **解析：** @@ -88,14 +92,16 @@ $代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1 #### **解析：** -记$A_i$为$A$中把除了第$i$列之外全部都换成$0$的矩阵，则将二次型展开为矩阵形式： +记$A_i$为$A$中除了第$i$行全都改为$0$的矩阵，$\boldsymbol{x}=\begin{bmatrix}x_1&x_2&\cdots&x_n\end{bmatrix}^T$。那么$$A_i\boldsymbol{x}=\begin{bmatrix}0 & 0 & \cdots & 0\\\vdots & \vdots & &\vdots\\a_{i1} & a_{i2} & \cdots &a_{in}\\\vdots & \vdots & &\vdots\\0 & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\\sum\limits_{j=1}^{n}a_{ij}x_j\\\vdots\\0\end{bmatrix}$$则$$(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2=(\sum\limits_{j=1}^{n}a_{ij}x_j)^2=(A_i\boldsymbol{x})^TA_i\boldsymbol{x}=\boldsymbol{x}^TA_i^TA_i\boldsymbol{x}$$将二次型用上式展开得： $$\begin{aligned} -f(\boldsymbol{x})&=\sum_{i=1}^{n}(<A_i\boldsymbol{x},A_i\boldsymbol{x}>)^2\\ -&=\sum\limits_{i=1}^n((A_i\boldsymbol{x})^TA_i\boldsymbol{x})\\ -&=\boldsymbol{x}^T\sum\limits_{i=i}^{n}A_i^T\sum\limits_{i=1}^nA_i\boldsymbol{x}\\ +f(\boldsymbol{x})&=\sum_{i=1}^{n}(\boldsymbol{x}^TA_i^TA_i\boldsymbol{x})\\ +&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^TA_i)\boldsymbol{x}\\ +&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i)\boldsymbol{x}\\ &=\boldsymbol{x}^T(A^T A)\boldsymbol{x} \end{aligned}$$ +其中因为$A_i^TA_j=\boldsymbol{0},$如果$i\neq j$，所以 +$$\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i=\sum\limits_{i=1}^nA_i^TA_i$$ • 二次型的矩阵需满足**对称性质**（即矩阵等于其转置），验证：$(A^T A)^T=A^T(A^T)^T=A^T A$，故 $A^T A$ 是对称矩阵，即为二次型的矩阵。 @@ -184,7 +190,7 @@ $$ ### **9. 证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$** -#### **步骤 1：建立递推公式** + **步骤 1：建立递推公式** 对 $D_n$ 按**第一行展开**（第一行元素为 $a_{11}=1,a_{12}=1$，其余 $a_{1j}=0$）： @@ -198,22 +204,20 @@ $D_n=a_{11}\times(-1)^{1+1}M_{11}+a_{12}\times(-1)^{1+2}M_{12}$ $$D_n=D_{n-1}+D_{n-2}$$ -#### **步骤 2：确定初始条件** + **步骤 2：确定初始条件** • 当 $n=1$ 时，$D_1=\begin{vmatrix}1\end{vmatrix}=1$； • 当 n=2 时，$D_2=\begin{vmatrix}1&1\\-1&1\end{vmatrix}=1\times1-1\times(-1)=2$。 +容易证明$n=1,n=2$时满足要证的式子 -#### **步骤 3：求解递推关系** - -递推公式 $D_n=D_{n-1}+D_{n-2}$ 是**斐波那契数列的变形**，斐波那契数列的通项公式（比内公式）为： - -$$F_k=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^k-\left(\frac{1-\sqrt{5}}{2}\right)^k\right]$$ - -对比初始条件：$D_1=1=F_2，D_2=2=F_3$，故 $D_n=F_{n+1}$（斐波那契数列的第 $n+1$ 项）。 - -将$F_{n+1}$代入比内公式，得： - -$$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$$ - -证毕。 \ No newline at end of file +**步骤3：运用数学归纳法** +假设当$n\le k$时，有$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。当$n=k+1$时，有$$ +\begin{aligned} +D_{k+1}&=D_k+D_{k-1}\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+1}-(\frac{1-\sqrt{5}}{2})^{k+1}+(\frac{1+\sqrt{5}}{2})^k-(\frac{1-\sqrt{5}}{2})^k)\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{3+\sqrt{5}}{2})-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2}))\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{1+\sqrt{5}}{2})^2-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2})^2)\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+2}-(\frac{1-\sqrt{5}}{2})^{k+2}) +\end{aligned} +$$满足条件，故由数学归纳法知，$\forall{n}\in\mathbb{N}_+,D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$. \ No newline at end of file -- 2.34.1 From 5887a7b71bf828fe5f32d85338d6e9b475c91ba9 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 17:12:22 +0800 Subject: [PATCH 221/274] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=BA=86=E6=9C=89?= =?UTF-8?q?=E5=85=B3=E7=89=B9=E5=BE=81=E5=80=BC=E7=9A=84=E7=B4=A0=E6=9D=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/特征值.md | 31 +++++++++++++++++++++++++++++++ 1 file changed, 31 insertions(+) create mode 100644 素材/特征值.md diff --git a/素材/特征值.md b/素材/特征值.md new file mode 100644 index 0000000..35ccf38 --- /dev/null +++ b/素材/特征值.md @@ -0,0 +1,31 @@ +>[!note] 定理 +>秩为$1$的矩阵$A\in\mathbb{R}^{n\times n}$的特征值有如下特征：（1）$0$为其特征值，且代数重数和几何重数均为$n-1$；（2）它的另一个特征值为$\mathrm{tr}(A)$. + +**证明：** +根据迹的定义，只需要证明（1）。 +因为$r(A)=1<n$,所以$|A|=0$，故$0$是$A$的一个特征值。考虑齐次线性方程组$A\boldsymbol{x}=\boldsymbol{0}$.由于$r(A)=1$，$\mathrm{dim}N(A)=n-1$，所以特征值$0$的几何重数为$n-1$。若$0$的代数重数为$n$，则$A\sim O$,而相似必等价，故$r(A)=0$，矛盾。又代数重数必定不小于几何重数，所以$0$的代数重数为$n-1$。 +特殊地，如果$A=\beta^T\alpha$,则$\mathrm{tr}(A)=\alpha\beta^T$. + +>[!example] 例1 +>设 $E$ 为 $3$ 阶单位矩阵，$\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $E-\alpha\alpha^T$ 的全部 $3$ 个特征值为$\underline{\qquad}$。 + +**解：** +设 $B=\alpha\alpha^T$，该矩阵为**秩 $1$ 矩阵**（因 $\alpha$ 是单位列向量，$\alpha^T\alpha=1$）。 + +• 秩 $1$ 矩阵的特征值性质：非零特征值为矩阵的迹 $\text{tr}(B)=\alpha^T\alpha=1$，其余 $n-1=2$ 个特征值为 $0$（秩 $1$ 矩阵的非零特征值个数等于秩）。 + +• 若 $B\boldsymbol{x}=\lambda\boldsymbol{x}$（$\boldsymbol{x}$为特征向量），则 $(E-B)\boldsymbol{x}=(1-\lambda)\boldsymbol{x}$，即$E-B$ 的特征值为 $1-\lambda$。 + +$代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1-0=1，即 1,1,0$。 + +>[!example] 例2 +>已知 $n(n\geq2)$维列向量 $\alpha,\beta$ 满足 $\beta^T\alpha=-3$，则方阵 $(\beta\alpha^T)^2$ 的非零特征值为$\underline{\qquad}$。 + +**解：** +设 $A=\beta\alpha^T$（秩 1 矩阵），计算 $A^2$： + +$A^2=(\beta\alpha^T)(\beta\alpha^T)=\beta(\alpha^T\beta)\alpha^T=(\alpha^T\beta)A$ + +• 注意：$\alpha^T\beta=(\beta^T\alpha)^T$（矩阵转置性质），而 $\beta^T\alpha=-3$（数，转置等于自身），故 $\alpha^T\beta=-3$，因此 $A^2=-3A$。 + +• 秩 $1$ 矩阵 $A$ 的非零特征值为 $\text{tr}(A)=\alpha^T\beta=-3$，设 $A\boldsymbol{x}=-3\boldsymbol{x}$，则 $A^2\boldsymbol{x}=(-3)A\boldsymbol{x}=(-3)^2\boldsymbol{x}=9\boldsymbol{x}$，即 $A^2$ 的非零特征值为 $9$。 \ No newline at end of file -- 2.34.1 From 2a5fc288a9b92c753100b9bf963519b4a208dc47 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 17:15:12 +0800 Subject: [PATCH 222/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E5=90=8D?= =?UTF-8?q?=E5=AD=97=E4=B8=AD=E7=9A=84=E6=97=A5=E6=9C=9F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1.13线代测试答案.md | 223 ++++++++++++++++++ .../1.13线代限时练.md | 66 ++++++ 2 files changed, 289 insertions(+) create mode 100644 编写小组/试卷/线代期末复习模拟/1.13线代测试答案.md create mode 100644 编写小组/试卷/线代期末复习模拟/1.13线代限时练.md diff --git a/编写小组/试卷/线代期末复习模拟/1.13线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.13线代测试答案.md new file mode 100644 index 0000000..4f0ed9d --- /dev/null +++ b/编写小组/试卷/线代期末复习模拟/1.13线代测试答案.md @@ -0,0 +1,223 @@ +# **1.13 线性代数限时练（题目 + 答案与解析）** + +## **第一部分：题目** + +### **1.** + +已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum_{i=1}^{3}\sum_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ + +### **2.（2013 秋 A）** + +$已知向量空间 V=\{(2a,2b,3b,3a)\mid a,b\in\mathbb{R}\}，则 V 的维数是\underline{\qquad}。$ + +### **3.（2018 秋 A）** + +$设 E 为 3 阶单位矩阵，\alpha 为一个 3 维单位列向量，则矩阵 E-\alpha\alpha^T 的全部 3 个特征值为\underline{\qquad}。$ + +### **4.（2018 秋 A）** + +$设 n 阶矩阵 A=[a_{ij}]_{n\times n}，则二次型 f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^{n}(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2 的矩阵为\underline{\qquad}。$ +### **5.（2018 秋 A）** + +$若 n 阶实对称矩阵 A 的特征值为 \lambda_i=(-1)^i（i=1,2,\cdots,n），则 A^{100}=\underline{\qquad}。$ + +### **6.（2022 秋 A・5）** + +$已知 n（n\geq2）维列向量 \alpha,\beta 满足 \beta^T\alpha=-3，则方阵 (\beta\alpha^T)^2 的非零特征值为\underline{\qquad}。$ + +### **7.（2022 秋 A）** + +$已知向量组 \alpha_1,\alpha_2,\alpha_3 线性无关（\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}^3），A 为 3 阶方阵，且满足：$ + +$$\begin{aligned} +A\alpha_1&=2\alpha_1-\alpha_2-\alpha_3,\\A\alpha_2&=\alpha_1+2\alpha_2+3\alpha_3,\\A\alpha_3&=2\alpha_1+4\alpha_2+\alpha_3 +\end{aligned} +$$ +(1) $证明 A\alpha_1,A\alpha_2,A\alpha_3 线性无关$； + +(2) $计算行列式 |E-A|（E 是 3 阶单位矩阵）$。 + +### **8.（2013 秋 A）** + +$求 n 阶方阵 A=\begin{bmatrix}1&1&1&\cdots&1\\1&0&1&\cdots&1\\1&1&0&\cdots&1\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&1&1&\cdots&0\end{bmatrix} 的逆矩阵。$ + +### **9.（2013 秋 A・三）** + +设 n 阶行列式： + +$$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$ + +证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。 + +## **第二部分：答案与解析** + +### 1. 答案：$\boxed{4c}$ + +#### **解析：** + +$设原矩阵为 A（即 |A|=c），全 1 矩阵 J=\boldsymbol{e}\boldsymbol{e}^T（其中 \boldsymbol{e}=(1,1,1)^T），需求解的行列式为 |A+J|。$ +由代数余子式性质：$\sum_{i,j}A_{ij}=\boldsymbol{e}^T A^*\boldsymbol{e}=3c$，且可逆矩阵的伴随矩阵满足$A^*=|A|A^{-1}=cA^{-1}$（因 $|A|=c\neq0，A$ 可逆），代入得 $\boldsymbol{e}^T A^{-1}\boldsymbol{e}=3$。 + +利用**Sherman-Morrison 行列式公式**：对可逆矩阵$A$ 和向量 $\boldsymbol{u},\boldsymbol{v}$，有 $|A+\boldsymbol{u}\boldsymbol{v}^T|=|A|(1+\boldsymbol{v}^T A^{-1}\boldsymbol{u})$。 + +此处 $\boldsymbol{u}=\boldsymbol{v}=\boldsymbol{e}$，代入得 $|A+J|=c(1+\boldsymbol{e}^T A^{-1}\boldsymbol{e})=c(1+3)=4c$。 + +### **2. 答案：$\boxed{2}$ + +#### **解析：** + +将向量空间 V 中的元素拆分为线性组合形式： + +$$(2a,2b,3b,3a)=a(2,0,0,3)+b(0,2,3,0)$$ + +设 $\boldsymbol{\alpha}=(2,0,0,3)，\boldsymbol{\beta}=(0,2,3,0)$，需验证 $\boldsymbol{\alpha},\boldsymbol{\beta}$ 线性无关： + +若 $k_1\boldsymbol{\alpha}+k_2\boldsymbol{\beta}=\boldsymbol{0}$（零向量），则 $\begin{cases}2k_1=0\\2k_2=0\\3k_2=0\\3k_1=0\end{cases}$，解得 $k_1=k_2=0$，故 $\boldsymbol{\alpha},\boldsymbol{\beta}$是 V 的一组基。 + +向量空间的维数等于基的个数，因此 $V$ 的维数为 $2$。 + +### **3. 答案：$\boxed{1,1,0}$ + +#### **解析：** + +设 $B=\alpha\alpha^T$，该矩阵为**秩 $1$ 矩阵**（因 $\alpha$ 是单位列向量，$\alpha^T\alpha=1$）。 + +• 秩 $1$ 矩阵的特征值性质：非零特征值为矩阵的迹 $\text{tr}(B)=\alpha^T\alpha=1$，其余 $n-1=2$ 个特征值为 $0$（秩 $1$ 矩阵的非零特征值个数等于秩）。 + +• 若 $B\boldsymbol{x}=\lambda\boldsymbol{x}$（$\boldsymbol{x}$为特征向量），则 $(E-B)\boldsymbol{x}=(1-\lambda)\boldsymbol{x}$，即$E-B$ 的特征值为 $1-\lambda$。 + +$代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1-0=1，即 1,1,0$。 + +### **4. 答案：$\boxed{A^T A}$** + +#### **解析：** + +记$A_i$为$A$中除了第$i$行全都改为$0$的矩阵，$\boldsymbol{x}=\begin{bmatrix}x_1&x_2&\cdots&x_n\end{bmatrix}^T$。那么$$A_i\boldsymbol{x}=\begin{bmatrix}0 & 0 & \cdots & 0\\\vdots & \vdots & &\vdots\\a_{i1} & a_{i2} & \cdots &a_{in}\\\vdots & \vdots & &\vdots\\0 & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\\sum\limits_{j=1}^{n}a_{ij}x_j\\\vdots\\0\end{bmatrix}$$则$$(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2=(\sum\limits_{j=1}^{n}a_{ij}x_j)^2=(A_i\boldsymbol{x})^TA_i\boldsymbol{x}=\boldsymbol{x}^TA_i^TA_i\boldsymbol{x}$$将二次型用上式展开得： + +$$\begin{aligned} +f(\boldsymbol{x})&=\sum_{i=1}^{n}(\boldsymbol{x}^TA_i^TA_i\boldsymbol{x})\\ +&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^TA_i)\boldsymbol{x}\\ +&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i)\boldsymbol{x}\\ +&=\boldsymbol{x}^T(A^T A)\boldsymbol{x} +\end{aligned}$$ +其中因为$A_i^TA_j=\boldsymbol{0},$如果$i\neq j$，所以 +$$\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i=\sum\limits_{i=1}^nA_i^TA_i$$ + +• 二次型的矩阵需满足**对称性质**（即矩阵等于其转置），验证：$(A^T A)^T=A^T(A^T)^T=A^T A$，故 $A^T A$ 是对称矩阵，即为二次型的矩阵。 + +### **5. 答案：$\boxed{E}$（单位矩阵）** + +#### **解析：** + +实对称矩阵可对角化，即存在可逆矩阵 P，使得 $P^{-1}AP=\Lambda$（$\Lambda$ 为对角矩阵，对角元为 A 的特征值 $\lambda_i=(-1)^i$）。 + +• 矩阵幂运算性质：$A^{100}=P\Lambda^{100}P^{-1}$。 + +• 计算 $\Lambda^{100}$：对角元为 $\lambda_i^{100}=[(-1)^i]^{100}=1$，故 $\Lambda^{100}=E$（单位矩阵）。 + +• 因此 $A^{100}=P E P^{-1}=P P^{-1}=E$。 + +### **6. 答案：$\boxed{9}$** + +#### **解析：** + +设 $A=\beta\alpha^T$（秩 1 矩阵），计算 $A^2$： + +$A^2=(\beta\alpha^T)(\beta\alpha^T)=\beta(\alpha^T\beta)\alpha^T=(\alpha^T\beta)A$ + +• 注意：$\alpha^T\beta=(\beta^T\alpha)^T$（矩阵转置性质），而 $\beta^T\alpha=-3$（数，转置等于自身），故 $\alpha^T\beta=-3$，因此 $A^2=-3A$。 + +• 秩 $1$ 矩阵 $A$ 的非零特征值为 $\text{tr}(A)=\alpha^T\beta=-3$，设 $A\boldsymbol{x}=-3\boldsymbol{x}$，则 $A^2\boldsymbol{x}=(-3)A\boldsymbol{x}=(-3)^2\boldsymbol{x}=9\boldsymbol{x}$，即 $A^2$ 的非零特征值为 $9$。 + +### **7. 答案：(1) 证明见解析；(2) $\boxed{20}$** + +#### **(1) 证明 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关** + +因 $\alpha_1,\alpha_2,\alpha_3$ 线性无关，故矩阵 $P=(\alpha_1,\alpha_2,\alpha_3)$ 可逆（列向量线性无关的矩阵可逆）。 + +由题设条件，将 $A$ 对 $\alpha_1,\alpha_2,\alpha_3$ 的作用表示为矩阵乘法： + +$$A(\alpha_1,\alpha_2,\alpha_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}=P C$$ + +其中 $C=\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}$，计算 $|C|$： + +$$\begin{aligned} +|C|&=2\times(2\times1-4\times3)-1\times(-1\times1-4\times(-1))+2\times(-1\times3-2\times(-1))\\ +&=2\times(-10)-1\times3+2\times(-1)\\ +&=-25\neq0 +\end{aligned} +$$ + +• 因 $|C|\neq0$，故 $C$ 可逆，$\text{rank}(C)=3$。 + +• 又 $P$ 可逆，故 $\text{rank}(A\alpha_1,A\alpha_2,A\alpha_3)=\text{rank}(P C)=\text{rank}(C)=3$，即 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关。 + +#### **(2) 计算 $|E-A|$** + +由 (1) 知$P^{-1}AP=C（A 与 C 相似）$，则 $E-A$ 与 $E-C$ 相似（相似矩阵的 “单位矩阵减矩阵” 仍相似），而**相似矩阵的行列式相等**，故 $|E-A|=|E-C|$。 + +计算 $E-C$： + +$$E-C=\begin{bmatrix}1-2&0-1&0-2\\0-(-1)&1-2&0-4\\0-(-1)&0-3&1-1\end{bmatrix}=\begin{bmatrix}-1&-1&-2\\1&-1&-4\\1&-3&0\end{bmatrix}$$ + +按第三行展开计算行列式： + +$$ +\begin{aligned} +|E-C|&=1\times\begin{vmatrix}-1&-2\\-1&-4\end{vmatrix}-(-3)\times\begin{vmatrix}-1&-2\\1&-4\end{vmatrix}+0\times(\text{余子式})\\ +&=1\times(4-2)+3\times(4+2)\\ +&=2+18=20 +\end{aligned} +$$ + +故 $|E-A|=20$。 + +### **8. 答案：** + +$$A^{-1}=\begin{bmatrix}-(n-2)&1&1&\cdots&1\\1&-1&0&\cdots&0\\1&0&-1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&0&0&\cdots&-1\end{bmatrix} +$$ +#### **解析：** + +通过 “行变换法” 或 “规律归纳” 推导： + +• 观察矩阵 A 的结构：第一行全为 1，其余行的对角元为 0，非对角元为 1。可先计算 n=2,3 时的逆矩阵，归纳规律： + +◦ $当 n=2 时，A=\begin{bmatrix}1&1\\1&0\end{bmatrix}，逆矩阵为 \begin{bmatrix}0&1\\1&-1\end{bmatrix}（符合上述形式，-(2-2)=0）$； + +◦ 当 $n=3$ 时，$A=\begin{bmatrix}1&1&1\\1&0&1\\1&1&0\end{bmatrix}$，逆矩阵为 $\begin{bmatrix}-1&1&1\\1&-1&0\\1&0&-1\end{bmatrix}$（符合上述形式，$-(3-2)=-1$）。 + +• 验证规律：对 n 阶矩阵，逆矩阵的第一行第一列元素为 -(n-2)，第一行其余元素为 1，第一列其余元素为 1，对角元（除第一行第一列）为 -1，非对角元（除第一行、第一列）为 0，即为上述形式。 + +### **9. 证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$** + + **步骤 1：建立递推公式** + +对 $D_n$ 按**第一行展开**（第一行元素为 $a_{11}=1,a_{12}=1$，其余 $a_{1j}=0$）： + +$D_n=a_{11}\times(-1)^{1+1}M_{11}+a_{12}\times(-1)^{1+2}M_{12}$ + +• $M_{11}$：去掉第一行第一列后的子式，即 $n-1$ 阶行列式 $D_{n-1}$（结构与 $D_n$ 一致）； + +• $M_{12}$：去掉第一行第二列后的子式，按第一列展开（第一列仅首元素为 $-1$），得 $-D_{n-2}$（符号需结合 $(-1)^{1+2}=-1$）。 + +因此递推公式为： + +$$D_n=D_{n-1}+D_{n-2}$$ + + **步骤 2：确定初始条件** + +• 当 $n=1$ 时，$D_1=\begin{vmatrix}1\end{vmatrix}=1$； + +• 当 n=2 时，$D_2=\begin{vmatrix}1&1\\-1&1\end{vmatrix}=1\times1-1\times(-1)=2$。 +容易证明$n=1,n=2$时满足要证的式子 + +**步骤3：运用数学归纳法** +假设当$n\le k$时，有$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。当$n=k+1$时，有$$ +\begin{aligned} +D_{k+1}&=D_k+D_{k-1}\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+1}-(\frac{1-\sqrt{5}}{2})^{k+1}+(\frac{1+\sqrt{5}}{2})^k-(\frac{1-\sqrt{5}}{2})^k)\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{3+\sqrt{5}}{2})-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2}))\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{1+\sqrt{5}}{2})^2-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2})^2)\\ +&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+2}-(\frac{1-\sqrt{5}}{2})^{k+2}) +\end{aligned} +$$满足条件，故由数学归纳法知，$\forall{n}\in\mathbb{N}_+,D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$. \ No newline at end of file diff --git a/编写小组/试卷/线代期末复习模拟/1.13线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.13线代限时练.md new file mode 100644 index 0000000..111f58e --- /dev/null +++ b/编写小组/试卷/线代期末复习模拟/1.13线代限时练.md @@ -0,0 +1,66 @@ +1. （[[线代2022秋B]]·1）已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. + +2. （[[线代2013秋A]]·4）已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. + +3. （[[线代2019秋A]]·9）设 $\boldsymbol{E}$ 为 $3$ 阶单位矩阵，$\boldsymbol\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $\boldsymbol E - \boldsymbol\alpha\boldsymbol\alpha^\mathrm{T}$ 的全部 $3$ 个特征值为\_\_\_\_\_\_\_. + +4. （[[线代2018秋A]]·12）设 $n$ 阶矩阵 $A=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^n(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2$ 的矩阵为\_\_\_\_\_\_\_. + +5. （[[线代2018秋A]]·11）若 $n$ 阶实对称矩阵 $\boldsymbol A$ 的特征值为 $\lambda_i=(-1)^i\ (i=1,2,\cdots,n)$，则 $\boldsymbol A^{100}=$ \_\_\_\_\_\_\_. + +6. ([[线代2022秋A]]·5）已知 $n(n\ge2)$ 维列向量 $\boldsymbol\alpha,\boldsymbol\beta$ 满足 $\boldsymbol\beta^\mathrm{T}\boldsymbol\alpha=-3$，则方阵 $(\boldsymbol{\beta\alpha}^\mathrm{T})^2$ 的非零特征值为\_\_\_\_\_\_\_\_\_\_. + +7. （[[线代2022秋A]]·七）已知向量组 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3$ 线性无关，其中 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3\in\mathbb{R}^3$，$\boldsymbol A$ 为 $3$ 阶方阵，且 +$$\boldsymbol{A\alpha_1}=2\boldsymbol\alpha_1-\boldsymbol\alpha_2-\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_2=\boldsymbol\alpha_1+2\boldsymbol\alpha_2+3\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_3=2\boldsymbol\alpha_1+4\boldsymbol\alpha_2+\boldsymbol\alpha_3.$$ +(1) 证明 $A\boldsymbol\alpha_1, A\boldsymbol\alpha_2, A\boldsymbol\alpha_3$ 线性无关； +(2) 计算行列式 $\boldsymbol E-\boldsymbol A$，其中 $\boldsymbol E$ 是 $3$ 阶单位矩阵. +```text + + + + + + + + + + + + + + + + + + + + + +``` + +8. （[[线代2013秋A]]）求 $n$ 阶方阵 $\boldsymbol{A} = \begin{bmatrix}1 & 1 & 1 & \dots & 1 \\1 & 0 & 1 & \dots & 1 \\1 & 1 & 0 & \dots & 1 \\\vdots & \vdots & \vdots & \ddots & \vdots \\1 & 1 & 1 & \dots & 0\end{bmatrix}$ 的逆。 +```text + + + + + + + + + + + + + + + + + + + + +``` + +9. （[[线代2013秋A]]·三）设 + $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$证明：$D_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]$ . \ No newline at end of file -- 2.34.1 From 8136d97a58a9afe55a2bcd8ec32b1cce781b310b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 17:15:32 +0800 Subject: [PATCH 223/274] vault backup: 2026-01-13 17:15:31 --- .../1.14线代测试答案.md | 223 ------------------ .../1.14线代限时练.md | 66 ------ 2 files changed, 289 deletions(-) delete mode 100644 编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md delete mode 100644 编写小组/试卷/线代期末复习模拟/1.14线代限时练.md diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md deleted file mode 100644 index 189ee67..0000000 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ /dev/null @@ -1,223 +0,0 @@ -# **1.14 线性代数限时练（题目 + 答案与解析）** - -## **第一部分：题目** - -### **1.** - -已知三阶行列式 $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{vmatrix}=c$，代数余子式之和 $\sum_{i=1}^{3}\sum_{j=1}^{3}A_{ij}=3c$，则行列式 $\begin{vmatrix}a_{11}+1&a_{12}+1&a_{13}+1\\a_{21}+1&a_{22}+1&a_{23}+1\\a_{31}+1&a_{32}+1&a_{33}+1\end{vmatrix}=$？ - -### **2.（2013 秋 A）** - -$已知向量空间 V=\{(2a,2b,3b,3a)\mid a,b\in\mathbb{R}\}，则 V 的维数是\underline{\qquad}。$ - -### **3.（2018 秋 A）** - -$设 E 为 3 阶单位矩阵，\alpha 为一个 3 维单位列向量，则矩阵 E-\alpha\alpha^T 的全部 3 个特征值为\underline{\qquad}。$ - -### **4.（2018 秋 A）** - -$设 n 阶矩阵 A=[a_{ij}]_{n\times n}，则二次型 f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^{n}(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2 的矩阵为\underline{\qquad}。$ -### **5.（2018 秋 A）** - -$若 n 阶实对称矩阵 A 的特征值为 \lambda_i=(-1)^i（i=1,2,\cdots,n），则 A^{100}=\underline{\qquad}。