【答案】B
【解析】
\(f^\prime(x)=e^{x^2}\sin x,f^{\prime\prime}(x)=2xe^{x^2}\sin x + e^{x^2}\cos x\)
\(f^\prime(0)=0,f^{\prime\prime}(0)=1\gt0\) 。
\(x = 0\) 是 \(f(x)\) 的极值点。

\(g^\prime(x)=e^{x^2}\sin^2 x + \sin2x\int_{0}^{x}e^{t^2}dt\) ，
\(g^{\prime\prime}(x)=e^{x^2}\sin2x + 2xe^{x^2}\sin^2 x + \sin2xe^{x^2} + 2\cos2x\int_{0}^{x}e^{t^2}dt\) 
\(g^\prime(0)=0,\ g^{\prime\prime}(0)=0,\ g^{\prime\prime\prime}(0)\gt0\) 。
\((0,0)\) 是 \(y = g(x)\) 的拐点。 

【答案】B 

【解析】 
\(\left|\sin\frac{n^{3}\pi}{n^{2}+1}\right|=\left|\sin\left(\frac{n^{3}\pi}{n^{2}+1}-n\pi\right)\right|=\left|\sin\frac{n}{n^{2}+1}\pi\right|\sim\frac{n}{n^{2}+1}\pi\sim\frac{1}{n}\pi\) . 
\(\sum_{n = 1}^{\infty}\frac{1}{n}\)发散  \(\therefore\)不是绝对收敛. 

\(\sin\frac{n^{3}\pi}{n^{2}+1}=(-1)^{n}\sin\left(\frac{n^{3}\pi}{n^{2}+1}-n\pi\right)=(-1)^{n}\sin\frac{n}{n^{2}+1}\pi\)，为交错级数. 

\(\sin\frac{n}{n^{2}+1}\pi\)递减，\(\therefore\)条件收敛. 

\(\sum_{n = 1}^{\infty}(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\) . 

\(\left|(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\right|=\left|-\frac{1}{3}\frac{1}{n^{\frac{2}{3}\times\frac{3}{2}}}+o\left(\frac{1}{n^{2}}\right)\right|\)（注：原解析此处\(n\)的次数书写可能有误，推测应为\(\left|-\frac{1}{3}\frac{1}{n^{\frac{4}{3}}}+o\left(\frac{1}{n^{\frac{4}{3}}}\right)\right|\) ，按你提供内容提取 ），实际准确是利用等价无穷小，当\(x\to0\)时，\(x - \tan x\sim-\frac{1}{3}x^{3}\)，令\(x = \frac{1}{\sqrt[3]{n^{2}}}=n^{-\frac{2}{3}}\)，则\(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\sim-\frac{1}{3}n^{-2}\)，所以\(\left|(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\right|\sim\frac{1}{3n^{2}}\)  ） 

\(\sum_{n = 1}^{\infty}\frac{1}{n^{2}}\)收敛 \(\therefore\sum_{n = 1}^{\infty}(-1)^{n}\left(\frac{1}{\sqrt[3]{n^{2}}}-\tan\frac{1}{\sqrt[3]{n^{2}}}\right)\) 绝对收敛 

【答案】D

【解析】A 错误，反例：
\( f(x) = \frac{\sin x^2}{x}, \lim\limits_{x \to +\infty} f(x) = 0 \)，但 \( \lim\limits_{x \to +\infty} f'(x) = \lim\limits_{x \to +\infty} \frac{2x^2 \cos x^2 - \sin x^2}{x^2} \)，极限不存在。

B 错误，反例：\( f(x) = \sqrt{x}, f'(x) = \frac{1}{2\sqrt{x}}, \lim\limits_{x \to +\infty} f'(x) = 0 \)，极限存在，但 \( \lim\limits_{x \to +\infty} f(x) \) 极限不存在。

C 错误，反例：
\( f(x) = \cos x \)，则 \( \lim\limits_{x \to +\infty} \frac{\int_{0}^{x} f(t)dt}{x} = \lim\limits_{x \to +\infty} \frac{\sin x}{x} \) 存在，但 \( \lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} \cos x \) 不存在。

