答案 C
解析 当$t>0$时，有$\begin{cases} x=3t, \\ y=t(\text{e}^t - 1), \end{cases}$ 可得
$\frac{\text{d}y}{\text{d}x} = \frac{\text{e}^t + t\text{e}^t - 1}{3}$.
当$t<0$时，有$\begin{cases} x=t, \\ y=-t(\text{e}^{-t} - 1), \end{cases}$ 可得
$\frac{\text{d}y}{\text{d}x} = \frac{-\text{e}^{-t} + t\text{e}^{-t} + 1}{1}$.
当$x=0$时，$t=0$.
$f'_+(0) = \lim\limits_{x \to 0^+} \frac{f(x) - f(0)}{x} = \lim\limits_{t \to 0^+} \frac{t(\text{e}^t - 1) - 0}{t} = 0$，
$f'_-(0) = \lim\limits_{x \to 0^-} \frac{f(x) - f(0)}{x} = \lim\limits_{t \to 0^-} \frac{-t(\text{e}^{-t} - 1)}{t} = 0$，
故$f'(0) = 0$.
又
$\lim\limits_{x \to 0^+} f'(x) = \lim\limits_{t \to 0^+} \frac{\text{e}^t + t\text{e}^t - 1}{3} = 0 = f'(0)$，
$\lim\limits_{x \to 0^-} f'(x) = \lim\limits_{t \to 0^-} \frac{-\text{e}^{-t} + t\text{e}^{-t} + 1}{1} = 0 = f'(0)$，
故$\lim\limits_{x \to 0} f'(x) = f'(0) = 0$.
选项C正确.

答案 D
解析 由
$[\text{e}^{2x}f(x)]'=\text{e}^{2x}[2f(x)+f'(x)]$，
$\lim\limits_{x\to+\infty}[2f(x)+f'(x)]=1$，
知
$\lim\limits_{x\to+\infty}[\text{e}^{2x}f(x)]'=+\infty$.
令$h(x)=\text{e}^{2x}f(x)$，由极限性质，知存在$x_0>1$.当$x>x_0$时，$h'(x)>1$；当$x>x_0$时，
应用拉格朗日中值定理，存在$\xi\in(x_0,x)$，使得
$h(x)-h(x_0)=h'(\xi)(x-x_0)>x-x_0$，
从而有
$h(x)>h(x_0)+x-x_0$.
上式两端取极限，得
$\lim\limits_{x\to+\infty}h(x)=+\infty$，
即
$\lim\limits_{x\to+\infty}\text{e}^{2x}f(x)=+\infty$（结论②正确）.
由洛必达法则，有
$\lim\limits_{x\to+\infty}f(x)=\lim\limits_{x\to+\infty}\frac{h(x)}{\text{e}^{2x}}=\lim\limits_{x\to+\infty}\frac{h'(x)}{2\text{e}^{2x}}$
$=\lim\limits_{x\to+\infty}\frac{2f(x)+f'(x)}{2}=\frac{1}{2}$，
故
$\lim\limits_{x\to+\infty}f'(x)=\lim\limits_{x\to+\infty}[2f(x)+f'(x)]-2\lim\limits_{x\to+\infty}f(x)$
$=1-2\lim\limits_{x\to+\infty}f(x)=1-2\times\frac{1}{2}=0$（结论④正确）.
选项D正确.

答案 B
解析 由已知，\( I = \int_{0}^{\frac{\pi}{4}} \frac{dx}{\sin^{p}x \cos^{q}x} + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^{p}x \cos^{q}x} \) 收敛.
由 \( \lim_{x \to 0^{+}} \frac{\frac{1}{\sin^{p}x \cos^{q}x}}{\frac{1}{x^{p}}} = 1 \)，知 \( \int_{0}^{\frac{\pi}{4}} \frac{dx}{\sin^{p}x \cos^{q}x} \) 与 \( \int_{0}^{\frac{\pi}{4}} \frac{1}{x^{p}}dx \) 的敛散性相同.
故当 \( p < 1 \) 时，\( \int_{0}^{\frac{\pi}{4}} \frac{dx}{\sin^{p}x \cos^{q}x} \) 收敛.
又 \( \lim_{x \to \frac{\pi}{2}^{-}} \frac{\frac{1}{\sin^{p}x \cos^{q}x}}{\frac{1}{(\frac{\pi}{2} - x)^{q}}} = \lim_{x \to \frac{\pi}{2}^{-}} \frac{\sin^{q}(\frac{\pi}{2} - x)}{\frac{1}{(\frac{\pi}{2} - x)^{q}}} = 1 \)，
所以 \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^{p}x \cos^{q}x} \) 与 \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{(\frac{\pi}{2} - x)^{q}} \) 的敛散性相同.
故当 \( q < 1 \) 时，\( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{dx}{\sin^{p}x \cos^{q}x} \) 收敛.
综上所述，当 \( p < 1 \) 且 \( q < 1 \) 时，原积分收敛.
选项 B 正确.

