【参考答案】B 

**参考解析**：
\[
\begin{align*}
k&=\lim_{x \to \infty}\frac{y}{x}=\lim_{x \to \infty}\frac{x\ln(e + \frac{1}{x - 1})}{x}=\lim_{x \to \infty}\ln(e + \frac{1}{x - 1}) = 1\\
b&=\lim_{x \to \infty}(y - kx)=\lim_{x \to \infty}[x\ln(e + \frac{1}{x - 1})-x]=\lim_{x \to \infty}x[\ln(e + \frac{1}{x - 1}) - 1]\\
&=\lim_{x \to \infty}x\ln[1 + \frac{1}{e(x - 1)}]=\lim_{x \to \infty}\frac{x}{e(x - 1)}=\frac{1}{e}
\end{align*}
\]
所以斜渐近线方程为 \(y = x + \frac{1}{e}\)。 

【答案】C.

【解析】二阶常系数齐次线性微分方程$y'' + ay' + by = 0$特征方程为：$\lambda^{2}+a\lambda + b = 0$

①当$\Delta = a^{2}-4b > 0$时，设其两不同特征值为$\lambda_{1},\lambda_{2}$，则其通解结构为：$y(x)=C_{1}e^{\lambda_{1}x}+C_{2}e^{\lambda_{2}x}$.

不管$\lambda_{1},\lambda_{2}$的正负如何，其在$x \to \infty$时，$y(x)\to\infty$，其必然为无界函数.舍去.

②当$\Delta = a^{2}-4b = 0$时，设其两相同特征值为$\lambda_{1}=\lambda_{2}$，则其通解结构为$y(x)=(C_{1}+C_{2}x)e^{\lambda_{1}x}$. 不管$\lambda_{1},\lambda_{2}$的正负如何，其在$x \to \infty$时，$y(x)\to\infty$，其必然为无界函数.舍去.

③当$\Delta = a^{2}-4b < 0$时，设其两特征值为$\lambda_{1,2}=\alpha\pm\beta i$，则其通解结构为

$y(x)=e^{\alpha x}(C_{1}\sin\beta x + C_{2}\cos\beta x)$.

若$\alpha\neq0$ ，则$x \to \infty$时，$y(x)=e^{\alpha x}(C_{1}\sin\beta x + C_{2}\cos\beta x)$为无界函数，从而$\alpha = 0$

即当$\alpha = 0$时，$y(x)=C_{1}\sin\beta x + C_{2}\cos\beta x$为有界函数.

此时$\Delta = a^{2}-4b < 0$，即$a = 0, b > 0$，故选（C）. 

【答案】C.

【解析】由参数方程可得

$f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3},&x\geq0\\ -x\sin x,&x < 0\end{cases}$

$\lim\limits_{x\to0^{+}}f(x)=\lim\limits_{x\to0^{-}}f(x)=0 = f(0)$，$f(x)$在$x = 0$处连续.

$f^{\prime}(x)=\begin{cases}\frac{1}{3}\sin\frac{x}{3}+\frac{x}{9}\cos\frac{x}{3},&x\geq0\\ -\sin x - x\cos x,&x < 0\end{cases}$，

$\lim\limits_{x\to0^{+}}f^{\prime}(x)=\lim\limits_{x\to0^{-}}f^{\prime}(x)=0 = f^{\prime}(0)$，$f^{\prime}(x)$在$x = 0$处连续.

$f_{+}^{\prime\prime}(0)=\lim\limits_{x\to0^{+}}\frac{f^{\prime}(x)-f^{\prime}(0)}{x}=\lim\limits_{x\to0^{+}}\frac{\frac{1}{3}\sin\frac{x}{3}+\frac{x}{9}\cos\frac{x}{3}}{x}=\frac{2}{9}$;

$f_{-}^{\prime\prime}(0)=\lim\limits_{x\to0^{-}}\frac{f^{\prime}(x)-f^{\prime}(0)}{x}=\lim\limits_{x\to0^{-}}\frac{-\sin x - x\cos x}{x}=-2$

$\lim\limits_{x\to0^{+}}f^{\prime\prime}(x)\neq\lim\limits_{x\to0^{-}}f^{\prime\prime}(x)$，$f^{\prime\prime}(x)$在$x = 0$处不可导，选C. 

