\(\int \frac{x}{e^{x}} d x=-(x+1) e^{-x}\) ,则 \(\int_{2}^{+\infty} \frac{x}{e^{x}} d x=-(x+1) e^{-x}|_{2} ^{+\infty}=3 e^{-2}-\lim _{x \to +\infty}(x+1) e^{-x}=3 e^{-2}\)

\(f(x)=\lim _{t \to 0}(1+\frac{\sin t}{x})^{\frac{x^{2}}{t}}=e^{\lim _{t \to 0} \frac{\sin t x^{2}}{x} t}=e^{x}\) ，\(x ≠0\) ，故 \(f(x)\) 有可去间断点 \(x=0\)

\(x<0\) 时, \(f'(x)=0\) ，\(f_{-}'(0)=0\)

\[f_{+}'(0)=\lim _{x \to 0^{+}} \frac{x^{\alpha} \cos \frac{1}{x^{\beta}}-0}{x}=\lim _{x \to 0^{+}} x^{\alpha-1} \cos \frac{1}{x^{\beta}}\]

\[

\begin{aligned}

& x>0 时, f'(x)=\alpha x^{\alpha-1} \cos \frac{1}{x^{\beta}}+(-1) x^{\alpha} \sin \frac{1}{x^{\beta}}(-\beta) \frac{1}{x^{\beta+1}} \\

& =\alpha x^{\alpha-1} \cos \frac{1}{x^{\beta}}+\beta x^{\alpha-\beta-1} \sin \frac{1}{x^{\beta}}

\end{aligned}

\]

\(f'(x)\) 在 \(x=0\) 处连续则: \(f_{-}'(0)=f_{+}'(0)=\lim _{x \to 0^{+}} x^{\alpha-1} \cos \frac{1}{x^{\beta}}=0\) 得 \(\alpha-1>0\)

\[f'(0)=\lim _{x \to 0^{+}} f'(x)=\lim _{x \to 0^{+}}\left(\alpha x^{\alpha-1} \cos \frac{1}{x^{\beta}}+\beta x^{\alpha-\beta-1} \sin \frac{1}{x^{\beta}}\right)=0\]

得: \(\alpha-\beta-1>0\) ,答案选择 A

根据图像观察存在两点,二阶导数变号.则拐点个数为2个。

此题考查二元复合函数偏导的求解。

令 \(u=x+y\) ，\(v=\frac{y}{x}\) ，则 \(x=\frac{u}{1+v}\) ，\(y=\frac{u v}{1+v}\) ，从而 \(f(x+y, \frac{y}{x})=x^{2}-y^{2}\) 变为

\[f(u, v)=\left(\frac{u}{1+v}\right)^{2}-\left(\frac{u v}{1+v}\right)^{2}=\frac{u^{2}(1-v)}{1+v}.\] 故 \(\frac{\partial f}{\partial u}=\frac{2 u(1-v)}{1+v}\) ，\(\frac{\partial f}{\partial v}=-\frac{2 u^{2}}{(1+v)^{2}}\) ，

因而 \(\left.\frac{\partial f}{\partial u}\right|_{\substack{u=1 \\ v=1}}=0\) ，\(\left.\frac{\partial f}{\partial v}\right|_{\substack{u=1 \\ v=1}}=-\frac{1}{2}\) ，故选(D)

根据图可得,在极坐标系下计算该二重积分的积分区域为

\[D=\left\{(r, \theta) | \frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}, \frac{1}{\sqrt{2 \sin 2 \theta}} \leq r \leq \frac{1}{\sqrt{\sin 2 \theta}}\right\}\]

所以

\[\iint_{D} f(x, y) d x d y=\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} d \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r d r\]

故选B.

\[(A, b)=\left(\begin{array}{cccc}1 & 1 & 1 & 1 \\ 1 & 2 & a & d \\ 1 & 4 & a^{2} & d^{2}\end{array}\right) \to\left(\begin{array}{cccc}1 & 1 & 1 & 1 \\ 0 & 1 & a-1 & d-1 \\ 0 & 0 & (a-1)(a-2) & (d-1)(d-2)\end{array}\right),\]