答案 B
解析 由已知，有$\frac{\partial f(x,y)}{\partial x} = -f(x,y)$，即$\frac{\partial f(x,y)}{\partial x} / f(x,y) = -1$.
上式两边对$x$积分，得
$\ln|f(x,y)| = -x + c_1(y)$，
解得
$f(x,y) = \pm \mathrm{e}^{c_1(y)} \cdot \mathrm{e}^{-x} = c(y) \cdot \mathrm{e}^{-x}$.
由$f(0,y) = 1 + y^2$，得$c(y) = 1 + y^2$. 故$f(x,y) = \mathrm{e}^{-x}(1 + y^2)$.
由
$\frac{\partial f}{\partial x}\bigg|_{(0,1)} = -\mathrm{e}^{-x}(1 + y^2)\bigg|_{(0,1)} = -2$，
$\frac{\partial f}{\partial y}\bigg|_{(0,1)} = \mathrm{e}^{-x} \cdot 2y\bigg|_{(0,1)} = 2$，
知
$\mathrm{d}f|_{(0,1)} = -2\mathrm{d}x + 2\mathrm{d}y$.
选项B正确.

答案 C
解析 \( I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d\theta \int_{0}^{\frac{1}{\sin\theta}} f(r) r dr \) 在 \( xOy \) 直角坐标系下的积分区域如下左图阴影部分所示，则有
\( I = \int_{0}^{1} dx \int_{x}^{1} f(\sqrt{x^2 + y^2}) dy \).
排除选项 A,B.
在 \( rO\theta \) 极坐标系下，如下右图阴影部分所示，则有
\( I = \int_{0}^{1} dr \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(r) r d\theta + \int_{1}^{\sqrt{2}} dr \int_{\frac{\pi}{4}}^{\arcsin\frac{1}{r}} f(r) r d\theta \).
选项 C 正确.

答案 D
解析 由$y'+2y=f(x)$，有
$y=y(x)=\mathrm{e}^{-\int 2\mathrm{d}x}\left(\int f(t)\mathrm{e}^{\int 2\mathrm{d}t}\mathrm{d}t + C\right)=C\mathrm{e}^{-2x} + \mathrm{e}^{-2x}\int_{0}^{x}\mathrm{e}^{2t}f(t)\mathrm{d}t$，
故
$\lim\limits_{x\to+\infty}y(x)=\lim\limits_{x\to+\infty}C\mathrm{e}^{-2x} + \lim\limits_{x\to+\infty}\frac{\int_{0}^{x}\mathrm{e}^{2t}f(t)\mathrm{d}t}{\mathrm{e}^{2x}}$
$=0 + \lim\limits_{x\to+\infty}\frac{\mathrm{e}^{2x}f(x)}{2\mathrm{e}^{2x}}$
$=\lim\limits_{x\to+\infty}\frac{1}{2}f(x)=\frac{1}{2}$.
选项D正确.

答案 B
解析 依题设，如图所示，用微元法.
任取细棒上\([x, x + dx]\)一段，则它到质点的距离为\( a - x \)。故质点与细棒之间的引力为
\( F = \int_{-l}^{0} \frac{G\rho m dx}{(a - x)^2} = G\rho m\left(\frac{1}{a} - \frac{1}{a + l}\right) = \frac{G\rho ml}{a(a + l)} \)。
选项B正确.

