答案 选(D).
解 若点$(x,y)$在曲线$y = y(x)$上，容易验证点$(-x,-y)$也在曲线$y = y(x)$上，因此$y(x)$为奇函数.
先考虑$x>0$时的情况，令$y=0$，则有$\frac{x}{2} = \arctan\frac{\sqrt{3}x}{2}$，令$x = \frac{2\tan\theta}{\sqrt{3}},\theta\in\left(0,\frac{\pi}{2}\right)$，则有$\frac{\tan\theta}{\sqrt{3}} = \theta$，即$\frac{\tan\theta}{\theta} = \sqrt{3}$.设$f(x) = \frac{\tan x}{x},x\in\left(0,\frac{\pi}{2}\right)$，则有
$f'(x) = \frac{x\sec^2 x - \tan x}{x^2} = \frac{2x - \sin2x}{2x^2\cos^2 x} > 0$，
又$f(0^+) = 1 < \sqrt{3},f\left(\frac{\pi}{2}^-\right) = +\infty$，因此方程$f(x) = \sqrt{3}$有唯一正根.由于$y(x)$为奇函数，根据对称性可知还有一个负零点，以及零点$x = 0$.综上可得$y(x)$的零点个数是3.
当$x \neq 0$时，方程可变形为$\frac{1}{2} + \frac{\sqrt{3}}{2}\frac{y}{x} = \frac{1}{x}\arctan\left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right)$，则$\lim\limits_{x\rightarrow\infty}\frac{y}{x} = -\frac{1}{\sqrt{3}}$.又有
$\frac{1}{\sqrt{3}}x + y = \frac{2}{\sqrt{3}}\arctan\left(\frac{\sqrt{3}}{2}x - \frac{1}{2}y\right)$，
而$\lim\limits_{x\rightarrow+\infty}y = -\infty,\lim\limits_{x\rightarrow-\infty}y = +\infty$，所以
$\lim\limits_{x\rightarrow+\infty}\left(\frac{1}{\sqrt{3}}x + y\right) = \frac{\pi}{\sqrt{3}}，\lim\limits_{x\rightarrow-\infty}\left(\frac{1}{\sqrt{3}}x + y\right) = -\frac{\pi}{\sqrt{3}}$，
因此有两条斜渐近线$y = \frac{\pi - x}{\sqrt{3}}$与$y = -\frac{\pi + x}{\sqrt{3}}$.

答案 选(C).
解 取$f(x) = 0,a_n = \frac{(-1)^n}{2}$，则$\lim\limits_{n\rightarrow\infty}f(a_n) = 0$，但是$\lim\limits_{n\rightarrow\infty}a_n$不存在，故(A)错误；
若$\lim\limits_{n\rightarrow\infty}a_n$存在，又因为$a_n \in (-1,1)$，不妨设$\lim\limits_{n\rightarrow\infty}a_n = b \in [-1,1]$，又函数$f(x)$在$[-1,1]$上连续，由复合函数极限运算法则知$\lim\limits_{n\rightarrow\infty}f(a_n) = f(b)$，故(B)错误；
取$f(x) = x^2,a_n = (-1)^n\frac{n}{n + 1}$，则$f(a_n) = \left(\frac{n}{n + 1}\right)^2$单调增，但是$\lim\limits_{n\rightarrow\infty}a_n$不存在，故(D)错误；
数列$\{a_n\}$单调，又因为$a_n \in (-1,1)$，故$\lim\limits_{n\rightarrow\infty}a_n$存在，不妨设$\lim\limits_{n\rightarrow\infty}a_n = b \in [-1,1]$，又函数$f(x)$在$[-1,1]$上连续，由复合函数极限运算法则知$\lim\limits_{n\rightarrow\infty}f(a_n) = f(b)$，故选(C).

答案 选(A).
解 点$x = 1,x = -1$及$x = \frac{1}{2}$均为间断点，且
$\lim\limits_{x\rightarrow-1}f(x) = 0，\lim\limits_{x\rightarrow1^-}f(x) = -\frac{2}{\sqrt{3}}\pi，\lim\limits_{x\rightarrow1^+}f(x) = \frac{2}{\sqrt{3}}\pi，\lim\limits_{x\rightarrow\frac{1}{2}}f(x) = \infty$，
所以(B),(C),(D)均正确.
由于当$x < -2$时$f(x)$没有定义，故由间断点的概念，点$x = -2$不是间断点.

