【答案】A
【解析】\(z+\ln z - \int_{y}^{x}e^{-t^{2}}dt = 0\)，分别对\(x\)，\(y\)求偏导，得：
\(\frac{\partial z}{\partial x}+\frac{1}{z}\frac{\partial z}{\partial x}-e^{-x^{2}} = 0\Rightarrow\frac{\partial z}{\partial x}=\frac{z}{z + 1}e^{-x^{2}}\)
\(\frac{\partial z}{\partial y}+\frac{1}{z}\frac{\partial z}{\partial y}+e^{-y^{2}} = 0.\Rightarrow\frac{\partial z}{\partial y}=\frac{-z}{z + 1}e^{-y^{2}}\)
\(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=\frac{z}{z + 1}(e^{-x^{2}}-e^{-y^{2}})\) 

【答案】B 
【解析】 
\(f^\prime(x)=e^{x^{2}}\sin x,f^{\prime\prime}(x)=2xe^{x^{2}}\sin x + e^{x^{2}}\cos x\) 
\(f^\prime(0)=0,f^{\prime\prime}(0)=1\gt0\). 
\(x = 0\)是\(f(x)\)的极值点. 
\(g^\prime(x)=e^{x^{2}}\sin^{2}x + \sin2x\int_{0}^{x}e^{t^{2}}dt\)， 
\(g^{\prime\prime}(x)=e^{x^{2}}\sin2x + 2xe^{x^{2}}\sin^{2}x + \sin2xe^{x^{2}} + 2\cos2x\int_{0}^{x}e^{t^{2}}dt\) 
\(g^\prime(0)=0,\ g^{\prime\prime}(0)=0,\ g^{\prime\prime\prime}(0)\gt0\). 
\((0,0)\)是\(y = g(x)\)的拐点. 

【答案】C
【解析】当 \( a = -2 \) 时，\( y'' + 4y' = 0 \)，通解：\( c_1 + c_2e^{-4t} \)，\( c \neq 0 \) 时，\( \int_{0}^{+\infty} (c_1 + c_2e^{-4x})dx \) 不收敛.
故B、D排除.
当 \( a = -\frac{1}{2} \) 时，\( y'' + y' + \frac{3}{2}y = 0 \)，通解：\( y(t) = e^{-\frac{1}{2}t}(a_1\cos(\frac{\sqrt{5}}{2}t) + B(\sin\frac{\sqrt{5}}{2}t)) \)
\( \int_{0}^{+\infty} y(x)dx \) 收敛.

【答案】C
【解析】由题易知，\(x \to 0\)时，\(f(x)\)是\(g(x)\)高阶无穷小.
则有\(\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = 0\)及\(\lim\limits_{x \to 0}f(x) = 0\)，\(\lim\limits_{x \to 0}g(x) = 0\).
又\(f(x)\)，\(g(x)\)在\(x = 0\)某去心邻域内有定义且不恒等于\(0\).
故对于A选项，等式两端同除\(g(x)\)得：
\(\frac{f(x)}{g(x)} + 1 = \frac{o[g(x)]}{g(x)}\)
取极限得
\(\lim\limits_{x \to 0}(\frac{f(x)}{g(x)} + 1) = \lim\limits_{x \to 0}\frac{o[g(x)]}{g(x)}\)
即\(0 + 1 = 0\)，显然A不成立.

对于B选项，等式两端同除\(f^2(x)\)得
\(\frac{g(x)}{f(x)} = \frac{o[f^2(x)]}{f^2(x)}\)
两端取极限得\(\lim\limits_{x \to 0}\frac{g(x)}{f(x)} = \lim\limits_{x \to 0}\frac{o[f^2(x)]}{f^2(x)}\)，即\(\infty = 0\)，显然不成立.

对于C选项，等式两端同除\(g(x)\)得
\(\frac{f(x)}{g(x)} = \frac{o[\mathrm{e}^{g(x)} - 1]}{g(x)}\)
取极限得\(\lim\limits_{x \to 0}\frac{f(x)}{g(x)} = \lim\limits_{x \to 0}\frac{o[\mathrm{e}^{g(x)} - 1]}{g(x)} = \lim\limits_{x \to 0}\frac{o[g(x)]}{g(x)}\)
显然有\(0 = 0\)，故C正确.