答案 B
解析 显然\( BX = 0 \)的解是\( ABX = 0 \)的解，故只要找出\( ABX = 0 \)的解是\( BX = 0 \)的解的充分必要条件即可.
对于选项A：\( \text{r}(A) = s \)，则\( AX = 0 \)只有零解，由\( ABX = 0 \)必有\( BX = 0 \)，即\( ABX = 0 \)的解是\( BX = 0 \)的解.故选项A是充分条件，并不必要.因为当\( \text{r}(A) < s \)时，\( ABX = 0 \)与\( BX = 0 \)仍可以同解.排除选项A.
对于选项B：当\( ABX = 0 \)与\( BX = 0 \)同解时，它们有相同的基础解系，从而\( \text{r}(AB) = \text{r}(B) \).反之，若\( \text{r}(AB) = \text{r}(B) \)，则\( ABX = 0 \)与\( BX = 0 \)的基础解系中解向量的个数相同，若\( \eta_1, \eta_2, \cdots, \eta_t \)是\( BX = 0 \)的基础解系，而\( BX = 0 \)的解必是\( ABX = 0 \)的解，故\( \eta_1, \eta_2, \cdots, \eta_t \)是\( ABX = 0 \)的线性无关解.因此\( \eta_1, \eta_2, \cdots, \eta_t \)必是\( ABX = 0 \)的基础解系.从而\( ABX = 0 \)与\( BX = 0 \)同解.
对于选项C：由\( \text{r}(A) = \text{r}(B) \)，不能证得\( \text{r}(AB) = \text{r}(B) \).
例如：
\( A = \begin{pmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{pmatrix} \)，\( B = \begin{pmatrix} 0&0&0 \\ 0&1&0 \\ 0&0&1 \end{pmatrix} \)，\( AB = \begin{pmatrix} 0&0&0 \\ 0&1&0 \\ 0&0&0 \end{pmatrix} \).
排除选项C.
对于选项D：由\( \text{r}(A) = m \)，不能证得\( \text{r}(AB) = \text{r}(B) \).
例如：
\( A = \begin{pmatrix} 1&0&0 \\ 0&0&1 \end{pmatrix} \)，\( B = \begin{pmatrix} 1&0 \\ 0&1 \\ 0&0 \end{pmatrix} \)，\( AB = \begin{pmatrix} 1&0 \\ 0&0 \end{pmatrix} \).
排除选项D.综上可知，选项B正确.

答案 A

解析：对A作初等列变换,相当于右乘相应的初等矩阵,对A作初等行变换,相当于左乘相应的初等矩阵.依题设,有
\[
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
A
\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{pmatrix}
=C,
\]
令\( Q = \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 1
\end{pmatrix} \)，则\( Q = \begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & 0
\end{pmatrix} \)为正交矩阵.
故\( Q^T A Q = Q^{-1} A Q = C \).
从而A与C相似,所以A与C有相同的特征值-1,-2,1.
故\( X^T C X \)在正交变换\( X = Q Y \)下的标准形为\( -y_1^2 + y_2^2 - 2y_3^2 \)，
选项A正确.

答案 C
解析 \( A = \begin{pmatrix} 1 - a & a \\ a & a \end{pmatrix} \)正定的充分必要条件是\( \begin{cases} 1 - a > 0, \\ (1 - a)a - a^2 > 0, \end{cases} \) 解得\( a \in (0, \frac{1}{2}) \).

\( f(x_1, x_2) = \begin{vmatrix} 1 - a & a & x_1 \\ a & a & x_2 \\ -x_1 & -x_2 & 0 \end{vmatrix} = x_1 \begin{vmatrix} a & a \\ -x_1 & -x_2 \end{vmatrix} - x_2 \begin{vmatrix} 1 - a & a \\ -x_1 & -x_2 \end{vmatrix} \)
\( = -a x_1 \begin{vmatrix} 1 & 1 \\ x_1 & x_2 \end{vmatrix} + x_2 \begin{vmatrix} 1 - a & a \\ x_1 & x_2 \end{vmatrix} \)
\( = a x_1 (x_1 - x_2) + x_2 [(1 - a)x_2 - a x_1] \)
\( = a x_1^2 + (1 - a)x_2^2 - 2a x_1 x_2 \)，
其矩阵\( B = \begin{pmatrix} a & -a \\ -a & 1 - a \end{pmatrix} \)，其正定的充分必要条件是\( \begin{cases} a > 0, \\ (1 - a)a - a^2 > 0, \end{cases} \) 解得\( a \in (0, \frac{1}{2}) \).因此\( A \)正定是\( f(x_1, x_2) \)正定的充分必要条件.选项 C 正确.

