数一2023

一、选择题: 1 10 小题, 每小题 5 分, 共 50 分. 下列每题给出的四个选项中, 只有一个选项是最符合题目要求的

(1)

曲线 $y=x \ln \left(\mathrm{e}+\frac{1}{x-1}\right)$ 的斜渐近线方程为 A. $y=x+\mathrm{e}$. B. $y=x+\frac{1}{\mathrm{e}}$. C. $y=x$. D. $y=x-\frac{1}{\mathrm{e}}$.

2025版

• 求斜渐近线的方法:$\displaystyle \begin{aligned} y = kx + b \end{aligned}$
  • $\displaystyle \begin{aligned} \lim_{x \to \infty} \left(\frac{y}{x}\right) = k \end{aligned}$
  • $\displaystyle \begin{aligned} \lim_{x \to \infty} (y - kx) = b \end{aligned}$
• $\displaystyle \begin{aligned} k = \lim_{x \to \infty} \frac{y}{x} \xlongequal[]{代入y} \lim_{x \to \infty} \frac{x \ln\left(e + \frac{1}{x-1}\right)}{x} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{消去x}\lim_{x \to \infty} \ln\left(e + \frac{1}{x-1}\right) \xrightarrow{\frac{1}{x-1} \to 0} \lim_{x \to \infty} \ln e = 1 \end{aligned}$
• $\displaystyle \begin{aligned} b \xlongequal[]{} \lim_{x \to \infty} (y - kx) \xlongequal[]{代入y和k}\lim_{x \to \infty} x \ln\left(e + \frac{1}{x-1}\right) - x \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{提取系数x}\lim_{x \to \infty} x \left[\ln\left(e + \frac{1}{x-1}\right) - 1\right] \xrightarrow{1 = \ln e} \lim_{x \to \infty} x \cdot \left[\ln\frac{e + \frac{1}{x-1}}{e}\right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{ln里面做除法} \lim_{x \to \infty} x \cdot \ln\left[1 + \frac{1}{e(x-1)}\right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow{\ln(1+x) \sim x} \lim_{x \to \infty} x \cdot \frac{1}{e(x-1)} = \frac{1}{e} \end{aligned}$
• 结合斜率 $k$ 和截距 $b$，得$y = kx + b= x + \frac{1}{e}$

gpt版

我们首先需要确定斜渐近线方程的两个关键元素：斜率 $k$ 和截距 $b$。这个问题涉及到求极限，以确定曲线 $y = x \ln \left(e + \frac{1}{x-1}\right)$ 的斜渐近线。

二叉树结构如下：

• 确定斜率 $k$
  • $\displaystyle k = \lim_{x \rightarrow \infty} \frac{y}{x} ==\lim _{x \rightarrow \infty} \frac{x \ln \left(e+\frac{1}{x-1}\right)}{x}= \lim_{x \rightarrow \infty} \ln\left(e + \frac{1}{x-1}\right) = \ln e = 1$
• 确定截距 $b$
  • 截距 $b$ 的定义
    • $\displaystyle b = \lim_{x \rightarrow \infty} (y - x)= \lim_{x \rightarrow \infty} \left[ x \ln\left(e + \frac{1}{x-1}\right) - x \right]$
      • $\displaystyle \xlongequal[]{提出x}\lim _{x \rightarrow \infty} x\left[\ln \left(e+\frac{1}{x-1}\right)-1\right]\xlongequal[]{1=\ln e}\lim _{x \rightarrow \infty} x\left[\ln \left(e+\frac{1}{x-1}\right)-\ln e\right]\xlongequal[]{减法变除法}\lim _{x \rightarrow \infty} x \ln \left(1+\frac{1}{\mathrm{e}(x-1)}\right)$
      • $\displaystyle \xlongequal[]{\ln (1+x～x)}\lim _{x \rightarrow \infty} \frac{x}{\mathrm{e}(x-1)}\xlongequal[]{分子分母的x消去}\frac{1}{\mathrm{e}}$
• 斜渐近线方程
  • 结合斜率 $k$ 和截距 $b$，得$y = kx + b= x + \frac{1}{e}$ 由此，我们得到了斜渐近线的方程 $y = x + \frac{1}{e}$，符合选项 B

答 应选 B.

(2)

若微分方程 $y^{\prime \prime}+a y^{\prime}+b y=0$ 的解在 $(-\infty,+\infty)$ 上有界, 则 A. $a<0, b>0$. B. $a>0, b>0$. C. $a=0, b>0$. D. $a=0, b<0$.

• 关于二阶常系数齐次线性微分方程 $\displaystyle \begin{aligned} y'' + p y' + q y = 0 \end{aligned}$ 通解的求法，具体步骤和解的形式如下：
  • 写出特征方程 $\displaystyle \begin{aligned} \lambda^2 + p\lambda + q = 0 \end{aligned}$。
  • 求出特征方程的两个根 $\displaystyle \begin{aligned} \lambda_1, \lambda_2 \end{aligned}$。
  • 根据特征方程两个根的不同情形，通解如下：

特征方程 $\displaystyle \begin{aligned} \lambda^2 + p\lambda + q = 0 \end{aligned}$ 的根	微分方程 $\displaystyle \begin{aligned} y'' + p y' + q y = 0 \end{aligned}$ 的通解
两个不同实根 $\displaystyle \begin{aligned} \lambda_1, \lambda_2 \end{aligned}$	$\displaystyle \begin{aligned} y = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} \end{aligned}$
两个相同实根 $\displaystyle \begin{aligned} \lambda_1 = \lambda_2 \end{aligned}$	$\displaystyle \begin{aligned} y = (C_1 + C_2 x) e^{\lambda_1 x} \end{aligned}$
一对共轭复根 $\displaystyle \begin{aligned} \lambda_{1, 2} = \alpha \pm \beta i \end{aligned}$	$\displaystyle \begin{aligned} y = e^{\alpha x}(C_1 \cos \beta x + C_2 \sin \beta x) \end{aligned}$

• 判断无界的条件
  • $\displaystyle \begin{aligned} & \lim _{x \rightarrow-\infty} e^{\lambda x}=\left\{\begin{array}{cc}0 & \lambda>0 \\+\infty & \lambda<0\end{array}\right. \end{aligned}$
  • $\displaystyle \begin{aligned} \lim _{x \rightarrow+\infty} e^{\lambda x}=\left\{\begin{array}{cl}+\infty & \lambda>0 \\0 & \lambda<0\end{array}\right. \end{aligned}$
  • 则$λ=0$
• 当微分方程 $y^{\prime \prime}+a y^{\prime}+b y=0$ 对应的特征方程 $r^2+a r+b=0$
  • 有实根时, $a^2-4 b \geqslant 0$, 设根为 $r_1, r_2$
    • 当$r_1≠r_2$时，则微分方程的解为 $y=\underbrace{C_1 \mathrm{e}^{r_1 x}}_{∞}+C_2 \mathrm{e}^{r_1 x}\left(r_1 \neq r_2\right)$，
      此时解在 $(-\infty$, $+\infty)$ 上无界.
      • 取 $\displaystyle \begin{aligned} C_1 = C_2 = 1 \end{aligned}$，$\displaystyle \begin{aligned} \lim_{x \to -\infty} (e^{\lambda_1 x} + e^{\lambda_2 x}) \end{aligned}$与$\displaystyle \begin{aligned} \lim_{x \to +\infty} (e^{\lambda_1 x} + e^{\lambda_2 x}) \end{aligned}$至少有一个为 $\displaystyle \begin{aligned} \infty \end{aligned}$。
    • 当$r_1=r_2$时，则微分方程的解为 $y=\left(C_1+C_2 x\right) \mathrm{e}^{r_1 x}\left(r_1=r_2\right)$,
      此时解在 $(-\infty$, $+\infty)$ 上无界.
      • 取 $\displaystyle \begin{aligned} C_1 = 0 \end{aligned}$，$\displaystyle \begin{aligned} C_2 = 1 \end{aligned}$，$\displaystyle \begin{aligned} \lim_{x \to -\infty} x e^{\lambda x} \end{aligned}$与$\displaystyle \begin{aligned} \lim_{x \to +\infty} x e^{\lambda x} \end{aligned}$至少有一个为 $\displaystyle \begin{aligned} \infty \end{aligned}$。
  • 当 $a^2-4 b<0$ 时, 设根为 $\alpha+\beta \mathrm{i}$, 则 $y=\left(C_1 \cos \beta x+C_2 \sin \beta x\right) \mathrm{e}^\alpha$,
    • 对于 $\displaystyle \begin{aligned} \alpha \neq 0 \end{aligned}$，取 $\displaystyle \begin{aligned} C_1 = 1 \end{aligned}$，$\displaystyle \begin{aligned} C_2 = 0 \end{aligned}$，$\displaystyle \begin{aligned} y = e^{\alpha x} \cos \beta x \end{aligned}$，在 $\displaystyle \begin{aligned} (-\infty, +\infty) \end{aligned}$ 上无界。
    • 对于 $\displaystyle \begin{aligned} \alpha = 0 \end{aligned}$，$\displaystyle \begin{aligned} y = C_1 \cos \beta x + C_2 \sin \beta x \end{aligned}$对任意常数 $\displaystyle \begin{aligned} C_1, C_2 \end{aligned}$，解在 $\displaystyle \begin{aligned} (-\infty, +\infty) \end{aligned}$ 上均有界。
      • 若想解在 $(-\infty,+\infty)$ 上有界, 则 $\alpha=0$, 又 $a=-\frac{\alpha}{2}$, 因此 $a=0$,
      • 结合 $a^2-4 b<0$ 可得 $b>0$. 故选 C.

(3)

设函数 $y=f(x)$ 由 $\displaystyle \left\{\begin{array}{l}x=2 t+|t|, \\ y=|t| \sin t\end{array}\right.$ 确定, 则 A. $f(x)$ 连续, $f^{\prime}(0)$ 不存在. B. $f^{\prime}(0)$ 存在, $f^{\prime}(x)$ 在 $x=0$ 处不连续. C. $f^{\prime}(x)$ 连续, $f^{\prime \prime}(0)$ 不存在. D. $f^{\prime \prime}(0)$ 存在, $f^{\prime \prime}(x)$ 在 $x=0$ 处不连续.