$ - -### **6.（2022 秋 A・5）** - -$已知 n（n\geq2）维列向量 \alpha,\beta 满足 \beta^T\alpha=-3，则方阵 (\beta\alpha^T)^2 的非零特征值为\underline{\qquad}。$ - -### **7.（2022 秋 A）** - -$已知向量组 \alpha_1,\alpha_2,\alpha_3 线性无关（\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}^3），A 为 3 阶方阵，且满足：$ - -$$\begin{aligned} -A\alpha_1&=2\alpha_1-\alpha_2-\alpha_3,\\A\alpha_2&=\alpha_1+2\alpha_2+3\alpha_3,\\A\alpha_3&=2\alpha_1+4\alpha_2+\alpha_3 -\end{aligned} -$$ -(1) $证明 A\alpha_1,A\alpha_2,A\alpha_3 线性无关$； - -(2) $计算行列式 |E-A|（E 是 3 阶单位矩阵）$。 - -### **8.（2013 秋 A）** - -$求 n 阶方阵 A=\begin{bmatrix}1&1&1&\cdots&1\\1&0&1&\cdots&1\\1&1&0&\cdots&1\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&1&1&\cdots&0\end{bmatrix} 的逆矩阵。$ - -### **9.（2013 秋 A・三）** - -设 n 阶行列式： - -$$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$ - -证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。 - -## **第二部分：答案与解析** - -### 1. 答案：$\boxed{4c}$ - -#### **解析：** - -$设原矩阵为 A（即 |A|=c），全 1 矩阵 J=\boldsymbol{e}\boldsymbol{e}^T（其中 \boldsymbol{e}=(1,1,1)^T），需求解的行列式为 |A+J|。$ -由代数余子式性质：$\sum_{i,j}A_{ij}=\boldsymbol{e}^T A^*\boldsymbol{e}=3c$，且可逆矩阵的伴随矩阵满足$A^*=|A|A^{-1}=cA^{-1}$（因 $|A|=c\neq0，A$ 可逆），代入得 $\boldsymbol{e}^T A^{-1}\boldsymbol{e}=3$。 - -利用**Sherman-Morrison 行列式公式**：对可逆矩阵$A$ 和向量 $\boldsymbol{u},\boldsymbol{v}$，有 $|A+\boldsymbol{u}\boldsymbol{v}^T|=|A|(1+\boldsymbol{v}^T A^{-1}\boldsymbol{u})$。 - -此处 $\boldsymbol{u}=\boldsymbol{v}=\boldsymbol{e}$，代入得 $|A+J|=c(1+\boldsymbol{e}^T A^{-1}\boldsymbol{e})=c(1+3)=4c$。 - -### **2. 答案：$\boxed{2}$ - -#### **解析：** - -将向量空间 V 中的元素拆分为线性组合形式： - -$$(2a,2b,3b,3a)=a(2,0,0,3)+b(0,2,3,0)$$ - -设 $\boldsymbol{\alpha}=(2,0,0,3)，\boldsymbol{\beta}=(0,2,3,0)$，需验证 $\boldsymbol{\alpha},\boldsymbol{\beta}$ 线性无关： - -若 $k_1\boldsymbol{\alpha}+k_2\boldsymbol{\beta}=\boldsymbol{0}$（零向量），则 $\begin{cases}2k_1=0\\2k_2=0\\3k_2=0\\3k_1=0\end{cases}$，解得 $k_1=k_2=0$，故 $\boldsymbol{\alpha},\boldsymbol{\beta}$是 V 的一组基。 - -向量空间的维数等于基的个数，因此 $V$ 的维数为 $2$。 - -### **3. 答案：$\boxed{1,1,0}$ - -#### **解析：** - -设 $B=\alpha\alpha^T$，该矩阵为**秩 $1$ 矩阵**（因 $\alpha$ 是单位列向量，$\alpha^T\alpha=1$）。 - -• 秩 $1$ 矩阵的特征值性质：非零特征值为矩阵的迹 $\text{tr}(B)=\alpha^T\alpha=1$，其余 $n-1=2$ 个特征值为 $0$（秩 $1$ 矩阵的非零特征值个数等于秩）。 - -• 若 $B\boldsymbol{x}=\lambda\boldsymbol{x}$（$\boldsymbol{x}$为特征向量），则 $(E-B)\boldsymbol{x}=(1-\lambda)\boldsymbol{x}$，即$E-B$ 的特征值为 $1-\lambda$。 - -$代入 B 的特征值 \lambda=1,0,0，得 E-B 的特征值为 1-1=0，1-0=1，1-0=1，即 1,1,0$。 - -### **4. 答案：$\boxed{A^T A}$** - -#### **解析：** - -记$A_i$为$A$中除了第$i$行全都改为$0$的矩阵，$\boldsymbol{x}=\begin{bmatrix}x_1&x_2&\cdots&x_n\end{bmatrix}^T$。那么$$A_i\boldsymbol{x}=\begin{bmatrix}0 & 0 & \cdots & 0\\\vdots & \vdots & &\vdots\\a_{i1} & a_{i2} & \cdots &a_{in}\\\vdots & \vdots & &\vdots\\0 & 0 & \cdots & 0\end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\\sum\limits_{j=1}^{n}a_{ij}x_j\\\vdots\\0\end{bmatrix}$$则$$(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2=(\sum\limits_{j=1}^{n}a_{ij}x_j)^2=(A_i\boldsymbol{x})^TA_i\boldsymbol{x}=\boldsymbol{x}^TA_i^TA_i\boldsymbol{x}$$将二次型用上式展开得： - -$$\begin{aligned} -f(\boldsymbol{x})&=\sum_{i=1}^{n}(\boldsymbol{x}^TA_i^TA_i\boldsymbol{x})\\ -&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^TA_i)\boldsymbol{x}\\ -&=\boldsymbol{x}^T(\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i)\boldsymbol{x}\\ -&=\boldsymbol{x}^T(A^T A)\boldsymbol{x} -\end{aligned}$$ -其中因为$A_i^TA_j=\boldsymbol{0},$如果$i\neq j$，所以 -$$\sum\limits_{i=1}^nA_i^T\sum\limits_{i=1}^nA_i=\sum\limits_{i=1}^nA_i^TA_i$$ - -• 二次型的矩阵需满足**对称性质**（即矩阵等于其转置），验证：$(A^T A)^T=A^T(A^T)^T=A^T A$，故 $A^T A$ 是对称矩阵，即为二次型的矩阵。 - -### **5. 答案：$\boxed{E}$（单位矩阵）** - -#### **解析：** - -实对称矩阵可对角化，即存在可逆矩阵 P，使得 $P^{-1}AP=\Lambda$（$\Lambda$ 为对角矩阵，对角元为 A 的特征值 $\lambda_i=(-1)^i$）。 - -• 矩阵幂运算性质：$A^{100}=P\Lambda^{100}P^{-1}$。 - -• 计算 $\Lambda^{100}$：对角元为 $\lambda_i^{100}=[(-1)^i]^{100}=1$，故 $\Lambda^{100}=E$（单位矩阵）。 - -• 因此 $A^{100}=P E P^{-1}=P P^{-1}=E$。 - -### **6. 答案：$\boxed{9}$** - -#### **解析：** - -设 $A=\beta\alpha^T$（秩 1 矩阵），计算 $A^2$： - -$A^2=(\beta\alpha^T)(\beta\alpha^T)=\beta(\alpha^T\beta)\alpha^T=(\alpha^T\beta)A$ - -• 注意：$\alpha^T\beta=(\beta^T\alpha)^T$（矩阵转置性质），而 $\beta^T\alpha=-3$（数，转置等于自身），故 $\alpha^T\beta=-3$，因此 $A^2=-3A$。 - -• 秩 $1$ 矩阵 $A$ 的非零特征值为 $\text{tr}(A)=\alpha^T\beta=-3$，设 $A\boldsymbol{x}=-3\boldsymbol{x}$，则 $A^2\boldsymbol{x}=(-3)A\boldsymbol{x}=(-3)^2\boldsymbol{x}=9\boldsymbol{x}$，即 $A^2$ 的非零特征值为 $9$。 - -### **7. 答案：(1) 证明见解析；(2) $\boxed{20}$** - -#### **(1) 证明 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关** - -因 $\alpha_1,\alpha_2,\alpha_3$ 线性无关，故矩阵 $P=(\alpha_1,\alpha_2,\alpha_3)$ 可逆（列向量线性无关的矩阵可逆）。 - -由题设条件，将 $A$ 对 $\alpha_1,\alpha_2,\alpha_3$ 的作用表示为矩阵乘法： - -$$A(\alpha_1,\alpha_2,\alpha_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}=P C$$ - -其中 $C=\begin{bmatrix}2&1&2\\-1&2&4\\-1&3&1\end{bmatrix}$，计算 $|C|$： - -$$\begin{aligned} -|C|&=2\times(2\times1-4\times3)-1\times(-1\times1-4\times(-1))+2\times(-1\times3-2\times(-1))\\ -&=2\times(-10)-1\times3+2\times(-1)\\ -&=-25\neq0 -\end{aligned} -$$ - -• 因 $|C|\neq0$，故 $C$ 可逆，$\text{rank}(C)=3$。 - -• 又 $P$ 可逆，故 $\text{rank}(A\alpha_1,A\alpha_2,A\alpha_3)=\text{rank}(P C)=\text{rank}(C)=3$，即 $A\alpha_1,A\alpha_2,A\alpha_3$ 线性无关。 - -#### **(2) 计算 $|E-A|$** - -由 (1) 知$P^{-1}AP=C（A 与 C 相似）$，则 $E-A$ 与 $E-C$ 相似（相似矩阵的 “单位矩阵减矩阵” 仍相似），而**相似矩阵的行列式相等**，故 $|E-A|=|E-C|$。 - -计算 $E-C$： - -$$E-C=\begin{bmatrix}1-2&0-1&0-2\\0-(-1)&1-2&0-4\\0-(-1)&0-3&1-1\end{bmatrix}=\begin{bmatrix}-1&-1&-2\\1&-1&-4\\1&-3&0\end{bmatrix}$$ - -按第三行展开计算行列式： - -$$ -\begin{aligned} -|E-C|&=1\times\begin{vmatrix}-1&-2\\-1&-4\end{vmatrix}-(-3)\times\begin{vmatrix}-1&-2\\1&-4\end{vmatrix}+0\times(\text{余子式})\\ -&=1\times(4-2)+3\times(4+2)\\ -&=2+18=20 -\end{aligned} -$$ - -故 $|E-A|=20$。 - -### **8. 答案：** - -$$A^{-1}=\begin{bmatrix}-(n-2)&1&1&\cdots&1\\1&-1&0&\cdots&0\\1&0&-1&\cdots&0\\\vdots&\vdots&\vdots&\ddots&\vdots\\1&0&0&\cdots&-1\end{bmatrix} -$$ -#### **解析：** - -通过 “行变换法” 或 “规律归纳” 推导： - -• 观察矩阵 A 的结构：第一行全为 1，其余行的对角元为 0，非对角元为 1。可先计算 n=2,3 时的逆矩阵，归纳规律： - -◦ $当 n=2 时，A=\begin{bmatrix}1&1\\1&0\end{bmatrix}，逆矩阵为 \begin{bmatrix}0&1\\1&-1\end{bmatrix}（符合上述形式，-(2-2)=0）$； - -◦ 当 $n=3$ 时，$A=\begin{bmatrix}1&1&1\\1&0&1\\1&1&0\end{bmatrix}$，逆矩阵为 $\begin{bmatrix}-1&1&1\\1&-1&0\\1&0&-1\end{bmatrix}$（符合上述形式，$-(3-2)=-1$）。 - -• 验证规律：对 n 阶矩阵，逆矩阵的第一行第一列元素为 -(n-2)，第一行其余元素为 1，第一列其余元素为 1，对角元（除第一行第一列）为 -1，非对角元（除第一行、第一列）为 0，即为上述形式。 - -### **9. 证明：$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$** - - **步骤 1：建立递推公式** - -对 $D_n$ 按**第一行展开**（第一行元素为 $a_{11}=1,a_{12}=1$，其余 $a_{1j}=0$）： - -$D_n=a_{11}\times(-1)^{1+1}M_{11}+a_{12}\times(-1)^{1+2}M_{12}$ - -• $M_{11}$：去掉第一行第一列后的子式，即 $n-1$ 阶行列式 $D_{n-1}$（结构与 $D_n$ 一致）； - -• $M_{12}$：去掉第一行第二列后的子式，按第一列展开（第一列仅首元素为 $-1$），得 $-D_{n-2}$（符号需结合 $(-1)^{1+2}=-1$）。 - -因此递推公式为： - -$$D_n=D_{n-1}+D_{n-2}$$ - - **步骤 2：确定初始条件** - -• 当 $n=1$ 时，$D_1=\begin{vmatrix}1\end{vmatrix}=1$； - -• 当 n=2 时，$D_2=\begin{vmatrix}1&1\\-1&1\end{vmatrix}=1\times1-1\times(-1)=2$。 -容易证明$n=1,n=2$时满足要证的式子 - -**步骤3：运用数学归纳法** -假设当$n\le k$时，有$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$。当$n=k+1$时，有$$ -\begin{aligned} -D_{k+1}&=D_k+D_{k-1}\\ -&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+1}-(\frac{1-\sqrt{5}}{2})^{k+1}+(\frac{1+\sqrt{5}}{2})^k-(\frac{1-\sqrt{5}}{2})^k)\\ -&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{3+\sqrt{5}}{2})-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2}))\\ -&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k}(\frac{1+\sqrt{5}}{2})^2-(\frac{1-\sqrt{5}}{2})^{k}(\frac{1-\sqrt{5}}{2})^2)\\ -&=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{k+2}-(\frac{1-\sqrt{5}}{2})^{k+2}) -\end{aligned} -$$满足条件，故由数学归纳法知，$\forall{n}\in\mathbb{N}_+,D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$. \ No newline at end of file diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md b/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md deleted file mode 100644 index 111f58e..0000000 --- a/编写小组/试卷/线代期末复习模拟/1.14线代限时练.md +++ /dev/null @@ -1,66 +0,0 @@ -1. （[[线代2022秋B]]·1）已知 $\begin{vmatrix}a_{11} & a_{12} & a_{13} \\a_{21} & a_{22} & a_{23} \\a_{31} & a_{32} & a_{33}\end{vmatrix} = c$ ，代数余子式之和 $\sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3} A_{ij} = 3c$ ，则 $\begin{vmatrix}a_{11}+1 & a_{12}+1 & a_{13}+1 \\a_{21}+1 & a_{22}+1 & a_{23}+1 \\a_{31}+1 & a_{32}+1 & a_{33}+1\end{vmatrix} =$ \_\_\_\_\_\_\_\_\_\_. - -2. （[[线代2013秋A]]·4）已知向量空间 $V=\{(2a,2b,3b,3a)|a,b\in\mathbb{R}\}$，则 $V$ 的维数是\_\_\_\_\_. - -3. （[[线代2019秋A]]·9）设 $\boldsymbol{E}$ 为 $3$ 阶单位矩阵，$\boldsymbol\alpha$ 为一个 $3$ 维单位列向量，则矩阵 $\boldsymbol E - \boldsymbol\alpha\boldsymbol\alpha^\mathrm{T}$ 的全部 $3$ 个特征值为\_\_\_\_\_\_\_. - -4. （[[线代2018秋A]]·12）设 $n$ 阶矩阵 $A=[a_{ij}]_{n\times n}$ ，则二次型 $f(x_1,x_2,\cdots,x_n)=\sum\limits_{i=1}^n(a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2$ 的矩阵为\_\_\_\_\_\_\_. - -5. （[[线代2018秋A]]·11）若 $n$ 阶实对称矩阵 $\boldsymbol A$ 的特征值为 $\lambda_i=(-1)^i\ (i=1,2,\cdots,n)$，则 $\boldsymbol A^{100}=$ \_\_\_\_\_\_\_. - -6. ([[线代2022秋A]]·5）已知 $n(n\ge2)$ 维列向量 $\boldsymbol\alpha,\boldsymbol\beta$ 满足 $\boldsymbol\beta^\mathrm{T}\boldsymbol\alpha=-3$，则方阵 $(\boldsymbol{\beta\alpha}^\mathrm{T})^2$ 的非零特征值为\_\_\_\_\_\_\_\_\_\_. - -7. （[[线代2022秋A]]·七）已知向量组 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3$ 线性无关，其中 $\boldsymbol\alpha_1,\boldsymbol\alpha_2,\boldsymbol\alpha_3\in\mathbb{R}^3$，$\boldsymbol A$ 为 $3$ 阶方阵，且 -$$\boldsymbol{A\alpha_1}=2\boldsymbol\alpha_1-\boldsymbol\alpha_2-\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_2=\boldsymbol\alpha_1+2\boldsymbol\alpha_2+3\boldsymbol\alpha_3, \boldsymbol A\boldsymbol\alpha_3=2\boldsymbol\alpha_1+4\boldsymbol\alpha_2+\boldsymbol\alpha_3.$$ -(1) 证明 $A\boldsymbol\alpha_1, A\boldsymbol\alpha_2, A\boldsymbol\alpha_3$ 线性无关； -(2) 计算行列式 $\boldsymbol E-\boldsymbol A$，其中 $\boldsymbol E$ 是 $3$ 阶单位矩阵. -```text - - - - - - - - - - - - - - - - - - - - - -``` - -8. （[[线代2013秋A]]）求 $n$ 阶方阵 $\boldsymbol{A} = \begin{bmatrix}1 & 1 & 1 & \dots & 1 \\1 & 0 & 1 & \dots & 1 \\1 & 1 & 0 & \dots & 1 \\\vdots & \vdots & \vdots & \ddots & \vdots \\1 & 1 & 1 & \dots & 0\end{bmatrix}$ 的逆。 -```text - - - - - - - - - - - - - - - - - - - - -``` - -9. （[[线代2013秋A]]·三）设 - $$D_n=\begin{vmatrix}1&1&0&\cdots&0&0\\-1&1&1&\cdots&0&0\\0&-1&1&\cdots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\cdots&1&1\\0&0&0&\cdots&-1&1\end{vmatrix}$$证明：$D_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^{n+1}-(\frac{1-\sqrt{5}}{2})^{n+1}]$ . \ No newline at end of file -- 2.34.1 From c7774cac8d3e471542a853cbaf23a2e844643c1c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Tue, 13 Jan 2026 18:44:13 +0800 Subject: [PATCH 224/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=A0=BC?= =?UTF-8?q?=E5=BC=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/微分中值定理.md | 62 +++++------- ...性方程组的系数矩阵与解关系.md | 94 +++++++------------ 2 files changed, 55 insertions(+), 101 deletions(-) diff --git a/素材/微分中值定理.md b/素材/微分中值定理.md index 6304f45..70e16dd 100644 --- a/素材/微分中值定理.md +++ b/素材/微分中值定理.md @@ -1,9 +1,6 @@ -### 例1 -设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $0 < a < b$，试证存在 $\xi, \eta \in (a, b)$，使得 -$$ -f'(\xi) = \frac{a + b}{2\eta} f'(\eta). -$$ +>[!example] 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $0 < a < b$，试证存在 $\xi, \eta \in (a, b)$，使得 $$f'(\xi) = \frac{a + b}{2\eta} f'(\eta).$$ **解析**： 本题结论中含有两个不同的中值 $\xi$ 和 $\eta$，且涉及两个不同的函数形式。可考虑分别对 $f(x)$ 和 $g(x)=x^2$ 在 $[a,b]$ 上应用柯西中值定理： @@ -23,7 +20,7 @@ $$ --- -### 例2 +>[!example] 例2 设函数 $f(x)$ 在 $[0,3]$ 上连续，在 $(0,3)$ 内可导，且 $f(0) + f(1) + f(2) = 3$，$f(3) = 1$，试证必存在 $\xi \in (0, 3)$，使 $f'(\xi) = 0$。 **解析**： @@ -31,10 +28,10 @@ $$ --- -### 例3 +>[!example] 例3 设 $f(x)$ 在区间 $[0,1]$ 上连续，在 $(0,1)$ 内可导，且 $f(0) = f(1) = 0$，$f(1/2) = 1$，试证： -1. 存在 $\eta \in (1/2, 1)$，使得 $f(\eta) = \eta$； -2. 对任意实数 $\lambda$，必存在 $\xi \in (0, \eta)$，使得 $f'(\xi) - \lambda [f(\xi) - \xi] = 1$。 +(1)存在 $\eta \in (1/2, 1)$，使得 $f(\eta) = \eta$； +(2)对任意实数 $\lambda$，必存在 $\xi \in (0, \eta)$，使得 $f'(\xi) - \lambda [f(\xi) - \xi] = 1$。 **解析**： (1) 令 $g(x)=f(x)-x$，则 $g(1/2)=1-1/2=1/2>0$，$g(1)=0-1=-1<0$，由零点定理，存在 $\eta \in (1/2,1)$，使 $g(\eta)=0$，即 $f(\eta)=\eta$。 @@ -48,12 +45,9 @@ $$ ## 5.2 微分中值定理及其应用 -### 例1 +>[!example] 例1 设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 -$$ -f(0) = 0, \quad |f'(x)| \leq 1, -$$ -证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 +$$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 **解析**： 对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 @@ -64,12 +58,9 @@ $$ --- -### 例2 +>[!example] 例2 设 $a_i \in \mathbb{R} (i = 0,1,2,\cdots,n)$，且满足 -$$ -a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} = 0， -$$ -证明：方程 $a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0$ 在 $(0,1)$ 内至少有一个实根。 +$$a_0 + \frac{a_1}{2} + \frac{a_2}{3} + \cdots + \frac{a_n}{n+1} = 0，$$证明：方程 $a_0 + a_1x + a_2x^2 + \cdots + a_nx^n = 0$ 在 $(0,1)$ 内至少有一个实根。 **解析**： 构造辅助函数 @@ -83,11 +74,9 @@ $$ --- -### 例3 +>[!example] 例3 设函数 $f(x)$ 在 $[a,b]$ 上可导，且 -$$ -f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0， -$$ +$$f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0，$$ 试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 **解析**： @@ -95,7 +84,7 @@ $$ --- -### 例4 +>[!example] 例4 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 **解析**： @@ -114,7 +103,7 @@ $$ --- -### 例5（柯西中值定理例） +>[!example] 例5（柯西中值定理例） 试证至少存在一点 $\xi \in (1, e)$，使 $\sin 1 = \cos \ln \xi$。 **解析**： @@ -133,7 +122,7 @@ $$ ## 练习 -### Ex3 +>[!example] Ex1 设 $f(x)$ 在 $(a, b)$ 内可导，且 $f'(x) \neq 1$。试证明 $f(x)$ 在 $(a, b)$ 内至多只有一个不动点，即方程 $f(x) = x$ 在 $(a, b)$ 内至多只有一个实根。 **解析**： @@ -145,14 +134,11 @@ $$ --- -### Ex4 +>[!example] Ex2 设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 -$$ -f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4} -$$ -证明： -1. 至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； -2. 若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 +$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： +(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； +(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 **解析**： (1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 @@ -160,7 +146,7 @@ $$ --- -### Ex5 +>[!example] Ex3 若 $f(x)$ 可导，试证在其两个零点间一定有 $f(x) + f'(x)$ 的零点。 **解析**： @@ -172,7 +158,7 @@ $$ --- -### Ex6 +>[!example] Ex4 设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 **解析**： @@ -184,11 +170,9 @@ $$ --- -### Ex7 +>[!example] Ex5 设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 -$$ -f(x_1 + x_2) < f(x_1) + f(x_2) -$$ +$$f(x_1 + x_2) < f(x_1) + f(x_2)$$ **解析**： 不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： diff --git a/素材/线性方程组的系数矩阵与解关系.md b/素材/线性方程组的系数矩阵与解关系.md index aaeb69a..4209625 100644 --- a/素材/线性方程组的系数矩阵与解关系.md +++ b/素材/线性方程组的系数矩阵与解关系.md @@ -1,22 +1,24 @@ 这是一个链接了方程组解空间与方程组系数秩的公式 >[!note] 解零度化定理： ->对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 -> $$\dim N(\boldsymbol{A})=n-r$$ +>对于齐次方程组 ${A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}{A}=r$，则 +> $$\dim N({A})=n-r$$ -已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. - $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ ->分析：在求解非齐次方程组通解的题目中，若是题目给出了特解与齐次方程组的解，那么大概率来说这个齐次方程组的解就可以拓展为齐次方程组通解（根据问题导向，不然写不出来了），那么如何由齐次方程组的解拓展为齐次方程组通解呢，那就要根据题目具体的条件进行分析了，这就要用到我们的解零度化定理来求齐次方程组解空间的维数 ->解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ ->故 $Ax=0$ 的解空间维数是 $1$ （5分） ->$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） ->$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） ->解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 +>[!example] 例1 +>已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $A\boldsymbol{x}=\beta$ 的通解. +**答案：** + $$\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$$ +**分析：** 在求解非齐次方程组通解的题目中，若是题目给出了特解与齐次方程组的解，那么大概率来说这个齐次方程组的解就可以拓展为齐次方程组通解（根据问题导向，不然写不出来了），那么如何由齐次方程组的解拓展为齐次方程组通解呢，那就要根据题目具体的条件进行分析了，这就要用到我们的解零度化定理来求齐次方程组解空间的维数 +**解析：** 由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ +故 $Ax=0$ 的解空间维数是 $1$ （5分） +$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） +$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） +解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 - 设 -$$ -A = \begin{bmatrix} + + >[!example] 例2 + >设 $$A = \begin{bmatrix} 1 & -1 & 0 & -1 \\ 1 & 1 & 0 & 3 \\ 2 & 1 & 2 & 6 @@ -28,87 +30,55 @@ B = \begin{bmatrix} \end{bmatrix}, \quad \alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad -\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} -$$ - +\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$$ (1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； - -(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 -4. (10分) 设 -$$ -A = \begin{bmatrix} -1 & -1 & 0 & -1 \\ -1 & 1 & 0 & 3 \\ -2 & 1 & 2 & 6 -\end{bmatrix}, \quad -B = \begin{bmatrix} -1 & 0 & 1 & 2 \\ -1 & -1 & a & a-1 \\ -2 & -3 & 2 & -2 -\end{bmatrix}, -$$ -向量 -$$ -\alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad -\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}. -$$ - -(1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； - -(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 - ---- +$\quad$ +(2) 若方程组 $A\boldsymbol{x} = \alpha$ 与方程组 $B\boldsymbol{x} = \beta$ 不同解，求 $a$ 的值。 **解：** - (1) 由于 $$ -\left( \begin{array}{c} +\begin{bmatrix} A \quad \alpha \\ B \quad \beta -\end{array} \right) = -\left( \begin{array}{ccccc} +\end{bmatrix} = +\begin{bmatrix} 1 & -1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 3 & 2 \\ 2 & 1 & 2 & 6 & 3 \\ 1 & 0 & 1 & 2 & 1 \\ 1 & -1 & a & a-1 & 0 \\ 2 & -3 & 2 & -2 & -1 -\end{array} \right) -$$ -$$ +\end{bmatrix} \rightarrow -\left( \begin{array}{ccccc} +\begin{bmatrix} 1 & -1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 -\end{array} \right), +\end{bmatrix}, $$ 故 $$ -R \left( \begin{array}{c} -A \quad \alpha \\ -B \quad \beta -\end{array} \right) = R(A, \alpha), +\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha], $$ 从而方程组 $$ \begin{cases} -Ax = \alpha, \\ -Bx = \beta +A\boldsymbol{x} = \alpha \\ +B\boldsymbol{x} = \beta \end{cases} $$ -与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 +与 $A\boldsymbol{x} = \alpha$ 同解，故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解。 -(2) 分析：不同解，却要可以求出a的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑解零度化定理 -由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 +(2) 分析：不同解，却要可以求出$a$的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑解零度化定理 +由于 $A\boldsymbol{x} = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $A\boldsymbol{x} = \alpha$ 与 $B\boldsymbol{x} = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $B\boldsymbol{x} = 0$ 的基础解系中解向量的个数，即 $$ -4 - R(A) < 4 - R(B), +4 - r(A) < 4 - r(B), $$ -故 $R(A) > R(B)$。又因 $R(A) = 3$，故 $R(B) < 3$，则 +故 $r(A) > r(B)$。又因 $r(A) = 3$，故 $r(B) < 3$，则 $$ \left| \begin{array}{ccc} 1 & 0 & 1 \\ -- 2.34.1 From 82353433d9a8d7c83cdbe4369f6b72560bb12ae3 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 13 Jan 2026 19:25:41 +0800 Subject: [PATCH 225/274] =?UTF-8?q?=E7=94=A8=E7=A7=A9=E7=9A=84=E4=B8=8D?= =?UTF-8?q?=E7=AD=89=E5=BC=8F=E2=80=9C=E5=A4=B9=E9=80=BC=E2=80=9D=E5=87=BA?= =?UTF-8?q?=E7=A1=AE=E5=88=87=E5=80=BC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...秩的不等式“夹逼”出确切值.md | 19 ++++++++++++++++++- 1 file changed, 18 insertions(+), 1 deletion(-) diff --git a/素材/用秩的不等式“夹逼”出确切值.md b/素材/用秩的不等式“夹逼”出确切值.md index 60acc1f..588d98e 100644 --- a/素材/用秩的不等式“夹逼”出确切值.md +++ b/素材/用秩的不等式“夹逼”出确切值.md @@ -6,4 +6,21 @@ >2. $\mathrm{rank}(\boldsymbol{A+B})<\mathrm{rank}\boldsymbol A+\mathrm{rank}\boldsymbol B$ >3. 矩阵加边不会减小秩； > ->特别的，在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ \ No newline at end of file +>特别的，在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ + +9. （20分）设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： + (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . + (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解. + 证明如下： +>(1)（10分） +> 对于方程组$Ax=0$ (a)和 $A^\mathrm{T}Ax=0$ (b)，b的解空间一定包含a的解空间（5分）； +>而方程b两边同时乘以$x^\mathrm{T}$，得 $x^\mathrm{T}A^\mathrm{T}Ax=0$ ，即 $(x^\mathrm{T}A^\mathrm{T})(Ax)=0 \to Ax=0$， +>所以a的解空间包含b的解空间（5分）， +>所以a,b同解，所以$\mathrm{rank}A=\mathrm{rank}{A^\mathrm{T}A}$ +>(2) （10分） +>$A^\mathrm{T}Ax=A^\mathrm{T}\beta \iff \mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ +>而由(1)的结论得等式左边 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}A$； +><span style="color:#ffff22;">关键步骤！</span> 等式右边 $\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})\ge \mathrm{rank}A^\mathrm{T}A=\mathrm{rank}A$ （5分） +><span style="color:#ffff22;">关键步骤！</span> 又$\mathrm{rank}(A^\mathrm{T}\begin{bmatrix}A&\beta\end{bmatrix})\le \min{(\mathrm{rank}A^\mathrm{T},\ \mathrm{rank}\begin{bmatrix}A&\beta\end{bmatrix})}=\mathrm{rank}A$ （5分） +>所以$\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})=\mathrm{rank}A$ +>故 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ 得证. \ No newline at end of file -- 2.34.1 From bb00842168a7f058e3838c3da5c593c703f26dbd Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Tue, 13 Jan 2026 19:28:37 +0800 Subject: [PATCH 226/274] =?UTF-8?q?=E7=A7=BB=E5=8A=A8=E4=BA=86=E8=AF=95?= =?UTF-8?q?=E5=8D=B7=E4=BD=8D=E7=BD=AE?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../高数期末真题/2017高数期末考试卷.md | 0 .../高数期末真题/2018高数期末考试卷.md | 0 .../高数期末真题/2019高数期末考试卷.md | 0 .../高数期末真题/2020高数期末考试卷.md | 0 4 files changed, 0 insertions(+), 0 deletions(-) rename {编写小组/试卷 => 试卷库}/高数期末真题/2017高数期末考试卷.md (100%) rename {编写小组/试卷 => 试卷库}/高数期末真题/2018高数期末考试卷.md (100%) rename {编写小组/试卷 => 试卷库}/高数期末真题/2019高数期末考试卷.md (100%) rename {编写小组/试卷 => 试卷库}/高数期末真题/2020高数期末考试卷.md (100%) diff --git a/编写小组/试卷/高数期末真题/2017高数期末考试卷.md b/试卷库/高数期末真题/2017高数期末考试卷.md similarity index 100% rename from 编写小组/试卷/高数期末真题/2017高数期末考试卷.md rename to 试卷库/高数期末真题/2017高数期末考试卷.md diff --git a/编写小组/试卷/高数期末真题/2018高数期末考试卷.md b/试卷库/高数期末真题/2018高数期末考试卷.md similarity index 100% rename from 编写小组/试卷/高数期末真题/2018高数期末考试卷.md rename to 试卷库/高数期末真题/2018高数期末考试卷.md diff --git a/编写小组/试卷/高数期末真题/2019高数期末考试卷.md b/试卷库/高数期末真题/2019高数期末考试卷.md similarity index 100% rename from 编写小组/试卷/高数期末真题/2019高数期末考试卷.md rename to 试卷库/高数期末真题/2019高数期末考试卷.md diff --git a/编写小组/试卷/高数期末真题/2020高数期末考试卷.md b/试卷库/高数期末真题/2020高数期末考试卷.md similarity index 100% rename from 编写小组/试卷/高数期末真题/2020高数期末考试卷.md rename to 试卷库/高数期末真题/2020高数期末考试卷.md -- 2.34.1 From 4c39673c9ca0d35d4ef9aef941e1eea20bdc9db6 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 00:38:29 +0800 Subject: [PATCH 227/274] vault backup: 2026-01-14 00:38:29 --- ...的解与秩的不等式（解析版）.md | 328 +++++++++++++++++- 1 file changed, 327 insertions(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index d5e4d4f..3cefce7 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -15,6 +15,11 @@ tags: 3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 +怎么理解： +1. 从线性方程组的角度，如果加上一列 $b$ 后的秩变大了，那么化为最简行阶梯型后下面一定多出来一行 $0=$某个常数 ，则必然无解。 +2. 秩等于 $n$ 就是有 $n$ 个无关的方程，则经过消元法后可以解出唯一解。 +3. 秩小于$n$ 就是方程不足 $n$ 个，消元消不完，也能解释为啥秩跟解空间维数的和为 $n$ + 把以上结论应用到齐次线性方程组，可得 推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 @@ -248,6 +253,96 @@ $$ 2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 +# 矩阵秩与线性方程组解的关系图解说明 + +--- + +## 概念回顾 + +- **rank[A]** 代表矩阵$A$的秩。 +- 秩的定义：矩阵列向量组中**极大线性无关组所含列向量的个数**。 +- 可以用“圆”或“空间”来表示矩阵列向量组张成的向量空间。 + +--- + +## 图解说明 + +假设 +$$ +A = [\alpha_1, \alpha_2, \alpha_3] +$$ +是$m \times 3$矩阵， +$$ +\beta_1, \beta_2 +$$ +是$m$维列向量。 + +### 1. 方程组有解的条件 + +方程组 +$$ +Ax = \beta_1 \quad \text{和} \quad Ax = \beta_2 +$$ +有解 +$\Leftrightarrow$$\beta_1, \beta_2$可由$A$的列向量线性表示。 + +在几何上，这表示： + +- 设$A$的列向量张成的空间为$S_A$。 +-$\beta_1, \beta_2 \in S_A$。 +- 即$S_A$“包含”$\beta_1, \beta_2$。 + +因此，$S_A$这个“圆”应当能够**覆盖**$\beta_1$和$\beta_2$。 + +--- + +### 2. 秩等价条件 + +已知： +$$ +\operatorname{rank}[A] = \operatorname{rank}[A \quad \beta_1 \quad \beta_2] +$$ +表示： +- 矩阵$A$的秩与增广矩阵$[A \mid \beta_1 \mid \beta_2]$的秩相等。 +- 这意味着$\beta_1, \beta_2$并没有“扩大”$A$的列空间。 + +因此： +$$ +\operatorname{rank}[A] = \operatorname{rank}[A \quad \beta_1 \quad \beta_2] +\quad\Leftrightarrow\quad +\beta_1, \beta_2 \in S_A +$$ +即$Ax = \beta_1$和$Ax = \beta_2$有解。 + +--- + +### 3. 等价写法 + +把$\beta_1, \beta_2$放在$A$的右侧构成一个更大的矩阵： +$$ +[A \quad \beta_1 \quad \beta_2] +$$ +其秩与$A$相同，说明： +1. 空间$S_{[A \ \beta_1 \ \beta_2]}$与$S_A$相同。 +2. 从初等行变换角度看：在行阶梯形中，$\beta_1, \beta_2$对应的列会被$A$的列线性表示，从而可化为零列（在解方程时体现为消去）。 +3. 存在$x_1, x_2$使得： + $$ + A(-x_1) = \beta_1, \quad A(-x_2) = \beta_2 + $$ + 这样在增广矩阵中可以通过列操作消去$\beta_1, \beta_2$，使其变为零列。 + +--- + +## 总结 + +- 秩相等 ⇔ 列空间相同 ⇔ 方程组有解。 +- 图示法：把$A$的列空间画成一个圆，$\beta_1, \beta_2$若落在圆内，则方程有解。 +- 矩阵的秩是判断线性方程组解的存在性的核心工具。 + +--- + +**注**：这里的“圆”是比喻，实际为**线性子空间**。 + # 秩的不等式 ### 1. 和的秩不超过秩的和 @@ -260,7 +355,7 @@ $$ \operatorname{rank}(A+B) \leq \operatorname{rank} A + \operatorname{rank} B $ 设 $A_{m \times n}, B_{n \times k}$，则 $$ \operatorname{rank}(AB) \leq \min\{\operatorname{rank} A, \operatorname{rank} B\} $$ -### 3. 重要不等式 +### 3. Sylvester(西尔维斯特)不等式 设 $A_{m \times n}, B_{n \times k}$，则 $$ \operatorname{rank}(AB) \geq \operatorname{rank} A + \operatorname{rank} B - n $$ @@ -311,6 +406,188 @@ B & B \end{bmatrix} \geq \text{rank } (A + B) $$ +注：（2）与（4）结合即第一个不等式的证明方法 + +> [!note] 证明1： +$$\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}$$ + +--- + +### 证明思路 +设 +$$ +A \in \mathbb{R}^{m \times n}, \quad B \in \mathbb{R}^{n \times p}, \quad C = AB \in \mathbb{R}^{m \times p}. +$$ + +--- + +#### 1. 先证$\operatorname{rank}(AB) \leq \operatorname{rank}(A)$ + +- 考虑$C$的列向量： + 设$B = [b_1, b_2, \dots, b_p]$，则 + $$ + C = [A b_1, A b_2, \dots, A b_p]. + $$ + 因此$C$的每一列都是$A$的列向量的线性组合。 +- 所以$C$的列空间是$A$的列空间的子空间，故 + $$ + \operatorname{rank}(C) \leq \operatorname{rank}(A)。 + $$ + +--- + +#### 2. 再证$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ + +- 考虑$C$的行向量： + 设$A = \begin{bmatrix} a_1^T \\ a_2^T \\ \vdots \\ a_m^T \end{bmatrix}$，则 + $$ + C = \begin{bmatrix} a_1^T B \\ a_2^T B \\ \vdots \\ a_m^T B \end{bmatrix}. + $$ + 因此$C$的每一行都是$B$的行向量的线性组合。 +- 所以$C$的行空间是$B$的行空间的子空间，故 + $$\operatorname{rank}(C) \leq \operatorname{rank}(B)$$ + +--- + +#### 3. 综合 +由 1 和 2 得 +$$ +\operatorname{rank}(AB) \leq \operatorname{rank}(A) \quad \text{且} \quad \operatorname{rank}(AB) \leq \operatorname{rank}(B), +$$ +即 +$$ +\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}. +$$ + +--- + +**证毕。** + +> [!note] 证明2 +> 证明 Sylvester 秩不等式： +$$\operatorname{rank}(AB) \ge \operatorname{rank}(A) + \operatorname{rank}(B) - n$$ +其中 +$A \in \mathbb{R}^{m \times n}, \; B \in \mathbb{R}^{n \times p}, \; AB \in \mathbb{R}^{m \times p}$。 + +--- + +### 证明思路 +设： +-$\operatorname{rank}(A) = r$ +-$\operatorname{rank}(B) = s$ +-$n$是矩阵乘法的中间维度，即$A$的列数、$B$的行数。 + +--- + +#### 1. 利用分块矩阵构造 +构造如下分块矩阵： +$$ +M = \begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} +\in \mathbb{R}^{(m+n) \times (n+p)} +$$ +其中$I_n$是$n \times n$单位矩阵，$O$是零矩阵。 + +--- + +#### 2. 对$M$进行初等变换 +从$M$的第二块行减去第一块行左乘某个矩阵（这里相当于对$M$做列初等变换），实际上我们可以对$M$做以下变换： +$$ +\begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} +\xrightarrow{\text{右乘 } \begin{bmatrix} I_n & -B \\ O & I_p \end{bmatrix}} +\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +$$ +初等变换不改变矩阵的秩，所以： +$$ +\operatorname{rank}(M) = \operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +$$ + +--- + +#### 3. 估计$\operatorname{rank}(M)$ +另一方面，由分块矩阵的秩不等式： +$$ +\operatorname{rank}(M) \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 + +更严格地，我们可以直接写： +$$ +\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} +A \\ +I_n +\end{bmatrix} + \operatorname{rank}\begin{bmatrix} +I_n & B +\end{bmatrix} - n +$$ +但更简单的常用方法是利用： +$$ +\operatorname{rank}\begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +因为$I_n$的存在使得两个子块的秩可以同时保持。 + +--- + +#### 4. 从变换后的矩阵得到下界 +观察变换后的矩阵： +$$ +\operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +\ge \operatorname{rank}\begin{bmatrix} +I_n & O +\end{bmatrix} + \operatorname{rank}([-AB]) +$$ +实际上更直接的方法是注意到： +$$ +\operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} += \operatorname{rank}\begin{bmatrix} +O & -AB \\ +I_n & O +\end{bmatrix} \quad (\text{列变换}) +$$ +即： +$$ += \operatorname{rank}\begin{bmatrix} +I_n & O \\ +O & AB +\end{bmatrix} = \operatorname{rank}(I_n) + \operatorname{rank}(AB) = n + \operatorname{rank}(AB) +$$ + +--- + +#### 5. 联立 +由初等变换保秩，得： +$$ +n + \operatorname{rank}(AB) = \operatorname{rank}(M) \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +整理得： +$$ +\operatorname{rank}(AB) \ge \operatorname{rank}(A) + \operatorname{rank}(B) - n +$$ + +--- + +## 重点思路 + >[!information] 思路1 >通过矩阵的秩的不等式，最大限度限制所求的表达式的取值范围，或者将其**限制到一个具体的值**. >在希望求一个矩阵的秩的确切值时，也可以考虑用不等式关系来“夹逼”，常见的不等式： @@ -322,4 +599,53 @@ $$ > [!note] 思路2 > 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ +> [!example] 例1 +> 设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： +> (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . +> (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解.（这一问用到这个方法） + +解析：证明如下： +>(1)（10分） +> 对于方程组$Ax=0$ (a)和 $A^\mathrm{T}Ax=0$ (b)，b的解空间一定包含a的解空间（5分）； +>而方程b两边同时乘以$x^\mathrm{T}$，得 $x^\mathrm{T}A^\mathrm{T}Ax=0$ ，即 $(x^\mathrm{T}A^\mathrm{T})(Ax)=0 \to Ax=0$， +>所以a的解空间包含b的解空间（5分）， +>所以a,b同解，所以$\mathrm{rank}A=\mathrm{rank}{A^\mathrm{T}A}$ +>(2) （10分） +>$A^\mathrm{T}Ax=A^\mathrm{T}\beta \iff \mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ +>而由(1)的结论得等式左边 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}A$； +>等式右边 $\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})\ge \mathrm{rank}A^\mathrm{T}A=\mathrm{rank}A$ （5分） +>又$\mathrm{rank}(A^\mathrm{T}\begin{bmatrix}A&\beta\end{bmatrix})\le \min{(\mathrm{rank}A^\mathrm{T},\ \mathrm{rank}\begin{bmatrix}A&\beta\end{bmatrix})}=\mathrm{rank}A$ （5分） +>所以$\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})=\mathrm{rank}A$ +>故 $\mathrm{rank}(A^\mathrm{T}A)=\mathrm{rank}(\begin{bmatrix}A^\mathrm{T}A&A^\mathrm{T}\beta\end{bmatrix})$ 得证. + +>[!example] 例2 +>已知$A, B, C, D$都是 4 阶非零矩阵，且$ABCD = O$，如果$|BC| \neq 0$，记$$r(A) + r(B) + r(C) + r(D) = r$$ +>则$r$的最大值是（ ）。 +>(A) 11 +>(B) 12 +>(C) 13 +>(D) 14 + +**解析思路**： +- 由$|BC| \neq 0$知$B, C$均可逆。 +- 由$ABCD = O$，故$r(AB) + r(CD) \le 4$，$r(A) + r(D) \le 4$ +- 又$B, C$满秩，即$r(B) = r(C) = 4$。 +- 于是$r = r(A) + r(B) + r(C) + r(D) \le 4 + 4 + 4 = 12$。 +- 存在构造使等号成立，故最大值为$12$。 +- 例如$$A = +\begin{pmatrix} +1 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 \\ +0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 +\end{pmatrix}, +\quad +D = +\begin{pmatrix} +0 & 0 & 0 & 0 \\ +0 & 0 & 0 & 0 \\ +1 & 0 & 0 & 0 \\ +0 & 1 & 0 & 0 +\end{pmatrix}$$ +**答案**： (B) 12 -- 2.34.1 From b53998a580e184012a7fca61f92f373716747ae2 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 00:49:27 +0800 Subject: [PATCH 228/274] vault backup: 2026-01-14 00:49:27 --- 编写小组/未命名 1.md | 0 编写小组/未命名.md | 0 ...线性方程组的解与秩的不等式.md | 627 ++++++++++++++++++ ...的解与秩的不等式（解析版）.md | 8 +- 4 files changed, 634 insertions(+), 1 deletion(-) create mode 100644 编写小组/未命名 1.md create mode 100644 编写小组/未命名.md create mode 100644 编写小组/讲义/线性方程组的解与秩的不等式.md diff --git a/编写小组/未命名 1.md b/编写小组/未命名 1.md new file mode 100644 index 0000000..e69de29 diff --git a/编写小组/未命名.md b/编写小组/未命名.md new file mode 100644 index 0000000..e69de29 diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md new file mode 100644 index 0000000..a8cb976 --- /dev/null +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -0,0 +1,627 @@ +--- +tags: + - 编写小组 +--- +**内部资料，禁止传播** +**编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 + +# 单方程组解的问题 + +### 线性方程组解的判定 + +对非齐次线性方程组 $A_{m\times n}x=b$, +1. 无解的充要条件是 $\text{rank}A < \text{rank}[A\ \ b]$； +2. 有唯一解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] = n$； +3. 有无穷多解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b] < n$。 +注:上述定理也说明非齐次线性方程组有解的充要条件是 $\text{rank}A = \text{rank}[A\ \ b]$。 + +怎么理解： +1. 从线性方程组的角度，如果加上一列 $b$ 后的秩变大了，那么化为最简行阶梯型后下面一定多出来一行 $0=$某个常数 ，则必然无解。 +2. 秩等于 $n$ 就是有 $n$ 个无关的方程，则经过消元法后可以解出唯一解。 +3. 秩小于$n$ 就是方程不足 $n$ 个，消元消不完，也能解释为啥秩跟解空间维数的和为 $n$ + +把以上结论应用到齐次线性方程组，可得 +推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 + +### 矩阵方程解的判定 + +本质上和线性方程组是一脉相承的，只是形式上更一般化。 +最常见的矩阵方程是 $\boldsymbol{AX} = \boldsymbol{B}$，其中$\boldsymbol{A}$是 $m\times n$ 矩阵，$\boldsymbol{B}$ 是 $m\times p$ 矩阵，$\boldsymbol{X}$ 是待求的$n\times p$矩阵。 +1. 有解的充要条件： +矩阵方程有解的充要条件是系数矩阵 $\boldsymbol{A}$ 的秩等于增广矩阵 $[\boldsymbol{A} \ \boldsymbol{B}]$的秩，即： +$$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}])$$ +这个结论和非齐次线性方程组有解的条件完全一致。 + +理解上可以将 $B$ 拆分成一列列 $b$ ，从而化归为上面的线性方程组问题 + +2. 解的结构： +- 唯一解：当$r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) = n$ 时，方程有唯一解。 +- 无穷多解：当 $r(\boldsymbol{A}) = r([\boldsymbol{A} \ \boldsymbol{B}]) < n$ 时，方程有无穷多解。 + +可逆矩阵 +- 当 $\boldsymbol{A}$ 是 n 阶可逆矩阵时，矩阵方程 $\boldsymbol{AX} = \boldsymbol{B}$有唯一解:$\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{B}$ + +其他形式的矩阵方程 +- 对于 $\boldsymbol{XA} = \boldsymbol{B}$ 形式的方程，可以转置为 $\boldsymbol{A}^T\boldsymbol{X}^T = \boldsymbol{B}^T$，再套用上述方法，或类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} =\text{rank}A$ +- 对于 $\boldsymbol{AXB} = \boldsymbol{C}$ 形式的方程，当 $\boldsymbol{A} 和 \boldsymbol{B}$ 都可逆时，有唯一解 $\boldsymbol{X} = \boldsymbol{A}^{-1}\boldsymbol{C}\boldsymbol{B}^{-1}$。 + +>[!example] **例1** +>设矩阵 +>$$A = \begin{bmatrix} +1 & a_1 & a_1^2 & a_1^3 \\ +1 & a_2 & a_2^2 & a_2^3 \\ +1 & a_3 & a_3^2 & a_3^3 \\ +1 & a_4 & a_4^2 & a_4^3 +\end{bmatrix}, +\quad +x = \begin{bmatrix} +x_1 \\ x_2 \\ x_3 \\ x_4 +\end{bmatrix}, +\quad +b = \begin{bmatrix} +1 \\ 1 \\ 1 \\ 1 +\end{bmatrix}$$ +其中常数 $a_1, a_2, a_3, a_4$ 互不相等，则线性方程组 $Ax = b$ 的解为 ______________。 + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +>[!example] **例2** +> 设 $A, B$ 均为 $n$ 阶方阵，满足$\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B,$ 且方程 $XA = B$ 有解。若 $\operatorname{rank} A = k$，则$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\underline{\hspace{3cm}}.$ + +# 多方程组的问题（线性方程组同解） + +## **$Ax=0$与$Bx=0$同解问题**： + +充要条件：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. +$Ax=\alpha$ 与$Bx=\beta$同解问题： +充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. + +如何理解（非严格证明，目的是便于理解）： +首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， +考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， +可以得到$N(A)\subset N(B)$,以及$N(B)\subset N(A)$. +$N(A)\subset N(B)$可以得到什么呢？ + +说明$Ax=0$的解比较少，$Bx=0$的解比较多，一个方程组解多就说明他的方程限制相对宽松，解少则说明方程要求比较严格。换言之，$Bx=0$的每个方程是由$Ax=0$的方程线性表示的,同理$N(B)\subset N(A)$ 可以得到 $Ax=0$ 的每个方程是由 $Bx=0$ 的方程线性表示的，进而说明这两个系数矩阵的行向量能够互相线性表示，即行向量组等价.用秩的语言表示：$rankA=rankB=rank\begin{bmatrix} A \\ B\end{bmatrix}$. + +另一个角度：这两个矩阵化成最简行阶梯型，是相同的，进行化简的时候只用到行变换，故它们的行向量组等价. + +需要注意的是,这个条件是充要的.非常的好用. +非齐次的时候同理. + +注意：由此，我们还能得到一些别的结论 +例如：$A$ 和 $B$ 等价（可以通过初等变换得到），并不能得到两方程同解，因为等价的初等变换可能包括初等列变换，而列变换可能改变两方程的解 + +>[!example] 例1 +>6. 已知方程组$\quad\begin{cases}x_1 + 2x_2 + 3x_3 = 0, \\2x_1 + 3x_2 + 5x_3 = 0, \\x_1 + x_2 + ax_3 = 0,\end{cases}$ 与$\quad\begin{cases}x_1 + bx_2 + cx_3 = 0, \\2x_1 + b^2x_2 + (c+1)x_3 = 0\end{cases}$同解，则 + (A) $a = 1, b = 0, c = 1$; + (B) $a = 1, b = 1, c = 2$; + (C) $a = 2, b = 0, c = 1$; + (D) $a = 2, b = 1, c = 2$. + +# 线性方程组的系数矩阵与解关系 + +在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： + +>[!note] 定理1： +>对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 +> $$\dim N(\boldsymbol{A})=n-r$$ + +> [!example] 例1 +> 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. + $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +> [!example] 例2 +> 设 $$ +A = \begin{bmatrix} +1 & -1 & 0 & -1 \\ +1 & 1 & 0 & 3 \\ +2 & 1 & 2 & 6 +\end{bmatrix}, \quad +B = \begin{bmatrix} +1 & 0 & 1 & 2 \\ +1 & -1 & a & a-1 \\ +2 & -3 & 2 & -2 +\end{bmatrix},$$ +向量 $$ +\alpha = \begin{bmatrix} 0 \\ 2 \\ 3 \end{bmatrix}, \quad +\beta = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.$$ +(1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； +(2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 + +```text + + + + + + + + + + + + + + + + + + + + + +``` +# 通过秩反过来得方程是否有解 + +>[!example] 例1 +>已知$\boldsymbol{A},\boldsymbol{B}$均为$m\times n$矩阵，$\beta_1,\beta_2$为$m$维列向量，则下列选项正确的有[ ] +(A)若$\mathrm{rank}\boldsymbol{A}=m$，则对于任意$m$维列向量$\boldsymbol{b},\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解. +(B)若$\boldsymbol{A}$与$\boldsymbol{B}$等价，则齐次线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$与$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$同解. +(C)矩阵方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，但$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解的充要条件是$$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ B}].$$ +(D)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解当且仅当$$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}].