答案 选(B)。
解 \(\lim\limits_{x \to 0^{+}}\frac{\int_{0}^{x}|\sin t|dt}{x^{\alpha}} = \lim\limits_{x \to 0^{+}}\frac{\int_{0}^{x}\sin tdt}{x^{\alpha}} = \lim\limits_{x \to 0^{+}}\frac{\sin x}{\alpha x^{\alpha - 1}} = \lim\limits_{x \to 0^{+}}\frac{1}{\alpha x^{\alpha - 2}}\)，由题意知，需极限存在，故\(\alpha \leq 2\)。
\(\lim\limits_{x \to +\infty}\frac{\int_{0}^{x}|\sin t|dt}{x^{\alpha}} = \lim\limits_{x \to +\infty}\frac{\int_{0}^{x}|\sin t|dt}{x} \cdot \frac{1}{x^{\alpha - 1}}\)，由周期函数的性质知 \(\lim\limits_{x \to +\infty}\frac{\int_{0}^{x}|\sin t|dt}{x} = \frac{2}{\pi}\)，
故若函数\(f(x)=\frac{\int_{0}^{x}|\sin t|dt}{x^{\alpha}}\)在\((0,+\infty)\)内有界，需\(\lim\limits_{x \to +\infty}\frac{\int_{0}^{x}|\sin t|dt}{x^{\alpha}}\)存在，即\(\alpha \geq 1\)。

答案 选(D).

解 \( \lim\limits_{x \to 0} \ln | 1 - e^{2x} | = -\infty \)，故直线 \( x = 0 \) 为垂直渐近线.

\( \lim\limits_{x \to +\infty} \frac{\ln | 1 - e^{2x} |}{x} = \lim\limits_{x \to +\infty} \frac{\ln (e^{2x} - 1)}{x} = 2 \).

\( \lim\limits_{x \to +\infty} (\ln | 1 - e^{2x} | - 2x) = \lim\limits_{x \to +\infty} \ln \frac{e^{2x} - 1}{e^{2x}} = 0 \)，故有一条斜渐近线 \( y = 2x \).

\( \lim\limits_{x \to -\infty} \ln | 1 - e^{2x} | = 0 \)，故直线 \( y = 0 \) 为水平渐近线.

答案 选(A)。

解 \(\frac{d}{dx}[(f(x) + 2)(1 + e^{-x})] = f'(x)(1 + e^{-x}) - (f(x) + 2)e^{-x}\)，
\(\lim\limits_{x \to 0^{+}} \frac{(f(x) + 2) \cdot x - 0}{x} = f(0) + 2\)，\(\lim\limits_{x \to 0^{-}} \frac{(f(x) + 2) \cdot (-x) - 0}{x} = -f(0) - 2\)。
又\(G'_{+}(0) = G'_{-}(0)\)，故\(f(0) + 2 = 0\)，即\(f(0) = -2\)，选(A)。

答案 选(A)

解 \( N = \int_{0}^{\frac{\pi}{2}} e^{\sin x}dx \stackrel{\sin x = t}{=} \int_{0}^{1} e^{t} \cdot \frac{1}{\sqrt{1 - t^{2}}}dt = \int_{0}^{1} \frac{e^{x}}{\sqrt{1 - x^{2}}}dx \)，

\( P = \int_{0}^{\frac{\pi}{4}} e^{\tan x}dx \stackrel{\tan x = t}{=} \int_{0}^{1} e^{t} \cdot \frac{1}{1 + t^{2}}dt = \int_{0}^{1} \frac{e^{x}}{1 + x^{2}}dx \)，

因为\( \frac{e^{x}}{1 + x^{2}} < e^{x} < \frac{e^{x}}{\sqrt{1 - x^{2}}} \)（\( 0 < x < 1 \)），所以\( P < M < N \)。故选(A)。

答案 选(C)。

解 \( f'_x(0,0) = \lim\limits_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = \lim\limits_{x \to 0} \frac{0 - 0}{x} = 0 \)，

当 \( (x,y) \neq (0,0) \) 时，\( f'_x(x,y) = \frac{y\cos(xy)(x^2+y^2) - x\sin(xy)}{(x^2+y^2)^{\frac{3}{2}}} \)，

\( f''_{xx}(0,0) = \lim\limits_{x \to 0} \frac{f'_x(x,0) - f'_x(0,0)}{x} = \lim\limits_{x \to 0} \frac{0 - 0}{x} = 0 \)，

\( f''_{xy}(0,0) = \lim\limits_{y \to 0} \frac{f'_x(0,y) - f'_x(0,0)}{y} = \lim\limits_{y \to 0} \frac{\frac{y}{|y|} - 0}{y} = \lim\limits_{y \to 0} \frac{1}{|y|} \) 不存在。故选(C)。

答案 选(B).

