答案 C
解析 因为
$\left[\frac{f(x)}{x}\right]' = \frac{x f'(x) - f(x)}{x^2} > 0$，$x \in (a,b)$，
所以$\frac{f(x)}{x}$在$[a,b]$上单调递增．
故$\frac{f(a)}{a} < \frac{f(x)}{x} < \frac{f(b)}{b}$，$\frac{f(a)}{ab} < \frac{f(x)}{x^2} < \frac{f(b)}{ab}$．
选项A，B，D分别改写为
$\frac{f(a)}{a} > \frac{f(b)}{b}$，$\frac{f(x)}{x^2} > \frac{f(b)}{ab}$，$\frac{f(x)}{x^2} < \frac{f(a)}{ab}$，
故选项A，B，D均不正确．
由正值函数$\frac{f(x)}{x}$在$[a,b]$上单调递增，知$x f(x) = \frac{f(x)}{x} \cdot x^2$在$[a,b]$上单调递增，故$x f(x) > a f(a)$．选项C正确．

答案 B
解析 当$x^2+y^2\neq0$时，有
$f'_x(x,y)=2x\sin\frac{1}{x^2+y^2}-\frac{2x}{x^2+y^2}\cos\frac{1}{x^2+y^2}$，
$f'_y(x,y)=2y\sin\frac{1}{x^2+y^2}-\frac{2y}{x^2+y^2}\cos\frac{1}{x^2+y^2}$.
当$x^2+y^2=0$时，有
$f'_x(0,0)=\lim\limits_{x\to0}\frac{f(x,0)-f(0,0)}{x}=\lim\limits_{x\to0}\frac{x^2\sin\frac{1}{x^2}}{x}=0$.
同理，$f'_y(0,0)=0$.
取$x_n=\frac{1}{2\sqrt{n\pi}}$，$y_n=\frac{1}{2\sqrt{n\pi}}$，当$n\to\infty$时，$x_n\to0$，$y_n\to0$.
$f'_x(x_n,y_n)=-2\sqrt{n\pi}\to-\infty$（$n\to\infty$），
故$f'_x(x,y)$在点$(0,0)$处间断.
又由$\lim\limits_{\substack{x\to0\\y\to0}}\frac{f(x,y)-f(0,0)-[f'_x(0,0)x+f'_y(0,0)y]}{\sqrt{x^2+y^2}}=\lim\limits_{\substack{x\to0\\y\to0}}\frac{(x^2+y^2)\sin\frac{1}{x^2+y^2}}{\sqrt{x^2+y^2}}$
$\xlongequal{\sqrt{x^2+y^2}=\rho}\lim\limits_{\rho\to0^+}\rho\sin\frac{1}{\rho^2}=0$，
可知$f(x,y)$在$(0,0)$处可微.选项B正确.

答案 B
解析 当$n\to\infty$时，有
$\frac{\sqrt{n+1}-\sqrt{n-1}}{n^{\alpha-1}}=\frac{2}{n^{\alpha-1}(\sqrt{n+1}+\sqrt{n-1})}=\frac{2}{n^{\alpha-\frac{1}{2}}(\sqrt{1+\frac{1}{n}}+\sqrt{1-\frac{1}{n}})}$
$\sim\frac{2}{n^{\alpha-\frac{1}{2}}\cdot2}=\frac{1}{n^{\alpha-\frac{1}{2}}}$.
当$\alpha-\frac{1}{2}>1$，即$\alpha>\frac{3}{2}$时，原级数绝对收敛.
由$\sum\limits_{n=1}^{\infty}(-1)^n\frac{1}{n^{2-\alpha}}$条件收敛，知$\sum\limits_{n=1}^{\infty}\frac{1}{n^{2-\alpha}}$发散，且$\frac{1}{n^{2-\alpha}}\to0(n\to\infty)$，故$0<2-\alpha\leqslant1$，于是$1\leqslant\alpha<2$.
综上所述，$\frac{3}{2}<\alpha<2$.选项B正确.

答案 D
解析 解法一，L的参数方程为\(\begin{cases}x = x, \\ y = f(x),\end{cases}\) \( x \)从\( a \)到\( 0 \)，则
原式\(=\frac{1}{2}\int_L -y\mathrm{d}x + x\mathrm{d}y = \frac{1}{2}\int_a^0 [-f(x) + x f'(x)]\mathrm{d}x\)
\(=\frac{1}{2}\int_0^a f(x)\mathrm{d}x - \frac{1}{2}\int_0^a x f'(x)\mathrm{d}x = \frac{1}{2}\int_0^a f(x)\mathrm{d}x - \frac{1}{2}\int_0^a x \mathrm{d}[f(x)]\)
\(=\frac{1}{2}\int_0^a f(x)\mathrm{d}x - \frac{1}{2}\left[ x f(x)\big|_0^a - \int_0^a f(x)\mathrm{d}x \right]\)
\(=\frac{1}{2}\int_0^a f(x)\mathrm{d}x - \frac{1}{2}\left[ a f(a) - \int_0^a f(x)\mathrm{d}x \right]\)
\(=\int_0^a f(x)\mathrm{d}x - \frac{1}{2}a f(a)\)
= 曲边三角形AOB的面积.

