因为

\[lim _{x \to 0} \frac{\int_{0}^{x^{2}}\left(e^{t^{3}}-1\right) d t}{x^{7}}=lim _{x \to 0} \frac{\int_{0}^{x^{2}} t^{3} d t}{x^{7}}=lim _{x \to 0} \frac{\frac{1}{4} t^{4}}{x^{7}}=lim _{x \to 0} \frac{\frac{1}{4} x^{8}}{x^{7}}=0,\]

故 \(x \to 0\) 时, \(\int_{0}^{x^{2}}(e^{t^{3}}-1) d t\) 是 \(x^{7}\) 的高阶无穷小。 故应选(C).

因 \(\lim _{x \to 0} f(x)=\lim _{x \to 0} \frac{e^{x}-1}{x}=\lim _{x \to 0} \frac{x}{x}=1=f(0)\)，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).

设圆柱体的底面半径为 \(r\) ,高为 \(h\) ,则圆柱体的体积 \(V=\pi r^{2} h\) ,圆柱体的表面积 \(S=2 \pi r^{2}+2 \pi r h\) ,从而

\[V'(t)=\pi\left[2 r \cdot r'(t) \cdot h+r^{2} \cdot h'(t)\right]\]

\[S'(t)=2 \pi \cdot 2 r \cdot r'(t)+2 \pi\left[r'(t) \cdot h+r \cdot h'(t)\right]\]

将 \(r=10\) , \(h=5\) , \(r'(t)=2\) , \(h'(t)=-3\) 分别代入中,解得

\[\left.V'(t)\right|_{\substack{r=10 \\ h=5}}=-100 \pi \mathrm{cm}^{3} / \mathrm{s},\]

\[\left.S'(t)\right|_{\substack{r=10 \\ h=5}}=40 \pi \mathrm{cm}^{2} / \mathrm{s},\]

故应选(C).

答案 A.
分析 本题主要考查函数性态的分析.
可以由$f(x)$的表达式分析$f(x)$的单调性,从而得到$f(x)$有2个零点需要满足的条件,最后确定$\frac{b}{a}$的取值范围;也可以将$f(x)$的零点问题先转化为曲线的交点问题,再讨论$\frac{b}{a}$的取值范围.

解（法一）$f(x)$的定义域为$(0,+∞)$.计算$f'(x)$得$f'(x)=a-\frac {b}{x}$.由于$a>0$,故若$b≤0,$则$f'(x)>0,f(x)$单调增加.此时$f(x)$不可能有2个零点.于是,$b>0.$
下面分析$f(x)$的单调性.
当$x=\frac {b}{a}$时,$f'(x)=0$;当$0<x<\frac {b}{a}$时,$f'(x)<0,f(x)$单调减少;当$x>\frac {b}{a}$时,$f'(x)>0,$$f(x)$单调增加.于是,$f(x)$在$(0,+∞)$上先单调减少,再单调增加.$f(\frac {b}{a})$为$f(x)$的最小值.
$f(x)$有2个零点等价于$f(\frac {b}{a})$小于零.否则$f(x)$恒大于等于零,不可能存在2个零点.
$f(\frac {b}{a})=a\cdot \frac {b}{a}-b\ln \frac {b}{a}=b - b\ln \frac {b}{a}=b(1-\ln \frac {b}{a})<0.$
又因为$b>0$,所以$b(1-\ln \frac {b}{a})<0$等价于$1-\ln \frac {b}{a}<0$,即$\ln \frac {b}{a}>1,\frac {b}{a}>e.$
因此,应选A.

（法二）同法一可知$b>0.$
$ax - b\ln x=0$等价于$\frac {x}{\ln x}=\frac {b}{a}$,故函数$f(x)=ax - b\ln x$有2个零点等价于曲线$y=\frac {x}{\ln x}$与直线$y=\frac {b}{a}$有2个交点.

由题意知 \(f(x)=\sec x=\frac{1}{\cos x}=1+a x+b x^{2}+o(x^{2})\)，故

\[\begin{aligned} 1 & =\cos \left(1+a x+b x^{2}+o(x^{2})\right) \\ & =\left(1-\frac{1}{2} x^{2}+o\left(x^{2}\right)\right)\left(1+a x+b x^{2}+o\left(x^{2}\right)\right) \\ & =1+a x+\left(b-\frac{1}{2}\right) x^{2}+o\left(x^{2}\right), \end{aligned}\]

由系数对应得 \(\begin{cases}a=0, \\ b-\frac{1}{2}=0\end{cases}\)，即 \(\begin{cases}a=0, \\ b=\frac{1}{2}.\end{cases}\) 故应选(D).

