# 1988 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 20 分, 每小题 4 分)

(1) 设 $f(x)=\left\{\begin{array}{ll}2 x+a, & x \leqslant 0 \\ \mathrm{e}^{x}(\sin x+\cos x), & x>0\end{array}\right.$ 在 $(-\infty,+\infty)$ 内连续, 则 $a=?$

$$\lim \limits_{x \rightarrow 0^{+}} e^{x}(\sin x+\cos x)=\lim \limits_{x \rightarrow 0}(2 x+a) \Rightarrow$$

$$a=1$$

(2) 设 $f(t)=\lim \limits_{x \rightarrow \infty} t\left(1+\frac{1}{x}\right)^{2 t x}$, 则 $f^{\prime}(t)=?$

$$f(t)=\lim \limits_{x \rightarrow \infty} t\left(1+\frac{1}{x}\right)^{2 t x} \Rightarrow$$

$$f(t)=t \cdot \lim \limits_{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x \cdot 2 t} \Rightarrow$$

$$f(t)=t e^{2 t} \Rightarrow f^{\prime}(t)=e^{2 t}+2 t e^{2 t}$$

(3) 设 $f(x)$ 连续, 且 $\int_{0}^{x^{3}-1} f(t) \mathrm{d} t=x$, 则 $f(7)=?$

$$\left[\int_{0}^{x^{3}-1} f(t) \mathrm{~ d} t=x\right]_{x}^{\prime} \Rightarrow 3 x^{2} f\left(x^{3}-1\right)=1 \Rightarrow$$

$$12 f(7)=1 \Rightarrow f(7)=\frac{1}{12}$$

(4) $\lim \limits_{x \rightarrow 0^{+}}\left(\frac{1}{\sqrt{x}}\right)^{\tan x}=?$

$$\lim \limits_{x \rightarrow 0^{+}}\left(\frac{1}{\sqrt{x}}\right)^{\tan x}=\lim \limits_{x \rightarrow 0^{+}} x^{\frac{-\tan x}{2}}=$$

$$\lim \limits_{x \rightarrow 0^{+}} e^{\frac{-\tan x}{2} \ln x}$$

$$\lim \limits_{x \rightarrow 0^{+}}\left(\frac{-\tan x}{2} \ln x\right)=$$

$$\lim \limits_{x \rightarrow 0^{+}} \frac{-\ln x}{2 / \tan x} \Rightarrow$$

$$\frac{-1 / x}{2 \cdot\left(\frac{-1}{\cos ^{2} x}\right) /(\tan x)^{2}} =$$

$$\frac{\frac{-1}{x}}{2 \cdot \frac{-1}{\cos^{2} x} \cdot \frac{\cos^{2} x}{\sin^{2} x}}$$

$$\frac{\frac{-1}{x}}{\frac{-2}{\sin ^{2} x}}=\frac{1}{x} \cdot \frac{\sin x}{2}=\frac{x}{2}$$

$$\lim \limits_{x \rightarrow 0^{+}} e^{ \frac{x}{2} } = e^{0} =1$$

(5) $\int_{0}^{4} e^{\sqrt{x}} \mathrm{~d} x=?$

$$t=\sqrt{x} \Rightarrow t \in(0,2) \Rightarrow x=t^{2} \Rightarrow \mathrm{~ d} x=2 t \mathrm{~ d} t$$

$$\int_{0}^{4} e^{\sqrt{x}} \mathrm{~ d} x=\int_{0}^{2} e^{t} 2 t \mathrm{~ d} t=2 \int_{0}^{2} t e^{t} \mathrm{~ d} t \Rightarrow$$

$$\left(t e^{t}-e^{t}\right)^{\prime}=e^{t}+t e^{t}-e^{t}=t e^{t} \Rightarrow$$

$$2 \int_{0}^{2} t e^{t} \mathrm{~ d} t=2 \cdot\left(t e^{t}-e^{t}\right) \Big|_{0} ^{2}=$$

$$2\left[\left(2 e^{2}-e^{2}\right)-(0-1)\right]=2\left(e^{2}+1\right)$$