D 正确，用 \( \lim\limits_{x \to +\infty} \frac{\int_{0}^{x} f(t)dt}{x} = \lim\limits_{x \to +\infty} \frac{f(x)}{1} = A \)，故选 D 。

【答案】A
【解析】由题易知，此二重积分积分区域为

\( D = \left\{(x,y)\vert 4 - x^2\leq y\leq 4, - 2\leq x\leq 2\right\} \)，对应图像为上图所示。
记 \( D_1 = \left\{(x,y)\vert 4 - x^2\leq y\leq 4, - 2\leq x\leq 0\right\} \)，\( D_2 = \left\{(x,y)\vert 4 - x^2\leq y\leq 4, 0\leq x\leq 2\right\} \)，且 \( I = \int_{-2}^{2}dx\int_{4 - x^2}^{4}f(x,y)dy \)，则 \( I = \iint_{D_1}f(x,y)d\sigma + \iint_{D_2}f(x,y)d\sigma \)，交换积分次序得
\( I = \int_{0}^{4}dy\int_{-2}^{-\sqrt{4 - y}}f(x,y)dx + \int_{0}^{4}dy\int_{\sqrt{4 - y}}^{2}f(x,y)dx \)
\( = \int_{0}^{4}\left[\int_{-2}^{-\sqrt{4 - y}}f(x,y)dx + \int_{\sqrt{4 - y}}^{2}f(x,y)dx\right]dy \)
故A正确。

【答案】B
【解析】

\( A = \begin{pmatrix}
1 & 1 & 1 \\
1 & 0 & 0 \\
1 & 0 & 0
\end{pmatrix} \)

\( (\lambda E - A) = \begin{pmatrix}
\lambda - 1 & -1 & -1 \\
-1 & \lambda & 0 \\
-1 & 0 & \lambda
\end{pmatrix} \to \begin{pmatrix}
\lambda - 1 & -1 & -1 \\
0 & \lambda & -\lambda \\
-1 & 0 & \lambda
\end{pmatrix} \to \lambda \begin{pmatrix}
\lambda - 1 & -1 & -1 \\
0 & 1 & -1 \\
-1 & 0 & \lambda
\end{pmatrix} \)

\( = \lambda[\lambda(\lambda - 1) - 1 - 1] \)

\( = \lambda(\lambda^2 - \lambda - 2) \)

\( = \lambda(\lambda - 2)(\lambda + 1) \)

解得

\( \lambda_1 = 0, \lambda_2 = 2, \lambda_3 = -1 \)

故正惯性指数为 1，选 B.

【答案】D

【解析】记\(A = (\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})\)，由\(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}\)线性无关，\(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}\)线性相关，可得\(r(A)=2\)。记\(\overline{A}=(A|\boldsymbol{\alpha}_{4}) = (\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3},\boldsymbol{\alpha}_{4})\)，再由\(\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{4}=\boldsymbol{0}\)，则\(r(\overline{A}) = 2\)。于是\(Ax = \boldsymbol{\alpha}_{4}\)有无穷多解。则\(x\boldsymbol{\alpha}_{1}+y\boldsymbol{\alpha}_{2}+z\boldsymbol{\alpha}_{3}=\boldsymbol{\alpha}_{4}\)等价于\((\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3})\cdot(x,y,z)^{\mathrm{T}}=\boldsymbol{\alpha}_{4}\)，即\(A\cdot(x,y,z)^{\mathrm{T}}=\boldsymbol{\alpha}_{4}\)。

若过原点，则\(\boldsymbol{\alpha}_{4}=\boldsymbol{0}\)与\(\boldsymbol{\alpha}_{1},\boldsymbol{\alpha}_{2}\)线性无关矛盾，故不过原点。

\(x\boldsymbol{\alpha}_{1}+y\boldsymbol{\alpha}_{2}+z\boldsymbol{\alpha}_{3}=\boldsymbol{\alpha}_{4}\Leftrightarrow\begin{cases}a_{11}x + a_{12}y + a_{13}z = a_{14}\\a_{21}x + a_{22}y + a_{23}z = a_{24}\\\cdots\\a_{n1}x + a_{n2}y + a_{n3}z = a_{n4}\end{cases}\)，由上述分析可知\(r(A)=r(\overline{A}) = 2\)，故两平面交于一条直线，且不过原点。故选D。