答案 $y = -x + 1$
解析 由$\int_{0}^{x}f(t - x)dt \stackrel{t - x = u}{=} \int_{-x}^{0}f(u)du$，得
$\int_{-x}^{0}f(u)du = x + \frac{1}{2}x^{2} + \ln|1 - x|$．
上式两边对$x$求导，得
$f(-x) = 1 + x + \frac{-1}{1 - x} = \frac{-x^{2}}{1 - x}$，
故
$f(x) = \frac{-x^{2}}{1 + x}$．
由$\lim_{x \to \infty}\frac{f(x)}{x} = -1$，$\lim_{x \to \infty}[f(x) + x] = 1$，知$y = f(x)$的斜渐近线方程为$y = -x + 1$．

答案 \( \frac{7\pi}{4} \)
解析 由积分式可知，积分区域
\( D = \left\{ (x, y) \mid 0 \leqslant y \leqslant 2, \sqrt{2y - y^2} \leqslant x \leqslant \sqrt{4 - y^2} \right\} \)，
如图阴影部分所示. 下面将积分式改用极坐标计算.

\( \int_{0}^{2} \mathrm{d}y \int_{\sqrt{2y - y^2}}^{\sqrt{4 - y^2}} (1 + x^2 + y^2) \mathrm{d}x = \int_{0}^{\frac{\pi}{2}} \mathrm{d}\theta \int_{2\sin\theta}^{2} (1 + r^2) r \mathrm{d}r \)
\( = \int_{0}^{\frac{\pi}{2}} \left. \left( \frac{1}{2}r^2 + \frac{1}{4}r^4 \right) \right|_{2\sin\theta}^{2} \mathrm{d}\theta \)
\( = \int_{0}^{\frac{\pi}{2}} \left[ 2(1 - \sin^2\theta) + 4(1 - \sin^4\theta) \right] \mathrm{d}\theta \)
\( = 2\int_{0}^{\frac{\pi}{2}} \cos^2\theta (3 + 2\sin^2\theta) \mathrm{d}\theta \)
\( = 2\int_{0}^{\frac{\pi}{2}} \cos^2\theta (5 - 2\cos^2\theta) \mathrm{d}\theta \)
\( = 10\int_{0}^{\frac{\pi}{2}} \cos^2\theta \mathrm{d}\theta - 4\int_{0}^{\frac{\pi}{2}} \cos^4\theta \mathrm{d}\theta = \frac{5\pi}{2} - \frac{3\pi}{4} = \frac{7\pi}{4} \).

答案 $\frac {x}{\sin x}$

解析
$\lim\limits _{t→0}[\frac {f(x+t\sin x)}{f(x)}]^{\frac {1}{t}}=e^{\lim\limits _{t→0}\frac {1}{t}\ln \frac {f(x+t\sin x)}{f(x)}}$．
又$\lim\limits _{t→0}\frac {1}{t}\ln \frac {f(x+t\sin x)}{f(x)}=\lim\limits _{t→0}\frac {\ln f(x+t\sin x)-\ln f(x)}{t\sin x}\cdot \sin x=[\ln f(x)]'\sin x$．
由已知，有
$e^{[\ln f(x)]'\sin x}=e^{\frac {\sin x - x\cos x}{x}}$，
即
$[\ln f(x)]'=\frac {\sin x - x\cos x}{x\sin x}$． ①
①式两边积分，得
$\ln f(x)=\int(\frac {1}{x}-\frac {\cos x}{\sin x})dx=\ln x - \ln \sin x + C_1$，
解得$f(x)=\frac {Cx}{\sin x}$．由$\lim\limits _{x→0^{+}}f(x)=1$，得$C=1$．故$f(x)=\frac {x}{\sin x}$．

答案 $\frac{\pi}{4} + \frac{1}{2}\ln 2$
解析
$\lim\limits_{n \to \infty} \sum\limits_{k=1}^{n} \int_{k}^{k+1} \frac{\arctan x}{x^2} dx = \lim\limits_{n \to \infty} \int_{1}^{n+1} \frac{\arctan x}{x^2} dx = \int_{1}^{+\infty} \frac{\arctan x}{x^2} dx$.
$\int_{1}^{+\infty} \frac{\arctan x}{x^2} dx = -\int_{1}^{+\infty} \arctan x d\left( \frac{1}{x} \right)$
$= -\frac{1}{x} \arctan x \bigg|_{1}^{+\infty} + \int_{1}^{+\infty} \frac{dx}{x(1 + x^2)}$
$= \frac{\pi}{4} + \int_{1}^{+\infty} \frac{dx}{x(1 + x^2)}$,
又 $\int_{1}^{+\infty} \frac{dx}{x(1 + x^2)} = \int_{1}^{+\infty} \left( \frac{1}{x} - \frac{x}{1 + x^2} \right) dx = \lim\limits_{b \to +\infty} \left( \ln x - \ln \sqrt{1 + x^2} \right) \bigg|_{1}^{b}$
$= \lim\limits_{b \to +\infty} \ln \frac{x}{\sqrt{1 + x^2}} \bigg|_{1}^{b} = \frac{1}{2}\ln 2$,
故 原式 $= \frac{\pi}{4} + \frac{1}{2}\ln 2$.