• • 连续与可导的关系
    • 连续不一定可导，可导必然连续
• 读题分别问你子孙三代
  • $f(x)$ 连续,
  • $f^{\prime}(0)$ 不存在， $f^{\prime}(0)$ 存在, $f^{\prime}(x)$ 在 $x=0$ 处不连续， $f^{\prime}(x)$ 连续
  • $f^{\prime \prime}(0)$ 不存在， $f^{\prime \prime}(0)$ 存在, $f^{\prime \prime}(x)$ 在 $x=0$ 处不连续.
• 函数 $y=f(x)$ 由 $\displaystyle \left\{\begin{array}{l}x=2 t+|t|, \\ y=|t| \sin t\end{array}\right.$ 确定
  • 当 $t \geqslant 0$ 时, 参数方程为 $\displaystyle \left\{\begin{array}{l}x=3 t, \\ y=t \sin t,\end{array}\right.$ 即 $x \geqslant 0, y=\frac{x}{3} \sin \frac{x}{3}$;
  • 当 $t<0$ 时,参数方程为 $\displaystyle \left\{\begin{array}{l}x=t, \\ y=-t \sin t,\end{array}\right.$ 即 $x<0, y=-x \sin x$.
• $f^{}(x)$ 是否连续
  存在函数值=左极限=右极限，则连续
  • $\displaystyle \begin{aligned} & f(x)= \begin{cases}\frac{x}{3} \sin \frac{x}{3}, & x \geqslant 0 \\-x \sin x, & x<0\end{cases} \end{aligned}$
  • $\displaystyle \begin{aligned} f\left(0^{-}\right)=f\left(0^{+}\right)=f(0)=0 \end{aligned}$
• $f^{\prime}(0)$ 是否存在
  定理1:可导 $\displaystyle \begin{aligned} \Leftrightarrow \end{aligned}$ 左右导数都存在且相等。
  • $\displaystyle \begin{aligned} f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{\frac{-\sin x}{x}-0}{x-0}=\lim _{x \rightarrow 0^{-}}-\sin x=0 \end{aligned}$
  • $\displaystyle \begin{aligned} f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x}=\lim _{x \rightarrow 0^{+}} \frac{\frac{x}{3} \sin \frac{x}{3}-0}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{1}{3} \sin \frac{x}{3}=0 \end{aligned}$
  • $\displaystyle \begin{aligned} f^{\prime}\left(0^{-}\right)=f^{\prime}\left(0^{+}\right)=0 \text { 。则 } f^{\prime}(0) \text { 存在 } \end{aligned}$
• 判断$f'(x)$是否连续
  (可导必然连续，这一步是否多余)
  • $\displaystyle \begin{aligned} x>0, f^{\prime}(x)=\frac{1}{3} \sin \frac{x}{3}+\frac{x}{9} \cos \frac{x}{3} \xrightarrow{x \rightarrow 0^{+}} \lim _{x \rightarrow 0^{+}} f^{\prime}(x)=0 \end{aligned}$
  • $\displaystyle \begin{aligned} x<0, f^{\prime}(x)=-\sin x-x \cos x \quad \xrightarrow{x \rightarrow 0^{-}} \lim _{x \rightarrow 0^{-}} f^{\prime}(x)=0 \end{aligned}$
  • $\displaystyle \begin{aligned} \lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{x \rightarrow 0^{-}} f^{\prime}(x)=f^{\prime}(0)=0 \end{aligned}$
• 判断$f''(x)$是否存在
  • $\displaystyle \begin{aligned} f^{\prime \prime}\left(0^{+}\right)=\lim _{x \rightarrow 0^{+}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{3} \sin \frac{x}{3}+\frac{x}{9} \cos \frac{x}{3}}{x} \end{aligned}$
    • $\displaystyle \begin{aligned} =\lim _{x \rightarrow 0^{+}} \frac{1}{3} \cdot \frac{\sin \frac{x}{3}}{x}+\frac{1}{9} \cdot \cos \frac{x}{3}=\lim _{x \rightarrow 0^{+}} \frac{1}{3} \cdot \frac{\frac{x}{3}}{x}+\frac{1}{9} . \end{aligned}$
    • $\displaystyle \begin{aligned} =\frac{1}{9}+\frac{1}{9}=\frac{2}{9} \end{aligned}$
  • $\displaystyle \begin{aligned} f^{\prime \prime}\left(0^{-}\right)=\lim _{x \rightarrow 0^{-}} \frac{f^{\prime}(x)-f^{\prime}(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{-\sin x-x \cos x}{x-0}=-2 \end{aligned}$
  • $\displaystyle \begin{aligned} f^{\prime \prime}\left(0^{+}\right) \neq f^{\prime \prime}\left(0^{-}\right) \text {、则 } f^{\prime \prime}(0) \text { 不存在 } \end{aligned}$ 答 应选 C.

(4)

已知 $a_{n}<b_{n}(n=1,2, \cdots)$. 若级数 $\displaystyle \sum_{n=1}^{\infty} a_{n}$ 与 $\displaystyle \sum_{n=1}^{\infty} b_{n}$ 均收敛, 则 “ $\displaystyle \sum_{n=1}^{\infty} a_{n}$ 绝对收敛” 是 “ $\displaystyle \sum_{n=1}^{\infty} b_{n}$ 绝对收敛” 的 A. 充分必要条件. B. 充分不必要条件. C. 必要不充分条件. D. 既不充分也不必要条件.

• 笠应选 A.
• 正向级数
  • 比较法
    • 一般形式
    • 极限形式
  • 比值法/根值法
• 交错级数
  • 莱布尼兹法则
• 由级数 $\displaystyle \sum_{n=1}^{\infty} a_n$ 与 $\displaystyle \sum_{n=1}^{\infty} b_n$ 均收敛且 $a_n<b_n$
  • 可以推出 $\displaystyle \sum^{\infty}\left|a_n-b_n\right|$ 收敛.
• 若 $\displaystyle \sum_{n=1}^{\infty} b_n$ 绝对收敛，
  • 由 $\left|a_n\right|=\left|a_n-b_n+b_n\right| \xrightarrow[]{|a \pm b| \leqslant|a|+|b|}\leqslant\left|a_n-b_n\right|+\left|b_n\right|$,
    • $\displaystyle \begin{aligned} \xrightarrow[]{大的收敛，小的也收敛}\end{aligned}$可知 $\displaystyle \sum_{n=1}^{\infty} a_n$ 绝对收敛.
• 若 $\displaystyle \sum_{n-1}^{\infty} a_n$ 绝对收敛,
  • 由 $\left|b_n\right|=\left|b_n-a_n+a_n\right|\xrightarrow[]{|a \pm b| \leqslant|a|+|b|} \leqslant\left|b_n-a_n\right|+\left|a_n\right|$，
  • $\displaystyle \begin{aligned} \xrightarrow[]{大的收敛，小的也收敛}\end{aligned}$可知 $\displaystyle \sum_{n=1}^{\infty} b_n$ 绝对收敛.
• 故选 A.

(5)

已知 $n$ 阶矩阵 $\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}$ 满足 $\boldsymbol{A B C}=\boldsymbol{O}, \boldsymbol{E}$ 为 $n$ 阶单位矩阵. 记矩阵 $\displaystyle \left(\begin{array}{cc}\boldsymbol{O} & \boldsymbol{A} \\ \boldsymbol{B} & \boldsymbol{E}\end{array}\right),\left(\begin{array}{cc}\boldsymbol{A} & \boldsymbol{C} \\ \boldsymbol{O} & \boldsymbol{E}\end{array}\right),\left(\begin{array}{cc}\boldsymbol{E} & \boldsymbol{A B} \\ \boldsymbol{A B} & \boldsymbol{O}\end{array}\right)$ 的秩分别为 $r_{1}$, $r_{2}, r_{3}$, 则 A. $r_{1} \leqslant r_{2} \leqslant r_{3}$. B. $r_{1} \leqslant r_{3} \leqslant r_{2}$. C. $r_{3} \leqslant r_{1} \leqslant r_{2}$. D. $r_{2} \leqslant r_{1} \leqslant r_{3}$.

• 答应选 B.
• 分块矩阵的重要公式
  • (1) $r(A \cdot A B)=r(A), r\binom{A}{B A}=r(A)$
  • (2) $\displaystyle r\left(\begin{array}{ll}A & 0 \\ 0 & B\end{array}\right)=r(A)+r(B)=r\left(\begin{array}{cc}A & A B \\ 0 & B\end{array}\right)=r\left(\begin{array}{cc}A & 0 \\ B A & B\end{array}\right)$
  • (3) $\displaystyle r\left(\begin{array}{ll}A & C \\ 0 & B\end{array}\right) \geqslant r(A)+r(B)$
• 矩阵的秩:越乘越小，越拼越大，分开加最大
  $\displaystyle \begin{aligned} & r(A B) \leqslant \begin{array}{l}r(A) \\ r(B)\end{array} \leq r(A, B) \leqslant r(A)+r(B) \\ \end{aligned}$
  • 越乘越小
    • $r(A B) \leqslant r(A)$
    • $r(A B) \leqslant r(B)$
  • 越拼越大:$\displaystyle \begin{aligned}r(A) \xrightarrow[]{上下拼}\leq r\binom{A}{B} \xrightarrow[]{左右拼}\leq r\left(\begin{array}{ll}A & C \\ B & 0\end{array}\right)\end{aligned}$
• 舒尔公式:用E去消0
  • $\displaystyle \begin{aligned} \left(\begin{array}{cc} 0 & A \\ BC & E \end{array}\right) \xrightarrow{r_1 + (-A) r_2} \left(\begin{array}{cc} -ABC & 0 \\ BC & E \end{array}\right) \xrightarrow{c_1 + c_2 \cdot (-BC)} \left(\begin{array}{cc} -ABC & 0 \\ 0 & E \end{array}\right) \end{aligned}$
    • $\displaystyle \begin{aligned} r(ABC) = 0, \, r = n \end{aligned}$
  • $\displaystyle \begin{aligned} \left(\begin{array}{ll} AB & C \\ 0 & E \end{array}\right) \xrightarrow{r_1 + (-C) r_2} \left(\begin{array}{ll} AB & 0 \\ 0 & E \end{array}\right) \end{aligned}$
    • $\displaystyle \begin{aligned} r_2 = r(AB) + r(E) = r(AB) + n \end{aligned}$
  • $\displaystyle \begin{aligned} \left(\begin{array}{l} EAB \\ AB 0 \end{array}\right) \xrightarrow{r_2 + (-AB) r_1} \left(\begin{array}{c} EAB \\ 0 \end{array}\right) \xrightarrow{c_2 + (-AB) c_1} \left(\begin{array}{c} E \\ 0 \end{array}\right) \end{aligned}$
    • $\displaystyle \begin{aligned} r_3 = n + r(ABAB) \end{aligned}$
• $\displaystyle \begin{aligned} \xrightarrow[]{矩阵越乘越小}r_3 < r_2 < r_1 \end{aligned}$

(6)

下列矩阵中不能相似于对角矩阵的是 A. $\displaystyle \left(\begin{array}{lll}1 & 1 & a \\ 0 & 2 & 2 \\ 0 & 0 & 3\end{array}\right)$. B. $\displaystyle \left(\begin{array}{lll}1 & 1 & a \\ 1 & 2 & 0 \\ a & 0 & 3\end{array}\right)$. C. $\displaystyle \left(\begin{array}{lll}1 & 1 & a \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right)$. D. $\displaystyle \left(\begin{array}{lll}1 & 1 & a \\ 0 & 2 & 2 \\ 0 & 0 & 2\end{array}\right)$.

• 答 应选 D.

• n阶方阵A可相似对角化的充分必要条件

  • A有n个线性无关的特征向量。
  • 对于A的每个特征值 $\displaystyle \begin{aligned} \lambda_i \end{aligned}$，线性无关的特征向量个数等于 $\displaystyle \begin{aligned} \lambda_i \end{aligned}$ 的重数。

• n阶方阵A可相似对角化的充分但不必要条件
  相似对角化推不出来的条件

  • A有n个不同的特征值。
  • A为实对称矩阵。

• $\mathrm{A}$ 中矩阵的特征值不同, 分别为 $1,2,3$

• $\mathrm{B}$中矩阵为实对称矩阵;

• $\mathrm{C}$ 中矩阵的特征值 2 为二重根, 其对应的线性无关特征向量个数为 2

  • $\displaystyle \begin{aligned} (C-2 E) x=0 \quad \Rightarrow \quad S=n-r(C-2 E)=3-1=2 . \end{aligned}$
  • $\displaystyle |c-2 E|=\left|\begin{array}{lll}1 & 1 & a \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right|$

• D 中矩阵的特征值 2 为二重根, 其对应的线性无关特征向量个数为 1 , 不可相似对角化,故选 D.