答案 选(C)
解 $\frac{\mathrm{d}y}{\mathrm{d}x} = mz^{m - 1}\frac{\mathrm{d}z}{\mathrm{d}x}$代入得
$mz^{m - 1}\frac{\mathrm{d}z}{\mathrm{d}x} = ax^\alpha + bz^{\beta m}$，即$m\frac{\mathrm{d}z}{\mathrm{d}x} = a\frac{x^\alpha}{z^{m - 1}} + bz^{\beta m - m + 1}$.
要使得此方程为齐次方程，则应有$\begin{cases}\alpha = m - 1,\\\beta m - m + 1 = 0.\end{cases}$消去$m$有$\alpha\beta - \alpha + \beta = 0$，即$\frac{1}{\alpha} - \frac{1}{\beta} = -1$.

答案 选(A).
解 $\frac{\partial z}{\partial x} = 2xe^{x - y} + (x^2 + y^2)e^{x - y}$，$\frac{\partial z}{\partial y} = 2ye^{x - y} - (x^2 + y^2)e^{x - y}$，
$\frac{\partial^2 z}{\partial x^2} = 2e^{x - y} + 4xe^{x - y} + (x^2 + y^2)e^{x - y}$，$\frac{\partial^2 z}{\partial y^2} = 2e^{x - y} - 4ye^{x - y} + (x^2 + y^2)e^{x - y}$，
所以$g(x,y) = \frac{\partial^2 z}{\partial x^2} - \frac{\partial^2 z}{\partial y^2} = 4(x + y)e^{x - y}$.
$g_x'(x,y) = 4(x + y + 1)e^{x - y}$，$g_y'(x,y) = 4(1 - x - y)e^{x - y}$，
对任意$(x,y) \in D$，有$g_x'(x,y) > 0,g_y'(x,y) > 0$，故
当$x_1 < x_2,y_1 < y_2$时，
$g(x_1,y_1) \leqslant g(x_2,y_1) \leqslant g(x_2,y_2)$ 或 $g(x_1,y_1) \leqslant g(x_1,y_2) \leqslant g(x_2,y_2)$.

答案 选(B).
解 积分区域如图所示，

$x = \sqrt{2 - y^2}$
1
$x = \sqrt{y}$
O 1 x
交换积分次序
$\int_{0}^{1}\mathrm{d}y\int_{0}^{\sqrt{y}}f(x,y)\mathrm{d}x + \int_{1}^{\sqrt{2}}\mathrm{d}y\int_{0}^{\sqrt{2 - y^2}}f(x,y)\mathrm{d}x = \int_{0}^{1}\mathrm{d}x\int_{x^2}^{\sqrt{2 - x^2}}f(x,y)\mathrm{d}y$.

答案 选(D).
解 由导数几何意义及参量函数求导可得
$\tan\alpha=\frac{\text{d}y}{\text{d}x}=\frac{\sin t}{1-\cos t}=\tan\frac{\pi-t}{2},$
故
$\frac{\cos\alpha}{\sqrt{y}}=\frac{\sin\frac{t}{2}}{\sqrt{1-\cos t}}=\frac{\sqrt{2}}{2}.$

答案 选(B).
解 $\boxed{\alpha}_{4}$可由$\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3}$表示，则$r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3})=r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})$，而$\boxed{\alpha}_{1}$不可由$\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4}$表示，则
$r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4}) \neq r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})$，从而
$r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3})=r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4}) \neq r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4}).$
$\begin{pmatrix} 1 & 1 & 1 & \vdots & 1 \\ 0 & 2 & -1 & \vdots & b \\ 1 & 1 & a & \vdots & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 1 & \vdots & 1 \\ 0 & 2 & -1 & \vdots & b \\ 0 & 0 & a-1 & \vdots & 0 \end{pmatrix}.$
① 若$a \neq 1$，则$r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3})=r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=3$，而
$(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & b \\ 1 & a & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 1 \\ 0 & -3 & b-2 \\ 0 & a-1 & 0 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & b-2 \end{pmatrix}.$
若$b \neq 2$，则$r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=3$，不满足题意；
若$b = 2$，则$r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=2$，满足题意.
② 若$a = 1$，则$r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3})=r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=2$，而
$(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=\begin{pmatrix} 1 & 1 & 1 \\ 2 & -1 & b \\ 1 & 1 & 1 \end{pmatrix}\sim\begin{pmatrix} 1 & 1 & 1 \\ 0 & -3 & b-2 \\ 0 & 0 & 0 \end{pmatrix},$
对任意$b \in \mathbb{R}$，则$r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})=2$，不满足题意.
综上，可知当$a \neq 1$且$b = 2$，$r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3})=r(\boxed{\alpha}_{1},\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4}) \neq r(\boxed{\alpha}_{2},\boxed{\alpha}_{3},\boxed{\alpha}_{4})$，故选(B).

