【解析】\(\lim _{x \to 0} f(x)=\lim _{x \to 0} \frac{e^{x}-1}{x}=\lim _{x \to 0} \frac{x}{x}=1=f(0)\)，故\(f(x)\)在\(x=0\)处连续。又\[f'(0)=\lim _{x \to 0} \frac{f(x)-f(0)}{x}=\lim _{x \to 0} \frac{\frac{e^{x}-1}{x}-1}{x}=\lim _{x \to 0} \frac{e^{x}-1-x}{x^{2}}=\lim _{x \to 0} \frac{1+x+\frac{1}{2} x^{2}+o\left(x^{2}\right)-1-x}{x^{2}}=\frac{1}{2},\]故应选(D)。

【解析】\[f_{x}'(x, y)=2 x \ln y, f_{y}'(x, y)=\frac{x^{2}}{y},\]则\(f_{x}'(1,1)=0\)，\(f_{y}'(1,1)=1\)，得\(df(1,1)=dy\)，故应选(C)。

【解析】由题意知，\(f(x)=\frac{\sin x}{1+x^{2}}=a x+b x^{2}+c x^{3}+o(x^{3})\)，故\[\begin{aligned} \sin x & =\left(1+x^{2}\right)\left(a x+b x^{2}+c x^{3}+o\left(x^{3}\right)\right) \\ & =a x+b x^{2}+(a+c) x^{3}+o\left(x^{3}\right), \end{aligned}\]又\(\sin x=x-\frac{1}{6} x^{3}+o(x^{3})\)，由泰勒展开的唯一性，比较系数，有\[\left\{\begin{array}{l} a=1, \\ b=0, \\ a+c=-\frac{1}{6}, \end{array}\right.\]解得\(\left\{\begin{array}{l} a=1, \\ b=0, \\ c=-\frac{7}{6}, \end{array}\right.\)故应选(A)。

【解析】将区间\([0,1]\)上均分成\(n\)份，取\(\xi_{k}\)为第\(k\)个小区间的中点，则\(\xi_{k}=\frac{\frac{k-1}{n}+\frac{k}{n}}{2}=\frac{2 k-1}{2 n}\)，\[\int_{0}^{1} f(x) d x=\lim _{n \to \infty} \sum_{k=1}^{n} f\left(\frac{2 k-1}{2 n}\right) \cdot \frac{1}{n},\]故应选(B)。

【解析】\[f=\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}-\left(x_{3}-x_{1}\right)^{2}=2 x_{2}^{2}+2 x_{1} x_{2}+2 x_{1} x_{3}+2 x_{2} x_{3},\]对应的矩阵\(A=\begin{pmatrix} 0 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 0 \end{pmatrix}\)，\[|\lambda E-A|=\left|\begin{array}{ccc} \lambda & -1 & -1 \\ -1 & \lambda-2 & -1 \\ -1 & -1 & \lambda \end{array}\right|=\lambda(\lambda-3)(\lambda+1),\] \(A\)的特征值为3，-1，0，故\(f\)的正、负惯性指数都是1，故应选(B)。