  • $\displaystyle \begin{aligned} (D-2 E) x=0 \quad \Rightarrow \quad S=n-r(D-2 E)=3-2=1 \end{aligned}$
  • $\displaystyle |D-2 E|=\left|\begin{array}{lll}1 & 1 & 9 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right|$

(7)

已知向量 $\displaystyle \boldsymbol{\alpha}_{1}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \boldsymbol{\alpha}_{2}=\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right), \boldsymbol{\beta}_{1}=\left(\begin{array}{l}2 \\ 5 \\ 9\end{array}\right), \boldsymbol{\beta}_{2}=\left(\begin{array}{l}1 \\ 0 \\ 1\end{array}\right)$. 若 $\boldsymbol{\gamma}$ 既可由 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}$ 线性表示, 也可由 $\boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}$ 线性表示, 则 $\boldsymbol{\gamma}=$ A. $\displaystyle k\left(\begin{array}{l}3 \\ 3 \\ 4\end{array}\right), k \in \mathbf{R}$. B. $\displaystyle k\left(\begin{array}{c}3 \\ 5 \\ 10\end{array}\right), k \in \mathbf{R}$. C. $\displaystyle k\left(\begin{array}{c}-1 \\ 1 \\ 2\end{array}\right), k \in \mathbf{R}$. D. $\displaystyle k\left(\begin{array}{l}1 \\ 5 \\ 8\end{array}\right), k \in \mathbf{R}$.

• 答 应选 D.

代入法

• $\displaystyle \left(\alpha_1, \alpha_2, y\right)=\left(\begin{array}{ll:l}1 & 2 & 1 \\ 2 & 1 & 5 \\ 3 & 1 & 8\end{array}\right) \xrightarrow[r_3-3 r_1]{r_2-2 r_1}\left(\begin{array}{cc:c}1 & 2 & 1 \\ 0 & -3 & 3 \\ 0 & -5 & 5\end{array}\right) \longrightarrow\left(\begin{array}{cc:c}1 & 2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0\end{array}\right)$
  • $\gamma$可由 $\alpha_1, \alpha_2$ 表示
• $\displaystyle \left(\beta_2, \beta_1, \gamma\right)=\left(\begin{array}{ll:l}1 & 2 & 1 \\ 0 & 5 & 5 \\ 1 & 9 & 8\end{array}\right) \rightarrow\left(\begin{array}{ll:l}1 & 2 & 1 \\ 0 & 5 & 5 \\ 0 & 7 & 7\end{array}\right) \rightarrow\left(\begin{array}{ll:l}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0\end{array}\right)$
  • $\gamma$可以由$\beta_2, \beta_1,$表示

直接算

• 若 $\boldsymbol{\gamma}$ 既可由 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}$ 线性表示, 也可由 $\boldsymbol{\beta}_{1}, \boldsymbol{\beta}_{2}$ 线性表示,
  • $\displaystyle \begin{aligned} & \exists x_1, x_2, x_3, x_4,使得 \gamma=x_1 \alpha_1+x_2 \alpha_2=x_3 \beta_1+x_4 \beta_2 \end{aligned}$
    • $\displaystyle \begin{aligned} \xrightarrow[]{移项整理}x_1\alpha_1+x_2 \alpha_2-x_3 \beta_1-x_4 \beta_2=0\end{aligned}$
    • $\displaystyle \begin{aligned}\xrightarrow[]{提出x} & \left(\alpha_1, \alpha_2,-\beta_1,-\beta_2\right) x=0 \quad \text { 有非0解 }\end{aligned}$
• $\displaystyle \left(\begin{array}{cccc}1 & 2 & -2 & -1 \\ 2 & 1 & -5 & 0 \\ 3 & 1 & -9 & -1\end{array}\right) \xrightarrow[]{\begin{aligned}& r_2-2 r_1 \\& r_3-3 r_1\end{aligned}}\left(\begin{array}{cccc}1 & 2 & -2 & -1 \\ 0 & -3 & -1 & 2 \\ 0 & -5 & -3 & 2\end{array}\right) \xrightarrow[r_3 \cdot\left(-\frac{3}{4}\right)]{\left\{\begin{array}{l} r_3-\frac{5}{3} r_2\\r_2 \cdot(-1) \\ \end{array}\right.}\left(\begin{array}{cccc}1 & 2 & -2 & -1 \\ 0 & 3 & 1 & -2 \\ 0 & 0 & 1 & 1\end{array}\right)$
  • $\displaystyle \left\{\begin{aligned}\xrightarrow[]{化为方程组求通解} x_1+2 x_2-2 x_3-x_4 & =0 \\ 3 x_2+x_3-2 x_4 & =0 \\ x_3+x_4 & =0\end{aligned} \right.$
    • $\displaystyle \xrightarrow[]{x_4为自由元}\left\{\begin{array}{l}x_1=-3 k \\ x_2=k \\ x_3=-k \\ x_4=k\end{array} \Rightarrow x=\left(\begin{array}{l}x_1 \\ x_2 \\ x_3 \\ x_4\end{array}\right)=\left(\begin{array}{c}-3 k \\ k \\ -k \\ k\end{array}\right) k \in R\right.$
• 将$\binom{x_1}{x_2}=\binom{-3 k}{k}$代回入$y=x_1 \alpha_1+x_2 \alpha_2$
  • $\displaystyle y=-3 k\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)+k\left(\begin{array}{l}2 \\ 1 \\ 1\end{array}\right)=\left(\begin{array}{c}-k \\ -5 k \\ -8 k\end{array}\right)=k\left(\begin{array}{l}-1 \\ -5 \\ -8\end{array}\right)=k\left(\begin{array}{l}1 \\ 5 \\ 8\end{array}\right)$

(8)

设随机变量 $X$ 服从参数为 1 的泊松分布, 则 $E(|X-E X|)=$ A. $\frac{1}{\mathrm{e}}$. B. $\frac{1}{2}$. C. $\frac{2}{\mathrm{e}}$. D. 1.

• 答 应选 C.

• 泊松分布的分布律与数字特征 若随机变量 $X$ 服从参数为 $\lambda$ 的泊松分布, $X \sim P(\lambda)$

  • 分布律$\displaystyle \begin{aligned}P\{X=k\}=\frac{\lambda^k \mathrm{e}^{-\lambda}}{k!}, k=0,1,2, \cdots\end{aligned}$, 其中 $\lambda>0$ 是常数,
  • 期望 $E(X)=\lambda$
  • 方差$D(X)=\lambda$

• $E(|X-E X|)\xlongequal[泊松分布]{EX=1}E|X-1|\xrightarrow[]{去掉绝对值}$

  • $\displaystyle \begin{aligned} E|X-1| \xlongequal[颗粒感:函数的期望]{P\{X=k\}=\frac{\lambda^k \mathrm{e}^{-\lambda}}{k!}}\sum_{k=0}^{\infty} |k-1| \cdot \frac{e^{-1}}{k!} \end{aligned}$
    • $\displaystyle |x-1|= \begin{cases}1-x, & x=0 \\ x-1 & x \geqslant 1\end{cases}$
  • $\displaystyle \begin{aligned} \xlongequal[]{把绝对值的0单独拆出来}\frac{e^{-1}}{0!} + \underbrace{\sum_{k=1}^{\infty} (k-1) \cdot \frac{e^{-1}}{k!}}_{计算方法:拆+期望}\end{aligned}$
  • $\displaystyle \begin{aligned} = \frac{e^{-1}}{0!} + \underbrace{\sum_{k=0}^{\infty} (k-1) \cdot \frac{e^{-1}}{k!} - (0-1) \cdot \frac{e^{-1}}{0!}}_{\sum_{k=1}^{\infty} }\end{aligned}$
  • $\displaystyle \begin{aligned} = e^{-1} + E(X-1) + e^{-1} \xlongequal[]{E(X-1)=E X-1=0}2e^{-1} \end{aligned}$

• 故选 C.

(9)

设 $X_{1}, X_{2}, \cdots, X_{n}$ 为来自总体 $\displaystyle N\left(\mu_{1}, \sigma^{2}\right)$ 的简单随机样本, $Y_{1}, Y_{2}, \cdots, Y_{m}$ 为来自总体 $\displaystyle N\left(\mu_{2}, 2 \sigma^{2}\right)$ 的简单随机样本,且两样本相互独立, 记 $\displaystyle \bar{X}=\frac{1}{n} \sum_{i=1}^{n} X_{i}, \bar{Y}=\frac{1}{m} \sum_{i=1}^{m} Y_{i}, S_{1}^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}, S_{2}^{2}=\frac{1}{m-1} \sum_{i=1}^{m}\left(Y_{i}-\bar{Y}\right)^{2}$, 则 A. $\frac{S_{1}^{2}}{S_{2}^{2}} \sim F(n, m)$. B. $\frac{S_{1}^{2}}{S_{2}^{2}} \sim F(n-1, m-1)$. C. $\frac{2 S_{1}^{2}}{S_{2}^{2}} \sim F(n, m)$. D. $\frac{2 S_{1}^{2}}{S_{2}^{2}} \sim F(n-1, m-1)$.

• 答 应选 D.

• 正态总体的抽样分布定理：

  • 设总体 $\displaystyle \begin{aligned} X \sim N(\mu, \sigma^2) \end{aligned}$。
    $\displaystyle \begin{aligned} X_1, X_2, \ldots, X_n \end{aligned}$ 是来自 $\displaystyle \begin{aligned} X \end{aligned}$ 的样本，
    样本均值 $\displaystyle \begin{aligned} \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \end{aligned}$，
    样本方差 $\displaystyle \begin{aligned} S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2 \end{aligned}$，则：
    • $\displaystyle \begin{aligned} \bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right) \end{aligned}$:属于正态分布
    • $\displaystyle \begin{aligned} \frac{(n-1) S^2}{\sigma^2} \sim \chi^2(n-1) \end{aligned}$:样本方差属于卡方分布
    • $\displaystyle \begin{aligned} \bar{X} \end{aligned}$ 与 $\displaystyle \begin{aligned} S^2 \end{aligned}$ 相互独立:样本均值与样本方差独立

• F分布：

  • 设 $\displaystyle \begin{aligned} U \sim \chi^2(n_1) \end{aligned}$，$\displaystyle \begin{aligned} V \sim \chi^2(n_2) \end{aligned}$，且 $\displaystyle \begin{aligned} U, V \end{aligned}$ 相互独立，
    则称随机变量 $\displaystyle \begin{aligned} F = \frac{U/n_1}{V/n_2} \end{aligned}$ 服从自由度为 $\displaystyle \begin{aligned} (n_1, n_2) \end{aligned}$ 的 F 分布，记为 $\displaystyle \begin{aligned} F \sim F(n_1, n_2) \end{aligned}$。

• $X_{1}, X_{2}, \cdots, X_{n}$ 为来自总体 $\displaystyle N\left(\mu_{1}, \sigma^{2}\right)$ 的简单随机样本

• $Y_{1}, Y_{2}, \cdots, Y_{m}$ 为来自总体 $\displaystyle N\left(\mu_{2}, 2 \sigma^{2}\right)$ 的简单随机样本

• 由卡方分布构造F分布

  • 样本方差构造卡方分布
    • $\displaystyle \begin{aligned} \frac{(n-1) S_{1}^{2}}{\sigma^2} \sim \chi^2(n-1) \end{aligned}$
    • $\displaystyle \begin{aligned} \frac{(m-1) S_{2}^{2}}{2\sigma^2} \sim \chi^2(m-1) \end{aligned}$
  • 卡方分布构造F分布
    • $\displaystyle \begin{aligned} F = \frac{U/n_1}{V/n_2} = \frac{(n-1) S_{1}^{2}/(n-1)}{(m-1) S_{2}^{2}/(m-1)} = \frac{2 S_{1}^{2}}{S_{2}^{2}} = F(n-1, m-1) \end{aligned}$

故选 D.

