答案 选(C)。

解 \( \lim_{x \to 0} \frac{x}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = \ln \frac{\pi}{2} \)，因此 \( x = 0 \) 为可去间断点。

\( \lim_{x \to \frac{\pi}{2}} \frac{x}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \frac{\frac{1}{x - \frac{\pi}{2}}}{\sec^2 x} = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \frac{\cos^2 x}{x - \frac{\pi}{2}} \)
\( = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \frac{-2\sin x \cos x}{1} = 0 \)。

或者 \( \lim_{x \to \frac{\pi}{2}} \frac{x}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \frac{1}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = \frac{\pi}{2} \lim_{x \to \frac{\pi}{2}} \cos x \ln\left| x - \frac{\pi}{2} \right| \)
令 \( \frac{\pi}{2} - x = t \)
\( = \frac{\pi}{2} \lim_{t \to 0} \sin t \ln |t| = 0 \)，

因此 \( x = \frac{\pi}{2} \) 为可去间断点。

\( \lim_{x \to -\frac{\pi}{2}} \frac{x}{\tan x} \ln\left| x - \frac{\pi}{2} \right| = 0 \)，因此 \( x = -\frac{\pi}{2} \) 为可去间断点。

答案 选(C).

解 \(F(x)=\arctan x\cdot\int_{0}^{x}f(t)dt - \int_{0}^{x}\left(\arctan t + \frac{t}{1 + t^{2}}\right)f(t)dt\)，
$F'(x)=\frac {1}{1+x^{2}}\int_{0}^{x}f(t)dt + f(x)\arctan x - \left(f(x)\arctan x + \frac {x}{1+x^{2}}f(x)\right)$
\(=\frac{1}{1 + x^{2}}\left(\int_{0}^{x}f(t)dt - xf(x)\right)\)

① 若\(f(x)\)为奇函数，则\(F^{\prime}(x)\)为偶函数，又\(F(0)=0\)，故\(F(x)\)为奇函数
② 若\(f(x)\)为偶函数，则\(F^{\prime}(x)\)为奇函数，故\(F(x)\)为偶函数
③ 若\(f(x)\)单增，则\(F^{\prime}(x)=\frac{1}{1 + x^{2}}\left(\int_{0}^{x}f(t)dt - xf(x)\right)\leqslant0\)，故\(F(x)\)单减

答案 选(A)。

解法一 令\( xt = u \)，\( f(x) = \begin{cases} \frac{1}{x} \int_{0}^{x} \sin u^2 du, & x \neq 0, \\ 0, & x = 0. \end{cases} \)

\( \int_{0}^{1} x f(x) dx = \int_{0}^{1} \left[ \int_{0}^{x} \sin u^2 du \right] dx = \left. \left( x \int_{0}^{x} \sin u^2 du \right) \right|_{0}^{1} - \int_{0}^{1} x \sin x^2 dx \)

\( = \int_{0}^{1} \sin u^2 du - \int_{0}^{1} x \sin x^2 dx = \int_{0}^{1} \sin x^2 dx - \int_{0}^{1} x \sin x^2 dx = \int_{0}^{1} (1 - x) \sin x^2 dx \)。

解法二 \( \int_{0}^{1} x f(x) dx = \int_{0}^{1} dx \int_{0}^{1} x \sin(x^2 t^2) dt \)

\( = \frac{1}{2} \int_{0}^{1} \frac{1 - \cos t^2}{t^2} dt = -\frac{1}{2} \int_{0}^{1} (1 - \cos t^2) d \frac{1}{t} = -\frac{1}{2}(1 - \cos 1) + \int_{0}^{1} \sin t^2 dt \)，

\( \int_{0}^{1} (1 - x) \sin x^2 dx = \int_{0}^{1} \sin x^2 dx - \int_{0}^{1} x \sin x^2 dx = \int_{0}^{1} \sin x^2 dx - \frac{1}{2}(1 - \cos 1) \)，

故选(A)。

答案 选(A).

解 由题意知 \( P(x) = \int_{0}^{1} |x - t| dt = \begin{cases} \frac{1}{2} - x, & x \leq 0, \\ x^2 - x + \frac{1}{2}, & 0 < x < 1, \\ x - \frac{1}{2}, & x \geq 1. \end{cases} \) 易得 \( P'(0) = -1 \)，

由 \( y' - P(x)y = \frac{1}{2}e^x - 1 \)，\( y(0) = 1 \)，进而可知 \( f'(0) = 0 \).