【答案】A

【解析】\(\boldsymbol{A} = \begin{pmatrix}1&0\\0&0\end{pmatrix},\boldsymbol{B} = \begin{pmatrix}0&0\\0&1\end{pmatrix},\boldsymbol{C} = \boldsymbol{E}\)，满足\(r(\boldsymbol{A}) + r(\boldsymbol{B}) + r(\boldsymbol{C}) = r(\boldsymbol{ABC}) + 2n\)，则\(r(\boldsymbol{A}) = 1,r(\boldsymbol{B}) = 1,r(\boldsymbol{C}) = 2\)，排除结论③④，故选A.

【答案】C 
【解析】 
\(D(aX + bY)=a^2DX + b^2DY + 2ab\rho\cdot1\cdot1\) 
\(= a^2 + b^2 + 2ab\rho = 1 + 2ab\rho = 1 + 2a\sqrt{1 - a^2}\rho = f(a)\) 
\(f^\prime(a)=\rho\left(2\sqrt{1 - a^2} + 2a\cdot\frac{-a}{\sqrt{1 - a^2}}\right)= 2\rho\left(\sqrt{1 - a^2} - \frac{a^2}{\sqrt{1 - a^2}}\right)= 0\) 
即\(2\rho\cdot\frac{1 - a^2 - a^2}{\sqrt{1 - a^2}} = 0\)，\(2a^2 = 1\Rightarrow a^2 = \frac{1}{2},b^2 = \frac{1}{2}\)，于是\(a = \pm\frac{1}{\sqrt{2}},b = \pm\frac{1}{\sqrt{2}}\)。所以最大值为\(1 + |\rho|\)，故选C。 

【答案】C 

【解析】由题意可知\(T\sim B(20,0.1).np = 20\times0.1 = 2\) 

\(P\{T\leq1\} = P\{T = 0\} + P\{T = 1\} = \frac{2^{0}}{0!}e^{-2} + \frac{2^{1}}{1!}e^{-2} = e^{-2} + 2e^{-2} = \frac{3}{e^{2}}\)

【答案】D

【解析】按题意需检验假设: \( H_0: \mu \leq 1 \)，\( H_1: \mu > 1 \)，这是右边检验问题，\( \sigma^2 = 2 \) 为已知，其拒绝域为 \( \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} > z_\alpha \)，解得 \( \bar{X} > \mu + \frac{\sigma}{\sqrt{n}}z_\alpha = 1 + \frac{\sqrt{2}}{\sqrt{n}}z_\alpha \)，故选 D.

【答案】$-1$ 
【解析】 
$\lim\limits_{x \to 0^{+}} \frac{e^{x\ln x} - 1}{-x\ln x} = \lim\limits_{x \to 0^{+}} \frac{x\ln x}{-x\ln x} = -1$

【答案】\(\frac{1}{8}\)

【解析】

\(S\left(-\frac{7}{2}\right)=S\left(-\frac{7}{2}+4\right)=S\left(\frac{1}{2}\right)=\frac{1}{8}\)  ．

【答案】1 

【解析】由题易知，\(\frac{\partial u}{\partial x}=y^{2}z^{3}\)，\(\frac{\partial u}{\partial y}=2xyz^{3}\)，\(\frac{\partial u}{\partial z}=3xy^{2}z^{2}\) 

则在\(x = 1,y = 1,z = 1\)处有\(\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)=(1,2,3)\) 

对于向量\(\boldsymbol{\bar{n}}=(2,2,-1)\)，归一化可得\(\boldsymbol{\bar{n}}_{0}=\left(\frac{2}{3},\frac{2}{3},-\frac{1}{3}\right)\) 

故\(\left. \frac{\partial u}{\partial \boldsymbol{n}}\right\vert _{(1,1,1)}=\left(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}\right)\cdot \boldsymbol{\bar{n}}_{0}=(1,2,3)\cdot \left(\frac{2}{3},\frac{2}{3},-\frac{1}{3}\right)=1\cdot \frac{2}{3}+2\cdot \frac{2}{3}+3\cdot \left(-\frac{1}{3}\right)=1\)

【答案】\(\boldsymbol{\frac{4}{3} - 2\sin1}\) 
【解析】由题易知可作曲线如右图所示. 

