【答案】 D
【分析】 当\( x \to 0 \)时，
\(\int_{0}^{\sin x} [(1+t)^t - 1]dt \sim \int_{0}^{\sin x} t^2 dt = \frac{1}{3}\sin^3 x \sim \frac{1}{3}x^3\)；
\(\int_{0}^{\sin x^2} (1+t)^{\frac{1}{t}} dt \sim \int_{0}^{\sin x^2} e dt = e\sin x^2 \sim e x^2\)；
\(\int_{0}^{e^{\sin x}} [e - (1+t)^{\frac{1}{t}}]dt \sim \int_{0}^{\sin x} \frac{e t}{2} dt = \frac{e}{4}\sin^2 x \sim \frac{e}{4}x^2\)；
\(\int_{0}^{\sin^2 x} (t e^t - t)dt \sim \int_{0}^{\sin^2 x} t^2 dt = \frac{1}{3}\sin^6 x \sim \frac{1}{3}x^6\)。
故选D.

【答案】 C
【分析】 \( D_1, D_2 \)是以原点为圆心、半径分别为\( R \)与\( \sqrt{2}R \)的圆，\( D_3 \)是正方形，边长为\( 2R \)，
如图所示.显然有

\( D_1 \subset D_3 \subset D_2 \).
又由于\( e^{-(x^2 + y^2)} \)恒正，因此C成立.

【答案】 B
【分析】
\( f^{(11)}(1) = \sum_{k=0}^{11} C_{11}^k [(x-1)^0]^{(k)} (\sin x)^{(11-k)} \big|_{x=1} \)
\( = C_{11}^{10} 10! \cdot \cos 1 = 11 \cdot \cos 1 \).

【答案】 D
【分析】 \( y = \sqrt[3]{x^3 - 3x^2} \)，则
\( a = \lim_{x \to \infty} \frac{y}{x} = \lim_{x \to \infty} \frac{\sqrt[3]{x^3 - 3x^2}}{x} = 1 \),
\( b = \lim_{x \to \infty} (y - x) = \lim_{x \to \infty} (\sqrt[3]{x^3 - 3x^2} - x) \)
\( \stackrel{t = \frac{1}{x}}{=} \lim_{t \to 0} \frac{\sqrt[3]{1 - 3t} - 1}{t} = -1 \),
故斜渐近线方程为\( y = x - 1 \).

【答案】 B
【分析】 令
\( F(x) = \int_{0}^{\int_{0}^{x} e^{-t^2} dt} f(u) du - \int_{0}^{x} f(t) e^{-x^2} dt, x \in [0,1] \),
则
\( F'(x) = f\left( \int_{0}^{x} e^{-t^2} dt \right) e^{-x^2} - f(x) e^{-x^2} = e^{-x^2} \left[ f\left( \int_{0}^{x} e^{-t^2} dt \right) - f(x) \right] \).
由\( 0 < e^{-x^2} \leq 1 \)知，\( \int_{0}^{x} e^{-t^2} dt \leq \int_{0}^{x} 1 dt = x \)，且\( f(x) \)单调递增，故\( F'(x) \leq 0 \)，又\( F(0) = 0 \)，故
\( F(1) \leq 0 \)，即\( \int_{0}^{\int_{0}^{1} e^{-t^2} dt} f(x) dx \leq \int_{0}^{1} f(x) e^{-x^2} dx \).故选项B正确.
因无法判断\( f(x) \)在\([0,1]\)上的正负，故选项C, D均不正确.