【解析】由已知，有\[\beta_{1}=\alpha_{1},\] \[\beta_{2}=\alpha_{2}-k \beta_{1}=\alpha_{2}-k \alpha_{1},\] \[\beta_{3}=\alpha_{3}-l_{1} \beta_{1}-l_{2} \beta_{2}=\alpha_{3}-l_{1} \alpha_{1}-l_{2}\left(\alpha_{2}-k \alpha_{1}\right),\]又\(\beta_{1}\)，\(\beta_{2}\)，\(\beta_{3}\)两两正交，故\((\beta_{1}, \beta_{2})=0\)，即\[\left(\alpha_{1}, \alpha_{2}-k \alpha_{1}\right)=\left(\alpha_{1}, \alpha_{2}\right)-k\left(\alpha_{1}, \alpha_{1}\right)=2-2 k=0,\]得\(k=1\)，此时\(\beta_{3}=\alpha_{3}-(l_{1}-l_{2}) \alpha_{1}-l_{2} \alpha_{2}\)。又\[\begin{aligned} \left(\beta_{1}, \beta_{3}\right) & =\left(\alpha_{1}, \alpha_{3}-\left(l_{1}-l_{2}\right) \alpha_{1}-l_{2} \alpha_{2}\right) \\ & =\left(\alpha_{1}, \alpha_{3}\right)-\left(l_{1}-l_{2}\right)\left(\alpha_{1}, \alpha_{1}\right)-l_{2}\left(\alpha_{1}, \alpha_{2}\right) \\ & =5-\left(l_{1}-l_{2}\right) \cdot 2-2 l_{2}=5-2 l_{1}=0, \end{aligned}\]得\(l_{1}=\frac{5}{2}\)。\(\beta_{2}=\alpha_{2}-\alpha_{1}=\begin{pmatrix} 0 \\ 2 \\ 0 \end{pmatrix}\)，又\[\begin{aligned} \left(\beta_{2}, \beta_{3}\right) & =\left(\beta_{2}, \alpha_{3}-\left(l_{1}-l_{2}\right) \alpha_{1}-l_{2} \alpha_{2}\right) \\ & =\left(\beta_{2}, \alpha_{3}\right)-\left(l_{1}-l_{2}\right)\left(\beta_{2}, \alpha_{1}\right)-l_{2}\left(\beta_{2}, \alpha_{2}\right) \\ & =2-\left(l_{1}-l_{2}\right) \cdot 0-l_{2} \cdot 4=2-4 l_{2}=0, \end{aligned}\]得\(l_{2}=\frac{1}{2}\)，故应选(A)。

【解析】对于(A)：\(r\left(\begin{array}{cc}A & O \\ O & A^{T} A\end{array}\right)=r(A)+r(A^{T} A)=r(A)+r(A)=2 r(A)\)；对于(B)：\(r\left(\begin{array}{cc}A & A B \\ O & A^{T}\end{array}\right)=r\left(\begin{array}{c}A \\ A^{T}\end{array}\right)+r(A B)\)，因\(\begin{pmatrix} E & B \end{pmatrix}\)行满秩，所以\(r\left(\begin{array}{ll}A & A B\end{array}\right)=r(A)\)，于是\(r\left(\begin{array}{cc}A & A B \\ O & A^{T}\end{array}\right)=r(A)+r(A^{T})=2 r(A)\)；对于(D)：\(r\left(\begin{array}{cc}A & O \\ B A & A^{T}\end{array}\right)=r(A)+r(A^{T})=2 r(A)\)；对于(C)，取\(A=\begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix}\)，\(B=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}\)，则\(\left(\begin{array}{cc}A & B A \\ O & A A^{T}\end{array}\right)=\begin{pmatrix} 1 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{pmatrix}\)，\(r\left(\begin{array}{cc}A & B A \\ O & A A^{T}\end{array}\right)=3 \neq 2 r(A)=2\)，故应选(C)。

【答案】(D).
【解】由\( P(A|B)=P(A) \)得\( P(AB)=P(A)P(B) \),即事件A,B独立,
于是\( P(A|\overline{B})=\frac{P(A\overline{B})}{P(\overline{B})}=\frac{P(A)P(\overline{B})}{P(\overline{B})}=P(A) \);
由\( P(A|B) > P(A) \)得\( P(AB) > P(A)P(B) \),
从而\( P(\overline{A}|\overline{B})=\frac{P(\overline{A}\overline{B})}{P(\overline{B})}=\frac{1 - P(A) - P(B) + P(AB)}{1 - P(B)} \)
\( > \frac{1 - P(A) - P(B) + P(A)P(B)}{1 - P(B)}=1 - P(A)=P(\overline{A}) \);
由\( P(A|B) > P(A|\overline{B}) \)得\( \frac{P(AB)}{P(B)} > \frac{P(A) - P(AB)}{1 - P(B)} \),整理得\( P(AB) > P(A)P(B) \),
则\( P(A|B)=\frac{P(AB)}{P(B)} > \frac{P(A)P(B)}{P(B)}=P(A) \),应选(D).

【答案】（C）.