解法二，用格林公式，令\( L_1 \)为从B到O的直线段，\( L_2 \)为从O到A的直线段，则
\(\frac{1}{2}\oint_{L + L_1 + L_2} -y\mathrm{d}x + x\mathrm{d}y = \iint_D \mathrm{d}x\mathrm{d}y = \)曲边三角形AOB的面积，
其中D表示曲边三角形AOB. 由于
\(\frac{1}{2}\int_{L_1} -y\mathrm{d}x + x\mathrm{d}y = 0\)，
\(\frac{1}{2}\int_{L_2} -y\mathrm{d}x + x\mathrm{d}y = \frac{1}{2}\int_0^a \left[ -\frac{f(a)}{a}x + x \cdot \frac{f(a)}{a} \right]\mathrm{d}x = 0\)，
故\(\frac{1}{2}\int_L -y\mathrm{d}x + x\mathrm{d}y = \frac{1}{2}\oint_{L + L_1 + L_2} -y\mathrm{d}x + x\mathrm{d}y - 0 - 0 = \)曲边三角形AOB的面积.
选项D正确.

答案 D
解析 由已知，$\boldsymbol{A}$是$(n-1)\times n$矩阵，且$\mathrm{r}(\boldsymbol{A})=\mathrm{r}(\boldsymbol{A}^{\mathrm{T}})=n-1$，故$\boldsymbol{AX}=\boldsymbol{0}$的基础解系有$n-\mathrm{r}(\boldsymbol{A})=1$个向量.
又因$\boldsymbol{\beta}_1,\boldsymbol{\beta}_2$与$\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\cdots,\boldsymbol{\alpha}_{n-1}$均正交，即$\boldsymbol{\alpha}_i^{\mathrm{T}}\boldsymbol{\beta}_1=0$，$\boldsymbol{\alpha}_i^{\mathrm{T}}\boldsymbol{\beta}_2=0(i=1,2,\cdots,n-1)$，故
$$\boldsymbol{A}\boldsymbol{\beta}_j=\begin{pmatrix}\boldsymbol{\alpha}_1^{\mathrm{T}}\\\boldsymbol{\alpha}_2^{\mathrm{T}}\\\vdots\\\boldsymbol{\alpha}_{n-1}^{\mathrm{T}}\end{pmatrix}\boldsymbol{\beta}_j=\begin{pmatrix}0\\0\\\vdots\\0\end{pmatrix}\quad (j=1,2),$$
即$\boldsymbol{\beta}_1,\boldsymbol{\beta}_2$是$\boldsymbol{AX}=\boldsymbol{0}$的两个不同解，从而$\boldsymbol{\beta}_1-\boldsymbol{\beta}_2$是$\boldsymbol{AX}=\boldsymbol{0}$的非零解. 故$\boldsymbol{AX}=\boldsymbol{0}$的通解为$k(\boldsymbol{\beta}_1-\boldsymbol{\beta}_2)$. 选项D正确.
由于$\boldsymbol{\beta}_1,\boldsymbol{\beta}_2$可能为零向量，$\boldsymbol{\beta}_1+\boldsymbol{\beta}_2$也可能是零向量，所以排除选项A,B,C.

答案 D
解析 对分块矩阵分别作初等变换.由于初等变换不改变矩阵的秩,所以有
$r_1 = r\begin{pmatrix} A&B \\ O&E \end{pmatrix} = r\begin{pmatrix} A&O \\ O&E \end{pmatrix} = n + r(A)$,
$r_2 = r\begin{pmatrix} BA&O \\ B&B \end{pmatrix} = r\begin{pmatrix} BA&O \\ O&B \end{pmatrix} = r(BA) + r(B)$,
$r_3 = r\begin{pmatrix} E&A \\ B&O \end{pmatrix} = r\begin{pmatrix} E&O \\ B&-BA \end{pmatrix} = r\begin{pmatrix} E&O \\ O&-BA \end{pmatrix}$
$= n + r(-BA) = n + r(BA)$.
由$r(B) \leqslant n$,$r(BA) \leqslant r(A)$,知$r_1 \geqslant r_3 \geqslant r_2$.选项D正确.