答案 选(C).
解 因为$a_{ij}=-A_{ij}$，所以$A=(a_{ij})=(-A_{ij})=-(A^{*})^{\text{T}}$，则$|A|=-|A^{*}|=-|A|^{2}$，从而$|A|=0$
或$-1$，又因为$A \neq O$，不妨假设$a_{11} \neq 0$，从而$|A|=a_{11}A_{11}+a_{12}A_{12}+a_{13}A_{13}=-(a_{11}^{2}+a_{12}^{2}+a_{13}^{2})<0$，所以
$|A|=-1$，故$AA^{\text{T}}=-AA^{*}=-|A|E=E$，即$A$为正交阵，所以$\boxed{\alpha}^{\text{T}}A^{\text{T}}A\boxed{\alpha}=\|\boxed{\alpha}\|^{2}$，故①正确.
$[A\boxed{\alpha},A\boxed{\beta}]=(A\boxed{\alpha})^{\text{T}}A\boxed{\beta}=\boxed{\alpha}^{\text{T}}A^{\text{T}}A\boxed{\beta}=\boxed{\alpha}^{\text{T}}\boxed{\beta}=[\boxed{\alpha},\boxed{\beta}]$，故②正确.
$|A+E|=|A+AA^{\text{T}}|=|A||E+A^{\text{T}}|=-|(E+A)^{\text{T}}|=-|A+E|$，所以$|A+E|=0$，从而
$(A+E)\boxed{x}=\boxed{0}$有非零解，故③正确.
$A$为正交阵，则$A$的实特征值为1或$-1$，又$|A|=-1$，所以$A$特征值$1,1,-1$或$-1,-1,-1$，从
而$\text{tr}(A)=1$或$\text{tr}(A)=-3$. 而$A^{2}$的特征值全为1，所以$\text{tr}(A^{2})=3$，所以$\text{tr}(A^{2}) \neq [\text{tr}(A)]^{2}$，故④
不正确.
取$A=\begin{pmatrix} & & 1 \\ & 1 & \\ 1 & & -1 \end{pmatrix}$满足题意，但$\text{tr}(A^{2})=3,[\text{tr}(A)]^{2}=1$，故④不正确.
综上所述，应选(C).

答案 选(D).
解 法一 设$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{\xi}=\boxed{0}$，则$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{\xi}=\begin{pmatrix} A\boxed{\xi} \\ BA\boxed{\xi} \end{pmatrix}=\boxed{0}$，故有$A\boxed{\xi}=\boxed{0}$，从而$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{x}=\boxed{0}$的解是$A\boxed{x}=\boxed{0}$
的解.
若$A\boxed{\xi}=\boxed{0}$，则$BA\boxed{\xi}=\boxed{0}$，从而$\begin{pmatrix} A\boxed{\xi} \\ BA\boxed{\xi} \end{pmatrix}=\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{\xi}=\boxed{0}$，即$A\boxed{x}=\boxed{0}$的解是$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{x}=\boxed{0}$的解，从而$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{x}=\boxed{0}$
与$A\boxed{x}=\boxed{0}$同解.
法二 $r\begin{pmatrix} A \\ BA \end{pmatrix}=r\begin{pmatrix} A \\ BA \end{pmatrix}=r\left[\begin{pmatrix} E \\ B \end{pmatrix}A\right]$，因为$3=r(E) \leqslant r\begin{pmatrix} E \\ B \end{pmatrix} \leqslant 3$，左乘列满秩矩阵，不改变矩阵的
秩，所以$r\left[\begin{pmatrix} E \\ B \end{pmatrix}A\right]=r(A)$，从而$r\begin{pmatrix} A \\ BA \end{pmatrix}=r\begin{pmatrix} A \\ BA \end{pmatrix}=r(A)$，故$\begin{pmatrix} A \\ BA \end{pmatrix}\boxed{x}=\boxed{0}$与$A\boxed{x}=\boxed{0}$同解.
选项(A)不正确，$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},B=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$，$\begin{pmatrix} A \\ B \end{pmatrix}\boxed{x}=\boxed{0}$仅有零解，所以不正确.
选项(B)不正确，$A=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$，$\begin{pmatrix} A \\ B \end{pmatrix}\boxed{x}=\boxed{0}$与$B\boxed{x}=\boxed{0}$同解，有公共非零解.
选项(C)不正确，$A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix},B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$，$\begin{pmatrix} B \\ BA \end{pmatrix}\boxed{x}=\boxed{0}$与$B\boxed{x}=\boxed{0}$不同解.