上式两边同时对 \( x \) 求导，得 \( f''(x) - P'(x)f(x) - P(x)f'(x) = \frac{1}{2}e^x \)，得 \( f''(0) = -\frac{1}{2} \). 于是 \( f(x) \) 在点 \( x = 0 \) 处取极大值，故选(A).

答案 选(D)。

解 建立复合函数：\( g(x,y) = u(s,t), s = x + y, t = x - y \)。

因为\( \frac{\partial g}{\partial y} = u_1'(s,t) - u_2'(s,t) = 2s\cos(s + t) = 2(x + y)\cos2x \)。

故\( g(x,y) = \int \frac{\partial g}{\partial y} dy = \int 2(x + y)\cos2x dy = (2xy + y^2)\cos2x + \phi(x) \)，又\( g(x,0) = 0 \)，从而\( g(x,y) = (2xy + y^2)\cos2x \)，即
\( u(x + y, x - y) = (2xy + y^2)\cos2x \).①

令\( s = x + y, t = x - y \)，则\( x = \frac{s + t}{2}, y = \frac{s - t}{2} \)，得
\( u(s,t) = \left[ 2 \cdot \frac{s + t}{2} \cdot \frac{s - t}{2} + \left( \frac{s - t}{2} \right)^2 \right] \cos\left( 2 \cdot \frac{s + t}{2} \right) = \frac{3s^2 - t^2 - 2st}{4} \cos(s + t) \)，
从而\( u(1,0) = \frac{3}{4}\cos1 \)。

【注】令\( x = y = \frac{1}{2} \)，代入①式，可得\( u(1,0) = \frac{3}{4}\cos1 \)。

答案 选(D).

解 由对称性可知 \( \iint_{D_{k}} xy^{2}e^{-x^{2}} d\sigma = 0 \)，故 \( I_{k} = \iint_{D_{k}} x^{2}\sin(x^{2} + y^{2}) d\sigma \)，
对任意 \( (x,y) \in D_{1} \subset D_{2} \)，且 \( x^{2}\sin(x^{2} + y^{2}) \geq 0 \)，\( (x,y) \in D_{2} \)，所以 \( 0 < I_{1} \leq I_{2} \).
由轮换对称性可知
\[
\begin{align*}
I_{3} &= \frac{1}{2} \iint_{D_{3}} (x^{2} + y^{2})\sin(x^{2} + y^{2}) d\sigma = \frac{1}{2} \int_{0}^{2\pi} d\theta \int_{0}^{\sqrt{2\pi}} r^{2}\sin r^{2} \cdot r dr \\
&= \frac{1}{4} \cdot 2\pi \int_{0}^{2\pi} t\sin t dt = -\pi^{2} < 0,
\end{align*}
\]
综上所述，\( I_{3} \leq I_{1} \leq I_{2} \)，故选(D).

答案 选（D）。

解 \( \lim\limits_{x \to 0} \frac{F(x)}{G(x)} = \lim\limits_{x \to 0} \frac{\int_{0}^{1} x^{2}f(xt)dt}{(1 + x)^{x} - 1} = \lim\limits_{x \to 0} \frac{x \int_{0}^{x} f(u)du}{e^{x \ln(1 + x)} - 1} = \lim\limits_{x \to 0} \frac{x \int_{0}^{x} f(u)du}{x^{2}} = \lim\limits_{x \to 0} \frac{\int_{0}^{x} f(u)du}{x} = \lim\limits_{x \to 0} \frac{f(x)}{1} = 1 \)，所以选（D）。

答案 选(D).

解 \( A\boxed{x} = \boxed{0} \)的基础解系为\( \boxed{\xi}_1 = (2, 1, 0)^T \)，\( \boxed{\xi}_2 = (1, 0, 1)^T \)，\( \boxed{\xi}_1, \boxed{\xi}_2 \)是\( A \)的对应于特征值\( \lambda = 0 \)的两线性无关的特征向量. 而\( \boxed{\eta} = (-1, 2, 1)^T \)是\( A\boxed{x} = \boxed{b} \)的解，即
\[
A\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ -3 \end{pmatrix} = -3\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix},
\]
故\( A \)的三个特征值分别为\( \lambda_1 = \lambda_2 = 0 \)，\( \lambda_3 = -3 \)，\( A + E \)的三个特征值分别为\( 1, 1, -2 \)，正负惯性指数分别为\( 2, 1 \)，故选(D).