由 \(r(A)=r(A, b)<3\) ,故 \(a=1\) 或 \(a=2\) ,同时 \(d=1\) 或 \(d=2\) 。故选(D)

由 \(x=P y\) ,故 \(f=x^{T} A x=y^{T}(P^{T} A P) y=2 y_{1}^{2}+y_{2}^{2}-y_{3}^{2}\) 且

\[P^{T} A P=\left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right)\]

\[Q=P\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0\end{array}\right)=P C\]

\[Q^{T} A Q=C^{T}\left(P^{T} A P\right) C=\left(\begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{array}\right)\]

所以 \(f=x^{T} A x=y^{T}(Q^{T} A Q) y=2 y_{1}^{2}-y_{2}^{2}+y_{3}^{2}\) 。选(A)

$\frac{d y}{d x}=\frac{d y}{d t} / \frac{d x}{d t}=\frac{3+3 t^{2}}{\frac{1}{1+t^{2}}}=3\left(1+t^{2}\right)^{2}$

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left[3\left(1+t^{2}\right)^{2}\right]=\frac{d\left[3\left(1+t^{2}\right)^{2}\right]}{d t} / \frac{d x}{d t}=\frac{12 t\left(1+t^{2}\right)}{\frac{1}{1+t^{2}}}=12 t\left(1+t^{2}\right)^{2}$

$\left.\frac{d^{2} y}{d x^{2}}\right|_{t=1}=48$

根据莱布尼茨公式得：

$f^{(n)}(0)=\left.C_{n}^{2} 2\left(2^{x}\right)^{(n-2)}\right|_{x=0}=\frac{n(n-1)}{2} 2(\ln 2)^{n-2}=n(n-1)(\ln 2)^{n-2}$

已知$\varphi(x)=x \int_{0}^{x^{2}} f(t) d t$，求导得$\varphi'(x)=\int_{0}^{x^{2}} f(t) d t+2 x^{2} f(x^{2})$，故有$\varphi(1)=\int_{0}^{1} f(t) d t=1$，$\varphi'(1)=1+2 f(1)=5$，则$f(1)=2$

由题意知：$y(0)=3$，$y'(0)=0$，由特征方程：$\lambda^{2}+\lambda-2=0$解得$\lambda_{1}=1$，$\lambda_{2}=-2$

所以微分方程的通解为：$y=C_{1} e^{x}+C_{2} e^{-2 x}$，代入$y(0)=3$，$y'(0)=0$解得：$C_{1}=2$，$C_{2}=1$

解得：$y=2 e^{x}+e^{-2 x}$

当$x=0$，$y=0$时$z=0$，两边对$x$求偏导可得$(3 e^{x+2 y+3 z}+x y) \frac{\partial z}{\partial x}=-y z-e^{x+2 y+3 z}$，对$y$求偏导可得$(3 e^{x+2 y+3 z}+x y) \frac{\partial z}{\partial y}=-x z-2 e^{x+2 y+3 z}$，将$(0,0,0)$代入得$\left.\frac{\partial z}{\partial x}\right|_{(0,0)}=-\frac{1}{3}$，$\left.\frac{\partial z}{\partial y}\right|_{(0,0)}=-\frac{2}{3}$，则可得$\left.d z\right|_{(0,0)}=-\frac{1}{3} d x-\frac{2}{3} d y=-\frac{1}{3}(d x+2 d y)$

$A$的所有特征值为2，-2，1，$B$的所有特征值为$2^{2}-2+1=3$，$(-2)^{2}-(-2)+1=7$，$1^{2}-1+1=1$，所以$|B|=3×7×1=21$

【答案】 \(a=-1, k=-\frac{1}{3}, b=-\frac{1}{2}\)