答案 A
解析 由\( A \sim B \)，知\( A \)与\( B \)有相同的特征值，\( B \)的特征值为\( 2,2,b \)。又
\[
|\lambda E - A|=\begin{vmatrix} \lambda - 1&-a&1 \\ -1&\lambda - 5&-1 \\ -4&-12&\lambda - 6 \end{vmatrix}=(\lambda - 2)(\lambda^2 - 10\lambda + 13 - a),
\]
由2是二重特征值知，若\( b = 2 \)，但\( \text{tr}(A) = 12 \neq \text{tr}(B) = 6 \)，\( A \)与\( B \)不相似，故\( \lambda^2 - 10\lambda + 13 - a \)含因子\( \lambda - 2 \)，即\( 2^2 - 10×2 + 13 - a = 0 \)，解得\( a = -3 \)。
又由\( \text{tr}(A) = \text{tr}(B) \)，即\( 1 + 5 + 6 = 2 + 2 + b \)，知\( b = 8 \)。
选项A正确。

答案 C

解析 由已知，有$\frac{X - \mu}{\sigma} \sim N(0,1)$，$\overline{X} \sim N(\mu,\frac{\sigma^{2}}{n})$，$\frac{\overline{X} - \mu}{\sigma}\sqrt{n} \sim N(0,1)$。

又

$P\{|X - \mu| < a\} = P\{\left|\frac{X - \mu}{\sigma}\right| < \frac{a}{\sigma}\}$，

$P\{|\overline{X} - \mu| < \pi\} = P\{\left|\frac{\sqrt{n}(\overline{X} - \mu)}{\sigma}\right| < \frac{\sqrt{n}\pi}{\sigma}\}$，

由

$P\{|X - \mu| < a\} = P\{|\overline{X} - \mu| < \pi\}$，

知$\frac{a}{\sigma} = \frac{\sqrt{n}\pi}{\sigma}$。故$a = \sqrt{n}\pi$。选项C正确。

答案 B
解析 由已知，有$X\sim N(1,2)$，$Y\sim N(1,2)$，$\rho_{XY}=0$，故$E(X)=E(Y)=1$，$D(X)=D(Y)=2$，且$X$与$Y$相互独立.
$\text{Cov}(U,V)=\text{Cov}(X+2Y,X-2Y)$
$=\text{Cov}(X,X)-2\text{Cov}(X,Y)+2\text{Cov}(Y,X)-4\text{Cov}(Y,Y)$
$=\text{Cov}(X,X)-4\text{Cov}(Y,Y)$
$=D(X)-4D(Y)$
$=2-4\times2=-6$，
$D(U)=D(X+2Y)=D(X)+4D(Y)=2+4\times2=10$，
$D(V)=D(X-2Y)=D(X)+4D(Y)=10$，
故$\rho_{UV}=\frac{\text{Cov}(U,V)}{\sqrt{D(U)}\sqrt{D(V)}}=\frac{-6}{\sqrt{10}\times\sqrt{10}}=-\frac{3}{5}$.
选项B正确.

答案 C
解析 \( \frac{1}{2n}\sum_{i=1}^{2n}X_i = \bar{X} \)，由已知\( \sigma^2 = 1 \)，知
\( \frac{(2n - 1)S^2}{\sigma^2} = (2n - 1) \cdot \frac{1}{2n - 1}\sum_{i=1}^{2n}(X_i - \bar{X})^2 = \sum_{i=1}^{2n}(X_i - \bar{X})^2 \sim \chi^2(2n - 1) \)，
故\( T_1 \sim \chi^2(2n - 1) \).
\( T_2 = \frac{1}{2}\sum_{i=1}^{2n}X_i^2 + \sum_{i=1}^{n}X_{2i-1}X_{2i} \)
\( = \frac{1}{2}(X_1^2 + X_2^2 + \cdots + X_{2n-1}^2 + X_{2n}^2) + X_1X_2 + X_3X_4 + \cdots + X_{2n-1}X_{2n} \)
\( = \frac{1}{2}(X_1 + X_2)^2 + \frac{1}{2}(X_3 + X_4)^2 + \cdots + \frac{1}{2}(X_{2n-1} + X_{2n})^2 \)
\( = \sum_{i=1}^{n}\left( \frac{X_{2i-1} + X_{2i}}{\sqrt{2}} \right)^2 \sim \chi^2(n) \).
这里\( \frac{X_{2i-1} + X_{2i}}{\sqrt{2}} \sim N(0,1) \).
故\( E(T_2) = n,D(T_1) = 2(2n - 1) \).
选项C正确.

答案 $\frac{(-1)^{n-1}}{n^2 2^n} \cdot (2n)!$
解析
$\ln(1+t)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}t^n}{n} \ (-1<t\leq1)$．
记$g(t)=\begin{cases} \frac{\ln(1+t)}{t}, & t\neq0, \\ 1, & t=0, \end{cases}$则
$g(t)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}t^{n-1}}{n}$，
故$f(x)=\int_{0}^{\frac{x^2}{2}} g(t)dt=\int_{0}^{\frac{x^2}{2}} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}t^{n-1}}{n} \right) dt=\sum_{n=1}^{\infty} \int_{0}^{\frac{x^2}{2}} \frac{(-1)^{n-1}t^{n-1}}{n} dt=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{2n}}{n^2 2^n}$，
$f^{(2n)}(0)=a_{2n} \cdot (2n)!=\frac{(-1)^{n-1}}{n^2 2^n} \cdot (2n)!$．