令 \(u=x+1, v=e^{x}\) ,则 \(f(u, v)=u^{2} \ln v\) ,从而 \(f(x, y)=x^{2} \ln y\) ,故

\[f_{x}'(x, y)=2 x \ln y, f_{y}'(x, y)=\frac{x^{2}}{y},\]

则 \(f_{x}'(1,1)=0\) , \(f_{y}'(1,1)=1\) ,得 \(d f(1,1)=d y\) ，故应选(C).

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 \(\xi_{k}=\frac{\frac{k-1}{n}+\frac{k}{n}}{2}=\frac{2 k-1}{2 n}\)，因此

\[\int_{0}^{1} f(x) d x=lim _{n \to \infty} \sum_{k=1}^{n} f\left(\frac{2 k-1}{2 n}\right) \cdot \frac{1}{n},\]

故应选(B).

答案 B.

分析 本题主要考查二次型的正、负惯性指数.
惯性定理 设二次型\( f = x^{T}Ax \)的秩为\( r \),且有两个可逆变换\( x = Cy \)及\( x = Pz \),使
\( f = k_{1}y_{1}^{2}+k_{2}y_{2}^{2}+\cdots +k_{r}y_{r}^{2} \) (\( k_{i}\neq 0 \))
及
\( f = \lambda_{1}z_{1}^{2}+\lambda_{2}z_{2}^{2}+\cdots +\lambda_{r}z_{r}^{2} \) (\( \lambda_{i}\neq 0 \)),
则\( k_{1},k_{2},\cdots,k_{r} \)中正数的个数与\( \lambda_{1},\lambda_{2},\cdots,\lambda_{r} \)中正数的个数相等.
二次型的标准形中正系数的个数称为二次型的正惯性指数,负系数的个数称为负惯性指数.
二次型通过可逆线性变换可化为标准形,在此过程中,二次型的正、负惯性指数不变.

解 (法一) 令\( \begin{cases} y_{1}=x_{1}+x_{2}, \\ y_{2}=x_{2}+x_{3}, \\ y_{3}=x_{3}, \end{cases} \)则\( x_{3}-x_{1}=y_{2}-y_{1} \),且该变换为可逆线性变换.在该变换下,二次型\( f(x_{1},x_{2},x_{3}) \)化为
\( g(y_{1},y_{2},y_{3})=y_{1}^{2}+y_{2}^{2}-(y_{2}-y_{1})^{2}=2y_{1}y_{2}. \)
再令\( \begin{cases} y_{1}=z_{1}+z_{2}, \\ y_{2}=z_{1}-z_{2}, \\ y_{3}=z_{3}, \end{cases} \)则\( g(y_{1},y_{2},y_{3}) \)化为
\( h(z_{1},z_{2},z_{3})=2(z_{1}+z_{2})(z_{1}-z_{2})=2z_{1}^{2}-2z_{2}^{2}. \)
因此,\( f(x_{1},x_{2},x_{3}) \)的一个标准形为\( 2z_{1}^{2}-2z_{2}^{2} \),其正、负惯性指数分别为1,1.应选B.

(法二) 将\( f(x_{1},x_{2},x_{3}) \)展开可得
\( f(x_{1},x_{2},x_{3})=2x_{2}^{2}+2x_{1}x_{2}+2x_{2}x_{3}+2x_{1}x_{3}. \)
该二次型对应的矩阵为\( A=\begin{pmatrix} 0&1&1 \\ 1&2&1 \\ 1&1&0 \end{pmatrix} \).不难发现,\( A \)的第二行为第一行与第三行的和,故\( r(A)\leq 2 \).又因为\( A \)有一个2阶非零子式\( \begin{vmatrix} 0&1 \\ 1&2 \end{vmatrix} \),所以\( r(A)\geq 2 \).于是,\( r(A)=2 \).

由于二次型的正、负惯性指数之和等于其对应矩阵的秩,而选项C、D的两数之和均为3,故可排除选项C、D.