【解析】

方法一:

\[因为 ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+o\left(x^{3}\right), sin x=x-\frac{x^{3}}{3 !}+o\left(x^{3}\right),\]

那么,

\[1=lim _{x \to 0} \frac{f(x)}{g(x)}=lim _{x \to 0} \frac{x+a ln (1+x)+b x sin x}{k x^{3}}=lim _{x \to 0} \frac{(1+a) x+\left(b-\frac{a}{2}\right) x^{2}+\frac{a}{3} x^{3}+o\left(x^{3}\right)}{k x^{3}},\]

\[可得: \left\{\begin{array}{l}1+a=0 \\ b-\frac{a}{2}=0, 所以, \left\{\begin{array}{l}a=-1 \\ b=-\frac{1}{2} \\ k=-\frac{1}{3}\end{array}\right.\end{array}\right.\]

方法二:

由题意得

\[1=lim _{x \to 0} \frac{f(x)}{g(x)}=lim _{x \to 0} \frac{x+a ln (1+x)+b x sin x}{k x^{3}}=lim _{x \to 0} \frac{1+\frac{a}{1+x}+b sin x+b x cos x}{3 k x^{2}}\]

由分母 \(\lim _{x \to 0} 3 k x^{2}=0\) ,得分子 \(\lim _{x \to 0}(1+\frac{a}{1+x}+b sin x+b x cos x)=\lim _{x \to 0}(1+a)=0\) ,求得

\[于是 1=lim _{x \to 0} \frac{f(x)}{g(x)}=lim _{x \to 0} \frac{1-\frac{1}{1+x}+b sin x+b x cos x}{3 k x^{2}}\]

\[\begin{aligned} & =lim _{x \to 0} \frac{x+b(1+x) sin x+b x(1+x) cos x}{3 k x^{2}(1+x)} \\ & =lim _{x \to 0} \frac{x+b(1+x) sin x+b x(1+x) cos x}{3 k x^{2}} \\ & =lim _{x \to 0} \frac{1+b sin x+b(1+x) cos x+b(1+x) cos x+b(1+x) cos x+b x cos x-b x(1+x) sin x}{6 k x} \end{aligned}\]

由分母 \(\lim _{x \to 0} 6 k x=0\) ,得分子

\[lim _{x \to 0}[1+b sin x+2 b(1+x) cos x+b x cos x-b x(1+x) sin x]=lim _{x \to 0}(1+2 b cos x)=0,\]

求得 \(b=-\frac{1}{2}\) 进一步, b 值代入原式

\[\begin{aligned} & 1=lim _{x \to 0} \frac{f(x)}{g(x)}=lim _{x \to 0} \frac{1-\frac{1}{2} sin x-(1+x) cos x-\frac{1}{2} x cos x+\frac{1}{2} x(1+x) sin x}{6 k x} \\ & =lim _{x \to 0} \frac{-\frac{1}{2} cos x-cos x+(1+x) sin x-\frac{1}{2} cos x+\frac{1}{2} x sin x+\frac{1}{2}(1+x) sin x+\frac{1}{2} x sin x+\frac{1}{2} x(1+x) cos x}{6 k} \\ & =\frac{-\frac{1}{2}}{6 k}, 求得 k=-\frac{1}{3} . \end{aligned}\]

【答案】 \(\frac{8}{\pi}\)

【解析】由旋转体的体积公式,得

\[V_{1}=\int_{0}^{\frac{\pi}{2}} \pi f^{2}(x) d x=\int_{0}^{\frac{\pi}{2}} \pi(A sin x)^{2} d x=\pi A^{2} \int_{0}^{\frac{\pi}{2}} \frac{1-cos 2 x}{2} d x=\frac{\pi^{2} A^{2}}{4}\]

\[V_{2}=\int_{0}^{\frac{\pi}{2}} 2 \pi x f(x) d x=-2 \pi A \int_{0}^{\frac{\pi}{2}} x d cos x=2 \pi A\]