答案 选(C).

解 由已知得\( A \neq O \)，\( \text{r}(A) \geq 1 \)，\( A^2 = AA = O \)，\( \text{r}(A) + \text{r}(A) \leq 3 \)，即有\( 1 \leq \text{r}(A) \leq \frac{3}{2} \)，可得\( \text{r}(A) = 1 \). 而非齐次线性方程组\( A^T A x = A^T b \)必有解，\( \text{r}(A^T A) = \text{r}(A^T A, A^T b) = \text{r}(A) = 1 \)，其线性无关的解向量的个数为\( 3 - \text{r}(A^T A) + 1 = 3 \). 故选(C).

答案 选(B).

解一 若存在不全为零的数\(k_1,k_2,k_3\)使
\[k_1(x_1\beta + \alpha_1) + k_2(x_2\beta + \alpha_2) + k_3(x_3\beta + \alpha_3) = 0,\]
此时\(x_1\beta + \alpha_1, x_2\beta + \alpha_2, x_3\beta + \alpha_3\)线性相关，而
\[k_1(x_1\beta + \alpha_1) + k_2(x_2\beta + \alpha_2) + k_3(x_3\beta + \alpha_3) = (k_1x_1 + k_2x_2 + k_3x_3)\beta + k_1\alpha_1 + k_2\alpha_2 + k_3\alpha_3,\]
由已知\(\alpha_1 - \alpha_2 + 2\alpha_3 = 0\)，可取\(k_1 = 1, k_2 = -1, k_3 = 2\)，只要满足\(x_1 - x_2 + 2x_3 = 0\)，此时\(k_1(x_1\beta + \alpha_1) + k_2(x_2\beta + \alpha_2) + k_3(x_3\beta + \alpha_3) = 0\)， 故选(B).

解二 (B)选项中，\((3\beta + \alpha_1, 5\beta + \alpha_2, \beta + \alpha_3) \stackrel{c_1 - c_2 + 2c_3}{\longrightarrow} (0, \alpha_2, \alpha_3)\)，故\(3\beta + \alpha_1, 5\beta + \alpha_2, \beta + \alpha_3\)必线性相关，选(B).

答案 填“$e$”.

解 原式$= \lim\limits_{x \to 0}\exp\{\frac{\ln(x + \sqrt{1 + x^2})}{e^x - 1}\} = \exp\{\lim\limits_{x \to 0}\frac{\ln(x + \sqrt{1 + x^2})}{x}\} = \exp\{\lim\limits_{x \to 0}\frac{\frac{1}{\sqrt{1 + x^2}}}{1}\} = e.$

答案 填“\( 20 - 6\pi \)”。

解 由 \( \begin{cases} x^2 + y^2 + z^2 = 2, \\ z^2 = 2x + 2y \end{cases} \Rightarrow \begin{cases} (x + 1)^2 + (y + 1)^2 = 4, \\ z^2 = 2x + 2y \end{cases} \)，又 \( \begin{cases} -1 \leq x \leq 1, \\ y \geq -1, \\ z \geq 0, \end{cases} \) 可设参数表达式
\( \begin{cases} x = -1 + 2\cos t, \\ y = -1 + 2\sin t, \\ z^2 = -4 + 4\cos t + 4\sin t, \end{cases} \) 其中 \( t \in [0, \frac{\pi}{2}] \)。

\( \int_{-1}^{1} y(x)z^2(x) \, dx = \int_{0}^{\frac{\pi}{2}} (2\sin t - 1)(4\cos t + 4\sin t - 4) \cdot 2\sin t \, dt \)
\( = 8 \int_{0}^{\frac{\pi}{2}} (\sin t - \sin t \cos t - 3\sin^2 t + 2\sin^2 t \cos t + 2\sin^3 t) \, dt = 20 - 6\pi \)。

