### 武忠祥2024决胜冲刺模拟卷一（数三）

$\text{A.}$ $f(x)$ 仅有 2 个间断点, $F(x)$ 为连续的偶函数. $\text{B.}$ $f(x)$ 仅有 2 个间断点, $F(x)$ 为连续的奇函数. $\text{C.}$ $f(x)$ 有 3 个间断点, $F(x)$ 有 3 个不可导点. $\text{D.}$ $f(x)$ 有 3 个间断点, $F(x)$ 有 2 个不可导点.

$\text{A.}$ $\left(3-2 x^2\right) \mathrm{e}^x+\mathrm{e}^{-x}$. $\text{B.}$ $3 \mathrm{e}^x+x \mathrm{e}^{-x}$. $\text{C.}$ $(3-2 x) \mathrm{e}^x+\mathrm{e}^{-x}$. $\text{D.}$ $\mathrm{e}^x+(3-2 x) \mathrm{e}^{-x}$.

$\text{A.}$ $g(0,0)=0$. $\text{B.}$ $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} g(x, y)$ 存在. $\text{C.}$ $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} g(x, y)$ 存在且 $g(0,0)=0$. $\text{D.}$ $g(x, y)$ 在点 $(0,0)$ 处连续, 且 $g(0,0)=0$.

$\text{A.}$ 1 $\text{B.}$ $x$. $\text{C.}$ $x^2$. $\text{D.}$ $\frac{x^3}{3}$.

$\text{A.}$ $\left(\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$. $\text{B.}$ $\left(\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$. $\text{C.}$ $\left(\begin{array}{lll}2 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$. $\text{D.}$ $\left(\begin{array}{lll}2 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)$.

$\text{A.}$ $r(\boldsymbol{A}, \boldsymbol{A B})=r(\boldsymbol{A})$. $\text{B.}$ $r(\boldsymbol{A}, \boldsymbol{B A})=r(\boldsymbol{A})$. $\text{C.}$ $r\left(\begin{array}{c}\boldsymbol{A} \\ \boldsymbol{A B}\end{array}\right)=r(\boldsymbol{A})$. $\text{D.}$ $r(\boldsymbol{A B})=r(\boldsymbol{B A})$.

$\text{A.}$ $\left(\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right)$. $\text{B.}$ $\left(\begin{array}{ll}2 & 0 \\ 0 & 0\end{array}\right)$. $\text{C.}$ $\left(\begin{array}{ll}3 & 0 \\ 0 & 0\end{array}\right)$. $\text{D.}$ $\left(\begin{array}{ll}-1 & -1 \\ -1 & -1\end{array}\right)$.

$\text{A.}$ $\frac{1}{4}$. $\text{B.}$ $\frac{1}{3}$. $\text{C.}$ $\frac{1}{2}$. $\text{D.}$ 1

$\text{A.}$ $a=n b$. $\text{B.}$ $b=n a$. $\text{C.}$ $a=\sqrt{n} b$. $\text{D.}$ $b=\sqrt{n} a$.

$\text{A.}$ $\frac{1}{n^2}$. $\text{B.}$ $\frac{2}{n^2}$. $\text{C.}$ $\frac{3}{n^2}$. $\text{D.}$ $\frac{4}{n^2}$.

(I) 存在 $c \in(a, b)$, 使得 $f(c)=\frac{a}{a+b}$;
(II) 存在两个不同的点 $\xi, \eta \in(a, b)$ 使得 $\frac{a}{f^{\prime}(\xi)}+\frac{b}{f^{\prime}(\eta)}=b^2-a^2$.

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