答案 $[1,+\infty)$
解析 由$\ln(1+x)-a\text{e}^x + 1 \leq \ln a$，有
$\ln(1+x) + \text{e}^{\ln a + 1} \leq \ln a + x + \text{e}^{\ln a + x}$. ①
令$g(x)=x + \text{e}^x$，则$g'(x)=1 + \text{e}^x > 0$，$g(x)$单调递增. ①式可表示为
$g(\ln(1+x)) \leq g(\ln a + x)$.
由此可知，$\ln(1+x) \leq \ln a + x$，即$\ln(1+x) - x \leq \ln a$.
下面求$h(x)=\ln(1+x) - x$的最大值.
$h'(x)=\frac{1}{1+x} - 1 = 0$，
当$x \in (-1,0)$时，$h'(x) > 0$；当$x \in (0,+\infty)$时，$h'(x) < 0$，故$h(x)$在$x=0$处取得极大值，也是最大值，即$h_{\max}=h(0)=0$. 因此$a \geq 1$，即$a \in [1,+\infty)$.

答案 1
解析 方法一:利用定积分的定义.
原极限$=\lim \limits_{n \to \infty} \dfrac{\sqrt{n}\left(1 + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}}\right)}{n} \bigg/ \dfrac{\sqrt{n}\left( \dfrac{1}{\sqrt{n + 1}} + \dfrac{1}{\sqrt{n + 2}} + \cdots + \dfrac{1}{\sqrt{n + 3n}}\right)}{n}$
$=\dfrac{\lim \limits_{n \to \infty} \dfrac{\sqrt{n}}{n}\left(1 + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}}\right)}{\lim \limits_{n \to \infty} \dfrac{\sqrt{n}}{n}\left( \dfrac{1}{\sqrt{n + 1}} + \dfrac{1}{\sqrt{n + 2}} + \cdots + \dfrac{1}{\sqrt{n + 3n}}\right)} \xlongequal{令} \dfrac{A}{B},$
其中 $A = \lim \limits_{n \to \infty} \dfrac{1}{n}\left( \dfrac{1}{\sqrt{\dfrac{1}{n}}} + \dfrac{1}{\sqrt{\dfrac{2}{n}}} + \cdots + \dfrac{1}{\sqrt{\dfrac{n}{n}}}\right) = \lim \limits_{n \to \infty} \dfrac{1}{n}\sum \limits_{i = 1}^{n} \dfrac{1}{\sqrt{\dfrac{i}{n}}} = \int_{0}^{1} \dfrac{1}{\sqrt{x}}\text{d}x = 2,$
$B = \lim \limits_{n \to \infty} \dfrac{\sqrt{n}}{n}\sum \limits_{i = 1}^{3n} \dfrac{1}{\sqrt{n + i}} = \lim \limits_{n \to \infty} \dfrac{3\sqrt{n}}{3n}\sum \limits_{i = 1}^{3n} \dfrac{1}{\sqrt{n + i}} = \sqrt{3}\lim \limits_{n \to \infty} \dfrac{1}{3n}\sum \limits_{i = 1}^{3n} \dfrac{\sqrt{3n}}{\sqrt{n + i}}$
$= \sqrt{3}\lim \limits_{n \to \infty} \dfrac{1}{3n}\sum \limits_{i = 1}^{3n} \dfrac{1}{\sqrt{\dfrac{1}{3} + \dfrac{i}{3n}}} = \sqrt{3}\int_{0}^{1} \dfrac{1}{\sqrt{x + \dfrac{1}{3}}}\text{d}x = 2.$
故原极限为 1.

方法二:利用夹逼准则.注意到分子、分母每一项都是$\dfrac{1}{\sqrt{k}}$的形式,故对$\dfrac{1}{\sqrt{k}}$进行放缩,有
$\dfrac{1}{\sqrt{k}} = \dfrac{2}{\sqrt{k} + \sqrt{k}} \leqslant \dfrac{2}{\sqrt{k - 1} + \sqrt{k}} = 2(\sqrt{k} - \sqrt{k - 1}),$
$\dfrac{1}{\sqrt{k}} \geqslant \dfrac{2}{\sqrt{k} + \sqrt{k + 1}} = 2(\sqrt{k + 1} - \sqrt{k}),$
因此
$1 + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} \leqslant 2(\sqrt{1} - \sqrt{0}) + 2(\sqrt{2} - \sqrt{1}) + \cdots + 2(\sqrt{n} - \sqrt{n - 1}) = 2\sqrt{n},$
$1 + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}} \geqslant 2(\sqrt{2} - \sqrt{1}) + 2(\sqrt{3} - \sqrt{2}) + \cdots + 2(\sqrt{n + 1} - \sqrt{n}) = 2\sqrt{n + 1} - 2,$
$\dfrac{1}{\sqrt{n + 1}} + \dfrac{1}{\sqrt{n + 2}} + \cdots + \dfrac{1}{\sqrt{4n}} \leqslant 2(\sqrt{n + 1} - \sqrt{n}) + 2(\sqrt{n + 2} - \sqrt{n + 1}) + \cdots +$
$2(\sqrt{4n} - \sqrt{4n - 1}) = 2(\sqrt{4n} - \sqrt{n}) = 2\sqrt{n},$
$\dfrac{1}{\sqrt{n + 1}} + \dfrac{1}{\sqrt{n + 2}} + \cdots + \dfrac{1}{\sqrt{4n}} \geqslant 2(\sqrt{n + 2} - \sqrt{n + 1}) + 2(\sqrt{n + 3} - \sqrt{n + 2}) + \cdots +$
$2(\sqrt{4n + 1} - \sqrt{4n}) = 2(\sqrt{4n + 1} - \sqrt{n + 1}),$
因此 $\dfrac{2\sqrt{n + 1} - 2}{2\sqrt{n}} \leqslant \dfrac{1 + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}}}{\dfrac{1}{\sqrt{n + 1}} + \dfrac{1}{\sqrt{n + 2}} + \cdots + \dfrac{1}{\sqrt{4n}}} \leqslant \dfrac{2\sqrt{n}}{2(\sqrt{4n + 1} - \sqrt{n + 1})},$
由夹逼准则可得原极限为 1.