【答案】A. 

【解析】

由条件可知\(\sum_{n = 1}^{\infty}(b_{n}-a_{n})\)为收敛的正项级数，进而绝对收敛；

设\(\sum_{n = 1}^{\infty}a_{n}\)绝对收敛，则由\(\vert b_{n}\vert=\vert b_{n}-a_{n}+a_{n}\vert\leq\vert b_{n}-a_{n}\vert+\vert a_{n}\vert\)与比较 ，得\(\sum_{n = 1}^{\infty}b_{n}\)绝对收敛；

设\(\sum_{n = 1}^{\infty}b_{n}\)绝对收敛，则由\(\vert a_{n}\vert=\vert a_{n}-b_{n}+b_{n}\vert\leq\vert a_{n}-b_{n}\vert+\vert b_{n}\vert\)与比较 ，得\(\sum_{n = 1}^{\infty}a_{n}\)绝对收敛；

故为充要条件。 

【答案】B.

【解析】

因为初等变换不改变矩阵的秩. 故

\(r_{1}=r\begin{pmatrix}O&A\\BC&E\end{pmatrix}=r\begin{pmatrix}-ABC&O\\BC&E\end{pmatrix}=r\begin{pmatrix}O&O\\BC&E\end{pmatrix}=n\)

\(r_{2}=r\begin{pmatrix}AB&C\\O&E\end{pmatrix}=r\begin{pmatrix}AB&O\\O&E\end{pmatrix}=n + r(AB)\)

\(r_{3}=r\begin{pmatrix}E&AB\\AB&O\end{pmatrix}=r\begin{pmatrix}E&O\\O&-ABAB\end{pmatrix}=n + r(ABAB)\)

又因为\(r(ABAB)\leq r(AB)\)故B正确. 

【答案】D. 

【解析】

选项 A 为上三角矩阵，特征值为主对角元素且 3 个特征值互不相同，故可相似对角化。

选项 B 为实对称矩阵，故可相似对角化

选项 C 的特征值为 1,2,2 ，二重特征值 2 的特征向量的基础解系含解向量个数为\(2 = 3 - r(C - 2E)\)所以可相似对角化.

故选 D. 

【答案】（D）.

【解析】

解析：由题意可知：设存在常数\(k_{1},k_{2},t_{1},t_{2}\)使得:

\(\gamma = k_{1}\alpha_{1}+k_{2}\alpha_{2};\gamma = t_{1}\beta_{1}+t_{2}\beta_{2}\)

即\(k_{1}\alpha_{1}+k_{2}\alpha_{2}=t_{1}\beta_{1}+t_{2}\beta_{2}\)，可得：\(\begin{cases}k_{1}+2k_{2}=2t_{1}+t_{2}\\2k_{1}+k_{2}=5t_{1}+0\cdot t_{2}\\3k_{1}+k_{2}=9t_{1}+t_{2}\end{cases}\)

解得：\(k_{1}=-3k_{2}\).

即\(\gamma = k_{1}\alpha_{1}+k_{2}\alpha_{2}=-3k_{2}\alpha_{1}+k_{2}\alpha_{2}=k_{2}\begin{bmatrix}-1\\-5\\-8\end{bmatrix}=k\begin{bmatrix}1\\5\\8\end{bmatrix},k\in R\)，选（D）. 

【答案】（C）. 

