答案 选(C).
解 $y'=\mathrm{e}^{\frac{1}{x}\ln(1+\mathrm{e}^{x})}\cdot\frac{\frac{x\mathrm{e}^{x}}{1+\mathrm{e}^{x}}-\ln(1+\mathrm{e}^{x})}{x^{2}}$,
当$x\lt 0$时,$y'\lt 0$,且$\lim\limits_{x\rightarrow-\infty}(1+\mathrm{e}^{x})^{\frac{1}{x}}=1$,$\lim\limits_{x\rightarrow0^{-}}(1+\mathrm{e}^{x})^{\frac{1}{x}}=0$,故$0\lt y\lt 1$.
当$x\gt 0$时,由$\frac{x\mathrm{e}^{x}}{1+\mathrm{e}^{x}}\lt x=\ln\mathrm{e}^{x}\lt \ln(1+\mathrm{e}^{x})$,故$y'\lt 0$,且
$\lim\limits_{x\rightarrow0^{+}}(1+\mathrm{e}^{x})^{\frac{1}{x}}=+\infty$, $\lim\limits_{x\rightarrow+\infty}(1+\mathrm{e}^{x})^{\frac{1}{x}}=\mathrm{e}^{\lim\limits_{x\rightarrow+\infty}\frac{\ln(1+\mathrm{e}^{x})}{x}}=\mathrm{e}$,
故$y\gt \mathrm{e}$.所以选(C).

答案 选(C).
解 若$x_{0}\neq0$,则$f''(x_{0})=\frac{\mathrm{e}^{x_{0}}+\mathrm{e}^{-x_{0}}-2}{x_{0}^{2}}\gt \frac{2-2}{x_{0}^{2}}=0$,故$f(x_{0})$是极小值.
若$x_{0}=0$,则当$x\neq0$时,$f''(x)-\frac{3f'(x)}{x}=\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}-2}{x^{2}}$,即
$\lim\limits_{x\rightarrow0}f''(x)-3\lim\limits_{x\rightarrow0}\frac{f'(x)-f'(0)}{x}=\lim\limits_{x\rightarrow0}\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}-2}{x^{2}}$,
得
$f''(0)-3f''(0)=\lim\limits_{x\rightarrow0}\frac{\mathrm{e}^{x}-\mathrm{e}^{-x}}{2x}=\lim\limits_{x\rightarrow0}\frac{\mathrm{e}^{x}+\mathrm{e}^{-x}}{2}=1$,
所以$f''(0)=-\frac{1}{2}\lt 0$,故$f(0)$为极大值.选(C).

答案 选(B).
解 （A）不正确.取$f(x)=x^{2}\cdot\alpha(x)$,其中$\alpha(x)=\begin{cases}0,\ \ x为有理数,\\x,\ \ x为无理数,\end{cases}$则$f(x)$在$x\neq0$的点均不连续,故$f(x)$在$x=0$点没有二阶导数.
（B）正确.$\lim\limits_{x\rightarrow a^{+}}\frac{|f(x)|}{x-a}=\lim\limits_{x\rightarrow a^{+}}\left|\frac{f(x)}{x-a}\right|$,$\lim\limits_{x\rightarrow a^{-}}\frac{|f(x)|}{x-a}=-\lim\limits_{x\rightarrow a^{-}}\left|\frac{f(x)}{x-a}\right|$,由$\lim\limits_{x\rightarrow a}\frac{|f(x)|}{x-a}$存在知$\lim\limits_{x\rightarrow a}\left|\frac{f(x)}{x-a}\right|=0$,故$\lim\limits_{x\rightarrow a}\frac{f(x)}{x-a}=0$,得$f(a)=0$,$\lim\limits_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)=0$.
（C）错误.取$f(x)=\arctan x$,则$\lim\limits_{x\rightarrow+\infty}f(x)=\frac{\pi}{2}$,$\lim\limits_{x\rightarrow-\infty}f(x)=-\frac{\pi}{2}$,故$\lim\limits_{x\rightarrow\infty}\arctan x$不存在,但曲线$y=\arctan x$有水平渐近线.
（D）错误.取$f(x)=x+\sin x$,有$\lim\limits_{x\rightarrow\infty}\frac{f(x)}{x}=\lim\limits_{x\rightarrow\infty}\frac{x+\sin x}{x}=1$,但$\lim\limits_{x\rightarrow\infty}[f(x)-x]=\lim\limits_{x\rightarrow\infty}\sin x$不存在,所以曲线$y=f(x)$没有斜渐近线.