答案 填“$\frac {π}{4}$”.
解 $∫_{0}^{1}\frac {\sqrt [n]{1+n\arctan x^{2}}}{1+x^{2}}dx≤∫_{0}^{1}\frac {1}{1+x^{2}}dx\cdot \sqrt [n]{1+n\arctan 1^{2}}≤\sqrt [n]{1+n}\cdot ∫_{0}^{1}\frac {1}{1+x^{2}}dx=\frac {π}{4}\sqrt [n]{1+n},$
$∫_{0}^{1}\frac {\sqrt [n]{1+n\arctan x^{2}}}{1+x^{2}}dx≥∫_{0}^{1}\frac {1}{1+x^{2}}dx=\frac {π}{4},$
由于 $\lim\limits _{n→∞}\sqrt [n]{1+n}=1$,由夹逼准则得 $\lim\limits _{n→∞}∫_{0}^{1}\frac {\sqrt [n]{1+n\arctan x^{2}}}{1+x^{2}}dx=\frac {π}{4}.$

答案 填“384”.
解 由二项式展开可得
$f(x)=\sum\limits _{k=0}^{8}C_{8}^{k}(\sqrt {x})^{k}+\sum\limits _{k=0}^{8}C_{8}^{k}(-\sqrt {x})^{k}=2\sum\limits _{k=0}^{4}C_{8}^{2k}(\sqrt {x})^{2k}=2\sum\limits _{k=0}^{4}C_{8}^{2k}x^{k},$
所以
$f^{(3)}(x)=2(\sum\limits _{k=0}^{4}C_{8}^{2k}x^{k})^{(3)}=2(C_{8}^{6}x^{3}+C_{8}^{8}x^{4})^{(3)}=56\cdot 3!+48x,$
故$f^{(3)}(1)=384.$

答案 填“1”.
解 $∫_{0}^{+∞}\frac {1}{\sqrt {x(x+2)^{3}}}dx=∫_{0}^{+∞}\frac {1}{(x+2)\sqrt {x(x+2)}}dx=∫_{0}^{+∞}\frac {1}{(x+2)\sqrt {(x+1)^{2}-1}}dx,$
令$x+1=\sec t,$
原式$=∫_{0}^{\frac {π}{2}}\frac {1}{(\sec t+1)\tan t}\sec t\tan tdt=∫_{0}^{\frac {π}{2}}\frac {\sec t}{(\sec t+1)}dt=∫_{0}^{\frac {π}{2}}\frac {1}{1+\cos t}dt$
$=∫_{0}^{\frac {π}{2}}\frac {1}{2\cos ^{2}\frac {t}{2}}dt=\tan \frac {t}{2}|_{0}^{\frac {π}{2}}=1.$

答案 填“$2f_{1}'(1,2,1)-8f_{2}'(1,2,1)+9f_{3}'(1,2,1)$”
解 $\frac {du}{dx}=2xf_{1}'+2f_{2}'\frac {dy}{dx}+3z^{2}f_{3}'\frac {dz}{dx}$,方程组每个方程两边同时对x求导,得
$\left\{\begin{array}{l} 2y\frac {dy}{dx}-2y-2x\frac {dy}{dx}+z^{2}\frac {dy}{dx}+2yz\frac {dz}{dx}=0,\\ 2x+2y\frac {dy}{dx}+2z\frac {dz}{dx}=0,\end{array}\right. $
当$x=1$时,解得$y=z=1$,代入上式,有
$\left\{\begin{array}{l} -2+\frac {dy}{dx}+2\frac {dz}{dx}=0,\\ 1+\frac {dy}{dx}+\frac {dz}{dx}=0,\end{array}\right. $
解得$\frac {dy}{dx}|_{(1,1,1)}=-4,\frac {dz}{dx}|_{(1,1,1)}=3$.所以
$\frac {du}{dx}=2f_{1}'(1,2,1)-8f_{2}'(1,2,1)+9f_{3}'(1,2,1).$