答案 $-\sqrt{2}\mathrm{e}^{-\frac{1}{2}}$

解析 由$f(x,y)=\int_{x^{2}+y^{2}}^{\frac{y}{x}} \mathrm{e}^{-t^{2}} \mathrm{d}t$，有
$\frac{\partial f}{\partial x}=\mathrm{e}^{-\left(\frac{y}{x}\right)^{2}} \cdot \left(-\frac{y}{x^{2}}\right) - \mathrm{e}^{-\left(x^{2}+y^{2}\right)^{2}} \cdot 2x$，
$\frac{\partial f}{\partial y}=\mathrm{e}^{-\left(\frac{y}{x}\right)^{2}} \cdot \frac{1}{x} - \mathrm{e}^{-\left(x^{2}+y^{2}\right)^{2}} \cdot 2y$，
故
$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = -2(x^{2}+y^{2})\mathrm{e}^{-\left(x^{2}+y^{2}\right)^{2}}$．
令$x^{2}+y^{2}=u(u>0)$，则$g(u)\stackrel{\text{记}}{=} -2u\mathrm{e}^{-u^{2}}$．
由$g'(u)=-2\mathrm{e}^{-u^{2}}(1 - 2u^{2})=0$，得$g(u)$的唯一驻点$u=\frac{1}{\sqrt{2}}$．
且当$0 < u < \frac{1}{\sqrt{2}}$时，$g'(u) < 0$；当$u > \frac{1}{\sqrt{2}}$时，$g'(u) > 0$．
故$g\left(\frac{1}{\sqrt{2}}\right)=-2 \cdot \frac{1}{\sqrt{2}}\mathrm{e}^{-\left(\frac{1}{\sqrt{2}}\right)^{2}} = -\sqrt{2}\mathrm{e}^{-\frac{1}{2}}$为极小值．

答案 10
解析 $(A+E)(A-2E)=O$即$A^2 - A - 2E = O$.
设A的任一特征值为$\lambda$，则$\lambda^2 - \lambda - 2 = 0$.故A可能的特征值为$-1,2$.
由$|A|=2$知，A的特征值只能为$-1,-1,2$，从而$A^{-1}$的特征值为$-1,-1,\frac{1}{2}$.
$|A^{-1}+A|=|(E+A^2)A^{-1}|=|E+A^2||A^{-1}|=|A+3E||A^{-1}|$
$=(-1+3)\cdot(-1+3)\cdot(2+3)\cdot(-1)\cdot(-1)\cdot\frac{1}{2}=10$.

解析 先求出$f(x)$．由$f(x+y)=f(x)+f(y)+1$，知$f(0)=-1$，故
$f'(x)=\lim\limits _{\Delta x→0}\dfrac {f(x+\Delta x)-f(x)}{\Delta x}=\lim\limits _{\Delta x→0}\dfrac {f(x)+f(\Delta x)+1-f(x)}{\Delta x}$
$=\lim\limits _{\Delta x→0}\dfrac {f(\Delta x)-f(0)}{\Delta x}=f'(0)=1$，
从而
$f(x)=f(0)+\int _{0}^{x}f'(t)\mathrm{d}t=x-1$．
于是
$\lim\limits _{x→0}\left[\dfrac {f(x)+1}{\sin x}\right]^{\frac {1}{x^{2}}}=\lim\limits _{x→0}\left(\dfrac {x}{\sin x}\right)^{\frac {1}{x^{2}}}=\mathrm{e}^{\lim\limits _{x→0}\frac {1}{x^{2}}\ln \frac {x}{\sin x}}$，
而
$\lim\limits _{x→0}\dfrac {1}{x^{2}}\ln \dfrac {x}{\sin x}=\lim\limits _{x→0}\dfrac {1}{x^{2}}\ln \left(\dfrac {x}{\sin x}+1-1\right)=\lim\limits _{x→0}\dfrac {1}{x^{2}}\cdot \left(\dfrac {x}{\sin x}-1\right)$
$=\lim\limits _{x→0}\dfrac {x-\left[x-\dfrac {x^{3}}{3!}+o(x^{3})\right]}{x^{3}}=\dfrac {1}{6}$．
故
原式$=\mathrm{e}^{\frac {1}{6}}$．