(10)

设 $X_{1}, X_{2}$ 为来自总体 $\displaystyle N\left(\mu, \sigma^{2}\right)$ 的简单随机样本, 其中 $\displaystyle \sigma(\sigma>0)$ 是末知参数. 若 $\displaystyle \hat{\sigma}=a\left|X_{1}-X_{2}\right|$ 为 $\displaystyle \sigma$ 的无偏估计,则 $a=$ A. $\frac{\sqrt{\pi}}{2}$. B. $\frac{\sqrt{2 \pi}}{2}$. C. $\sqrt{\pi}$. D. $\sqrt{2 \pi}$.

• 一维正态分布

  • (1) 服从正态分布 $\displaystyle N\left(\mu, \sigma^2\right)$ 的随机变量 $X$ 的概率密度为
    $\displaystyle f(x)=\frac{1}{\sqrt{2 \pi} \sigma} \mathrm{e}^{-\frac{(x-\mu)^2}{2 \sigma^2}}, \quad-\infty<x<+\infty .$
  • (2) 若 $\displaystyle X_1 \sim N\left(\mu_1, \sigma_1^2\right), \quad X_2 \sim N\left(\mu_2, \sigma_2^2\right)$, 且 $X_1, X_2$ 相互独立,
    则对于任意的实数 $a_1, a_2$，有$\displaystyle a_1 X_1+a_2 X_2 \sim N\left(a_1 \mu_1+a_2 \mu_2, a_1^2 \sigma_1^2+a_2^2 \sigma_2^2\right) .$

• 无偏估计

  • 设 $\hat{\theta}$ 是 $\theta$ 的估计量, 若 $E(\hat{\theta})=\theta$, 则称 $\hat{\theta}$ 为 $\theta$ 的无偏估计量.
    • 估计量的数学期望等于参数本身

• $\displaystyle \begin{aligned} X \sim N\left(\mu, \sigma_{1}^{2}\right), \quad Y \sim N\left(\mu_{1}, \sigma_{2}^{2}\right) \quad \text{、} X \text{与} Y \text{独立} \end{aligned}$

  • $\displaystyle \begin{aligned} \xrightarrow[]{线性组合}aX + bY \sim N\left(a\mu_{1} + b\mu_{2}, a^{2}\sigma_{1}^{2} + b^{2}\sigma_{2}^{2}\right) \end{aligned}$

• 本题

  • $\displaystyle \begin{aligned} Y = X_{1} - X_{2} \sim N\left(0, 2\sigma^{2}\right) \end{aligned}$
  • 求正态的概率密度$\displaystyle \begin{aligned} f_{Y(y)} = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}} \xrightarrow{\mu=0, \sigma=\sqrt{2}\sigma} \frac{1}{2\sqrt{2}\sigma} e^{\frac{-y^{2}}{4\sigma^{2}}} \end{aligned}$

• 用定义求期望:
  $\displaystyle \begin{aligned} E|Y| \xlongequal[何处算积分]{何处求概率} \int_{-\infty}^{+\infty} |y| f_{Y(y)} dy \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[]{代入f(y)} \int_{-\infty}^{+\infty} |y| \cdot \frac{1}{2\sqrt{2}\sigma} \cdot e^{\frac{-y^{2}}{4\sigma^{2}}} dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{去绝对值} 2\int_{0}^{+\infty} y \cdot \frac{1}{2\sqrt{2}\sigma} \cdot e^{\frac{-y^{2}}{4\sigma^{2}}} dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{凑e的指数的微分的系数}2\int_{0}^{+\infty} \frac{2y}{4\sigma^{2}} \cdot \frac{\sigma}{\sqrt{\pi}} \cdot e^{\frac{-y^{2}}{4\sigma^{2}}} dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{凑微分}2\int_{0}^{+\infty} \frac{\sigma}{\sqrt{\pi}} \cdot e^{\frac{-y^{2}}{4\sigma^{2}}} d\left(\frac{y^{2}}{4\sigma^{2}}\right) \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\int e^{-x} d x=-e^{-x}} -2 \cdot \frac{\sigma}{\sqrt{\pi}} \cdot e^{\frac{-y^{2}}{4\sigma^{2}}} \Big|_{0}^{+\infty} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{e^{-\infty}=0, e^0=1} -2 \cdot \frac{\sigma}{\sqrt{\pi}} \cdot (0 - 1) \xlongequal[]{负负得正}\frac{2\sigma}{\sqrt{\pi}} \end{aligned}$

• $\displaystyle \begin{aligned} a \cdot E|X_1 - X_2| = a \cdot E|Y| = a \cdot \frac{2\sigma}{\sqrt{\pi}} = 6 \end{aligned}$

  • $\displaystyle \begin{aligned} a = \frac{\sqrt{\pi}}{2} \end{aligned}$

答应选 A. 解 $\displaystyle E\left(a\left|X_1-X_2\right|\right)=a E\left(\left|X_1-X_2\right|\right)=a \cdot \frac{2 \sigma}{\sqrt{\pi}}=\sigma$, 得 $a=\frac{\sqrt{\pi}}{2}$, 其中, $\displaystyle X_1-X_2 \sim N\left(0,2 \sigma^2\right)$, 令 $Z=X_1-X_2$, 有

• $\displaystyle \begin{aligned} E\left(\left|X_1-X_2\right|\right) & =\int_{-\infty}^{+\infty}|z| \cdot \frac{1}{\sqrt{2 \pi} \sqrt{2} \sigma} \cdot \mathrm{e}^{-\frac{z^2}{\omega^2}} \mathrm{~d} z=2 \int_0^{+\infty} \frac{z}{2 \sqrt{\pi} \sigma} \mathrm{e}^{-\frac{x^2}{\omega^2}} \mathrm{~d} z \end{aligned}$
• $\displaystyle \begin{aligned} =2 \frac{1}{2 \sqrt{\pi} \sigma}\left(-2 \sigma^2\right) \int_0^{+\infty} \mathrm{e}^{-\frac{x^2}{4^2}} \mathrm{~d}\left(-\frac{z^2}{4 \sigma^2}\right)=-\left.\frac{2 \sigma}{\sqrt{\pi}} \mathrm{e}^{-\frac{z^2}{\omega^2}}\right|_0 ^{+\infty}=\frac{2 \sigma}{\sqrt{\pi}} . \end{aligned}$

故选 A. 二、填空题: $11 \sim 16$ 小题, 每小题 5 分, 共 30 分.

(11)

当 $x \rightarrow 0$ 时, 函数 $f(x)=a x+b x^{2}+\ln (1+x)$ 与 $g(x)=\mathrm{e}^{x^{2}}-\cos x$ 是等价无穷小, 则 $a b=$

答 应填 -2 .

• 整体思想：$\displaystyle \lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=1 \Rightarrow a \cdot b$
• 由$f(x)=a x+b x^{2}$，可以考虑将ln,e,cos都展开到二阶
  • $\displaystyle \begin{aligned} f(x)=a x+b x^2+x-\frac{1}{2} x^2+o\left(x^2\right)=(a+1) x+\left(b-\frac{1}{2}\right) x^2+o\left(x^2\right) . \end{aligned}$
  • $\displaystyle \begin{aligned} g(x)=1+x^2+o\left(x^2\right)-\left[1-\frac{1}{2} x^2+o\left(x^2\right)\right]=\frac{3}{2} x^2+o\left(x^2\right) \end{aligned}$
• $\displaystyle \begin{aligned} 1=\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{(a+1) x+\left(b-\frac{1}{2}\right) x^2+o\left(x^2\right)}{\frac{3}{2} x^2+o\left(x^2\right)} \end{aligned}$
• $\displaystyle \begin{aligned} \Rightarrow a+1=0, \quad b-\frac{1}{2}=\frac{3}{2} \Rightarrow a=-1, b=2 . \end{aligned}$

• (12)

曲面 $z=x+2 y+\ln \left(1+x^{2}+y^{2}\right)$ 在点 $(0,0,0)$ 处的切平面方程为

• 答 应填 $x+2 y-z=0$.
• 切平面方程的公式：$\left.\frac{\partial F}{\partial x}\right|_{\left(x_0, y_0, z_0\right)} \cdot\left(x-x_0\right)+\left.\frac{\partial F}{\partial y}\right|_{\left(x_0, y_0, z_0\right)} \cdot\left(y-y_0\right)+\left.\frac{\partial F}{\partial z}\right|_{\left(x_0, y_0, z_0\right)} \cdot\left(z-z_0\right)=0$.
  • 若曲面的方程为 $F(x, y, z)=0$,
  • 则该曲面在点 $P\left(x_0, y_0, z_0\right)$ 的一个法向量为 $\left.\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)\right|_{\left(x_0, y_0, z_0\right)}$.
  • 由于曲面在一点的切平面与曲面在该点的法向量垂直, 故曲面在点 $P$ 的切平面方程为
    $\left.\left(x-x_0, y-y_0, z-z_0\right) \cdot\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)\right|_{\left(x_0, y_0, z_0\right)}=0$
• 令 $F(x, y, z)=x+2 y+\ln \left(1+x^2+y^2\right)-z$, 有
  • $\left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)=\left(1+\frac{2 x}{1+x^2+y^2}, 2+\frac{2 y}{1+x^2+y^2},-1\right)$
  • $(0,0,0)$ 处的法向量为 $(1,2,-1)$
• $1 \cdot(x-0)+2(y-0)+(-1)(z-0)=0$
• 切平面方程： $x+2 y-z=0$.

(13)

设 $f(x)$ 是周期为 2 的周期函数, 且 $f(x)=1-x, x \in[0,1]$. 若 $\displaystyle f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty} a_{n} \cos n \pi x$,则 $\displaystyle \sum_{n=1}^{\infty} a_{2 n}=$