解 \(r = \frac{1}{\sin\theta}\)得\(r\sin\theta = 1\)，故\(y = 1\)，\(r = 2\sin\theta\)得\(r^{2} = 2r\sin\theta\)，故\(x^{2} + y^{2} = 2y\)，故

\(D_{1} = \{(x,y) | 0 \leq x \leq 1, x \leq y \leq 1\}\)，\(D_{2} = \{(x,y) | 0 \leq x \leq 1, 1 \leq y \leq 1 + \sqrt{1 - x^{2}}\}\)，

所以\(D = D_{1} + D_{2} = \{(x,y) | 0 \leq x \leq 1, x \leq y \leq 1 + \sqrt{1 - x^{2}}\}\)，故选(B).

答案 选(D).

解 由 \( \int_{0}^{x} f(t)dt \) 是周期函数的充要条件为 \( \int_{0}^{T} f(x)dx = 0 \) 知 ① 、② 均正确.
\( \int_{0}^{+\infty} f(x)dx = 0 \Rightarrow f(x) \equiv 0 \Rightarrow \int_{0}^{x} f(t)dt = 0 \)，故 ③ 正确.
\( F(x+T) - F(x) = \int_{x}^{x+T} f(t)dt - \int_{0}^{T} f(t)dt = 0 \)，故 ④ 正确.

答案 选(B)。

解 \( B = (\alpha_1, \alpha_2, \alpha_3)\begin{pmatrix} 1 & 0 & -2 \\ -1 & 1 & 0 \\ 0 & \lambda & 1 \end{pmatrix} = A\begin{pmatrix} 1 & 0 & -2 \\ -1 & 1 & 0 \\ 0 & \lambda & 1 \end{pmatrix} \)，

因为\( \alpha_1, \alpha_2, \alpha_3 \)两两正交，所以\( \alpha_1, \alpha_2, \alpha_3 \)线性无关，故\( |A| \neq 0 \)，所以\( |A| = |B| \)的充分必要条件为

\( \begin{vmatrix} 1 & 0 & -2 \\ -1 & 1 & 0 \\ 0 & \lambda & 1 \end{vmatrix} = 2\lambda + 1 = 1 \)，

即\( \lambda = 0 \)，故选(B)。

答案 选(D).

解 \( \begin{pmatrix} A & B \\ O & C \end{pmatrix} \stackrel{r_1 \times A^{-1}}{\sim} \begin{pmatrix} E & B \\ O & C \end{pmatrix} \stackrel{r_2 - r_1 \times B}{\sim} \begin{pmatrix} E & O \\ O & C \end{pmatrix} \stackrel{c_1 \times A}{\sim} \begin{pmatrix} A & O \\ O & C \end{pmatrix} \)，所以\( r\begin{pmatrix} A & B \\ O & C \end{pmatrix} = r\begin{pmatrix} A & O \\ O & C \end{pmatrix} \)；

\( \begin{pmatrix} A & B \\ O & C \end{pmatrix} \sim \begin{pmatrix} E & O \\ O & C \end{pmatrix} \stackrel{c_2 + A \times c_1}{\sim} \begin{pmatrix} E & O \\ A & C \end{pmatrix} \)，所以\( r\begin{pmatrix} A & B \\ O & C \end{pmatrix} = r\begin{pmatrix} E & O \\ A & C \end{pmatrix} \)；

\( \begin{pmatrix} A & B \\ O & C \end{pmatrix} \sim \begin{pmatrix} A & O \\ O & C \end{pmatrix} \stackrel{r_1 + B \times r_2}{\sim} \begin{pmatrix} A & BC \\ O & C \end{pmatrix} \)，所以\( r\begin{pmatrix} A & B \\ O & C \end{pmatrix} = r\begin{pmatrix} A & BC \\ O & C \end{pmatrix} \)；