【解析】

由\(X\sim P(1)\)，知：\(P(X = k)=\frac{\lambda^{k}}{k!}e^{-\lambda}=\frac{e^{-1}}{k!}(k = 0,1,2,\cdots)\)，\(EX = 1\)

则

\[
\begin{aligned}
E(|X - EX|) 
&= \sum_{k = 0}^{+\infty}|x_{k}-EX|P_{k} \\
&= |0 - 1|\cdot e^{-1} + |1 - 1|\cdot e^{-1} + |2 - 1|\cdot\frac{e^{-1}}{2!} + \cdots + (k - 1)\frac{1}{k!}e^{-1} + \cdots \\
&= e^{-1} + e^{-1}\sum_{k = 2}^{\infty}(k - 1)\cdot\frac{1}{k!} \\
&= e^{-1} + e^{-1}\sum_{k = 2}^{\infty}\frac{1}{(k - 1)!} - e^{-1}\sum_{k = 2}^{\infty}\frac{1}{k!} \\
&= e^{-1} + e^{-1}(e - 1) - e^{-1}(e - 1 - 1) \\
&= \frac{2}{e}
\end{aligned}
\]

选（C）. 

【答案】（D）.

【解析】：由\(\frac{(n - 1)S^{2}}{\sigma^{2}}\sim\chi^{2}(n - 1)\)知：

\(\frac{(n - 1)S_{1}^{2}}{\sigma^{2}}\sim\chi^{2}(n - 1);\frac{(m - 1)S_{1}^{2}}{2\sigma^{2}}\sim\chi^{2}(m - 1)\)

又两样本之间相互独立，故

\(\frac{2S_{1}^{2}}{S_{2}^{2}}=\frac{\frac{(n - 1)S_{1}^{2}}{\sigma^{2}}/n - 1}{\frac{(m - 1)S_{1}^{2}}{2\sigma^{2}}/m - 1}\sim F(n - 1,m - 1)\) 

选（D）. 

【答案】（A）.

【解析】：因为\(X_{1},X_{2}\)为来自总体\(N(\mu,\sigma^{2})\)的简单随机样本，故\((X_{1}-X_{2})\sim N(0,2\sigma^{2})\).

令\(Y = X_{1}-X_{2}\)，则随机变量\(Y\)的概率密度为：

\(f(y)=\frac{1}{\sqrt{2\pi}\cdot\sqrt{2}\sigma}e^{-\frac{y^{2}}{4\sigma^{2}}},-\infty<y<+\infty\)

则

\[
\begin{aligned}
E(\hat{\sigma}) 
&= a\int_{-\infty}^{+\infty}|y|\frac{1}{2\sqrt{\pi}\cdot\sigma}e^{-\frac{y^{2}}{4\sigma^{2}}}dy \\
&= \frac{a}{\sqrt{\pi}\sigma}\int_{0}^{+\infty}ye^{-\frac{y^{2}}{4\sigma^{2}}}dy \\
&= -\frac{2a\sigma}{\sqrt{\pi}}\int_{0}^{+\infty}e^{-\frac{y^{2}}{4\sigma^{2}}}d\left(-\frac{y^{2}}{4\sigma^{2}}\right) \\
&= \frac{2a\sigma}{\sqrt{\pi}}
\end{aligned}
\]

因\(E(\hat{\sigma})=\sigma\)，则\(a = \frac{\sqrt{\pi}}{2}\)，选（A）. 

【答案】\(ab = - 2\).

【解析】

\[ \begin{aligned} &\lim_{x\to0}\frac{ax + bx^{2}+\ln(1 + x)}{e^{x^{2}}-\cos x} \\ &= \lim_{x\to0}\frac{ax + bx^{2}+x-\frac{1}{2}x^{2}+\frac{1}{3}x^{3}-\frac{1}{4}x^{4}+o(x^{4})}{1 + x^{2}+\frac{1}{2}(x^{4})-(1-\frac{x^{2}}{2}+\frac{1}{4!}x^{4})+o(x^{4})} \\ &= \lim_{x\to0}\frac{(a + 1)x+(b-\frac{1}{2})x^{2}+o(x^{2})}{\frac{3}{2}x^{2}+o(x^{2})} \end{aligned} \]

\(a + 1 = 0\)，\(a=-1\)，\(b-\frac{1}{2}=\frac{3}{2}\)，\(b = 2\)，

\(ab=-2\). 

【答案】\(x + 2y - z = 0\)

【解析】\(F(x,y,z)=x + 2y+\ln(1 + x^{2}+y^{2})-z\).