答案 选(D).
解 $I_{1}-I_{2}=\int_{0}^{\frac{\pi}{2}}\frac{\sin x-\cos x}{1+x^{2}}\mathrm{d}x=\int_{0}^{\frac{\pi}{4}}\frac{\sin x-\cos x}{1+x^{2}}\mathrm{d}x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin x-\cos x}{1+x^{2}}\mathrm{d}x$
$=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x-\sin x}{1+\left(\frac{\pi}{2}-x\right)^{2}}\mathrm{d}x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin x-\cos x}{1+x^{2}}\mathrm{d}x=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{\cos x-\sin x}{1+\left(\frac{\pi}{2}-x\right)^{2}}-\frac{\cos x-\sin x}{1+x^{2}}\right]\mathrm{d}x$
$=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\cos x-\sin x)\left[\frac{1}{1+\left(\frac{\pi}{2}-x\right)^{2}}-\frac{1}{1+x^{2}}\right]\mathrm{d}x\lt 0$,
故$I_{1}\lt I_{2}$;

$J_{1}-J_{2}=\int_{0}^{1}\mathrm{e}^{x}\sqrt{1-x}\mathrm{d}x-\int_{0}^{1}\mathrm{e}^{x}\frac{x}{2\sqrt{1-x}}\mathrm{d}x=\int_{0}^{1}\mathrm{e}^{x}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x$
$=\int_{0}^{\frac{2}{3}}\mathrm{e}^{x}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x+\int_{\frac{2}{3}}^{1}\mathrm{e}^{x}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x\lt \int_{0}^{\frac{2}{3}}\mathrm{e}^{\frac{2}{3}}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x+\int_{0}^{1}\mathrm{e}^{\frac{2}{3}}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x$
$=\mathrm{e}^{\frac{2}{3}}\int_{0}^{1}\frac{2-3x}{2\sqrt{1-x}}\mathrm{d}x\stackrel{\mathrm{令}\sqrt{1-x}=t}{=}\mathrm{e}^{\frac{2}{3}}\int_{0}^{1}(3t^{2}-1)\mathrm{d}t=0$,
故$J_{1}\lt J_{2}$.所以选(D).

答案 选(D).
解 因为$\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}\frac{f(x,y)}{\ln(1+|x|+|y|)}=0$且$f(x,y)$在点$(0,0)$处连续,所以$f(0,0)=\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}f(x,y)=0$.
$f'_{x}(0,0)=\lim\limits_{x\rightarrow0}\frac{f(x,0)-f(0,0)}{x}=\lim\limits_{x\rightarrow0}\frac{f(x,0)}{x}=\lim\limits_{\substack{x\rightarrow0\\y=0}}\frac{f(x,y)}{\ln(1+|x|+|y|)}\cdot\frac{\ln(1+|x|+|y|)}{x}$
$=0\cdot(\pm1)=0$,
由对称性可知,$f'_{y}(0,0)=0$.

$\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}\frac{f(x,y)-f(0,0)-f'_{x}(0,0)x-f'_{y}(0,0)y}{\sqrt{x^{2}+y^{2}}}=\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}\frac{f(x,y)}{\sqrt{x^{2}+y^{2}}}$
$=\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}\frac{f(x,y)}{\ln(1+|x|+|y|)}\cdot\frac{\ln(1+|x|+|y|)}{\sqrt{x^{2}+y^{2}}}$
又因为$\left|\frac{\ln(1+|x|+|y|)}{\sqrt{x^{2}+y^{2}}}\right|\leqslant\frac{|x|+|y|}{\sqrt{x^{2}+y^{2}}}\leqslant2$,故$\frac{\ln(1+|x|+|y|)}{\sqrt{x^{2}+y^{2}}}$为有界函数,从而
$\lim\limits_{\substack{x\rightarrow0\\y\rightarrow0}}\frac{f(x,y)-f(0,0)-f'_{x}(0,0)x-f'_{y}(0,0)y}{\sqrt{x^{2}+y^{2}}}=0$,
所以函数$f(x,y)$在点$(0,0)$处可微.

答案 选(A)
解 原极限 $= \lim\limits_{n \to \infty} \sum_{i=1}^{n} \sum_{j=n-i}^{n+i} \ln \left( \frac{1 + \frac{i}{n}}{1 + \frac{j}{n}} \right) \frac{1}{n^2}$.
令 $x = \frac{i}{n}, y = \frac{j}{n}$, 由于 $1 \leqslant i \leqslant n$, 从而 $\frac{1}{n} \leqslant \frac{i}{n} \leqslant 1$, 故当 $n \to \infty$ 时, $0 \leqslant x \leqslant 1$. 同理,由于 $n - i \leqslant j \leqslant n + i$, 从而 $1 - \frac{i}{n} \leqslant \frac{j}{n} \leqslant 1 + \frac{i}{n}$, 故 $1 - x \leqslant y \leqslant 1 + x$.
记 $D = \{(x,y) \mid 0 \leqslant x \leqslant 1,1 - x \leqslant y \leqslant 1 + x\}$, 则
原极限 $= \iint_{D} \ln \frac{1+x}{1+y} \mathrm{d}\sigma = \int_{0}^{1} \mathrm{d}x \int_{1-x}^{1+x} \ln \frac{1+x}{1+y} \mathrm{d}y$
$= \int_{0}^{1} \mathrm{d}y \int_{1-y}^{1} \ln \frac{1+x}{1+y} \mathrm{d}x + \int_{1}^{2} \mathrm{d}y \int_{y-1}^{1} \ln \frac{1+x}{1+y} \mathrm{d}x$,
故选(A).