若取\( A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)，\( B = \begin{pmatrix} 1 & -1 \\ 1 & -1 \end{pmatrix} \)，\( C = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \)，则\( BC = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \)，于是有

\( r\begin{pmatrix} A & B \\ O & C \end{pmatrix} = 3 \)，\( r\begin{pmatrix} A & O \\ O & BC \end{pmatrix} = r(A) = 2 \)，

故\( r\begin{pmatrix} A & B \\ O & C \end{pmatrix} \neq r\begin{pmatrix} A & O \\ O & BC \end{pmatrix} \).

【注】此题也可以利用列变换和行变换的方法直接得到答案.

答案 选（A）。

解 由\( x^T A x \stackrel{x=Py}{=} y^T P^T A P y = y^T B y \)，有\( B = P^T A P \)，\( P \)可逆，所以\( A,B \)等价，且合同。

又\( P^T = P^* \)，所以\( |P| = |P^T| = |P^*| = |P|^2 \)，即\( |P|(1 - |P|) = 0 \)。

再由\( |P| \neq 0 \)，得\( |P| = 1 \)，从而\( P^T = P^* = |P| P^{-1} = P^{-1} \)，即\( P^T P = E \)，从而\( P \)为正交阵，故\( B = P^{-1} A P \)，故\( A,B \)相似，所以选（A）。

答案 填“\(-\frac{5}{6}\)”．

解一 由题意\(\lim\limits_{x \to 0} \frac{x\cos x - a\tan x}{x\ln(1 + bx^2)} = 1\)，即\(\lim\limits_{x \to 0} \frac{x\cos x - x + x - a\tan x}{bx^3} = 1\)．

因为\(\lim\limits_{x \to 0} \frac{x(\cos x - 1)}{bx^3} = -\frac{1}{2b}\)，所以\(\lim\limits_{x \to 0} \frac{x - a\tan x}{bx^3} = 1 + \frac{1}{2b}\)，于是\( a = 1 \)，得\(\lim\limits_{x \to 0} \frac{x - \tan x}{bx^3} = -\frac{1}{3b} = 1 + \frac{1}{2b}\)，故\( b = -\frac{5}{6} \)，所以\( ab = -\frac{5}{6} \)．

解二 由题意\(\lim\limits_{x \to 0} \frac{x\cos x - a\tan x}{x\ln(1 + bx^2)} = \lim\limits_{x \to 0} \frac{x\left(1 - \frac{1}{2}x^2\right) - a\left(x + \frac{1}{3}x^3\right) + o(x^3)}{bx^3}\)

\( = \lim\limits_{x \to 0} \frac{(1 - a)x - \left(\frac{1}{2} + \frac{a}{3}\right)x^3 + o(x^3)}{bx^3} = 1\)，故\( a = 1 \)，\( b = -\frac{5}{6} \)，所以\( ab = -\frac{5}{6} \)．

答案 填“$-\frac{1}{2}$”。

解 $\frac{\text{d}y}{\text{d}x} = \frac{1 - \frac{\text{e}^t}{1 + \text{e}^{2t}}}{\frac{2\text{e}^{2t}}{1 + \text{e}^{2t}}} = \frac{1 + \text{e}^{2t} - \text{e}^t}{2\text{e}^{2t}},$
$\frac{\text{d}^2 y}{\text{d}x^2} = \frac{(2\text{e}^{2t} - \text{e}^t) \cdot 2\text{e}^{2t} - (1 + \text{e}^{2t} - \text{e}^t) \cdot 4\text{e}^{2t}}{4\text{e}^{4t}} \cdot \frac{1 + \text{e}^{2t}}{2\text{e}^{2t}} = \frac{(\text{e}^t - 2)(1 + \text{e}^{2t})}{4\text{e}^{4t}},$
$\frac{\text{d}^2 y}{\text{d}x^2}\big|_{t=0} = -\frac{1}{2}.$