【解】$\overline{X} \sim N(\mu_{1},\frac{\sigma_{1}^{2}}{n})$，$\overline{Y} \sim N(\mu_{2},\frac{\sigma_{2}^{2}}{n})$，

则$E(\hat{\theta}) = E(\overline{X}) - E(\overline{Y}) = \mu_{1} - \mu_{2} = \theta$；

$D(\hat{\theta}) = D(\overline{X} - \overline{Y}) = D(\overline{X}) + D(\overline{Y}) - 2\text{Cov}(\overline{X},\overline{Y})$

$= \frac{\sigma_{1}^{2}}{n} + \frac{\sigma_{2}^{2}}{n} - \frac{2}{n}[\text{Cov}(X_{1},\overline{Y}) + \text{Cov}(X_{2},\overline{Y}) + \cdots + \text{Cov}(X_{n},\overline{Y})]$

$= \frac{\sigma_{1}^{2}}{n} + \frac{\sigma_{2}^{2}}{n} - \frac{2}{n^{2}}[\text{Cov}(X_{1},Y_{1}) + \text{Cov}(X_{2},Y_{2}) + \cdots + \text{Cov}(X_{n},Y_{n})]$

$= \frac{\sigma_{1}^{2}}{n} + \frac{\sigma_{2}^{2}}{n} - \frac{2}{n^{2}} \cdot n\rho\sigma_{1}\sigma_{2} = \frac{\sigma_{1}^{2} + \sigma_{2}^{2} - 2\rho\sigma_{1}\sigma_{2}}{n}$，应选（C）.

【答案】（B）。
【解】 由题$\overline{X} \sim N(11.5, \frac{1}{4})$，或$\frac{\overline{X} - 11.5}{\frac{1}{2}} \sim N(0,1)$，
犯第二类错误的概率为
$P\{\overline{X} < 11\} = P\{\frac{\overline{X} - 11.5}{\frac{1}{2}} < -1\} = \Phi(-1) = 1 - \Phi(1)$，
应选（B）。

【解析】

\[\begin{aligned} \int_{0}^{+\infty} \frac{1}{x^{2}+2 x+2} d x & =\int_{0}^{+\infty} \frac{1}{(x+1)^{2}+1} d x=\left.arctan (x+1)\right|_{0} ^{+\infty} \\ & =lim _{x \to+\infty} arctan (x+1)-arctan (1) \\ & =\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4} . \end{aligned}\]

【答案】$\frac{2}{3}$。
【解】$\frac{\text{d}y}{\text{d}x} = \frac{\text{d}y/\text{d}t}{\text{d}x/\text{d}t} = \frac{4t\text{e}^t + 2t}{2\text{e}^t + 1} = 2t$，$\frac{\text{d}^2 y}{\text{d}x^2} = \frac{\text{d}(2t)/\text{d}t}{\text{d}x/\text{d}t} = \frac{2}{2\text{e}^t + 1}$，
则$\frac{\text{d}^2 y}{\text{d}x^2}\big|_{t=0} = \frac{2}{3}$。

【答案】\( x^{2} \)。
【解】令 \( x = e^{t} \)，\( D = \frac{d}{dt} \)，则
\( xy' = Dy \)，\( x^{2}y'' = D(D - 1)y \)，
代入欧拉方程得
\( \frac{d^{2}y}{dt^{2}} - 4y = 0 \)，
特征方程为 \( \lambda^{2} - 4 = 0 \)，特征根为 \( \lambda_{1} = -2 \)，\( \lambda_{2} = 2 \)，
\( \frac{d^{2}y}{dt^{2}} - 4y = 0 \) 的通解为 \( y = C_{1}e^{-2t} + C_{2}e^{2t} \)，原方程的通解为
\( y = \frac{C_{1}}{x^{2}} + C_{2}x^{2} \)，
由 \( y(1) = 1 \)，\( y'(1) = 2 \) 得 \( C_{1} + C_{2} = 1 \)，\( -2C_{1} + 2C_{2} = 2 \)，解得 \( C_{1} = 0 \)，\( C_{2} = 1 \)，
故 \( y = x^{2} \)。