【答案】 A
【分析】 设细杆的质量为\( M \)，长为\( l \)，建立如图所示的坐标系.设质点距杆左端(原点)的距离为\( x_0(x_0 > 1) \)，则引力微元大小为

\( dF = G \cdot \frac{M \cdot \frac{dx}{l} \cdot a}{(x_0 - x)^2} = Ga \frac{1}{(x_0 - x)^2} dx \),
则此瞬时引力大小为
\( F = \int_{0}^{1} dF = Ga \int_{0}^{1} \frac{1}{(x_0 - x)^2} dx \)
\( = Ga \frac{1}{x_0 - x} \big|_{x=0}^{x=1} = Ga \left( \frac{1}{x_0 - 1} - \frac{1}{x_0} \right) \)
\( = Ga \frac{1}{x_0(x_0 - 1)} \).
故所求引力做功大小为
\( W = \left| \int_{\frac{3}{2}}^{\frac{4}{3}} Ga \frac{dx_0}{x_0(x_0 - 1)} \right| = \left| Ga \ln \frac{x_0 - 1}{x_0} \big|_{\frac{3}{2}}^{\frac{4}{3}} \right| \)
\( = \left| -Ga \ln \frac{4}{3} \right| = Ga \ln \frac{4}{3} \).

【答案】 A
【分析】 对于①，可举反例\( f(x) = \frac{\sin x^2}{x} \)，\( \lim_{x \to +\infty} f(x) \)存在，但\( \lim_{x \to +\infty} f'(x) \)不存在.
对于②，可举反例\( f(x) = x \)，\( \lim_{x \to +\infty} f'(x) \)存在，但\( \lim_{x \to +\infty} f(x) \)不存在.
对于③，不妨设\( a > 0 \)，则存在\( X > 0 \)，当\( x > X \)时，\( f'(x) > \frac{a}{2} \)，故
\( f(x) = f(X) + \int_{X}^{x} f'(t)dt > f(X) + \frac{a}{2}(x - X) \),
当\( x \to +\infty \)时，\( f(x) \to +\infty \).故③正确.
对于④，可举反例\( f(x) = \sqrt{x} \)，\( \lim_{x \to +\infty} f'(x) = 0 \)，但\( f(x) \)在\( x \to +\infty \)时无界.
故选A.

【答案】 B
【分析】 记\( \alpha_1 = \begin{pmatrix} a \\ 1 \\ 1 \end{pmatrix} \)，\( \alpha_2 = \begin{pmatrix} 1 \\ a \\ 1 \end{pmatrix} \)，\( \alpha_3 = \begin{pmatrix} 1 \\ a \\ a \end{pmatrix} \)，\( \alpha_4 = \begin{pmatrix} 1 \\ 1 \\ a \end{pmatrix} \)，由题意，向量组\( \alpha_1, \alpha_2, \alpha_4 \)可由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示，可知\( \alpha_4 \)可由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示，所以方程组\( x_1\alpha_1 + x_2\alpha_2 + x_3\alpha_3 = \alpha_4 \)有解，其系数行列式
\( |A| = |\alpha_1, \alpha_2, \alpha_3| = \begin{vmatrix} a & 1 & 1 \\ 1 & a & a \\ 1 & 1 & a \end{vmatrix} = (a + 1)(a - 1)^2 \).
当\( |\alpha_1, \alpha_2, \alpha_3| = (a + 1)(a - 1)^2 \neq 0 \)，即\( a \neq \pm 1 \)时，\( \alpha_4 \)可由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示；
当\( a = 1 \)时，有\( \alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \)，可知\( \alpha_4 \)仍可由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示；
当\( a = -1 \)时，对增广矩阵\( \overline{A} \)作初等行变换，则
\( \overline{A} = (\alpha_1, \alpha_2, \alpha_3 \mid \alpha_4) = \begin{pmatrix} -1 & 1 & 1 & 1 \\ 1 & -1 & -1 & 1 \\ 1 & 1 & -1 & -1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 & 1 \\ -1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \end{pmatrix} \)
\( \to \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 0 & 0 & 2 \\ 0 & 2 & 0 & -2 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & -1 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 1 \end{pmatrix} \),
可知\( r(\overline{A}) = 3, r(A) = 2 \)，此时\( \alpha_4 \)不能由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示.

由题设知，\( A \)可经初等列变换化成\( B \)，则\( B \)也可经初等列变换化成\( A \)，即\( \alpha_3 \)可由\( \alpha_1, \alpha_2, \alpha_4 \)线性表示.同理可得\( a \neq -2 \).
综上，\( a \neq -1 \)且\( a \neq -2 \).故选B.