另一方面,若\( f \)的负惯性指数为0,则\( f \)非负,即对任意\( (x_{1},x_{2},x_{3}) \),均有\( f(x_{1},x_{2},x_{3})\geq 0 \).但是,\( f(1,0,-1)=1+1-4<0 \),矛盾.因此,选项A不正确.

根据排除法,应选B.

答案 D.
分析 本题主要考查向量组之间的线性表示与方程组的解.
若矩阵\( P \)有\( n \)行,则将\( P \)写成分块的形式\( \begin{pmatrix}\alpha_1^{\text{T}}\\\alpha_2^{\text{T}}\\\vdots\\\alpha_n^{\text{T}}\end{pmatrix} \),从而\( Px = 0 \)等价于\( \begin{pmatrix}\alpha_1^{\text{T}}x\\\alpha_2^{\text{T}}x\\\vdots\\\alpha_n^{\text{T}}x\end{pmatrix} = 0 \).方程组\( Px = 0 \)的解与\( P \)的行向量均正交.
解 (法一) 由已知条件可知\( A \)的列向量组可由\( B \)的列向量组线性表示,故\( A^{\text{T}} \)的行向量组可由\( B^{\text{T}} \)的行向量组线性表示.若\( x_0 \)为\( B^{\text{T}}x = 0 \)的解,则\( x_0 \)与\( B^{\text{T}} \)的行向量均正交,从而与\( A^{\text{T}} \)的行向量均正交.根据分析中的结论,可得\( A^{\text{T}}x_0 = 0 \).
因此,\( B^{\text{T}}x = 0 \)的解均为\( A^{\text{T}}x = 0 \)的解.应选D.
(法二) 由于向量组\( \alpha_1,\alpha_2,\alpha_3 \)可由向量组\( \beta_1,\beta_2,\beta_3 \)线性表示,故存在矩阵\( P \),使得\( A = BP \).
由\( A = BP \)可得\( A^{\text{T}} = P^{\text{T}}B^{\text{T}} \).于是,若\( B^{\text{T}}x = 0 \),则\( A^{\text{T}}x = P^{\text{T}}B^{\text{T}}x = P^{\text{T}}(B^{\text{T}}x) = 0 \),即\( B^{\text{T}}x = 0 \)的解均为\( A^{\text{T}}x = 0 \)的解.
因此,应选D.

法1. 验证 \(\begin{pmatrix}1 & 0 & 0 \\ 2 & -1 & 0 \\ -3 & 2 & 1\end{pmatrix} A \begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{pmatrix}=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\)，故应选(C).

法2. 对 \((A, E)\) 作行变换：

\[\begin{aligned} (A, E) & =\left(\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & 0 & 0 \\ 2 & -1 & 1 & 0 & 1 & 0 \\ -1 & 2 & -5 & 0 & 0 & 1 \end{array}\right) \to\left(\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & -1 & 3 & -2 & 1 & 0 \\ 0 & 2 & -6 & 1 & 0 & 1 \end{array}\right) \\ & \to\left(\begin{array}{ccc|ccc} 1 & 0 & -1 & 1 & 0 & 0 \\ 0 & 1 & -3 & 2 & -1 & 0 \\ 0 & 0 & 0 & -3 & 2 & 1 \end{array}\right), \end{aligned}\]

得 \(P=\begin{pmatrix}1 & 0 & 0 \\ 2 & -1 & 0 \\ -3 & 2 & 1\end{pmatrix}\)，此时 \(PA=B=\begin{pmatrix}1 & 0 & -1 \\ 0 & 1 & -3 \\ 0 & 0 & 0\end{pmatrix}\)。

对 \((B, E)\) 作列变换得 \(Q=\begin{pmatrix}1 & 0 & 1 \\ 0 & 1 & 3 \\ 0 & 0 & 1\end{pmatrix}\)，使 \(BQ=\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\)，即 \(PAQ\) 为对角矩阵，故应选(C).