答案 填“480”。

解 由莱布尼茨公式，得
\( (x^2 e^{2x})^{(n)} = (e^{2x})^{(n)} \cdot x^2 + n(e^{2x})^{(n - 1)} \cdot 2x + \frac{n(n - 1)}{2} \cdot (e^{2x})^{(n - 2)} \cdot 2 = 2^{n - 2}[4x^2 + 4nx + n(n - 1)]e^{2x} \)，
\( \left( \frac{2x}{1 - x^2} \right)^{(n)} = \left( \frac{1}{1 - x} \right)^{(n)} - \left( \frac{1}{1 + x} \right)^{(n)} = \frac{n!}{(1 - x)^{n + 1}} + \frac{(-1)^{n - 1}n!}{(1 + x)^{n + 1}} \)，
或者因为\( \frac{2x}{1 - x^2} \)是奇函数，所以\( \left. \left( \frac{2x}{1 - x^2} \right)^{(6)} \right|_{x = 0} = 0 \)，故\( f^{(6)}(0) = 480 \)。

答案 填“\( (x - 1)y + 2y\text{e}^{-x} \)”。

解 \( f(x,y) = \text{e}^{-\int 1\text{d}x} \left[ \int xy\text{e}^{\int 1\text{d}x} \text{d}x + \phi(y) \right] = \text{e}^{-x} \left[ \int xy\text{e}^x \text{d}x + \phi(y) \right] = \text{e}^{-x} \left[ y(x\text{e}^x - \text{e}^x) + \phi(y) \right] \)
\( = (x - 1)y + \phi(y)\text{e}^{-x} \)。

由 \( f(0,y) = y \) 代入得 \( \phi(y) = 2y \)，故 \( f(x,y) = (x - 1)y + 2y\text{e}^{-x} \)。

答案 填“\(-x\ln x\)”。

解 \( \frac{\partial z}{\partial x} = 2x f\left(\frac{y}{x}\right) - y f'\left(\frac{y}{x}\right) \)，\( \frac{\partial^2 z}{\partial x^2} = 2 f\left(\frac{y}{x}\right) - 2 \frac{y}{x} f'\left(\frac{y}{x}\right) + \frac{y^2}{x^2} f''\left(\frac{y}{x}\right) \)，

\( \frac{\partial z}{\partial y} = x f'\left(\frac{y}{x}\right) \)，\( \frac{\partial^2 z}{\partial y^2} = f''\left(\frac{y}{x}\right) \)，

因为 \( \frac{\partial^2 z}{\partial x^2} - \frac{y^2}{x^2} \frac{\partial^2 z}{\partial y^2} = 2 \frac{y}{x} \)，所以

\( 2 f\left(\frac{y}{x}\right) - 2 \frac{y}{x} f'\left(\frac{y}{x}\right) = 2 \frac{y}{x} \)，即 \( f(t) - t f'(t) = t \)，

解得 \( f(t) = \mathrm{e}^{\int \frac{1}{t} \mathrm{d}t} \left[ \int -1 \cdot \mathrm{e}^{-\int \frac{1}{t} \mathrm{d}t} \mathrm{d}t + C \right] = t(C - \ln t) \)，

由 \( f(1) = 0 \)，解得 \( C = 0 \)，故 \( f(x) = -x \ln x \)。

答案 填“3”

解 由题意知\( A \xrightarrow[ r_1 \leftrightarrow r_2 ]{4r_2, -2r_3} C \)，故\( E(1,2) \cdot E(3(-2)) \cdot E(2(4)) \cdot A = C \)，则
\( C = \begin{pmatrix} 0 & 4 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -2 \end{pmatrix} A = PA \)，其中\( P = \begin{pmatrix} 0 & 4 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & -2 \end{pmatrix} \)，由\( AB = C \)，可得\( AB = PA \)，A可逆，故\( B = A^{-1}PA \)，B与P相似，由\( |P - \lambda E| = -(\lambda - 2)(\lambda + 2)^2 = 0 \)，可得B的三个特征值分别为2，-2，-2，\( B + E \)的三个特征值分别为3，-1，-1，
从而\( |B + E| = 3 \)。

