### 李永乐全国硕士研究生招生考试模拟一试卷2024版（数二）

$\text{A.}$ $f(0)=1$. $\text{B.}$ $\lim _{x \rightarrow 0} f(x)=1$. $\text{C.}$ $f^{\prime}(0)=-1$. $\text{D.}$ $f^{\prime}(0)=1$.

$\text{A.}$ $x=-1$ 为 $f(x)$ 的第一类间断点. $\text{B.}$ $x=1$ 为 $f(x)$ 的第一类间断点. $\text{C.}$ $x=-1$ 为 $f(x)$ 的第二类间断点. $\text{D.}$ $x=1$ 为 $f(x)$ 的第二类间断点.

$\text{A.}$ $f(x, y)$ 在 $(0,0)$ 点连续. $\text{B.}$ $f_x^{\prime}(0,0)=f_y^{\prime}(0,0)=0$. $\text{C.}$ $f(x, y)$ 在 $(0,0)$ 处可微. $\text{D.}$ $f(x, y)$ 在点 $(0,0)$ 处取极大值.

$\text{A.}$ 0 $\text{B.}$ 1 $\text{C.}$ 2 $\text{D.}$ 3

$\text{A.}$ 0 $\text{B.}$ $\frac{1}{2}$. $\text{C.}$ 10 $\text{D.}$ 20

$\text{A.}$ $g(0,0)=0$. $\text{B.}$ $\lim _{\substack{x \rightarrow 0 \\ y \rightarrow 0}} g(x, y)$ 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 $g(0,0)=0$.

$\text{A.}$ $\int_1^4 \mathrm{~d} y \int_{\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x$. $\text{B.}$ $\int_1^4 \mathrm{~d} y \int_{2-y}^{\sqrt{y}} f(x, y) \mathrm{d} x$. $\text{C.}$ $\int_0^1 \mathrm{~d} y \int_{\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x+\int_1^4 \mathrm{~d} y \int_{2-y}^{\sqrt{y}} f(x, y) \mathrm{d} x$. $\text{D.}$ $\int_0^1 \mathrm{~d} y \int_{\sqrt{y}}^{\sqrt{y}} f(x, y) \mathrm{d} x+\int_1^4 \mathrm{~d} y \int_{-\sqrt{y}}^{2-y} f(x, y) \mathrm{d} x$.

$\text{A.}$ 0 $\text{B.}$ 1 $\text{C.}$ 2 $\text{D.}$ 3

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

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

$\lim _{n \rightarrow \infty} n\left[\mathrm{e}\left(1+\frac{1}{n}\right)^{-n}-1\right]=$

$\int_0^1 \ln (1+\sqrt{x}) \mathrm{d} x=$ ________ .

( I ) $\frac{1}{4} \leqslant \theta(x) < \frac{1}{2} ;$
(II) $\lim _{x \rightarrow 0^{+}} \theta(x)=\frac{1}{4}, \lim _{x \rightarrow+\infty} \theta(x)=\frac{1}{2}$.

(I) 求 $a$ 的值;
(II) 矩阵 $\boldsymbol{A}$ 能否对角化? 若能, 求可逆矩阵 $\boldsymbol{P}$, 使得 $\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}$ 为对角矩阵.

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