### 2024年全国硕士研究生招生考试（数学一）模拟考试

$\text{A.}$ 连续但不可导. $\text{B.}$ 可导但 $f^{\prime}(0) \neq 0$. $\text{C.}$ 极限存在但不连续. $\text{D.}$ 可微且 $\left.\mathrm{d} f(x)\right|_{x=0}=0$.

\begin{aligned} & I_1=\int_0^{2 \pi} \mathrm{d} \theta \int_0^1(\cos r+r \cos \theta+r \sin \theta) r \mathrm{~d} r, \\ & I_2=\int_0^{2 \pi} \mathrm{d} \theta \int_0^1\left(\cos r^2-r \cos \theta+r \sin \theta\right) r \mathrm{~d} r, \\ & I_3=\int_0^{2 \pi} \mathrm{d} \theta \int_0^1\left(\cos r^4-r \cos \theta-r \sin \theta\right) r \mathrm{~d} r, \end{aligned}
$\text{A.}$ $I_1 < I_2 < I_3$. $\text{B.}$ $I_2 < I_1 < I_3$. $\text{C.}$ $I_3 < I_2 < I_1$. $\text{D.}$ $I_3 < I_1 < I_2$.

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

$\text{A.}$ $\alpha>\frac{5}{2}$. $\text{B.}$ $2 < \alpha < 3$. $\text{C.}$ $\frac{1}{2} < \alpha < 1$. $\text{D.}$ $\alpha < 3$.

$\text{A.}$ 11 $\text{B.}$ 12 $\text{C.}$ 13 $\text{D.}$ 14

$$\left\{\begin{array}{l} x_1+x_2+a x_3=4 \\ x_1-x_2+2 x_3=-4 \\ -x_1+a x_2+x_3=a^2 \end{array}\right.$$

$\text{A.}$ $\text{B.}$ $\text{C.}$ $\text{D.}$

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

$\text{A.}$ $p_1 \leqslant p_2$. $\text{B.}$ $p_1 \geqslant p_2$. $\text{C.}$ $f_1(x) \leqslant f_2(x)$. $\text{D.}$ $f_1(x) \geqslant f_2(x)$.

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

$\text{A.}$ $f_Y(y)=f(y)+f(-y)$. $\text{B.}$ $f_Y(y)=\frac{f(y)+f(-y)}{2}$. $\text{C.}$ $f_Y(y)=\left\{\begin{array}{cc}f(y)+f(-y), & y>0, \\ 0, & y \leqslant 0 .\end{array}\right.$ $\text{D.}$ $f_Y(y)=\left\{\begin{array}{cc}\frac{f(y)+f(-y)}{2}, & y>0, \\ 0, & y \leqslant 0 .\end{array}\right.$

$$I=\iint_{\Sigma} 2(1-x y) \mathrm{d} y \mathrm{~d} z+(x+1) y \mathrm{~d} z \mathrm{~d} x-4 y z^2 \mathrm{~d} x \mathrm{~d} y,$$

(1) 求 $f\left(x_1, x_2, x_3\right)=0$ 的解;
(2) 设二次型 $f\left(x_1, x_2, x_3\right)$ 的规范形为 $z_1^2+z_2^2$, 求正交变换 $x=Q y$, 使得二次型 $f\left(x_1, x_2, x_3\right)$ 化为标准形.

(1) 求参数 $\theta$ 的矩估计量;
(2) 设 $U=\min \left\{X_1, X_2, \cdots, X_n\right\}$, 求 $E(U)$.

$$\lim _{x \rightarrow 0}\left(\frac{-\cot x}{\mathrm{e}^{-x}}+\frac{1}{\mathrm{e}^{-2 x} \sin ^2 x}+\frac{1}{x^2}\right)$$

• 无限看试题

• 下载试题

• 组卷