答案 0

解析
\(\text{grad}\ u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial u}{\partial z} \right)\)，
\[
\text{rot}(\text{grad}\ u) = \begin{vmatrix}
\boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z}
\end{vmatrix}
= \left( \frac{\partial^2 u}{\partial z \partial y} - \frac{\partial^2 u}{\partial y \partial z} \right)\boldsymbol{i} - \left( \frac{\partial^2 u}{\partial z \partial x} - \frac{\partial^2 u}{\partial x \partial z} \right)\boldsymbol{j} + \left( \frac{\partial^2 u}{\partial y \partial x} - \frac{\partial^2 u}{\partial x \partial y} \right)\boldsymbol{k} = 0.
\]

答案 $(1,1,1)^{\mathrm{T}}$
解析 依题设，设从基$\alpha_1,\alpha_2,\alpha_3$到基$\beta_1,\beta_2,\beta_3$的过渡矩阵为$P$，则
$$(\beta_1,\beta_2,\beta_3)=(\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}=(\alpha_1,\alpha_2,\alpha_3)P,$$
故$(\alpha_1,\alpha_2,\alpha_3)=(\beta_1,\beta_2,\beta_3)P^{-1}$. 又由于
$$\alpha=(\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}=(\beta_1,\beta_2,\beta_3)P^{-1}\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$$
$$=(\beta_1,\beta_2,\beta_3)\begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}=(\beta_1,\beta_2,\beta_3)\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$$
故$\alpha$在基$\beta_1,\beta_2,\beta_3$下的坐标为$(1,1,1)^{\mathrm{T}}$.

答案 $R_{\alpha}=\left\{\frac{\overline{X}-\mu_{0}}{\frac{S}{\sqrt{n}}}<-t_{\alpha}(n-1)\right\}$

解析：方差$\sigma^{2}$未知，检验均值$\mu$，应用$t$检验法，因$H_{1}$是单边的，故是单边检验。

在$H_{0}$成立的条件下，$T=\frac{\overline{X}-\mu_{0}}{\frac{S}{\sqrt{n}}}\sim t(n-1)$。
对于给定的检验水平$\alpha$，使$P\{T<-t_{\alpha}(n-1)\}=\alpha$，故拒绝域为
$R_{\alpha}=\{T<-t_{\alpha}(n-1)\}=\left\{\frac{\overline{X}-\mu_{0}}{\frac{S}{\sqrt{n}}}<-t_{\alpha}(n-1)\right\}$。

解析

积分区域\( D_t \)如图阴影部分所示，故
\( F(t) = \iint_{D_t} f_{yx}''(x,y) \, dxdy \)
\( = \int_{0}^{2t} dx \int_{0}^{t} f_{yx}''(x,y) dy \)
\( = \int_{0}^{2t} dx \left[ d f_x'(x,y) \right]_{0}^{t} = \int_{0}^{2t} \left. f_x'(x,y) \right|_{0}^{t} dx \)
\( = \int_{0}^{2t} \left[ f_x'(x,t) - f_x'(x,0) \right] dx \)
\( = \int_{0}^{2t} d [f(x,t)] - \int_{0}^{2t} d [f(x,0)] \)
\( = \left. f(x,t) \right|_{0}^{2t} - \left. f(x,0) \right|_{0}^{2t} \)
\( = f(2t,t) - f(0,t) - [f(2t,0) - f(0,0)] \)，
\( \lim\limits_{t \to 0^+} \frac{F(t)}{t} = \lim\limits_{t \to 0^+} \frac{f(2t,t) - f(2t,0) - [f(0,t) - f(0,0)]}{t} \)
\( \stackrel{\text{洛必达法则}}{=} \lim\limits_{t \to 0^+} \frac{f_1'(2t,t) \cdot 2 + f_2'(2t,t) \cdot 1 - f_1'(2t,0) \cdot 2}{1} - \lim\limits_{t \to 0^+} \frac{f(0,t) - f(0,0)}{t} \)
\( = 2f_1'(0,0) + f_2'(0,0) - 2f_1'(0,0) - f_2'(0,0) = 0 \)