答案 填“\(\frac{1}{\sqrt{\pi}}\)”。

解一
\[
\begin{align*}
\int_{0}^{+\infty} [1 - \varphi(x)]dx &= \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty} \left[\int_{0}^{+\infty} e^{-t^2} dt - \int_{0}^{x} e^{-t^2} dt\right]dx \\
&= \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty} \left[\int_{x}^{+\infty} e^{-t^2} dt\right]dx = \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty} dt \int_{0}^{t} e^{-t^2} dx \\
&= \frac{2}{\sqrt{\pi}}\int_{0}^{+\infty} t e^{-t^2} dt = -\frac{1}{\sqrt{\pi}} e^{-t^2} \bigg|_{0}^{+\infty} = \frac{1}{\sqrt{\pi}}.
\end{align*}
\]

解二
\[
\begin{align*}
\int_{0}^{+\infty} [1 - \varphi(x)]dx &= x[1 - \varphi(x)] \bigg|_{0}^{+\infty} + \int_{0}^{+\infty} x \frac{2}{\sqrt{\pi}} e^{-x^2} dx \\
&= \lim_{x \to +\infty} \frac{1 - \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} dt}{\frac{1}{x}} + \frac{1}{\sqrt{\pi}}(-e^{-x^2}) \bigg|_{0}^{+\infty} \\
&= \lim_{x \to +\infty} \frac{-\frac{2}{\sqrt{\pi}} e^{-x^2}}{-\frac{1}{x^2}} + \frac{1}{\sqrt{\pi}} = -\frac{2}{\sqrt{\pi}} \lim_{x \to +\infty} \frac{x^2}{e^{x^2}} + \frac{1}{\sqrt{\pi}} = \frac{1}{\sqrt{\pi}}.
\end{align*}
\]

答案 填“\( y(x) = \frac{1}{12}\left( x\mathrm{e}^{3x} - \frac{1}{3}\mathrm{e}^{3x} \right) \)”。

解 \( r^3 - 2r^2 - 3r = r(r - 3)(r + 1) = 0 \)，解得 \( r_1 = 0 \)，\( r_2 = 3 \)，\( r_3 = -1 \)。

令 \( p = y' \)，则 \( p'' - 2p' - 3p = \mathrm{e}^{3x} \)。

令 \( p^* = \mathrm{e}^{3x} \cdot a \cdot x \) 代入得 \( a = \frac{1}{4} \)，

又 \( \int \frac{x}{4}\mathrm{e}^{3x} \mathrm{d}x = \frac{1}{12}\int x\mathrm{d}\mathrm{e}^{3x} = \frac{x}{12}\mathrm{e}^{3x} - \frac{1}{12}\int \mathrm{e}^{3x} \mathrm{d}x = \frac{x}{12}\mathrm{e}^{3x} - \frac{1}{36}\mathrm{e}^{3x} + C \)，

取 \( y^* = \frac{1}{12}\left( x\mathrm{e}^{3x} - \frac{1}{3}\mathrm{e}^{3x} \right) \)，故原方程通解为

\( y = C_1 + C_2\mathrm{e}^{3x} + C_3\mathrm{e}^{-x} + \frac{1}{12}\left( x\mathrm{e}^{3x} - \frac{1}{3}\mathrm{e}^{3x} \right) \)。

将 \( y(0) = -\frac{1}{36} \)，\( y'(0) = 0 \)，\( y''(0) = \frac{1}{4} \) 代入得 \( C_1 = 0 \)，\( C_2 = 0 \)，\( C_3 = 0 \)，故 \( y(x) = \frac{1}{12}\left( x\mathrm{e}^{3x} - \frac{1}{3}\mathrm{e}^{3x} \right) \)。

答案 填“\(\frac{1}{4}e - \frac{5}{84}\)”。

解 令\(P(x,y) = axy + e^{y^2}\)，\(Q(x,y) = 2x^2 + bxye^{y^2}\)，显然\(P(x,y)\)与\(Q(x,y)\)有一阶的连续偏导数，即函数\(u(x,y)\)的二阶偏导数连续，从而\(\frac{\partial^2 u}{\partial x\partial y} = \frac{\partial^2 u}{\partial y\partial x}\)，得\(ax + 2ye^{y^2} = 4x + bye^{y^2}\)，所以\(a = 4\)，\(b = 2\)。即\(du = (4xy + e^{y^2})dx + (2x^2 + 2xye^{y^2})dy\)。

