From 20ca449ab992b896daba7fbaf5aad9d6930e93ff Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Thu, 1 Jan 2026 17:49:42 +0800 Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86=E9=83=A8=E5=88=86?= =?UTF-8?q?=E6=A0=BC=E5=BC=8F=EF=BC=8C=E5=85=84=E5=BC=9F=E5=A7=90=E5=A6=B9?= =?UTF-8?q?=E4=BB=AC=E6=B3=A8=E6=84=8F=E8=BE=93=E5=85=A5=E8=A7=84=E8=8C=83?= =?UTF-8?q?=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。