解析 （Ⅰ）由已知，设注入\( t \)秒后液面高度为\( y \)米，则此时液体的体积为
\( V = \pi \int_{0}^{y} x^2 dy = \pi \int_{0}^{y} (y + \sin y)^2 dy \)，
故
\( \frac{dV}{dt} = \frac{dV}{dy} \cdot \frac{dy}{dt} = \pi (y + \sin y)^2 \cdot \frac{dy}{dt} \)。
由\( \frac{dV}{dt} = \pi \)，知\( \frac{dy}{dt} = \frac{1}{(y + \sin y)^2} \)，故所求液面上升的速率为
\( \frac{dy}{dt}\big|_{y = \frac{\pi}{4}} = \frac{1}{(\frac{\pi}{4} + \frac{\sqrt{2}}{2})^2} = \frac{16}{(\pi + 2\sqrt{2})^2} (m/s) \)。
（Ⅱ）设需要\( t \)秒的时间将容器内注满液体，则
\( \pi \int_{0}^{\frac{\pi}{2}} (y + \sin y)^2 dy = \pi t \)。
\( t = \int_{0}^{\frac{\pi}{2}} (y + \sin y)^2 dy = \int_{0}^{\frac{\pi}{2}} (y^2 + 2y\sin y + \sin^2 y) dy \)
\( = \frac{1}{3}y^3\big|_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} 2y\sin y dy + \int_{0}^{\frac{\pi}{2}} \sin^2 y dy \)
\( = \frac{\pi^3}{24} + \frac{\pi}{4} + 2 (s) \)。

解析 （Ⅰ）\( (2 + x^2)y' + 2xy = 2 \)变形为
\( y' + \frac{2x}{2 + x^2}y = \frac{2}{2 + x^2} \)，
解得
\( y = \mathrm{e}^{-\int\frac{2x}{2 + x^2}\mathrm{d}x}\left( \int\frac{2}{2 + x^2}\mathrm{e}^{\int\frac{2x}{2 + x^2}\mathrm{d}x}\mathrm{d}x + C \right) = (2 + x^2)^{-1}(2x + C) \)．
由\( f(0) = \frac{1}{2} \)，得\( C = 1 \)．故\( y = f(x) = \frac{1 + 2x}{2 + x^2} \)．

（Ⅱ）由\( f'(x) = \frac{2(x + 2)(1 - x)}{(2 + x^2)^2} = 0 \)，得\( x = -2 \)，\( x = 1 \)，且
\( \lim\limits_{x\to -\infty}f(x) = 0 \)，\( \lim\limits_{x\to +\infty}f(x) = 0 \)，\( f\left(-\frac{1}{2}\right) = 0 \)．

列表如下：

| \( x \)          | \( (-\infty, -2) \) | \( -2 \)  | \( \left(-2, -\frac{1}{2}\right) \) | \( -\frac{1}{2} \) | \( \left(-\frac{1}{2}, 1\right) \) | \( 1 \)  | \( (1, +\infty) \) |
|------------------|---------------------|-----------|------------------------------------|--------------------|------------------------------------|----------|---------------------|
| \( f'(x) \)      | \( - \)             | \( 0 \)   | \( + \)                            | \( + \)            | \( + \)                            | \( 0 \)  | \( - \)             |
| \( f(x) \)       | \( \searrow \)      | 极小值\( -\frac{1}{2} \) | \( \nearrow \)                      | \( 0 \)            | \( \nearrow \)                    | 极大值\( 1 \) | \( \searrow \)      |

\( f(x) \)在区间\( [t, +\infty) \)上的最小值与最大值分别记为\( m(t) \)与\( M(t) \)．由上表可知：
① 当\( t \leqslant -2 \)时，\( m(t) = f(-2) = -\frac{1}{2} \)，\( M(t) = f(1) = 1 \)；
② 当\( -2 \lt t \leqslant -\frac{1}{2} \)时，\( m(t) = f(t) = \frac{1 + 2t}{2 + t^2} \)，\( M(t) = f(1) = 1 \)；
③ 当\( -\frac{1}{2} \lt t \leqslant 1 \)时，无最小值，\( M(t) = f(1) = 1 \)；
④ 当\( t \gt 1 \)时，无最小值，\( M(t) = f(t) = \frac{1 + 2t}{2 + t^2} \)．