由题 \(V_{1}=V_{2}\) 求得 \(A=\frac{8}{\pi}\)

【答案】极小值 \(f(0,-1)=-1\)

【解析】 \(f_{x y}^{\prime \prime}(x, y)=2(y+1) e^{x}\) 两边对 y 积分,得

\[f_{x}'(x, y)=2\left(\frac{1}{2} y^{2}+y\right) e^{x}+\varphi(x)=\left(y^{2}+2 y\right) e^{x}+\varphi(x),\]

故 \(f_{x}'(x, 0)=\varphi(x)=(x+1) e^{x}\) ,

\[求得 \varphi(x)=e^{x}(x+1),\]

故 \(f_{x}'(x, y)=(y^{2}+2 y) e^{x}+e^{x}(1+x)\) ,两边关于 x 积分,得

\[\begin{aligned} f(x, y) & =\left(y^{2}+2 y\right) e^{x}+\int e^{x}(1+x) d x \\ & =\left(y^{2}+2 y\right) e^{x}+\int(1+x) d e^{x} \\ & =\left(y^{2}+2 y\right) e^{x}+(1+x) e^{x}-\int e^{x} d x \\ & =\left(y^{2}+2 y\right) e^{x}+(1+x) e^{x}-e^{x}+C \\ & =\left(y^{2}+2 y\right) e^{x}+x e^{x}+C \end{aligned}\]

由 \(f(0, y)=y^{2}+2 y+C=y^{2}+2 y\) ,求得 \(C=0\) 所以 \(f(x, y)=(y^{2}+2 y) e^{x}+x e^{x}\)

\[令 \left\{\begin{array}{c}f_{x}'=\left(y^{2}+2 y\right) e^{x}+e^{x}+x e^{x}=0 \\ f_{y}'=(2 y+2) e^{x}=0\end{array}\right., 求得 \left\{\begin{array}{c}x=0 \\ y=-1\end{array}\right..\]

\[又 f_{x x}''=\left(y^{2}+2 y\right) e^{x}+2 e^{x}+x e^{x},\]

\[f_{x y}''=2(y+1) e^{x}, f_{y y}''=2 e^{x},\]

当 \(x=0\) , \(y=-1\) 时, \(A=f_{x x}^{\prime \prime}(0,-1)=1\) , \(B=f_{x y}^{\prime \prime}(0,-1)=0\) , \(C=f_{y y}^{\prime \prime}(0,-1)=2\) , \(A C-B^{2}>0\) , \(f(0,-1)=-1\) 为极小值.

【答案】 \(\frac{\pi}{4}-\frac{2}{5}\)

\[\begin{aligned} 【解析】 \iint_{D} x(x+y) d x d y & =\iint_{D} x^{2} d x d y \\ & =2 \int_{0}^{1} d x \int_{x^{2}}^{\sqrt{2-x^{2}}} x^{2} d y \\ & =2 \int_{0}^{1} x^{2}\left(\sqrt{2-x^{2}}-x^{2}\right) d x \\ & =2 \int_{0}^{1} x^{2} \sqrt{2-x^{2}} d x-\frac{2}{5}=\sqrt{2} \sin t 2 \int_{0}^{\frac{\pi}{4}} 2 \sin ^{2} t 2 \cos ^{2} t d t-\frac{2}{5} \\ & =2 \int_{0}^{\frac{\pi}{4}} \sin ^{2} 2 t d t-\frac{2}{5}=\int_{0}^{\frac{\pi}{2}} \sin ^{2} u d u-\frac{2}{5}=\frac{\pi}{4}-\frac{2}{5} . \end{aligned}\]

【答案】2个

\[【解析】 f'(x)=-\sqrt{1+x^{2}}+2 x \sqrt{1+x^{2}}=\sqrt{1+x^{2}}(2 x-1)\]

令 \(f'(x)=0\) ,得驻点为 \(x=\frac{1}{2}\)