【解析】

\[\begin{aligned} \int_{-\infty}^{+\infty}|x| 3^{-x^{2}} d x & =2 \int_{0}^{+\infty} x 3^{-x^{2}} d x=\int_{0}^{+\infty} 3^{-x^{2}} d\left(x^{2}\right) \\ & =-\left.\frac{3^{-x_{0}^{2}}}{ln 3}\right|_{0} ^{+\infty}=-\frac{1}{ln 3}\left(lim _{x \to \infty} 3^{-x^{2}}-1\right)=\frac{1}{ln 3} . \end{aligned}\]

【解析】

\[\frac{d y}{ d x}=\frac{\frac{d y}{ d t}}{\frac{d x}{ d t}}=\frac{4\left[e^{t}+(t-1) e^{t}\right]+2 t}{2 e^{t}+1}=\frac{4 t e^{t}+2 t}{2 e^{t}+1}=2 t,\]

\[\frac{d^{2} y}{ d x^{2}}=\frac{d}{d x}\left(\frac{d y}{ d x}\right)=\frac{d}{d x}(2 t)=2 \cdot \frac{1}{2 e'+1},\]

\[\left.\frac{d^{2} y}{ d x^{2}}\right|_{t=0}=\frac{2}{3} .\]

【解析】在等式\((x+1) z+y \ln z-\arctan (2 x y)=1\)两边对x求偏导有

\[z+(x+1) \cdot \frac{\partial z}{\partial x}+y \cdot \frac{1}{z} \cdot \frac{\partial z}{\partial x}-\frac{2 y}{1+(2 x y)^{2}}=0\]

将\(x=0\)，\(y=2\)代入原方程得\(z+2 \ln z=1\)，易知\(z=1\)，再将\(x=0\)、\(y=2\)、\(z=1\)代入偏导方程，解得\(\frac{\partial z}{\partial x}|_{(0,2)}=1\)

【解析】

\[f(t)=\int_{1}^{t^{2}} d x \int_{-\infty}^{t} \sin \frac{x}{y} d y=\int_{1}^{t} \left[-\left.y \cos \frac{x}{y}\right|_{1} ^{y^{2}}\right] d y=\int_{1}^{t}\left(-t \cos t + t \cos \frac{1}{t}\right) d y\]

从而\(f'(t)=-t \cos t + t \cos \frac{1}{t}\)，故\(f'(\frac{\pi}{2})=\frac{\pi}{2} \cos \frac{2}{\pi}\)

【解析】特征方程为\(\lambda^{3}-1=0\)，解得\(\lambda=1\)（三重根），三个线性无关特解为\(e^{x}\)，\(x e^{x}\)，\(x^{2} e^{x}\)，故通解为

【解析】四阶行列式中含\(x^{3}\)的项为\((-1)^{\tau(2134)} a_{12} a_{21} a_{33} a_{44}=-x^{3}\)和\((-1)^{\tau(4231)} a_{14} a_{22} a_{33} a_{41}=-4 x^{3}\)，故系数为\(-1-4=-5\)

【解析】

\[
\begin{aligned}
\lim _{x \to 0}\left(\frac{1+\int_{0}^{x} e^{t^{2}} d t}{e^{x}-1}-\frac{1}{\sin x}\right) &= \lim _{x \to 0} \frac{\sin x + \sin x \cdot \int_{0}^{x} e^{t^{2}} d t - e^{x} + 1}{(e^{x}-1)\sin x} \\
&= \lim _{x \to 0} \frac{\sin x - e^{x} + 1}{x^{2}} + \lim _{x \to 0} \frac{\sin x \cdot \int_{0}^{x} e^{t^{2}} d t}{x^{2}} \\
&= \lim _{x \to 0} \frac{x + o(x^{2}) - \left(1 + x + \frac{1}{2}x^{2} + o(x^{2})\right) + 1}{x^{2}} + \lim _{x \to 0} \frac{\int_{0}^{x} e^{t^{2}} d t}{x} \\
&= -\frac{1}{2} + \lim _{x \to 0} e^{x^{2}} = -\frac{1}{2} + 1 = \frac{1}{2} .
\end{aligned}
\]

【解析】因为

\[
f(x)=\frac{x|x|}{1 + x}=
\begin{cases} 
\frac{x^2}{1 + x}, & x>0, \\
0, & x = 0, \\
-\frac{x^2}{1 + x}, & -1<x<0, \\
\text{无定义}, & x=-1, \\
-\frac{x^2}{1 + x}, & x<-1.
\end{cases}
\]