• 答 应填 0 .
• 正弦级数与余弦级数 ：设 $f(x)$ 是周期为 $2 l$ 的周期函数. 记 $C=\left\{x \left\lvert\, f(x)=\frac{1}{2}\left[f\left(x^{-}\right)+f\left(x^{+}\right)\right]\right.\right\}$.
  • 当 $f(x)$ 为奇函数时, $f(x)$ 的傅里叶级数是只含正弦项的正弦级数
    • $\displaystyle f(x)=\sum_{n=1}^{\infty} b_n \sin \frac{n \pi x}{l} \quad(x \in C)$
    • 其中 $\displaystyle b_n=\frac{2}{l} \int_0^l f(x) \sin \frac{n \pi x}{l} \mathrm{~d} x \quad(n=1,2, \cdots)$.
  • 当 $f(x)$ 为偶函数时, $f(x)$ 的傅里叶级数是只含常数项和余弦项的余弦级数
    • $\displaystyle f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos \frac{n \pi x}{l} \quad(x \in C),$
    • 其中 $\displaystyle a_n=\frac{2}{l} \int_0^l f(x) \cos \frac{n \pi x}{l} \mathrm{~d} x \quad(n=0,1,2, \cdots)$.
• $\displaystyle \begin{aligned} &\text { 当 } n \geqslant 1 \text { 时} \end{aligned}$
  • $\displaystyle \begin{aligned}a_n & =2 \int_0^1(1-x) \cos n \pi x d x\xlongequal[]{分布}2\left[\int_0^1 \cos n \pi x d x-\int_0^1 x \cos n \pi x d x\right]\end{aligned}$
    • $\displaystyle \begin{aligned} =2\left[\underbrace{\left.\frac{1}{n \pi} \sin n \pi x\right|_0 ^1}_{0-0=0}-\underbrace{\frac{1}{n \pi} \int_0^1 x d(\sin n \pi x)}_{凑微分}\right] \end{aligned}$
  • $\displaystyle \begin{aligned} =-\frac{2}{n \pi} \int_0^1 x d(\sin n \pi x)\xlongequal[]{分布}-\frac{2}{n \pi}\left(\underbrace{\left.x \sin n \pi x\right|_0 ^1}_{0-0=0}-\int_0^1 \sin n \pi x d x\right) \end{aligned}$
    • $\displaystyle \begin{aligned} =\frac{2}{n \pi} \int_0^1 \sin n \pi x d x\xlongequal[]{凑微分}\left.\frac{2}{n \pi} \cdot\left(-\frac{\cos n \pi x}{n \pi}\right)\right|_0 ^1 \end{aligned}$
  • $\displaystyle \begin{aligned} =-\frac{2}{n^2 \pi^2}(\cos n \pi-1) . \end{aligned}$
• $\displaystyle \begin{aligned}\xrightarrow[]{将n换成2n} a_{2 n} & =-\frac{2}{(2 n)^2 \pi^2}(\cos 2 n \pi-1)=0 \xrightarrow[]{\cos 2 n \pi=1}\sum_{n=1}^{\infty} a_{2 n}=0 \end{aligned}$
• 类似题目
  • 将函数 $f(x)=2+|x|(-1 \leq x \leq 1)$ 展开成以 2 为周期的傅里叶级数，并由此求级数 $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}$ 的和. (1991 年数学一试题)
  • 设函数 $f(x)=\pi x+x^2(-\pi<x<\pi)$ 的傅里叶级数展开式为 $\displaystyle \frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n \cos n x+b_n \sin n x\right)$, 则其中系数 $b_3$ 的值为 $\qquad$ (1993 年数学一试题)
  • 将函数 $f(x)=x-1(0 \leq x \leq 2)$ 展开成周期为 4 的余弦级数。(1995 年数学一试题)
  • 设 $\displaystyle x^2=\sum_{n=0}^{\infty} a_n \cos n x(-\pi \leq x \leq \pi)$, 则 $a_2=$ $\qquad$ (2003 年数学一试题)
  • 将函数 $f(x)=1-x^2(0 \leq x \leq \pi)$ 展开成余弦级数, 并 $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2}$ 的和.(2008 年数学一试题)

(14)

设连续函数 $f(x)$ 满足: $\displaystyle f(x+2)-f(x)=x, \int_{0}^{2} f(x) \mathrm{d} x=0$, 则 $\displaystyle \int_{1}^{3} f(x) \mathrm{d} x=$

• 答 应填 $\frac{1}{2}$. 解
• $\displaystyle \begin{aligned} \int_1^3 f(x) d x & =\int_0^3 f(x) d x-\int_0^1 f(x) d x \end{aligned}$
• $\displaystyle \begin{aligned} \xlongequal[]{\int_0^3 \rightarrow \int_0^2+\int_2^3}\int_0^2 f(x) d x+\int_2^3 f(x) d x-\int_0^1 f(x) d x \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\int_{0}^{2} f(x) \mathrm{d} x=0}\int_2^3 f(x) d x-\int_0^1 f(x) d x \end{aligned}$
• $\displaystyle \begin{aligned} \xlongequal[]{\int_0^1 f(x+2) d x\xlongequal[]{t=x+2}\int_2^3 f(x) d x }\int_0^1 f(x+2) d x-\int_0^1 f(x) d x\xlongequal[]{f(x+2)-f(x)=x}\int_0^1 x d x \xlongequal[]{牛莱}\left.\frac{1}{2} x^2\right|_0 ^1=\frac{1}{2} . \end{aligned}$

(15)

已知向量 $\displaystyle \boldsymbol{\alpha}_{1}=\left(\begin{array}{l}1 \\ 0 \\ 1 \\ 1\end{array}\right), \boldsymbol{\alpha}_{2}=\left(\begin{array}{c}-1 \\ -1 \\ 0 \\ 1\end{array}\right), \boldsymbol{\alpha}_{3}=\left(\begin{array}{c}0 \\ 1 \\ -1 \\ 1\end{array}\right), \boldsymbol{\beta}=\left(\begin{array}{c}1 \\ 1 \\ 1 \\ -1\end{array}\right), \boldsymbol{\gamma}=k_{1} \boldsymbol{\alpha}_{1}+k_{2} \boldsymbol{\alpha}_{2}+k_{3} \boldsymbol{\alpha}_{3}$. 若 $\boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_{i}=\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_{i}(i=1,2,3)$,则 $k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=$

• 答 应填 $\frac{11}{9}$.

• 向量内积的概念

  • 设有 $n$ 维向量$\displaystyle \boldsymbol{x}=\left(\begin{array}{c}x_1 \\x_2 \\\vdots \\x_n\end{array}\right), \quad \boldsymbol{y}=\left(\begin{array}{c}y_1 \\y_2 \\\vdots \\y_n\end{array}\right) \text {, }$
  • 令$(\boldsymbol{x}, \boldsymbol{y})=x_1 y_1+x_2 y_2+\cdots+x_n y_n,$则$(\boldsymbol{x}, \boldsymbol{y})$ 称为向量 $\boldsymbol{x}$ 与 $\boldsymbol{y}$ 的内积.
    • $(\boldsymbol{x}, \boldsymbol{y})=\boldsymbol{x}^{\mathrm{T}} \boldsymbol{y}=\boldsymbol{y}^{\mathrm{T}} \boldsymbol{x},$
      • 当 $(\boldsymbol{x}, \boldsymbol{y})=0$ 时,称向量 $\boldsymbol{x}$ 与 $\boldsymbol{y}$ 正交.
    • $(\boldsymbol{x}, \boldsymbol{x})=\|\boldsymbol{x}\|^2$,即 $\boldsymbol{x}$ 的模长的平方.

• 解：注意到 $\boldsymbol{\alpha}_1^{\mathrm{T}} \boldsymbol{\alpha}_2=\boldsymbol{\alpha}_2^{\mathrm{T}} \boldsymbol{\alpha}_3=\boldsymbol{\alpha}_3^{\mathrm{T}} \boldsymbol{\alpha}_1=0$, 故 $\boldsymbol{\alpha}_1, \boldsymbol{\alpha}_2, \boldsymbol{\alpha}_3$ 相互正交, 从而

• 由$\boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_{i}=\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_{i}(i=1,2,3)$

  • 计算$\boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_{i}$
    • $\displaystyle \begin{aligned} & \boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_1=\left(k_1 \boldsymbol{\alpha}_1+k_2 \boldsymbol{\alpha}_2+k_3 \boldsymbol{\alpha}_3\right)^{\mathrm{T}} \boldsymbol{\alpha}_1=k_1 \alpha_1^{\top} \alpha_1\underbrace{+k_2 \alpha_2^{\top} \alpha_1+k_3 \alpha_3^{\top} \alpha_1}_{\boldsymbol{\alpha}_1^{\mathrm{T}} \boldsymbol{\alpha}_2=\boldsymbol{\alpha}_2^{\mathrm{T}} \boldsymbol{\alpha}_3=\boldsymbol{\alpha}_3^{\mathrm{T}} \boldsymbol{\alpha}_1=0}=k_1\left\|\boldsymbol{\alpha}_1\right\|^2 \end{aligned}$
    • $\displaystyle \begin{aligned} \boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_2=\left(k_1 \boldsymbol{\alpha}_1+k_2 \boldsymbol{\alpha}_2+k_3 \boldsymbol{\alpha}_3\right)^{\mathrm{T}} \boldsymbol{\alpha}_2=k_2\left\|\boldsymbol{\alpha}_2\right\|^2 \end{aligned}$
    • $\displaystyle \begin{aligned} \boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_3=\left(k_1 \boldsymbol{\alpha}_1+k_2 \boldsymbol{\alpha}_2+k_3 \boldsymbol{\alpha}_3\right)^{\mathrm{T}} \boldsymbol{\alpha}_3=k_3\left\|\boldsymbol{\alpha}_3\right\|^2\end{aligned}$
  • 计算$\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}$
    • $\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_1=1, \quad \boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_2=-3, \quad \boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_3=-1$
  • 由$\boldsymbol{\gamma}^{\mathrm{T}} \boldsymbol{\alpha}_{i}=\boldsymbol{\beta}^{\mathrm{T}} \boldsymbol{\alpha}_{i}(i=1,2,3)$
    • $k_1\left\|\alpha_1\right\|^2=1， \quad k_2\left\|\alpha_2\right\|^2=-3， \quad k_3\left\|\alpha_3\right\|^2=-1$
      • 即$3 k_1=1, \quad 3 k_2=-3, \quad 3 k_3=-1 .$
      • $k_1=\frac{1}{3}, k_2=-1, k_3=-\frac{1}{3}$.

• 因此, $k_1^2+k_2^2+k_3^2=\frac{1}{9}+1+\frac{1}{9}=\frac{11}{9}$.

(16)

设随机变量 $X$ 与 $Y$ 相互独立, 且 $X \sim B\left(1, \frac{1}{3}\right), Y \sim B\left(2, \frac{1}{2}\right)$, 则 $P\{X=Y\}=$

• 答 应填 $\frac{1}{3}$.
• $C_n^k=\frac{n!}{k!(n-k)!}$，比如：$C_5^3=\frac{5!}{3!\cdot 2!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \cdot(2 \times 1)}=10$
• $\displaystyle \begin{array}{l|l|l}x & 0 & 1 \\\hline p & \frac{2}{3} & \frac{1}{3}\end{array}$，$\displaystyle \begin{array}{c|c|c|c}Y & 0 & 1 & 2 \\\hline p & \frac{1}{4} & \frac{1}{2} & \frac{1}{4}\end{array}$
  • $\displaystyle \begin{aligned} P\{X=Y\} \xlongequal[]{分析X=Y}P\{X=0, Y=0\}+P\{X=1, Y=1\} \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{独立性}P\{X=0\} P\{Y=0\}+P\{X=1\} P\{Y=1\} \end{aligned}$
    • $\displaystyle \begin{aligned} \begin{array}{lll}X \sim B\left(1, \frac{1}{3}\right) & P\{X=0\}=C_1^0\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^1=\frac{2}{3} & P\{X=1\}=C_1^1\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^0=\frac{1}{3} \\ Y \sim B\left(2, \frac{1}{2}\right) & P\{Y=0\}=C_2^0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^2=\frac{1}{4} & P\{Y=1\}=C_2^1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^1=\frac{1}{2}\end{array}\end{aligned}$
  • $\displaystyle =\begin{aligned} \frac{2}{3} \times \frac{1}{4}+\frac{1}{3} \times \frac{1}{2} \end{aligned}\begin{aligned} =\frac{1}{6}+\frac{1}{6}=\frac{1}{3} . \end{aligned}$

三、解答题: $17 \sim 22$ 小题, 共 70 分. 解答应写出文字说明、证明过程或演算步聚.

(17)

(本题满分 10 分)

设曲线 $y=y(x)(x>0)$ 经过点 $(1,2)$, 该曲线上任一点 $P(x, y)$ 到 $y$ 轴的距离等于该点处的切线在 $y$ 轴上的截距.

(1) 求 $y(x)$;

(2) 求函数 $\displaystyle f(x)=\int_{1}^{x} y(t) \mathrm{d} t$ 在 $(0,+\infty)$ 上的最大值.