【注】 本题实质上是\( \alpha_4 \)可由\( \alpha_1, \alpha_2, \alpha_3 \)线性表示且\( \alpha_3 \)可由\( \alpha_1, \alpha_2, \alpha_4 \)线性表示，进而转化成非齐次线性方程组有解的问题.

【答案】 C
【分析】 设二次型矩阵为\( A \)，则
\( P^{-1}AP = P^T AP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -2 \end{pmatrix} \),
则\( e_1, e_2, e_3 \)分别是\( A \)的对应于特征值1,1,-2的特征向量，于是\( -e_3 \)也是\( A \)的对应于特征值-2的特征向量.因此
\( Q^{-1}AQ = Q^T AQ = \begin{pmatrix} -2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \),
从而\( f(x_1, x_2, x_3) \)在正交变换\( x = Qy \)下的标准形为\( -2y_1^2 + y_2^2 + y_3^2 \).故选C.

【答案】 C
【分析】 必要性( \( \Leftarrow \) ).若\( A = CB, BC = E_r, r(C) = r, \) 则\( A^2 = CBCB = CE_r B = A \),
成立.
充分性( \( \Rightarrow \) ).由\( A^2 = A \)得\( A(A - E) = O \)，故\( r(A) + r(A - E) \leq n \).又\( r(A) + r(A - E) \geq r(A - (A - E)) = r(E) = n, \) 可得\( r(A) + r(A - E) = n \).由\( A^2 = A \)可得矩阵\( A \)的特征值只能为0或1，而\( A \)属于0的线性无关的特征向量个数为\( n - r(A) \)，属于1的线性无关的特征向量个数为\( n - r(A - E) \).故\( A \)的线性无关的特征向量个数为
\( n - r(A) + n - r(A - E) = n \)，则\( A \)可相似对角化.又\( r(A) = r, \) 则存在可逆矩阵\( P \),使
\( A = P^{-1} \begin{pmatrix} E_r & O \\ O & O \end{pmatrix} P \)，即\( A = P^{-1} \begin{pmatrix} E_r \\ O \end{pmatrix} (E_r, O) P \stackrel{记}{=} CB \)，其中
\( C = P^{-1} \begin{pmatrix} E_r \\ O \end{pmatrix}, r(C) = r, B = (E_r, O) P \),
且\( BC = (E_r, O) PP^{-1} \begin{pmatrix} E_r \\ O \end{pmatrix} = E_r \), 成立.
故选C.

【答案】 2
【分析】原式$=\lim\limits _{x→0}\frac {|x|^{x}\cdot x^{2}\cdot (\sqrt {1+x^{2}}+1)}{x^{2}}=\lim\limits _{x→0}|x|^{x}\cdot \lim\limits _{x→0}(\sqrt {1+x^{2}}+1)=1\cdot 2=2$．

【答案】 $\frac {3}{8}π-1$
【分析】原式$=∫_{0}^{1}\arcsin(1-x)d(\frac {x^{2}}{2})=\frac {x^{2}}{2}\arcsin(1-x)|_{0}^{1}+\frac {1}{2}∫_{0}^{1}\frac {x^{2}}{\sqrt {1-(1-x)^{2}}}dx$
$\frac {1-x=t}{} \frac {1}{2}∫_{0}^{1}\frac {(1-t)^{2}}{\sqrt {1-t^{2}}}dt=\frac {1}{2}∫_{0}^{1}\frac {t^{2}-1-2t+2}{\sqrt {1-t^{2}}}dt$
$=-\frac {1}{2}∫_{0}^{1}\sqrt {1-t^{2}}dt-∫_{0}^{1}\frac {t}{\sqrt {1-t^{2}}}dt+∫_{0}^{1}\frac {1}{\sqrt {1-t^{2}}}dt$
$=-\frac {1}{2}\cdot \frac {π}{4}+\sqrt {1-t^{2}}|_{0}^{1}+\arcsin t|_{0}^{1}=\frac {3}{8}π-1$．