\(f(x)\)在\(x = 0\)连续，又

\[
f^\prime(x)=
\begin{cases} 
\frac{2x + x^2}{(1 + x)^2}, & x>0, \\
-\frac{2x + x^2}{(1 + x)^2}, & -1<x<0, \\
-\frac{2x + x^2}{(1 + x)^2}, & x<-1.
\end{cases}
\]

因为\(\lim\limits_{x \to 0^+}f^\prime(x)=\lim\limits_{x \to 0^-}f^\prime(x)=0\)，故\(f_+^\prime(0)=f_-^\prime(0)=0\)，即\(f^\prime(0)=0\)，从而

\[
f^\prime(x)=
\begin{cases} 
\frac{2x + x^2}{(1 + x)^2}, & x\geq0, \\
-\frac{2x + x^2}{(1 + x)^2}, & -1<x<0, \\
-\frac{2x + x^2}{(1 + x)^2}, & x<-1.
\end{cases}
\]

\(f^\prime(x)\)在\(x = 0\)处连续，又

\[
f^{\prime\prime}(x)=
\begin{cases} 
\frac{(2 + 2x)(1 + x)^2 - (2x + x^2)\cdot2(1 + x)}{(1 + x)^4}=\frac{2}{1 + x}-\frac{4x + 2x^2}{(1 + x)^3}=\frac{2}{(1 + x)^3}, & x>0, \\
-\frac{2}{(1 + x)^3}, & x<0 \text{且} x\neq -1.
\end{cases}
\]

因为\(\lim\limits_{x \to 0^+}f^{\prime\prime}(x)=2,\lim\limits_{x \to 0^-}f^{\prime\prime}(x)=-2\)，故\(f_+^{\prime\prime}(0)\neq f_-^{\prime\prime}(0)\)，从而\(f^{\prime\prime}(0)\)不存在，故

\[
f^{\prime\prime}(x)=
\begin{cases} 
\frac{2}{(1 + x)^3}, & x>0, \\
-\frac{2}{(1 + x)^3}, & x<0 \text{且} x\neq -1.
\end{cases}
\]

当\(x>0\)时，\(f^{\prime\prime}(x)>0\)，当\(-1<x<0\)时，\(f^{\prime\prime}(x)<0\)，

当\(x<-1\)时，\(f^{\prime\prime}(x)>0\)，从而曲线\(y = f(x)\)的凹区间为\((-\infty,-1)\)和\((0,+\infty)\)，凸区间为\((-1,0)\)。

因\(\lim\limits_{x \to -1}f(x)=\infty\)，故\(x = -1\)为曲线\(y = f(x)\)的一条铅垂渐近线；

因\(\lim\limits_{x \to \infty}f(x)=\lim\limits_{x \to \infty}\frac{x}{1 + x}|x|=+\infty\)，从而曲线\(y = f(x)\)无水平渐近线；

因

\[
k_1=\lim\limits_{x \to +\infty}\frac{f(x)}{x}=\lim\limits_{x \to +\infty}\frac{x^2}{(1 + x)x}=\lim\limits_{x \to +\infty}\frac{x}{1 + x}=1,
\]

\[
b_1=\lim\limits_{x \to +\infty}[f(x)-x]=\lim\limits_{x \to +\infty}(\frac{x^2}{1 + x}-x)=\lim\limits_{x \to +\infty}\frac{x^2 - x - x^2}{1 + x}=-1,
\]

故\(y = x - 1\)为\(x \to +\infty\)时曲线\(y = f(x)\)的一条斜渐近线。

因

\[
k_2=\lim\limits_{x \to -\infty}\frac{f(x)}{x}=\lim\limits_{x \to -\infty}\frac{-x^2}{(1 + x)x}=-\lim\limits_{x \to -\infty}\frac{x}{1 + x}=-1,
\]

\[
b_2=\lim\limits_{x \to -\infty}(f(x)+x)=\lim\limits_{x \to -\infty}(-\frac{x^2}{1 + x}+x)=\lim\limits_{x \to -\infty}\frac{x + x^2 - x^2}{1 + x}=1,
\]

故\(y = -x + 1\)为\(x \to -\infty\)时曲线\(y = f(x)\)的一条斜渐近线。

【解析】在等式\(\int\frac{f(x)}{\sqrt{x}}dx = \frac{1}{6}x^2 - x + C\)的两端对\(x\)求导，有\(\frac{f(x)}{\sqrt{x}} = \frac{1}{3}x - 1\)，故