答案 选(A).
解 $\int_{0}^{+\infty} \frac{\ln x}{\sqrt[3]{x^a - x^b}} \mathrm{d}x = \int_{0}^{1} \frac{\ln x}{\sqrt[3]{x^a - x^b}} \mathrm{d}x + \int_{1}^{+\infty} \frac{\ln x}{\sqrt[3]{x^a - x^b}} \mathrm{d}x$, $\lim\limits_{x \to +\infty} \frac{\sqrt[3]{x^a - x^b}}{\frac{1}{x^{\frac{a}{3}}}} = 1$, 由 $\int_{1}^{+\infty} \frac{\ln x}{\sqrt[3]{x^a - x^b}} \mathrm{d}x$ 收敛,故 $\frac{a}{3} > 1$, 即 $a > 3$.
$\lim\limits_{x \to 1} \frac{\ln x}{\sqrt[3]{x^a - x^b}} = 0, \lim\limits_{x \to 0^+} \frac{\frac{1}{\sqrt[3]{x^a - x^b}}}{\frac{1}{x^{\frac{b}{3}}}} = 1$, 故 $\frac{b}{3} < 1$, 即 $b < 3$, 故选(A).

答案 选(A).
解 $\boldsymbol{N}=\begin{pmatrix} \boldsymbol{E} & \boldsymbol{B} \\ \boldsymbol{A} & \boldsymbol{O} \end{pmatrix} \xrightarrow{c_2 - c_1 \cdot \boldsymbol{B}} \begin{pmatrix} \boldsymbol{E} & \boldsymbol{O} \\ \boldsymbol{A} & -\boldsymbol{AB} \end{pmatrix} \xrightarrow{r_2 - \boldsymbol{A} \cdot r_1} \begin{pmatrix} \boldsymbol{E} & \boldsymbol{O} \\ \boldsymbol{O} & -\boldsymbol{AB} \end{pmatrix}$,所以 $\mathrm{r}(\boldsymbol{N}) = n + \mathrm{r}(\boldsymbol{AB})$.
$\boldsymbol{M}=\begin{pmatrix} \boldsymbol{A} & \boldsymbol{AB} \\ \boldsymbol{CA} & \boldsymbol{O} \end{pmatrix} \xrightarrow{r_2 - \boldsymbol{C} \cdot r_1} \begin{pmatrix} \boldsymbol{A} & \boldsymbol{AB} \\ \boldsymbol{O} & -\boldsymbol{CAB} \end{pmatrix} \xrightarrow{c_2 - c_1 \cdot \boldsymbol{B}} \begin{pmatrix} \boldsymbol{A} & \boldsymbol{O} \\ \boldsymbol{O} & -\boldsymbol{CAB} \end{pmatrix}$,
$\mathrm{r}(\boldsymbol{M}) = \mathrm{r}(\boldsymbol{A}) + \mathrm{r}(\boldsymbol{CAB})$, 因为 $\mathrm{r}(\boldsymbol{M}) = \mathrm{r}(\boldsymbol{N})$, 则 $n + \mathrm{r}(\boldsymbol{AB}) = \mathrm{r}(\boldsymbol{A}) + \mathrm{r}(\boldsymbol{CAB})$, 若 $\mathrm{r}(\boldsymbol{C}) = n$, 则 $\mathrm{r}(\boldsymbol{CAB}) = \mathrm{r}(\boldsymbol{AB})$, 从而 $\mathrm{r}(\boldsymbol{A}) = n$, 故选(A).
当 $m = n$ 时, 若 $\boldsymbol{A} = \boldsymbol{C} = \boldsymbol{E}, \boldsymbol{B} = \boldsymbol{O}$ 时, 显然满足题意, 故选项(C)(D)不正确.
取 $m = 3, n = 2, \boldsymbol{A} = \boldsymbol{E}, \boldsymbol{B} = \boldsymbol{O}, \boldsymbol{C}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$, 满足题意, 故(B)不正确.

答案 选(C)
解 因为 $\boldsymbol{\beta}$ 可由 $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3$ 线性表示, 所以 $\mathrm{r}(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3) = \mathrm{r}(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3,\boldsymbol{\beta})$.
$\begin{pmatrix} 1 & 1 & -1 & \vdots & 3 \\ 2 & 0 & a & \vdots & 2 \\ -1 & 1 & 3 & \vdots & b \end{pmatrix} \sim \begin{pmatrix} 1 & 1 & -1 & \vdots & 3 \\ 0 & -2 & a + 2 & \vdots & -4 \\ 0 & 0 & a + 4 & \vdots & b - 1 \end{pmatrix}$,
故有 $a \neq -4$ 或 $a = -4, b = 1$.
又因为 $\boldsymbol{\beta}$ 不能由 $\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2$ 表示, 所以 $\mathrm{r}(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\beta}) \neq \mathrm{r}(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2) = 2$, 即 $b \neq 1$. 综上所得 $a \neq -4$ 且 $b \neq 1$. 故应选(C).

