1990 年考研数二真题解析

前言

在本文中,荒原之梦网从考试实战的角度出发,详细解析了考研数学二 1990 年的真题。

注意事项:
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 15 分, 每小题 3 分)

(1) 曲线 $\left\{\begin{array}{l}x=\cos ^{3} t, \\ y=\sin ^{3} t\end{array}\right.$ 上对应于 $t=\frac{\pi}{6}$ 处的法线方程是

注意:本题说的法线方程不是切线方程。

$$
\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} t}=3 \sin ^{2} t \cdot \cos t
$$

$$
\frac{\mathrm{d} x}{\mathrm{d} t}=3 \cos ^{2} t(-\sin t) \Rightarrow
$$

$$
t=\frac{\pi}{6} \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} t}=3 \cdot\left(\frac{1}{2}\right)^{2} \cdot\left(\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3}}{8}
$$

$$
t=\frac{\pi}{6} \Rightarrow \frac{\mathrm{d} x}{\mathrm{d} t}=3 \cdot\left(\frac{\sqrt{3}}{2}\right)^{2}\left(-\frac{1}{2}\right)=\frac{-9}{8} \Rightarrow
$$

又:

$$
\frac{\mathrm{d} y}{\mathrm{d} x}=k=\frac{3 \sqrt{3}}{8} \cdot \frac{8}{-9}=\frac{-\sqrt{3}}{3}
$$

$$
t=\frac{\pi}{6} \Rightarrow x=\left(\frac{\sqrt{3}}{2}\right)^{3}=\frac{3 \sqrt{3}}{8}
$$

$$
t=\frac{\pi}{6} \Rightarrow y=\left(\frac{1}{2}\right)^{3}=\frac{1}{8} \Rightarrow
$$

于是:

$$
\frac{-\sqrt{3}}{3} k^{\prime}=1 \Rightarrow k^{\prime}=\frac{3}{\sqrt{3}}=\sqrt{3} \Rightarrow
$$

因此:

$$
y-\frac{1}{8}=\sqrt{3}\left(x-\frac{3 \sqrt{3}}{8}\right) \Rightarrow
$$

$$
y=\sqrt{3} x-\frac{9}{8}+\frac{1}{8} \Rightarrow y=\sqrt{3} x-1
$$

(2) 设 $y=\mathrm{e}^{\tan \frac{1}{x}} \cdot \sin \frac{1}{x}$, 则 $y^{\prime}=$

$$
y^{\prime}=\left(e^{\tan \frac{1}{x}}\right)^{\prime} \cdot \sin \frac{1}{x}+e^{\tan \frac{1}{x}} \cdot\left(\sin \frac{1}{x}\right)^{\prime} \Rightarrow
$$

$$
y^{\prime}=\left(\tan \frac{1}{x}\right)^{\prime} \cdot e^{\tan \frac{1}{x}} \cdot \sin \frac{1}{x}+
$$

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

$$
y^{\prime}=\frac{1}{\cos ^{2} \frac{1}{x}} \cdot\left(\frac{-1}{x^{2}}\right) \cdot e^{\tan \frac{1}{x}} \cdot \sin \frac{1}{x}+
$$

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

$$
y^{\prime}=\frac{-e^{\tan \frac{1}{x}}}{x^{2}}\left[\frac{\sin \frac{1}{x}}{\cos ^{2} \frac{1}{x}}+\cos \frac{1}{x}\right] \Rightarrow
$$

$$
y^{\prime}=\frac{-e^{\tan \frac{1}{x}}}{x^{2}}\left[\sec ^{2} \frac{1}{x} \cdot \sin \frac{1}{x}+\cos \frac{1}{x}\right]
$$

(3) $\int_{0}^{1} x \sqrt{1-x} \mathrm{~d} x=$

$$
t=\sqrt{1-x} \Rightarrow t^{2}=1-x \Rightarrow x=1-t^{2} \Rightarrow
$$

$$
\mathrm{d} x=-2 t \mathrm{d} t \Rightarrow x \in(0,1) \Rightarrow 1-x \in(1,0) \Rightarrow
$$

$$
\sqrt{1-x} \in(1,0) \Rightarrow t \in(1,0) \Rightarrow
$$

$$
\int_{0}^{1} x \sqrt{1-x} \mathrm{d} x=\int_{1}^{0}\left(1-t^{2}\right) \cdot t(-2 t) \mathrm{d} t=
$$

$$
2 \int_{0}^{1}\left(1-t^{2}\right) t^{2} \mathrm{d} t=2 \int_{0}^{1}\left(t^{2}-t^{4}\right) \mathrm{d} t=
$$

$$
2\left[\left.\frac{1}{3} t^{3}\right|_{0} ^{1}-\left.\frac{1}{5} t^{5}\right|_{0} ^{1}\right]=\frac{4}{15}
$$

(4) 下列两个积分大小关系式: $\int_{-2}^{-1} \mathrm{e}^{-x^{3}} \mathrm{~d} x \longrightarrow \int_{-2}^{-1} \mathrm{e}^{x^{3}} \mathrm{~d} x$.

$$
t=-x \Rightarrow x=-t \Rightarrow t \in(2,1) \Rightarrow
$$

$$
\int_{-2}^{-1} e^{-x^{3}} \mathrm{d} x=\int_{2}^{1} e^{t^{3}}(-1) \mathrm{d} t=\int_{1}^{2} e^{t^{3}} \mathrm{d} t
$$

$$
\int_{-1}^{-2} e^{x^{3}} \mathrm{d} x>0 \Rightarrow \int_{-1}^{-2} e^{x^{3}} \mathrm{d} x<\int_{1}^{2} e^{x^{3}} \mathrm{d} x \Rightarrow
$$

$$
\int_{-2}^{-1} e^{x^{3}} \mathrm{d} x<0 \Rightarrow \int_{-2}^{-1} e^{x^{3}} \mathrm{d} x<\int_{1}^{2} e^{x^{3}} \mathrm{d} x
$$

或者(积分区间相同的时候,比较被积函数):

$$
x \in(-2,-1) \Rightarrow-x^{3}>x^{3} \Rightarrow e^{-x^{3}}>e^{x^{3}} \Rightarrow
$$

$$
\int_{-2}^{-1} e^{-x^{3}} \mathrm{d} x>\int_{-2}^{-1} e^{-x^{3}} \mathrm{d} x
$$

(5) 设函数 $f(x)=\left\{\begin{array}{l}1,|x| \leqslant 1, \\ 0,|x|>1,\end{array}\right.$ 则函数 $f[f(x)]=$

$$
|f(x)| \leqslant 1 \Rightarrow f[f(x)] \leq 1 = 1
$$