解析 （Ⅰ）因为方向导数取得最大值的方向为梯度方向，所以$f(x,y)$在$P(2,1)$处的梯度与$\boxed{l}=(0,1)$方向相同，由
$f'_{x}(2,1)=(2x + ay)|_{(2,1)}=4 + a$，
$f'_{y}(2,1)=(ax + 2by)|_{(2,1)}=2a + 2b$，
可知$\text{grad }f(2,1)=(4 + a, 2a + 2b)$与$\boxed{l}=(0,1)$同向，故
$4 + a = 0$，$2a + 2b > 0$.
又$\frac{\partial f}{\partial \boxed{l}}\big|_{(2,1)} = 2 = \|\text{grad }f(2,1)\| = |2a + 2b|$，故$a = -4$，$b = 5$.

（Ⅱ）由（Ⅰ）知，$f(x,y)=1$，即$x^{2} - 4xy + 5y^{2} = 1$，曲线上任一点$(x,y)$到$O(0,0)$的距离为
$d = \sqrt{x^{2} + y^{2}}$.
利用拉格朗日乘数法，求$x^{2} + y^{2}$在条件$x^{2} - 4xy + 5y^{2} = 1$下的最值.

令$L = x^{2} + y^{2} + \lambda(x^{2} - 4xy + 5y^{2} - 1)$，则
$\begin{cases}
L'_{x} = 2x + 2\lambda x - 4\lambda y = 0, & ①\\
L'_{y} = 2y - 4\lambda x + 10\lambda y = 0, & ②\\
L'_{\lambda} = x^{2} - 4xy + 5y^{2} - 1 = 0. & ③
\end{cases}$

由①$\cdot \frac{x}{2} +$②$\cdot \frac{y}{2}$可得
$x^{2} + y^{2} + \lambda(x^{2} - 4xy + 5y^{2}) = 0$.
因由$x^{2} - 4xy + 5y^{2} = 1$，知$x^{2} + y^{2} = -\lambda$，故只需求$\lambda$即可.

由①式与②式，得方程组
$\begin{cases}
(1 + \lambda)x - 2\lambda y = 0, \\
-2\lambda x + (1 + 5\lambda)y = 0.
\end{cases}$（*）

该方程组有非零解的充分必要条件是
$\begin{vmatrix}
1 + \lambda & -2\lambda \\
-2\lambda & 1 + 5\lambda
\end{vmatrix} = \lambda^{2} + 6\lambda + 1 = 0$，

解得$-\lambda = 3 \pm 2\sqrt{2} = (\sqrt{2} \pm 1)^{2}$，故$d = \sqrt{-\lambda} = \sqrt{2} \pm 1$，即所求最大值为$d = \sqrt{2} + 1$，最小值为$d = \sqrt{2} - 1$.

解析 （Ⅰ）$(a+x)f'(x)+f(x)=a$变形为$f'(x)+\frac{1}{a+x}f(x)=\frac{a}{a+x}$，
$f(x)=\mathrm{e}^{-\int \frac{1}{a+x}\mathrm{d}x}\left(\int \frac{a}{a+x}\mathrm{e}^{\int \frac{1}{a+x}\mathrm{d}x}\mathrm{d}x+C\right)=\frac{ax+C}{a+x}$，
由$f(0)=1$，得$C=a$，故$f(x)=\frac{a(1+x)}{a+x}(x\geqslant 0)$．
（Ⅱ）由（Ⅰ）及已知条件知，$x_{n+1}=f(x_n)=\frac{a(1+x_n)}{a+x_n}>0(n=1,2,\cdots)$．
又由
$f'(x)=\frac{a(a-1)}{(a+x)^2}\begin{cases}>0,\quad a>1,\\<0,\quad \frac{1}{2}\leqslant a<1,\end{cases}$
知当$a>1$时，$f(x)$单调递增．故$\{x_n\}$必单调．又由$f(x)$在$[0,+\infty)$上连续，且$\lim \limits_{x\rightarrow +\infty}f(x)=a$，知$f(x)$在$[0,+\infty)$上有界，从而$\{x_n\}$有界．由单调有界准则，知$\lim \limits_{n\rightarrow \infty}x_n$存在．记$\lim \limits_{n\rightarrow \infty}x_n=A$，$x_{n+1}=\frac{a(1+x_n)}{a+x_n}$两边取极限$(n\rightarrow \infty)$，得$A=\frac{a(1+A)}{a+A}$，解得$A=\sqrt{a}$．当$\frac{1}{2}\leqslant a<1$时，$f(x)$在$x>0$内单调递减，此时$\{x_n\}$不单调．由于
$\vert x_{n+1}-\sqrt{a}\vert=\left\vert \frac{a(1+x_n)}{a+x_n}-\frac{a(1+\sqrt{a})}{a+\sqrt{a}}\right\vert=\frac{a(1-a)}{(a+x_n)(a+\sqrt{a})}\vert x_n-\sqrt{a}\vert$
$\leqslant \frac{1-a}{2a}\vert x_n-\sqrt{a}\vert\leqslant \frac{1}{2}\vert x_n-\sqrt{a}\vert$
$\leqslant \left(\frac{1}{2}\right)^2\vert x_{n-1}-\sqrt{a}\vert\leqslant \cdots\leqslant \left(\frac{1}{2}\right)^n\vert x_1-\sqrt{a}\vert$，
故$\lim \limits_{n\rightarrow \infty}x_n=\sqrt{a}$．