• • $\displaystyle \begin{aligned} & P(x, y) \text { 处的切线方程 } y-y_0=y^{\prime}\left(x_0\right)\left(x-x_0\right) \end{aligned}$

• 该曲线上任一点 $P(x, y)$ 到 $y$ 轴的距离等于该点处的切线在 $y$ 轴上的截距

  • 求截距:$\displaystyle \begin{aligned} 令 x=0 \text { 、得 }|O A|=y=y_0-y^{\prime}\left(x_0\right) x_0 \end{aligned}$
  • 该点处的切线在 $y$ 轴上的截距:$\displaystyle \begin{aligned} |P H|=x_0 \end{aligned}$

• $\displaystyle \begin{aligned} |O A|=|P H| \end{aligned}$，得$x_0=y_0-y^{\prime}\left(x_0\right) x_0$

  • $\displaystyle \begin{aligned} \xrightarrow[]{整理}y-x y^{\prime}=x \xrightarrow{\text { 同除 } x} y^{\prime}-\frac{1}{x} y=-1\end{aligned}$
    • $\displaystyle \begin{aligned} \xrightarrow[]{ p=-\frac{1}{x}, q=- 1}y=e^{\int \frac{1}{x} d x}\left[\int(-1) e^{-\int \frac{1}{x} d x} d x+C\right] \end{aligned}$
    • $\displaystyle \begin{aligned} =e^{\ln x}\left[\int(-1) e^{-\ln x} d x+C\right] \end{aligned}$
    • $\displaystyle \begin{aligned} =x\left[\int(-1) \frac{1}{x} d x+C\right] \end{aligned}$
    • $\displaystyle \begin{aligned} =x[-\ln x+C] \end{aligned}$

• $\displaystyle \begin{aligned} y(1)=2\end{aligned}$，代入得$2=1 \cdot[-\ln 1+C] \text { ，得 } C=2$

  • 从而y的方程:$\displaystyle \begin{aligned} y=x(2-\ln x) \end{aligned}$

• (2) 求函数 $\displaystyle f(x)=\int_{1}^{x} y(t) \mathrm{d} t$ 在 $(0,+\infty)$ 上的最大值.

• $\displaystyle \begin{aligned} f(x)=\int_1^x y(t) d t=\int_1^x t \cdot(2-\ln t) d t \end{aligned}$

• 一阶导求极值:$\displaystyle \begin{aligned} f^{\prime}(x)=x \cdot(2-\ln x)=0 \end{aligned}$，得 $x=0 \text { 或 } x=e^2$

  • $\displaystyle \begin{aligned} \text { 因 } x>0 \text {, 则驻点 } x=e^2 \end{aligned}$
  • 通过增减性判断是否是最值
    • $\displaystyle \begin{aligned} & \text { 当 } 0<x<e^2, f^{\prime}(x)>0,↗ \end{aligned}$
    • $\displaystyle \begin{aligned} \text { 当 } x>e^2, f^{\prime}(x)<0, \text { ↘ } \end{aligned}$
    • 则 $\displaystyle \begin{aligned} x = e^2 \end{aligned}$ 为 $\displaystyle \begin{aligned} f(x) \end{aligned}$ 在 $\displaystyle \begin{aligned} (0, +\infty) \end{aligned}$ 上的最大值点。

• $\displaystyle \begin{aligned} \xrightarrow[]{x = e^2 }f_{\text {max }}=f\left(e^2\right) & =\int_1^{e^2} t(2-\ln t) d t \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[]{拆开}\int_1^{e^2} 2 t d t-\int_1^{e^2} t \ln t d t \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{凑微分+分布}\left.t^2\right|_1 ^{e^2}-\frac{1}{2} \int_1^{e^2} \ln t d\left(t^2\right) \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{脆弱坚强}e^4-1-\frac{1}{2}\left[\left.t^2 \cdot \ln t\right|_1 ^{e^2}-\int_1^{e^2} t^2 \cdot d(\ln t)\right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{(\ln t)^{\prime}=\frac{1}{t}}e^4-1-\frac{1}{2}\left[2 e^4-0-\int_1^{e^2} t^2 \cdot \frac{1}{t} d t\right] \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{牛莱}e^4-e^4-1+\left.\frac{1}{2} \cdot \frac{t^2}{2}\right|_1 ^{e^2} \end{aligned}$
  • $\displaystyle \begin{aligned} =-1+\frac{1}{2}\left(\frac{e^4}{2}-\frac{1}{2}\right)=\frac{e^4}{4}-\frac{5}{4} \end{aligned}$

(18)

(本题满分 12 分)

求函数 $f(x, y)=\left(y-x^{2}\right)\left(y-x^{3}\right)$ 的极值.

• • 求条件极值的方法
    • ① $\displaystyle \begin{aligned} f_x = f_y = 0 \end{aligned}$
    • ② $\displaystyle \begin{aligned} A = f_{xx}, B = f_{xy}, C = f_{yy} \end{aligned}$
    • $\displaystyle \begin{aligned} AC - B^2 \begin{cases} > 0 & \begin{cases} A > 0: \text{极小} \\ A < 0: \text{极大} \end{cases} \\ < 0: \text{非极值} \\ = 0: \text{进一步讨论} \end{cases} \end{aligned}$

• $\displaystyle \begin{aligned} f(x, y) = (y - x^2)(y - x^3) \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[]{乘法分配} y^2 - x^3y - x^2y + x^5 \end{aligned}$

• 求可能的极值点，也就是驻点

  • $\displaystyle \begin{aligned} f'_x = -3x^2y - 2xy + 5x^4 \end{aligned}$
  • $\displaystyle \begin{aligned} f'_y = 2y - x^3 - x^2 \end{aligned}$
  • 令 $\displaystyle \begin{aligned} \begin{cases} f'_x = 0 \\ f'_y = 0 \end{cases} \end{aligned}$，得 $\displaystyle \begin{aligned} \begin{cases} -3x^2y - 2xy + 5x^4 = 0 \\ 2y - x^3 - x^2 = 0 \end{cases} \end{aligned}$
    • $\displaystyle \begin{aligned} y = \frac{x^3 + x^2}{2} \xrightarrow[]{代入消元法}\end{aligned}$，得$\displaystyle \begin{aligned} -3x^2 \cdot \frac{x^3 + x^2}{2} - 2x \cdot \frac{x^3 + x^2}{2} + 5x^4 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[]{整理}-\frac{3}{2} \cdot (x^5 + x^4) - (x^4 + x^3) + 5x^4 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[]{合并同类项}-\frac{3}{2}x^5 + \left(-\frac{3}{2} - 1 + 5\right)x^4 - x^3 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[]{整理}-\frac{3}{2}x^5 + \frac{5}{2}x^4 - x^3 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[]{同除x^3}-\frac{3}{2}x^2 + \frac{5}{2}x - 1 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[]{同乘2}3x^2 - 5x + 2 = 0 \end{aligned}$
      • $\displaystyle \begin{aligned} \xrightarrow[因式分解]{\begin{array}{lr}1 & -1 \\3 & 2\end{array}}(x - 1)(3x - 2) = 0 \end{aligned}$
    • 得 $\displaystyle \begin{aligned} x = 0, x = 1, x = \frac{2}{3} \end{aligned}$
      • $\displaystyle \begin{aligned} (x, y) \end{aligned}$ 为 $\displaystyle \begin{aligned} \left(0, 0\right), \left(1, 1\right), \left(\frac{2}{3}, \frac{10}{27}\right) \end{aligned}$

• 求$A,B,C$

  • 由一阶导
    • $\displaystyle \begin{aligned} f'_x = -3x^2y - 2xy + 5x^4 \end{aligned}$
    • $\displaystyle \begin{aligned} f'_y = 2y - x^3 - x^2 \end{aligned}$
  • 求二阶导
    • $\displaystyle \begin{aligned} \begin{cases} A = f''_{xx} = 20x^3 - 6xy - 2y \\ B = f''_{xy} = -3x^2 - 2x \\ C = f''_{yy} = 2 \end{cases} \end{aligned}$

• 分别讨论三种情况$\displaystyle \begin{aligned} (x, y) \end{aligned}$ 为 $\displaystyle \begin{aligned} \left(0, 0\right), \left(1, 1\right), \left(\frac{2}{3}, \frac{10}{27}\right) \end{aligned}$

• 当 $\displaystyle \begin{aligned} x = 0, y = 0 \end{aligned}$ 时,

  • $A,B,C$
    • $\displaystyle \begin{aligned} A = 0 - 0 - 0 = 0 \end{aligned}$
    • $\displaystyle \begin{aligned} B = 0 - 0 = 0 \end{aligned}$
    • $\displaystyle \begin{aligned} C = 2 \end{aligned}$
  • $\displaystyle \begin{aligned} AC - B^2 = 0 \cdot 2 - 0 = 0 \end{aligned}$，可能存在极值点
  • $\displaystyle \begin{aligned} f(x, y) = (y - x^2)(y - x^3) \end{aligned}$
    • 取 $\displaystyle \begin{aligned} y = 2x^2 \end{aligned}$
      • $\displaystyle \begin{aligned} f(x, y) = (2x^2 - x^2)(x^2 - x^3) = x^2 \cdot (x^2 - x^3) = x^4 \cdot (1 - x) > 0 \end{aligned}$
    • 取 $\displaystyle \begin{aligned} y = 2x^3 \end{aligned}$
      • $\displaystyle \begin{aligned} f(x, y) = (2x^3 - x^2)(2x^3 - x^3) = (2x^3 - x^2) \cdot x^3 = x^5(2x - 1) < 0 \end{aligned}$
    • 有的点大、有的点小、所以点 $\displaystyle \begin{aligned} (0,0) \end{aligned}$ 不是极值点
    • 
      

• $\displaystyle \begin{aligned} (x, y) = (1,1) \end{aligned}$

  • $A,B,C$
    • $\displaystyle \begin{aligned} A = 20 - 6 - 2 = 12 \end{aligned}$
    • $\displaystyle \begin{aligned} B = -5 \end{aligned}$
    • $\displaystyle \begin{aligned} C = 2 \end{aligned}$
  • $\displaystyle \begin{aligned} AC - B^2 = 24 - 25 = -1 < 0 \end{aligned}$
    • $\displaystyle \begin{aligned} (x, y) = (1,1) \end{aligned}$ 不是极值点

• $\displaystyle \begin{aligned} (x, y) = \left(\frac{2}{3}, \frac{10}{27}\right) \end{aligned}$ 时

  • 求$A,B,C$
    • $\displaystyle \begin{aligned} A = f''_{xx} = 20x^3 - 6xy - 2y = 20\left(\frac{2}{3}\right)^3 - 6\cdot\frac{2}{3}\cdot\frac{10}{27} - 2\cdot\frac{10}{27} \end{aligned}$
      • $\displaystyle \begin{aligned} = 20\cdot\frac{8}{27} - 4\cdot\frac{10}{27} - 2\cdot\frac{10}{27} = \frac{100}{27} \end{aligned}$
    • $\displaystyle \begin{aligned} B = f''_{xy} = -3x^2 - 2x = -3\left(\frac{2}{3}\right)^2 - 2\cdot\frac{2}{3} = -3\cdot\frac{4}{9} - \frac{4}{3} = -\frac{8}{3} \end{aligned}$
    • $\displaystyle \begin{aligned} C = f''_{yy} = 2 \end{aligned}$
  • 判断是否是极值点
    • $\displaystyle \begin{aligned} AC - B^2 = \frac{100}{27} \times 2 - \frac{64}{9} > 0 \end{aligned}$
    • $\displaystyle \begin{aligned} A = \frac{100}{27} > 0 \end{aligned}$
    • 则 $\displaystyle \begin{aligned} \left(\frac{2}{3}, \frac{10}{27}\right) \end{aligned}$ 是极小值点。
      • $\displaystyle \begin{aligned} f\left(\frac{2}{3}, \frac{10}{27}\right) = (y - x^2)(y - x^3) \end{aligned}$
        • $\displaystyle \begin{aligned} = \left(\frac{10}{27} - \frac{4}{9}\right)\left(\frac{10}{27} - \frac{8}{27}\right) \end{aligned}$
        • $\displaystyle \begin{aligned} = -\frac{2}{27} \cdot \frac{2}{27} = -\frac{4}{27^2} \end{aligned}$