【答案】$4\pi$．
【解】 设Σ所围成的几何体为$\Omega$，由高斯公式得
$I = \iint_{\Sigma} x^2 \mathrm{d}y\mathrm{d}z + y^2 \mathrm{d}z\mathrm{d}x + z \mathrm{d}x\mathrm{d}y = \iiint_{\Omega} (2x + 2y + 1)\mathrm{d}v$，
由积分的奇偶性得
$I = \iiint_{\Omega} 1\mathrm{d}v = 2 \iint_{D_{xy}} \mathrm{d}x\mathrm{d}y = 2 \cdot \pi \cdot 1 \cdot 2 = 4\pi$．

【解析】【分析】求 \(A_{11}+A_{21}+A_{31}\) ，相当于求 \(A^{*}\) 的第1行元素之和 已知 A 的每行元素之和均为2，故

\[A\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\left(\begin{array}{l} 2 \\ 2 \\ 2 \end{array}\right)=2\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) .\]

又 \(A^{*} A=|A|E=3E\) ，故

\[A^{*}\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)=\frac{3}{2}\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right),\]

故 \(A_{11}+A_{21}+A_{31}=\frac{3}{2}\)

【解析】由题意有

\[P\{X=0\}=P\{X=1\}=\frac{1}{2},\]

\[\begin{aligned} P\{Y=0\} & =P\{Y=0 | X=0\} P\{X=0\}+P\{Y=0 | X=1\} P\{X=1\} \\ & =\frac{3}{5} \cdot \frac{1}{2}+\frac{2}{5} \cdot \frac{1}{2}=\frac{1}{2}, \end{aligned}\]

\[P\{Y=1\}=1-P\{Y=0\}=\frac{1}{2},\]

\[\begin{aligned} P\{X=0, Y=0\} & =P\{Y=0 | X=0\} P\{X=0\} \\ & =\frac{3}{5} \cdot \frac{1}{2}=\frac{3}{10}, \end{aligned}\]

故 \((X, Y)\) 的概率分布为

| $Y$ $X$ | $0$ | $1$ |

| --- | --- | --- |

| $0$ | 3/10 | 2/10 |

| $1$ | 2/10 | 3/10 |

从而 \(\rho_{X Y}=\frac{cov(X, Y)}{\sqrt{D X} \cdot \sqrt{D Y}}=\frac{E(X Y)-E(X) E(Y)}{\sqrt{D X} \cdot \sqrt{D Y}}=\frac{\frac{3}{10}-\frac{1}{2} × \frac{1}{2}}{\frac{1}{2} × \frac{1}{2}}=\frac{1}{5}.\)

**方法一**
$\lim\limits _{x→0}\left(\frac {1+\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)=\lim\limits _{x→0}\frac {\left(1+\int_{0}^{x}\text{e}^{t^{2}}dt\right)\sin x-\text{e}^{x}+1}{(\text{e}^{x}-1)\sin x}$
$=\lim\limits _{x→0}\frac {\left(1+\int_{0}^{x}\text{e}^{t^{2}}dt\right)\sin x-\text{e}^{x}+1}{x^{2}}$
$=\lim\limits _{x→0}\left(\frac {\sin x-x}{x^{2}}+\frac {\int_{0}^{x}\text{e}^{t^{2}}dt\cdot \sin x-\text{e}^{x}+1+x}{x^{2}}\right)$
$=\lim\limits _{x→0}\frac {\int_{0}^{x}\text{e}^{t^{2}}dt\cdot \sin x-\text{e}^{x}+1+x}{x^{2}}$
$=\lim\limits _{x→0}\frac {\sin x}{x}\cdot \frac {\int_{0}^{x}\text{e}^{t^{2}}dt}{x}-\lim\limits _{x→0}\frac {\text{e}^{x}-1-x}{x^{2}}$
$=\lim\limits _{x→0}\text{e}^{x^{2}}-\lim\limits _{x→0}\frac {\text{e}^{x}-1}{2x}=1-\frac {1}{2}=\frac {1}{2}$。