记\( L_0 \)是从\( x = -1 \)到\( x = 1 \)的直线， 
并记曲线积\( I = \int_{L} (y + \cos x)dx + (2x + \cos y)dy \) 
则在\( L_0 \)与\( L \)所围的封闭区域可用格林公式 
即\( I_1 = \oint_{L_0 + L} (y + \cos x)dx + (2x + \cos y)dy \) 
\( = \iint_{D} 2 - 1d\sigma = \int_{-1}^{1} dx\int_{0}^{1 - x^2} dy = \int_{-1}^{1} (1 - x^2)dx = \frac{4}{3} \) 

又\( I_2 = \int_{L_0} (y + \cos x)dx + (2x + \cos y)dy = \int_{-1}^{1} \cos xdx = 2\sin1 \)，故\( I = \frac{4}{3} - 2\sin1 \)

【答案】\(-4\)

【解析】由题知，\(\boldsymbol{A}=\begin{pmatrix}4&2&-3\\a&3&-4\\b&5&-7\end{pmatrix}\)，若\(\boldsymbol{A}^{2}\boldsymbol{x}=\boldsymbol{0}\)与\(\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}\)同解，则三秩相同，即\(r(\boldsymbol{A}) = r(\boldsymbol{A}^{2}) = r\begin{pmatrix}\boldsymbol{A}\\\boldsymbol{A}^{2}\end{pmatrix}\)。如果\(\boldsymbol{A}\)可逆，三秩显然相同，则\(\boldsymbol{A}^{2}\boldsymbol{x}=\boldsymbol{0}\)与\(\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}\)同解，于是要想\(\boldsymbol{A}^{2}\boldsymbol{x}=\boldsymbol{0}\)与\(\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}\)不同解，即\(\boldsymbol{A}\)不可逆，于是\(\vert\boldsymbol{A}\vert = 0\)。根据行列式的倍加性质易得
\(\vert\boldsymbol{A}\vert=\begin{vmatrix}4&2&-3\\a&3&-4\\b&5&-7\end{vmatrix}=\begin{vmatrix}4&2&-1\\a&3&-1\\b&5&-2\end{vmatrix}=\begin{vmatrix}4&0&-1\\a&1&-1\\b&1&-2\end{vmatrix}= 4(-2 + 1)-(a - b)\)，令\(\vert\boldsymbol{A}\vert = 0\)，有\(a - b = -4\)。

【答案】 \(\frac{4}{5}\)

【解析】\(P(A\overline{B}) + P(\overline{A}B)=P(A) - P(AB)+P(B) - P(AB)\)

\(= P(A) + P(B) - 2P(A)P(B)\)

\(P(A\cup B)=P(A) + P(B) - P(AB)=\frac{5}{8}\)，\(\Rightarrow 3P(B) - 2P^{2}(B)=\frac{5}{8}\)

\(24P(B) - 16P^{2}(B)=5\)，\(16P^{2}(B) - 24P(B) + 5 = 0\)

\(\Rightarrow(4P(B) - 1)(4P(B) - 5)=0\)

\(P(B)=\frac{1}{4}\)，\(P(A)=\frac{1}{2}\)

\(P(A\overline{B}) + P(\overline{A}B)=P(A) - P(AB)+P(B) - P(AB)=P(A) + P(B) - 2P(A)P(B)\)

\(=\frac{1}{2}+\frac{1}{4}-2\times\frac{1}{4}\times\frac{1}{2}=\frac{1}{2}\)

\(P = \frac{\frac{1}{2}}{\frac{5}{8}}=\frac{1}{2}\times\frac{8}{5}=\frac{4}{5}\)