(19)

(本题满分 12 分)

设空间有界区域 $\Omega$ 由柱面 $x^{2}+y^{2}=1$ 与平面 $z=0$ 和 $x+z=1$ 围成, $\displaystyle \Sigma$ 为 $\Omega$ 边界面的外侧,计算曲面积分$\displaystyle I=\iint_{\Sigma} 2 x z \mathrm{~d} y \mathrm{~d} z+x z \cos y \mathrm{~d} z \mathrm{~d} x+3 y z \sin x \mathrm{~d} x \mathrm{~d} y$

• • 
    

• $\displaystyle \begin{aligned} I \xlongequal[]{原式} \iint_{\Sigma} 2xz \, dy \, dz + xz\cos y \, dz \, dx + 3yz\sin x \, dx \, dy \end{aligned}$

  • $\displaystyle \begin{aligned} = \iiint_{\Omega} \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \, dv \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{高斯}\iiint_{\Omega} 2z - xz\sin y + 3y\sin x \, dv \end{aligned}$

• $\displaystyle \begin{aligned} \xrightarrow[]{x \geq \sin y 与3 y \sin x是奇函数} \iiint_{\Omega} 2z \, dv \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[]{柱线法}\iint_{\Sigma} dx \, dy \int_{0}^{1-x} 2z \, dz \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{\left.z^2\right|_0 ^{1-x}}\iint_{x^2 + y^2 \leq 1} (1-x)^2 \, dx \, dy \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{完全平方}\iint_{x^2 + y^2 \leq 1} (1 + x^2 - 2x) \, dx \, dy \end{aligned}$
  • $\displaystyle \begin{aligned} = \iint_{D} (1 + x^2 - 2x) \, dx \, dy \end{aligned}$
    

• $\displaystyle \begin{aligned} \xlongequal[]{x是奇函数}\iint_{D} dx \, dy + \iint_{D} x^2 \, dx \, dy \xrightarrow[]{\iint_D y^2 d x d y}\end{aligned}$

• $\displaystyle \begin{aligned} \xlongequal[然后极坐标求解]{轮转对称性}\pi + \frac{1}{2} \iint_{D} (x^2 + y^2) \, dx \, dy \end{aligned}$

  • $\displaystyle \begin{aligned} \xlongequal[x^2 + y^2=r^2]{化为极坐标}\pi + \frac{1}{2} \int_{0}^{2\pi} d\theta \int_{0}^{1} r^2 \cdot r \, dr \end{aligned}$
  • $\displaystyle \begin{aligned} = \pi + \frac{1}{2} \cdot 2\pi \cdot \frac{1^4}{4} = \frac{5\pi}{4} \end{aligned}$

(20)

(本题满分 12 分)

设函数 $f(x)$ 在 $[-a, a]$ 上具有 2 阶连续导数. 证明:

(1) 若 $f(0)=0$, 则存在 $\xi \in(-a, a)$, 使得 $f^{\prime \prime}(\xi)=\frac{1}{a^{2}}[f(a)+f(-a)]$;

(2) 若 $f(x)$ 在 $(-a, a)$ 内取得极值, 则存在 $\eta \in(-a, a)$, 使得

• $\left|f^{\prime \prime}(\eta)\right| \geqslant \frac{1}{2 a^{2}}|f(a)-f(-a)| .$

• • 带拉格朗日型余项的一阶泰勒公式
    • 设函数 $f(x)$ 在 $[a, b]$ 上具有二阶导数, $c$ 为 $(a, b)$ 内一点, 则 $f(x)$ 在 $x=c$ 处的带拉格朗日型余项的一阶泰勒公式为
      • $f(x)=f(c)+f^{\prime}(c)(x-c)+\frac{f^{\prime \prime}(\xi)}{2!}(x-c)^2,$
      • 其中 $\xi$ 介于 $x$ 与 $c$ 之间.
    • 展开位置:中间点(左右界限相加除以二)，左右端点，题中给的特殊点

• 解 (1) 由题设, 根据一阶带拉格朗日余项的泰勒公式, 存在 $x_1 \in(-a, 0), x_2 \in(0, a)$, 使得

  • 在0点展开:$\displaystyle \begin{aligned} f(x) = f(0) + f'(0)x + \frac{1}{2} f''(\xi)x^2 \end{aligned}$
    • $\displaystyle \begin{aligned} f(-a)=f(0)-f^{\prime}(0) a+\frac{1}{2} f^{\prime \prime}\left(x_1\right) a^2\xlongequal[]{f(0)=0}-f^{\prime}(0) a+\frac{1}{2} f^{\prime \prime}\left(x_1\right) a^2, \end{aligned}$
    • $\displaystyle \begin{aligned} f(a)=f(0)+f^{\prime}(0) a+\frac{1}{2} f^{\prime \prime}\left(x_2\right) a^2\xlongequal[]{f(0)=0}f^{\prime}(0) a+\frac{1}{2} f^{\prime \prime}\left(x_2\right) a^2 . \end{aligned}$
  • 所以 $f(-a)+f(a)=\frac{a^2}{2}\left[f^{\prime \prime}\left(x_1\right)+f^{\prime \prime}\left(x_2\right)\right]$.
    • 因为 $f^{\prime \prime}(x)$ 连续, 所以由介值定理存在 $\xi \in\left[x_1, x_2\right] \subset(-a, a)$, 使得
    • $f^{\prime \prime}(\xi)\xlongequal[]{平均值}\frac{1}{2}\left[f^{\prime \prime}\left(x_1\right)+f^{\prime \prime}\left(x_2\right)\right] .$
      • 没有点，就创造一个
  • 综上,存在 $\xi \in(-a, a)$, 使得$f^{\prime \prime}(\xi)=\frac{1}{a^2}[f(a)+f(-a)] .$

(2)

• (2) 设 $f(x)$ 在 $x_0 \in(-a, a)$ 取得极值, 则 $\xrightarrow[]{极值点处导数为0}f^{\prime}\left(x_0\right)=0$
  • 根据一阶带拉格朗日余项的泰勒公式, 存在 $b \in\left(-a, x_0\right), c \in\left(x_0, a\right)$, 使得
    $\displaystyle \begin{aligned} f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{1}{2} f''(\eta)(x - x_0)^2, \quad \eta \text{在} x \text{与} x_0 \text{之间}. \end{aligned}$
    • $\displaystyle \begin{aligned} f(-a)=f\left(x_0\right)-f^{\prime}\left(x_0\right)\left(a+x_0\right)+\frac{1}{2} f^{\prime \prime}(b)\left(a+x_0\right)^2=f\left(x_0\right)+\frac{1}{2} f^{\prime \prime}(b)\left(a+x_0\right)^2, \end{aligned}$
    • $\displaystyle \begin{aligned} f(a)=f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(a-x_0\right)+\frac{1}{2} f^{\prime \prime}(c)\left(a-x_0\right)^2=f\left(x_0\right)+\frac{1}{2} f^{\prime \prime}(c)\left(a-x_0\right)^2 . \end{aligned}$
  • 所以 $f(a)-f(-a)=\frac{1}{2}\left[f^{\prime \prime}(c)\left(a-x_0\right)^2-f^{\prime \prime}(b)\left(a+x_0\right)^2\right]$.
  • 不妨设 $\left|f^{\prime \prime}(b)\right| \leqslant\left|f^{\prime \prime}(c)\right|$, 记 $\eta=c$, 则
    • $|f(a)-f(-a)| \leqslant \frac{1}{2}\left|f^{\prime \prime}(\eta)\right|\left[\left(a-x_0\right)^2+\left(a+x_0\right)^2\right] \leqslant 2 a^2\left|f^{\prime \prime}(\eta)\right| .$
• 综上, 存在 $\eta \in(-a, a)$, 使得$\left|f^{\prime \prime}(\eta)\right| \geqslant \frac{1}{2 a^2}|f(a)-f(-a)| .$

(2)高昆仑版

(2) 设 $f(x)$ 在 $x_0 \in(-a, a)$ 处职得极值, 于是 $f^{\prime}\left(x_0\right)=0$.

• $\displaystyle \begin{aligned} f(x) & =f\left(x_0\right)+f^{\prime}\left(x_0\right)\left(x-x_0\right)+\frac{1}{2} f^{\prime \prime}(\tau)\left(x-x_0\right)^2 \end{aligned}$
  • $\displaystyle \begin{aligned} \xlongequal[]{f^{\prime}\left(x_0\right)=0}f\left(x_0\right)+\frac{1}{2} f^{\prime \prime}(\tau)\left(x-x_0\right)^2 . \end{aligned}$
    • 取 $\displaystyle \begin{aligned} x = -a \end{aligned}$，有$\displaystyle \begin{aligned} f(-a) = f(x_0) + \frac{1}{2} f''(\eta_1)(-a - x_0)^2. \end{aligned}$
    • 取 $\displaystyle \begin{aligned} x = a \end{aligned}$，有$\displaystyle \begin{aligned} f(a) = f(x_0) + \frac{1}{2} f''(\eta_2)(a - x_0)^2. \end{aligned}$
      • 以上两式相减，得$\displaystyle \begin{aligned} f(-a) - f(a) = \frac{1}{2} f''(\eta_1)(a + x_0)^2 - \frac{1}{2} f''(\eta_2)(a - x_0)^2. \end{aligned}$
• 于是$\displaystyle \begin{aligned} |f(-a) - f(a)| = \left| \frac{1}{2} f''(\eta_1)(a + x_0)^2 - \frac{1}{2} f''(\eta_2)(a - x_0)^2 \right| \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{|a \pm b| \leqslant|a|+|b|}\leq \left| \frac{1}{2} f''(\eta_1)(a + x_0)^2 \right| + \left| \frac{1}{2} f''(\eta_2)(a - x_0)^2 \right|. \end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{\left|f''(\eta)\right| = \max\left\{\left|f''(\eta_1)\right|, \left|f''(\eta_2)\right|\right\} }\leq \frac{1}{2} \left|f''(\eta)\right| \cdot \left[(a + x_0)^2 + (a - x_0)^2\right] \end{aligned}$
  • $\displaystyle \begin{aligned}\xlongequal[]{完全平方展开后整理}\frac{1}{2} \left|f''(\eta)\right| \cdot (2a^2 + 2x_0^2)\end{aligned}$
  • $\displaystyle \begin{aligned} \xrightarrow[]{x_0^2<a^2}< \frac{1}{2} \left|f''(\eta)\right| \cdot 4a^2 = 2a^2 \left|f''(\eta)\right|. \end{aligned}$
• 所以$\displaystyle \begin{aligned} \left|f''(\eta)\right| > \frac{\left|f(-a) - f(a)\right|}{2a^2}. \end{aligned}$

(21)

(本题满分 12 分)

已知二次型

• $\displaystyle \begin{aligned} f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}+2 x_{2}^{2}+2 x_{3}^{2}+2 x_{1} x_{2}-2 x_{1} x_{3}, \end{aligned}$

• $\displaystyle \begin{aligned} g\left(y_{1}, y_{2}, y_{3}\right)=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2 y_{2} y_{3} . \end{aligned}$ (1) 求可逆变换 $\boldsymbol{x}=\boldsymbol{P} \boldsymbol{y}$ 将 $f\left(x_{1}, x_{2}, x_{3}\right)$ 化成 $g\left(y_{1}, y_{2}, y_{3}\right)$; (2) 是否存在正交变换 $\boldsymbol{x}=\boldsymbol{Q y}$ 将 $f\left(x_{1}, x_{2}, x_{3}\right)$ 化成 $g\left(y_{1}, y_{2}, y_{3}\right)$ ?