对于D等式两端同除\(g(x)\)得
\(\frac{f(x)}{g^2(x)} = \frac{o[g^2(x)]}{g^2(x)}\)
取极限得\(\lim\limits_{x \to 0}\frac{f(x)}{g^2(x)} = \lim\limits_{x \to 0}\frac{o[g^2(x)]}{g^2(x)}\)，显然不成立.

综上选C. 

【答案】A

【解析】由题易知，此二重积分积分区域为 

$D = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, -2 \leq x \leq 2\right\}$，对应图像为上图所示。

记$D_1 = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, -2 \leq x \leq 0\right\}$，$D_2 = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, 0 \leq x \leq 2\right\}$，且 $I = \int_{-2}^{2}dx\int_{4 - x^2}^{4}f(x,y)dy$，则$I = \iint\limits_{D_1} f(x,y)d\sigma + \iint\limits_{D_2} f(x,y)d\sigma$，交换积分次序得 

$I = \int_{0}^{4}dy\int_{-2}^{-\sqrt{4 - y}} f(x,y)dx + \int_{0}^{4}dy\int_{\sqrt{4 - y}}^{2} f(x,y)dx$ 

$= \int_{0}^{4}\left[\int_{-2}^{-\sqrt{4 - y}} f(x,y)dx + \int_{\sqrt{4 - y}}^{2} f(x,y)dx\right]dy$ 

故 A 正确。

【答案】A 

【解析】由题可知，其对应如图所示. 

单位质点\(P\)与单位质点\(Q\)之间的引力为 

\(F = G\frac{1\cdot1}{r^{2}}\) 

其中\(r\)为两质点间的距离. 

且由图可知 \(r^{2}=x^{2}+1\) 

又引力\(F\)在\(x\)方向上的力投影为 \(F_{x}=F\cos\theta=\frac{x}{\sqrt{1 + x^{2}}}F\). 

故克服引力做功为： 

\(W = \int_{0}^{1}F_{x}\mathrm{d}x=\int_{0}^{1}\frac{G}{(x^{2}+1)}\frac{x}{\sqrt{1 + x^{2}}}\mathrm{d}x=\int_{0}^{1}\frac{Gx}{(1 + x^{2})^{\frac{3}{2}}}\mathrm{d}x\) 

【答案】B

【解析】①\(\lim\limits_{x \to 0}\frac{|f(x)| - f(0)}{x} = A\).
\(\Rightarrow |f(0)| - f(0) = 0 \Rightarrow f(0) = |f(0)| \Rightarrow f(0) = 0\). 或\(f(0)\)为正.

若\(f(0)\)为正，则当\(x \to 0\)时. 有 \(f(x) > 0\).则\(\lim\limits_{x \to 0}\frac{|f(x) - f(0)|}{x} = \lim\limits_{x \to 0}\frac{f(x) - f(0)}{x} = A \Rightarrow f'(0) = A\)
若\(f(0) = 0\)，由\(\lim\limits_{x \to 0}\frac{|f(x)| - f(0)}{x} = \lim\limits_{x \to 0}\frac{|f(x)|}{x} = A\)
\(\Rightarrow \begin{cases}\lim\limits_{x \to 0^+}\left|\frac{f(x)}{x}\right| = A \\ \lim\limits_{x \to 0^-}\left|-\frac{f(x)}{x}\right| = A \end{cases} \Rightarrow\)若\(A = 0\)，则\(f'(0)\)存在，若\(A \neq 0\)，则\(f'(0)\)不存在.

②由\(\lim\limits_{x \to 0}\frac{f(x) - |f(0)|}{x} = A \Rightarrow f(0) = |f(0)|\) ，则\(f(0) = 0\)或\(f(0) > 0\)
若\(f(0) = 0\)，则\(A = \lim\limits_{x \to 0}\frac{f(x) - |f(0)|}{x} = \lim\limits_{x \to 0}\frac{f(x)}{x} = f'(0)\)
若\(f(0) > 0\)，则\(A = \lim\limits_{x \to 0}\frac{f(x) - f(0)}{x} = f'(0)\) 从而②成立.