答案 选(D).
解 $\boldsymbol{A}$ 为 $2 \times 3$ 矩阵, 则 $\boldsymbol{AA}^{\mathrm{T}},\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 分别为 2 阶,3 阶方阵. 由 $|\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A} - 2\boldsymbol{E}| = 0$, 可知 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 的一个特征值为 2, 从而 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}\boldsymbol{x} = 2\boldsymbol{x}\ (\boldsymbol{x} \neq \boldsymbol{0})$, 因 $\boldsymbol{x} \neq \boldsymbol{0}$, 所以 $\boldsymbol{A}\boldsymbol{x} \neq \boldsymbol{0}$, 故 $\boldsymbol{AA}^{\mathrm{T}}(\boldsymbol{A}\boldsymbol{x}) = 2(\boldsymbol{A}\boldsymbol{x})$, 2 也为 $\boldsymbol{AA}^{\mathrm{T}}$ 的特征值; 又 $|\boldsymbol{AA}^{\mathrm{T}}| = 2, \boldsymbol{AA}^{\mathrm{T}}$ 两个特征值的乘积为 2, 易知 $\boldsymbol{AA}^{\mathrm{T}}$ 两个特征值分别为 1 和 2. 同理可得 1 也为 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 的特征值, 又由 $|\boldsymbol{AA}^{\mathrm{T}}| = 2, \boldsymbol{AA}^{\mathrm{T}}$ 为满秩阵, $\mathrm{r}(\boldsymbol{AA}^{\mathrm{T}}) = \mathrm{r}(\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}) = 2$, 三阶矩阵 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 为降秩矩阵, 0 必为 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 的特征值, 从而 $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A}$ 的三个特征值分别为 2,1,0, $\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A} + 2\boldsymbol{E}$ 的三个特征值分别为 4,3,2, 故 $|\boldsymbol{A}^{\mathrm{T}}\boldsymbol{A} + 2\boldsymbol{E}| = 4 \cdot 3 \cdot 2 = 24$. 故选(D).

答案 填“$\frac {1}{e}$”.
解 原式$=\lim\limits _{x→+∞}e^{\frac {1}{\ln x}\cdot \ln (\frac {π}{2}-arctanx)}=e^{\lim\limits _{x→+∞}\frac {\frac {1}{\frac {π}{2}-arctanx}\cdot \frac {-1}{1+x^{2}}}{\frac {1}{x}}}=e^{\lim\limits _{x→+∞}\frac {-x}{(\frac {π}{2}-arctanx)(1+x^{2})}}=e^{\lim\limits _{x→+∞}\frac {-\frac {1}{x}\cdot \frac {x^{2}}{1+x^{2}}}{\frac {π}{2}-arctanx}}$
$=e^{\lim\limits _{x→+∞}\frac {-\frac {1}{x}}{\frac {π}{2}-arctanx}}=e^{\lim\limits _{x→+∞}\frac {-\frac {1}{x^{2}}}{-\frac {1}{1+x^{2}}}}=e^{\lim\limits _{x→+∞}\frac {-(1+x^{2})}{-x^{2}}}=e^{-1}.$

答案 填“$\frac {\sqrt {6}}{4}$”.
解 $∫_{0}^{x}f(x-t)dt\frac {令x-t=u}{}∫_{0}^{x}f(u)du=∫_{0}^{x}f(t)dt.$
令$F(x)=∫_{0}^{x}f(t)dt$,则$F'(x)\cdot F(x)=xcos^{4}x$,两边同时在$[0,π]$上积分得
$∫_{0}^{π}F'(x)\cdot F(x)dx=∫_{0}^{π}xcos^{4}xdx,$
即
$\frac {1}{2}F^{2}(x)|^{π}_{0}=\frac {π}{2}∫_{0}^{π}cos^{4}xdx=\frac {π}{2}\cdot 2\cdot \frac {3}{4}\cdot \frac {1}{2}\cdot \frac {π}{2}=\frac {3}{16}π^{2},$
故$F^{2}(π)=\frac {3}{8}π^{2}$,所以$F(π)=\frac {\sqrt {6}}{4}π,F(π)=-\frac {\sqrt {6}}{4}π$(舍去),得
$\frac {1}{π}∫_{0}^{π}f(x)dx=\frac {1}{π}F(π)=\frac {\sqrt {6}}{4}.$

答案 填“$f_{1}'+f_{2}'-(x+y)f_{3}'$”.
解 $\frac {\partial u}{\partial x}=f_{1}'+f_{3}'\frac {\partial z}{\partial x},\frac {\partial u}{\partial y}=f_{2}'+f_{3}'\frac {\partial z}{\partial y}.$
令$G(x,y,z)=∫_{0}^{xy+z}e^{-t^{2}}dt-1$,则$G'_{x}=e^{-(xy+z)^{2}}y,G'_{y}=e^{-(xy+z)^{2}}x,G'_{z}=e^{-(xy+z)^{2}}$,故
$\frac {\partial z}{\partial x}=-\frac {G'_{x}}{G'_{z}}=-y,\frac {\partial z}{\partial y}=-\frac {G'_{y}}{G'_{z}}=-x,$
于是
$\frac {\partial u}{\partial x}+\frac {\partial u}{\partial y}=f_{1}'-yf_{3}'+f_{2}'-xf_{3}'=f_{1}'+f_{2}'-(x+y)f_{3}'.$