【答案】 $\{ 1\}$
【分析】 首先，$a>0$，若不然，当$x>0$时，$e^{ax}≤1$，此时$e^{ax}≥1+x$不成立．
令$f(x)=e^{ax}-1-x$，则$f'(x)=ae^{ax}-1$，令$f'(x)=0$，得$e^{ax}=\frac {1}{a}$，即$ax=-\ln a$，解得$x=-\frac {\ln a}{a}$．因为
$f''(x)=a^{2}e^{ax}>0$，
故
$f_{\min}=f\left(-\frac {\ln a}{a}\right)=e^{-\ln a}-1+\frac {\ln a}{a}=\frac {1}{a}-1+\frac {\ln a}{a}≥0$，
即$1-a+\ln a≥0$，又$\ln a≤a-1$，故$\ln a=a-1$，得$a=1$．
综上，$a$的取值范围是$\{ 1\}$．

【注】 考生熟知对任意的$x∈R$，有$e^{x}≥1+x$，但本题的解答显然不是那么轻易的，要明确本题是含参不等式问题，才能走上正确的解题之路．

【答案】 $2π$
【分析】 平面区域如图所示，则面积为

$S=2∫_{-∞}^{+∞}\frac {1}{1+x^{2}}dx=4\arctan x|_{0}^{+∞}=4\cdot \frac {π}{2}=2π$．

【答案】 $\frac {\sqrt {3}}{3}\ln 2dx-\frac {\sqrt {3}}{6}dy$
【分析】
$\frac {\partial z}{\partial x}=\frac {y^{x}\ln y}{\sqrt {1-(y^{x})^{2}}}$，
$\frac {\partial z}{\partial y}=\frac {xy^{x-1}}{\sqrt {1-(y^{x})^{2}}}$，
故
$\frac {\partial z}{\partial x}|_{(-1,2)}=\frac {\frac {1}{2}\ln 2}{\frac {\sqrt {3}}{2}}=\frac {\sqrt {3}}{3}\ln 2$，
$\frac {\partial z}{\partial y}|_{(-1,2)}=\frac {-1×\frac {1}{4}}{\frac {\sqrt {3}}{2}}=-\frac {\sqrt {3}}{6}$，
于是
$dz|_{(-1,2)}=\frac {\sqrt {3}}{3}\ln 2dx-\frac {\sqrt {3}}{6}dy$．

【答案】 $\{ k|k∈R且k≠-1\}$
【分析】 方程组(Ⅱ)的通解为
$k_{1}(2,-1,k+2,1)^{T}+k_{2}(-1,2,4,k+8)^{T}$
$=(2k_{1}-k_{2},-k_{1}+2k_{2},k_{1}(k+2)+4k_{2},k_{1}+k_{2}(k+8))^{T}$，
其中$k_{1},k_{2}$为任意常数．将通解代入方程组(Ⅰ)并整理得$\begin{cases} -(k+1)k_{1}=0, \\ (k+1)k_{1}-(k+1)k_{2}=0, \end{cases}$则
方程组(Ⅰ)与(Ⅱ)有非零公共解
$\Leftrightarrow$关于$k_{1},k_{2}$的方程组$\begin{cases} -(k+1)k_{1}=0, \\ (k+1)k_{1}-(k+1)k_{2}=0 \end{cases}$有非零解
$\Leftrightarrow \begin{vmatrix} -(k+1) & 0 \\ k+1 & -(k+1) \end{vmatrix}=0$
$\Leftrightarrow k=-1$．

