From 445f632860fa8b3e1584fdccc42a9abd903241ff Mon Sep 17 00:00:00 2001 From: Cym10x Date: Wed, 31 Dec 2025 21:43:47 +0800 Subject: [PATCH] =?UTF-8?q?=E8=B0=83=E6=95=B4=E8=A7=A3=E6=9E=90=E6=A0=BC?= =?UTF-8?q?=E5=BC=8F+=E9=AB=98=E6=95=B0=E6=A8=A1=E6=8B=9F=E8=AF=95?= =?UTF-8?q?=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.