答案 填“$\frac {2}{5}$”.
解 转化为直角坐标系下二次积分为
$∫_{\frac {π}{4}}^{\frac {3π}{4}}dθ∫_{0}^{\sqrt {\frac {3}{1+sin^{2}θ}}}2r^{3}sinθcosθ(1+rcosθ)dr=∫_{-1}^{1}dx∫_{|x|}^{\sqrt {\frac {3-x^{2}}{2}}}2(x^{2}y+xy)dy,$
由对称性可知，$∫_{-1}^{1}dx∫_{|x|}^{\sqrt {\frac {3-x^{2}}{2}}}2xydy=0,$
原积分$=2∫_{0}^{1}dx∫_{x}^{\sqrt {\frac {3-x^{2}}{2}}}2x^{2}ydy=2∫_{0}^{1}x^{2}(\frac {3-x^{2}}{2}-x^{2})dx=3∫_{0}^{1}(x^{2}-x^{4})dx=\frac {2}{5}.$

答案 填“$\sqrt {x}$”.
解 $\frac {dx}{dy}+\frac {1}{y}x=3y$,故
$x=e^{-∫\frac {1}{y}dy}[∫3ye^{∫\frac {1}{y}dy}dy+C]=\frac {1}{y}(y^{3}+C)=y^{2}+\frac {C}{y},$
由$y(1)=1$得$C=0$.故$y^{2}=x$,又$y(1)=1$,所以$y=\sqrt {x}.$

答案 填“3”.
解 因为$Ax=0$与$B^{T}x=0$同解,所以$r(A)=r(B^{T})=r\begin{pmatrix} A\\ B^{T}\end{pmatrix} =2.$
$(A^{T},B)=\begin{pmatrix} 1&a&|&1&1\\ -1&1&|&1&2\\ 1&2&|&1&b\end{pmatrix} \sim \begin{pmatrix} 1&a&|&1&1\\ 0&1+a&|&2&3\\ 0&2-a&|&0&b-1\end{pmatrix},$
因为$r(A^{T})=r(A^{T},B)=2$,所以$a=2,b=1$,从而$a+b=3.$

解 （Ⅰ）等式两边同时对$x$求偏导得$yf(xy) = y^2f(x) + 2x\int_{1}^{y}f(t)\mathrm{d}t$.
令$x = 1$得$yf(y) = y^2f(1) + 2\int_{1}^{y}f(t)\mathrm{d}t$，又$f(1) = 3$，等式两边同时对$y$求导得
$f(y) + yf'(y) = 6y + 2f(y)$，即$f'(y) - \frac{1}{y}f(y) = 6$.
故$f(y) = \text{e}^{\int\frac{1}{y}\mathrm{d}y}\left[\int 6\text{e}^{-\int\frac{1}{y}\mathrm{d}y}\mathrm{d}y + C\right] = y(6\ln y + C) = 6y\ln y + Cy$.
又$f(1) = 3$得$C = 3$，所以$f(y) = 6y\ln y + 3y$，即$f(x) = 6x\ln x + 3x$.

（Ⅱ）$V = \int_{1}^{\text{e}}2\pi(x - 1)(6x\ln x + 3x)\mathrm{d}x = 6\pi\int_{1}^{\text{e}}(2x^2\ln x - 2x\ln x + x^2 - x)\mathrm{d}x$
$= \left(\frac{7}{9}\text{e}^3 - \frac{1}{9} - \text{e}^2\right)6\pi$.

证明 （Ⅰ）$F(t + T) - F(t) = \int_{0}^{t+T}f(x)\mathrm{d}x - \frac{t+T}{T}\int_{0}^{T}f(x)\mathrm{d}x - \int_{0}^{t}f(x)\mathrm{d}x + \frac{t}{T}\int_{0}^{T}f(x)\mathrm{d}x$
$= \int_{t}^{t+T}f(x)\mathrm{d}x - \int_{0}^{T}f(x)\mathrm{d}x = 0$，
故$F(t)$是以$T$为周期的函数.

（Ⅱ）因为$F(t)$是$(-\infty,+\infty)$内连续的周期函数，故$F(t)$在$[0,T]$上有界，所以$F(t)$在$(-\infty,+\infty)$内有界.得
$\lim\limits_{x\rightarrow+\infty}\frac{F(x)}{x} = \lim\limits_{x\rightarrow+\infty}\left(\frac{1}{x}\int_{0}^{x}f(t)\mathrm{d}t - \frac{1}{T}\int_{0}^{T}f(t)\mathrm{d}t\right) = 0$，
即$\lim\limits_{x\rightarrow+\infty}\frac{\int_{0}^{x}f(t)\mathrm{d}t}{x} = \frac{1}{T}\int_{0}^{T}f(t)\mathrm{d}t$.