则方程组(Ⅰ)与(Ⅱ)没有非零公共解⇔k≠-1，即k的取值范围为$\{ k|k\in \mathbf{R}且k\neq -1\}$。

【解】当$x \to 0$时，$f(x) > 0$，有
$$
\begin{align*}
f(x)^{f(x)} - f(0)^{f(x)} &= f(0)^{f(x)} \left\{ \left[ \frac{f(x)}{f(0)} \right]^{f(x)} - 1 \right\} \\
&= f(0)^{f(x)} \left[ \mathrm{e}^{f(x) \ln \frac{f(x)}{f(0)}} - 1 \right] \sim f(0)^{f(x)} f(x) \ln \frac{f(x)}{f(0)},
\end{align*}
$$
故
$$
\begin{align*}
\text{原式} &= \lim_{x \to 0} \frac{f(0)^{f(x)} f(x) \ln \frac{f(x)}{f(0)}}{x} \\
&= \lim_{x \to 0} f(0)^{f(x)} f(x) \lim_{x \to 0} \frac{\ln f(x) - \ln f(0)}{x - 0} \\
&= f(0)^{f(0)} f(0) \cdot \lim_{x \to 0} \frac{\ln f(x) - \ln f(0)}{f(x) - f(0)} \cdot \frac{f(x) - f(0)}{x} \\
&= f(0)^{f(0)} f(0) \cdot \frac{f'(0)}{f(0)} \\
&= f'(0) f(0)^{f(0)} \\
&= ba^a.
\end{align*}
$$

【解】(1) $x^2 y' + (x^2 - 3)y^2 = 0$整理得$y' = \left( \frac{3}{x^2} - 1 \right) y^2$，即$y^{-2} y' = \frac{3}{x^2} - 1$，此为可分离变量型微分方程，即$\frac{\mathrm{d}y}{y^2} = \left( \frac{3}{x^2} - 1 \right) \mathrm{d}x$，积分得$-\frac{1}{y} = -\frac{3}{x} - x + C$，即$y = \frac{x}{x^2 - Cx + 3}$。由$y(1) = 1$，知$\frac{1}{4 - C} = 1$，有$C = 3$，故
$$
y(x) = \frac{x}{x^2 - 3x + 3}.
$$
(2)由(1)知，
$$
\int_0^3 y^2(x) \mathrm{d}x = \int_0^3 \frac{x^2}{(x^2 - 3x + 3)^2} \mathrm{d}x.
$$
因
$$
x^2 - 3x + 3 = \left( x - \frac{3}{2} \right)^2 + \frac{3}{4},
$$
令$x - \frac{3}{2} = \frac{\sqrt{3}}{2} \tan u \left( |u| < \frac{\pi}{2} \right)$，则
$$
(x^2 - 3x + 3)^2 = \left( \frac{3}{4} \sec^2 u \right)^2 = \frac{9}{16} \sec^4 u, \mathrm{d}x = \frac{\sqrt{3}}{2} \sec^2 u \mathrm{d}u.
$$
当$x = 0$时，$u = -\frac{\pi}{3}$；当$x = 3$时，$u = \frac{\pi}{3}$。于是
$$
\begin{align*}
\int_0^3 \frac{x^2}{(x^2 - 3x + 3)^2} \mathrm{d}x &= \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( \frac{3}{4} \tan^2 u + \frac{3\sqrt{3}}{2} \tan u + \frac{9}{4} \right) \cdot \frac{16}{9} \cdot \frac{\sqrt{3}}{2} \cos^2 u \mathrm{d}u \\
&= \frac{8}{3\sqrt{3}} \cdot 2 \int_0^{\frac{\pi}{3}} \left( \frac{3}{4} \tan^2 u + \frac{9}{4} \right) \cos^2 u \mathrm{d}u \\
&= \frac{4}{\sqrt{3}} \int_0^{\frac{\pi}{3}} (\sin^2 u + 3\cos^2 u) \mathrm{d}u = \frac{4}{\sqrt{3}} \int_0^{\frac{\pi}{3}} (2 + \cos 2u) \mathrm{d}u \\
&= \frac{4}{\sqrt{3}} \left( 2u + \frac{1}{2} \sin 2u \right) \bigg|_0^{\frac{\pi}{3}} = \frac{8\sqrt{3}\pi}{9} + 1.
\end{align*}
$$