\(f(x) = \frac{1}{3}x\sqrt{x} - \sqrt{x}\)，

\(f^\prime(x) = \frac{1}{2}\sqrt{x} - \frac{1}{2\sqrt{x}} = \frac{1}{2}(\sqrt{x} - \frac{1}{\sqrt{x}}) = \frac{1}{2}\frac{x - 1}{\sqrt{x}}\)，

则

\(S = \int_{4}^{9}\sqrt{1 + [f^\prime(x)]^2}dx = \int_{4}^{9}\frac{(x + 1)}{2\sqrt{x}}dx = \int_{4}^{9}(x + 1)d(\sqrt{x})\)

\( = \int_{2}^{3}(x^2 + 1)dx = \frac{1}{3}x^3\big|_{2}^{3} + 1\)

\( = \frac{1}{3}(27 - 8) + 1 = \frac{22}{3}\)，

\(A = 2\pi\int_{4}^{9}f(x)\sqrt{1 + [f^\prime(x)]^2}dx = 2\pi\int_{4}^{9}(\frac{1}{3}x\sqrt{x} - \sqrt{x})\frac{(x + 1)}{2\sqrt{x}}dx\)

\( = 2\pi\int_{4}^{9}(\frac{1}{6}x^2 - \frac{1}{3}x - \frac{1}{2})dx\)

\( = 2\pi\cdot\frac{425}{18} = \frac{425}{9}\pi\)。 

【解析】（Ⅰ）因\(xy' - 6y = - 6\)，故\(y' - \frac{6}{x}y = - \frac{6}{x}\)，从而

\[
\begin{align*}
y&=e^{\int\frac{6}{x}dx}(\int - \frac{6}{x}\cdot\frac{1}{x^6}dx + C)\\
&=x^6(x^{- 6} + C) = 1 + Cx^6
\end{align*}
\]

又\(y(\sqrt{3}) = 10\)，故\(C = \frac{1}{3}\)，从而\(y(x) = 1 + \frac{1}{3}x^6\)。

（Ⅱ）曲线\(y = y(x)\)在点\(P\)的法线方程为\(Y - (1 + \frac{1}{3}x^6) = - \frac{1}{2x^5}(X - x)\)，令\(X = 0\)，则\(Y = 1 + \frac{1}{3}x^6 + \frac{1}{2x^4}\)。

令\(g(x) = 1 + \frac{1}{3}x^6 + \frac{1}{2x^4},x\gt0\)，则

\[
\begin{align*}
g'(x)&= 2x^5 + \frac{1}{2}\cdot(- 4)x^{- 5}= 2(x^5 - x^{- 5}) = 2\cdot\frac{x^{10} - 1}{x^5}
\end{align*}
\]

令\(g'(x) = 0\)，解得\(x = 1\)，当\(x\gt1\)时，\(g'(x)\gt0\)；当\(0\lt x\lt1\)时，\(g'(x)\lt0\)，故\(x = 1\)为唯一的极小值点也一定是最小值点，此时点\(P\)的坐标为\((1,\frac{4}{3})\) 。 

【解析】

\[
\begin{aligned}
\iint_{D} xydxdy &= \int_{0}^{\frac{\pi}{4}} d\theta\int_{0}^{\sqrt{\cos2\theta}} r\cos\theta r\sin\theta rdr \\
&= \int_{0}^{\frac{\pi}{4}} \cos\theta\sin\theta\int_{0}^{\sqrt{\cos2\theta}} r^{3}dr \\
&= \int_{0}^{\frac{\pi}{4}} \sin\theta\cos\theta\frac{1}{4}\cos^{2}2\theta d\theta \\
&= \frac{1}{4}\int_{0}^{\frac{\pi}{4}}\frac{1}{2}\sin2\theta\cos^{2}2\theta d\theta \\
&= -\frac{1}{16}\int_{0}^{\frac{\pi}{4}} \cos^{2}2\theta d(\cos2\theta) \\
&= -\frac{1}{16}\cdot\frac{1}{3}\cdot\cos^{3}2\theta\bigg|_{0}^{\frac{\pi}{4}} \\
&= \frac{1}{48}
\end{aligned}
\]

【解析】

\(\vert\lambda\boldsymbol{E}-\boldsymbol{A}\vert=\begin{vmatrix}\lambda - 2&-1&0\\-1&\lambda - 2&0\\-1&-a&\lambda - b\end{vmatrix}=(\lambda - b)[(\lambda - 2)^2 - 1]\)

\(=(\lambda - b)(\lambda^2 - 4\lambda + 3)=(\lambda - b)(\lambda - 1)(\lambda - 3)\)

故\(\boldsymbol{A}\)的特征值为\(\lambda_1 = b\)，\(\lambda_2 = 1\)，\(\lambda_3 = 3\).