解析 积分区间$D_t$如图阴影部分所示，故

$F(t) = \iint_{D_t} f_{yx}''(x,y)dxdy$
$= \int_{0}^{2t} dx \int_{0}^{t} f_{yx}''(x,y)dy$
$= \int_{0}^{2t} dx \int_{0}^{t} d[f_x'(x,y)] = \int_{0}^{2t} f_x'(x,y)|_{0}^{t} dx$
$= \int_{0}^{2t} [f_x'(x,t) - f_x'(x,0)] dx$
$= \int_{0}^{2t} d[f(x,t)] - \int_{0}^{2t} d[f(x,0)]$
$= f(x,t)|_{0}^{2t} - f(x,0)|_{0}^{2t}$
$= f(2t,t) - f(0,t) - [f(2t,0) - f(0,0)]$。
$\lim_{t \to 0^+} \frac{F(t)}{t} = \lim_{t \to 0^+} \frac{f(2t,t) - f(2t,0) - [f(0,t) - f(0,0)]}{t}$
$\stackrel{洛必达法则}{=} \lim_{t \to 0^+} \frac{f_1'(2t,t) \cdot 2 + f_2'(2t,t) \cdot 1 - f_1'(2t,0) \cdot 2}{1} - \lim_{t \to 0^+} \frac{f(0,t) - f(0,0)}{t}$
$= 2f_1'(0,0) + f_2'(0,0) - 2f_1'(0,0) - f_2'(0,0) = 0$。

解析 （Ⅰ）令$F(x) = \int_{0}^{x}f(t)dt$，则$F'(x) = f(x)$，$F''(x) = f'(x)$，$F(0) = 0$，$F(1) = 0$。
对$F(x)$应用泰勒公式。对$\forall a\in[0,1]$，有
$F(x) = F(a) + F'(a)(x - a) + \frac{F''(\xi)}{2}(x - a)^2$， ①
其中$\xi$介于$a$与$x$之间。将$x = 1$，$x = 0$分别代入①式，有
$0 = F(1) = F(a) + F'(a)(1 - a) + \frac{1}{2}F''(\xi_1)(1 - a)^2$， ②
$0 = F(0) = F(a) + F'(a)(0 - a) + \frac{1}{2}F''(\xi_2)(0 - a)^2$， ③
其中$\xi_1$介于$1$与$a$之间，$\xi_2$介于$0$与$a$之间。②式$-$③式，得
$f(a) = F'(a) = -\frac{1}{2}[F''(\xi_1)(1 - a)^2 - F''(\xi_2)a^2]$，
故
$|f(a)| = |F'(a)| \leq \frac{1}{2}[|F''(\xi_1)|(1 - a)^2 + |F''(\xi_2)|a^2]$
$= \frac{1}{2}[|f'(\xi_1)|(1 - a)^2 + |f'(\xi_2)|a^2]$。
令$|f'(\xi)| = \max\{|f'(\xi_1)|, |f'(\xi_2)|\}$，$\xi\in(0,1)$，则
$|f(a)| = |F'(a)| \leq \frac{1}{2}|f'(\xi)|[(1 - a)^2 + a^2] \leq \frac{1}{2}|f'(\xi)| \leq \frac{1}{2}M$。
由$a$是$[0,1]$上任一点，知$|f(x)| \leq \frac{1}{2}M$，$x\in[0,1]$。

（Ⅱ）由于
$\iint\limits_{D}f(y)dxdy = \int_{0}^{1}dy\int_{y}^{1}f(y)dx = \int_{0}^{1}(1 - y)f(y)dy$
$= \int_{0}^{1}\left(\frac{1}{2} - y\right)f(y)dy + \frac{1}{2}\int_{0}^{1}f(y)dy$
$= \int_{0}^{1}\left(\frac{1}{2} - y\right)f(y)dy$，
故
$\left|\iint\limits_{D}f(y)dxdy\right| \leq \frac{M}{2}\int_{0}^{1}\left|\frac{1}{2} - y\right|dy \stackrel{t = y - \frac{1}{2}}{=} \frac{M}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}|t|dt = M\int_{0}^{\frac{1}{2}}tdt = \frac{M}{8}$。
命题证毕。

解析 （Ⅰ）由\( A \)与\( B \)相似，知\( r(A) = r(B) \)，而\( r(B) = 2(a \neq 0) \). 从而
\[
|A| = \begin{vmatrix} a & 0 & 1 \\ 0 & -a & 0 \\ 1 & 0 & a \end{vmatrix} = -a(a^2 - 1) = 0,
\]
得\( a = \pm 1 \).
当\( a = 1 \)时，\( \text{tr} A = 1 \)，\( \text{tr} B = -1 \)，\( A \)与\( B \)不相似，故\( a = -1 \).