答案 填“$π$”
解法一 $∫_{D_{t}}^{0}e^{-\sqrt {x^{2}+y^{2}}}dσ=∫_{0}^{2π}dθ∫_{0}^{t}e^{-r}rdr=2π\cdot [1-(t+1)e^{-t}]$,所以
原极限$=\lim\limits _{t→0^{+}}\frac {1}{\sqrt {1+2t^{2}}-1}∫_{D_{t}}^{0}e^{-\sqrt {x^{2}+y^{2}}}dσ=\lim\limits _{t→0^{+}}\frac {2π\cdot [1-(t+1)e^{-t}]}{\sqrt {1+2t^{2}}-1}$
$=2π\lim\limits _{t→0^{+}}\frac {[1-(t+1)e^{-t}]}{t^{2}}=2π\lim\limits _{t→0^{+}}\frac {te^{-t}}{2t}=π.$
解法二 根据积分中值定理,存在$(ξ,η)∈D_{t}$,使得
$∫_{D_{t}}^{0}e^{-\sqrt {x^{2}+y^{2}}}dσ=e^{-\sqrt {ξ^{2}+η^{2}}}πt^{2},$
所以原极限$=\lim\limits _{t→0^{+}}\frac {e^{-\sqrt {ξ^{2}+η^{2}}}πt^{2}}{\sqrt {1+2t^{2}}-1}=π\lim\limits _{t→0^{+}}\frac {e^{-\sqrt {ξ^{2}+η^{2}}}t^{2}}{t^{2}}=π\lim\limits _{ξ→0}e^{-\sqrt {ξ^{2}+η^{2}}}=π.$

答案 填“$a>0$”.
解 $A=\begin{pmatrix} 0&a&2\\ a&0&-2\\ 2&-2&0\end{pmatrix},$
$|A-λE|=| -λ&a&2\\ a&-λ&-2\\ 2&-2&-λ| =(a-λ)| 1&1&0\\ a&-λ&-2\\ 2&-2&-λ| =(a-λ)(λ^{2}+aλ-8),$
则$λ_{1}=a,λ_{2}\cdot λ_{3}=-8$,所以$λ_{2},λ_{3}$一正一负.
又因为正惯性指数为2,所以A的特征值只能为正、正、负,故$λ_{1}=a>0.$
注:此题二次型中不含平方项,配方法不易.

解法一 原式\(=\int \sqrt{\mathrm{e}^{x}-1} \cdot \frac{1}{\left(\mathrm{e}^{x}+3\right)^{2}} \mathrm{~d}\left(\mathrm{e}^{x}+3\right)=-\int \sqrt{\mathrm{e}^{x}-1} \mathrm{~d} \frac{1}{\mathrm{e}^{x}+3}\)
\(=-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+\int \frac{1}{\mathrm{e}^{x}+3} \cdot \frac{\mathrm{e}^{x}}{2 \sqrt{\mathrm{e}^{x}-1}} \mathrm{~d} x\).

令\(\sqrt{\mathrm{e}^{x}-1}=t, \mathrm{e}^{x}=t^{2}+1, x=\ln \left(t^{2}+1\right)\)
\(\int \frac{\mathrm{e}^{x}}{\left(\mathrm{e}^{x}+3\right) \cdot 2 \sqrt{\mathrm{e}^{x}-1}} \mathrm{~d} x=\int \frac{t^{2}+1}{\left(t^{2}+4\right) \cdot 2 t} \cdot \frac{2 t}{t^{2}+1} \mathrm{~d} t=\int \frac{1}{t^{2}+4} \mathrm{~d} t=\frac{1}{2} \arctan \frac{t}{2}+C\)
\(=\frac{1}{2} \arctan \frac{\sqrt{\mathrm{e}^{x}-1}}{2}+C\),

故原式\(=-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+\frac{1}{2} \arctan \frac{\sqrt{\mathrm{e}^{x}-1}}{2}+C\).