**方法二**
$\lim\limits _{x→0}\left(\frac {1+\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)=\lim\limits _{x→0}\left(\frac {\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}+\frac {1}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)$，
由$\lim\limits _{x→0}\frac {\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}=\lim\limits _{x→0}\frac {\text{e}^{x^{2}}}{\text{e}^{x}}=1$，
$\lim\limits _{x→0}\left(\frac {1}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)=\lim\limits _{x→0}\frac {\sin x-\text{e}^{x}+1}{(\text{e}^{x}-1)\sin x}=\lim\limits _{x→0}\frac {\sin x-\text{e}^{x}+1}{x^{2}}$
$=\frac {1}{2}\lim\limits _{x→0}\frac {\cos x-\text{e}^{x}}{x}=\frac {1}{2}\lim\limits _{x→0}(-\sin x-\text{e}^{x})=-\frac {1}{2}$，
得$\lim\limits _{x→0}\left(\frac {1+\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)=1-\frac {1}{2}=\frac {1}{2}$。

**方法三**
由泰勒公式得$\text{e}^{t^{2}}=1+t^{2}+o(t^{2})$，
从而$\int_{0}^{x}\text{e}^{t^{2}}dt=x+\frac {x^{3}}{3}+o(x^{3})$，于是有
$\lim\limits _{x→0}\left(\frac {1+\int_{0}^{x}\text{e}^{t^{2}}dt}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)=\lim\limits _{x→0}\left[\frac {1+x+\frac {x^{3}}{3}+o(x^{3})}{\text{e}^{x}-1}-\frac {1}{\sin x}\right]=\lim\limits _{x→0}\left(\frac {1+x}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)$
$=\lim\limits _{x→0}\frac {x}{\text{e}^{x}-1}+\lim\limits _{x→0}\left(\frac {1}{\text{e}^{x}-1}-\frac {1}{\sin x}\right)$
$=1+\lim\limits _{x→0}\frac {\sin x-\text{e}^{x}+1}{(\text{e}^{x}-1)\sin x}=1+\lim\limits _{x→0}\frac {\sin x-\text{e}^{x}+1}{x^{2}}$
$=1+\lim\limits _{x→0}\frac {\cos x-\text{e}^{x}}{2x}=1+\lim\limits _{x→0}\frac {-\sin x-\text{e}^{x}}{2}=\frac {1}{2}$。

【解】\( \sum_{n=1}^{\infty}u_{n}(x)=\sum_{n=1}^{\infty}\mathrm{e}^{-nx}+\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)} \)，
当\( \lim_{n\to\infty}\dfrac{\mathrm{e}^{-(n+1)x}}{\mathrm{e}^{-nx}}=\mathrm{e}^{-x}<1 \)即\( x > 0 \)时，\( \sum_{n=1}^{\infty}\mathrm{e}^{-nx} \)收敛；
再由\( \lim_{n\to\infty}\dfrac{\dfrac{1}{(n+1)(n+2)}}{\dfrac{1}{n(n+1)}}=1 \)得\( \sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)} \)的收敛半径为\( R = 1 \)，
当\( x=\pm1 \)时，\( \sum_{n=1}^{\infty}\dfrac{(\pm1)^{n+1}}{n(n+1)}|=\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}=1 \)，故\( \sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)} \)的收敛域为\( [-1,1] \)，
故级数\( \sum_{n=1}^{\infty}u_{n}(x) \)的收敛域为\( (0,1] \).

令\( S(x)=\sum_{n=1}^{\infty}u_{n}(x)=\sum_{n=1}^{\infty}\mathrm{e}^{-nx}+\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)}=S_{1}(x)+S_{2}(x) \)，
且\( S_{1}(x)=\sum_{n=1}^{\infty}\mathrm{e}^{-nx}=\dfrac{\mathrm{e}^{-x}}{1-\mathrm{e}^{-x}}=\dfrac{1}{\mathrm{e}^{x}-1} \)；
\( S_{2}(x)=\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n(n+1)}=\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n}-\sum_{n=1}^{\infty}\dfrac{x^{n+1}}{n+1}=x\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}-\sum_{n=1}^{\infty}\dfrac{x^{n}}{n}+x \)
\( =(1 - x)\ln(1 - x) + x(0 < x < 1) \)，
当\( x = 1 \)时，由\( S_{1}(1)=\dfrac{1}{\mathrm{e}-1} \)，\( S_{2}(1)=1 \)得\( S(1)=\dfrac{1}{\mathrm{e}-1}+1=\dfrac{\mathrm{e}}{\mathrm{e}-1} \)，
故
\[
S(x)=
\begin{cases}
\dfrac{1}{\mathrm{e}^{x}-1}+(1 - x)\ln(1 - x)+x, & 0 < x < 1, \\
\dfrac{\mathrm{e}}{\mathrm{e}-1}, & x = 1.
\end{cases}
\]