③由\(\lim\limits_{x \to 0}\frac{|f(x)|}{x}\)存在，则 \(|f(0)| = 0 \Rightarrow f(0) = 0\).  则同①，从而③错.
④ \(\lim\limits_{x \to 0}\frac{|f(x)| - |f(0)|}{x}\)存在，则\(|f(x)|\)在\(x = 0\)处可导\(\Rightarrow f(x)\)在\(x = 0\)处可导，④正确.

从而①③错误，②④正确，选B. 

【答案】D 
【解析】 

令\(A = \begin{pmatrix}1&2&0\\2&a&0\\0&0&b\end{pmatrix}\)，为实对称矩阵，对应二次型为\(f(x_1,x_2,x_3)=x_1^2 + ax_2^2 + bx_3^2 + 4x_1x_2\)，则用配方法将其化为标准型，\(f(x_1,x_2,x_3)=(x_1 + 2x_2)^2 + (a - 4)x_2^2 + bx_3^2\)。已知\(A\)有一正两负特征值，则\(\begin{cases}a - 4 \lt 0 \\ b \lt 0 \end{cases} \Rightarrow \begin{cases}a \lt 4 \\ b \lt 0 \end{cases}\)，故选D.

【答案】B

【解析】

A选项：\(\begin{pmatrix}1&1&0&1\\1&2&1&3\\2&3&1&4\end{pmatrix} \to \begin{pmatrix}1&1&0&1\\0&1&1&2\\0&1&1&2\end{pmatrix} \to \begin{pmatrix}1&1&0&1\\0&1&1&2\\0&0&0&0\end{pmatrix}\) 

B选项：\(\begin{pmatrix}1&1&0&1\\1&1&2&5\\1&1&1&3\end{pmatrix} \to \begin{pmatrix}1&1&0&1\\0&0&2&4\\0&0&1&2\end{pmatrix} \to \begin{pmatrix}1&1&0&1\\0&0&1&2\\0&0&0&0\end{pmatrix}\) 

C选项：\(\begin{pmatrix}1&0&0&1\\0&1&0&3\\0&1&0&0\end{pmatrix} \to \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\end{pmatrix}\) 

D选项：\(\begin{pmatrix}1&1&2&3\\1&2&2&3\\2&3&4&6\end{pmatrix} \to \begin{pmatrix}1&1&2&3\\0&1&0&0\\0&1&0&0\end{pmatrix} \to \begin{pmatrix}1&1&2&3\\0&1&0&0\\0&0&0&0\end{pmatrix}\)

【答案】D
【解析】
取\(\boldsymbol{A}=\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\)，\(\boldsymbol{B}=\begin{pmatrix}1&0& - 1\\1&0& - 1\\1&0& - 1\end{pmatrix}\)，则\(\boldsymbol{AB}=\begin{pmatrix}3&0& - 3\\3&0& - 3\\3&0&3\end{pmatrix}\)，\(r(\boldsymbol{AB}) = 1\).
\(\boldsymbol{BA}=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\)，\(r(\boldsymbol{BA}) = 0\).排除B，C.
\(\boldsymbol{A + B}=\begin{pmatrix}2&1&0\\2&1&0\\2&1&0\end{pmatrix}\)，\(r(\boldsymbol{A + B}) = 1\)排除A,故选D.

【解析】\(a = 2\) 。
原式
 \(=\int_{1}^{+\infty}\frac{a}{x(2x + a)}dx=\ln x-\ln(2x + a)\big|_{1}^{+\infty}=\int_{1}^{+\infty}\frac{a}{x(2x + a)}dx=\ln x-\ln(2x + a)\big|_{1}^{+\infty}\)
\(=\lim\limits_{x\rightarrow+\infty}\ln\frac{x}{2x + a}-\ln\frac{1}{2 + a}=\lim\limits_{x\rightarrow+\infty}\frac{x}{x + a}+\ln(2 + a)\)
\(\therefore\ln\frac{1}{2}+\ln(2 + a)=\ln2\)
\(\ln(2 + a)=2\ln2\Rightarrow a = 2\) 