$$ + +**解：** +(A)一方面$\mathrm{rank}[\boldsymbol{A\ b}]\ge \mathrm{rank}\boldsymbol{A}=m$,另一方面矩阵$[\boldsymbol{A\ b}]$只有$m$行，所以它的秩必然不大于$m$，所以$\mathrm{rank}[\boldsymbol{A\ b}]=m=\mathrm{rank}\boldsymbol{A}$，即方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{b}$总有解。 + +(B)等价的矩阵只需要是经过初等变换可以变成同一个矩阵就行了，但齐次线性方程组同解需要只经过初等行变换就能变成同一个矩阵才行，后一个条件明显更强，所以后一种更“难”达成，B就不对。 + +(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出C正确。 + +(D)我们同样有两种方法去解这道题，一种是形式化的、严谨的，另一种是理解性的、直观的。 +1)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解$\Leftrightarrow\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]=\mathrm{rank}[\boldsymbol{A\ \beta_2}]$,故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。 +2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 +3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 + +# 矩阵秩与线性方程组解的关系图解说明 + +--- + +## 概念回顾 + +- **rank[A]** 代表矩阵$A$的秩。 +- 秩的定义：矩阵列向量组中**极大线性无关组所含列向量的个数**。 +- 可以用“圆”或“空间”来表示矩阵列向量组张成的向量空间。 + +--- + +## 图解说明 + +假设 +$$ +A = [\alpha_1, \alpha_2, \alpha_3] +$$ +是$m \times 3$矩阵， +$$ +\beta_1, \beta_2 +$$ +是$m$维列向量。 + +### 1. 方程组有解的条件 + +方程组 +$$ +Ax = \beta_1 \quad \text{和} \quad Ax = \beta_2 +$$ +有解 +$\Leftrightarrow$$\beta_1, \beta_2$可由$A$的列向量线性表示。 + +在几何上，这表示： + +- 设$A$的列向量张成的空间为$S_A$。 +-$\beta_1, \beta_2 \in S_A$。 +- 即$S_A$“包含”$\beta_1, \beta_2$。 + +因此，$S_A$这个“圆”应当能够**覆盖**$\beta_1$和$\beta_2$。 + +--- + +### 2. 秩等价条件 + +已知： +$$ +\operatorname{rank}[A] = \operatorname{rank}[A \quad \beta_1 \quad \beta_2] +$$ +表示： +- 矩阵$A$的秩与增广矩阵$[A \mid \beta_1 \mid \beta_2]$的秩相等。 +- 这意味着$\beta_1, \beta_2$并没有“扩大”$A$的列空间。 + +因此： +$$ +\operatorname{rank}[A] = \operatorname{rank}[A \quad \beta_1 \quad \beta_2] +\quad\Leftrightarrow\quad +\beta_1, \beta_2 \in S_A +$$ +即$Ax = \beta_1$和$Ax = \beta_2$有解。 + +--- + +### 3. 等价写法 + +把$\beta_1, \beta_2$放在$A$的右侧构成一个更大的矩阵： +$$ +[A \quad \beta_1 \quad \beta_2] +$$ +其秩与$A$相同，说明： +1. 空间$S_{[A \ \beta_1 \ \beta_2]}$与$S_A$相同。 +2. 从初等行变换角度看：在行阶梯形中，$\beta_1, \beta_2$对应的列会被$A$的列线性表示，从而可化为零列（在解方程时体现为消去）。 +3. 存在$x_1, x_2$使得： + $$ + A(-x_1) = \beta_1, \quad A(-x_2) = \beta_2 + $$ + 这样在增广矩阵中可以通过列操作消去$\beta_1, \beta_2$，使其变为零列。 + +--- + +## 总结 + +- 秩相等 ⇔ 列空间相同 ⇔ 方程组有解。 +- 图示法：把$A$的列空间画成一个圆，$\beta_1, \beta_2$若落在圆内，则方程有解。 +- 矩阵的秩是判断线性方程组解的存在性的核心工具。 + +--- + +**注**：这里的“圆”是比喻，实际为**线性子空间**。 + +# 秩的不等式 + +### 1. 和的秩不超过秩的和 + +设 $A, B$ 为同型矩阵，则 +$$ \operatorname{rank}(A+B) \leq \operatorname{rank} A + \operatorname{rank} B $$ + +### 2. 积的秩不超过任何因子的秩 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \leq \min\{\operatorname{rank} A, \operatorname{rank} B\} $$ + +### 3. Sylvester(西尔维斯特)不等式 + +设 $A_{m \times n}, B_{n \times k}$，则 +$$ \operatorname{rank}(AB) \geq \operatorname{rank} A + \operatorname{rank} B - n $$ +特别地，当 $AB = 0$ 时，有 $\operatorname{rank} A + \operatorname{rank} B \leq n$。 + +### 4. 分块式 + +设 $A_{n \times n}$， $B_{n \times n}$，则 + +$$(1)\ \mathrm{rank} + +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } A, \quad \text{rank } +\begin{bmatrix} +A \\ +B +\end{bmatrix} \geq \text{rank } B +$$ + +$$(2)\ \mathrm{rank} +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } A + \text{rank } B +$$ + +$$(3)\ \mathrm{rank} +\begin{bmatrix} +A & E_n \\ +0 & B +\end{bmatrix} \geq \text{rank } A + \text{rank } B +$$ + +$$(4)\ \mathrm{rank} +\begin{bmatrix} +A & 0 \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A & B \\ +0 & B +\end{bmatrix} = \text{rank } +\begin{bmatrix} +A + B & B \\ +B & B +\end{bmatrix} \geq \text{rank } (A + B) +$$ + +注：（2）与（4）结合即第一个不等式的证明方法 + +> [!note] 证明1： +$$\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}$$ + +--- + +### 证明思路 +设 +$$ +A \in \mathbb{R}^{m \times n}, \quad B \in \mathbb{R}^{n \times p}, \quad C = AB \in \mathbb{R}^{m \times p}. +$$ + +--- + +#### 1. 先证$\operatorname{rank}(AB) \leq \operatorname{rank}(A)$ + +- 考虑$C$的列向量： + 设$B = [b_1, b_2, \dots, b_p]$，则 + $$ + C = [A b_1, A b_2, \dots, A b_p]. + $$ + 因此$C$的每一列都是$A$的列向量的线性组合。 +- 所以$C$的列空间是$A$的列空间的子空间，故 + $$ + \operatorname{rank}(C) \leq \operatorname{rank}(A)。 + $$ + +--- + +#### 2. 再证$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ + +- 考虑$C$的行向量： + 设$A = \begin{bmatrix} a_1^T \\ a_2^T \\ \vdots \\ a_m^T \end{bmatrix}$，则 + $$ + C = \begin{bmatrix} a_1^T B \\ a_2^T B \\ \vdots \\ a_m^T B \end{bmatrix}. + $$ + 因此$C$的每一行都是$B$的行向量的线性组合。 +- 所以$C$的行空间是$B$的行空间的子空间，故 + $$\operatorname{rank}(C) \leq \operatorname{rank}(B)$$ + +--- + +#### 3. 综合 +由 1 和 2 得 +$$ +\operatorname{rank}(AB) \leq \operatorname{rank}(A) \quad \text{且} \quad \operatorname{rank}(AB) \leq \operatorname{rank}(B), +$$ +即 +$$ +\operatorname{rank}(AB) \leq \min\{\operatorname{rank}(A), \operatorname{rank}(B)\}. +$$ + +--- + +**证毕。** + +> [!note] 证明2 +> 证明 Sylvester 秩不等式： +$$\operatorname{rank}(AB) \ge \operatorname{rank}(A) + \operatorname{rank}(B) - n$$ +其中 +$A \in \mathbb{R}^{m \times n}, \; B \in \mathbb{R}^{n \times p}, \; AB \in \mathbb{R}^{m \times p}$。 + +--- + +### 证明思路 +设： +-$\operatorname{rank}(A) = r$ +-$\operatorname{rank}(B) = s$ +-$n$是矩阵乘法的中间维度，即$A$的列数、$B$的行数。 + +--- + +#### 1. 利用分块矩阵构造 +构造如下分块矩阵： +$$ +M = \begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} +\in \mathbb{R}^{(m+n) \times (n+p)} +$$ +其中$I_n$是$n \times n$单位矩阵，$O$是零矩阵。 + +--- + +#### 2. 对$M$进行初等变换 +从$M$的第二块行减去第一块行左乘某个矩阵（这里相当于对$M$做列初等变换），实际上我们可以对$M$做以下变换： +$$ +\begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} +\xrightarrow{\text{右乘 } \begin{bmatrix} I_n & -B \\ O & I_p \end{bmatrix}} +\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +$$ +初等变换不改变矩阵的秩，所以： +$$ +\operatorname{rank}(M) = \operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +$$ + +--- + +#### 3. 估计$\operatorname{rank}(M)$ +另一方面，由分块矩阵的秩不等式： +$$ +\operatorname{rank}(M) \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 + +更严格地，我们可以直接写： +$$ +\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} +A \\ +I_n +\end{bmatrix} + \operatorname{rank}\begin{bmatrix} +I_n & B +\end{bmatrix} - n +$$ +但更简单的常用方法是利用： +$$ +\operatorname{rank}\begin{bmatrix} +A & O \\ +I_n & B +\end{bmatrix} \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +因为$I_n$的存在使得两个子块的秩可以同时保持。 + +--- + +#### 4. 从变换后的矩阵得到下界 +观察变换后的矩阵： +$$ +\operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} +\ge \operatorname{rank}\begin{bmatrix} +I_n & O +\end{bmatrix} + \operatorname{rank}([-AB]) +$$ +实际上更直接的方法是注意到： +$$ +\operatorname{rank}\begin{bmatrix} +A & -AB \\ +I_n & O +\end{bmatrix} += \operatorname{rank}\begin{bmatrix} +O & -AB \\ +I_n & O +\end{bmatrix} \quad (\text{列变换}) +$$ +即： +$$ += \operatorname{rank}\begin{bmatrix} +I_n & O \\ +O & AB +\end{bmatrix} = \operatorname{rank}(I_n) + \operatorname{rank}(AB) = n + \operatorname{rank}(AB) +$$ + +--- + +#### 5. 联立 +由初等变换保秩，得： +$$ +n + \operatorname{rank}(AB) = \operatorname{rank}(M) \ge \operatorname{rank}(A) + \operatorname{rank}(B) +$$ +整理得： +$$ +\operatorname{rank}(AB) \ge \operatorname{rank}(A) + \operatorname{rank}(B) - n +$$ + +--- + +## 重点思路 + +>[!information] 思路1 +>通过矩阵的秩的不等式，最大限度限制所求的表达式的取值范围，或者将其**限制到一个具体的值**. +>在希望求一个矩阵的秩的确切值时，也可以考虑用不等式关系来“夹逼”，常见的不等式： +>1. $\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB})\le\min{\{\mathrm{rank}\boldsymbol{A}, \mathrm{rank}\boldsymbol{B}\}}$ +>2. $\mathrm{rank}(\boldsymbol{A+B})<\mathrm{rank}\boldsymbol A+\mathrm{rank}\boldsymbol B$ +>3. 矩阵加边不会减小秩； +> + +> [!note] 思路2 +> 在遇到诸如 $AB=O$ 的情况，务必要想到$\mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}-n\le\mathrm{rank}(\boldsymbol{AB}) \Rightarrow \mathrm{rank}\boldsymbol{A}+\mathrm{rank}\boldsymbol{B}\le n$ + +> [!example] 例1 +> 设 $A$ 是 $m\times n$ 实矩阵， $\beta \neq 0$ 是 $m$ 维实列向量，证明： +> (1) $\mathrm{rank}A=\mathrm{rank}(A^\mathrm{T}A)$ . +> (2) 线性方程组 $A^\mathrm{T}Ax = A^\mathrm{T}\beta$ 有解.（这一问用到这个方法） + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +>[!example] 例2 +>已知$A, B, C, D$都是 4 阶非零矩阵，且$ABCD = O$，如果$|BC| \neq 0$，记$$r(A) + r(B) + r(C) + r(D) = r$$ +>则$r$的最大值是（ ）。 +>(A) 11 +>(B) 12 +>(C) 13 +>(D) 14 + +```text + + + + + + + + + + + + + + + + + + + + + +``` diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 3cefce7..5605d81 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -135,7 +135,13 @@ $N(A)\subset N(B)$可以得到什么呢？ (C) $a = 2, b = 0, c = 1$; (D) $a = 2, b = 1, c = 2$. -解析：类似于方程 $AX = B$ 有解的充要条件是$\text{rank} \begin{bmatrix} A & B \end{bmatrix} = \text{rank}A$，由方程 $XA = B$ 有解可知 $\text{rank} \begin{bmatrix} A \\ B \end{bmatrix} = \text{rank}B=\text{rank}A=k$，由初等变换不改变秩得$$\text{rank} \begin{bmatrix} B & O \\ A & E \end{bmatrix} =\text{rank} \begin{bmatrix} B & O \\ O & E \end{bmatrix}=n+k$$ +>答案：**D** +>解析：$\text{(I)}:\begin{bmatrix}1&2&3\\2&3&5\\1&1&a\end{bmatrix}x=0$，$\text{(II)}: \begin{bmatrix}1&b&c\\2&b^2&c+1\end{bmatrix}x=0$，对方程做初等行变换： +>$\text{(I)}:\begin{bmatrix}1&0&1\\0&1&1\\0&0&a-2\end{bmatrix}x=0$，$\text{(II)}: \begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}x=0$，记系数矩阵分别为$A,B$ +>因为方程(I),(II)同解，所以$\text{rank}A=\text{rank}B$，而$\text{rank}A\ge 2,\text{rank}B\le 2$，故$\text{rank}A=\text{rank}B=2$，故$a-2=0 \to a=2$；所以方程组(I)的解为$x=k(1,1,-1)^T$； +>令$k=1,x=(1,1,-1)^T$代入方程组(II)得$\begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}\begin{bmatrix}1\\1\\-1\end{bmatrix}=\begin{bmatrix}1+b-c\\b^2-2b+c-1\end{bmatrix}=0$，解得$\begin{cases}b=0\\c=1\end{cases}$或$\begin{cases}b=1\\c=2\end{cases}$；然而，当$\begin{cases}b=0\\c=1\end{cases}$时，$\begin{bmatrix}1&b&c\\0&b^2-2b&1-c\end{bmatrix}=\begin{bmatrix}1&0&1\\0&0&0\end{bmatrix}$，不符合$\text{rank}B=2$的约束，故舍去； +>综上，$\begin{cases}a=2\\b=1\\c=2\end{cases}$ + # 线性方程组的系数矩阵与解关系 在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： -- 2.34.1 From 67a8889fd54cac576dca4abec87b65533d175146 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 10:48:30 +0800 Subject: [PATCH 229/274] vault backup: 2026-01-14 10:48:29 --- 编写小组/讲义/线性方程组的解与秩的不等式.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index a8cb976..2032607 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -18,7 +18,7 @@ tags: 怎么理解： 1. 从线性方程组的角度，如果加上一列 $b$ 后的秩变大了，那么化为最简行阶梯型后下面一定多出来一行 $0=$某个常数 ，则必然无解。 2. 秩等于 $n$ 就是有 $n$ 个无关的方程，则经过消元法后可以解出唯一解。 -3. 秩小于$n$ 就是方程不足 $n$ 个，消元消不完，也能解释为啥秩跟解空间维数的和为 $n$ +3. 秩小于 $n$ 就是方程不足 $n$ 个，消元消不完，也能解释为啥秩跟解空间维数的和为 $n$ 把以上结论应用到齐次线性方程组，可得 推论 齐次线性方程组 $A_{m\times n}x=0$ 有非零解（无穷多解）的充要条件是 $\text{rank}A < n$，即系数矩阵的秩小于未知数个数。 -- 2.34.1 From c57e5c088fb92d01d912866c202524227907ca54 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 10:49:06 +0800 Subject: [PATCH 230/274] vault backup: 2026-01-14 10:49:06 --- 素材/特征值.md | 1 + 1 file changed, 1 insertion(+) diff --git a/素材/特征值.md b/素材/特征值.md index 35ccf38..a0d0906 100644 --- a/素材/特征值.md +++ b/素材/特征值.md @@ -1,3 +1,4 @@ + >[!note] 定理 >秩为$1$的矩阵$A\in\mathbb{R}^{n\times n}$的特征值有如下特征：（1）$0$为其特征值，且代数重数和几何重数均为$n-1$；（2）它的另一个特征值为$\mathrm{tr}(A)$. -- 2.34.1 From 071ab75d0d66a635a75ab7f0a1e2aa4695b757d1 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 10:55:36 +0800 Subject: [PATCH 231/274] vault backup: 2026-01-14 10:55:36 --- 素材/特征值与相似.md | 1 + ...性方程组的系数矩阵与解关系.md | 3 +- ...的解与秩的不等式（解析版）.md | 52 +++++++++---------- 3 files changed, 27 insertions(+), 29 deletions(-) diff --git a/素材/特征值与相似.md b/素材/特征值与相似.md index 2ce056b..d08044a 100644 --- a/素材/特征值与相似.md +++ b/素材/特征值与相似.md @@ -1 +1,2 @@ + $设 E 为 3 阶单位矩阵，\alpha 为一个 3 维单位列向量，则矩阵 E-\alpha\alpha^T 的全部 3 个特征值为\underline{\qquad}。$ diff --git a/素材/线性方程组的系数矩阵与解关系.md b/素材/线性方程组的系数矩阵与解关系.md index 4209625..cde46ac 100644 --- a/素材/线性方程组的系数矩阵与解关系.md +++ b/素材/线性方程组的系数矩阵与解关系.md @@ -1,5 +1,6 @@ 这是一个链接了方程组解空间与方程组系数秩的公式 ->[!note] 解零度化定理： + +>[!note] 秩零化度定理： >对于齐次方程组 ${A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}{A}=r$，则 > $$\dim N({A})=n-r$$ diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 5605d81..08a5383 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -144,9 +144,9 @@ $N(A)\subset N(B)$可以得到什么呢？ # 线性方程组的系数矩阵与解关系 -在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： +在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系，下面是一个链接了方程组解空间与方程组系数秩的公式。 ->[!note] 定理1： +>[!note] 秩零化度定理： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ @@ -154,11 +154,14 @@ $N(A)\subset N(B)$可以得到什么呢？ > 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ ->解析：由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ ->故 $Ax=0$ 的解空间维数是 $1$ （5分） ->$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） ->$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） ->解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 +**答案：** + $$\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$$ +**分析：** 在求解非齐次方程组通解的题目中，若是题目给出了特解与齐次方程组的解，那么大概率来说这个齐次方程组的解就可以拓展为齐次方程组通解（根据问题导向，不然写不出来了），那么如何由齐次方程组的解拓展为齐次方程组通解呢，那就要根据题目具体的条件进行分析了，这就要用到我们的秩零化度定理来求齐次方程组解空间的维数 +**解析：** 由 $\alpha_3=2\alpha_1+\alpha_2$ 可得 $A$ 的列向量组线性相关， $|A|=0$；又因为 $A$ 的三个特征值各不相同，故 $A$ 有两个不为零的特征值 $\lambda_1,\lambda_2$，且 $A$ 可相似对角化，即 $A=P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P$，$\mathrm{rank}A=\mathrm{rank}(P^{-1}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}P)=\mathrm{rank}\begin{bmatrix}\lambda_1&&\\&\lambda_2&\\&&0\end{bmatrix}=2$ +故 $Ax=0$ 的解空间维数是 $1$ （5分） +$\beta=\alpha_1+3\alpha_2+4\alpha_3$，所以 $(1,3,4)^\mathrm{T}$ 为特解；（5分） +$\alpha_3=2\alpha_1+\alpha_2$，所以$A\begin{bmatrix}2k\\k\\-k\end{bmatrix}=2\alpha_1+\alpha_2-\alpha_3=0$，所以 $(2,1,-1)^\mathrm{T}$ 为基础解系；（10分） +解空间维数是 $1$ ，方程的解 $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ 维数是 $1$，该解完备 > [!example] 例2 > 设 $$ @@ -178,57 +181,51 @@ B = \begin{bmatrix} (1) 证明：方程组 $Ax = \alpha$ 的解均为方程组 $Bx = \beta$ 的解； (2) 若方程组 $Ax = \alpha$ 与方程组 $Bx = \beta$ 不同解，求 $a$ 的值。 ---- **解：** - (1) 由于 $$ -\left( \begin{array}{c} +\begin{bmatrix} A \quad \alpha \\ B \quad \beta -\end{array} \right) = -\left( \begin{array}{ccccc} +\end{bmatrix} = +\begin{bmatrix} 1 & -1 & 0 & -1 & 0 \\ 1 & 1 & 0 & 3 & 2 \\ 2 & 1 & 2 & 6 & 3 \\ 1 & 0 & 1 & 2 & 1 \\ 1 & -1 & a & a-1 & 0 \\ 2 & -3 & 2 & -2 & -1 -\end{array} \right) -$$ -$$ +\end{bmatrix} \rightarrow -\left( \begin{array}{ccccc} +\begin{bmatrix} 1 & -1 & 0 & -1 & 0 \\ 0 & 1 & 0 & 2 & 1 \\ 0 & 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 -\end{array} \right), +\end{bmatrix}, $$ 故 $$ -R \left( \begin{array}{c} -A \quad \alpha \\ -B \quad \beta -\end{array} \right) = R(A, \alpha), +\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha], $$ 从而方程组 $$ \begin{cases} -Ax = \alpha, \\ -Bx = \beta +A\boldsymbol{x} = \alpha \\ +B\boldsymbol{x} = \beta \end{cases} $$ -与 $Ax = \alpha$ 同解，故 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解。 +与 $A\boldsymbol{x} = \alpha$ 同解，故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解。 -(2) 由于 $Ax = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $Ax = \alpha$ 与 $Bx = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $Bx = 0$ 的基础解系中解向量的个数，即 +(2) 分析：不同解，却要可以求出$a$的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑秩零化度定理 +由于 $A\boldsymbol{x} = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $A\boldsymbol{x} = \alpha$ 与 $B\boldsymbol{x} = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $B\boldsymbol{x} = 0$ 的基础解系中解向量的个数，即 $$ -4 - R(A) < 4 - R(B), +4 - r(A) < 4 - r(B), $$ -故 $R(A) > R(B)$。又因 $R(A) = 3$，故 $R(B) < 3$，则 +故 $r(A) > r(B)$。又因 $r(A) = 3$，故 $r(B) < 3$，则 $$ \left| \begin{array}{ccc} 1 & 0 & 1 \\ @@ -237,7 +234,6 @@ $$ \end{array} \right| = 0, $$ 解得 $a = 1$。 - # 通过秩反过来得方程是否有解 >[!example] 例1 -- 2.34.1 From f860d4d04cad256ff509cde8f6636a42bd21a34a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 10:56:17 +0800 Subject: [PATCH 232/274] vault backup: 2026-01-14 10:56:16 --- .../讲义/线性方程组的解与秩的不等式.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 2032607..0c106ee 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -123,9 +123,9 @@ $N(A)\subset N(B)$可以得到什么呢？ # 线性方程组的系数矩阵与解关系 -在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系： +在研究线性方程组的解的性质（例如维数）时，我们通常要与其系数矩阵本身的性质产生联系，下面是一个链接了方程组解空间与方程组系数秩的公式。 ->[!note] 定理1： +>[!note] 秩零化度定理： >对于齐次方程组 $\boldsymbol{A}_{m \times n}\boldsymbol{x}=\boldsymbol{0}$，设$\mathrm{rank}\boldsymbol{A}=r$，则 > $$\dim N(\boldsymbol{A})=n-r$$ @@ -198,6 +198,7 @@ B = \begin{bmatrix} ``` + # 通过秩反过来得方程是否有解 >[!example] 例1 -- 2.34.1 From df6f648c06537928ee2e822615d309503b9e5919 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 11:19:49 +0800 Subject: [PATCH 233/274] vault backup: 2026-01-14 11:19:49 --- .../线性方程组的解与秩的不等式（解析版）.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 08a5383..308ba3f 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -152,8 +152,7 @@ $N(A)\subset N(B)$可以得到什么呢？ > [!example] 例1 > 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. - $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ - + **答案：** $$\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$$ **分析：** 在求解非齐次方程组通解的题目中，若是题目给出了特解与齐次方程组的解，那么大概率来说这个齐次方程组的解就可以拓展为齐次方程组通解（根据问题导向，不然写不出来了），那么如何由齐次方程组的解拓展为齐次方程组通解呢，那就要根据题目具体的条件进行分析了，这就要用到我们的秩零化度定理来求齐次方程组解空间的维数 -- 2.34.1 From 6bc2b639014fbee568eedb9470bc60cc8dcd75b7 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 11:20:15 +0800 Subject: [PATCH 234/274] vault backup: 2026-01-14 11:20:15 --- 编写小组/讲义/线性方程组的解与秩的不等式.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 0c106ee..1b9a78b 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -131,8 +131,7 @@ $N(A)\subset N(B)$可以得到什么呢？ > [!example] 例1 > 已知三阶方阵 $A=\begin{bmatrix}\alpha_1&\alpha_2&\alpha_3\end{bmatrix}$ 有三个不同的特征值，其中$\alpha_3=2\alpha_1+\alpha_2$，若 $\beta=\alpha_1+3\alpha_2+4\alpha_3$ ，求线性方程组 $Ax=\beta$ 的通解. - $\begin{bmatrix}1\\3\\4\end{bmatrix}+k\begin{bmatrix}2\\1\\-1\end{bmatrix}, k\in\mathbb{R}$ - + ```text -- 2.34.1 From 305ec0e93d4bac3929c4aefbdfacdc4cdbaf21ae Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 11:28:18 +0800 Subject: [PATCH 235/274] vault backup: 2026-01-14 11:28:18 --- .../线性方程组的解与秩的不等式（解析版）.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 308ba3f..3490ddf 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -208,7 +208,7 @@ B \quad \beta $$ 故 $$ -\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha], +\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha]=\mathrm{rank} \begin{bmatrix}A&\alpha\\A&\alpha\\B&\beta\end{bmatrix}, $$ 从而方程组 $$ @@ -217,7 +217,7 @@ A\boldsymbol{x} = \alpha \\ B\boldsymbol{x} = \beta \end{cases} $$ -与 $A\boldsymbol{x} = \alpha$ 同解，故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解。 +与 $A\boldsymbol{x} = \alpha$ 同解，故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解（取交集为其中之一：包含关系） (2) 分析：不同解，却要可以求出$a$的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑秩零化度定理 由于 $A\boldsymbol{x} = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $A\boldsymbol{x} = \alpha$ 与 $B\boldsymbol{x} = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $B\boldsymbol{x} = 0$ 的基础解系中解向量的个数，即 -- 2.34.1 From 73773fd5daacf5e058e2a87c12ae14ada5839c43 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 11:36:05 +0800 Subject: [PATCH 236/274] vault backup: 2026-01-14 11:36:05 --- ...组的解与秩的不等式（解析版）.md | 14 ++++---------- 1 file changed, 4 insertions(+), 10 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 3490ddf..fa15bf4 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -112,6 +112,8 @@ $$ $Ax=\alpha$ 与$Bx=\beta$同解问题： 充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. +解包含的关系：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix} \Leftrightarrow A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解 + 如何理解（非严格证明，目的是便于理解）： 首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， 考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， @@ -208,16 +210,8 @@ B \quad \beta $$ 故 $$ -\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha]=\mathrm{rank} \begin{bmatrix}A&\alpha\\A&\alpha\\B&\beta\end{bmatrix}, -$$ -从而方程组 -$$ -\begin{cases} -A\boldsymbol{x} = \alpha \\ -B\boldsymbol{x} = \beta -\end{cases} -$$ -与 $A\boldsymbol{x} = \alpha$ 同解，故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解（取交集为其中之一：包含关系） +\mathrm{rank} \begin{bmatrix}A&\alpha\\B&\beta\end{bmatrix} = \mathrm{rank}[A\ \alpha]$$ +故 $A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解。 (2) 分析：不同解，却要可以求出$a$的具体值，说明这是一个与秩相关的题，而与解相关的秩的问题我们就可以考虑秩零化度定理 由于 $A\boldsymbol{x} = \alpha$ 的解均为 $Bx = \beta$ 的解，若 $A\boldsymbol{x} = \alpha$ 与 $B\boldsymbol{x} = \beta$ 同解，则与题意矛盾，故 $Ax = \alpha$ 的解是 $Bx = \beta$ 解的真子集。于是 $Ax = 0$ 的基础解系中解向量的个数小于 $B\boldsymbol{x} = 0$ 的基础解系中解向量的个数，即 -- 2.34.1 From 600d95654e907d4f889c039b8cde383d487fb60c Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 11:36:28 +0800 Subject: [PATCH 237/274] vault backup: 2026-01-14 11:36:28 --- 编写小组/讲义/线性方程组的解与秩的不等式.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 1b9a78b..bfed081 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -98,6 +98,8 @@ b = \begin{bmatrix} $Ax=\alpha$ 与$Bx=\beta$同解问题： 充要条件：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} B &\beta\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix}$. +解包含的关系：$rank\begin{bmatrix} A & \alpha\end{bmatrix}=rank\begin{bmatrix} A &\alpha\\ B&\beta\end{bmatrix} \Leftrightarrow A\boldsymbol{x} = \alpha$ 的解均为 $B\boldsymbol{x} = \beta$ 的解 + 如何理解（非严格证明，目的是便于理解）： 首先，为了简化问题，我们只考虑齐次线性方程组同解问题，对于$Ax=0$与$Bx=0$， 考虑这两个齐次线性方程组的解空间，分别记为$N(A)$,$N(B)$,这两个集合是完全相同的， -- 2.34.1 From 19c92df270a76cb985d7497935f80fcfaf4b0cfd Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 12:48:03 +0800 Subject: [PATCH 238/274] vault backup: 2026-01-14 12:48:03 --- 素材/微分中值定理.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/微分中值定理.md b/素材/微分中值定理.md index 70e16dd..5b1bf51 100644 --- a/素材/微分中值定理.md +++ b/素材/微分中值定理.md @@ -141,7 +141,7 @@ $$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： (2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 **解析**： -(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 (2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 --- -- 2.34.1 From 51b584367093b4e4831ac8b772f62b9933c40e6a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 12:48:37 +0800 Subject: [PATCH 239/274] vault backup: 2026-01-14 12:48:37 --- 编写小组/未命名 1.md | 0 编写小组/未命名.md | 0 2 files changed, 0 insertions(+), 0 deletions(-) delete mode 100644 编写小组/未命名 1.md delete mode 100644 编写小组/未命名.md diff --git a/编写小组/未命名 1.md b/编写小组/未命名 1.md deleted file mode 100644 index e69de29..0000000 diff --git a/编写小组/未命名.md b/编写小组/未命名.md deleted file mode 100644 index e69de29..0000000 -- 2.34.1 From 891241ca52a8962f2bd5c3ae3723be386bb2385f Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 13:00:37 +0800 Subject: [PATCH 240/274] vault backup: 2026-01-14 13:00:37 --- 笔记分享/LaTeX（KaTeX）输入规范.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md index 9674991..c88bffc 100644 --- a/笔记分享/LaTeX（KaTeX）输入规范.md +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -4,4 +4,4 @@ 3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ 4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ 5. \$\$双美元符号之间不要打回车！除非你有\begin{...}\end{...}\$\$ -6. 矩阵和向量要加粗，用\boldsymbol{}，比如$\boldsymbol{A},\boldsymbol{x}$。 \ No newline at end of file +6. 矩阵不用加粗，但向量要加粗，用\boldsymbol{}，比如$A,\boldsymbol{x}$。 \ No newline at end of file -- 2.34.1 From 371e03f45e4f53d6d7882253a5d03e1da6c022ee Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 14 Jan 2026 13:23:12 +0800 Subject: [PATCH 241/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 微分中值定理的不等式问题.md | 72 +++++++++++++++++++++++++ 1 file changed, 72 insertions(+) create mode 100644 微分中值定理的不等式问题.md diff --git a/微分中值定理的不等式问题.md b/微分中值定理的不等式问题.md new file mode 100644 index 0000000..ad98b77 --- /dev/null +++ b/微分中值定理的不等式问题.md @@ -0,0 +1,72 @@ +## 例一 +设 $e < a < b < e^2$，证明： +$$ +\ln^2 b - \ln^2 a > \frac{4}{e^2}(b-a). +$$ + +**证明**： +考虑函数 $f(x) = \ln^2 x$，则 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导。由拉格朗日中值定理，存在 $\xi \in (a,b)$，使得 +$$ +\frac{\ln^2 b - \ln^2 a}{b-a} = f'(\xi) = \frac{2\ln \xi}{\xi}. +$$ +令 $g(x) = \dfrac{2\ln x}{x}$，求导得 +$$ +g'(x) = \frac{2(1-\ln x)}{x^2}. +$$ +当 $x > e$ 时，$\ln x > 1$，故 $g'(x) < 0$，即 $g(x)$ 在 $(e, +\infty)$ 上单调递减。 +由于 $e < a < \xi < b < e^2$，所以 +$$ +g(\xi) > g(e^2) = \frac{2\ln e^2}{e^2} = \frac{4}{e^2}. +$$ +因此 +$$ +\frac{\ln^2 b - \ln^2 a}{b-a} > \frac{4}{e^2}, +$$ +即 +$$ +\ln^2 b - \ln^2 a > \frac{4}{e^2}(b-a). +$$ +证毕。 +## 例2 +设 $a > e$，$0 < x < y < \dfrac{\pi}{2}$，证明： +$$ +a^y - a^x > (\cos x - \cos y) \cdot a^x \ln a. +$$ + +**证明**： +令 $f(t) = a^t$，则 $f(t)$ 在 $[x, y]$ 上连续，在 $(x, y)$ 内可导。由拉格朗日中值定理，存在 $\xi \in (x, y)$，使得 +$$ +\frac{a^y - a^x}{\cos x - \cos y} = \frac{a^\xi \ln a}{\sin \xi } +$$ + $$\frac{a^\xi \ln a}{\sin \xi }>a^\xi \ln a>a^x \ln a$$ + 证毕 +## 例3 +证明：当 $x>0$ 时， +$$ +\frac{\arctan x}{\ln(1+x)} \leq \frac{1+\sqrt{2}}{2}. +$$ + +**证明**： +考虑函数 $f(t) = \arctan t$ 与 $g(t) = \ln(1+t)$，两者在 $[0, x]$ 上连续，在 $(0, x)$ 内可导，且 $g'(t) = \frac{1}{1+t} \neq 0$。由柯西中值定理，存在 $\xi \in (0, x)$，使得 +$$ +\frac{\arctan x}{\ln(1+x)} = \frac{f(x) - f(0)}{g(x) - g(0)} = \frac{f'(\xi)}{g'(\xi)} = \frac{1/(1+\xi^2)}{1/(1+\xi)} = \frac{1+\xi}{1+\xi^2}. +$$ +令 $\phi(\xi) = \dfrac{1+\xi}{1+\xi^2}$，则 +$$ +\phi'(\xi) = \frac{(1+\xi^2) - (1+\xi) \cdot 2\xi}{(1+\xi^2)^2} = \frac{1 - 2\xi - \xi^2}{(1+\xi^2)^2} = \frac{2 - (1+\xi)^2}{(1+\xi^2)^2}. +$$ +令 $\phi'(\xi) = 0$，得 $(1+\xi)^2 = 2$，因 $\xi > 0$，故 $\xi = \sqrt{2} - 1$。 +当 $0 < \xi < \sqrt{2} - 1$ 时，$\phi'(\xi) > 0$；当 $\xi > \sqrt{2} - 1$ 时，$\phi'(\xi) < 0$。 +因此 $\phi(\xi)$ 在 $\xi = \sqrt{2} - 1$ 处取得最大值： +$$ +\phi(\sqrt{2} - 1) = \frac{1 + (\sqrt{2} - 1)}{1 + (\sqrt{2} - 1)^2} = \frac{\sqrt{2}}{1 + (3 - 2\sqrt{2})} = \frac{\sqrt{2}}{4 - 2\sqrt{2}} = \frac{\sqrt{2}}{2(2 - \sqrt{2})}. +$$ +化简： +$$ +\frac{\sqrt{2}}{2(2 - \sqrt{2})} = \frac{\sqrt{2}}{2} \cdot \frac{1}{2 - \sqrt{2}} = \frac{\sqrt{2}}{2} \cdot \frac{2 + \sqrt{2}}{2} = \frac{\sqrt{2}(2 + \sqrt{2})}{4} = \frac{2\sqrt{2} + 2}{4} = \frac{1 + \sqrt{2}}{2}. +$$ +于是对任意 $\xi > 0$，有 $\phi(\xi) \leq \dfrac{1+\sqrt{2}}{2}$，从而 +$$ +\frac{\arctan x}{\ln(1+x)} \leq \frac{1+\sqrt{2}}{2}, \quad x > 0. +$$ +等号在 $\xi = \sqrt{2} - 1$ 时成立，即存在 $x > 0$ 使等号成立。证毕。 \ No newline at end of file -- 2.34.1 From a4e0d896be48ec4f5d74f7617b7acafc6eb71558 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 14 Jan 2026 13:23:33 +0800 Subject: [PATCH 242/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../微分中值定理的不等式问题.md | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 微分中值定理的不等式问题.md => 素材/微分中值定理的不等式问题.md (100%) diff --git a/微分中值定理的不等式问题.md b/素材/微分中值定理的不等式问题.md similarity index 100% rename from 微分中值定理的不等式问题.md rename to 素材/微分中值定理的不等式问题.md -- 2.34.1 From 743fbcfeac89e7050ad30e88102744d5e424f1f3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 14 Jan 2026 13:28:20 +0800 Subject: [PATCH 243/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 素材/微分中值定理的不等式问题.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/素材/微分中值定理的不等式问题.md b/素材/微分中值定理的不等式问题.md index ad98b77..b7c2c97 100644 --- a/素材/微分中值定理的不等式问题.md +++ b/素材/微分中值定理的不等式问题.md @@ -1,3 +1,5 @@ + +当看到多元不等式问题时候，我们可以考虑用微分中值定理来解决,(慎用，因为微分中值定理的放缩精度并不高，但是一旦可以用就非常巧妙) ## 例一 设 $e < a < b < e^2$，证明： $$ -- 2.34.1 From dc545dd1fe28ab77ac43c3a77309f7ef04282505 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 13:49:41 +0800 Subject: [PATCH 244/274] vault backup: 2026-01-14 13:49:41 --- 编写小组/黑马试卷/1.16黑马试卷.md | 126 ++++++++++++++++++ 1 file changed, 126 insertions(+) create mode 100644 编写小组/黑马试卷/1.16黑马试卷.md diff --git a/编写小组/黑马试卷/1.16黑马试卷.md b/编写小组/黑马试卷/1.16黑马试卷.md new file mode 100644 index 0000000..bfb7100 --- /dev/null +++ b/编写小组/黑马试卷/1.16黑马试卷.md @@ -0,0 +1,126 @@ +### 一、选择题（共3小题，每小题2分，共10分） + +1. 函数 +$$ + f(x) = \begin{cases} + \sqrt{x} \sin \frac{1}{x}, & x > 0, \\ + 0, & x \leq 0 + \end{cases} +$$ + 在点$x = 0$处（ ）。 + (A) 不连续 + (B) 连续但不可导 + (C) 可导且$f'(0) = 0$ + (D) 可导且$f'(0) \neq 0$ + +2. 数列极限$\lim_{n \to \infty} (e^{-n} + \pi^{-n})^{\frac{1}{n}}$的值为（ ）。 + (A)$e$ + (B)$\pi$ + (C)$\frac{1}{e}$ + (D)$\frac{1}{\pi}$ + +3. 曲线$y = \frac{x^3 - x^2}{2 + x^2}$的渐近线为（ ）。 + (A)$y = x - 1$ + (B)$y = x + 1$ + (C)$y = x$ + (D)$y = \frac{1}{2}x$ + +--- + +### 二、填空题（共3小题，每小题2分，共10分） + +4. 函数$f(x) = xe^{-x^2}$在$(-\infty,+\infty)$上的最大值为 $\underline{\qquad}$。 + +5. 曲线$C: x = \frac{1}{2} \cos t, y = \sin t, t \in [0,2\pi]$在点$(0,-1)$处的曲率为 $\underline{\qquad}$。 + +6. 不定积分$\int \frac{1}{x(1+2\ln x)} dx = \underline{\qquad}$。 + +--- + +### 三、解答题（共4小题，共80分） + +7. 设$y(x)$是由曲线方程$\sin x + y + e^x = 2$确定的隐函数，试计算$\frac{dy}{dx} \bigg|_{x=0}$的值，并求该曲线在点$P(0,1)$处的切线方程。（6分） + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +8. 计算不定积分 +$$ + \int \frac{x}{1+\sqrt{1-x^2}} dx。 +$$ + （6分） + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +9. 设曲线$f(x) = x^3 + ax^2 + 18x$（$a$为大于零的常数）的拐点正好位于$x$轴上，试求$a$的值及曲线$y = f(x)$的拐点坐标。（6分） + +```text + + + + + + + + + + + + + + + + + + + + + +``` + +10. 计算极限 + $$\lim_{x \to +\infty} \left[ x + x^2 \ln \left( 1 - \frac{1}{x} \right) \right]。$$ + （6分） -- 2.34.1 From a1216ac1465940fe09c99f8545b5ebf4620ff003 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 13:55:37 +0800 Subject: [PATCH 245/274] vault backup: 2026-01-14 13:55:37 --- 笔记分享/达布定理.md | 4 ++++ 1 file changed, 4 insertions(+) create mode 100644 笔记分享/达布定理.md diff --git a/笔记分享/达布定理.md b/笔记分享/达布定理.md new file mode 100644 index 0000000..8c1c529 --- /dev/null +++ b/笔记分享/达布定理.md @@ -0,0 +1,4 @@ +>[!note] 定理： +>如果函数$f(x)$在区间$[a,b]$上可导，则其导函数$f'(x)$在$[a,b]$上有介值性质，即若$f(x)$在$[a,b]$上的值域为$[m,M]$,则$\forall \xi\in[m,M]$，总$\exists \eta\in[a,b],$有$\xi=f'(\eta)$. + +**证明**：若$m=M$，结论显然成立.若$m<M$,设$f'(x_1)=m,f'(x_2)=M$,不妨设$x_1<x_2$.任取$\xi\in(m,M)$,令$$g(x)=f(x)-\xi x,x\in[a,b].$$于是$g(x)$在$[a,b]$上可导，且$$g'(x_1)=f'(x_1)-\xi<0,g'(x_2)=f'(x_2)-\xi>0.$$由零值定理，$\exists \eta\in(x_1,x_2) \subset(a,b),g'(\eta)=0\implies f'(\eta)=\xi$，证毕. \ No newline at end of file -- 2.34.1 From 91fc5d07d1b834695f75233457e8b6242b489398 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 14 Jan 2026 15:06:32 +0800 Subject: [PATCH 246/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../微分中值定理的不等式问题.md | 39 ++++++++++++++++++- 1 file changed, 38 insertions(+), 1 deletion(-) diff --git a/素材/微分中值定理的不等式问题.md b/素材/微分中值定理的不等式问题.md index b7c2c97..f916ba4 100644 --- a/素材/微分中值定理的不等式问题.md +++ b/素材/微分中值定理的不等式问题.md @@ -1,5 +1,42 @@ -当看到多元不等式问题时候，我们可以考虑用微分中值定理来解决,(慎用，因为微分中值定理的放缩精度并不高，但是一旦可以用就非常巧妙) +## 微分中值定理证明不等式的要点归纳 + +### 1. **识别不等式结构** + - 若不等式形如 $f(b) - f(a)$ 与 $b-a$ 的关系，或含有函数值差与自变量差之商，可考虑**拉格朗日中值定理**。 + - 若不等式涉及两个不同函数值差的比值，可考虑**柯西中值定理**。 + - 若结论中出现高阶导数（如二阶导），可能需用**泰勒公式**。 + +### 2. **选择合适定理与辅助函数** + - **拉格朗日定理**：常用于"单函数"差值型不等式，构造 $f(x)$ 使 $f'(\xi)$ 出现在不等式中。 + - **柯西定理**：适用于"双函数"比值型不等式，构造 $f(x), g(x)$ 使 $\frac{f'(\xi)}{g'(\xi)}$ 出现。 + - **辅助函数构造**：常借助常见函数如 $\ln x, e^x, x^n, \arctan x, \sin x, \cos x$ 等，通过求导形式匹配目标。 + +### 3. **利用导数单调性估计中值** + - 应用中值定理得到含 $\xi$ 的表达式后，需估计 $f'(\xi)$ 的范围。 + - 若 $f'(x)$ 单调，则根据 $\xi$ 所属区间确定 $f'(\xi)$ 的上下界，从而导出不等式。 + +### 4. **处理多中值与多次应用** + - 若结论含两个及以上中值，可能需要**多次应用中值定理**（如先在子区间上用拉格朗日，再对导数用罗尔或柯西）。 + - 有时需**结合不同定理**，例如先用柯西得到比值，再用拉格朗日简化。 + +### 5. **验证定理条件** + - 确保函数在闭区间连续、开区间可导，且分母函数导数不为零（柯西定理）。 + +### 6. **结合其他技巧** + - **放大缩小**：对得到的中值表达式进行适当放缩。 + - **函数最值**：若中值表达式为某函数值，可求该函数在区间上的最值。 + - **反证法**：假设不等式不成立，推出矛盾。 + +### 7. **常见题型模式** + - **单中值不等式**：直接构造辅助函数用拉格朗日，利用 $f'(\xi)$ 的范围证明。 + - **双函数比值不等式**：用柯西定理化为导数比，再分析导数比的取值范围。 + - **含参数的不等式**：将参数视为变量，构造含参函数应用中值定理。 + +### 8. **书写规范** + - 清晰写出所构造的函数、使用的区间、定理名称。 + - 明确中值 $\xi$ 的存在范围，并利用该范围进行不等推导。 + +掌握以上要点，可系统解决大多数与微分中值定理相关的不等式证明题。 ## 例一 设 $e < a < b < e^2$，证明： $$ -- 2.34.1 From 937861c613514e7e868f0f3d7099188cf701f13c Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E9=83=91=E5=93=B2=E8=88=AA?= <zzh18070015413@qq.com> Date: Wed, 14 Jan 2026 16:45:00 +0800 Subject: [PATCH 247/274] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E6=96=87?= =?UTF-8?q?=E4=BB=B6=EF=BC=9A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../微分中值定理的不等式问题.md | 75 ++++++++++++------- 1 file changed, 49 insertions(+), 26 deletions(-) diff --git a/素材/微分中值定理的不等式问题.md b/素材/微分中值定理的不等式问题.md index f916ba4..730737f 100644 --- a/素材/微分中值定理的不等式问题.md +++ b/素材/微分中值定理的不等式问题.md @@ -3,8 +3,7 @@ ### 1. **识别不等式结构** - 若不等式形如 $f(b) - f(a)$ 与 $b-a$ 的关系，或含有函数值差与自变量差之商，可考虑**拉格朗日中值定理**。 - - 若不等式涉及两个不同函数值差的比值，可考虑**柯西中值定理**。 - - 若结论中出现高阶导数（如二阶导），可能需用**泰勒公式**。 + ### 2. **选择合适定理与辅助函数** - **拉格朗日定理**：常用于"单函数"差值型不等式，构造 $f(x)$ 使 $f'(\xi)$ 出现在不等式中。 @@ -12,31 +11,9 @@ - **辅助函数构造**：常借助常见函数如 $\ln x, e^x, x^n, \arctan x, \sin x, \cos x$ 等，通过求导形式匹配目标。 ### 3. **利用导数单调性估计中值** - - 应用中值定理得到含 $\xi$ 的表达式后，需估计 $f'(\xi)$ 的范围。 - - 若 $f'(x)$ 单调，则根据 $\xi$ 所属区间确定 $f'(\xi)$ 的上下界，从而导出不等式。 + - 应用中值定理得到含 $\xi$ 的表达式后，可以通过函数极值的求法求出其最大最小值进行比较 -### 4. **处理多中值与多次应用** - - 若结论含两个及以上中值，可能需要**多次应用中值定理**（如先在子区间上用拉格朗日，再对导数用罗尔或柯西）。 - - 有时需**结合不同定理**，例如先用柯西得到比值，再用拉格朗日简化。 -### 5. **验证定理条件** - - 确保函数在闭区间连续、开区间可导，且分母函数导数不为零（柯西定理）。 - -### 6. **结合其他技巧** - - **放大缩小**：对得到的中值表达式进行适当放缩。 - - **函数最值**：若中值表达式为某函数值，可求该函数在区间上的最值。 - - **反证法**：假设不等式不成立，推出矛盾。 - -### 7. **常见题型模式** - - **单中值不等式**：直接构造辅助函数用拉格朗日，利用 $f'(\xi)$ 的范围证明。 - - **双函数比值不等式**：用柯西定理化为导数比，再分析导数比的取值范围。 - - **含参数的不等式**：将参数视为变量，构造含参函数应用中值定理。 - -### 8. **书写规范** - - 清晰写出所构造的函数、使用的区间、定理名称。 - - 明确中值 $\xi$ 的存在范围，并利用该范围进行不等推导。 - -掌握以上要点，可系统解决大多数与微分中值定理相关的不等式证明题。 ## 例一 设 $e < a < b < e^2$，证明： $$ @@ -108,4 +85,50 @@ $$ $$ \frac{\arctan x}{\ln(1+x)} \leq \frac{1+\sqrt{2}}{2}, \quad x > 0. $$ -等号在 $\xi = \sqrt{2} - 1$ 时成立，即存在 $x > 0$ 使等号成立。证毕。 \ No newline at end of file +等号在 $\xi = \sqrt{2} - 1$ 时成立，即存在 $x > 0$ 使等号成立。证毕。 + +## 例3 + +(1) 证明：存在 $\theta \in (0, 1)$ 使得 $\ln(1+x) - \ln\left(1+\frac{x}{2}\right) = \frac{x}{2+(1+\theta)x}, \, x > 0$; +(2) 证明不等式 +$$ +\left(1+\frac{1}{n}\right)^{n+1} < e\left(1+\frac{1}{2n}\right), +$$ +其中 $n$ 为正整数。 + +## 解答 + +**证明** +（1）对 $x > 0$ 定义函数 $f(t) = \ln(1+t), t \in \left[\frac{x}{2}, x\right]$, +由拉格朗日中值定理知：存在 $\theta \in (0, 1)$ 使得 + +$$ +\begin{aligned} +f(x) - f\left(\frac{x}{2}\right) &= \ln(1+x) - \ln\left(1+\frac{x}{2}\right) \\ +&= \frac{1}{1+\frac{x}{2}+\theta} \cdot \frac{x}{2} \\ +&= \frac{x}{2+(1+\theta)x}. +\end{aligned} +$$ + +（2）不等式两边取对数，可知仅证明 $(n+1)\ln\left(1+\frac{1}{n}\right) < 1 + \ln\left(1+\frac{1}{2n}\right)$ 即可。 + +令 $F(x) = x + x\ln\left(1+\frac{x}{2}\right) - (x+1)\ln(1+x), x \geq 0$，则由（1）知 + +$$ +\begin{aligned} +F'(x) &= 1 + \frac{\frac{x}{2}}{1+\frac{x}{2}} + \ln\left(1+\frac{x}{2}\right) - 1 - \ln(1+x) \\ +&= \frac{\frac{x}{2}}{1+\frac{x}{2}} - \left[\ln(1+x) - \ln\left(1+\frac{x}{2}\right)\right] \\ +&= \frac{\frac{x}{2}}{1+\frac{x}{2}} - \frac{\frac{x}{2}}{1+(1+\theta)\frac{x}{2}} \\ +&= \frac{\frac{x}{2}}{1+\frac{x}{2}} - \frac{\frac{x}{2}}{1+\frac{x}{2}} = 0. +\end{aligned} +$$ + +因此 $F(x) > F(0) = 0, x > 0$。即 $(x+1)\ln(1+x) < x + x\ln\left(1+\frac{x}{2}\right), x > 0$。 + +令 $x = \frac{1}{n}$，则有 $(n+1)\ln\left(1+\frac{1}{n}\right) < 1 + \ln\left(1+\frac{1}{2n}\right)$。因此对任意正整数 $n$ 有不等式 + +$$ +\left(1+\frac{1}{n}\right)^{n+1} < e\left(1+\frac{1}{2n}\right) +$$ + +成立。 \ No newline at end of file -- 2.34.1 From 152ab205d95a1e933a6f4a9e84e0f32f4a29bbe0 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 17:00:48 +0800 Subject: [PATCH 248/274] vault backup: 2026-01-14 17:00:48 --- ...方程组的解与秩的不等式（解析版）.md | 10 ++++++---- 1 file changed, 6 insertions(+), 4 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index fa15bf4..395f157 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -241,12 +241,14 @@ $$ (B)等价的矩阵只需要是经过初等变换可以变成同一个矩阵就行了，但齐次线性方程组同解需要只经过初等行变换就能变成同一个矩阵才行，后一个条件明显更强，所以后一种更“难”达成，B就不对。 -(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出C正确。 +(C)方程$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解$\Leftrightarrow\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}\boldsymbol{A}$,方程$\boldsymbol{B}\boldsymbol{Y}=\boldsymbol{A}$无解$\Leftrightarrow\mathrm{rank}\boldsymbol{B}<\mathrm{rank}[\boldsymbol{B\ A}]$,而$\mathrm{rank}[\boldsymbol{A\ B}]=\mathrm{rank}[\boldsymbol{B\ A}]$,故C正确。 +这是纯形式化的解答，不过当然是正确的。但是怎么理解这个结果呢？$\boldsymbol{A}\boldsymbol{X}=\boldsymbol{B}$有解，就是说我们可以用矩阵$\boldsymbol{A}$表示矩阵$\boldsymbol{B}$,也就是说，$\boldsymbol{A}$中包含了$\boldsymbol{B}$中的所有信息，也就是$\mathrm{rank}\boldsymbol{A}\ge\mathrm{rank}\boldsymbol{B}$；另一方面，$\boldsymbol{BY}=\boldsymbol{A}$无解说明我们无法用矩阵$\boldsymbol{B}$表示矩阵$\boldsymbol{A}$，也就是说，$\boldsymbol{B}$中没有包含$\boldsymbol{A}$中的所有信息，那么$\mathrm{rank}\boldsymbol{B}<\mathrm{rank}\boldsymbol{A}$；再加上有解的充要条件得出正向是正确的。 +反过来，如果$\mathrm{rank}{A}=\mathrm{rank}[{A\ B}]$，说明$A$中已经包含了矩阵$[A\ B]$的所有信息，所以$A$中就也包含了$B$中的所有信息，所以方程组$AX=B$有解；而如果又有$\mathrm{rank}B<\mathrm{rank}A$，则$B$没有完全包含$A$中的所有信息，或者说，$B$的信息真包含于$A$的信息，所以无法用$B$表示$A$，即$BY=A$无解。 (D)我们同样有两种方法去解这道题，一种是形式化的、严谨的，另一种是理解性的、直观的。 1)线性方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解$\Leftrightarrow\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]=\mathrm{rank}[\boldsymbol{A\ \beta_2}]$,故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。 -2)也可以从初等变换的角度来理解，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 -3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$,反过来也是一样的。这就说明D是正确的。 +2)也可以从初等变换的角度来解答，方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解说明$\boldsymbol{\beta_1}$可以用$\boldsymbol{A}$的列向量线性表示，从而$[\boldsymbol{A\ \beta_1}]$可以通过初等列变换变成$[\boldsymbol{A\ O}]$，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1}]$；同理可以得出关于$\boldsymbol{\beta_2}$的结论。 +3)同样，怎么直观地理解？我们一样用信息量的观点去看。方程$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$有解，意味着$\boldsymbol{A}$中包含了$\boldsymbol{\beta_1}$中的所有信息，同理，$\boldsymbol{A}$中也包含了$\boldsymbol{\beta_2}$中的所有信息，这就意味着矩阵$[\boldsymbol{A\ \beta_1\ \beta_2}]$中所有的信息其实只需要用$\boldsymbol{A}$就可以表示，故$\mathrm{rank}\boldsymbol{A}=\mathrm{rank}[\boldsymbol{A\ \beta_1\ \beta_2}]$。反过来也一样，如果有$\mathrm{rank}{A}=\mathrm{rank}[{A\ \boldsymbol{\beta_1\ \beta_2}}]$，那么$A$中就包含了$\boldsymbol{\beta_1},\boldsymbol{\beta_2}$中的所有的信息，所以可以用$A$去表示向量$\boldsymbol{\beta_1}$和$\boldsymbol{\beta_2}$，也就是方程组$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_1}$与$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{\beta_2}$同时有解。这就说明D是正确的。 # 矩阵秩与线性方程组解的关系图解说明 @@ -254,7 +256,7 @@ $$ ## 概念回顾 -- **rank[A]** 代表矩阵$A$的秩。 +- __$\mathrm{rank}A$__ 代表矩阵$A$的秩。 - 秩的定义：矩阵列向量组中**极大线性无关组所含列向量的个数**。 - 可以用“圆”或“空间”来表示矩阵列向量组张成的向量空间。 -- 2.34.1 From 58a1b08da16a3ed809536ce17a5fb1ccd259bcbd Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:05:57 +0800 Subject: [PATCH 249/274] vault backup: 2026-01-14 17:05:57 --- 素材/罗尔定理与拉格朗日定理.md | 108 ++++++++++++++++++++ 1 file changed, 108 insertions(+) create mode 100644 素材/罗尔定理与拉格朗日定理.md diff --git a/素材/罗尔定理与拉格朗日定理.md b/素材/罗尔定理与拉格朗日定理.md new file mode 100644 index 0000000..bee8105 --- /dev/null +++ b/素材/罗尔定理与拉格朗日定理.md @@ -0,0 +1,108 @@ +## **罗尔定理** +### **原理** +若函数 f(x) 满足以下三个条件： +在闭区间 $[a,b]$ 上连续； +在开区间 $(a,b)$ 内可导； +区间端点函数值相等，即 $f(a)=f(b)$； +则在 $(a,b)$ 内至少存在一点 $\xi$，使得 $f'(\xi)=0$。 +罗尔定理的几何意义为：满足条件的函数曲线在区间内至少有一条水平切线。 +它是拉格朗日中值定理（$f(b)-f(a)=f'(\xi)(b-a)$）当 $f(a)=f(b)$ 时的特例。 + +### **适用条件** +罗尔定理的核心适用题型是证明导函数方程 $f'(\xi)=0$ 在区间 $(a,b)$ 内有根以及衍生的相关证明题。 +具体可分为以下几类： +1.直接证明 $f'(\xi)$=0 存在根 +题目给出函数 f(x) 在 $[a,b]$ 上的连续性、$(a,b)$ 内的可导性，且满足 $f(a)=f(b)$，直接应用罗尔定理证明存在 $\xi\in(a,b)$ 使得 $f'(\xi)=0$。 +2.构造辅助函数证明导函数相关方程有根 +对于形如 $f'(\xi)+g(\xi)f(\xi)=0$、$f''(\xi)=0$ 等方程，需构造满足罗尔定理条件的辅助函数 $F(x)$，通过 $F(a)=F(b)$ 推导 $F'(\xi)=0$，进而等价转化为目标方程。 +3.结合多次罗尔定理证明高阶导数零点存在 +若函数 f(x) 有 n+1 个点的函数值相等，可多次应用罗尔定理，证明其 n 阶导数 $f^{(n)}(\xi)=0$ 在对应区间内有根。 +4.证明函数恒为常数（反证法结合罗尔定理） +若 $f'(x)\equiv0$ 在区间内成立，可通过反证法假设存在两点函数值不等，结合罗尔定理推出矛盾，进而证明函数为常数。 + +### **例题** +>[!example] 例1 +设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 + +**解析**： +设辅助函数 $\varphi(x) = x^n f(x)$，则 $\varphi(0)=0$，$\varphi(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使得 $\varphi'(\xi)=0$即 +$$ +n\xi^{n-1} f(\xi) + \xi^n f'(\xi) = 0 +$$ +两边除以 $\xi^{n-1}$ ($\xi>0$)，得 $nf(\xi) + \xi f'(\xi) = 0$。 + + + +>[!example] 例2 +设函数 $f(x)$ 在 $[a,b]$ 上可导，且 +$$f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0，$$ +试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 + + +**解析**： +由导数极限定理及 $f'_+(a)f'_-(b) > 0$，知在 $a$ 右侧和 $b$ 左侧，$f(x)$ 的符号相同，不妨设 $f'_+(a)>0$，$f'_-(b)>0$。则在 $a$ 右侧附近 $f(x)>0$，在 $b$ 左侧附近 $f(x)>0$。由于 $f(a)=f(b)=0$，由极值点的费马定理，$f(x)$ 在 $(a,b)$ 内至少有一个极大值点，该点处导数为零。又因为 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导，且 $f(a)=f(b)$，由罗尔定理至少存在一点 $c \in (a,b)$ 使 $f'(c)=0$。结合极大值点处的导数零点，可知至少有两个导数为零的点。 + + + +>[!example] 例3 +设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 +$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： +(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； +(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 + +**解析**： +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 +(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 + + + + +## **拉格朗日中值定理** +### **原理** +若函数 f(x) 满足两个条件: +在闭区间 $[a,b]$ 上连续； +在开区间 $(a,b)$ 内可导； +则在 $(a,b)$ 内至少存在一点 $\xi$，使得 +$f(b)-f(a)=f'(\xi)(b-a)$ +也可写成等价形式 $f'(\xi)=\dfrac{f(b)-f(a)}{b-a}$。 +是罗尔定理的推广，同时也是柯西中值定理的特例。其几何意义为：满足条件的函数曲线在区间 (a,b) 内，至少存在一点的切线与连接端点 (a,f(a)) 和 (b,f(b)) 的弦平行。 + +### **适用条件** +拉格朗日中值定理的核心适用题型是建立函数增量与导数的关联，进行不等式的证明，这是最常见的题型。通过对目标函数在指定区间上应用拉格朗日中值定理，得到 $f(b)-f(a)=f'(\xi)(b-a)$，再利用导数 $f'(\xi)$ 的取值范围（有界性、正负性）放大或缩小式子，推导不等式。 + +### **例题** +>[!example] 例1 +设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 +$$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 + +**解析**： +对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 +$$ +f(x) - f(0) = f'(\xi)(x-0) +$$ +即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 + + + + +>[!example] 例2 +设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 +$$f(x_1 + x_2) < f(x_1) + f(x_2)$$ + +**解析**： +不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： +$$ +f(x_1+x_2)-f(x_2) = f'(\xi_1)x_1, \quad \xi_1 \in (x_2, x_1+x_2) +$$ +$$ +f(x_1)-f(0) = f'(\xi_2)x_1, \quad \xi_2 \in (0, x_1) +$$ +于是 +$$ +f(x_1+x_2)-f(x_2)-f(x_1) = [f'(\xi_1)-f'(\xi_2)]x_1 +$$ +对 $f'(x)$ 在 $[\xi_2,\xi_1]$ 上应用拉格朗日中值定理，存在 $\xi \in (\xi_2,\xi_1)$，使 +$$ +f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 +$$ +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。、 \ No newline at end of file -- 2.34.1 From dc5c7e60efcaceb107c99747546c588d81c4cd2a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:06:14 +0800 Subject: [PATCH 250/274] vault backup: 2026-01-14 17:06:14 --- ...朗日定理.md => 罗尔定理与拉格朗日中值定理.md} | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 素材/{罗尔定理与拉格朗日定理.md => 罗尔定理与拉格朗日中值定理.md} (100%) diff --git a/素材/罗尔定理与拉格朗日定理.md b/素材/罗尔定理与拉格朗日中值定理.md similarity index 100% rename from 素材/罗尔定理与拉格朗日定理.md rename to 素材/罗尔定理与拉格朗日中值定理.md -- 2.34.1 From 036148cc494372bcf3d4d797ba389d794a0c7198 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:12:07 +0800 Subject: [PATCH 251/274] vault backup: 2026-01-14 17:12:07 --- 素材/罗尔定理与拉格朗日中值定理.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index bee8105..f39b7ab 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -20,6 +20,8 @@ 4.证明函数恒为常数（反证法结合罗尔定理） 若 $f'(x)\equiv0$ 在区间内成立，可通过反证法假设存在两点函数值不等，结合罗尔定理推出矛盾，进而证明函数为常数。 +罗尔定理针对于一个函数，不同于柯西中值定理针对于两个函数 + ### **例题** >[!example] 例1 设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 -- 2.34.1 From 6ed260ff9808c8893a7661d25c4b3e9cc3ed9e2b Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:19:50 +0800 Subject: [PATCH 252/274] vault backup: 2026-01-14 17:19:50 --- 素材/罗尔定理与拉格朗日中值定理.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index f39b7ab..9e2bf6f 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -73,6 +73,7 @@ $f(b)-f(a)=f'(\xi)(b-a)$ 拉格朗日中值定理的核心适用题型是建立函数增量与导数的关联，进行不等式的证明，这是最常见的题型。通过对目标函数在指定区间上应用拉格朗日中值定理，得到 $f(b)-f(a)=f'(\xi)(b-a)$，再利用导数 $f'(\xi)$ 的取值范围（有界性、正负性）放大或缩小式子，推导不等式。 ### **例题** + >[!example] 例1 设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 $$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 @@ -85,8 +86,6 @@ $$ 即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 - - >[!example] 例2 设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 $$f(x_1 + x_2) < f(x_1) + f(x_2)$$ -- 2.34.1 From c10444c59543e50146dcc05fd3a96eaff7172f09 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:20:00 +0800 Subject: [PATCH 253/274] vault backup: 2026-01-14 17:20:00 --- 素材/罗尔定理与拉格朗日中值定理.md | 3 --- 1 file changed, 3 deletions(-) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index 9e2bf6f..7b03460 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -56,9 +56,6 @@ $$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： (1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 (2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 - - - ## **拉格朗日中值定理** ### **原理** 若函数 f(x) 满足两个条件: -- 2.34.1 From ac994fb6ab7b20aa16ad47ac9db641846035b8fc Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:21:39 +0800 Subject: [PATCH 254/274] vault backup: 2026-01-14 17:21:39 --- 素材/罗尔定理与拉格朗日中值定理.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index 7b03460..3b19eb2 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -103,4 +103,4 @@ $$ $$ f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 $$ -故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。、 \ No newline at end of file +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 \ No newline at end of file -- 2.34.1 From 43cf299c7893ab313d4388ebd6ab8a910b8e7498 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:28:01 +0800 Subject: [PATCH 255/274] vault backup: 2026-01-14 17:28:01 --- ...的解与秩的不等式（解析版）.md | 31 +++++++++++++++++++ 1 file changed, 31 insertions(+) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 395f157..d2e0955 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -646,3 +646,34 @@ D = \end{pmatrix}$$ **答案**： (B) 12 +>[!example] 例3 + 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 +（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 + $$f(x) = f(0) + x f'(x\theta(x));$$ + （2）求 + $$\lim_{x \to 0} \theta(x).$$ + + + +解： +1. 证** 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 + +$$ +f(x)=f(0)+x f'(x\theta(x)). +$$ + +如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 + +2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 + +$$ +\begin{aligned} +\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} +&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ +&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ +&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ +&= \frac{f''(0)}{2}, +\end{aligned} +$$ + +所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 \ No newline at end of file -- 2.34.1 From 84459d2a5ee164ed2eb283a0ad1e532a67fb0b3a Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:28:42 +0800 Subject: [PATCH 256/274] vault backup: 2026-01-14 17:28:42 --- .../线性方程组的解与秩的不等式（解析版）.md | 4 +--- 1 file changed, 1 insertion(+), 3 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index d2e0955..bb93886 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -653,10 +653,8 @@ D = （2）求 $$\lim_{x \to 0} \theta(x).$$ - - 解： -1. 证** 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 +1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 $$ f(x)=f(0)+x f'(x\theta(x)). -- 2.34.1 From 94d32c445fc6555bdb786586478640e14cff0ff8 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:38:36 +0800 Subject: [PATCH 257/274] vault backup: 2026-01-14 17:38:36 --- .../线性方程组的解与秩的不等式（解析版）.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index bb93886..6bccef1 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -436,9 +436,9 @@ $$ #### 2. 再证$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ - 考虑$C$的行向量： - 设$A = \begin{bmatrix} a_1^T \\ a_2^T \\ \vdots \\ a_m^T \end{bmatrix}$，则 + 设$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$，则 $$ - C = \begin{bmatrix} a_1^T B \\ a_2^T B \\ \vdots \\ a_m^T B \end{bmatrix}. + C = \begin{bmatrix} a_1 B \\ a_2 B \\ \vdots \\ a_m B \end{bmatrix}. $$ 因此$C$的每一行都是$B$的行向量的线性组合。 - 所以$C$的行空间是$B$的行空间的子空间，故 -- 2.34.1 From cfe519ddf3c78d3efb3a4f2eab406e8be41ebd3a Mon Sep 17 00:00:00 2001 From: pjokerx <1433560268@qq.com> Date: Wed, 14 Jan 2026 17:38:43 +0800 Subject: [PATCH 258/274] vault backup: 2026-01-14 17:38:43 --- ...理与定积分中值定理的综合运用：.md | 12 ++++++++++++ 1 file changed, 12 insertions(+) create mode 100644 微分中值定理与定积分中值定理的综合运用：.md diff --git a/微分中值定理与定积分中值定理的综合运用：.md b/微分中值定理与定积分中值定理的综合运用：.md new file mode 100644 index 0000000..9301c2d --- /dev/null +++ b/微分中值定理与定积分中值定理的综合运用：.md @@ -0,0 +1,12 @@ +经过对近十年的期末测试题的观察，微分中值定理通常不会单独出题，而是与积分中值定理一起出，本模块旨在通过几道经典的题目，让同学们熟悉微分中值与定积分中值的综合运用。 +首先我们来回顾定积分中值定理： +>[!note] 定理 +>如果函数$f(x)$在闭区间$[a,b]$上连续，则在积分区间$[a,b]$上至少有一点$\xi$，使 +>$$\int_{a}^{b}f(x) \mathrm{d} x=f(\xi)(b-a)\qquad(a\le\xi\le b)$$ + +>[!example] 例题1 +>已知函数$f(x)$在$[0,2]$上可导，且$f(0)=0$，$\large{\int}_{1}^{2}f(x)\mathrm{d}x=0$. 证明：至少存在$\xi\in(0,2)$，使得$f'(\xi)=2022f(\xi)$ + +>[!example] 例题2 +>设函数$f(x)$在闭区间$[0,2]$上可导，且$\large{\int}_{0}^{1}f(x)\mathrm{d}x=0$.证明：至少存在一点$\xi\in(0,2)$，使得$f'(\xi)=\frac{2}{2-\xi}f(\xi)$ + -- 2.34.1 From b04df6f4f651fc081740b216fc6b4ce94b108978 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:44:49 +0800 Subject: [PATCH 259/274] vault backup: 2026-01-14 17:44:49 --- ...线性方程组的解与秩的不等式.md | 26 ++-------- ...的解与秩的不等式（解析版）.md | 52 +------------------ 2 files changed, 6 insertions(+), 72 deletions(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index bfed081..673e8d9 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -407,9 +407,9 @@ $$ #### 2. 再证$\operatorname{rank}(AB) \leq \operatorname{rank}(B)$ - 考虑$C$的行向量： - 设$A = \begin{bmatrix} a_1^T \\ a_2^T \\ \vdots \\ a_m^T \end{bmatrix}$，则 + 设$A = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_m \end{bmatrix}$，则 $$ - C = \begin{bmatrix} a_1^T B \\ a_2^T B \\ \vdots \\ a_m^T B \end{bmatrix}. + C = \begin{bmatrix} a_1 B \\ a_2 B \\ \vdots \\ a_m B \end{bmatrix}. $$ 因此$C$的每一行都是$B$的行向量的线性组合。 - 所以$C$的行空间是$B$的行空间的子空间，故 @@ -490,15 +490,6 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -更严格地，我们可以直接写： -$$ -\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} -A \\ -I_n -\end{bmatrix} + \operatorname{rank}\begin{bmatrix} -I_n & B -\end{bmatrix} - n -$$ 但更简单的常用方法是利用： $$ \operatorname{rank}\begin{bmatrix} @@ -511,16 +502,7 @@ $$ --- #### 4. 从变换后的矩阵得到下界 -观察变换后的矩阵： -$$ -\operatorname{rank}\begin{bmatrix} -A & -AB \\ -I_n & O -\end{bmatrix} -\ge \operatorname{rank}\begin{bmatrix} -I_n & O -\end{bmatrix} + \operatorname{rank}([-AB]) -$$ + 实际上更直接的方法是注意到： $$ \operatorname{rank}\begin{bmatrix} @@ -530,7 +512,7 @@ I_n & O = \operatorname{rank}\begin{bmatrix} O & -AB \\ I_n & O -\end{bmatrix} \quad (\text{列变换}) +\end{bmatrix} \quad (\text{行变换}) $$ 即： $$ diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index 6bccef1..f413cce 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -519,15 +519,6 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -更严格地，我们可以直接写： -$$ -\operatorname{rank}(M) \ge \operatorname{rank}\begin{bmatrix} -A \\ -I_n -\end{bmatrix} + \operatorname{rank}\begin{bmatrix} -I_n & B -\end{bmatrix} - n -$$ 但更简单的常用方法是利用： $$ \operatorname{rank}\begin{bmatrix} @@ -540,16 +531,7 @@ $$ --- #### 4. 从变换后的矩阵得到下界 -观察变换后的矩阵： -$$ -\operatorname{rank}\begin{bmatrix} -A & -AB \\ -I_n & O -\end{bmatrix} -\ge \operatorname{rank}\begin{bmatrix} -I_n & O -\end{bmatrix} + \operatorname{rank}([-AB]) -$$ + 实际上更直接的方法是注意到： $$ \operatorname{rank}\begin{bmatrix} @@ -559,7 +541,7 @@ I_n & O = \operatorname{rank}\begin{bmatrix} O & -AB \\ I_n & O -\end{bmatrix} \quad (\text{列变换}) +\end{bmatrix} \quad (\text{行变换}) $$ 即： $$ @@ -645,33 +627,3 @@ D = 0 & 1 & 0 & 0 \end{pmatrix}$$ **答案**： (B) 12 - ->[!example] 例3 - 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 -（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 - $$f(x) = f(0) + x f'(x\theta(x));$$ - （2）求 - $$\lim_{x \to 0} \theta(x).$$ - -解： -1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 - -$$ -f(x)=f(0)+x f'(x\theta(x)). -$$ - -如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 - -2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 - -$$ -\begin{aligned} -\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} -&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ -&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ -&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ -&= \frac{f''(0)}{2}, -\end{aligned} -$$ - -所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 \ No newline at end of file -- 2.34.1 From 850c10194729d9a417ddbaa0d2ccaa3eb68f8fb2 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 17:45:46 +0800 Subject: [PATCH 260/274] vault backup: 2026-01-14 17:45:46 --- .../微分中值定理（解析版）.md | 74 +++++++++++++++++++ 1 file changed, 74 insertions(+) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index b52eaf9..e9dd243 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -5,3 +5,77 @@ tags: **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 +### 多次运用中值定理 +多次运用中值定理一般有如下特征： +1. 有多个中值（如$\xi,\eta$两个中值）； +2. 有二阶导出现。 +这种题目一般会比较难，而且通常要结合其他方法，比如反证法、构造函数等。 + +多次用中值定理又分为两种： +1. 在同一区间（或一个区间包含另一个区间）上用不同的中值定理； +2. 在相邻区间上对原函数和导函数用同一个（也可能是不同的）中值定理。 + +>[!example] 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $0 < a < b$，试证存在 $\xi, \eta \in (a, b)$，使得 $$f'(\xi) = \frac{a + b}{2\eta} f'(\eta).$$ + +**分析：** + 首先注意到题目要求我们证明的式子中有两个中值（即在区间内的某个点及其函数值），故想到可能会用多次中值定理。右边有$\frac{f'(\eta)}{\eta}$的形式，一般会想到用拉格朗日或者柯西，又题中没有哪个值让$f$等于$0$，所以大概率就是柯西中值定理。 + 令$g(x)=x^2$，则有$$\exists\eta\in(a,b),\frac{f'(\eta)}{g'(\eta)}=\frac{f'(\eta)}{2\eta}=\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f(b)-f(a)}{b^2-a^2}$$ + 再对比要求的式子，知道再用一次拉格朗日中值定理就行。 + +**解**： +对 $f(x)$ 和 $g(x)=x^2$ 在 $[a,b]$ 上应用柯西中值定理，得存在 $\eta \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b^2-a^2} = \frac{f'(\eta)}{2\eta} +$$ +整理得 +$$ +\frac{f(b)-f(a)}{b-a} = \frac{a+b}{2\eta} f'(\eta) +$$ +再对 $f(x)$ 在 $[a,b]$ 上应用拉格朗日中值定理，存在 $\xi \in (a,b)$，使得 +$$ +\frac{f(b)-f(a)}{b-a} = f'(\xi) +$$ +比较两式即得结论。 + +>[!example] 例2 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 + +**分析：** +![[Pasted image 20260114164542.png]] + 二阶导的零点就是图像的拐点，从图中能直观地看出来，函数图像的凹凸性确实发生了改变。现在的问题就是如何证明。 + 首先可以很直观地看到，函数图像应当有两条与直线$AB$平行的切线，由拉格朗日中值定理也可以证明这一点。这样，$f'(x)$就在不同地方取到了相同的函数值，这就想到用罗尔定理，从而可以证明题中结论。 + +**解**： +弦 $AB$ 的方程为 +$$ +y = f(a) + \frac{f(b)-f(a)}{b-a}(x-a) +$$ +由条件，$f(c) = f(a) + \frac{f(b)-f(a)}{b-a}(c-a)$。分别对 $f(x)$ 在 $[a,c]$ 和 $[c,b]$ 上应用拉格朗日中值定理，存在 $\xi_1 \in (a,c)$，$\xi_2 \in (c,b)$，使得 +$$ +f'(\xi_1) = \frac{f(c)-f(a)}{c-a} = \frac{f(b)-f(a)}{b-a} +$$ +$$ +f'(\xi_2) = \frac{f(b)-f(c)}{b-c} = \frac{f(b)-f(a)}{b-a} +$$ +故 $f'(\xi_1)=f'(\xi_2)$。再对 $f'(x)$ 在 $[\xi_1,\xi_2]$ 上应用罗尔定理，存在 $\xi \in (\xi_1,\xi_2) \subset (a,b)$，使 $f''(\xi)=0$。 + +>[!example] 例3 +设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 +$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： +(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； +(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 + +**分析：** + 初看没有什么思路，涉及到二阶导一般会用泰勒展开或者对导数用中值定理。我们对一阶导和二阶导的性质一概不清楚，所以应该考虑对导数用中值定理。 + 直接对$f(x)$用中值定理吗？不是，这样子我们得不出任何的结论。或许我们应该构造一个新的函数，让我们更容易研究一些。观察要证的式子，由于我们完全不知道导数的性质，所以考虑函数$g(x)=f(x)-x^2$，有$g'(x)=f'(x)-2x,g''(x)=f''(x)-2$，且$g(0)=0,g(1)=0,g(\frac{1}{2})>0$.看到这些$0$就舒服很多了。注意这里应该是多次运用中值定理中的第2种情况，因为题目很明显给了我们分割区间的一个点：$\large{\frac{1}{2}}$. + +**解**： +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。