解法二 原式\(=-\int \sqrt{\mathrm{e}^{x}-1} \mathrm{~d} \frac{1}{\mathrm{e}^{x}+3}=-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+\int \frac{1}{\mathrm{e}^{x}+3} \mathrm{~d} \sqrt{\mathrm{e}^{x}-1}\)
\(=-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+\int \frac{1}{\left(\mathrm{e}^{x}-1\right)+4} \mathrm{~d} \sqrt{\mathrm{e}^{x}-1}=-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+\frac{1}{2} \arctan \frac{\sqrt{\mathrm{e}^{x}-1}}{2}+C\)

解法三 令\(\sqrt{\mathrm{e}^{x}-1}=t, \mathrm{e}^{x}=t^{2}+1\),则
原式\(=\int \frac{\left(t^{2}+1\right) \cdot t}{\left(t^{2}+4\right)^{2}} \cdot \frac{2 t}{t^{2}+1} \mathrm{~d} t=2 \int \frac{t^{2}}{\left(t^{2}+4\right)^{2}} \mathrm{~d} t\).

令\(t=2 \tan u\),
原式\(=2 \int \frac{4 \tan ^{2} u}{16 \sec ^{4} u} \cdot 2 \sec ^{2} u \mathrm{~d} u=\int \sin ^{2} u \mathrm{~d} u=\int \frac{1-\cos 2 u}{2} \mathrm{~d} u=\frac{u}{2}-\frac{1}{2} \sin u \cos u+C\)
\(=\frac{1}{2} \arctan \frac{t}{2}-\frac{t}{t^{2}+4}+C=\frac{1}{2} \arctan \frac{\sqrt{\mathrm{e}^{x}-1}}{2}-\frac{\sqrt{\mathrm{e}^{x}-1}}{\mathrm{e}^{x}+3}+C\).

证明 因为\(f(x)\)为\([a,b]\)上的连续,所以\(\int_{a}^{x} f(t) \mathrm{d} t\)在\([a,b]\)上可导,

（Ⅰ）因为\(\lim _{x \rightarrow a^{+}} F(x)=\lim _{x \rightarrow a^{+}} \frac{\int_{a}^{x} f(t) \mathrm{d} t}{x-a}=\lim _{x \rightarrow a^{+}} f(x)=f(a)=F(a)\),即\(F(x)\)在\(x=a\)右连续,又
\(F(x)\)在\((a,b]\)上连续,所以\(F(x)\)为\([a,b]\)上的连续函数.

（Ⅱ）因为\(f(x)\)为\([a,b]\)上的连续单调增加函,故对任意\(x \in(a, b)\),当\(a \leqslant t \leqslant x\)时,\(f(x)-f(t)\)
\(\geqslant 0\),从而\(\int_{a}^{x}[f(x)-f(t)] \mathrm{d} t>0\),故
\(F^{\prime}(x)=\frac{f(x)(x-a)-\int_{a}^{x} f(t) \mathrm{d} t}{(x-a)^{2}}=\frac{\int_{a}^{x}[f(x)-f(t)] \mathrm{d} t}{(x-a)^{2}}>0\),
又\(F(x)\)在\([a,b]\)上连续,所以\(F(x)\)在\([a,b]\)上单调增加.

解 \(f_{x}^{\prime}=3 x^{2}+3 y-3, f_{y}^{\prime}(x, y)=3 y^{2}+3 x-3\),令\(f_{x}^{\prime}=0, f_{y}^{\prime}(x, y)=0\),整理得
\(\begin{cases}
x^{2}+y-1=0, & ① \\
y^{2}+x-1=0, & ②
\end{cases}\)
由①−②,得\((x-y)(x+y-1)=0\),得驻点
\((1,0)\), \((0,1)\), \(\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)\), \(\left(\frac{-1-\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}\right)\).