\(u(x,y) = \int (4xy + e^{y^2})dx = 2x^2y + xe^{y^2} + \phi(y)\)，

又由\(\frac{\partial u}{\partial y} = 2x^2 + 2xye^{y^2} = 2x^2 + 2xye^{y^2} + \phi'(y)\)，解得\(\phi'(y) = 0\)，即\(\phi(y) = C\)。

于是有 \(u(x,y) = 2x^2y + xe^{y^2} + C\)，代入\(u(0,0) = 0\)，得\(C = 0\)。则\(u(x,y) = 2x^2y + xe^{y^2}\)。

\(\iint\limits_D u(x,y)d\sigma = \int_0^1 dy \int_0^{\sqrt{y}} (2x^2y + xe^{y^2}) dx = \int_0^1 \left( \frac{1}{2}y e^{y^2} + \frac{2}{3}y^{\frac{5}{2}} \right) dy = \frac{1}{4}e - \frac{5}{84}\)。

答案 填“\(k(-1,-1,2,-1)^{\mathrm{T}},k\in\mathbb{R}\)”。

解 \(\begin{pmatrix}A\\B\end{pmatrix}x = 0\)的解\(\alpha\)是\(Ax = 0\)与\(Bx = 0\)的公共解\(\alpha\)，故\(\alpha = k_{1}\xi_{1}+k_{2}\xi_{2}=l_{1}\eta_{1}+l_{2}\eta_{2}\)，得方程组
\(k_{1}\xi_{1}+k_{2}\xi_{2}-l_{1}\eta_{1}-l_{2}\eta_{2}=0\)，\((\xi_{1},\xi_{2},-\eta_{1},-\eta_{2})\begin{pmatrix}k_{1}\\k_{2}\\l_{1}\\l_{2}\end{pmatrix}=0\)，即\(\begin{pmatrix}1&-1&-1&0\\0&1&-2&-1\\-1&0&1&-1\\0&1&-2&-1\end{pmatrix}\begin{pmatrix}k_{1}\\k_{2}\\l_{1}\\l_{2}\end{pmatrix}=0\)，
\(\begin{pmatrix}1&-1&-1&0\\0&1&-2&-1\\-1&0&1&-1\\0&1&-2&-1\end{pmatrix}\sim\begin{pmatrix}1&0&0&2\\0&1&0&1\\0&0&1&1\\0&0&0&0\end{pmatrix}\)。
令\(\begin{pmatrix}k_{1}\\k_{2}\\l_{1}\\l_{2}\end{pmatrix}=\begin{pmatrix}-2k\\-k\\-k\\k\end{pmatrix}\)，故\(\alpha=-2k\xi_{1}-k\xi_{2}=-k\eta_{1}+k\eta_{2}=k\begin{pmatrix}-1\\-1\\2\\-1\end{pmatrix},k\in\mathbb{R}\)。

解 由题意\( f(1) = 0 \)，\( f'(1) = 1 \)。令\( u = 1 + e^{x^2} - e^t \)，则\( du = -e^t dt \)，那么

原式\( = \lim\limits_{x \to 0} \frac{\int_{1}^{e^{x^2}} f(u) du}{x^2 \cdot \ln(1 + \cos x - 1)} = \lim\limits_{x \to 0} \frac{\int_{1}^{e^{x^2}} f(u) du}{x^2 (\cos x - 1)} = \lim\limits_{x \to 0} \frac{\int_{1}^{e^{x^2}} f(u) du}{-\frac{1}{2} x^4} \)

\( = \lim\limits_{x \to 0} \frac{f(e^{x^2}) \cdot e^{x^2} \cdot 2x}{-2x^3} = -\lim\limits_{x \to 0} \frac{f(e^{x^2})}{x^2} = -\lim\limits_{x \to 0} \frac{f(e^{x^2}) - f(1)}{x^2} \)

\( = -\lim\limits_{x \to 0} \frac{f(e^{x^2}) - f(1)}{e^{x^2} - 1} \cdot \frac{e^{x^2} - 1}{x^2} = -f'(1) = -1 \)

解 原式\(=\frac{1}{2}\int \ln(x + \sqrt{1 + x^2})d\left(\frac{1}{1 - x^2}\right) = \frac{1}{2}\frac{\ln(x + \sqrt{1 + x^2})}{1 - x^2} + \frac{1}{2}\int \frac{1}{(x^2 - 1)\sqrt{1 + x^2}}dx\)，