（Ⅱ）当\( a = -1 \)时，由
\[
|\lambda E - A| = \begin{vmatrix} \lambda + 1 & 0 & -1 \\ 0 & \lambda - 1 & 0 \\ -1 & 0 & \lambda + 1 \end{vmatrix} = \lambda(\lambda - 1)(\lambda + 2),
\]
得\( A \)的特征值为\( 0,1,-2 \). 由\( A \)与\( B \)相似，知\( B \)的特征值也是\( 0,1,-2 \).
由
\[
0E - A = \begin{pmatrix} 1 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -1 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \alpha_1 = (1,0,1)^T \).

由
\[
E - A = \begin{pmatrix} 2 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 2 \end{pmatrix} \to \begin{pmatrix} -1 & 0 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \alpha_2 = (0,1,0)^T \).

由
\[
-2E - A = \begin{pmatrix} -1 & 0 & -1 \\ 0 & -3 & 0 \\ -1 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \alpha_3 = (-1,0,1)^T \). 将\( \alpha_1, \alpha_2, \alpha_3 \)单位化，得
\[
\gamma_1 = \frac{1}{\sqrt{2}}(1,0,1)^T, \quad \gamma_2 = (0,1,0)^T, \quad \gamma_3 = \frac{1}{\sqrt{2}}(-1,0,1)^T
\]

令\( Q_1 = (\gamma_1, \gamma_2, \gamma_3) \)，则\( Q_1^{-1}AQ_1 = \Lambda = \text{diag}(0,1,-2) \).

由
\[
0E - B = \begin{pmatrix} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \beta_1 = (1,1,0)^T \).

由
\[
E - B = \begin{pmatrix} 2 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} -1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \beta_2 = (0,0,1)^T \).

由
\[
-2E - B = \begin{pmatrix} -1 & -1 & 0 \\ -1 & -1 & 0 \\ 0 & 0 & -3 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},
\]
得特征向量\( \beta_3 = (-1,1,0)^T \). 将\( \beta_1, \beta_2, \beta_3 \)单位化，得
\[
\xi_1 = \frac{1}{\sqrt{2}}(1,1,0)^T, \quad \xi_2 = (0,0,1)^T, \quad \xi_3 = \frac{1}{\sqrt{2}}(-1,1,0)^T.
\]

令\( Q_2 = (\xi_1, \xi_2, \xi_3) \)，则\( Q_2^{-1}BQ_2 = \Lambda = \text{diag}(0,1,-2) \).

故\( Q_1^{-1}AQ_1 = Q_2^{-1}BQ_2 \)，即\( (Q_1Q_2^{-1})^{-1}A(Q_1Q_2^{-1}) = B \).

令\( Q = Q_1Q_2^{-1} \)，则\( Q^{-1}AQ = B \)，其中
\[
Q = Q_1Q_2^{-1} = Q_1Q_2^T = \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\ 0 & 1 & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ 0 & 0 & 1 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}.
\]

（Ⅲ）由（Ⅱ）知\( Q^{-1}AQ = B \)，得\( AQ = QB \). 令\( AQ = QB = P \)，则
\[
A = PQ^{-1} = PQ^T, \quad B = Q^{-1}P = Q^T P,
\]
故
\[
AB = PQ^T \cdot Q^T P = P(Q^T)^2 P.
\]
又
\[
(Q^T)^2 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}^2 = E,
\]
故
\[
AB = P(Q^T)^2 P = P^2,
\]
其中\( P = AQ = \begin{pmatrix} -1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix} \)为所求矩阵.

【注】 第（Ⅲ）问也可以采用如下方法计算：
由\( Q^{-1}AQ = B \)，有
\[
AB = A(Q^{-1}AQ) = AQ^{-1}(AQ) = AQ^T(AQ).
\]
由\( Q = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix} \)，知\( Q^T = Q \).
故\( AB = (AQ)(AQ) \).
令\( AQ = P \)，则\( AB = P^2 \).