（Ⅲ）$\int_{0}^{\pi}\frac{1}{1+\cos^2t}\mathrm{d}t = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+\cos^2t}\mathrm{d}t = 2\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\cos^2t}\mathrm{d}t = 2\int_{0}^{\frac{\pi}{2}}\frac{1}{\sec^2t + 1}\mathrm{d}(\tan t)$
$= 2\int_{0}^{\frac{\pi}{2}}\frac{1}{\tan^2t + 2}\mathrm{d}(\tan t) = \frac{2}{\sqrt{2}}\arctan\frac{\tan t}{\sqrt{2}}\bigg|_{0}^{\frac{\pi}{2}} = \frac{\sqrt{2}}{2}\pi$，
$\int_{0}^{\pi}\frac{\sin^2t}{1+\cos^2t}\mathrm{d}t = \int_{0}^{\pi}\frac{1 - \cos^2t}{1+\cos^2t}\mathrm{d}t = 2\int_{0}^{\pi}\frac{1}{1+\cos^2t}\mathrm{d}t - \pi = (\sqrt{2} - 1)\pi$，
$\lim\limits_{x\rightarrow+\infty}\frac{\int_{0}^{x}\frac{\sin^2t}{1+\cos^2t}\mathrm{d}t}{\int_{0}^{x}\frac{1}{1+\cos^2t}\mathrm{d}t} = \lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}\int_{0}^{x}\frac{\sin^2t}{1+\cos^2t}\mathrm{d}t}{\frac{1}{x}\int_{0}^{x}\frac{1}{1+\cos^2t}\mathrm{d}t} = \frac{\frac{1}{\pi}\int_{0}^{\pi}\frac{\sin^2t}{1+\cos^2t}\mathrm{d}t}{\frac{1}{\pi}\int_{0}^{\pi}\frac{1}{1+\cos^2t}\mathrm{d}t} = \frac{\sqrt{2} - 1}{\frac{\sqrt{2}}{2}} = 2 - \sqrt{2}$.

解 （Ⅰ）由题意可知，
$f'_x(x,y) = 3(x - y)^2$，$f'_y(x,y) = -3(x^2 - 2xy + 2y^2 - 2)$，
故$f(x,y) = \int 3(x - y)^2\mathrm{d}x = (x - y)^3 + C(y)$，又
$f'_y(x,y) = -3(x - y)^2 + C'(y) = -3(x^2 - 2xy + 2y^2 - 2)$，
故$C'(y) = 6 - 3y^2$，解得$C(y) = 6y - y^3 + C$，从而$f(x,y) = (x - y)^3 - y^3 + 6y + C$.代入$f(0,0) = 0$，有$f(x,y) = (x - y)^3 - y^3 + 6y$.

（Ⅱ）① 区域内：令$\begin{cases}f'_x(x,y) = 3(x - y)^2 = 0,\\f'_y(x,y) = -3(x^2 - 2xy + 2y^2 - 2) = 0,\end{cases}$
解得驻点$(\sqrt{2},\sqrt{2})\in D,(-\sqrt{2},-\sqrt{2})\notin D$.

② 区域边界上：$L_1:x = 0(0\leqslant y\leqslant3),f(0,y) = -2y^3 + 6y,f'(0,y) = -6y^2 + 6$，解得驻点$(0,1)\in L_1,(0,-1)\notin L_1$.

$L_2:y = 0(0\leqslant x\leqslant3),f(x,0) = x^3,f'(x,0) = 3x^2$，解得驻点$(0,0)\in L_2$.

$L_3:y = 3 - x(0\leqslant x\leqslant3),f(x,3 - x) = (2x - 3)^3 - (3 - x)^3 + 6(3 - x),f' = 3(3x - 5)^2$，解得驻点$\left(\frac{5}{3},\frac{4}{3}\right)\in L_3$.

③ 边界曲线的交点为$(0,0),(0,3),(3,0)$.

④ 比较以上点的函数值，有$f(\sqrt{2},\sqrt{2}) = 4\sqrt{2},f(0,0) = 0,f(0,1) = 4,f\left(\frac{5}{3},\frac{4}{3}\right) = \frac{17}{3},f(3,0) = 27,f(0,3) = -36$，所以$f(3,0) = 27$为最大值，$f(0,3) = -36$为最小值.

解 利用$y = x$将区域$D$分割为$D_1,D_2$，如图所示，

$\iint_{D}f(x,y)\mathrm{d}\sigma = \iint_{D_1}\frac{(y + 1)\text{e}^y}{(1 + x)^2}\mathrm{d}\sigma + \iint_{D_2}(x^2 + y^2)\mathrm{d}\sigma$.