\(\vec{n}=(F_{x}',F_{y}',F_{z}')=(1+\frac{2x}{1 + x^{2}+y^{2}},2+\frac{2y}{1 + x^{2}+y^{2}},-1)\)

故在点\((0,0,0)\)处的法向量为\(\vec{n}=(1,2,-1)\)即切平面方程为 \(x + 2y - z = 0\). 

【答案】\(0\)

【解析】\(a_{n}=2\int_{0}^{1}(1 - x)\cos n\pi xdx = 2(\int_{0}^{1}\cos n\pi xdx-\int_{0}^{1}x\cos n\pi xdx)\)

\(=-\frac{2}{n\pi}\int_{0}^{1}xd(\sin n\pi x)\)

\(=\frac{-2}{n^{2}\pi^{2}}(\cos n\pi - 1)\)，

故\(\sum_{n = 1}^{\infty}a_{2n}=\frac{-1}{2n^{2}\pi^{2}}(\cos 2n\pi - 1)=0\) 

【答案】\(\frac{1}{2}\).

【解析】

\[ \begin{aligned} \int_{1}^{3}f(x)dx&=\int_{1}^{2}f(x)dx+\int_{2}^{3}f(x)dx=\int_{1}^{2}f(x)dx+\int_{0}^{1}f(x + 2)dx\\ &=\int_{1}^{2}f(x)dx+\int_{0}^{1}[f(x)+x]dx=\int_{1}^{2}f(x)dx+\int_{0}^{1}f(x)dx+\int_{0}^{1}xdx\\ &=\int_{0}^{2}f(x)dx+\int_{0}^{1}xdx=0+\frac{1}{2}=\frac{1}{2}. \end{aligned} \]

【答案】\(\frac{11}{9}\).

【解析】

\(\gamma^{T}\alpha_{1}=\beta^{T}\alpha_{1}=1\Rightarrow k_{1}\alpha_{1}^{T}\alpha_{1}+k_{2}\alpha_{2}^{T}\alpha_{2}+k_{3}\alpha_{3}^{T}\alpha_{3}=1\Rightarrow k_{1}=\frac{1}{3}\)

同理可得\(k_{2}=-1,k_{3}=-\frac{1}{3}\)

\(k_{1}^{2}+k_{2}^{2}+k_{3}^{2}=\frac{11}{9}\) 

【答案】\(\frac{1}{3}\).

【解析】

因为\(X\sim B(1,\frac{1}{3})\)，\(Y\sim B(2,\frac{1}{2})\)所以\(X\)取值为\(0,1\)，\(Y\)的取值为\(0,1,2\)

又因为\(X\)与\(Y\)相互独立，所以

\[ \begin{aligned} P\{X = Y\} &= P\{X = 0, Y = 0\} + P\{X = 1, Y = 1\} \\ &= \frac{2}{3}C_{2}^{0}\left(\frac{1}{2}\right)^2 + \frac{1}{3}C_{2}^{1}\left(\frac{1}{2}\right)^2 = \frac{1}{3} \end{aligned} \]

（1）曲线\(L\)在点\(P(x,y)\)处的切线方程为\(Y - y = y'(X - x)\)，令\(X = 0\)， 则切线在\(y\)轴上的截距为\(Y = y - xy'\)，则\(x = y - xy'\)，即\(y'-\frac{1}{x}y=-1\)，解得

\(y(x)=x(C - \ln x)\)，其中\(C\)为任意常数.

又\(y(1)=2\)，则\(C = 2\)，故\(y(x)=x(2 - \ln x)\).

（2）由（1）可知，\(f(x)=\int_{1}^{x}t(2 - \ln t)dt\)，故\(f'(x)=x(2 - \ln x)=0\). 则驻点为\(x = e^{2}\).

当\(0<x<e^{2}\)时，\(f'(x)>0\)；当\(x>e^{2}\)时，\(f'(x)<0\)故\(f(x)\)在\(x = e^{2}\)处取得极大值，同时也是最大值，且最大值为\(f(e^{2})=\frac{1}{4}e^{4}-\frac{5}{4}\). 