【证】必要性$(\Rightarrow)$.
由题设，对任意的$x \in \mathbf{R}$，$f(x) = f(x + T)$，则
$$
\int_a^{a+T} f(x) \mathrm{d}x = \int_a^0 f(x) \mathrm{d}x + \int_0^T f(x) \mathrm{d}x + \int_T^{a+T} f(x) \mathrm{d}x,
$$
其中
$$
\int_T^{a+T} f(x) \mathrm{d}x \stackrel{x = t + T}{=} \int_0^a f(t + T) \mathrm{d}t = \int_0^a f(t) \mathrm{d}t,
$$
故$\int_a^{a+T} f(x) \mathrm{d}x = \int_a^0 f(x) \mathrm{d}x + \int_0^a f(x) \mathrm{d}x + \int_0^T f(x) \mathrm{d}x = \int_0^T f(x) \mathrm{d}x$，即$\int_a^{a+T} f(x) \mathrm{d}x$为常数.

充分性$(\Leftarrow)$.
因$\int_a^{a+T} f(x) \mathrm{d}x \stackrel{\text{记}}{=} C$，则$\left[ \int_a^{a+T} f(x) \mathrm{d}x \right]_a' = 0$，即$f(a + T) - f(a) = 0$，也即任给常数$a \in \mathbf{R}$，有$f(a + T) = f(a)$，故$f(x)$以$T$为周期.

证毕.

【解】积分区域如图所示，则

$$
\begin{align*}
\text{原式} &= -\int_0^1 \mathrm{d}x \int_x^1 (\mathrm{e}^{-y^2} + \mathrm{e}^y \sin y) \mathrm{d}y \\
&= -\int_0^1 \mathrm{d}y \int_0^y (\mathrm{e}^{-y^2} + \mathrm{e}^y \sin y) \mathrm{d}x \\
&= -\int_0^1 y \mathrm{e}^{-y^2} \mathrm{d}y - \int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y \\
&= \frac{1}{2} \mathrm{e}^{-y^2} \bigg|_0^1 - \int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y \\
&= \frac{1}{2} \mathrm{e}^{-1} - \frac{1}{2} - \int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y.
\end{align*}
$$
其中，
$$
\begin{align*}
\int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y &= \int_0^1 y \sin y \mathrm{d}(\mathrm{e}^y) \\
&= y \sin y \cdot \mathrm{e}^y \bigg|_0^1 - \int_0^1 \mathrm{e}^y (\sin y + y \cos y) \mathrm{d}y \\
&= \mathrm{e} \sin 1 - \int_0^1 \mathrm{e}^y \sin y \mathrm{d}y - \int_0^1 y \cos y \mathrm{d}(\mathrm{e}^y) \\
&= \mathrm{e} \sin 1 - \int_0^1 \mathrm{e}^y \sin y \mathrm{d}y - y \cos y \cdot \mathrm{e}^y \bigg|_0^1 + \int_0^1 \mathrm{e}^y (\cos y - y \sin y) \mathrm{d}y \\
&= \mathrm{e}(\sin 1 - \cos 1) + \mathrm{e}^y \cos y \bigg|_0^1 - \int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y \\
&= \mathrm{e} \sin 1 - 1 - \int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y,
\end{align*}
$$
则
$$
\int_0^1 y \mathrm{e}^y \sin y \mathrm{d}y = \frac{1}{2}(\mathrm{e} \sin 1 - 1).
$$
故
$$
\begin{align*}
\text{原式} &= \frac{1}{2} \mathrm{e}^{-1} - \frac{1}{2} - \frac{1}{2}(\mathrm{e} \sin 1 - 1) \\
&= \frac{\mathrm{e}^{-1} - \mathrm{e} \sin 1}{2}.
\end{align*}
$$