解 由题意，\( \lim_{x \to +\infty} \frac{f(x)}{x} = \lim_{x \to +\infty} \frac{x^{a+x-1}}{(b+x)^x} = \lim_{x \to +\infty} \frac{x^{a-1}}{(1 + \frac{b}{x})^x} = \frac{1}{e^2} \)。又\( \lim_{x \to +\infty} (1 + \frac{b}{x})^x = e^b \)，故\( a = 1, b = 2 \)。

\( \lim_{x \to +\infty} \left( \frac{x^{x+1}}{(2+x)^x} - \frac{x}{e^2} \right) = \lim_{x \to +\infty} x \left( \frac{1}{(1 + \frac{2}{x})^x} - \frac{1}{e^2} \right) = \lim_{x \to +\infty} x \cdot \frac{e^2 - (1 + \frac{2}{x})^x}{e^4} \)

\( = \frac{1}{e^4} \lim_{x \to +\infty} x \left( e^2 - e^{x \ln(1 + \frac{2}{x})} \right) = \frac{1}{e^2} \lim_{x \to +\infty} x \left[ 2 - x \ln(1 + \frac{2}{x}) \right] \stackrel{t = \frac{1}{x}}{=} \frac{1}{e^2} \lim_{t \to 0^+} \frac{2t - \ln(1 + 2t)}{t^2} \)

\( = \frac{1}{e^2} \lim_{t \to 0^+} \frac{\frac{1}{2}(2t)^2}{t^2} = \frac{2}{e^2} \)，故\( c = \frac{2}{e^2} \)。

解 (Ⅰ)① 当\( x > 0 \)时，
\( f(x) = \lim\limits_{n \to \infty} \frac{6x}{1 + x^2} \cdot \frac{\text{e}^{nx} - \sin(\text{e}^{nx})}{\text{e}^{3nx}} = \frac{6x}{1 + x^2} \lim\limits_{n \to \infty} \frac{1 - \text{e}^{-nx} \sin(\text{e}^{nx})}{\text{e}^{2nx}} = 0 \)。

② 当\( x = 0 \)时，\( f(0) = 0 \)。

③ 当\( x < 0 \)时，
\( f(x) = \frac{6x}{1 + x^2} \lim\limits_{n \to \infty} \frac{\text{e}^{nx} - \sin(\text{e}^{nx})}{\text{e}^{3nx}} = \frac{6x}{1 + x^2} \lim\limits_{n \to \infty} \frac{\frac{1}{6}\text{e}^{3nx}}{\text{e}^{3nx}} = \frac{x}{1 + x^2} \)。

故\( f(x) = \begin{cases} 0, & x \geq 0, \\ \frac{x}{1 + x^2}, & x < 0. \end{cases} \)

(Ⅱ)\( V_y = \int_{0}^{1} 2\pi x \cdot \frac{x}{2} \, dx + \int_{-1}^{0} 2\pi |x| \cdot \frac{|x|}{1 + x^2} \, dx - \int_{-1}^{0} 2\pi |x| \cdot \frac{|x|}{2} \, dx \)
\( = \frac{\pi}{3} + 2\pi \int_{-1}^{0} \frac{x^2}{1 + x^2} \, dx - \pi \int_{-1}^{0} x^2 \, dx = 2\pi - 2\pi \arctan x \big|_{-1}^{0} = 2\pi - \frac{\pi^2}{2} \)。

解 令\( F(x,y,z) = x^3 + 2y^2 + z^2 - 8yz - 3x + 23 \)，

\( \frac{\partial z}{\partial x} = \frac{3x^2 - 3}{8y - 2z} \)，\( \frac{\partial z}{\partial y} = \frac{4y - 8z}{8y - 2z} \)，令\( \frac{\partial z}{\partial x} = 0 \)，\( \frac{\partial z}{\partial y} = 0 \)，得\( \begin{cases} x^2 = 1, \\ y = 2z, \\ x^3 + 2y^2 + z^2 - 8yz - 3x + 23 = 0. \end{cases} \)

解得驻点\( (1, 2\sqrt{3}) \)，\( (1, -2\sqrt{3}) \)，\( \left(-1, \frac{10\sqrt{7}}{7}\right) \)，\( \left(-1, -\frac{10\sqrt{7}}{7}\right) \)。