在$[0,\frac{1}{2}],[\frac{1}{2},1]$上分别用拉格朗日中值定理得$\exists\eta_1\in(0,\frac{1}{2}),\eta_2\in(\frac{1}{2},1)$，使得$$g'(\eta_1)=\frac{g(1/2)-g(0)}{1/2}>0,g'(\eta_2)=\frac{g(1)-g(1/2)}{1/2}<0$$对$g'(x)$在区间$[\eta_1,\eta_2]$上用拉格朗日中值定理得，$\exists\xi\in(\eta_1,\eta_2)$，使得$$g''(\xi)=\frac{g'(\eta_2)-g'(\eta_1)}{\eta_2-\eta_1}<0,$$即$$f''(\xi)<2.$$ +(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，故$g(x_0)\le0$。由$g(0)=0,g(1)=0,g(1/2)>0,\exists\alpha\in(0,1),g(\alpha)=0$.故由罗尔中值定理，$$\exists\beta_1\in(0,\alpha),\beta_2\in(\alpha,1),g'(\beta_1)=g'(\beta_2)=0,$$从而$$\exists\beta\in(\beta_1,\beta_2),g''(\beta)=0$$这与$f''(x)\neq2$，即$g''(x)\neq0$矛盾，故结论成立。 + +**题后总结：** + 1. 注意观察要证等式的形式，如果需要用到中值定理，就看看应该是哪个中值定理，比如等于某个确定的值一般就是罗尔，有多个函数（有些比较荫蔽，尤其是幂函数）一般是柯西，其他一些奇奇怪怪的情况基本上是拉格朗日 + 2. 注意分析和综合相结合的方法，从结论向前推一推，推不动了再从条件往后推一推，这是证明题的很重要的思路 + 3. 数形结合可以给我们很大的信心并提供思路，但是也要小心用 + -- 2.34.1 From 91148e072c8e4c90e7919fe09657e1b45a83567c Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:45:59 +0800 Subject: [PATCH 261/274] vault backup: 2026-01-14 17:45:59 --- ...罗尔定理与拉格朗日中值定理.md | 33 ++++++++++++++++++- 1 file changed, 32 insertions(+), 1 deletion(-) diff --git a/素材/罗尔定理与拉格朗日中值定理.md b/素材/罗尔定理与拉格朗日中值定理.md index 3b19eb2..b657011 100644 --- a/素材/罗尔定理与拉格朗日中值定理.md +++ b/素材/罗尔定理与拉格朗日中值定理.md @@ -103,4 +103,35 @@ $$ $$ f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 $$ -故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 \ No newline at end of file +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 + + +>[!example] 例3 + 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 +（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 + $$f(x) = f(0) + x f'(x\theta(x));$$ + （2）求 + $$\lim_{x \to 0} \theta(x).$$ + +解： +1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 + +$$ +f(x)=f(0)+x f'(x\theta(x)). +$$ + +如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 + +2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 + +$$ +\begin{aligned} +\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} +&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ +&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ +&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ +&= \frac{f''(0)}{2}, +\end{aligned} +$$ + +所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 \ No newline at end of file -- 2.34.1 From 1c0748ad13bd8ba782b1daa000c9bfbb342648de Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:47:42 +0800 Subject: [PATCH 262/274] vault backup: 2026-01-14 17:47:42 --- .../线性方程组的解与秩的不等式（解析版）.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md index f413cce..6ab1770 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式（解析版）.md @@ -519,7 +519,7 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -但更简单的常用方法是利用： +利用： $$ \operatorname{rank}\begin{bmatrix} A & O \\ -- 2.34.1 From d997c228c4a5741f5f5d3b82aa0ff1a255c9f267 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Wed, 14 Jan 2026 17:48:06 +0800 Subject: [PATCH 263/274] vault backup: 2026-01-14 17:48:06 --- 编写小组/讲义/线性方程组的解与秩的不等式.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 673e8d9..6d5f365 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -490,7 +490,7 @@ $$ $$ 这是因为$M$左上块为$A$，右下块为$B$，中间有单位矩阵，所以$A$和$B$的秩可以同时取到。 -但更简单的常用方法是利用： +利用： $$ \operatorname{rank}\begin{bmatrix} A & O \\ -- 2.34.1 From 4e5fa271417f2d2e3a6995229b2cdf15a7d3de72 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Wed, 14 Jan 2026 17:50:53 +0800 Subject: [PATCH 264/274] vault backup: 2026-01-14 17:50:53 --- .../讲义/图片/微分中值定理图.png | Bin 0 -> 35034 bytes .../讲义/微分中值定理（解析版）.md | 3 +-- 2 files changed, 1 insertion(+), 2 deletions(-) create mode 100644 编写小组/讲义/图片/微分中值定理图.png diff --git a/编写小组/讲义/图片/微分中值定理图.png b/编写小组/讲义/图片/微分中值定理图.png new file mode 100644 index 0000000000000000000000000000000000000000..4ccc9f8faa47ec0b57db70a849bf9c7f33677a1f GIT binary patch literal 35034 zcmag`1yojR7c~suCJ2b!AYIZ*w=~kJNT;+SAPs_q2#DOYpoCIV0#ec?Azgx$pwcB> zQUb!aHs^Vs^S<Bz|KoRzGtTgwvG2Y2bzN(%x#pZ}2j9J;a2baJ2ZcgiR#KGJK%ve( zM4`@jVqb*+@+i*hIsEI4i-v+Us-TBz5&m-CO6s;03RM(|dt`P2{(i|(QQrlH!f!(U zIn(BlZHYo*9F%0Gv^<Pg$K16BhwCnEh;wg8V6M?)Mq2!9_eg3?2d^@i-ek5>4eM); 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zzBC)@$CCkJLYJ>TT)EbIo#y7lgmQ|Au9eAR{3m_J=u&`2)A_I1*y7$F_T})ODVh2{ m^#009CZhyjF8K4-XIA%y-u@nUge#N*KkCZ6m2N3moccfD1;vv9 literal 0 HcmV?d00001 diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index e9dd243..53c82e6 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -41,8 +41,7 @@ $$ >[!example] 例2 设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内二阶可导，又若 $f(x)$ 的图形与联结 $A(a, f(a))$，$B(b, f(b))$ 两点的弦交于点 $C(c, f(c))$ ($a \leq c \leq b$)，证明在 $(a, b)$ 内至少存在一点 $\xi$，使得 $f''(\xi) = 0$。 -**分析：** -![[Pasted image 20260114164542.png]] +**分析：![[微分中值定理图.png]]** 二阶导的零点就是图像的拐点，从图中能直观地看出来，函数图像的凹凸性确实发生了改变。现在的问题就是如何证明。 首先可以很直观地看到，函数图像应当有两条与直线$AB$平行的切线，由拉格朗日中值定理也可以证明这一点。这样，$f'(x)$就在不同地方取到了相同的函数值，这就想到用罗尔定理，从而可以证明题中结论。 -- 2.34.1 From dab8187a9db5c8e54817852c0cf8c3008477c92e Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Thu, 15 Jan 2026 23:42:01 +0800 Subject: [PATCH 265/274] vault backup: 2026-01-15 23:42:01 --- .../1.17高数限时练.md | 137 +++++++++ .../1.17高数限时练（解析版）.md | 288 ++++++++++++++++++ 2 files changed, 425 insertions(+) create mode 100644 编写小组/试卷/高数期末复习模拟/1.17高数限时练.md create mode 100644 编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md diff --git a/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md new file mode 100644 index 0000000..0dcd9a8 --- /dev/null +++ b/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md @@ -0,0 +1,137 @@ +--- +tags: + - 高数复习模拟 +--- +### 一、选择题（共3小题，每小题6分，共18分） + +1. 下列级数发散的是（　）。 + + (A) $\sum_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ + (B) $\sum_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ + (C) $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ + (D) $\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ + +2. 当 $x \to 0$时，与 $x - \sin x$同阶的无穷小是（　）。 + + (A) $x + \tan x$ + (B) $x \tan x$ + (C) $x^2 + \tan x$ + (D) $x^2 \tan x$ + +3. 已知函数 $f(x)$在 $x = 1$的某邻域内三阶可导，且 + $$ + \lim_{x \to 1} \frac{f(x) - 2}{(x - 1)^2 \ln x} = \frac{1}{3}, + $$ + 则（　）。 + (A) $x = 1$为函数 $f(x)$的极大值点 + (B) 点 $(1, 2)$为曲线 $y = f(x)$的拐点 + (C) $x = 1$为函数 $f(x)$的极小值点 + (D) 点 $(1, 0)$为曲线 $y = f(x)$的拐点 + +--- + +### 二、填空题（共3小题，每小题6分，共18分） + +4. 极限 $$\lim_{n \to \infty} \frac{1}{n} \left( \tan \frac{\pi}{4n} + \tan \frac{2\pi}{4n} + \cdots + \tan \frac{n\pi}{4n} \right)$$ 的值为______。 + +5. 已知 $\int f(x^2) dx = x \ln x + C，$则 $f'(2) = \underline{\quad }。$ + +6. 极坐标曲线 $\rho = \theta$ 在点 $(\rho, \theta) = (\pi, \pi)$ 处的切线的直角坐标方程为______ + +--- + +### 三、解答题（共4小题，共64分） + +7. 计算数列极限 + $$ + \lim_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}。 + $$ + （12分） + +```text + + + + + + + + + + + + + + + +``` + +7. (16分)已知函数 $f(x)$ 在 $[0,1]$上连续，在 $(0,1)$内可导，且 $f(0)=0$ ，$f(1)=1$ 。试证明： + （1）（6分）存在 $\xi \in (0,1)$，使得 + $$ + f'(\xi) = 2\xi。 + $$ + （2）（10分）对任意正数 $a,b$，在 $(0,1)$内存在相异的两点 $x_1, x_2$，使得 + $$ + \frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = a+b。 + $$ + +```text + + + + + + + + + + + + + + + +``` + +7. （16分）已知当 $x \to 0$ 时，函数 $f(x) = \sqrt{a + bx^2} - \cos x$ 与 $x^2$ 是等价无穷小。 + （1）求参数 $a, b$ 的值；（6分） + （2）计算极限 $\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） + +```text + + + + + + + + + + + + + + + +``` + +8. （20分）设函数 $f(x)$在 $[0,1]$上可导，$\int_{0}^{\frac{1}{2}} f(x) dx = 0$。证明存在 $\xi \in (0,1)$，使得 +$$f(\xi) = (1 - \xi) f'(\xi)。$$ +```text + + + + + + + + + + + + + + + +``` \ No newline at end of file diff --git a/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md b/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md new file mode 100644 index 0000000..7bc6e37 --- /dev/null +++ b/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md @@ -0,0 +1,288 @@ +--- +tags: + - 高数复习模拟 +--- +### 一、选择题（共3小题，每小题6分，共18分） + +1. 下列级数发散的是（　）。 + + (A) $\sum_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ + (B) $\sum_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ + (C) $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ + (D) $\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ + +2. 当 $x \to 0$时，与 $x - \sin x$同阶的无穷小是（　）。 + + (A) $x + \tan x$ + (B) $x \tan x$ + (C) $x^2 + \tan x$ + (D) $x^2 \tan x$ + +3. 已知函数 $f(x)$在 $x = 1$的某邻域内三阶可导，且 + $$ + \lim_{x \to 1} \frac{f(x) - 2}{(x - 1)^2 \ln x} = \frac{1}{3}, + $$ + 则（　）。 + (A) $x = 1$为函数 $f(x)$的极大值点 + (B) 点 $(1, 2)$为曲线 $y = f(x)$的拐点 + (C) $x = 1$为函数 $f(x)$的极小值点 + (D) 点 $(1, 0)$为曲线 $y = f(x)$的拐点 + +--- + +### 二、填空题（共3小题，每小题6分，共18分） + +4. 极限 $$\lim_{n \to \infty} \frac{1}{n} \left( \tan \frac{\pi}{4n} + \tan \frac{2\pi}{4n} + \cdots + \tan \frac{n\pi}{4n} \right)$$ 的值为______。 + +5. 已知 $\int f(x^2) dx = x \ln x + C，$则 $f'(2) = \underline{\quad }。$ + +6. 极坐标曲线 $\rho = \theta$ 在点 $(\rho, \theta) = (\pi, \pi)$ 处的切线的直角坐标方程为______ + +--- + +### 三、解答题（共4小题，共64分） + +7. 计算数列极限 + $$ + \lim_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}。 + $$ + （12分） + +```text + + + + + + + + + + + + + + + +``` + +7. (16分)已知函数 $f(x)$ 在 $[0,1]$上连续，在 $(0,1)$内可导，且 $f(0)=0$ ，$f(1)=1$ 。试证明： + （1）（6分）存在 $\xi \in (0,1)$，使得 + $$ + f'(\xi) = 2\xi。 + $$ + （2）（10分）对任意正数 $a,b$，在 $(0,1)$内存在相异的两点 $x_1, x_2$，使得 + $$ + \frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = a+b。 + $$ + +```text + + + + + + + + + + + + + + + +``` + +7. （16分）已知当 $x \to 0$ 时，函数 $f(x) = \sqrt{a + bx^2} - \cos x$ 与 $x^2$ 是等价无穷小。 + （1）求参数 $a, b$ 的值；（6分） + （2）计算极限 $\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） + +```text + + + + + + + + + + + + + + + +``` + +8. （20分）设函数 $f(x)$在 $[0,1]$上可导，$\int_{0}^{\frac{1}{2}} f(x) dx = 0$。证明存在 $\xi \in (0,1)$，使得 +$$f(\xi) = (1 - \xi) f'(\xi)。$$ +```text + + + + + + + + + + + + + + + +``` + +# 高等数学（Ⅰ）考试试卷（A）卷 大题解答 + +## 7. 计算极限 +$$ +\lim_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}} +$$ + +**解：** +令$L = \lim\limits_{x \to +\infty} (x^2 + 2^x)^{\frac{1}{x}}$，取对数得： +$$ +\ln L = \lim_{x \to +\infty} \frac{\ln(x^2 + 2^x)}{x} +$$ + +当$x \to +\infty$时，$2^x \gg x^2$，故： +$$ +x^2 + 2^x \sim 2^x +$$ +$$ +\ln(x^2 + 2^x) = x\ln 2 + \ln\left(1 + \frac{x^2}{2^x}\right) \sim x\ln 2 +$$ +因此： +$$ +\ln L = \lim_{x \to +\infty} \frac{x\ln 2}{x} = \ln 2 +$$ +$$ +L = e^{\ln 2} = 2 +$$ + +**答案：**$\boxed{2}$ + +--- + +## 8. 中值定理证明题 + +已知$f(x)$在$[0,1]$上连续，$(0,1)$内可导，$f(0)=0$，$f(1)=1$。 + +**(1)** 证明存在$\xi \in (0,1)$使$f'(\xi) = 2\xi$ + +**证：** +构造$g(x) = f(x) - x^2$ +-$g(0) = f(0) - 0 = 0$ +-$g(1) = f(1) - 1 = 0$ + +由罗尔定理，存在$\xi \in (0,1)$使$g'(\xi) = 0$，即： +$$ +f'(\xi) - 2\xi = 0 \quad \Rightarrow \quad f'(\xi) = 2\xi +$$ + +--- + +**(2)** 对任意$a,b > 0$，存在相异$x_1, x_2 \in (0,1)$使： +$$ +\frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = a+b +$$ + +**证：** +由介值定理，存在$c \in (0,1)$使$f(c) = \dfrac{a}{a+b}$（因$0 < \frac{a}{a+b} < 1$） + +1. 在$[0,c]$上用拉格朗日中值定理： +$$ +f'(x_1) = \frac{f(c)-f(0)}{c-0} = \frac{\frac{a}{a+b}}{c}, \quad x_1 \in (0,c) +$$ +$$ +\frac{a}{f'(x_1)} = a \cdot \frac{c}{\frac{a}{a+b}} = c(a+b) +$$ + +2. 在$[c,1]$上用拉格朗日中值定理： +$$ +f'(x_2) = \frac{f(1)-f(c)}{1-c} = \frac{1 - \frac{a}{a+b}}{1-c} = \frac{\frac{b}{a+b}}{1-c}, \quad x_2 \in (c,1) +$$ +$$ +\frac{b}{f'(x_2)} = b \cdot \frac{1-c}{\frac{b}{a+b}} = (1-c)(a+b) +$$ + +相加得： +$$ +\frac{a}{f'(x_1)} + \frac{b}{f'(x_2)} = c(a+b) + (1-c)(a+b) = a+b +$$ +且$x_1 < c < x_2$，故$x_1 \neq x_2$。 + +--- + +## 9. 等价无穷小与参数确定 + +已知$f(x) = \sqrt{a + bx^2} - \cos x \sim x^2 \ (x\to 0)$ + +**(1)** 求$a, b$ + +**解：** +展开至$x^4$： +$$ +\sqrt{a+bx^2} = \sqrt{a} + \frac{b\sqrt{a}}{2a}x^2 - \frac{b^2\sqrt{a}}{8a^2}x^4 + o(x^4) +$$ +$$ +\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + o(x^4) +$$ +代入$f(x)$： +$$ +f(x) = (\sqrt{a}-1) + \left(\frac{b}{2\sqrt{a}} + \frac{1}{2}\right)x^2 + \left(-\frac{b^2}{8a^{3/2}} - \frac{1}{24}\right)x^4 + o(x^4) +$$ + +由$f(x) \sim x^2$得： +1.$\sqrt{a}-1 = 0 \Rightarrow a = 1$ +2.$\frac{b}{2} + \frac{1}{2} = 1 \Rightarrow b = 1$ + +**(2)** 计算$\lim\limits_{x\to 0} \dfrac{f(x) - x^2}{x^4}$ + +**解：** +$a=1, b=1$时： +$$ +f(x) = \sqrt{1+x^2} - \cos x = x^2 - \frac{1}{6}x^4 + o(x^4) +$$ +$$ +f(x) - x^2 = -\frac{1}{6}x^4 + o(x^4) +$$ +$$ +\lim_{x\to 0} \frac{f(x) - x^2}{x^4} = -\frac{1}{6} +$$ + +**答案：** +(1)$a=1, b=1$ +(2)$-\dfrac{1}{6}$ + +--- + +## 10. 积分条件与微分方程形式的中值定理 + +已知$f(x)$在$[0,1]$上可导，$\int_0^{1/2} f(x) dx = 0$ +证明存在$\xi \in (0,1)$使$f(\xi) = (1-\xi)f'(\xi)$ + +**证：** +令$g(x) = (1-x)f(x)$，则： +$$ +g'(x) = -f(x) + (1-x)f'(x) +$$ +要证等式等价于$g'(\xi) = 0$ + +设$h(x) = \int_0^x f(t)dt$，由已知$h(1/2) = 0$，且$h(0)=0$ +由罗尔定理，存在$c \in (0,1/2)$使$h'(c) = f(c) = 0$ + +于是： +$$ +g(c) = (1-c)f(c) = 0, \quad g(1) = (1-1)f(1) = 0 +$$ +在$[c,1]$上对$g(x)$应用罗尔定理，存在$\xi \in (c,1) \subset (0,1)$使： +$$ +g'(\xi) = 0 \quad \Rightarrow \quad f(\xi) = (1-\xi)f'(\xi) +$$ + +证毕。 \ No newline at end of file -- 2.34.1 From c8f1325a6f2f48244803c42be6a2a751c24b6622 Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Fri, 16 Jan 2026 08:45:26 +0800 Subject: [PATCH 266/274] vault backup: 2026-01-16 08:45:25 --- ...柯西中值定理与常见辅助函数.md | 229 ++++++++++++++++++ 1 file changed, 229 insertions(+) create mode 100644 素材/柯西中值定理与常见辅助函数.md diff --git a/素材/柯西中值定理与常见辅助函数.md b/素材/柯西中值定理与常见辅助函数.md new file mode 100644 index 0000000..cbf664f --- /dev/null +++ b/素材/柯西中值定理与常见辅助函数.md @@ -0,0 +1,229 @@ +## **柯西中值定理** + +### **原理** + +设函数 $f(x)$ 和 $g(x)$ 满足以下条件： +1. 在闭区间 $[a, b]$ 上连续； +2. 在开区间 $(a, b)$ 内可导； +3. 对任意 $x \in (a, b)$，有 $g'(x) \neq 0$； +则在 $(a, b)$ 内至少存在一点 $\xi$，使得： + +$$ +\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(\xi)}{g'(\xi)} +$$ + +柯西中值定理的几何意义为：由参数方程 $(g(t), f(t))$ 表示的曲线，在两点间的割线斜率等于曲线上某点切线的斜率。 + +它与拉格朗日中值定理的关系为：当 $g(x) = x$ 时，柯西中值定理退化为拉格朗日中值定理。它是处理两个函数之间微分中值关系的通用形式。 + +### **适用条件** + +柯西中值定理的核心适用题型是**证明形如 $\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)}$ 的等式成立**，以及处理**涉及两个中值点 $\xi, \eta$ 的问题**。 + +常见应用方向包括： +1. 直接证明存在性等式； +2. 通过函数配对，将目标等式转化为柯西中值定理的标准形式； +3. 处理“双中值问题”，常与拉格朗日中值定理结合使用。 + +### **例题** + +>[!example] 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $a>0$。证明存在 $\xi \in (a, b)$，使得： +$$\frac{f(b)-f(a)}{b-a} = \xi f'(\xi) \cdot \frac{\ln(b/a)}{b-a}$$ + +**解析**： +将等式变形为： +$$ +\frac{f(b)-f(a)}{\ln b - \ln a} = \xi f'(\xi) +$$ +取 $g(x) = \ln x$，则 $g'(x) = \frac{1}{x} \neq 0$ 在 $(a, b)$ 内成立。 +对 $f(x)$ 与 $g(x)$ 应用柯西中值定理，存在 $\xi \in (a, b)$ 使得： +$$ +\frac{f(b)-f(a)}{\ln b - \ln a} = \frac{f'(\xi)}{1/\xi} = \xi f'(\xi) +$$ +整理即得所求。 + +--- + +>[!example] 例2 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，证明存在不同的 $\xi, \eta \in (a, b)$，使得： +$$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ + +**解析**： +1. 对 $f(x)$ 与 $g(x) = \frac{x^2}{2}$ 应用柯西中值定理，存在 $\eta \in (a, b)$ 使得： + $$ + \frac{f(b)-f(a)}{(b^2 - a^2)/2} = \frac{f'(\eta)}{\eta} + $$ + 整理得： + $$ + f(b)-f(a) = \frac{b^2 - a^2}{2\eta} f'(\eta) + $$ + +2. 对 $f(x)$ 应用拉格朗日中值定理，存在 $\xi \in (a, b)$ 使得： + $$ + f(b)-f(a) = (b-a) f'(\xi) + $$ + +3. 联立两式，消去 $f(b)-f(a)$ 得： + $$ + (b-a) f'(\xi) = \frac{(b-a)(a+b)}{2\eta} f'(\eta) + $$ + 由于 $b-a \neq 0$，约去后即得： + $$ + f'(\xi) = \frac{a+b}{2\eta} f'(\eta) + $$ + +--- + +>[!example] 例3 +设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： +$$f(b)-f(a) = \frac{3\xi^2}{a^2+ab+b^2} f'(\xi)(b-a)$$ + +**解析**： +将等式变形为： +$$ +\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} +$$ +取 $g(x) = x^3$，则 $g'(x) = 3x^2 \neq 0$ 在 $(a, b)$ 内成立。 +由柯西中值定理，存在 $\xi \in (a, b)$ 使得： +$$ +\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} +$$ +整理后即得所求。 + +--- + +## **辅助函数的构造方法** + +### **原理** + +在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导后的结果**。 + +### **常见构造类型** + +#### 1. 乘积型与商型 +若结论形如： +$$ +f'(\xi)g(\xi) + f(\xi)g'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = f(x)g(x) +$$ +若结论形如： +$$ +f'(\xi)g(\xi) - f(\xi)g'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = \frac{f(x)}{g(x)} \quad (g(x) \neq 0) +$$ + +#### 2. 含幂函数因子 +若结论形如： +$$ +n f(\xi) + \xi f'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = x^n f(x) +$$ + +#### 3. 一阶线性微分结构 +若结论形如： +$$ +f'(\xi) + P(\xi)f(\xi) = 0 +$$ +可构造积分因子： +$$ +\mu(x) = e^{\int P(x)dx} +$$ +并设辅助函数： +$$ +F(x) = \mu(x) f(x) +$$ + +#### 4. 对数型 +若结论形如： +$$ +\frac{f'(\xi)}{f(\xi)} = k +$$ +可构造辅助函数： +$$ +F(x) = \ln|f(x)| - kx +$$ + +#### 5. 常数变易法 +若结论形如： +$$ +f'(\xi) = \lambda f(\xi) +$$ +可构造辅助函数： +$$ +F(x) = e^{-\lambda x} f(x) +$$ + +--- + +### **例题** + +>[!example] 例1 +设 $f(x)$ 在 $[0, 1]$ 上连续，在 $(0, 1)$ 内可导，且 $f(0)=0$，$f(1)=1$。 +证明：存在 $\xi \in (0, 1)$ 使得$f'(\xi) = 2\xi f(\xi)$ + +**解析**： +将结论改写为： +$$ +f'(\xi) - 2\xi f(\xi) = 0 +$$ +属于一阶线性微分结构，其中 $P(x) = -2x$。 +积分因子为： +$$ +\mu(x) = e^{\int (-2x)dx} = e^{-x^2} +$$ +构造辅助函数： +$$ +F(x) = e^{-x^2} f(x) +$$ +则 $F(0) = 0$，$F(1) = e^{-1}$。 +需进一步寻找另一个点 $c$ 使 $F(c)=0$，才可应用罗尔定理。通常需结合题目其他条件（如积分中值定理、零点定理等）找出该点。 + +--- + +>[!example] 例2 +设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) = 0$。 +证明：存在 $\xi \in (a, b)$ 使得：$f'''(\xi) + k f''(\xi) = 0$ + +**解析**： +结论可写为： +$$ +\bigl[ e^{kx} f''(x) \bigr]' \big|_{x=\xi} = 0 +$$ +因此构造辅助函数： +$$ +H(x) = e^{kx} f''(x) +$$ +由条件可推知存在 $\eta_1, \eta_2 \in (a, b)$ 使 $f''(\eta_1) = f''(\eta_2) = 0$，从而 $H(\eta_1)=H(\eta_2)=0$。 +对 $H(x)$ 应用罗尔定理即得证。 + +--- + +>[!example] 例3 +设 $f(x)$ 在 $[0, 1]$ 上可导，且$f(1) = 2\int_0^{1/2} e^{1-x} f(x) dx$ +证明：存在 $\xi \in (0, 1)$ 使得：$f'(\xi) = (1-\xi) f(\xi)$ + +**解析**： +结论化为： +$$ +f'(\xi) - (1-\xi) f(\xi) = 0 +$$ +积分因子为： +$$ +\mu(x) = e^{\int (x-1) dx} = e^{\frac{x^2}{2} - x} +$$ +构造辅助函数： +$$ +F(x) = e^{\frac{x^2}{2} - x} f(x) +$$ +利用题设积分条件与积分中值定理，可找到 $\eta \in (0, \frac{1}{2})$ 使 $F(\eta) = F(1)$，再对 $F(x)$ 应用罗尔定理即证。 + -- 2.34.1 From 81daaf6593dd7492f994ce82cfc033522e4973ac Mon Sep 17 00:00:00 2001 From: unknown <18951088369@163.com> Date: Fri, 16 Jan 2026 08:51:00 +0800 Subject: [PATCH 267/274] vault backup: 2026-01-16 08:51:00 --- 素材/柯西中值定理与常见辅助函数.md | 9 +++++---- 1 file changed, 5 insertions(+), 4 deletions(-) diff --git a/素材/柯西中值定理与常见辅助函数.md b/素材/柯西中值定理与常见辅助函数.md index cbf664f..76332e8 100644 --- a/素材/柯西中值定理与常见辅助函数.md +++ b/素材/柯西中值定理与常见辅助函数.md @@ -91,7 +91,8 @@ $$ $$ 整理后即得所求。 ---- + + ## **辅助函数的构造方法** @@ -136,7 +137,7 @@ f'(\xi) + P(\xi)f(\xi) = 0 $$ 可构造积分因子： $$ -\mu(x) = e^{\int P(x)dx} +\mu(x) = e^{\int P(x)\mathrm{d}x} $$ 并设辅助函数： $$ @@ -179,7 +180,7 @@ $$ 属于一阶线性微分结构，其中 $P(x) = -2x$。 