【解】 设$M(x,y,z)\in C$，点$M$到$xOy$坐标面的距离$d=|z|$，
令$F=z^{2}+\lambda(x^{2}+2y^{2}-z-6)+\mu(4x+2y+z-30)$，
由$\begin{cases} F'_{x}=2\lambda x+4\mu=0, \\ F'_{y}=4\lambda y+2\mu=0, \\ F'_{z}=2z-\lambda+\mu=0, \\ F'_{\lambda}=x^{2}+2y^{2}-z-6=0, \\ F'_{\mu}=4x+2y+z-30=0 \end{cases}$得$\begin{cases} x=4, \\ y=1, \\ z=12, \end{cases}$或$\begin{cases} x=-8, \\ y=-2, \\ z=66, \end{cases}$
故$C$上的点$(-8,-2,66)$到$xOy$面的距离最大为$66$.

【解】（Ⅰ）显然\( I(D) = \iint_D (4 - x^2 - y^2) \, dxdy \)取最大值的区域为\( 4 - x^2 - y^2 \geq 0 \)，
即\( D_1 = \{(x, y) \mid x^2 + y^2 \leq 4\} \)，则
\( I(D_1) = \iint_{D_1} (4 - x^2 - y^2) \, dxdy = 2\pi \int_0^2 r(4 - r^2) \, dr \)
\( = 2\pi \int_0^2 (4r - r^3) \, dr = 2\pi (8 - 4) = 8\pi \)；

（Ⅱ）令\( L_0: x^2 + 4y^2 = r^2 (r > 0, L_0 \)在\( L \)内，取逆时针），设\( \partial D_1 \)与\( L_0^- \)所围成的区域为\( D_0 \)，
\( L_0 \)围成的区域为\( D_2 \)，则
\( \int_{\partial D_1} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} \)
\( = \oint_{\partial D_1 + L_0^-} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} + \)
\( \int_{L_0} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} \)，
而\( \oint_{\partial D_1 + L_0^-} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} = \iint_{D_0} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dxdy = 0 \)，
\( \int_{L_0} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} \)
\( = \frac{1}{r^2} \int_{L_0} (x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy \)
\( = \frac{1}{r^2} \iint_{D_2} (8xy \mathrm{e}^{x^2 + 4y^2} - 1 - 8xy \mathrm{e}^{x^2 + 4y^2} - 1) dxdy \)
\( = \frac{-2}{r^2} \iint_{D_2} dxdy = \frac{-2}{r^2} \cdot \pi \cdot r \cdot \frac{r}{2} = -\pi \)。
故\( \int_{\partial D_1} \frac{(x \mathrm{e}^{x^2 + 4y^2} + y)dx + (4y \mathrm{e}^{x^2 + 4y^2} - x)dy}{x^2 + 4y^2} = -\pi \)。

【解】(Ⅰ)由
\( |\lambda E - A| = \begin{vmatrix} \lambda - a&-1&1 \\ -1&\lambda - a&1 \\ 1&1&\lambda - a \end{vmatrix} = \begin{vmatrix} \lambda - a + 1&-(\lambda - a + 1)&0 \\ -1&\lambda - a&1 \\ 1&1&\lambda - a \end{vmatrix} \)
\( = (\lambda - a + 1)\begin{vmatrix} 1&-1&0 \\ -1&\lambda - a&1 \\ 1&1&\lambda - a \end{vmatrix} \)
\( = (\lambda - a + 1)\begin{vmatrix} 1&0&0 \\ -1&\lambda - a - 1&1 \\ 1&2&\lambda - a \end{vmatrix} \)
\( = (\lambda - a + 1)^2(\lambda - a - 2) = 0 \)，
得\( \lambda_1 = \lambda_2 = a - 1 \)，\( \lambda_3 = a + 2 \)，
由\( (a - 1)E - A = \begin{pmatrix} -1&-1&1 \\ -1&-1&1 \\ 1&1&-1 \end{pmatrix} \to \begin{pmatrix} 1&1&-1 \\ 0&0&0 \\ 0&0&0 \end{pmatrix} \)得\( \lambda_1 = \lambda_2 = a - 1 \)对应的线性无关的特征向量为\( \alpha_1 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \)，\( \alpha_2 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \)；

