2025 年考研数学二真题解析

2025 年全国硕士研究生招生考试

数学（二）

科目代码：302

考生注意事项

1. 答题前，考生须在答题卡指定位置上填写考生编号和考生姓名；在答题卡指定位置上填写报考单位、考生姓名和考生编号，并涂写考生编号信息点.
2. 选择题的答案必须涂写在答题卡相应题号的选项上，非选择题的答案必须书写在答题卡指定位置的边框区域内.超出答题区域书写的答案无效；在草稿纸、试题册上答题无效.
3. 填（书）写部分必须使用黑色字迹签字笔书写，字迹工整、笔记清楚；涂写部分必须使用 2B 铅笔填涂.
4. 考试结束，将答题卡和试题册按规定交回.

一、选择题

第 1～10 小题，每小题 5 分，共 50 分.下列每题给出的四个选项中，只有一个选项是符合题目要求的.

1

设函数 \(z=z\left( x,y \right)\) 由 \(z+\ln z-\int_{y}^{x} e^{-t^{2}} \mathrm{~d} t=0\) 确定，则 \(\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=\)

A. \(\dfrac{z}{z+1}\left( e^{-x^{2}}-e^{-y^{2}} \right)\)

B. \(\dfrac{z}{z+1}\left( e^{-x^{2}}+e^{-y^{2}} \right)\)

C. \(-\dfrac{z}{z+1}\left( e^{-x^{2}}-e^{-y^{2}} \right)\)

D. \(-\dfrac{z}{z+1}\left( e^{-x^{2}}+e^{-y^{2}} \right)\)

答案：A

解析：

原式两边对 \(x\) 求导得 \(\dfrac{\partial z}{\partial x}+\dfrac{1}{z}\dfrac{\partial z}{\partial x}-e^{-x^{2}}=0\).

原式两边对 \(y\) 求导得 \(\dfrac{\partial z}{\partial y}+\dfrac{1}{z}\dfrac{\partial z}{\partial y}+e^{-y^{2}}=0\).

两式相加得 \(\dfrac{\partial z}{\partial x}+\dfrac{\partial z}{\partial y}=\dfrac{z}{z+1}\left( e^{-x^{2}}-e^{-y^{2}} \right)\).

2

已知函数 \(f\left( x \right)=\int_{0}^{x} e^{t^{2}}\sin t \mathrm{~d} t\)，\(g\left( x \right)=\int_{0}^{x} e^{t^{2}} \mathrm{~d} t \cdot \sin^{2} x\)，则

A. \(x=0\) 是 \(f\left( x \right)\) 的极值点，也是 \(g\left( x \right)\) 的极值点.

B. \(x=0\) 是 \(f\left( x \right)\) 的极值点，\(\left( 0,0 \right)\) 是曲线 \(y=g\left( x \right)\) 的拐点.

C. \(x=0\) 是 \(f\left( x \right)\) 的极值点，\(\left( 0,0 \right)\) 是曲线 \(y=f\left( x \right)\) 的拐点.

D. \(\left( 0,0 \right)\) 是曲线 \(y=f\left( x \right)\) 的拐点，也是曲线 \(y=g\left( x \right)\) 的拐点.

答案：B

解析：

\(f^{\prime}\left( x \right)=e^{x^{2}}\sin x\)，\(f^{\prime\prime}\left( x \right)=e^{x^{2}}\cdot 2x\sin x+e^{x^{2}}\cos x\).

\(g^{\prime}\left( x \right)=e^{x^{2}}\sin^{3} x+\int_{0}^{x} e^{t^{2}} \mathrm{~d} t \cdot 2\sin x\cos x\).

\(g^{\prime\prime}\left( x \right)=e^{x^{2}}\cdot 2x\sin^{3} x+e^{x^{2}}\cdot 3\sin^{2} x\cos x+2e^{x^{2}}\sin x\cos x+\int_{0}^{x} e^{t^{2}} \mathrm{~d} t \cdot 2\cos 2x\).

\(f^{\prime}\left( 0 \right)=0\)，\(f^{\prime\prime}\left( 0 \right)=1>0\)，所以 \(x=0\) 是 \(f\left( x \right)\) 的极值点，但不是拐点.

\(g^{\prime}\left( 0 \right)=0\)，\(g^{\prime\prime}\left( 0 \right)=0\).

继续求导可得 \(g^{\prime\prime\prime}\left( x \right)=2e^{x^{2}}\sin^{3} x+x\left( 2e^{x^{2}}\sin^{3} x \right)+4\cos 2x , e^{x^{2}}+2\sin 2x\left( e^{x^{2}} \right)-4\sin 2x\int_{0}^{x} e^{t^{2}} \mathrm{~d} t+2\cos 2x , e^{x^{2}}\).