【解析】\(y = x - 1\)
可得无水平渐近线、铅直渐近线，故求斜渐近线即可．

\(k = \lim\limits_{x \to \infty} \frac{\sqrt[3]{x^3 - 3x^2 + 1}}{x} = \lim\limits_{x \to \infty} \sqrt[3]{\frac{x^3 - 3x^2 + 1}{x^3}} = 1\)

\(\begin{align}b &= \lim\limits_{x \to \infty} \sqrt[3]{x^3 - 3x^2 + 1} - x \\&= \lim\limits_{x \to \infty} x \cdot \left( \left( 1 + \frac{1 - 3x^2}{x^3} \right)^{\frac{1}{3}} - 1 \right) \\&= \lim\limits_{x \to \infty} x \cdot \frac{1}{3} \cdot \frac{1 - 3x^2}{x^3} = -1\end{align}\)

故\(y = x - 1\)．

【解析】

$\lim\limits_{n \to \infty} \frac{1}{n^2} \left[ \ln \frac{1}{n} + 2\ln \frac{2}{n} + \cdots + (n - 1)\ln \frac{n - 1}{n} \right]$

$= \lim\limits_{n \to \infty} \left( \frac{1}{n}\ln \frac{1}{n} + \frac{2}{n}\ln \frac{2}{n} + \cdots + \frac{n - 1}{n}\ln \frac{n - 1}{n} \right) \cdot \frac{1}{n}$

$= \lim\limits_{n \to \infty} \left( \frac{1}{n}\ln \frac{1}{n} + \frac{2}{n}\ln \frac{2}{n} + \cdots + \frac{n - 1}{n}\ln \frac{n - 1}{n} + \frac{n}{n}\ln \frac{n}{n} \right) \cdot \frac{1}{n}$

$= \lim\limits_{n \to \infty} \sum_{i = 1}^{n} \frac{i}{n}\ln \frac{i}{n} \cdot \frac{1}{n}$

$= \int_{0}^{1} x\ln x dx$

$= \frac{1}{2} \int_{0}^{1} \ln x dx^2 = \frac{1}{2} \left( x^2\ln x \big|_{0}^{1} - \int_{0}^{1} x^2 \cdot \frac{1}{x} dx \right)$

$= -\frac{1}{2} \int_{0}^{1} x dx = -\frac{1}{4}$

【解析】

\( \begin{cases}x = \ln(1 + 2t) ①\\2t - \int_{1}^{y + t^2} e^{-u^2} du = 0 ②\end{cases} \)

由②两边关于 \( t \) 求导，则 \( 2 - e^{-(y + t^2)^2} \cdot \left( \frac{dy}{dt} + 2t \right) = 0 \)。

当 \( t = 0 \) 时，\( y = 1 \)，\( 2 - e^{-1} \cdot \frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = 2e \)。

则 \( \frac{dy}{dx}\big|_{t = 0} = \frac{\frac{dy}{dt}\big|_{t = 0}}{\frac{dx}{dt}\big|_{t = 0}} = \frac{2e}{2} = e \) 。

【答案】\(3x^{2}-4xy + 5y^{2}=4\) 
【解析】 
解： 
\[
\begin{aligned}
&(2y - 3x)dx + (2x - 5y)dy = 0 \\
\Rightarrow \quad &2ydx + 2xdy - 3xdx - 5ydy = 0 \\
\Rightarrow \quad &d(2xy) - d\left(\frac{3}{2}x^{2}\right) - d\left(\frac{5}{2}y^{2}\right) = 0
\end{aligned}
\]
即 
\[ \begin{aligned} &d\left(2xy-\frac{3}{2}x^{2}-\frac{5}{2}y^{2}\right)= 0 \\ \Rightarrow &2xy-\frac{3}{2}x^{2}-\frac{5}{2}y^{2}= c \end{aligned} \]

又因为
\[y(1) = 1\]
则
\[2 - \frac{3}{2} - \frac{5}{2} = c,\Rightarrow c = -2\]
即
\[2xy - \frac{3}{2}x^2 - \frac{5}{2}y^2 = -2\]
\[3x^2 - 4xy + 5y^2 = 4\]
则所求方程为\(y = \frac{2x + \sqrt{20 - 11x^2}}{5}\)  。