而\(\int \frac{1}{(x^2 - 1)\sqrt{1 + x^2}}dx \stackrel{x = \tan t}{=} \int \frac{\cos t}{\sin^2 t - \cos^2 t}dt = \frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}\sin t - 1}{\sqrt{2}\sin t + 1}\right| + C = \frac{1}{2\sqrt{2}}\ln\left|\frac{\sqrt{2}x - \sqrt{1 + x^2}}{\sqrt{2}x + \sqrt{1 + x^2}}\right| + C\)，

因而原式\(=\frac{1}{2}\frac{\ln(x + \sqrt{1 + x^2})}{1 - x^2} + \frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}x - \sqrt{1 + x^2}}{\sqrt{2}x + \sqrt{1 + x^2}}\right| + C\)。

解 \( \frac{\partial f}{\partial x} = 2x(1 - x^2 - y^2)e^{-(x^2 + 2y^2)} \)，\( \frac{\partial f}{\partial y} = 2y(1 - 2x^2 - 2y^2)e^{-(x^2 + 2y^2)} \)，

令\( \frac{\partial f}{\partial x} = 0 \)，\( \frac{\partial f}{\partial y} = 0 \)，即\( \begin{cases} 2x(1 - x^2 - y^2) = 0, \\ 2y(1 - 2x^2 - 2y^2) = 0, \end{cases} \)

解得驻点\( (0,0) \)，\( (-1,0) \)，\( (1,0) \)，\( (0,\frac{1}{\sqrt{2}}) \)，\( (0,-\frac{1}{\sqrt{2}}) \).

\( \frac{\partial^2 f}{\partial x^2} = 2(2x^4 + 2x^2y^2 - 5x^2 - y^2 + 1)e^{-(x^2 + 2y^2)} \)，\( \frac{\partial^2 f}{\partial x\partial y} = 4xy(2x^2 + 2y^2 - 3)e^{-(x^2 + 2y^2)} \)，

\( \frac{\partial^2 f}{\partial y^2} = 2(8y^4 + 8x^2y^2 - 2x^2 - 10y^2 + 1)e^{-(x^2 + 2y^2)} \)，

因为\( f(x,y) \)关于\( yOz \)，\( xOz \)坐标面对称，只要代入驻点\( (0,0) \)，\( (1,0) \)，\( (0,\frac{1}{\sqrt{2}}) \)即可.

①\( (0,0) \)处，\( A = 2 > 0 \)，\( B = 0 \)，\( C = 2 \)，\( \Delta = B^2 - AC = -4 < 0 \)，故\( f(0,0) = 0 \)为极小值；

② 点\( (1,0) \)处，\( A = -4e^{-1} < 0 \)，\( B = 0 \)，\( C = -2e^{-1} \)，\( \Delta = B^2 - AC = -8e^{-2} < 0 \)，故\( f(1,0) = e^{-1} \)为极大值，由对称性可知\( f(-1,0) = e^{-1} \)也为极大值；

③ 点\( (0,\frac{1}{\sqrt{2}}) \)处，\( A = e^{-1} \)，\( B = 0 \)，\( C = -4e^{-1} \)，\( \Delta = B^2 - AC = 4e^{-2} > 0 \)，故\( (0,\frac{1}{\sqrt{2}}) \)不是极值点，

由对称性可知\( (0,-\frac{1}{\sqrt{2}}) \)也不是极值点.

解 引入曲线\(x = \frac{1}{\sqrt{1 + y^{2}}}\)，将区域\(D\)分割成\(D_1\)，\(D_2\)，
\(D_1 = \left\{(x,y)\mid 0 \leqslant y \leqslant 1, 0 \leqslant x \leqslant \frac{1}{\sqrt{1 + y^{2}}}\right\}\)，
\(D_2 = \left\{(x,y)\mid \frac{1}{\sqrt{2}} \leqslant x \leqslant 1, \frac{\sqrt{1 - x^{2}}}{x} \leqslant y \leqslant 1\right\}\)，
则\(\iint_{D} f(x,y)d\sigma = \iint_{D_1} x\arctan^{2}y d\sigma + \iint_{D_2} \frac{x^{2}y}{\sqrt{2}x + 1} d\sigma\)。