解：
\[
\begin{flalign}
\int_{0}^{1}\frac{1}{(x + 1)(x^2 - 2x + 2)}dx 
&= \int_{0}^{1}\left(\frac{A}{x + 1}+\frac{Bx + C}{x^2 - 2x + 2}\right)dx \nonumber \\
&= \int_{0}^{1}\left(\frac{\frac{1}{5}}{x + 1}+\frac{-\frac{1}{5}x+\frac{3}{5}}{x^2 - 2x + 2}\right)dx \nonumber \\
&= \frac{1}{5}\ln|1 + x|\bigg|_{0}^{1} 
- \frac{1}{10}\ln|x^2 - 2x + 2|\bigg|_{0}^{1} 
+ \frac{2}{5}\arctan(x - 1)\bigg|_{0}^{1} \nonumber \\
&= \frac{3}{10}\ln2 + \frac{1}{10}\pi \nonumber
\end{flalign}
\]

解：

令\( u = \frac{x}{y} \)，则\( \frac{\partial g}{\partial x} = f'(u)\frac{1}{y} \)，\( \frac{\partial g}{\partial y} = f'(u)\left(-\frac{x}{y^2}\right) \)

又\( g(x, x) = f\left(\frac{x}{x}\right) = f(1) = 1 \)，\(\left.\frac{\partial g}{\partial x}\right\rvert_{(x, x)} = f'(1)\frac{1}{x} = \frac{2}{x} \)，故\( f'(1) = 2 \)

\( \frac{\partial^2 g}{\partial x^2} = \left(f''(u)\frac{1}{y}\right)\frac{1}{y} = f''(u)\frac{1}{y^2} \)  ......(1)

\( \frac{\partial^2 g}{\partial x\partial y} = f''(u)\left(-\frac{x}{y^2}\right)\frac{1}{y} + f'(u)\left(-\frac{1}{y^2}\right) = -\frac{x}{y^3}f''(u) - \frac{1}{y^2}f'(u) \)  ..(2)

\( \frac{\partial^2 g}{\partial y^2} = f''(u)\left(-\frac{x}{y^2}\right)\frac{x}{y} + f'(u)\left(\frac{2x}{y^2}\right) = \frac{x^2}{y^4}f''(u) + \frac{2x}{y^3}f'(u) \)  ...(3)

将（1）（2）（3）代入\( x^2\frac{\partial^2 g}{\partial x^2} + xy\frac{\partial^2 g}{\partial x\partial y} + y^2\frac{\partial^2 g}{\partial y^2} = 1 \)化简得：

\( u^2f''(u) + uf'(u) = 1 \)，即\( f''(u) + \frac{1}{u^2}f'(u) = \frac{1}{u^2} \)。

令\( p' = f'(u) \)则\( p'' + \frac{1}{u}p' = \frac{1}{u^2} \)

\( p = e^{-\int \frac{1}{u}du}\left[\int \frac{1}{u^2}e^{\int \frac{1}{u}du}du + C\right] = \frac{1}{u}\left[\int \frac{1}{u}du + C\right] = \frac{\ln u}{u} + \frac{C}{u} \)

又\(\left.p'\right\rvert_{u = 1} = C = 2\)，故\( p = \frac{\ln u}{u} + \frac{2}{u} \)

因此，\( \frac{\partial f}{\partial u} = \frac{\ln u}{u} + \frac{2}{u} \)，积分得\( f(u) = \int \left(\frac{\ln u}{u} + \frac{2}{u}\right)du = \frac{1}{2}\ln^2 u + 2\ln u + C \)，

又\( f(1) = C = 1 \)，故\( f(u) = \frac{1}{2}\ln^2 u + 2\ln u + 1 \) 。 

解：充分性：若对\((a,b)\)内任意的\(x_1,x_2,x_3\)，当\(x_1\lt x_2\lt x_3\)时，都有
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
则在\((a,b)\)内取任意的\(x_1\lt x_2\lt x_3\lt x_4\lt x_5\)，有
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}\lt\frac{f(x_4) - f(x_3)}{x_4 - x_3}\lt\frac{f(x_5) - f(x_4)}{x_5 - x_4}
\]
在\(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}\)两边同时令\(x_2 \to x_1^+\)，得
\[
f_+'(x_1)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}
\]
两边同时令\(x_2 \to x_3^-\)，得\(\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_3)\)，即
\[
f_+'(x_1)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_3)
\]
同理可得\(f_+'(x_3)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_5)\) .因为
\[
\frac{f(x_3) - f(x_1)}{x_3 - x_1}\lt\frac{f(x_5) - f(x_3)}{x_5 - x_3}
\]
所以\(f_+'(x_1)\leq f_-'(x_5)\) .由\(x_1,x_5\)的任意性，可得\(f'(x)\)在\((a,b)\)内严格单调递增，充分性得证。