【答案】\(k\begin{pmatrix}1\\1\\ - 1\\ - 1\end{pmatrix}+\begin{pmatrix}1\\0\\0\\4\end{pmatrix}\)，\(k\)为任意常数 
【解析】由于\(\alpha_1 + \alpha_2 = \alpha_3 + \alpha_4\)，故\(\alpha_1, \alpha_2, \alpha_3, \alpha_4\)线性相关；且已知\(\alpha_1, \alpha_2, \alpha_3\)线性无关，则\(r(A)=3\) 。那么\(Ax = \alpha_1 + 4\alpha_4\)等价于\((\alpha_1, \alpha_2, \alpha_3, \alpha_4)\cdot\begin{pmatrix}1\\0\\0\\4\end{pmatrix}=\alpha_1 + 4\alpha_4\)，故\(\begin{pmatrix}1\\0\\0\\4\end{pmatrix}\)是\(Ax = \alpha_1 + 4\alpha_4\)的一个特解。又因为\(\alpha_1 + \alpha_2 = \alpha_3 + \alpha_4\)，即\(\alpha_1 + \alpha_2 - \alpha_3 - \alpha_4 = 0\)，则\((\alpha_1, \alpha_2, \alpha_3, \alpha_4)\cdot\begin{pmatrix}1\\1\\ - 1\\ - 1\end{pmatrix}= 0\)。由\(s = n - r(A)=4 - 3 = 1\)可得，\(\begin{pmatrix}1\\1\\ - 1\\ - 1\end{pmatrix}\)是\(Ax = 0\)的一个基础解系，故\(Ax = \alpha_1 + 4\alpha_4\)的通解为\(k\begin{pmatrix}1\\1\\ - 1\\ - 1\end{pmatrix}+\begin{pmatrix}1\\0\\0\\4\end{pmatrix}\)，其中\(k\)为任意常数。 

解：
\[
\begin{flalign}
\int_{0}^{1}\frac{1}{(x + 1)(x^2 - 2x + 2)}dx 
&= \int_{0}^{1}\left(\frac{A}{x + 1}+\frac{Bx + C}{x^2 - 2x + 2}\right)dx \nonumber \\
&= \int_{0}^{1}\left(\frac{\frac{1}{5}}{x + 1}+\frac{-\frac{1}{5}x+\frac{3}{5}}{x^2 - 2x + 2}\right)dx \nonumber \\
&= \frac{1}{5}\ln|1 + x|\bigg|_{0}^{1} 
- \frac{1}{10}\ln|x^2 - 2x + 2|\bigg|_{0}^{1} 
+ \frac{2}{5}\arctan(x - 1)\bigg|_{0}^{1} \nonumber \\
&= \frac{3}{10}\ln2 + \frac{1}{10}\pi \nonumber
\end{flalign}
\]

解：
已知\(\ln(1 + x) + \ln(1 - x)=x - \frac{1}{2}x^2 + o(x^2) - x - \frac{1}{2}x^2 + o(x^2)= - x^2 + o(x^2)\)
\(e^{2\sin x}=1 + 2\sin x + \frac{1}{2}(2\sin x)^2 + o(x^2)=1 + 2\sin x + 2\sin^2 x + o(x^2)\)
因此，\(-3 = \lim\limits_{x \to 0}\frac{xf(x) - e^{2\sin x} + 1}{-x^2}=\lim\limits_{x \to 0}\frac{xf(x) - (1 + 2\sin x + 2\sin^2 x + o(x^2)) + 1}{-x^2}\)
\(=\lim\limits_{x \to 0}\frac{xf(x) - 2\sin x}{-x^2} + \lim\limits_{x \to 0}\frac{-2\sin^2 x}{-x^2}\)
可以得出\(\lim\limits_{x \to 0}\frac{xf(x) - 2\sin x}{-x^2} = - 5\)，\(\lim\limits_{x \to 0}\frac{xf(x) - 2\left(x - \frac{1}{6}x^3 + o(x^3)\right)}{x^2} = 5\)，
进一步可以得出\(\lim\limits_{x \to 0}\frac{xf(x) - 2x}{x^2} = 5\)，以及\(\lim\limits_{x \to 0}\frac{f(x) - 2}{x} = 5\)，
\(\lim\limits_{x \to 0}[f(x) - 2] = 0\)，可得\(\lim\limits_{x \to 0}f(x) = 2 = f(0)\)，故\(f^\prime(0) = \lim\limits_{x \to 0}\frac{f(x) - 2}{x} = 5\) 。 