\(\iint_{D_1} x\arctan^{2}y d\sigma = \int_{0}^{1} dy \int_{0}^{\frac{1}{\sqrt{1 + y^{2}}}} x\arctan^{2}y dx = \frac{1}{2} \int_{0}^{1} \frac{1}{1 + y^{2}} \arctan^{2}y dy\)
\(= \frac{1}{2} \int_{0}^{1} \arctan^{2}y d\arctan y = \frac{1}{6} \arctan^{3}y \big|_{0}^{1} = \frac{\pi^{3}}{384}\)。

\(\iint_{D_2} \frac{x^{2}y}{\sqrt{2}x + 1} d\sigma = \int_{\frac{1}{\sqrt{2}}}^{1} dx \int_{\frac{\sqrt{1 - x^{2}}}{x}}^{1} \frac{x^{2}y}{\sqrt{2}x + 1} dy = \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{x^{2}}{\sqrt{2}x + 1} \left(\frac{2x^{2} - 1}{x^{2}}\right) dx\)
\(= \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} (\sqrt{2}x - 1) dx = \frac{3\sqrt{2}}{8} - \frac{1}{2}\)。

所以\(\iint_{D} f(x,y)d\sigma = \frac{\pi^{3}}{384} + \frac{3\sqrt{2}}{8} - \frac{1}{2}\)。

证 (Ⅰ) 假设$f(x) \leq 0$，$x \in [a,b]$，又$f(x)$在$[a,b]$上连续，且不恒等于零。则$\int_{a}^{b}f(x)dx \lt 0$，与条件矛盾，故假设不成立，即存在$c \in (a,b)$使得$f(c) \gt 0$，又$f(a) \lt 0$，$f(b) \lt 0$，由零点定理知存在$\xi_1 \in (a,c)$，$\xi_2 \in (c,b)$，使得$f(\xi_1) = 0$，$f(\xi_2) = 0$。

(Ⅱ) 令$F(x) = f(x) \cdot e^{\int_{a}^{x - 1}(t)dt}$。由(Ⅰ)知$F(\xi_1) = F(\xi_2) = 0$，由罗尔定理知，存在$\eta \in (a,b)$使得$F'(\eta) = 0$，即$f'(\eta) + [f(\eta)]^k = 0$。