解析 $yy''-y'^2=1$为不显含$x$的可降阶方程．
令$y'=p$，则$y''=p\dfrac{\mathrm{d}p}{\mathrm{d}y}$，代入原方程，得
$\dfrac{p\mathrm{d}p}{1+p^2}=\dfrac{\mathrm{d}y}{y}$．
上式两边积分，得
$\sqrt{1+p^2}=C_1y$．
由$p=y'(1)=0$，得$C_1=1$，故$y=\sqrt{1+p^2}$．
由$yy''=1+y'^2$，知当$y>0$时，$y''>0$．再由$y'(1)=0$，知当$x>1$时，$y'>0$；当$x<1$时，$y'<0$．故由$y=\sqrt{1+p^2}$，得$p=y'=\pm\sqrt{y^2 - 1}$，即
$\dfrac{\mathrm{d}y}{\sqrt{y^2 - 1}}=\pm\mathrm{d}x$．
上式两边积分，并注意到$y(1)=1$，解得
$\ln(y+\sqrt{y^2 - 1})=\pm(x - 1)$，
即
$y+\sqrt{y^2 - 1}=\mathrm{e}^{\pm(x - 1)}$．  ①
上式有理化，得
$y - \sqrt{y^2 - 1}=\mathrm{e}^{\mp(x - 1)}$．  ②
①式$+$②式，得
$y=\dfrac{1}{2}\left[\mathrm{e}^{x - 1}+\mathrm{e}^{-(x - 1)}\right]$．

（Ⅱ）由（Ⅰ）知，$y=f(x)=\dfrac{1}{2}\left[\mathrm{e}^{x - 1}+\mathrm{e}^{-(x - 1)}\right]$关于直线$x=1$对称，曲线$\begin{cases}y=f(x),\\z=0\end{cases}$绕直线$\begin{cases}x=1,\\z=0\end{cases}$旋转一周所得曲面$\Sigma_1$及$\Sigma$如图所示．

利用高斯公式，有
$I=\iiint_V \dfrac{1}{\sqrt{y^2 - 1}}\mathrm{d}x\mathrm{d}y\mathrm{d}z$，
其中$V$为$\Sigma$所围区域．采用“先二后一”计算，有
$I=\int_{\sqrt{2}}^{\mathrm{e}} \dfrac{1}{\sqrt{y^2 - 1}}\mathrm{d}y\iint_{D_y} \mathrm{d}z\mathrm{d}x$，
其中$D_y$为$(\sqrt{2},\mathrm{e})$内任一点$y$作平行于$xOz$面的平面与$\Sigma$所截得的圆形区域，其面积为
$\pi(x - 1)^2=\pi\ln^2(y+\sqrt{y^2 - 1})$，

故
$I=\int_{\sqrt{2}}^{\mathrm{e}} \dfrac{1}{\sqrt{y^2 - 1}}\cdot\pi\ln^2(y+\sqrt{y^2 - 1})\mathrm{d}y$
$=\pi\int_{\sqrt{2}}^{\mathrm{e}} \ln^2(y+\sqrt{y^2 - 1})\mathrm{d}[\ln(y+\sqrt{y^2 - 1})]$
$=\dfrac{\pi}{3}\ln^3(y+\sqrt{y^2 - 1})\bigg|_{\sqrt{2}}^{\mathrm{e}}$
$=\dfrac{\pi}{3}\left[\ln^3(\mathrm{e}+\sqrt{\mathrm{e}^2 - 1}) - \ln^3(\sqrt{2}+1)\right]$．