\( \frac{\partial^2 z}{\partial x^2} = \frac{6x(8y - 2z) - (3x^2 - 3)\frac{\partial (8y - 2z)}{\partial x}}{(8y - 2z)^2} \)，\( \frac{\partial^2 z}{\partial x \partial y} = \frac{ - (3x^2 - 3)\frac{\partial (8y - 2z)}{\partial y} }{(8y - 2z)^2} \)，

\( \frac{\partial^2 z}{\partial y^2} = \frac{ \left(4 - 8\frac{\partial z}{\partial y}\right)(8y - 2z) - (4y - 8z)\frac{\partial (8y - 2z)}{\partial y} }{(8y - 2z)^2} \)。

① 点\( (1, 2\sqrt{3}) \)处，\( z = \sqrt{3} \)，\( A = \frac{\sqrt{3}}{7} > 0 \)，\( B = 0 \)，\( C = \frac{2\sqrt{3}}{21} \)，\( B^2 - AC = -\frac{6}{147} < 0 \)，\( z(1, 2\sqrt{3}) = \sqrt{3} \)为极小值。

② 点\( (1, -2\sqrt{3}) \)处，\( z = -\sqrt{3} \)，\( A = -\frac{\sqrt{3}}{7} \)，\( B = 0 \)，\( C = -\frac{2\sqrt{3}}{21} \)，\( B^2 - AC = -\frac{6}{147} < 0 \)，\( z(1, -2\sqrt{3}) = -\sqrt{3} \)为极大值。

③ 点\( \left(-1, \frac{10\sqrt{7}}{7}\right) \)处，\( z = \frac{5\sqrt{7}}{7} \)，\( A = -\frac{3}{5\sqrt{7}} \)，\( B = 0 \)，\( C = \frac{2}{5\sqrt{7}} \)，\( B^2 - AC > 0 \)，所以\( \left(-1, \frac{10\sqrt{7}}{7}\right) \)不是极值点。

④ 点\( \left(-1, -\frac{10\sqrt{7}}{7}\right) \)处，\( z = -\frac{5\sqrt{7}}{7} \)，\( A = \frac{3}{5\sqrt{7}} \)，\( B = 0 \)，\( C = -\frac{2}{5\sqrt{7}} \)，\( B^2 - AC > 0 \)，所以\( \left(-1, -\frac{10\sqrt{7}}{7}\right) \)也不是极值点。

解 由轮换对称性，
\[
\iint\limits_{D} \frac{(1 + x^2)(x^2 + y^2 - xy)}{2 + x^2 + y^2} d\sigma = \iint\limits_{D} \frac{(1 + y^2)(x^2 + y^2 - xy)}{2 + x^2 + y^2} d\sigma
\]
所以
\[
I = \frac{1}{2} \iint\limits_{D} (x^2 + y^2 - xy) d\sigma
\]

令 \( \begin{cases} x = 1 + r\cos\theta, \\ y = 1 + r\sin\theta, \end{cases} \) 则
\[
\begin{align*}
I &= \frac{1}{2} \iint\limits_{D} [(x - y)^2 + xy] d\sigma \\
&= \frac{1}{2} \int_{0}^{2\pi} d\theta \int_{0}^{1} \left[ r^2 - r^2 \sin\theta\cos\theta + 1 + r\cos\theta + r\sin\theta \right] \cdot r dr \\
&= \frac{1}{2} \int_{0}^{2\pi} \left( \frac{1}{4} - \frac{1}{4} \sin\theta\cos\theta + \frac{1}{2} + \frac{1}{3} \cos\theta + \frac{1}{3} \sin\theta \right) d\theta \\
&= \frac{1}{2} \left( \frac{1}{4} \cdot 2\pi + \frac{1}{2} \cdot 2\pi \right) = \frac{3}{4}\pi
\end{align*}
\]

证 令\( F(u) = \int_{0}^{u}f(x)dx \)，则由题设知\( F(u) \)有三阶连续导数，将\( F(a) \)，\( f(-a) \)分别在\( x = 0 \)处展开得

\( \int_{0}^{a}f(x)dx = F(0) + F'(0)a + \frac{F''(0)}{2}a^{2} + \frac{F'''(\xi_{1})}{3!}a^{3} = f(0)a + \frac{f'(0)}{2}a^{2} + \frac{f''(\xi_{1})}{6}a^{3} \)，