解 因为\(\vert A\vert = 0\)，所以\(\lambda = 0\)是\(A\)的特征值。若\(\lambda = 0\)是二重特征值，则\(\lambda_{2} = \lambda_{3} = 0\)，\(\lambda_{1} = \mathrm{tr}(A)-\lambda_{2}-\lambda_{3} = 4\)。若\(\lambda = 0\)是单特征值，则\(\lambda_{1} = 0\)，\(\lambda_{2} = \lambda_{3} = \frac{\mathrm{tr}(A)-\lambda_{1}}{2} = 2\)。
设\(\boldsymbol{\alpha}_{1} = \boldsymbol{\alpha} = (1, -1, 1)^{\mathrm{T}}\)为属于单特征值\(\lambda_{1}\)的特征向量，设\(\boldsymbol{\alpha}_{2} = (x_{1}, x_{2}, x_{3})^{\mathrm{T}}\)是属于\(\lambda_{2} = \lambda_{3}\)的特征向量，因为\(\boldsymbol{\alpha}_{1}\)，\(\boldsymbol{\alpha}_{2}\)正交，所以\(x_{1}-x_{2} + x_{3} = 0\)，取\(\boldsymbol{\alpha}_{2} = (1, 1, 0)^{\mathrm{T}}\)，设\(\boldsymbol{\alpha}_{3} = (x_{1}, x_{2}, x_{3})^{\mathrm{T}}\)是属于\(\lambda_{2} = \lambda_{3}\)的另一个特征向量，令\(\boldsymbol{\alpha}_{3}\)与\(\boldsymbol{\alpha}_{1}\)，\(\boldsymbol{\alpha}_{2}\)均正交，即\(\begin{cases}x_{1}-x_{2} + x_{3} = 0 \\ x_{1} + x_{2} = 0\end{cases}\)
取\(\boldsymbol{\alpha}_{3} = (-1, 1, 2)^{\mathrm{T}}\)，因\(\boldsymbol{\alpha}_{1}\)，\(\boldsymbol{\alpha}_{2}\)，\(\boldsymbol{\alpha}_{3}\)已两两正交，将\(\boldsymbol{\alpha}_{1}\)，\(\boldsymbol{\alpha}_{2}\)，\(\boldsymbol{\alpha}_{3}\)单位化，得
\(\boldsymbol{\xi}_{1} = \left(\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)^{\mathrm{T}}\)，\(\boldsymbol{\xi}_{2} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0\right)^{\mathrm{T}}\)，\(\boldsymbol{\xi}_{3} = \left(\frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}\right)^{\mathrm{T}}\)。
（Ⅰ）\(A\)的单特征值不等于 0，即\(\lambda_{1} = 4\)，\(\lambda_{2} = \lambda_{3} = 0\)，
令\(P = (\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \boldsymbol{\xi}_{3}) = \begin{pmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}}\end{pmatrix}\)，\(\boldsymbol{\Lambda} = \begin{pmatrix}4 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\)，则
\(A = P\boldsymbol{\Lambda}P^{-1} = P\boldsymbol{\Lambda}P^{\mathrm{T}} = \frac{4}{3}\begin{pmatrix}1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1\end{pmatrix}\)。
\(A = \frac{4}{3}\begin{pmatrix}1 & -1 & 1 \\ -1 & 1 & -1 \\ 1 & -1 & 1\end{pmatrix} \sim \begin{pmatrix}1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\)，\(Ax = 0\)与\(x_{1}-x_{2} + x_{3} = 0\)同解，
基础解系为\(\boldsymbol{\eta}_{1} = (1, 1, 0)^{\mathrm{T}}\)，\(\boldsymbol{\eta}_{2} = (-1, 0, 1)^{\mathrm{T}}\)，故通解为\(x = k_{1}\boldsymbol{\eta}_{1} + k_{2}\boldsymbol{\eta}_{2}\)，\(k_{1}, k_{2} \in \mathbb{R}\)。
【注】基础解系也可以取\(\boldsymbol{\alpha}_{2}\)，\(\boldsymbol{\alpha}_{3}\)或\(\boldsymbol{\xi}_{2}\)，\(\boldsymbol{\xi}_{3}\)。
（Ⅱ）\(A\)的单特征值等于 0，即\(\lambda_{1} = 0\)，\(\lambda_{2} = \lambda_{3} = 2\)，由上可知，令\(P = (\boldsymbol{\xi}_{1}, \boldsymbol{\xi}_{2}, \boldsymbol{\xi}_{3})\)，有\(A = P\boldsymbol{\Lambda}P^{\mathrm{T}}\)，所以
\(A + 2E = P(\boldsymbol{\Lambda} + 2E)P^{\mathrm{T}} = P\begin{pmatrix}2 & & \\ & 4 & \\ & & 4\end{pmatrix}P^{\mathrm{T}}\)
\(= P\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}P^{\mathrm{T}} = P\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}P^{\mathrm{T}}P\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}P^{\mathrm{T}}\)。
令\(B = P\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}P^{\mathrm{T}}\)，因为\(B\)的特征值为\(\sqrt{2}\)，\(2\)，\(2\)，且\(B\)为对称阵，故\(B\)正定，且
\(B = P\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}P^{\mathrm{T}} = \begin{pmatrix}\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{6}} \\ \frac{-1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{6}}\end{pmatrix}\begin{pmatrix}\sqrt{2} & & \\ & 2 & \\ & & 2\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{-1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{2}{\sqrt{6}}\end{pmatrix} = \begin{pmatrix}\frac{4 + \sqrt{2}}{3} & \frac{2-\sqrt{2}}{3} & \frac{\sqrt{2}-2}{3} \\ \frac{2-\sqrt{2}}{3} & \frac{4 + \sqrt{2}}{3} & \frac{2-\sqrt{2}}{3} \\ \frac{\sqrt{2}-2}{3} & \frac{2-\sqrt{2}}{3} & \frac{4 + \sqrt{2}}{3}\end{pmatrix}\)，
有\(A = B^{2}-2E\)。