由\( (a + 2)E - A = \begin{pmatrix} 2&-1&1 \\ -1&2&1 \\ 1&1&2 \end{pmatrix} \to \begin{pmatrix} 1&-2&-1 \\ 0&1&1 \\ 0&0&0 \end{pmatrix} \to \begin{pmatrix} 1&0&1 \\ 0&1&1 \\ 0&0&0 \end{pmatrix} \)得\( \lambda_3 = a + 2 \)对应的特征向量为\( \alpha_3 = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \)，
令\( \beta_1 = \alpha_1 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \)，\( \beta_2 = \alpha_2 - \frac{(\alpha_2, \beta_1)}{(\beta_1, \beta_1)}\beta_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + \frac{1}{2}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)，\( \beta_3 = \alpha_3 = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \)，
再令\( \gamma_1 = \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} \)，\( \gamma_2 = \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \)，\( \gamma_3 = \frac{1}{\sqrt{3}}\begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \)，
得正交矩阵\( P = \begin{pmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}} \\ 0&\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}} \end{pmatrix} \)，
使得\( P^TAP = \begin{pmatrix} a - 1&0&0 \\ 0&a - 1&0 \\ 0&0&a + 2 \end{pmatrix} \)。

(Ⅱ)由\( P^T[(a + 3)E - A]P = \begin{pmatrix} 4&0&0 \\ 0&4&0 \\ 0&0&1 \end{pmatrix} \)得
\( (a + 3)E - A = P\begin{pmatrix} 4&0&0 \\ 0&4&0 \\ 0&0&1 \end{pmatrix}P^T = P\begin{pmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&1 \end{pmatrix}P^T \cdot P\begin{pmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&1 \end{pmatrix}P^T \)，
令\( C = P\begin{pmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&1 \end{pmatrix}P^T \)，
\( = \begin{pmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}} \\ 0&\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}} \end{pmatrix}\begin{pmatrix} 2&0&0 \\ 0&2&0 \\ 0&0&1 \end{pmatrix}\begin{pmatrix} -\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}&0 \\ \frac{1}{\sqrt{6}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}}&-\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{3}} \end{pmatrix} \)
\( = \frac{1}{3}\begin{pmatrix} 5&-1&1 \\ -1&5&1 \\ 1&1&5 \end{pmatrix} \)。
则\( C^2 = (a + 3)E - A \)。

【解】(Ⅰ)X的密度函数为
\[ f_X(x)=\begin{cases}
1, & 0<x<1, \\
0, & 其他.
\end{cases} \]
(Ⅱ)由\( Y = 2 - X \)得\( Z = \frac{2 - X}{X} \),
\( F_Z(z) = P\{Z \leq z\} = P\left\{ \frac{2}{X} - 1 \leq z \right\} \),
当\( z < 1 \)时,\( F_Z(z) = 0 \);
当\( z \geq 1 \)时,\( F_Z(z) = P\left\{ X \geq \frac{2}{z + 1} \right\} = \int_{\frac{2}{z + 1}}^1 1\text{d}x = 1 - \frac{2}{z + 1} = \frac{z - 1}{z + 1} \),
即
\[ F_Z(z)=\begin{cases}
0, & z < 1, \\
\frac{z - 1}{z + 1}, & z \geq 1,
\end{cases} \]
故Z的密度函数为
\[ f_Z(z)=\begin{cases}
0, & z \leq 1, \\
\frac{2}{(z + 1)^2}, & z > 1.
\end{cases} \]
(Ⅲ)\( E\left( \frac{X}{Y} \right) = E\left( \frac{X}{2 - X} \right) = \int_0^1 \frac{x}{2 - x}\text{d}x = 2\ln 2 - 1 \).