\(g^{\prime\prime\prime}\left( 0 \right)=6>0\)，故过 \(x=0\) 为 \(g\left( x \right)\) 的拐点.

3

如果对微分方程 \(y^{\prime\prime}-2ay^{\prime}+\left( a+2 \right)y=0\) 的任一解 \(y\left( x \right)\)，反常积分 \(\int_{0}^{+\infty} y\left( x \right) \mathrm{~d} x\) 均收敛，那么 \(a\) 的取值范围是

A. \(\left( -2,-1 \right]\)

B. \(\left( -\infty,-1 \right]\)

C. \(\left( -2,0 \right)\)

D. \(\left( -\infty,0 \right)\)

答案：C

解析：

特征方程为 \(r^{2}-2ar+\left( a+2 \right)=0\).

要使任一解对应的反常积分 \(\int_{0}^{+\infty} y\left( x \right) \mathrm{~d} x\) 均收敛，须有两个特征根都小于 \(0\)，即 \(\dfrac{a+2}{1}>0\)，\(a<0\)，\(\dfrac{-2a}{1}<0\).

故 \(-2<a<0\).

4

设函数 \(f\left( x \right)\)，\(g\left( x \right)\) 在 \(x=0\) 的某去心邻域内有定义且恒不为零.若当 \(x\to 0\) 时，\(f\left( x \right)\) 是 \(g\left( x \right)\) 的高阶无穷小，则当 \(x\to 0\) 时，

A. \(f\left( x \right)+g\left( x \right)=o\left( g\left( x \right) \right)\)

B. \(f\left( x \right)g\left( x \right)=o\left( f^{2}\left( x \right) \right)\)

C. \(f\left( x \right)=o\left( e^{g\left( x \right)}-1 \right)\)

D. \(f\left( x \right)=o\left( g^{2}\left( x \right) \right)\)

答案：C

解析：

由题意知 \(f\left( x \right)=o\left( g\left( x \right) \right)\)，即 \(\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=0\).

对于 A，\(\lim\limits_{x\to 0}\dfrac{f\left( x \right)+g\left( x \right)}{g\left( x \right)}=0+1=1\)，不是无穷小.

对于 B，\(\lim\limits_{x\to 0}\dfrac{f\left( x \right)g\left( x \right)}{f^{2}\left( x \right)}=\lim\limits_{x\to 0}\dfrac{g\left( x \right)}{f\left( x \right)}=\infty\).

对于 C，\(\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{e^{g\left( x \right)}-1}=\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{g\left( x \right)}=0\).

对于 D，\(\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{g^{2}\left( x \right)}=\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{g\left( x \right)}\cdot \dfrac{1}{g\left( x \right)}\)，不是未定式，不能推出结论.

5

设函数 \(f\left( x,y \right)\) 连续，则 \(\int_{-2}^{2} \mathrm{d} x \int_{4-x^{2}}^{2} f\left( x,y \right)\mathrm{d} y=\)

A. \(\int_{0}^{4}\left[ \int_{-2}^{-\sqrt{4-y}} f\left( x,y \right)\mathrm{d} x+\int_{\sqrt{4-y}}^{2} f\left( x,y \right)\mathrm{d} x \right]\mathrm{d} y\)

B. \(\int_{0}^{4}\left[ \int_{-2}^{\sqrt{4-y}} f\left( x,y \right)\mathrm{d} x+\int_{\sqrt{4-y}}^{2} f\left( x,y \right)\mathrm{d} x \right]\mathrm{d} y\)

C. \(\int_{0}^{4}\left[ \int_{-2}^{-\sqrt{4-y}} f\left( x,y \right)\mathrm{d} x+\int_{2}^{\sqrt{4-y}} f\left( x,y \right)\mathrm{d} x \right]\mathrm{d} y\)

D. \(2\int_{0}^{4} \mathrm{d} y \int_{\sqrt{4-y}}^{2} f\left( x,y \right)\mathrm{d} x\)

答案：A

解析：

原式表示的区域为 \(-2\le x\le 2\)，\(4-x^{2}\le y\le 2\).交换积分次序后可得

\(\int_{-2}^{2} \mathrm{d} x \int_{4-x^{2}}^{2} f\left( x,y \right)\mathrm{d} y=\int_{0}^{4}\left[ \int_{-2}^{-\sqrt{4-y}} f\left( x,y \right)\mathrm{d} x+\int_{\sqrt{4-y}}^{2} f\left( x,y \right)\mathrm{d} x \right]\mathrm{d} y\).