积分因子为： $$ -\mu(x) = e^{\int (-2x)dx} = e^{-x^2} +\mu(x) = e^{\int (-2x)\mathrm{d}x} = e^{-x^2} $$ 构造辅助函数： $$ @@ -219,7 +220,7 @@ f'(\xi) - (1-\xi) f(\xi) = 0 $$ 积分因子为： $$ -\mu(x) = e^{\int (x-1) dx} = e^{\frac{x^2}{2} - x} +\mu(x) = e^{\int (x-1) \mathrm{d}x} = e^{\frac{x^2}{2} - x} $$ 构造辅助函数： $$ -- 2.34.1 From 6b522c4ede4a8741ba4cff3e840561e807b7cc76 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Fri, 16 Jan 2026 09:07:15 +0800 Subject: [PATCH 268/274] vault backup: 2026-01-16 09:07:15 --- ...分中值定理.md => 微分中值定理部分题目汇总.md} | 0 1 file changed, 0 insertions(+), 0 deletions(-) rename 素材/{微分中值定理.md => 微分中值定理部分题目汇总.md} (100%) diff --git a/素材/微分中值定理.md b/素材/微分中值定理部分题目汇总.md similarity index 100% rename from 素材/微分中值定理.md rename to 素材/微分中值定理部分题目汇总.md -- 2.34.1 From 4330697ce5281b441fa3586b2a057e851063a409 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 16 Jan 2026 09:27:19 +0800 Subject: [PATCH 269/274] vault backup: 2026-01-16 09:27:19 --- .../讲义/微分中值定理（解析版）.md | 2 +- .../1.17高数限时练.md | 12 ++++++------ .../1.17高数限时练（解析版）.md | 14 +++++++------- 3 files changed, 14 insertions(+), 14 deletions(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 53c82e6..0720e7d 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -70,7 +70,7 @@ $$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： 直接对$f(x)$用中值定理吗？不是，这样子我们得不出任何的结论。或许我们应该构造一个新的函数，让我们更容易研究一些。观察要证的式子，由于我们完全不知道导数的性质，所以考虑函数$g(x)=f(x)-x^2$，有$g'(x)=f'(x)-2x,g''(x)=f''(x)-2$，且$g(0)=0,g(1)=0,g(\frac{1}{2})>0$.看到这些$0$就舒服很多了。注意这里应该是多次运用中值定理中的第2种情况，因为题目很明显给了我们分割区间的一个点：$\large{\frac{1}{2}}$. **解**： -(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。在$[0,\frac{1}{2}],[\frac{1}{2},1]$上分别用拉格朗日中值定理得$\exists\eta_1\in(0,\frac{1}{2}),\eta_2\in(\frac{1}{2},1)$，使得$$g'(\eta_1)=\frac{g(1/2)-g(0)}{1/2}>0,g'(\eta_2)=\frac{g(1)-g(1/2)}{1/2}<0$$对$g'(x)$在区间$[\eta_1,\eta_2]$上用拉格朗日中值定理得，$\exists\xi\in(\eta_1,\eta_2)$，使得$$g''(\xi)=\frac{g'(\eta_2)-g'(\eta_1)}{\eta_2-\eta_1}<0,$$即$$f''(\xi)<2.$$ +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。在$[0,\frac{1}{2}],[\frac{1}{2},1]$上分别用拉格朗日中值定理得，$\exists\eta_1\in(0,\frac{1}{2}),\eta_2\in(\frac{1}{2},1)$，使得$$g'(\eta_1)=\frac{g(1/2)-g(0)}{1/2}>0,g'(\eta_2)=\frac{g(1)-g(1/2)}{1/2}<0$$对$g'(x)$在区间$[\eta_1,\eta_2]$上用拉格朗日中值定理得，$\exists\xi\in(\eta_1,\eta_2)$，使得$$g''(\xi)=\frac{g'(\eta_2)-g'(\eta_1)}{\eta_2-\eta_1}<0,$$即$$f''(\xi)<2.$$ (2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，故$g(x_0)\le0$。由$g(0)=0,g(1)=0,g(1/2)>0,\exists\alpha\in(0,1),g(\alpha)=0$.故由罗尔中值定理，$$\exists\beta_1\in(0,\alpha),\beta_2\in(\alpha,1),g'(\beta_1)=g'(\beta_2)=0,$$从而$$\exists\beta\in(\beta_1,\beta_2),g''(\beta)=0$$这与$f''(x)\neq2$，即$g''(x)\neq0$矛盾，故结论成立。 **题后总结：** diff --git a/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md b/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md index 0dcd9a8..24e7993 100644 --- a/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md +++ b/编写小组/试卷/高数期末复习模拟/1.17高数限时练.md @@ -6,10 +6,10 @@ tags: 1. 下列级数发散的是（　）。 - (A) $\sum_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ - (B) $\sum_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ - (C) $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ - (D) $\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ + (A) $\sum\limits_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ + (B) $\sum\limits_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ + (C) $\sum\limits_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ + (D) $\sum\limits_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ 2. 当 $x \to 0$时，与 $x - \sin x$同阶的无穷小是（　）。 @@ -96,7 +96,7 @@ tags: 7. （16分）已知当 $x \to 0$ 时，函数 $f(x) = \sqrt{a + bx^2} - \cos x$ 与 $x^2$ 是等价无穷小。 （1）求参数 $a, b$ 的值；（6分） - （2）计算极限 $\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） + （2）计算极限 $\lim\limits_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） ```text @@ -116,7 +116,7 @@ tags: ``` -8. （20分）设函数 $f(x)$在 $[0,1]$上可导，$\int_{0}^{\frac{1}{2}} f(x) dx = 0$。证明存在 $\xi \in (0,1)$，使得 +8. （20分）设函数 $f(x)$在 $[0,1]$上可导，$\large{\int_{0}^{\frac{1}{2}}} f(x) dx = 0$。证明存在 $\xi \in (0,1)$，使得 $$f(\xi) = (1 - \xi) f'(\xi)。$$ ```text diff --git a/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md b/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md index 7bc6e37..9d6c9e1 100644 --- a/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md +++ b/编写小组/试卷/高数期末复习模拟/1.17高数限时练（解析版）.md @@ -6,10 +6,10 @@ tags: 1. 下列级数发散的是（　）。 - (A) $\sum_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ - (B) $\sum_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ - (C) $\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ - (D) $\sum_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ + (A) $\sum\limits_{n=1}^\infty \frac{4^n + 1}{n^4 + 1}$ + (B) $\sum\limits_{n=1}^\infty \frac{n^4 + 1}{4^n + 1}$ + (C) $\sum\limits_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n} + 1}$ + (D) $\sum\limits_{n=1}^\infty (-1)^n \frac{\sqrt{n}}{n + 1}$ 2. 当 $x \to 0$时，与 $x - \sin x$同阶的无穷小是（　）。 @@ -96,7 +96,7 @@ tags: 7. （16分）已知当 $x \to 0$ 时，函数 $f(x) = \sqrt{a + bx^2} - \cos x$ 与 $x^2$ 是等价无穷小。 （1）求参数 $a, b$ 的值；（6分） - （2）计算极限 $\lim_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） + （2）计算极限 $\lim\limits_{x \to 0} \frac{f(x) - x^2}{x^4}$的值。（10分） ```text @@ -263,7 +263,7 @@ $$ ## 10. 积分条件与微分方程形式的中值定理 -已知$f(x)$在$[0,1]$上可导，$\int_0^{1/2} f(x) dx = 0$ +已知$f(x)$在$[0,1]$上可导，$\large{\int}_0^{1/2} f(x) dx = 0$ 证明存在$\xi \in (0,1)$使$f(\xi) = (1-\xi)f'(\xi)$ **证：** @@ -273,7 +273,7 @@ g'(x) = -f(x) + (1-x)f'(x) $$ 要证等式等价于$g'(\xi) = 0$ -设$h(x) = \int_0^x f(t)dt$，由已知$h(1/2) = 0$，且$h(0)=0$ +设$h(x) = \large{\int_0^x }f(t)dt$，由已知$h(1/2) = 0$，且$h(0)=0$ 由罗尔定理，存在$c \in (0,1/2)$使$h'(c) = f(c) = 0$ 于是： -- 2.34.1 From da1cb74f50f104e4ebae583fa822e6729f23f890 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= <wkn339224@qq.com> Date: Fri, 16 Jan 2026 09:30:04 +0800 Subject: [PATCH 270/274] vault backup: 2026-01-16 09:30:04 --- ...分中值定理与定积分中值定理的综合运用：.md | 0 素材/{解的问题.md => 线性方程组解的问题.md} | 0 编写小组/讲义/线性方程组的解与秩的不等式.md | 2 +- 3 files changed, 1 insertion(+), 1 deletion(-) rename 微分中值定理与定积分中值定理的综合运用：.md => 素材/微分中值定理与定积分中值定理的综合运用：.md (100%) rename 素材/{解的问题.md => 线性方程组解的问题.md} (100%) diff --git a/微分中值定理与定积分中值定理的综合运用：.md b/素材/微分中值定理与定积分中值定理的综合运用：.md similarity index 100% rename from 微分中值定理与定积分中值定理的综合运用：.md rename to 素材/微分中值定理与定积分中值定理的综合运用：.md diff --git a/素材/解的问题.md b/素材/线性方程组解的问题.md similarity index 100% rename from 素材/解的问题.md rename to 素材/线性方程组解的问题.md diff --git a/编写小组/讲义/线性方程组的解与秩的不等式.md b/编写小组/讲义/线性方程组的解与秩的不等式.md index 6d5f365..3b466c3 100644 --- a/编写小组/讲义/线性方程组的解与秩的不等式.md +++ b/编写小组/讲义/线性方程组的解与秩的不等式.md @@ -3,7 +3,7 @@ tags: - 编写小组 --- **内部资料，禁止传播** -**编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 +**编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁** # 单方程组解的问题 -- 2.34.1 From 277ec22cecadf53e002f83f44dbe88860d823b3b Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Fri, 16 Jan 2026 11:22:43 +0800 Subject: [PATCH 271/274] vault backup: 2026-01-16 11:22:43 --- .../微分中值定理（解析版）.md | 276 +++++++++++++++++- 1 file changed, 275 insertions(+), 1 deletion(-) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 0720e7d..907b88d 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -5,7 +5,281 @@ tags: **内部资料，禁止传播** **编委会（不分先后，姓氏首字母顺序）：陈峰华 陈玉阶 程奕铭 韩魏 刘柯妤 卢吉辚 王嘉兴 王轲楠 彭靖翔 郑哲航 钟宇哲 支宝宁 -### 多次运用中值定理 +## **辅助函数的构造方法** + +### **原理** + +在证明与导数相关的等式或不等式时，常通过构造辅助函数，将原问题转化为对某个函数应用中值定理（如罗尔定理、拉格朗日定理等）。构造辅助函数的核心思想是：**将待证等式视为某个函数求导后的结果**。 + +### **常见构造类型** + +#### 1. 乘积型与商型 +若结论形如： +$$ +f'(\xi)g(\xi) + f(\xi)g'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = f(x)g(x) +$$ +若结论形如： +$$ +f'(\xi)g(\xi) - f(\xi)g'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = \frac{f(x)}{g(x)} \quad (g(x) \neq 0) +$$ + +#### 2. 含幂函数因子 +若结论形如： +$$ +n f(\xi) + \xi f'(\xi) = 0 +$$ +可构造辅助函数： +$$ +F(x) = x^n f(x) +$$ + +#### 3. 一阶线性微分结构 +若结论形如： +$$ +f'(\xi) + P(\xi)f(\xi) = 0 +$$ +可构造积分因子： +$$ +\mu(x) = e^{\int P(x)\mathrm{d}x} +$$ +并设辅助函数： +$$ +F(x) = \mu(x) f(x) +$$ + +#### 4. 对数型 +若结论形如： +$$ +\frac{f'(\xi)}{f(\xi)} = k +$$ +可构造辅助函数： +$$ +F(x) = \ln|f(x)| - kx +$$ + +#### 5. 常数变易法 +若结论形如： +$$ +f'(\xi) = \lambda f(\xi) +$$ +可构造辅助函数： +$$ +F(x) = e^{-\lambda x} f(x) +$$ + +--- + +### **例题**（先看完后面的知识再做这个） + +>[!example] 例1 +设 $f(x)$ 在 $[0, 1]$ 上连续，在 $(0, 1)$ 内可导，且 $f(0)=0$，$f(1)=1$。 +证明：存在 $\xi \in (0, 1)$ 使得$f'(\xi) = 2\xi f(\xi)$ + +**解析**： +将结论改写为： +$$ +f'(\xi) - 2\xi f(\xi) = 0 +$$ +属于一阶线性微分结构，其中 $P(x) = -2x$。 +积分因子为： +$$ +\mu(x) = e^{\int (-2x)\mathrm{d}x} = e^{-x^2} +$$ +构造辅助函数： +$$ +F(x) = e^{-x^2} f(x) +$$ +则 $F(0) = 0$，$F(1) = e^{-1}$。 +需进一步寻找另一个点 $c$ 使 $F(c)=0$，才可应用罗尔定理。通常需结合题目其他条件（如积分中值定理、零点定理等）找出该点。 + +--- + +>[!example] 例2 +设 $f(x)$ 在 $[a, b]$ 上三阶可导，且 $f(a) = f'(a) = f(b) = 0$。 +证明：存在 $\xi \in (a, b)$ 使得：$f'''(\xi) + k f''(\xi) = 0$ + +**解析**： +结论可写为： +$$ +\bigl[ e^{kx} f''(x) \bigr]' \big|_{x=\xi} = 0 +$$ +因此构造辅助函数： +$$ +H(x) = e^{kx} f''(x) +$$ +由条件可推知存在 $\eta_1, \eta_2 \in (a, b)$ 使 $f''(\eta_1) = f''(\eta_2) = 0$，从而 $H(\eta_1)=H(\eta_2)=0$。 +对 $H(x)$ 应用罗尔定理即得证。 + +--- + +>[!example] 例3 +设 $f(x)$ 在 $[0, 1]$ 上可导，且$f(1) = 2\int_0^{1/2} e^{1-x} f(x) dx$ +证明：存在 $\xi \in (0, 1)$ 使得：$f'(\xi) = (1-\xi) f(\xi)$ + +**解析**： +结论化为： +$$ +f'(\xi) - (1-\xi) f(\xi) = 0 +$$ +积分因子为： +$$ +\mu(x) = e^{\int (x-1) \mathrm{d}x} = e^{\frac{x^2}{2} - x} +$$ +构造辅助函数： +$$ +F(x) = e^{\frac{x^2}{2} - x} f(x) +$$ +利用题设积分条件与积分中值定理，可找到 $\eta \in (0, \frac{1}{2})$ 使 $F(\eta) = F(1)$，再对 $F(x)$ 应用罗尔定理即证。 + +## **罗尔定理** + +### **原理** +若函数 f(x) 满足以下三个条件： +在闭区间 $[a,b]$ 上连续； +在开区间 $(a,b)$ 内可导； +区间端点函数值相等，即 $f(a)=f(b)$； +则在 $(a,b)$ 内至少存在一点 $\xi$，使得 $f'(\xi)=0$。 +罗尔定理的几何意义为：满足条件的函数曲线在区间内至少有一条水平切线。 +它是拉格朗日中值定理（$f(b)-f(a)=f'(\xi)(b-a)$）当 $f(a)=f(b)$ 时的特例。 + +### **适用条件** +罗尔定理的核心适用题型是证明导函数方程 $f'(\xi)=0$ 在区间 $(a,b)$ 内有根以及衍生的相关证明题。 +具体可分为以下几类： +1.直接证明 $f'(\xi)$=0 存在根 +题目给出函数 f(x) 在 $[a,b]$ 上的连续性、$(a,b)$ 内的可导性，且满足 $f(a)=f(b)$，直接应用罗尔定理证明存在 $\xi\in(a,b)$ 使得 $f'(\xi)=0$。 +2.构造辅助函数证明导函数相关方程有根 +对于形如 $f'(\xi)+g(\xi)f(\xi)=0$、$f''(\xi)=0$ 等方程，需构造满足罗尔定理条件的辅助函数 $F(x)$，通过 $F(a)=F(b)$ 推导 $F'(\xi)=0$，进而等价转化为目标方程。 +3.结合多次罗尔定理证明高阶导数零点存在 +若函数 f(x) 有 n+1 个点的函数值相等，可多次应用罗尔定理，证明其 n 阶导数 $f^{(n)}(\xi)=0$ 在对应区间内有根。 +4.证明函数恒为常数（反证法结合罗尔定理） +若 $f'(x)\equiv0$ 在区间内成立，可通过反证法假设存在两点函数值不等，结合罗尔定理推出矛盾，进而证明函数为常数。 + +罗尔定理针对于一个函数，不同于柯西中值定理针对于两个函数 + +### **例题** +>[!example] 例1 +设 $f(x)$ 在 $[0,1]$ 连续，$(0,1)$ 可导，且 $f(1) = 0$，求证存在 $\xi \in (0,1)$ 使得 $nf(\xi) + \xi f'(\xi) = 0$。 + +**解析**： +设辅助函数 $\varphi(x) = x^n f(x)$，则 $\varphi(0)=0$，$\varphi(1)=0$。由罗尔定理，存在 $\xi \in (0,1)$，使得 $\varphi'(\xi)=0$即 +$$ +n\xi^{n-1} f(\xi) + \xi^n f'(\xi) = 0 +$$ +两边除以 $\xi^{n-1}$ ($\xi>0$)，得 $nf(\xi) + \xi f'(\xi) = 0$。 + + + +>[!example] 例2 +设函数 $f(x)$ 在 $[a,b]$ 上可导，且 +$$f(a) = f(b) = 0，\quad f'_+(a)f'_-(b) > 0，$$ +试证明 $f'(x) = 0$ 在 $(a,b)$ 内至少有两个根。 + + +**解析**： +由导数极限定理及 $f'_+(a)f'_-(b) > 0$，知在 $a$ 右侧和 $b$ 左侧，$f(x)$ 的符号相同，不妨设 $f'_+(a)>0$，$f'_-(b)>0$。则在 $a$ 右侧附近 $f(x)>0$，在 $b$ 左侧附近 $f(x)>0$。由于 $f(a)=f(b)=0$，由极值点的费马定理，$f(x)$ 在 $(a,b)$ 内至少有一个极大值点，该点处导数为零。又因为 $f(x)$ 在 $[a,b]$ 上连续，在 $(a,b)$ 内可导，且 $f(a)=f(b)$，由罗尔定理至少存在一点 $c \in (a,b)$ 使 $f'(c)=0$。结合极大值点处的导数零点，可知至少有两个导数为零的点。 + + + +>[!example] 例3 +设 $f(x)$ 在 $[0, 1]$ 上具有二阶导数，且满足 +$$f(0) = 0, \, f(1) = 1, \, f\left(\frac{1}{2}\right) > \frac{1}{4}$$证明： +(1)至少存在一点 $\xi \in (0, 1)$，使得 $f''(\xi) < 2$； +(2)若对一切 $x \in (0, 1)$，有 $f''(x) \neq 2$，则当 $x \in (0, 1)$ 时，恒有 $f(x) > x^2$。 + +**解析**： +(1) 考虑函数 $g(x)=f(x)-x^2$，则 $g(0)=0$，$g(1)=0$，$g(1/2)=f(1/2)-1/4>0$。由极值点的费马定理及罗尔定理，$g(x)$ 在 $(0,1)$ 内存在极大值点 $\eta$，且 $g'(\eta)=0$，$g''(\eta) \leq 0$。即 $f'(\eta)=2\eta$，$f''(\eta) \leq 2$。若 $f''(\eta) < 2$，则取 $\xi=\eta$ 即可；若 $f''(\eta)=2$，则考虑在 $\eta$ 两侧应用拉格朗日中值定理，可找到另一个点 $\xi$ 使得 $f''(\xi)<2$。 +(2) 用反证法。假设存在 $x_0 \in (0,1)$ 使 $f(x_0) \leq x_0^2$，结合 $f(0)=0$，$f(1)=1$ 和 $f(1/2)>1/4$，利用连续性及中值定理可推出存在 $\xi$ 使 $f''(\xi)=2$，矛盾。 + +## **拉格朗日中值定理** +### **原理** +若函数 f(x) 满足两个条件: +在闭区间 $[a,b]$ 上连续； +在开区间 $(a,b)$ 内可导； +则在 $(a,b)$ 内至少存在一点 $\xi$，使得 +$f(b)-f(a)=f'(\xi)(b-a)$ +也可写成等价形式 $f'(\xi)=\dfrac{f(b)-f(a)}{b-a}$。 +是罗尔定理的推广，同时也是柯西中值定理的特例。其几何意义为：满足条件的函数曲线在区间 (a,b) 内，至少存在一点的切线与连接端点 (a,f(a)) 和 (b,f(b)) 的弦平行。 + +### **适用条件** +拉格朗日中值定理的核心适用题型是建立函数增量与导数的关联，进行不等式的证明，这是最常见的题型。通过对目标函数在指定区间上应用拉格朗日中值定理，得到 $f(b)-f(a)=f'(\xi)(b-a)$，再利用导数 $f'(\xi)$ 的取值范围（有界性、正负性）放大或缩小式子，推导不等式。 + +### **例题** + +>[!example] 例1 +设函数 $f(x)$ 在 $(-1,1)$ 内可微，且 +$$f(0) = 0, \quad |f'(x)| \leq 1,$$证明：在 $(-1,1)$ 内，$|f(x)| < 1$。 + +**解析**： +对任意 $x \in (-1,1)$，由拉格朗日中值定理，存在 $\xi$ 介于 $0$ 与 $x$ 之间，使得 +$$ +f(x) - f(0) = f'(\xi)(x-0) +$$ +即 $f(x) = f'(\xi) x$。由于 $|f'(\xi)| \leq 1$，$|x| < 1$，故 $|f(x)| = |f'(\xi)| \cdot |x| < 1$。 + + +>[!example] 例2 +设 $f''(x) < 0$，$f(0) = 0$，证明对任意 $x_1 > 0, x_2 > 0$ 有 +$$f(x_1 + x_2) < f(x_1) + f(x_2)$$ + +**解析**： +不妨设 $0 < x_1 < x_2$。由拉格朗日中值定理： +$$ +f(x_1+x_2)-f(x_2) = f'(\xi_1)x_1, \quad \xi_1 \in (x_2, x_1+x_2) +$$ +$$ +f(x_1)-f(0) = f'(\xi_2)x_1, \quad \xi_2 \in (0, x_1) +$$ +于是 +$$ +f(x_1+x_2)-f(x_2)-f(x_1) = [f'(\xi_1)-f'(\xi_2)]x_1 +$$ +对 $f'(x)$ 在 $[\xi_2,\xi_1]$ 上应用拉格朗日中值定理，存在 $\xi \in (\xi_2,\xi_1)$，使 +$$ +f'(\xi_1)-f'(\xi_2) = f''(\xi)(\xi_1-\xi_2) < 0 +$$ +故 $f(x_1+x_2)-f(x_2)-f(x_1) < 0$，即 $f(x_1+x_2) < f(x_1)+f(x_2)$。 + + +>[!example] 例3 + 设 $f(x)$ 在 $(-\infty, +\infty)$ 内二阶可导，且 $f''(x) \neq 0$。 +（1）证明：对于任何非零实数 $x$，存在唯一的 $\theta(x)$ ($0<\theta(x)<1$)，使得 + $$f(x) = f(0) + x f'(x\theta(x));$$ + （2）求 + $$\lim_{x \to 0} \theta(x).$$ + +解： +1. 证： 对于任何非零实数 $x$，由中值定理，存在 $\theta(x)$ $(0<\theta(x)<1)$，使得 + +$$ +f(x)=f(0)+x f'(x\theta(x)). +$$ + +如果这样的 $\theta(x)$ 不唯一，则存在 $\theta_{1}(x)$ 与 $\theta_{2}(x)$ $(\theta_{1}(x)<\theta_{2}(x))$，使得 $f'(x\theta_{1}(x))=f'(x\theta_{2}(x))$，由罗尔定理，存在一点 $\xi$，使得 $f''(\xi)=0$，这与 $f''(x)\neq 0$ 矛盾。所以 $\theta(x)$ 是唯一的。 + +2. 解 注意到 $f''(0)=\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x\theta(x)}$，又知 + +$$ +\begin{aligned} +\lim_{x\rightarrow 0} \frac{f'(x\theta(x))-f'(0)}{x} +&= \lim_{x\rightarrow 0} \frac{\frac{f(x)-f(0)}{x}-f'(0)}{x} \\ +&= \lim_{x\rightarrow 0} \frac{f(x)-f(0)-x f'(0)}{x^{2}} \\ +&= \lim_{x\rightarrow 0} \frac{f'(x)-f'(0)}{2x} \\ +&= \frac{f''(0)}{2}, +\end{aligned} +$$ + +所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 + +## 多次运用中值定理 + 多次运用中值定理一般有如下特征： 1. 有多个中值（如$\xi,\eta$两个中值）； 2. 有二阶导出现。 -- 2.34.1 From 12235f60881783f76e96162d8d04936e23c3962e Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Fri, 16 Jan 2026 11:24:34 +0800 Subject: [PATCH 272/274] vault backup: 2026-01-16 11:24:34 --- .../微分中值定理（解析版）.md | 97 +++++++++++++++++++ 1 file changed, 97 insertions(+) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 907b88d..3c65454 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -278,6 +278,103 @@ $$ 所以 $\lim_{x\rightarrow 0} \theta(x)=\frac{1}{2}$。 +## **柯西中值定理** + +### **原理** + +设函数 $f(x)$ 和 $g(x)$ 满足以下条件： +1. 在闭区间 $[a, b]$ 上连续； +2. 在开区间 $(a, b)$ 内可导； +3. 对任意 $x \in (a, b)$，有 $g'(x) \neq 0$； +则在 $(a, b)$ 内至少存在一点 $\xi$，使得： + +$$ +\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(\xi)}{g'(\xi)} +$$ + +柯西中值定理的几何意义为：由参数方程 $(g(t), f(t))$ 表示的曲线，在两点间的割线斜率等于曲线上某点切线的斜率。 + +它与拉格朗日中值定理的关系为：当 $g(x) = x$ 时，柯西中值定理退化为拉格朗日中值定理。它是处理两个函数之间微分中值关系的通用形式。 + +### **适用条件** + +柯西中值定理的核心适用题型是**证明形如 $\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(\xi)}{g'(\xi)}$ 的等式成立**，以及处理**涉及两个中值点 $\xi, \eta$ 的问题**。 + +常见应用方向包括： +1. 直接证明存在性等式； +2. 通过函数配对，将目标等式转化为柯西中值定理的标准形式； +3. 处理“双中值问题”，常与拉格朗日中值定理结合使用。 + +### **例题** + +>[!example] 例1 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，且 $a>0$。证明存在 $\xi \in (a, b)$，使得： +$$\frac{f(b)-f(a)}{b-a} = \xi f'(\xi) \cdot \frac{\ln(b/a)}{b-a}$$ + +**解析**： +将等式变形为： +$$ +\frac{f(b)-f(a)}{\ln b - \ln a} = \xi f'(\xi) +$$ +取 $g(x) = \ln x$，则 $g'(x) = \frac{1}{x} \neq 0$ 在 $(a, b)$ 内成立。 +对 $f(x)$ 与 $g(x)$ 应用柯西中值定理，存在 $\xi \in (a, b)$ 使得： +$$ +\frac{f(b)-f(a)}{\ln b - \ln a} = \frac{f'(\xi)}{1/\xi} = \xi f'(\xi) +$$ +整理即得所求。 + +--- + +>[!example] 例2 +设 $f(x)$ 在 $[a, b]$ 上连续，在 $(a, b)$ 内可导，证明存在不同的 $\xi, \eta \in (a, b)$，使得： +$$f'(\xi) = \frac{a+b}{2\eta} f'(\eta)$$ + +**解析**： +1. 对 $f(x)$ 与 $g(x) = \frac{x^2}{2}$ 应用柯西中值定理，存在 $\eta \in (a, b)$ 使得： + $$ + \frac{f(b)-f(a)}{(b^2 - a^2)/2} = \frac{f'(\eta)}{\eta} + $$ + 整理得： + $$ + f(b)-f(a) = \frac{b^2 - a^2}{2\eta} f'(\eta) + $$ + +2. 对 $f(x)$ 应用拉格朗日中值定理，存在 $\xi \in (a, b)$ 使得： + $$ + f(b)-f(a) = (b-a) f'(\xi) + $$ + +3. 联立两式，消去 $f(b)-f(a)$ 得： + $$ + (b-a) f'(\xi) = \frac{(b-a)(a+b)}{2\eta} f'(\eta) + $$ + 由于 $b-a \neq 0$，约去后即得： + $$ + f'(\xi) = \frac{a+b}{2\eta} f'(\eta) + $$ + +--- + +>[!example] 例3 +设 $0 < a < b$，证明存在 $\xi \in (a, b)$，使得： +$$f(b)-f(a) = \frac{3\xi^2}{a^2+ab+b^2} f'(\xi)(b-a)$$ + +**解析**： +将等式变形为： +$$ +\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} +$$ +取 $g(x) = x^3$，则 $g'(x) = 3x^2 \neq 0$ 在 $(a, b)$ 内成立。 +由柯西中值定理，存在 $\xi \in (a, b)$ 使得： +$$ +\frac{f(b)-f(a)}{b^3 - a^3} = \frac{f'(\xi)}{3\xi^2} +$$ +整理后即得所求。 + + + + + ## 多次运用中值定理 多次运用中值定理一般有如下特征： -- 2.34.1 From 18c312d426dd0c9cff5605bf3321f8f571578747 Mon Sep 17 00:00:00 2001 From: Cym10x <Cym_em10x@outlook.com> Date: Fri, 16 Jan 2026 11:37:37 +0800 Subject: [PATCH 273/274] =?UTF-8?q?LaTeX=EF=BC=88KaTeX=EF=BC=89=E8=BE=93?= =?UTF-8?q?=E5=85=A5=E8=A7=84=E8=8C=83?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 笔记分享/LaTeX（KaTeX）输入规范.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/笔记分享/LaTeX（KaTeX）输入规范.md b/笔记分享/LaTeX（KaTeX）输入规范.md index c88bffc..6f6d9c0 100644 --- a/笔记分享/LaTeX（KaTeX）输入规范.md +++ b/笔记分享/LaTeX（KaTeX）输入规范.md @@ -4,4 +4,4 @@ 3. 微分算子d应当用正体，被微分的表达式用正常的斜体：$\mathrm{d}f(x)=f'(x)\mathrm{d}x$ 4. 极限和求和求积符号用\limits，如$\lim\limits_{x\to0}$和$\sum\limits_{n=0}^{\infty}$ 5. \$\$双美元符号之间不要打回车！除非你有\begin{...}\end{...}\$\$ -6. 矩阵不用加粗，但向量要加粗，用\boldsymbol{}，比如$A,\boldsymbol{x}$。 \ No newline at end of file +6. 不对矩阵作加粗要求，但向量一定要加粗，用\boldsymbol{}，比如$A,\boldsymbol{x}$。 \ No newline at end of file -- 2.34.1 From 2e3107bff8656cac77e230346542a8c43992c962 Mon Sep 17 00:00:00 2001 From: idealist999 <2974730459@qq.com> Date: Fri, 16 Jan 2026 11:44:56 +0800 Subject: [PATCH 274/274] vault backup: 2026-01-16 11:44:56 --- 编写小组/讲义/微分中值定理（解析版）.md | 8 ++++++++ 1 file changed, 8 insertions(+) diff --git a/编写小组/讲义/微分中值定理（解析版）.md b/编写小组/讲义/微分中值定理（解析版）.md index 3c65454..0c81049 100644 --- a/编写小组/讲义/微分中值定理（解析版）.md +++ b/编写小组/讲义/微分中值定理（解析版）.md @@ -74,6 +74,14 @@ $$ $$ F(x) = e^{-\lambda x} f(x) $$ +或者写成： +$$ +f'(\xi) + \lambda f(\xi) = 0 +$$ +则构造辅助函数： +$$ +F(x) = e^{\lambda x} f(x) +$$ --- -- 2.34.1