### 2022年全国硕士研究生招生统一考试数学试题及详细参考解答(数三)

(1) 若 $\alpha(x) \sim \beta(x)$, 则 $\alpha^2(x) \sim \beta^2(x)$ ；
(2) 若 $\alpha^2(x) \sim \beta^2(x)$, 则 $\alpha(x) \sim \beta(x)$
(3) 若 $\alpha(x) \sim \beta(x)$ ，则 $\alpha(x)-\beta(x) \sim o(\alpha(x))$ ；
(4) 若 $\alpha(x)-\beta(x) \sim o(\alpha(x))$, 则 $\alpha(x) \sim \beta(x)$.

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

$$F(x, y)=\int_0^{x-y}(x-y-t) f(t) \mathrm{d} t$$

$\text{A.}$ $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}, \frac{\partial^2 F}{\partial x^2}=\frac{\partial^2 F}{\partial y^2}$ $\text{B.}$ $\frac{\partial F}{\partial x}=\frac{\partial F}{\partial y}, \frac{\partial^2 F}{\partial x^2}=-\frac{\partial^2 F}{\partial y^2}$ $\text{C.}$ $\frac{\partial \boldsymbol{F}}{\partial x}=-\frac{\partial \boldsymbol{F}}{\partial y}, \frac{\partial^2 \boldsymbol{F}}{\partial x^2}=\frac{\partial^2 F}{\partial y^2}$ $\text{D.}$ $\frac{\partial F}{\partial x}=-\frac{\partial F}{\partial y}, \frac{\partial^2 F}{\partial x^2}=-\frac{\partial^2 F}{\partial y^2}$

$\text{A.}$ $I_1 < I_2 < I_3$ $\text{B.}$ $I_2 < I_1 < I_3$ $\text{C.}$ $I_1 < I_3 < I_2$ $\text{D.}$ $I_3 < I_2 < I_1$

$\text{A.}$ 存在可逆矩阵 $\boldsymbol{P}, \boldsymbol{Q}$ ，使得 $\boldsymbol{A}=\boldsymbol{P} \boldsymbol{\Lambda} \boldsymbol{Q}$ $\text{B.}$ 存在可逆矩阵 $\boldsymbol{P}$ ，使得 $\boldsymbol{A}=\boldsymbol{P} \boldsymbol{\Lambda} \boldsymbol{P}^{-1}$ $\text{C.}$ 存在正交矩阵 $Q$ ，使得 $\boldsymbol{A}=\boldsymbol{Q} \boldsymbol{\Lambda} Q^{-1}$ $\text{D.}$ 存在可逆矩阵 $\boldsymbol{P}$ ，使得 $\boldsymbol{A}=\boldsymbol{P} \boldsymbol{\Lambda} \boldsymbol{P}^T$

$\text{A.}$ 无解 $\text{B.}$ 有解 $\text{C.}$ 有无穷多解或无解 $\text{D.}$ 有唯一解或无解

$\text{A.}$ $\{0,1\}$ $\text{B.}$ $\{\lambda \mid \lambda \in R, \lambda \neq-2\}$ $\text{C.}$ $\{\lambda \mid \lambda \in R$ 且 $\lambda \neq-1, \lambda \neq-2\}$ $\text{D.}$ $\{\lambda \mid \lambda \in R$ 且 $\lambda \neq-1\}$

$\text{A.}$ 2 $\text{B.}$ 4 $\text{C.}$ 6 $\text{D.}$ 10

$\text{A.}$ $\frac{1}{8}$ $\text{B.}$ $\frac{1}{6}$ $\text{C.}$ $\frac{1}{3}$ $\text{D.}$ $\frac{1}{2}$

$\text{A.}$ -0.6 $\text{B.}$ -0.36 $\text{C.}$ 0 $\text{D.}$ 0.48

$\int_0^2 \frac{2 x-4}{x^2+2 x+4} \mathrm{~d} x=$

$\int_{-\infty}^{+\infty} \mathrm{d} x \int_{-\infty}^{+\infty} f(x) f(y-x) \mathrm{d} y=$

$$P(A)=P(B)=P(C)=\frac{1}{3}$$

$$y^{\prime}+\frac{1}{2 \sqrt{x}} y=2+\sqrt{x}, y(1)=3$$

$y=\sqrt{4-x^2}$ 以及 $x$ 轴所围成的区域。

$$f\left(x_1, x_2, x_3\right)=3 x_1^2+4 x_2^2+3 x_3^2+2 x_1 x_3 .$$

(1) 求正交变换 $X=Q Y$ 将 $f\left(x_1, x_2, x_3\right)$ 化为标准形；
(2) 证明: $\min _{x \neq 0} \frac{f(x)}{x^T x}=2$.

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