在 \((-\infty, \frac{1}{2})\) , \(f(x)\) 单调递减，在 \((\frac{1}{2},+\infty)\) , \(f(x)\) 单调递增

故 \(f(\frac{1}{2})\) 为唯一的极小值,也是最小值。

\[而 \begin{aligned} f\left(\frac{1}{2}\right) & =\int_{\frac{1}{2}}^{1} \sqrt{1+t^{2}} d t+\int_{1}^{\frac{1}{4}} \sqrt{1+t} d t=\int_{\frac{1}{2}}^{1} \sqrt{1+t^{2}} d t-\int_{\frac{1}{4}}^{1} \sqrt{1+t} d t \\ & =\int_{\frac{1}{2}}^{1} \sqrt{1+t^{2}} d t-\int_{\frac{1}{2}}^{1} \sqrt{1+t} d t-\int_{\frac{1}{4}}^{\frac{1}{2}} \sqrt{1+t} d t \end{aligned}\]

在 \((\frac{1}{2}, 1)\) \(\sqrt{1+t^{2}}<\sqrt{1+t}\) 故 \(\int_{\frac{1}{2}}^{1} \sqrt{1+t^{2}} d t-\int_{\frac{1}{2}}^{1} \sqrt{1+t} d t<0\) 从而有 \(f(\frac{1}{2})<0\)

\[lim _{x \to-\infty} f(x)=lim _{x \to-\infty}\left[\int_{x}^{1} \sqrt{1+t^{2}} d t+\int_{1}^{x^{2}} \sqrt{1+t} d t\right]=+\infty\]

\[lim _{x \to+\infty} f(x)=lim _{x \to+\infty}\left[\int_{x}^{1} \sqrt{1+t^{2}} d t+\int_{1}^{x^{2}} \sqrt{1+t} d t\right]=lim _{x \to+\infty}\left[\int_{1}^{x^{2}} \sqrt{1+t} d t-\int_{1}^{x} \sqrt{1+t^{2}} d t\right]\]

而 \(\lim _{x \to +\infty} \frac{\int_{1}^{x^{2}} \sqrt{1+t} d t}{\int_{1}^{x} \sqrt{1+t^{2}} d t}=\lim _{x \to +\infty} \frac{2 x \sqrt{1+x^{2}}}{\sqrt{1+x^{2}}}=+\infty\) 因此 \(\lim _{x \to +\infty} f(x)=+\infty\) 所以函数 \(f(x)\) 在 \((-\infty, \frac{1}{2})\) 及 \((\frac{1}{2},+\infty)\) 上各有一个零点,所以零点个数为2

【答案】30min

【解析】设 t 时刻物体温度为 \(x(t)\) ,比例常数为 \(k(>0)\) ,介质温度为 m ,则 \(\frac{d x}{d t}=-k(x-m)\) .从而 \(x(t)=C e^{-k t}+m\) \(x(0)=120\) , \(m=20\) ,所以 \(C=100\) ,即 \(x(t)=100 e^{-k t}+20\) 又 \(x(\frac{1}{2})=30\) 所以 \(k=2 \ln 10\) ,所以 \(x(t)=\frac{1}{100^{t-1}}+20\) 当 \(x=21\) 时, \(t=1\) ,所以还需要冷却30min

【证明】根据题意得点 \((b, f(b))\) 处的切线方程为 \(y-f(b)=f'(b)(x-b)\) 令 \(y=0\) 得 \(x_{0}=b-\frac{f(b)}{f'(b)}\)
因为 \(f'(x)>0\) 所以 \(f(x)\) 单调递增,又因为 \(f(a)=0\) 所以 \(f(b)>0\) ,又因为 \(f'(b)>0\) 
\[所以 x_{0}=b-\frac{f(b)}{f'(b)}<b\]
又因为 \(x_{0}-a=b-a-\frac{f(b)}{f'(b)}\) 在区间 \((a,b)\) 上应用拉格朗日中值定理有 
\[\frac{f( b)-f(a)}{b-a}=f'(\xi), \xi \in(a, b)\]

