### 2015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(数二)

$\text{A.}$ $\int_2^{+\infty} \frac{1}{\sqrt{x}} \mathrm{~d} x$ $\text{B.}$ $\int_2^{+\infty} \frac{\ln x}{x} \mathrm{~d} x$ $\text{C.}$ $\int_2^{+\infty} \frac{1}{x \ln x} \mathrm{~d} x$ $\text{D.}$ $\int_2^{+\infty} \frac{x}{e^x} \mathrm{~d} x$

$\text{A.}$ 连续 $\text{B.}$ 有可去间断点 $\text{C.}$ 有跳跃间断点 $\text{D.}$ 有无穷间断点

$\text{A.}$ $\alpha-\beta>1$ $\text{B.}$ $0 < \alpha-\beta \leq 1$ $\text{C.}$ $\alpha-\beta>2$ $\text{D.}$ $0 < \alpha-\beta \leq 2$

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

$\text{A.}$ $\frac{1}{2}, 0$ $\text{B.}$ $0, \frac{1}{2}$ $\text{C.}$ $-\frac{1}{2}, 0$ $\text{D.}$ $0,-\frac{1}{2}$

$\text{A.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$ $\text{B.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$ $\text{C.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$ $\text{D.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$

$\text{A.}$ $a \notin \Omega, d \notin \Omega$ $\text{B.}$ $a \notin \Omega, d \in \Omega$ $\text{C.}$ $a \in \Omega, d \notin \Omega$ $\text{D.}$ $a \in \Omega, d \in \Omega$

$\text{A.}$ $2 y_1^2-y_2^2+y_3^2$ $\text{B.}$ $2 y_1^2+y_2^2-y_3^2$ $\text{C.}$ $2 y_1^2-y_2^2-y_3^2$ $\text{D.}$ $2 y_1^2+y_2^2+y_3^2$

$$f_x^{\prime}(x, 0)=(x+1) e^x, f(0, y)=y^2+2 y ，$$

$$D=\left\{(x, y) \mid x^2+y^2 \leq 2, y \geq x^2\right\} .$$

$$f(a)=0, f^{\prime}(x)>0, f^{\prime \prime}(x)>0 .$$

(1) 求 $a$ 的值;
(2) 若矩阵 $\boldsymbol{X}$ 满足 $\boldsymbol{X}-\boldsymbol{X} \boldsymbol{A}^2-\boldsymbol{A} \boldsymbol{X}+\boldsymbol{A} \boldsymbol{X} \boldsymbol{A}^2=E$ ，其中 $E$ 为 3 阶单位矩阵，求 $X$

$$B=\left(\begin{array}{ccc} 1 & -2 & 0 \\ 0 & b & 0 \\ 0 & 3 & 1 \end{array}\right) \text {. }$$
(1) 求 $a, b$ 的值;
(2) 求可逆矩阵 $P$ ，使得 $P^{-1} A P$ 为对角矩阵

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