①若\(b = 1\)，则\(\lambda_1 = \lambda_2 = 1\)是\(\boldsymbol{A}\)的二重特征值，\(\lambda_3 = 3\)，因\(\boldsymbol{A}\)可相似对角化，故\(\text{r}(\boldsymbol{E}-\boldsymbol{A}) = 1\).

又\(\boldsymbol{E}-\boldsymbol{A}=\begin{pmatrix}-1&-1&0\\-1&-1&0\\-1&-a&0\end{pmatrix}\to\begin{pmatrix}1&1&0\\0&1 - a&0\\0&0&0\end{pmatrix}\)，

得\(a = 1\).

可求得\(\boldsymbol{A}\)属于特征值\(1\)的两个线性无关的特征向量\(\boldsymbol{\alpha}_1 = (-1,1,0)^\text{T}\)，\(\boldsymbol{\alpha}_2 = (0,0,1)^\text{T}\).

对于\(\lambda_3 = 3\)，求解\((3\boldsymbol{E}-\boldsymbol{A})\boldsymbol{x}=\boldsymbol{0}\)，

\(3\boldsymbol{E}-\boldsymbol{A}=\begin{pmatrix}1&-1&0\\-1&1&0\\-1&-1&2\end{pmatrix}\to\begin{pmatrix}1&-1&0\\0&1&-1\\0&0&0\end{pmatrix}\)

求得\(\boldsymbol{A}\)属于特征值\(3\)的线性无关的特征向量为\(\boldsymbol{\alpha}_3 = (1,1,1)^\text{T}\).

可取\(\boldsymbol{P}=(\boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\boldsymbol{\alpha}_3)\)，则\(\boldsymbol{P}\)是可逆矩阵，满足\(\boldsymbol{P}^{-1}\boldsymbol{AP}=\begin{pmatrix}1&&\\&1&\\&&3\end{pmatrix}\).

②若\(b = 3\)，则\(\lambda_1 = \lambda_3 = 3\)是\(\boldsymbol{A}\)的二重特征值，\(\lambda_2 = 1\)，因\(\boldsymbol{A}\)可相似对角化，故\(\text{r}(3\boldsymbol{E}-\boldsymbol{A}) = 1\).

又\(3\boldsymbol{E}-\boldsymbol{A}=\begin{pmatrix}1&-1&0\\-1&1&0\\-1&-a&0\end{pmatrix}\to\begin{pmatrix}1&-1&0\\0&a + 1&0\\0&0&0\end{pmatrix}\)

得\(a = -1\).

可求得\(\boldsymbol{A}\)属于特征值\(3\)的两个线性无关的特征向量\(\boldsymbol{\beta}_1 = (1,1,0)^\text{T},\boldsymbol{\beta}_2 = (0,0,1)^\text{T}\).

对于\(\lambda_2 = 1\)，求解\((\boldsymbol{E}-\boldsymbol{A})\boldsymbol{x}=\boldsymbol{0}\)，

\(\boldsymbol{E}-\boldsymbol{A}=\begin{pmatrix}-1&-1&0\\-1&-1&0\\-1&1&-2\end{pmatrix}\to\begin{pmatrix}1&1&0\\0&1&-1\\0&0&0\end{pmatrix}\)

求得\(\boldsymbol{A}\)属于特征值\(1\)的一个线性无关的特征向量为\(\boldsymbol{\beta}_3 = (-1,1,1)^\text{T}\).

可取\(\boldsymbol{P}=(\boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\boldsymbol{\beta}_3)\)，则\(\boldsymbol{P}\)是可逆矩阵，满足\(\boldsymbol{P}^{-1}\boldsymbol{AP}=\begin{pmatrix}3&&\\&3&\\&&1\end{pmatrix}\)