$\iint_{D_2}(x^2 + y^2)\mathrm{d}\sigma = \int_{0}^{1}\mathrm{d}y\int_{0}^{y}(x^2 + y^2)\mathrm{d}x = \frac{4}{3}\int_{0}^{1}y^3\mathrm{d}y = \frac{1}{3}$.

$\iint_{D_1}\frac{(y + 1)\text{e}^y}{(1 + x)^2}\mathrm{d}\sigma = \int_{0}^{1}\mathrm{d}x\int_{0}^{x}\frac{(y + 1)\text{e}^y}{(1 + x)^2}\mathrm{d}y = \int_{0}^{1}\frac{y\text{e}^y}{(1 + x)^2}\bigg|_{0}^{x}\mathrm{d}x = \int_{0}^{1}\frac{x\text{e}^x}{(1 + x)^2}\mathrm{d}x$
$= -\int_{0}^{1}x\text{e}^x\mathrm{d}\frac{1}{1 + x} = -\frac{x\text{e}^x}{1 + x}\bigg|_{0}^{1} + \int_{0}^{1}\frac{1}{1 + x}\mathrm{d}x\text{e}^x$
$= -\frac{\text{e}}{2} + \int_{0}^{1}\text{e}^x\mathrm{d}x = -\frac{\text{e}}{2} + \text{e} - 1 = \frac{\text{e}}{2} - 1$.

所以$\iint_{D}f(x,y)\mathrm{d}\sigma = \frac{1}{3} + \frac{\text{e}}{2} - 1 = \frac{\text{e}}{2} - \frac{2}{3}$.

证明一 （Ⅰ）$\int_{0}^{a}f(x)\mathrm{d}x-\frac{1}{2}a^{2}f(a)=\int_{0}^{a}[f(x)-f(a)x]\mathrm{d}x=0$，
由积分中值定理知，存在$c\in(0,a)$有$[f(c)-f(a)c]\cdot a=0$.
令$F(x)=f(x)-f(a)x$，故$F(c)=0$，又$f(x)$为奇函数，得$F(x)$在$[-a,a]$为奇函数，所以$F(0)=0$，$F(-c)=-F(c)=0$.
由$F(c)=0=F(0)$，故存在$\xi\in(0,a)$，有$F'(\xi)=0$，即$f'(\xi)=f(a)$.
（Ⅱ）由（Ⅰ）知$F(-c)=F(0)$，故存在$x_{0}\in(-a,0)$，有$F'(x_{0})=0$，即$f'(x_{0})=f(a)$.
令$G(x)=(f'(x)-f(a))e^{-\lambda x}$，得$G(x_{0})=0$，$G(\xi)=0$，由罗尔定理知存在$\eta\in(-a,a)$，有$G'(\eta)=0$，即$f''(\eta)-\lambda f'(\eta)+\lambda f(a)=0$.

证明二 （Ⅰ）令$F(t)=\int_{0}^{t}f(x)\mathrm{d}x-\frac{1}{2}t^{2}f(a)$，显然$F(t)$是偶函数，则$F(a)=0$，$F(0)=0$，$F(-a)=0$，由三次罗尔定理可得存在$\xi\in(0,a)$，有$F''(\xi)=0$，即$f'(\xi)=f(a)$.
（Ⅱ）因为$f(x)$为奇函数，所以$f'(x)$是偶函数，故$f'(-\xi)=f(a)$.
令$G(x)=(f'(x)-f(a))e^{-\lambda x}$，得$G(-\xi)=0$，$G(\xi)=0$，由罗尔定理知存在$\eta\in(-a,a)$，有$G'(\eta)=0$，即$f''(\eta)-\lambda f'(\eta)+\lambda f(a)=0$.

解 $\boldsymbol{A}=(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})=\begin{pmatrix}a&0&1\\0&-a&0\\1&0&a\end{pmatrix}$，$\boldsymbol{B}=(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=\begin{pmatrix}a&a&a^{2}\\0&1&0\\1&1&a\end{pmatrix}\stackrel{r}{\to}\begin{pmatrix}1&1&a\\0&1&0\\0&0&0\end{pmatrix}$.

（Ⅰ）由题设知向量组（i）与（ii）等价，
$\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})=\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3},\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=\mathrm{r}(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})$，而$\mathrm{r}(\boldsymbol{B})=2$，
从而$\mathrm{r}(\boldsymbol{A})=2$，$|\boldsymbol{A}|=\begin{vmatrix}a&0&1\\0&-a&0\\1&0&a\end{vmatrix}=-a(a^{2}-1)=0$，又已知$a\neq0$，可得$a=\pm1$；

当$a=1$时，
$(\boldsymbol{A},\boldsymbol{B})=\begin{pmatrix}1&0&1&1&1&1\\0&-1&0&0&1&0\\1&0&1&1&1&1\end{pmatrix}\stackrel{r}{\to}\begin{pmatrix}1&0&1&1&1&1\\0&-1&0&0&1&0\\0&0&0&0&0&0\end{pmatrix}$.
此时，$\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})=\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3},\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=\mathrm{r}(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=2$.