【解】令
\[
\begin{cases}
f'_x = 12x^2 - 16xy + 16y^2 - 33 = 0, \\
f'_y = -8x^2 + 32xy - 48 = 0,
\end{cases}
\]
解得
\[
\begin{cases}
x = 2, \\
y = \frac{5}{4}
\end{cases}
\]
(此处只保留\(0 \leq y \leq x \leq 3\)的解),则\(f\left(2, \frac{5}{4}\right) = -84\)。
又在边界\(y = 0(0 \leq x \leq 3)\)上，
\[
f(x, 0) = 4x^3 - 33x,
\]
令\(f'_x(x, 0) = 12x^2 - 33 = 0\)，得\(x = \frac{\sqrt{11}}{2}\)，得点\(\left(\frac{\sqrt{11}}{2}, 0\right)\)，\(f\left(\frac{\sqrt{11}}{2}, 0\right) = -11\sqrt{11}\)。
在边界\(x = 3(0 \leq y \leq 3)\)上，
\[
f(3, y) = 9 - 120y + 48y^2,
\]
令\(f'_y(3, y) = -120 + 96y = 0\)，得\(y = \frac{5}{4}\)，得点\(\left(3, \frac{5}{4}\right)\)，\(f\left(3, \frac{5}{4}\right) = -66\)。
在边界\(y = x(0 \leq x \leq 3)\)上，
\[
f(x, x) = 12x^3 - 81x,
\]
令\(f'(x, x) = 36x^2 - 81 = 0\)，得\(x = \frac{3}{2}\)，得点\(\left(\frac{3}{2}, \frac{3}{2}\right)\)，\(f\left(\frac{3}{2}, \frac{3}{2}\right) = -81\)。
且\(f(0, 0) = 0\)，\(f(3, 0) = 9\)，\(f(3, 3) = 81\)。
综上，\(f(x, y)\)在\(D\)上的取值范围是\([-84, 81]\)。

【解】(1)由\(A\)与\(B\)相似，可知\(|A| = |B|\)，即\(-6 - 2a = 2\)，解得\(a = -4\)。
由\(Ax = x + (b, -b, 2b)^T\)的一个解为\((0, -1, 1)^T\)，可得\((A - E)\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} b \\ -b \\ 2b \end{pmatrix}\)，又因\((A - E)\begin{pmatrix} 0 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}\)，故\(b = 1\)。
(2)设可逆矩阵\(P = (\xi_1, \xi_2, \xi_3)\)，使得\(P^{-1}AP = B\)，即\(AP = PB\)，也即
\[
A(\xi_1, \xi_2, \xi_3) = (\xi_1, \xi_2, \xi_3)\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix},
\]
于是有\((A\xi_1, A\xi_2, A\xi_3) = (\xi_1, \xi_1 + \xi_2, 2\xi_3)\)，则
\[
\begin{cases}
A\xi_1 = \xi_1, \\
A\xi_2 = \xi_1 + \xi_2, \\
A\xi_3 = 2\xi_3.
\end{cases}
\]
由\(A\xi_1 = \xi_1\)，得\((E - A)\xi_1 = 0\)，解\((E - A)x = 0\)得\(\xi_1 = (1, -1, 2)^T\)；
由\(A\xi_2 = \xi_1 + \xi_2\)得\((E - A)\xi_2 = -\xi_1\)，解\((E - A)x = (-1, 1, -2)^T\)得\(\xi_2 = (0, -1, 1)^T\)；
由\(A\xi_3 = 2\xi_3\)，得\((2E - A)\xi_3 = 0\)，解\((2E - A)x = 0\)得\(\xi_3 = (0, 1, 0)^T\)。
故\(P = \begin{pmatrix} 1 & 0 & 0 \\ -1 & -1 & 1 \\ 2 & 1 & 0 \end{pmatrix}\)，使得\(P^{-1}AP = B\)，于是\(A = PBP^{-1}\)，所以
\[
A^{100} = PB^{100}P^{-1} = \begin{pmatrix} 1 & 0 & 0 \\ -1 & -1 & 1 \\ 2 & 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 100 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2^{100} \end{pmatrix}\begin{pmatrix} 1 & 0 & 0 \\ -2 & 0 & 1 \\ -1 & 1 & 1 \end{pmatrix}
\]
\[
= \begin{pmatrix} -199 & 0 & 100 \\ 201 - 2^{100} & 2^{100} & -101 + 2^{100} \\ -400 & 0 & 201 \end{pmatrix}.
\]