单选题 （共 15 题 ），每题只有一个选项正确
$\int_0^1 \frac{\arcsin \sqrt{x}}{\sqrt{x(1-x)}} d x=$
$\text{A.}$ $\frac{\pi^2}{4}$.
$\text{B.}$ $\frac{\pi^2}{8}$.
$\text{C.}$ $\frac{\pi}{4}$.
$\text{D.}$ $\frac{\pi}{8}$.
设 $f(x, y)$ 具有一阶偏导数, 且在任意的 $(x, y)$ 都有 $\frac{\partial f(x, y)}{\partial x}>0, \frac{\partial f(x, y)}{\partial y} < 0$,则
$\text{A.}$ $f(0,0)>f(1,1)$.
$\text{B.}$ $f(0,0) < f(1,1)$.
$\text{C.}$ $f(0,1)>f(1,0)$.
$\text{D.}$ $f(0,1) < f(1,0)$.
二元函数 $z=x y(3-x-y)$ 的极值点是
$\text{A.}$ $(0,0)$.
$\text{B.}$ $(0,3)$.
$\text{C.}$ $(3,0)$.
$\text{D.}$ $(1,1)$.
设 $z=z(x, y)$ 由 $\left\{\begin{array}{l}x=u e ^v, \\ y=u v,(u>0, v>1) \\ z=v\end{array}\right.$ 所确定, 则 $\frac{\partial^2 z}{\partial x \partial y}=$
$\text{A.}$ $\frac{x y}{z(1-z)^3}$.
$\text{B.}$ $\frac{x y}{z(z-1)^3}$.
$\text{C.}$ $\frac{z}{x y(1-z)^3}$.
$\text{D.}$ $\frac{z}{x y(z-1)^3}$.
设 $f^{\prime}\left(x_0\right)$ 存在, 则 $\lim _{\Delta x \rightarrow 0} \frac{f\left(x_0-\Delta x\right)-f\left(x_0\right)}{\Delta x}=$.
$\text{A.}$ $f^{\prime}\left(x_0\right)$
$\text{B.}$ $-f^{\prime}\left(x_0\right)$
$\text{C.}$ $2 f^{\prime}\left(x_0\right)$
$\text{D.}$ 不存在
(1) 设 $f(x)$ 满足 $\lim _{x \rightarrow 0} \frac{\sqrt{1+f(x) \sin 2 x}-1}{e^{x^2}-1}=1$, 则( )
$\text{A.}$ $f(0)=0$
$\text{B.}$ $\lim _{x \rightarrow 0} f(x)=0$
$\text{C.}$ $f^{\prime}(0)=1$
$\text{D.}$ $\lim _{x \rightarrow 0} f^{\prime}(x)=1$
已知 $f(x)=\max \left\{1, x^2\right\}$, 则 $\int f(x) d x=$
$\text{A.}$ $\begin{cases}\frac{x^3}{3}-\frac{2}{3}+C, & x < -1 \\ x+C, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+\frac{2}{3}+C, & x>1\end{cases}$
$\text{B.}$ $\left\{\begin{array}{cc}\frac{x^3}{3}+C, & x < -1 \\ x+C, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+C, & x>1\end{array}\right.$
$\text{C.}$ $\left\{\begin{array}{cc}\frac{x^3}{3}+C_1, & x < -1 \\ x+C_2, & -1 \leq x \leq 1 \\ \frac{x^3}{3}+C_3, & x>1\end{array}\right.$
$\text{D.}$ $\left\{\begin{array}{l}\frac{x^3}{3}-\frac{4}{3}+C, \quad x < -1 \\ x+C, \quad-1 \leq x \leq 1 \\ \frac{x^3}{3}+\frac{2}{3}+C, \quad x>1\end{array}\right.$
设连续函数 $f(x, y)$ 满足 $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} \frac{f(x, y)-x-2 y-4}{x^2+y^2}=-1$, 则 $\lim _{h \rightarrow 0} \frac{f(2 h, 0)-f(0,-h)}{h}=($ )
$\text{A.}$ -1
$\text{B.}$ 2
$\text{C.}$ 3
$\text{D.}$ 4
设 $f(x)=\int_{-1}^x t \cos t d t, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, 则曲线 $y=f(x)$ 与 $x$ 轴所围图形的面积为 $(\quad)$
$\text{A.}$ $2 \int_0^1 x \sin x d x$.
$\text{B.}$ $2 \int_0^1 x^2 \sin x d x$.
$\text{C.}$ $2 \int_0^1 x \cos x d x$.
$\text{D.}$ $2 \int_0^1 x^2 \cos x d x$.
$f(x)$ 当 $x \rightarrow x_0$ 时的右极限 $f\left(x_0^{+}\right)$和左极限 $f\left(x_0^{-}\right)$存在且相等是 $\lim _{x \rightarrow x_0} f(x)$ 存在的 $\qquad$条件
$\text{A.}$ 必要
$\text{B.}$ 充分
$\text{C.}$ 充要
$\text{D.}$ 充分不必要
若 $f\left(x, x^2\right)=x^2 e ^{-x},\left.f_x^{\prime}(x, y)\right|_{y=x^2}=-x^2 e ^{-x}$, 则当 $x \neq 0$ 时, $\left.f_y^{\prime}(x, y)\right|_{y=x^2}=$ ( ).
$\text{A.}$ $2 x e ^{-x}$
$\text{B.}$ $\left(-x^2+2 x\right) e ^{-x}$
$\text{C.}$ $e^{-x}$
$\text{D.}$ $(2 x-1) e ^{-x}$
设 $f(x)=\int_0^x\left( e ^{\cos t} \cos t-k\right) d t$, 若积分 $\int_a^{a+2 \pi} f(x) d x$ 的值与 $a$ 无关, 则 $k=(\quad)$.
$\text{A.}$ $\int_0^{2 \pi} e ^{\cos x} \cos x d x$
$\text{B.}$ $\frac{1}{2 \pi} \int_0^{2 \pi} e ^{\cos x} \cos x d x$
$\text{C.}$ $\int_0^\pi e^{\cos x} \cos x d x$
$\text{D.}$ 0
设 $f$ 为二元可微函数, $z=y f\left(\frac{y}{x}, x y\right)$, 则 $\frac{x}{y} \cdot \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=$
$\text{A.}$ $f+2 \frac{x}{y} \cdot f_1^{\prime}$
$\text{B.}$ $f-2 \frac{x}{y} \cdot f_1^{\prime}$
$\text{C.}$ $f+2 x y f_2^{\prime}$
$\text{D.}$ $f-2 x y f_2^{\prime}$
设 $f(x)=\frac{(x+1) \sin (x-1)}{x(x-1)^2}$, 则 $x=1$ 是 $f(x)$ 的 ( ).
$\text{A.}$ 跳跃间断点
$\text{B.}$ 连续点
$\text{C.}$ 可去间断点
$\text{D.}$ 无穷间断点
设 $f(x)=\frac{1+e^{-x^2}}{1-e^{-x^2}}$, 则曲线 $f(x)$ ().
$\text{A.}$ 仅有水平渐近线
$\text{B.}$ 仅有铅直渐近线
$\text{C.}$ 既有水平渐近线又有铅直渐近线
$\text{D.}$ 没有渐近线