再证必要性，即已知\(f'(x)\)单调递增，在\([x_1,x_2],[x_2,x_3]\)上分别使用拉格朗日中值定理，知存在\(\xi_1 \in (x_1,x_2),\xi_2 \in (x_2,x_3)\)，使
\[
f'(\xi_1)=\frac{f(x_2) - f(x_1)}{x_2 - x_1}, \quad f'(\xi_2)=\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
又由\(f'(x)\)单调递增，且\(\xi_1\lt\xi_2\)知，\(f'(\xi_1)\lt f'(\xi_2)\)，即
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
必要性得证。

综上所述，充要条件得证。

【解析】直线 $\begin{cases} x=0, \\ y=0 \end{cases}$ 即为 $z$ 轴，故旋转曲面为经过三个坐标轴的圆锥面，$\Sigma_1$ 是以原点为顶点，底面为平面 $x + y + z = 1$ 与锥面的相交的圆 $\Gamma$ 的圆锥面的侧面. 故圆 $\Gamma$ 经过 $(0,0,1)$ 点，它的圆心为直线 $\begin{cases} x=t, \\ y=t, \\ z=t \end{cases}$ 与平面 $x + y + z = 1$ 的交点，代入得 $t = \frac{1}{3}$，即圆心为 $\left( \frac{1}{3}, \frac{1}{3}, \frac{1}{3} \right)$，故圆的半径 $r = \sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3}$. 平面 $x + y + z = 1$ 的单位法向量为 $\vec{n}^\circ = \frac{1}{\sqrt{3}}(1,1,1)$. 记 $\Sigma_1$ 与 $\Gamma$ 围成的圆域 $S$（方向向上）围成的圆锥体为 $\Omega$，则 $\Omega$ 的高为 $h = \frac{\sqrt{3}}{3}$. 于是由高斯公式，得
$$\iint_{\Sigma_1} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y$$
$$= \oiint_{\Sigma_1 + S} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y - \iint_{S} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y$$
$$= \iiint_{\Omega} 3\text{d}V - \iint_{S} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y$$

圆锥体的体积为 $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \cdot \frac{2}{3} \cdot \frac{\sqrt{3}}{3} = \frac{2\pi}{9\sqrt{3}}$，故 $\iiint_{\Omega} 3\text{d}V = \frac{2\sqrt{3}\pi}{9}$. 由两类曲面积分之间关系，可得
$$\iint_{S} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y = \iint_{S} [x \cos\alpha + (y + 1) \cos\beta + (z + 2) \cos\gamma]\text{d}S$$
$$= \frac{1}{\sqrt{3}} \iint_{S} (x + y + z + 3)\text{d}S$$
$$= \frac{4}{\sqrt{3}} \iint_{S} \text{d}S = \frac{4}{\sqrt{3}} \cdot \pi \cdot \frac{2}{3} = \frac{8\sqrt{3}\pi}{9}.$$

故
$$I = \iint_{\Sigma_1} x\text{d}y\text{d}z + (y + 1)\text{d}z\text{d}x + (z + 2)\text{d}x\text{d}y = \frac{2\sqrt{3}\pi}{9} - \frac{8\sqrt{3}\pi}{9} = -\frac{2\sqrt{3}\pi}{3}$$

