# 1989 年考研数二真题解析

## 前言

1. 按照原试卷结构，每页一类题，点击页码可以切换；
2. 蓝色部分为题干。

## 一、填空题 (本题满分 21 分, 每小题 3 分)

(1) $\lim \limits_{x \rightarrow 0} x \cot 2 x=$

$$\lim \limits_{x \rightarrow 0} x \cos 2 x=\lim \limits_{x \rightarrow 0} x \cdot \frac{1}{\tan 2 x}=\lim \limits_{x \rightarrow 0} \frac{x}{2 x}=\frac{1}{2}$$

(2) $\int_{0}^{\pi} t \sin t \mathrm{~d} t=$

$$\int_{0}^{\pi} t \sin t \mathrm{~d} t=t \cdot 2 \int_{0}^{\frac{\pi}{2}} \sin t \mathrm{~d} t=\left.t \cdot 2 \cdot(-\cos t)\right|_{0} ^{\frac{\pi}{2}}$$

$$-\left.2 t \cos t\right|_{0} ^{\frac{\pi}{2}}=-(0-\pi)=\pi$$

$$\int_{0}^{\pi} t \sin t \mathrm{~d} t=-\int_{0}^{\pi} t \mathrm{~d} (\cos t)=$$

$$-\left[\left.t \cos t\right|_{0} ^{\pi}-\int_{0}^{\pi} \cos t \mathrm{~d} t\right]=$$

$$-\left[-\pi-\left.\sin t\right|_{0} ^{\pi}\right]=\pi$$

(3) 曲线 $y=\int_{0}^{x}(t-1)(t-2) \mathrm{d} t$ 在点 $(0,0)$ 处的切线方程是

$$y^{\prime}=(x-1)(x-2) \Rightarrow x=0 \Rightarrow k=y^{\prime}=2 \Rightarrow$$

$$y-0=2(x-0)=y=2 x$$

(4) 设 $f(x)=x(x+1)(x+2) \cdots(x+n)$, 则 $f^{\prime}(0)=$

$$g(x)=(x+1)(x+2) \cdots(x+n) \Rightarrow$$

$$f(x)=x g(x) \Rightarrow f^{\prime}(x)=g(x)+x g^{\prime}(x) \Rightarrow$$

$$f^{\prime}(0)=g(0) \Rightarrow g(0)=1 \cdot 2 \cdot 3 \cdots n \Rightarrow$$

$$f^{\prime}(0)=n !$$

(5) 设 $f(x)$ 是连续函数, 且 $f(x)=x+2 \int_{0}^{1} f(t) \mathrm{d} t$, 则 $f(x)=$

$$f(x)=x+2 \int_{0}^{1} f(t) \mathrm{~d} t \Rightarrow A=\int_{0}^{1} f(t) \mathrm{~d} t \Rightarrow$$

$$f(x)=x+2 A \Rightarrow \int_{0}^{1} f(x) \mathrm{~d} x=\int_{0}^{1} x \mathrm{~d} x+2 A \int_{0}^{1} \mathrm{~d} x$$

$$A=\left.\frac{1}{2} x^{2}\right|_{0} ^{1}+2 A \Rightarrow-A=\frac{1}{2} \Rightarrow A=\frac{-1}{2} \Rightarrow$$

$$f(x)=x+2 \cdot\left(\frac{-1}{2}\right) \Rightarrow f(x)=x-1$$

(6) 设 $f(x)=\left\{\begin{array}{ll}a+b x^{2}, & x \leqslant 0, \\ \frac{\sin b x}{x}, & x>0\end{array}\right.$ 在 $x=0$ 处连续, 则常数 $a$ 与 $b$ 应满足的关系是

$$\lim \limits_{x \rightarrow 0}\left(a+b x^{2}\right)=\lim \limits_{x \rightarrow 0^{+}} \frac{\sin b x}{x} \Rightarrow$$

$$a=\frac{b x}{x} \Rightarrow a=b$$

(7) 设 $\tan y=x+y$, 则 $\mathrm{d} y=$

$$y(x) \Rightarrow$$

$$\tan y=x+y \Rightarrow \frac{1}{\cos ^{2} y} \cdot \frac{\mathrm{~d} y}{\mathrm{~d} x}=1+\frac{\mathrm{~d} y}{\mathrm{~d} x}$$

$$\frac{\mathrm{~d} y}{\mathrm{~d} x}\left(\frac{1}{\cos ^{2} y}-1\right)=1 \Rightarrow$$

$$\frac{\mathrm{~d} y}{\mathrm{~d} x} \cdot \frac{1-\cos ^{2} y}{\cos ^{2} y}=1 \Rightarrow$$

$$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\cos ^{2} y}{\sin ^{2} y} \Rightarrow$$

$$\frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{\cos ^{2} y}{\sin ^{2} y} \mathrm{~d} x=\cos ^{2} t \mathrm{~d} x$$