解析 （Ⅰ）由\( |\lambda E - A| = \begin{vmatrix} \lambda - 1&-1&1\\ 0&\lambda - 1&-1\\ 0&0&\lambda - 1\end{vmatrix} = (\lambda - 1)^3 \)，
得\( A \)的特征值为\( \lambda_1 = \lambda_2 = \lambda_3 = 1 \).
由\( E - A = \begin{pmatrix} 0&-1&1\\ 0&0&-1\\ 0&0&0\end{pmatrix} \)，得\( \lambda = 1 \)的特征向量为\( \alpha_1 = \begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} \).
解方程组\( (E - A)\alpha_2 = \alpha_1 \).
由\( \begin{pmatrix} 0&-1&1&\vdots&1\\ 0&0&-1&\vdots&0\\ 0&0&0&\vdots&0\end{pmatrix} \)，得\( \alpha_2 = k_1\begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} + \begin{pmatrix} 0\\ -1\\ 0\end{pmatrix} = \begin{pmatrix} k_1\\ -1\\ 0\end{pmatrix} \)（\( k_1 \)为任意常数）.
解方程组\( (E - A)\alpha_3 = \alpha_2 \).
由\( \begin{pmatrix} 0&-1&1&\vdots&k_1\\ 0&0&-1&\vdots&-1\\ 0&0&0&\vdots&0\end{pmatrix} \)，得\( \alpha_3 = k_2\begin{pmatrix} 1\\ 0\\ 0\end{pmatrix} + \begin{pmatrix} 0\\ 1 - k_1\\ 1\end{pmatrix} = \begin{pmatrix} k_2\\ 1 - k_1\\ 1\end{pmatrix} \)（\( k_2 \)为任意常数）.
\( P = (\alpha_1, \alpha_2, \alpha_3) = \begin{pmatrix} 1&k_1&k_2\\ 0&-1&1 - k_1\\ 0&0&1\end{pmatrix} \)，
由\( |P| = -1 \neq 0 \)，知\( P \)可逆.

（Ⅱ）由已知及（Ⅰ），有
\( A\alpha_1 = \alpha_1 \)，\( A\alpha_2 = \alpha_2 - \alpha_1 \)，\( A\alpha_3 = \alpha_3 - \alpha_2 \)，
则
\( A(\alpha_1, \alpha_2, \alpha_3) = (\alpha_1, \alpha_2 - \alpha_1, \alpha_3 - \alpha_2) \)
\( = (\alpha_1, \alpha_2, \alpha_3)\begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1\end{pmatrix} \)，
故
\( P^{-1}AP = \begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1\end{pmatrix} \)， 且 \( |A| = 1 \).

又由于
\( P^{-1}A^*P = P^{-1}(|A|A^{-1})P = P^{-1}A^{-1}P = (P^{-1}AP)^{-1} \)
\( = \begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1\end{pmatrix}^{-1} = \begin{pmatrix} 1&1&1\\ 0&1&1\\ 0&0&1\end{pmatrix} \)，
故
\( P^{-1}(A + A^*)P = P^{-1}AP + P^{-1}A^*P \)
\( = \begin{pmatrix} 1&-1&0\\ 0&1&-1\\ 0&0&1\end{pmatrix} + \begin{pmatrix} 1&1&1\\ 0&1&1\\ 0&0&1\end{pmatrix} \)
\( = \begin{pmatrix} 2&0&1\\ 0&2&0\\ 0&0&2\end{pmatrix} \).

解析 （Ⅰ）依题设，有
$f_X(x)=\begin{cases}1, & 1<x<2, \\ 0, & 其他,\end{cases}$
$f_{Y|X}(y|x)=\begin{cases}x\text{e}^{-xy}, & y>0, \\ 0, & y\leq0,\end{cases}$
故(X,Y)的概率密度为
$f(x,y)=f_X(x)\cdot f_{Y|X}(y|x)=\begin{cases}x\text{e}^{-xy}, & 1<x<2,y>0, \\ 0, & 其他.\end{cases}$
由卷积公式，有
$f_Z(z)=\int_{-\infty}^{+\infty}f\left(x,\frac{z}{x}\right)\cdot\frac{1}{|x|}\text{d}x.$
当$z\leq0$时，$f_Z(z)=0$.
当$z>0$时，$f_Z(z)=\int_{1}^{2}\text{e}^{-z}\text{d}x=\text{e}^{-z}$.
故
$f_Z(z)=\begin{cases}\text{e}^{-z}, & z>0, \\ 0, & z\leq0,\end{cases}$
即$Z=XY$服从参数为1的指数分布.
（Ⅱ）由已知及（Ⅰ），有
$E(X)=\frac{1+2}{2}=\frac{3}{2}$，$E(Z)=1$，
$\text{Cov}(X,Z)=E(XZ)-E(X)E(Z)$.
又
$E(XZ)=E(X^2Y)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}x^2y\cdot f(x,y)\text{d}x\text{d}y$
$=\int_{1}^{2}x^2\text{d}x\int_{0}^{+\infty}y\cdot x\text{e}^{-xy}\text{d}y$
$=\int_{1}^{2}x^2\cdot\frac{1}{x}\text{d}x=\int_{1}^{2}x\text{d}x=\frac{3}{2}$，
故$\text{Cov}(X,Z)=\frac{3}{2}-\frac{3}{2}\times1=0$，X与Z不相关.

【注】$\int_{0}^{+\infty}y\cdot x\text{e}^{-xy}\text{d}y=\frac{1}{x}$是参数$\lambda=x$的指数分布的数学期望.