• • 合同变换
    • 设 $\boldsymbol{A}$ 为 $n$ 阶矩阵， $\boldsymbol{P}$ 为 $n$ 阶可逆矩阵， $\boldsymbol{P}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{P}$ 为对 $\boldsymbol{A}$ 作合同变换所得矩阵, 所得矩阵 $\boldsymbol{B}=\boldsymbol{P}^{\mathrm{T}} \boldsymbol{A P}$ 与矩阵 $\boldsymbol{A}$ 合同.
    • $f=x^{\top} A x \xlongequal[]{x=p y} y^{\top} p^{\top} A p y=g$

方法1

• f和g具有相同的规范型

  • $\displaystyle \begin{aligned} f \xrightarrow{x=P_1z} h \quad P_1^TAP_1 = \Lambda \end{aligned}$
  • $\displaystyle \begin{aligned} g \xrightarrow{y=P_2z} h \quad P_2^TBP_2 = \Lambda \end{aligned}$
  • $\displaystyle \begin{aligned} B = (P_2^T)^{-1} \Lambda P_2^{-1} = (P_2^{-1})^T P_1^T A P_1 P_2^{-1} \end{aligned}$
    • 记 $\displaystyle \begin{aligned} P = P_1 P_2^{-1} \end{aligned}$，则 $\displaystyle \begin{aligned} B = P^T A P \end{aligned}$

• $\displaystyle \begin{aligned} & A=\left(\begin{array}{ccc}1 & 1 & -1 \\1 & 2 & 0 \\-1 & 0 & 2\end{array}\right) \quad B=\left(\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 1 \\0 & 1 & 1\end{array}\right) \end{aligned}$

• $\displaystyle \begin{aligned} A=\left(\begin{array}{ccc}1 & 1 & -1 \\1 & 2 & 0 \\-1 & 0 & 2\end{array}\right) \xrightarrow{r_2-r_1}\left(\begin{array}{ccc}1 & 1 & -1 \\0 & 1 & 1 \\-1 & 0 & 2\end{array}\right) \xrightarrow{C_2-C_1}\left(\begin{array}{ccc}1 & 0 & -1 \\0 & 1 & 1 \\-1 & 1 & 2\end{array}\right) \end{aligned}$

  • $\displaystyle \begin{aligned}\xrightarrow{r_3+r_1} \left(\begin{array}{ccc}1 & 0 & -1 \\0 & 1 & 1 \\0 & 1 & 1\end{array}\right) \xrightarrow{C_3+C_1}\left(\begin{array}{ccc}1 & 0 & 0 \\0 & 1 & 1 \\0 & 1 & 1\end{array}\right)\end{aligned}$

• $\displaystyle \begin{aligned} P_1=\left(\begin{array}{ccc}1 & -1 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right) \quad P_2=\left(\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right) \quad \end{aligned}$

  • $\displaystyle \begin{aligned}P=P_1 P_2=\left(\begin{array}{ccc}1 & -1 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)\left(\begin{array}{lll}1 & 0 & 1 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)=\left(\begin{array}{ccc}1 & -1 & 1 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right)\end{aligned}$

方法2

解：(1)

• $\displaystyle \begin{aligned} f(x_1, x_2, x_3) = x_1^2 + 2x_2^2 + 2x_3^2 + 2x_1x_2 - 2x_1x_3 \end{aligned}$
  • $\displaystyle \begin{aligned} = (x_1 + x_2 - x_3)^2 + x_2^2 + x_3^2 + 2x_2x_3 \end{aligned}$
  • $\displaystyle \begin{aligned} = (x_1 + x_2 - x_3)^2 + (x_2 + x_3)^3 + 0x_3^2. \end{aligned}$
  • $\displaystyle \begin{aligned} \begin{array}{l}\left\{\begin{array}{l}x_{1} + x_{2} - x_{3} = z_{1} \\x_{2} + x_{3} = z_{2} \\x_{3} = z_{3}\end{array}\right.\end{array} \end{aligned}$
• \displaystyle \begin{aligned} \begin{align*}& g(y_1, y_2, y_3) = y_1^2 + y_2^2 + y_3^2 + 2y_2y_3 = y_1^2 + (y_2 + y_3)^2 + 0y_3^2.\end{align*} \end{aligned}
  • $\displaystyle \begin{aligned} \begin{array}{l}\left\{\begin{array}{l}y_{1} = z_{1} \\y_{2} + y_{3} = z_{2}\end{array}\right.\end{array} \end{aligned}$
  • 则 $\displaystyle \begin{aligned} g(y_{1}, y_{2}, y_{3}) = z_{1}^{2} + z_{2}^{2} + 0z_{3}^{2} \end{aligned}$。
• (1) =(2)，得$\displaystyle \begin{aligned} \left\{\begin{array}{l}x_{1} + x_{2} - x_{3} = y_{1} \\x_{2} + x_{3} = y_{2} + y_{3} \\x_{3} = y_{3}\end{array}\right. \end{aligned}$
  • 即$\displaystyle \begin{aligned} \left\{\begin{array}{l}x_{1} = y_{1} - y_{2} + y_{3} \\x_{2} = y_{2} \\x_{3} = y_{3}\end{array}\right. \end{aligned}$
  • 记 $\displaystyle \begin{aligned} P = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$，则在 $\displaystyle \begin{aligned} x = P y \end{aligned}$ 下，有 $\displaystyle \begin{aligned} f(x_1, x_2, x_3) = g(y_1, y_2, y_3) \end{aligned}$。

(2)

二次型 $\displaystyle \begin{aligned} f(x_1, x_2, x_3) \end{aligned}$ 与 $\displaystyle \begin{aligned} g(y_1, y_2, y_3) \end{aligned}$ 对应的矩阵分别为

• $\displaystyle \begin{aligned} A = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 2 & 0 \\ -1 & 0 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 1 & 1 \end{pmatrix}. \end{aligned}$

• 是否存在正交变换 $\displaystyle \begin{aligned} x = Q y \end{aligned}$ 将 $\displaystyle \begin{aligned} f \end{aligned}$ 化为 $\displaystyle \begin{aligned} g \end{aligned}$

  • $\displaystyle \begin{aligned} \xrightarrow[]{等价} \text{是否存在正交矩阵 } Q, \text{使得 } Q^T A Q = B \text{ 或 } Q^{-1} A Q = B \text{ (相似)}. \end{aligned}$

• 由于 $\displaystyle \begin{aligned} \operatorname{tr}(A) \neq \operatorname{tr}(B) \end{aligned}$，所以矩阵 $\displaystyle \begin{aligned} A \end{aligned}$ 与 $\displaystyle \begin{aligned} B \end{aligned}$ 不相似，

  • 故，不存在正交变换 $\displaystyle \begin{aligned} x = Q y \end{aligned}$ 将 $\displaystyle \begin{aligned} f \end{aligned}$ 化为 $\displaystyle \begin{aligned} g \end{aligned}$。

(22)

(本题满分 12 分)

设二维随机变量 $(X, Y)$ 的概率密度为

• $\displaystyle f(x, y)= \begin{cases}\frac{2}{\pi}\left(x^{2}+y^{2}\right), & x^{2}+y^{2} \leqslant 1, \\ 0, & \text { 其他. }\end{cases}$

(1) 求 $X$ 与 $Y$ 的协方差; (2) $X$ 与 $Y$ 是否相互独立? (3) 求 $Z=X^{2}+Y^{2}$ 的概率密度.

• • 解 (1) 因为$\operatorname{Cov}(X, Y)=E(X Y)-E X E Y$
  • $\displaystyle \begin{aligned} E X=\iint_{x^{\prime}+y^{\prime}<1} x \cdot \frac{2}{\pi}\left(x^2+y^2\right) \mathrm{d} x \mathrm{~d} y=0, \end{aligned}$
  • $\displaystyle \begin{aligned} E Y=\iint_{x^{\prime}+y^{\prime}<1} y \cdot \frac{2}{\pi}\left(x^2+y^2\right) \mathrm{d} x \mathrm{~d} y=0, \end{aligned}$
  • $\displaystyle \begin{aligned} E(X Y)=\iint_{x^{\prime}+y^{\prime}<1} x y \cdot \frac{2}{\pi}\left(x^2+y^2\right) \mathrm{d} x \mathrm{~d} y=0, \end{aligned}$

• 所以 $X$ 与 $Y$ 的协方差为

  • $\operatorname{Cov}(X, Y)=E(X Y)-E X E Y=0 .$

• (2) 对 $\displaystyle \begin{aligned} \forall x, y \end{aligned}$，有 $\displaystyle \begin{aligned} f(x, y) = f_X(x) \cdot f_Y(y) \end{aligned}$，则 $\displaystyle \begin{aligned} X \end{aligned}$ 与 $\displaystyle \begin{aligned} Y \end{aligned}$ 相互独立。

  • 若 $\displaystyle \begin{aligned} f(x, y) \end{aligned}$（二元函数）不能分解成两个一元函数 $\displaystyle \begin{aligned} f_X(x) \end{aligned}$ 和 $\displaystyle \begin{aligned} f_Y(y) \end{aligned}$ 的乘积，则 $\displaystyle \begin{aligned} X \end{aligned}$ 与 $\displaystyle \begin{aligned} Y \end{aligned}$ 不相互独立。
  • 若 $\displaystyle \begin{aligned} f(x, y) \end{aligned}$ 的非零区域不是矩形，则 $\displaystyle \begin{aligned} X \end{aligned}$ 与 $\displaystyle \begin{aligned} Y \end{aligned}$ 不相互独立。

• (3) 记 $Z$ 的分布函数为 $F_Z(z)$.

  • $\displaystyle \begin{aligned} F_Z(z) = P\{Z \leq z\} = P\{X^2 + Y^2 \leq z\} \end{aligned}$，其中 $\displaystyle \begin{aligned} 0 \leq Z \leq 1 \end{aligned}$，$\displaystyle \begin{aligned} -\infty < z < +\infty \end{aligned}$。
    • 若 $\displaystyle \begin{aligned} z < 0 \end{aligned}$，则 $\displaystyle \begin{aligned} F_Z(z) = 0 \end{aligned}$。
    • 若 $\displaystyle \begin{aligned} z \geq 1 \end{aligned}$，则 $\displaystyle \begin{aligned} F_Z(z) = 1 \end{aligned}$。
    • 若 $\displaystyle \begin{aligned} 0 \leq z < 1 \end{aligned}$，则
      $\displaystyle \begin{aligned} F_Z(z) = P\{X^2 + Y^2 \leq z\} = \iint\limits_{x^2+y^2 \leq z} \frac{2}{\pi} (x^2 + y^2) \, dx \, dy \end{aligned}$
      • $\displaystyle \begin{aligned}\int_0^{2 \pi} \mathrm{~d} \theta \int_0^{\sqrt{z}} \frac{2}{\pi} r^3 \mathrm{~d} r=z^2 .\end{aligned}$
  • $\displaystyle \begin{aligned} f_Z(z) = \begin{cases} 2z, & 0 < z < 1, \\0, & \text{其他}.\end{cases} \end{aligned}$