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。

(1) 曲线 ${\begin{cases} x = \cos^{3} t, \\ y = \sin^{3} t \end{cases}$ 上对应于 $t = \frac{π}{6}$ 处的法线方程是

注意：本题说的法线方程不是切线方程。

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

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

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

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

又：

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

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

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

于是：

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

因此：

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

或者（积分区间相同的时候，比较被积函数）：

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

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

(5) 设函数 $f (x) = {\begin{cases} 1, | x | ⩽ 1, \\ 0, | x | > 1, \end{cases}$ 则函数 $f [f (x)] =$

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

页码： 12345678