当$a=-1$时，
$(\boldsymbol{A},\boldsymbol{B})=\begin{pmatrix}-1&0&1&-1&-1&1\\0&1&0&0&1&0\\1&0&-1&1&1&-1\end{pmatrix}\stackrel{r}{\to}\begin{pmatrix}-1&0&1&-1&-1&1\\0&1&0&0&1&0\\0&0&0&0&0&0\end{pmatrix}$.
此时有$\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})=\mathrm{r}(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3},\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=\mathrm{r}(\boldsymbol{\beta}_{1},\boldsymbol{\beta}_{2},\boldsymbol{\beta}_{3})=2$，从而$a=\pm1$.

（Ⅱ）若矩阵$\boldsymbol{A}$与$\boldsymbol{B}$相似，则$\boldsymbol{A}$与$\boldsymbol{B}$必等价，于是$\mathrm{r}(\boldsymbol{A})=\mathrm{r}(\boldsymbol{B})=2$. 若$\boldsymbol{A}$与$\boldsymbol{B}$不等价，则$\boldsymbol{A}$与$\boldsymbol{B}$一定不相似，故仅讨论$a=\pm1$两种情况.

当$a=1$时，$\mathrm{tr}(\boldsymbol{A})\neq\mathrm{tr}(\boldsymbol{B})$，故不存在可逆矩阵$\boldsymbol{P}$，使得$\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P}=\boldsymbol{B}$；

当$a=-1$时，$\boldsymbol{A}=\begin{pmatrix}-1&0&1\\0&1&0\\1&0&-1\end{pmatrix}$，$\boldsymbol{B}=\begin{pmatrix}-1&-1&1\\0&1&0\\1&1&-1\end{pmatrix}$

由$|\boldsymbol{A}-\lambda\boldsymbol{E}|=|\boldsymbol{B}-\lambda\boldsymbol{E}|=\lambda(1-\lambda)(\lambda+2)$，矩阵$\boldsymbol{A},\boldsymbol{B}$有相同的特征值：$\lambda_{1}=0$，$\lambda_{2}=1$，$\lambda_{3}=-2$. 三个特征值互异，因此$\boldsymbol{A},\boldsymbol{B}$与同一个对角阵$\boldsymbol{\Lambda}=\begin{pmatrix}0&&\\&1&\\&&-2\end{pmatrix}$相似，故存在可逆矩阵$\boldsymbol{P}$，使得$\boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P}=\boldsymbol{B}$，下面求$\boldsymbol{P}$.

当$\lambda_{1}=0$时，求$\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\xi}_{1}=(1,0,1)^{\mathrm{T}}$，$\boldsymbol{B}\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\eta}_{1}=(1,0,1)^{\mathrm{T}}$.

当$\lambda_{2}=1$时，求$(\boldsymbol{A}-\boldsymbol{E})\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\xi}_{2}=(0,1,0)^{\mathrm{T}}$，$(\boldsymbol{B}-\boldsymbol{E})\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\eta}_{2}=(-1,3,1)^{\mathrm{T}}$.

当$\lambda_{3}=-2$时，求$(\boldsymbol{A}+2\boldsymbol{E})\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\xi}_{3}=(-1,0,1)^{\mathrm{T}}$，$(\boldsymbol{B}+2\boldsymbol{E})\boldsymbol{x}=\boldsymbol{0}$，可得$\boldsymbol{\eta}_{3}=(-1,0,1)^{\mathrm{T}}$.

令$\boldsymbol{P}_{1}=(\boldsymbol{\xi}_{1},\boldsymbol{\xi}_{2},\boldsymbol{\xi}_{3})$，$\boldsymbol{P}_{2}=(\boldsymbol{\eta}_{1},\boldsymbol{\eta}_{2},\boldsymbol{\eta}_{3})$，$\boldsymbol{P}_{1}^{-1}\boldsymbol{A}\boldsymbol{P}_{1}=\boldsymbol{\Lambda}=\boldsymbol{P}_{2}^{-1}\boldsymbol{B}\boldsymbol{P}_{2}$，即$(\boldsymbol{P}_{1}\boldsymbol{P}_{2}^{-1})^{-1}\boldsymbol{A}(\boldsymbol{P}_{1}\boldsymbol{P}_{2}^{-1})=\boldsymbol{B}$，所求可逆矩阵
$\boldsymbol{P}=\boldsymbol{P}_{1}\boldsymbol{P}_{2}^{-1}=\begin{pmatrix}1&0&-1\\0&1&0\\1&0&1\end{pmatrix}\begin{pmatrix}1&-1&-1\\0&3&0\\1&1&1\end{pmatrix}^{-1}=\frac{1}{6}\begin{pmatrix}1&0&-1\\0&1&0\\1&0&1\end{pmatrix}\begin{pmatrix}3&0&3\\0&2&0\\-3&-2&3\end{pmatrix}=\frac{1}{6}\begin{pmatrix}6&2&0\\0&2&0\\0&-2&6\end{pmatrix}$.