【解析】

\(\begin{cases}f_{x}'=-2(2y + 3xy - 5x^{3}) = 0\\f_{y}'=2y - x^{2}-x^{3}=0\end{cases}\)，得驻点为\((0,0),(1,1),(\frac{2}{3},\frac{10}{27})\)

\(\begin{cases}f_{xx}''=-(2y + 3xy - 5x^{3})-x(3y - 15x^{2})\\f_{xy}''=-2x - 3x^{2}\\f_{yy}''=2\end{cases}\)

代入点\((0,0)\)可得\(\begin{cases}A = 0\\B = 0\\C = 2\end{cases}\)，则\(AC - B^{2}=0\)判别法失效，故当\(x\to0\)时，取\(y = x^{2}+kx^{3}\)可得\(f(x,y)=kx^{5}\)故附近值既有大于\(0\)也有小于\(0\)。因此\((0,0)\)不是极值点

代入点\((1,1)\)可得\(\begin{cases}A = 12\\B = -5\\C = 2\end{cases}\)，则\(AC - B^{2}<0\)故\((1,1)\)不是极值点

代入点\((\frac{2}{3},\frac{10}{27})\)可得\(\begin{cases}A=\frac{100}{27}\\B = -\frac{8}{3}\\C = 2\end{cases}\)，则\(AC - B^{2}>0\)且\(A>0\)故\((\frac{2}{3},\frac{10}{27})\)是极小值点. 故\(f(\frac{2}{3},\frac{10}{27})=-\frac{4}{729}\)为极小值. 

【解析】

由高斯公式可得：

\(I=\iint_{\Sigma}2xz\mathrm{d}y\mathrm{d}z+xz\cos y\mathrm{d}z\mathrm{d}x + 3yz\sin x\mathrm{d}x\mathrm{d}y=\iiint_{\Omega}(2z - xz\sin y + 3y\sin x)dV\)

由对称性知\(\iiint_{\Omega}(xz\sin y + 3y\sin x)dV = 0\)，则

\[ \begin{aligned} I &= \iiint_{\Omega} 2z \,dV = \iint_{D_{xy}} dx \, dy \int_{0}^{1 - x} 2z \, dz = \iint_{D_{xy}} (1 - x)^2 dx \, dy \\ &= \iint_{D_{xy}} (1 - 2x + x^2) dx \, dy = \pi + \frac{1}{2} \int_{0}^{2\pi} d\theta \int_{0}^{1} r^3 \, dr = \frac{5\pi}{4} \end{aligned} \]

【解析】

（Ⅰ）证明：由泰勒中值定理

\(f(x)=f(0)+f'(0)x+\frac{f''(\eta)}{2!}x^{2}=f'(0)x+\frac{f''(\eta)}{2!}x^{2}\)，\(\eta\)介于\(0\)与\(x\)之间，则

\(f(a)=f'(0)a+\frac{f''(\eta_{1})}{2!}a^{2}\)，\(0<\eta_{1}<a\) ①

\(f(-a)=f'(0)(-a)+\frac{f''(\eta_{2})}{2!}a^{2}\)，\(-a<\eta_{2}<0\) ②

①+②得：\(f(a)+f(-a)=\frac{a^{2}}{2}[f''(\eta_{1})+f''(\eta_{2})]\) ③

又\(f''(x)\)在\([\eta_{2},\eta_{1}]\)上连续，则必有最大值\(M\)与最小值\(m\)，即 \(m\leq f''(\eta_{1})\leq M\)；\(m\leq f''(\eta_{2})\leq M\)；从而\(m\leq\frac{f''(\eta_{1}) + f''(\eta_{2})}{2}\leq M\)；

由介值定理得：存在\(\xi\in[\eta_{2},\eta_{1}]\subset(-a,a)\)，有\(\frac{f''(\eta_{1}) + f''(\eta_{2})}{2}=f''(\xi)\)，代入③得：

\(f(a)+f(-a)=a^{2}f''(\xi)\)，\(f''(\xi)=\frac{f(a)+f(-a)}{a^{2}}\)

（Ⅱ）证明：设\(f(x)\)在\(x = x_{0}\in(-a,a)\)取极值，且\(f(x)\)在\(x = x_{0}\)可导，则\(f'(x_{0}) = 0\).