\[所以 x_{0}-a=b-a-\frac{f(b)}{f'(b)}=\frac{f(b)}{f'(\xi)}-\frac{f(b)}{f'(b)}=f(b) \frac{f'(b)-f'(\xi)}{f'(b) f'(\xi)}\]
因为 \(f^{\prime \prime}(x)>0\) 所以 \(f'(x)\) 单调递增 所以 \(f'(b)>f'(\xi)\) 所以 \(x_{0}-a>0\) ,即 \(x_{0}>a\) ,所以 \(a<x_{0}<b\) ,结论得证.

(1) \(A^{3}=O \Rightarrow |A|=0 \Rightarrow \begin{vmatrix}a & 1 & 0 \\ 1 & a & -1 \\ 0 & 1 & a\end{vmatrix}=\begin{vmatrix}0 & 1 & 0 \\ 1 - a^{2} & a & -1 \\ -a & 1 & a\end{vmatrix}=a^{3}=0 \Rightarrow a=0\)

(2)由题意知

\[

\begin{aligned}

& X - XA^{2} - AX + AXA^{2} = E \Rightarrow X(E - A^{2}) - AX(E - A^{2}) = E \\

& \Rightarrow (E - A)X(E - A^{2}) = E \Rightarrow X = (E - A)^{-1}(E - A^{2})^{-1} = [(E - A^{2})(E - A)]^{-1} \\

& \Rightarrow X = (E - A^{2} - A)^{-1}

\end{aligned}

\]

\[E - A^{2} - A = \begin{pmatrix}0 & -1 & 1 \\ -1 & 1 & 1 \\ -1 & -1 & 2\end{pmatrix}\]，通过初等行变换求逆可得

\[

\therefore X = \begin{pmatrix}2 & 0 & -1 \\ -1 & 1 & -1 \\ 2 & 1 & -1\end{pmatrix}

\]

【解析】(I) \(A\sim B\Rightarrow tr(A)=tr(B)\Rightarrow 3 + a = 1 + b + 1\)
\(\vert A\vert=\vert B\vert\Rightarrow\begin{vmatrix}0&2&-3\\-1&3&-3\\1&-2&a\end{vmatrix}=\begin{vmatrix}1&-2&0\\0&b&0\\0&3&1\end{vmatrix}\)
\(\therefore\begin{cases}a - b = -1\\2a - b = 3\end{cases}\Rightarrow\begin{cases}a = 4\\b = 5\end{cases}\)

(II)
\(A=\begin{pmatrix}0&2&-3\\-1&3&-3\\1&-2&3\end{pmatrix}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}+\begin{pmatrix}-1&2&-3\\-1&2&-3\\1&-2&3\end{pmatrix}=E + C\)
\(C=\begin{pmatrix}-1&2&-3\\-1&2&-3\\1&-2&3\end{pmatrix}=\begin{pmatrix}-1\\-1\\1\end{pmatrix}(1\ -2\ 3)\)
\(C\)的特征值\(\lambda_1 = \lambda_2 = 0,\lambda_3 = 4\)
\(\lambda = 0\)时\((0E - C)x = 0\)的基础解系为\(\xi_1=(2,1,0)^T;\xi_2=(-3,0,1)^T\) 
\(\lambda = 5\)时\((4E - C)x = 0\)的基础解系为\(\xi_3=(-1,-1,1)^T\) 
\(A\)的特征值\(\lambda_A = 1 + \lambda_C:1,1,5\) 
令\(P = (\xi_1,\xi_2,\xi_3)=\begin{pmatrix}2&-3&-1\\1&0&-1\\0&1&1\end{pmatrix}\), 
\(\therefore P^{-1}AP=\begin{pmatrix}1&&\\&1&\\&&5\end{pmatrix}\)