解：
\(\frac{\partial f}{\partial x} = -2x\mathrm{e}^{-y} \Rightarrow f(x,y) = -x^2\mathrm{e}^{-y} + \varphi(y) \)。则\(\frac{\partial f}{\partial y} = x^2\mathrm{e}^{-y} + \varphi^\prime(y) = \mathrm{e}^{-y}x^2 - (y + 1)\mathrm{e}^{-y}\)
则\(\varphi^\prime(y) = -(y + 1)\mathrm{e}^{-y}\)
\(\Rightarrow \varphi(y) = (y + 2)\mathrm{e}^{-y} + C\)
则
\(f(x,y) = -x^2\mathrm{e}^{-y} + (y + 2)\mathrm{e}^{-y} + C\)
又\(f(0,0) = 2\)，\(\Rightarrow C = 0\)

则\(f(x,y) = -x^2\mathrm{e}^{-y} + (y + 2)\mathrm{e}^{-y} \)。

\(\begin{cases}
\frac{\partial f}{\partial x} = -2x\mathrm{e}^{-y} = 0 \\
\frac{\partial f}{\partial y} = \mathrm{e}^{-y}(x^2 - y - 1) = 0
\end{cases} \Rightarrow \begin{cases}
x = 0 \\
y = -1
\end{cases}\)

则驻点\((0, -1)\)

\(f_{xx} = -2\mathrm{e}^{-y}, f_{xy} = 2x\mathrm{e}^{-y}, f_{yy} = \mathrm{e}^{-y}(x^2 - y - 1) - \mathrm{e}^{-y} = \mathrm{e}^{-y}(x^2 - y) \)

在点\((0, -1)\)处\(A = -2\mathrm{e}, B = 0, C = -\mathrm{e}\)

则\(AC - B^2 > 0, A < 0\)，从而\( f(x,y) \)在\((0, -1)\)处有极大值，且极大值为
\(f(0, -1) = \mathrm{e}\) 

【答案】 \(12\pi - \frac{16}{3}\)

由题可知，积分区域\(D =\{(x,y)|(x - 2)^2 + y^2 \leq 2^2, x^2 + (y - 2)^2 \leq 2^2\}\)

对应图形为右图所示，

显然积分区域关于\(y = x\)对称

记\(D_1 =\{(x,y)| x^2 + (y - 2)^2 \leq 2^2, y \leq x\}\)

\(D_2 =\{(x,y)|(x - 2)^2 + y^2 \leq 2^2, y \geq x\}\)

故\(I = \iint_D (x - y)^2 dxdy = \iint_{D_1} (x - y)^2 dxdy + \iint_{D_2} (x - y)^2 dxdy\)

\(= 2\iint_{D_1} (x - y)^2 dxdy\)

令\(x = r\cos\theta, y = r\sin\theta\) 则在积分区域\(D_1\)上有 \(\begin{cases}0 \leq r \leq 4\cos\theta \\ 0 \leq \theta \leq \frac{\pi}{4} \end{cases}\)

则\(\iint_{D_1} (x - y)^2 dxdy = \iint_{D_1} (x^2 + y^2 - 2xy) dxdy = \int_{0}^{\frac{\pi}{4}} d\theta \int_{0}^{4\cos\theta} (r^2 - 2r^2 \cos\theta\sin\theta) dr\)

\(= \int_{0}^{\frac{\pi}{4}} d\theta \int_{0}^{4\cos\theta} (r^3 - 2r^3 \cos\theta\sin\theta) dr = 6\pi - \frac{8}{3}\)