\(f_{x x}^{\prime \prime}(x, y)=6 x, \quad f_{x y}^{\prime \prime}(x, y)=3, \quad f_{y y}^{\prime \prime}(x, y)=6 y\).

| 驻点 | \(A\) | \(B\) | \(C\) | \(B^2-AC\) | 极值性 |
| --- | --- | --- | --- | --- | --- |
| \((1,0)\) | \(6\) | \(3\) | \(0\) | \(9>0\) | 不取极值 |
| \((0,1)\) | \(0\) | \(3\) | \(6\) | \(9>0\) | 不取极值 |
| \(\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)\) | \(3(\sqrt{5}-1)\) | \(3\) | \(3(\sqrt{5}-1)\) | \(9-9(\sqrt{5}-1)^2<0\) | 极小值 |
| \(\left(\frac{-1-\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}\right)\) | \(-3(\sqrt{5}+1)\) | \(3\) | \(-3(\sqrt{5}+1)\) | \(9-9(\sqrt{5}+1)^2<0\) | 极大值 |

综上可知,\(f\left(\frac{-1+\sqrt{5}}{2}, \frac{-1+\sqrt{5}}{2}\right)=\frac{7-5 \sqrt{5}}{2}\)为极小值,\(f\left(\frac{-1-\sqrt{5}}{2}, \frac{-1-\sqrt{5}}{2}\right)=\frac{7+5 \sqrt{5}}{2}\)为极大值.

解 积分区域如图所示.

\(\iint_{D} \frac{\ln \left(1+\tan \sqrt{x^{2}+y^{2}}\right)}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} \sigma=\int_{0}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{0}^{\frac{\pi}{4}} \frac{\ln (1+\tan r)}{r} \cdot r \mathrm{~d} r\)
\(=\frac{\pi}{2} \int_{0}^{\frac{\pi}{4}} \ln (1+\tan r) \mathrm{d} r=\frac{\pi}{2} \int_{0}^{\frac{\pi}{4}}[\ln (\cos r+\sin r)-\ln \cos r] \mathrm{d} r\)
\(=\frac{\pi}{2} \int_{0}^{\frac{\pi}{4}}\left[\ln \left[\sqrt{2} \cos \left(r-\frac{\pi}{4}\right)\right]-\ln \cos r\right] \mathrm{d} r\)
\(=\frac{\pi}{2}\left[\int_{0}^{\frac{\pi}{4}} \ln \sqrt{2} \mathrm{~d} r+\int_{0}^{\frac{\pi}{4}} \ln \cos \left(r-\frac{\pi}{4}\right) \mathrm{d} r-\int_{0}^{\frac{\pi}{4}} \ln \cos r \mathrm{~d} r\right]\),

由于\(\int_{0}^{\frac{\pi}{4}} \ln \cos \left(r-\frac{\pi}{4}\right) \mathrm{d} r \stackrel{t=\frac{\pi}{4}-r}{=} \int_{\frac{\pi}{4}}^{0} \ln \cos (-t)(-\mathrm{d} t)=\int_{0}^{\frac{\pi}{4}} \ln \cos t \mathrm{~d} t\),所以
\(\iint_{D} \frac{\ln \left(1+\tan \sqrt{x^{2}+y^{2}}\right)}{\sqrt{x^{2}+y^{2}}} \mathrm{~d} \sigma=\frac{\ln 2}{16} \pi^{2}+\frac{\pi}{2}\left(\int_{0}^{\frac{\pi}{4}} \ln \cos t \mathrm{~d} t-\int_{0}^{\frac{\pi}{4}} \ln \cos r \mathrm{~d} r\right)=\frac{\ln 2}{16} \pi^{2}\).

（Ⅰ）证明 必要性
$$f(x+h)=f(x)+f'(x)h+\frac{f''(\xi_1)}{2}h^2,f(x-h)=f(x)-f'(x)h+\frac{f''(\xi_2)}{2}h^2,$$
两式相加得
$$f(x+h)+f(x-h)=2f(x)+\frac{h^2}{2}\left[f''(\xi_1)+f''(\xi_2)\right].$$
因为$f''(x)\geqslant 0$，故$f(x+h)-2f(x)+f(x-h)\geqslant 0$.

充分性
$$\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=\lim_{h\to 0}\frac{f'(x+h)-f'(x-h)}{2h}$$
$$=\lim_{h\to 0}\frac{f'(x+h)-f'(x)-\left[f'(x-h)-f'(x)\right]}{2h}$$
$$=\lim_{h\to 0}\frac{f'(x+h)-f'(x)}{2h}-\lim_{h\to 0}\frac{f'(x-h)-f'(x)}{2h}=f''(x).$$

由$\frac{f(x+h)-2f(x)+f(x-h)}{h^2}\geqslant 0$，根据极限保号性得$f''(x)\geqslant 0$.