又\(f(x)=f(x_{0})+f'(x_{0})(x - x_{0})+\frac{f''(\gamma)}{2!}(x - x_{0})^{2}=f(x_{0})+\frac{f''(\gamma)}{2!}(x - x_{0})^{2}\)，\(\gamma\)介于\(0\)与\(x\)之间，则\(f(-a)=f(x_{0})+\frac{f''(\gamma_{1})}{2!}(-a - x_{0})^{2}\)，\(-a<\gamma_{1}<0\)

\(f(a)=f(x_{0})+\frac{f''(\gamma_{2})}{2!}(a - x_{0})^{2}\)，\(0<\gamma_{2}<a\)

从而\(\vert f(a)-f(-a)\vert=\vert\frac{1}{2}(a - x_{0})^{2}f''(\gamma_{2})-\frac{1}{2}(a + x_{0})^{2}f''(\gamma_{1})\vert\)

\(\leq\frac{1}{2}\vert(a - x_{0})^{2}f''(\gamma_{2})\vert+\frac{1}{2}\vert(a + x_{0})^{2}f''(\gamma_{1})\vert\)

又\(\vert f''(x)\vert\)连续，设\(M = \max\{\vert f''(\gamma_{1})\vert,\vert f''(\gamma_{2})\vert\}\)，则

\(\vert f(a)-f(-a)\vert\leq\frac{1}{2}M(a + x_{0})^{2}+\frac{1}{2}M(a - x_{0})^{2}\leq M(a^{2}+x_{0}^{2})\)

又\(x_{0}\in(-a,a)\)，则\(\vert f(a)-f(-a)\vert\leq M(a^{2}+x_{0}^{2})\leq2Ma^{2}\)，则\(M\geq\frac{1}{2a^{2}}\vert f(a)-f(-a)\vert\)，

即存在\(\eta=\gamma_{1}\)或\(\eta=\gamma_{2}\in(-a,a)\)，有

\(\vert f''(\eta)\vert\geq\frac{1}{2a^{2}}\vert f(a)-f(-a)\vert\). 

【解析】

（1）利用配方法将\(f(x_{1},x_{2},x_{3})\)化为规范形

\(f(x_{1},x_{2},x_{3})=x_{1}^{2}+2x_{2}^{2}+2x_{3}^{2}+2x_{1}x_{2}-2x_{1}x_{3}=(x_{1}+x_{2}-x_{3})^{2}+(x_{2}+x_{3})^{2}\)

令\(\begin{cases}z_{1}=x_{1}+x_{2}-x_{3}\\z_{2}=x_{2}+x_{3}\\z_{3}=x_{3}\end{cases}\)，即\(\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}1&1& - 1\\0&1&1\\0&0&1\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}\)，则\(f(x_{1},x_{2},x_{3})=z_{1}^{2}+z_{2}^{2}\)

再将\(g(y_{1},y_{2},y_{3})\)化为规范形

\(g(y_{1},y_{2},y_{3})=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2y_{2}y_{3}=y_{1}^{2}+(y_{2}+y_{3})^{2}\)

令\(\begin{cases}z_{1}=y_{1}\\z_{2}=y_{2}+y_{3}\\z_{3}=y_{3}\end{cases}\)则\(g(y_{1},y_{2},y_{3})=z_{1}^{2}+z_{2}^{2}\)

即\(\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}\)使得\(g(y_{1},y_{2},y_{3})=z_{1}^{2}+z_{2}^{2}\)

从而有\(\begin{pmatrix}z_{1}\\z_{2}\\z_{3}\end{pmatrix}=\begin{pmatrix}1&1& - 1\\0&1&1\\0&0&1\end{pmatrix}\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}\)