解：（1）\(f(\lambda)=\vert A - \lambda E\vert=(1 - \lambda)[(\lambda - a)(\lambda + 1)+4]\)
可得\((1 - a)(1 + 1)+4 = 0\)，\(a = 3\)
（2）由（1）可知\(A=\begin{pmatrix}0&-1&2\\-1&0&2\\-1&-1&3\end{pmatrix}\)，\(\vert A - \lambda E\vert = 0\)，得\(A\)中\(\lambda_1=\lambda_2=\lambda_3 = 1\)，
\(A\alpha=\alpha + \beta\)，\(A^{2}\alpha=\alpha + 2\beta\Rightarrow(A - E)\alpha=\beta\)，\((A^{2}-E)\alpha = 2\beta = 2(A - E)\alpha\)
\((A - E)^{2}\alpha = 0\)，其中\((A - E)^{2}=0\)，故\(\alpha\)为任意的非零向量，\(\alpha=(a_1,a_2,a_3)^{\mathrm{T}}\)，\(a_1a_2a_3\neq0\)
\(\Rightarrow\beta=(A - E)\alpha=\begin{pmatrix}-1&-1&2\\-1&-1&2\\-1&-1&2\end{pmatrix}\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}=\begin{pmatrix}2a_3 - a_1 - a_2\\2a_3 - a_1 - a_2\\2a_3 - a_1 - a_2\end{pmatrix}\)
其中\(a_1 + a_2\neq2a_3\)
则综上\(\alpha=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\)，\(\beta=\begin{pmatrix}2a_3 - a_1 - a_2\\2a_3 - a_1 - a_2\\2a_3 - a_1 - a_2\end{pmatrix}\)，\((a_1a_2a_3\neq0,a_1 + a_2\neq2a_3)\)

【解析】(1) \( P\{Y > 0\} = P\{X > 100\} = \int_{100}^{+\infty} \frac{2 \times 100^2}{(100 + x)^3} \, dx = \frac{1}{4} \),
\[
\begin{align*}
EY &= \int_{100}^{+\infty} (x - 100) \cdot \frac{2 \times 100^2}{(100 + x)^3} dx \\
&= 2 \times 100^2 \int_{100}^{+\infty} \frac{x - 100}{(100 + x)^3} dx \\
&= 2 \times 100^2 \int_{100}^{+\infty} \left[ \frac{1}{(x + 100)^2} - \frac{200}{(x + 100)^3} \right] dx \\
&= 2 \times 100^2 \int_{100}^{+\infty} \left[ -\frac{1}{x + 100} + \frac{100}{(x + 100)^2} \right] dx \\
&= 2 \times 100^2 \left( \frac{1}{200} - \frac{100}{200^2} \right) = 50.
\end{align*}
\]

(2) 由题设可知 \( N \sim P(8) \)，\( P\{M = k|N = n\} = C_n^k \left( \frac{1}{4} \right)^k \left( \frac{3}{4} \right)^{n - k} \)，于是可得
\[
\begin{align*}
P\{M = k\} &= \sum_{n=k}^{\infty} P\{N = n\} \cdot P\{M = k|N = n) \\
&= \sum_{n=k}^{\infty} \frac{8^n}{n!} \mathrm{e}^{-8} \cdot C_n^k \cdot \left( \frac{1}{4} \right)^k \cdot \left( \frac{3}{4} \right)^{n - k} \\
&= \mathrm{e}^{-8} \sum_{n=k}^{\infty} \frac{C_n^k \left( \frac{1}{4} \right)^k \cdot 8^k \cdot \left( \frac{3}{4} \right)^{n - k} \cdot 8^{n - k}}{n!} = \mathrm{e}^{-8} \sum_{n=k}^{\infty} \frac{C_n^k \cdot 2^k \cdot 6^{n - k}}{n!} \\
&= \mathrm{e}^{-8} \cdot 2^k \sum_{n=k}^{\infty} \frac{6^{n - k}}{k!(n - k)!} = \mathrm{e}^{-8} \cdot \frac{2^k}{k!} \sum_{n=k}^{\infty} \frac{6^{n - k}}{(n - k)!} \\
&= \mathrm{e}^{-8} \cdot \frac{2^k}{k!} \sum_{n=0}^{\infty} \frac{6^n}{n!} = \mathrm{e}^{-8} \cdot \frac{2^k}{k!} \cdot \mathrm{e}^6 = \frac{2^k}{k!} \mathrm{e}^{-2}
\end{align*}
\]
即 \( P\{M = k\} = \frac{2^k}{k!} \mathrm{e}^{-2} \)．所以 \( M \) 服从参数为 \( \lambda = 2 \) 的泊松分布．