（Ⅱ）因为$(x_0,f(x_0))$是曲线的拐点，不妨设存在$\delta>0$，当$x\in(x_0-\delta,x_0)$时，曲线为凹曲线，故由（Ⅰ）知，$f''(x)\geqslant 0$；当$x\in(x_0,x_0+\delta)$时，曲线为凸曲线，故由（Ⅰ）知，$f''(x)\leqslant 0$.

因为函数$f(x)$在$(-\infty,+\infty)$上二阶可导，故函数$f'(x)$在$(-\infty,+\infty)$上一阶可导，且当$x\in(x_0-\delta,x_0+\delta)$时，有$f'(x_0)\geqslant f'(x)$，则根据费马引理知$f''(x_0)=0$.
 

解 （Ⅰ）$\boldsymbol{A}=\begin{pmatrix}2&-1&1\\-1&2&-1\\1&-1&a\end{pmatrix}$，由$\boldsymbol{A}\boldsymbol{\alpha}=\lambda\boldsymbol{\alpha}$，代入$\boldsymbol{\alpha}=\begin{pmatrix}1\\b\\1\end{pmatrix}$，整理得$\begin{pmatrix}3-b\\2b-2\\1-b+a\end{pmatrix}=\begin{pmatrix}\lambda\\\lambda b\\\lambda\end{pmatrix}$，所以$a=2$，$b=-1,\lambda=4$或者$a=2,b=2,\lambda=1$，因为$b<0$，所以$a=2,b=-1,\lambda=4$.

（Ⅱ）$|\boldsymbol{A}-\lambda\boldsymbol{E}|=\begin{vmatrix}2-\lambda&-1&1\\-1&2-\lambda&-1\\1&-1&2-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}1&1&0\\-1&2-\lambda&-1\\1&-1&2-\lambda\end{vmatrix}=(1-\lambda)^2(4-\lambda)$，故$\boldsymbol{A}$的特征值$\lambda_1=\lambda_2=1,\lambda_3=4$.

当$\lambda_1=\lambda_2=1$时，$\boldsymbol{A}-\boldsymbol{E}\sim\begin{pmatrix}1&-1&1\\0&0&0\\0&0&0\end{pmatrix}$，得线性无关的特征向量为$\boldsymbol{\xi}_1=\begin{pmatrix}1\\1\\0\end{pmatrix},\boldsymbol{\xi}_2=\begin{pmatrix}-1\\0\\1\end{pmatrix}$，正交单位化得$\boldsymbol{\eta}_1=\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}\\0\end{pmatrix}$，$\boldsymbol{\eta}_2=\begin{pmatrix}-\frac{1}{\sqrt{6}}\\\frac{1}{\sqrt{6}}\\\frac{2}{\sqrt{6}}\end{pmatrix}$；由题意可知，对应$\lambda_3=4$的特征向量为$\boldsymbol{\xi}_3=\boldsymbol{\alpha}=\begin{pmatrix}1\\-1\\1\end{pmatrix}$，单位化得$\boldsymbol{\eta}_3=\begin{pmatrix}\frac{1}{\sqrt{3}}\\-\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{3}}\end{pmatrix}$.令$\boldsymbol{x}=\boldsymbol{Qy}=(\boldsymbol{\eta}_1,\boldsymbol{\eta}_2,\boldsymbol{\eta}_3)\boldsymbol{y}=\begin{pmatrix}\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{6}}&-\frac{1}{\sqrt{3}}\\0&\frac{2}{\sqrt{6}}&\frac{1}{\sqrt{3}}\end{pmatrix}\boldsymbol{y}$，得标准型$f=y_1^2+y_2^2+4y_3^2$.

（Ⅲ）由$\boldsymbol{Q}$为正交矩阵，得$\boldsymbol{Q}^\text{T}\boldsymbol{Q}=\boldsymbol{E}$，故$\boldsymbol{x}^\text{T}\boldsymbol{x}=(\boldsymbol{Qy})^\text{T}\boldsymbol{Qy}=\boldsymbol{y}^\text{T}\boldsymbol{Q}^\text{T}\boldsymbol{Qy}=\boldsymbol{y}^\text{T}\boldsymbol{y}=1$，所以
$$f=y_1^2+y_2^2+4y_3^2=\boldsymbol{y}^\text{T}\boldsymbol{y}+3y_3^2=1+3y_3^2,$$
因此当$y_3=0$，$f$取最小值为1.