于是可得\(\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}=P\begin{pmatrix}y_{1}\\y_{2}\\y_{3}\end{pmatrix}\) 其中\(P=\begin{pmatrix}1&1& - 1\\0&1&1\\0&0&1\end{pmatrix}^{-1}\begin{pmatrix}1&0&0\\0&1&1\\0&0&1\end{pmatrix}=\begin{pmatrix}1& - 1&1\\0&1&0\\0&0&1\end{pmatrix}\)为所求矩阵，可将\(f(x_{1},x_{2},x_{3})\)化成\(g(y_{1},y_{2},y_{3})\)

（2）二次型\(f(x_{1},x_{2},x_{3})=x_{1}^{2}+2x_{2}^{2}+2x_{3}^{2}+2x_{1}x_{2}-2x_{1}x_{3}\)的矩阵\(A=\begin{pmatrix}1&1& - 1\\1&2&0\\ - 1&0&2\end{pmatrix}\)

二次型\(g(y_{1},y_{2},y_{3})=y_{1}^{2}+y_{2}^{2}+y_{3}^{2}+2y_{2}y_{3}\)的矩阵\(B=\begin{pmatrix}1&0&0\\0&1&1\\0&1&1\end{pmatrix}\)

若存在正交变换\(x = Qy\)，使得\(Q^{T}AQ=Q^{-1}AQ = B\)可得\(A\)与\(B\)相似. 但有\(tr(A)=5\)，\(tr(B)=3\) ，故\(A\)与\(B\)不相似，因此不存在正交变换\(x = Qy\)使得\(f(x_{1},x_{2},x_{3})\)化成\(g(y_{1},y_{2},y_{3})\)。 

【解析】

（1）\(Cov(X,Y)=E(XY)-E(X)E(Y)\)

\(E(XY)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}xyf(x,y)dxdy=\iint_{x^{2}+y^{2}\leq1}xy\frac{2}{\pi}(x^{2}+y^{2})dxdy = 0\)

\(E(X)=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}xf(x,y)dxdy=\iint_{x^{2}+y^{2}\leq1}x\frac{2}{\pi}(x^{2}+y^{2})dxdy = 0\)

所以\(Cov(X,Y)=0\).

（2）当\(-1<x<1\)时

\[ \begin{aligned} f_{X}(x) &= \int_{-\infty}^{+\infty} f(x,y) \, dy = \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} \frac{2}{\pi}(x^2 + y^2) \, dy = 2 \int_{0}^{\sqrt{1 - x^2}} \frac{2}{\pi}(x^2 + y^2) \, dy \\ &= \frac{4}{\pi} \left[ x^2 \sqrt{1 - x^2} + \frac{1}{3}(1 - x^2)^{\frac{3}{2}} \right] \end{aligned} \]

\[ \therefore f_{X}(x) = \begin{cases} \frac{4}{\pi} \left[ x^2 \sqrt{1 - x^2} + \frac{1}{3}(1 - x^2)^{\frac{3}{2}} \right], & -1 < x < 1 \\ 0, & \text{其他} \end{cases} \]

同理可得\(\therefore f_{Y}(y)\)=

\[
\begin{cases}
\frac{4}{\pi} \left[ y^2 \sqrt{1 - y^2} + \frac{1}{3}(1 - y^2)^{\frac{3}{2}} \right], & -1 < y < 1 \\
0, & \text{其他}
\end{cases}
\]

\(f(x,y)\neq f_{X}(x)f_{Y}(y)\)，\(X\)与\(Y\)不相互独立；

（3）\(F_{Z}(z)=P\{Z\leq z\}=P\{X^{2}+Y^{2}\leq z\}\) \(z<0\)时，\(F_{Z}(z)=0\)；

\(0\leq z<1\)时，\(F_{Z}(z)=P\{X^{2}+Y^{2}\leq z\}=\int_{0}^{2\pi}d\theta\int_{0}^{\sqrt{z}}\frac{2}{\pi}r^{3}dr = z^{2}\)；

\(z\geq1\)时，\(F_{Z}(z)=1\)。

\[
\therefore f_{Z}(z) = 
\begin{cases}
2z, & 0 < z < 1 \\
0, & \text{其他}
\end{cases}
\]