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

其中\( \xi_{1} \in (0,a) \)，\( \xi_{2} \in (-a,0) \)，由此知

\( 2\int_{0}^{a}f(x)dx = 2f(0)a + f'(0)a^{2} + \frac{f''(\xi_{1})}{3}a^{3} \)，

\( af(-a) = f(0)a - f'(0)a^{2} + \frac{f''(\xi_{2})}{2}a^{3} \)，

两式相加并整理得

\( \int_{0}^{a}f(x)dx = \frac{a}{2}[3f(0) - f(-a)] + \frac{a^{3}}{12}[2f''(\xi_{1}) + 3f''(\xi_{2})] \)。

由于\( f''(x) \)在\( [-a,a] \)上连续，故存在\( m,M \)，使\( m \leq f''(x) \leq M \)，从而\( 5m \leq 2f''(\xi_{1}) + 3f''(\xi_{2}) \leq 5M \)。由闭区间上连续函数性质知，存在\( \xi \in [-a,a] \)，使\( 2f''(\xi_{1}) + 3f''(\xi_{2}) = 5f''(\xi) \)，因此存在\( \xi \in [-a,a] \)使

\( \int_{0}^{a}f(x)dx = \frac{a}{2}[3f(0) - f(-a)] + \frac{5}{12}f''(\xi)a^{3} \)。

解(Ⅰ)由题设知
\[
A(\alpha_1,\alpha_2,\alpha_3) = (2\alpha_1 + \alpha_2 + \alpha_3, \alpha_1 + 2\alpha_2 - \alpha_3, 3\alpha_3) = (\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & -1 & 3 \end{pmatrix}.
\]
令\( P = (\alpha_1,\alpha_2,\alpha_3) \)，\( B = \begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & -1 & 3 \end{pmatrix} \)，由\( \alpha_1,\alpha_2,\alpha_3 \)线性无关，可知\( P \)可逆，\( P^{-1}AP = B \)，\( A \)与\( B \)相似。由\( |A - \lambda E| = |B - \lambda E| = \begin{vmatrix} 2 - \lambda & 1 & 0 \\ 1 & 2 - \lambda & 0 \\ 1 & -1 & 3 - \lambda \end{vmatrix} = 0 \)，可得\( A \)的三个特征值为\( \lambda_1 = \lambda_2 = 3 \)，\( \lambda_3 = 1 \)。

对于矩阵\( B \)，当\( \lambda_1 = \lambda_2 = 3 \)时，求\( (B - 3E)x = 0 \)得线性无关的两个特征向量为\( \eta_1 = (1, 1, 0)^T \)，\( \eta_2 = (0, 0, 1)^T \)；当\( \lambda_3 = 1 \)时，求\( (B - E)x = 0 \)得线性无关的特征向量为\( \eta_3 = (1, -1, -1)^T \)。

令\( Q = (\eta_1,\eta_2,\eta_3) \)，则\( Q^{-1}BQ = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)，

而\( B = P^{-1}AP \)，\( Q^{-1}BQ = Q^{-1}P^{-1}APQ = (PQ)^{-1}A(PQ) = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)，令\( R = PQ = (\alpha_1,\alpha_2,\alpha_3)\begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & -1 \end{pmatrix} = (\alpha_1 + \alpha_2, \alpha_3, \alpha_1 - \alpha_2 - \alpha_3) \)， 则\( R^{-1}AR = \begin{pmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1 \end{pmatrix} \)，从而\( A \)的特征值为\( \lambda_1 = \lambda_2 = 3 \)，对应的线性无关的特征向量为\( \alpha_1 + \alpha_2, \alpha_3 \)；特征值\( \lambda_3 = 1 \)，对应的线性无关的特征向量为\( \alpha_1 - \alpha_2 - \alpha_3 \)。

(Ⅱ)由\( A \)的三个特征值为\( 3,3,1 \)，可得\( (A^2 - 4A + E)^{-1} \)的三个特征值为\( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \)。从而\( R^{-1}(A^2 - 4A + E)^{-1}R = -\frac{1}{2}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = -\frac{1}{2}E \)，故\( (A^2 - 4A + E)^{-1} = -\frac{1}{2}E \)。