故\(I = 2(6\pi - \frac{8}{3}) = 12\pi - \frac{16}{3}\) 。

解：充分性：若对\((a,b)\)内任意的\(x_1,x_2,x_3\)，当\(x_1\lt x_2\lt x_3\)时，都有
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
则在\((a,b)\)内取任意的\(x_1\lt x_2\lt x_3\lt x_4\lt x_5\)，有
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}\lt\frac{f(x_4) - f(x_3)}{x_4 - x_3}\lt\frac{f(x_5) - f(x_4)}{x_5 - x_4}
\]
在\(\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}\)两边同时令\(x_2 \to x_1^+\)，得
\[
f_+'(x_1)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}
\]
两边同时令\(x_2 \to x_3^-\)，得\(\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_3)\)，即
\[
f_+'(x_1)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_3)
\]
同理可得\(f_+'(x_3)\leq\frac{f(x_3) - f(x_1)}{x_3 - x_1}\leq f_-'(x_5)\) .因为
\[
\frac{f(x_3) - f(x_1)}{x_3 - x_1}\lt\frac{f(x_5) - f(x_3)}{x_5 - x_3}
\]
所以\(f_+'(x_1)\leq f_-'(x_5)\) .由\(x_1,x_5\)的任意性，可得\(f'(x)\)在\((a,b)\)内严格单调递增，充分性得证。

再证必要性，即已知\(f'(x)\)单调递增，在\([x_1,x_2],[x_2,x_3]\)上分别使用拉格朗日中值定理，知存在\(\xi_1 \in (x_1,x_2),\xi_2 \in (x_2,x_3)\)，使
\[
f'(\xi_1)=\frac{f(x_2) - f(x_1)}{x_2 - x_1}, \quad f'(\xi_2)=\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
又由\(f'(x)\)单调递增，且\(\xi_1\lt\xi_2\)知，\(f'(\xi_1)\lt f'(\xi_2)\)，即
\[
\frac{f(x_2) - f(x_1)}{x_2 - x_1}\lt\frac{f(x_3) - f(x_2)}{x_3 - x_2}
\]
必要性得证。

综上所述，充要条件得证。

（1）\(\boldsymbol{A}\)与\(\boldsymbol{B}\)合同知，二者有相同的正负惯性指数
显然，\(\boldsymbol{B}\)的特征值为\(k\)、\(6\)、\(0\)，故\(\boldsymbol{A}\)有特征值\(0\).故\(\vert\boldsymbol{A}\vert = 0\).
计算得\(\vert\boldsymbol{A}\vert=- 3(a - 4)\)，即有\(a = 4\).
此时\(\vert\lambda\boldsymbol{E}-\boldsymbol{A}\vert=\lambda(\lambda - 3)(\lambda - 6)\).
知\(\boldsymbol{A}\)的特征值为\(3\)、\(6\)、\(0\). \(P = 2\). 故\(k\gt0\).
(2)由\(\boldsymbol{Q}^{\text{T}}\boldsymbol{A}\boldsymbol{Q}=\boldsymbol{B}\)，知\(\boldsymbol{Q}^{-1}\boldsymbol{A}\boldsymbol{Q}=\boldsymbol{B}\).
故\(\boldsymbol{A}\)、\(\boldsymbol{B}\)相似，特征值相同，故\(k = 3\).对\(\boldsymbol{A}:\lambda = 3\)时，解\((3\boldsymbol{E}-\boldsymbol{A})\boldsymbol{X}=\boldsymbol{0}\)，
得\(\boldsymbol{c}_{1}=\begin{bmatrix}1\\1\\1\end{bmatrix}\)，\(\lambda = 6\)时，解\((6\boldsymbol{E}-\boldsymbol{A})\boldsymbol{x}=\boldsymbol{0}\)，得\(\boldsymbol{c}_{2}=\begin{bmatrix}-1\\0\\1\end{bmatrix}\)
\(\lambda = 0\)时，解\((0\cdot\boldsymbol{E}-\boldsymbol{A})\boldsymbol{x}=\boldsymbol{0}\)得\(\boldsymbol{c}_{3}=\begin{bmatrix}1\\-2\\1\end{bmatrix}\)
再单位化得：\(\boldsymbol{Q}=\begin{bmatrix}\frac{1}{\sqrt{3}}&-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{3}}&0&-\frac{2}{\sqrt{6}}\\\frac{1}{\sqrt{3}}&\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}\end{bmatrix}\)