6

设单位质点 \(P\)，\(Q\) 分别位于点 \(\left( 0,0 \right)\) 和 \(\left( 0,1 \right)\) 处，\(P\) 从点 \(\left( 0,0 \right)\) 出发沿 \(x\) 轴正向移动，记 \(G\) 为引力常量，则当质点 \(P\) 移动到点 \(\left( 1,0 \right)\) 时，克服质点 \(Q\) 的引力所做的功为

A. \(\int_{0}^{1}\dfrac{G}{x^{2}+1}\mathrm{d} x\)

B. \(\int_{0}^{1}\dfrac{Gx}{\left( x^{2}+1 \right)^{\frac{3}{2}}}\mathrm{d} x\)

C. \(\int_{0}^{1}\dfrac{G}{\left( x^{2}+1 \right)^{\frac{3}{2}}}\mathrm{d} x\)

D. \(\int_{0}^{1}\dfrac{G\left( x+1 \right)}{\left( x^{2}+1 \right)^{\frac{3}{2}}}\mathrm{d} x\)

答案：B

解析：

当 \(P\) 运动到 \(\left( x,0 \right)\) 时，\(\overrightarrow{QP}=\left( x,-1 \right)\)，两点间距离为 \(\sqrt{x^{2}+1}\).

引力沿 \(x\) 方向的分量为 \(\dfrac{Gx}{\left( x^{2}+1 \right)^{\frac{3}{2}}}\)，故克服引力所做的功为

\(\int_{0}^{1}\dfrac{Gx}{\left( x^{2}+1 \right)^{\frac{3}{2}}}\mathrm{d} x\).

7

设函数 \(f\left( x \right)\) 连续，给出下列四个条件：

① \(\lim\limits_{x\to 0}\dfrac{\left| f\left( x \right) \right|-f\left( 0 \right)}{x}\) 存在；

② \(\lim\limits_{x\to 0}\dfrac{f\left( x \right)-\left| f\left( 0 \right) \right|}{x}\) 存在；

③ \(\lim\limits_{x\to 0}\dfrac{\left| f\left( x \right) \right|}{x}\) 存在；

④ \(\lim\limits_{x\to 0}\dfrac{\left| f\left( x \right) \right|-\left| f\left( 0 \right) \right|}{x}\) 存在.

其中能得到“\(f\left( x \right)\) 在 \(x=0\) 处可导”的条件的个数是

A. 1

B. 2

C. 3

D. 4

答案：D

解析：

① 若 \(\lim\limits_{x\to 0}\dfrac{\left| f\left( x \right) \right|-f\left( 0 \right)}{x}\) 存在，则分子极限为 \(0\).由连续性得 \(\lim\limits_{x\to 0}\left| f\left( x \right) \right|=\left| f\left( 0 \right) \right|\)，故 \(\left| f\left( 0 \right) \right|=f\left( 0 \right)\)，从而 \(f\left( 0 \right)\ge 0\).若 \(f\left( 0 \right)>0\)，则在 \(0\) 的某邻域内 \(f\left( x \right)>0\)，故原极限就是 \(\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}\).若 \(f\left( 0 \right)=0\)，则由左右极限比较可得导数为 \(0\).故 ① 正确.

② 由极限存在及连续性得 \(f\left( 0 \right)=\left| f\left( 0 \right) \right|\ge 0\)，再将其化为与 ① 类似情形，可得 \(f\left( x \right)\) 在 \(x=0\) 处可导.故 ② 正确.

③ 设 \(\lim\limits_{x\to 0}\dfrac{\left| f\left( x \right) \right|}{x}=A\)，则左右极限存在.比较左右极限可得 \(A=-A=0\)，从而 \(\lim\limits_{x\to 0}\dfrac{f\left( x \right)}{x}=0\)，故可导.故 ③ 正确.

④ 当 \(f\left( 0 \right)>0\) 时，与 ① 类似；当 \(f\left( 0 \right)=0\) 时，与 ③ 类似；当 \(f\left( 0 \right)<0\) 时，在 \(0\) 的某邻域内 \(f\left( x \right)<0\)，原式与 \(\dfrac{-f\left( x \right)+f\left( 0 \right)}{x}\) 只差一个负号，仍可推出可导.故 ④ 正确.

综上，选 D.

8

设矩阵 \(\left( \begin{array}{ccc} 1 & 2 & 0 \ 2 & a & 0 \ 0 & 0 & b \end{array} \right)\) 有一个正特征值和两个负特征值，则

A. \(a>4\)，\(b>0\)

B. \(a<4\)，\(b>0\)

C. \(a>4\)，\(b<0\)

D. \(a<4\)，\(b<0\)

答案：D

解析：

由题意，

\(\left| \begin{array}{ccc} 1 & 2 & 0 \\ 2 & a & 0 \\ 0 & 0 & b \end{array} \right|=b\left( a-4 \right)>0\).

故有 \(a>4,b>0\) 或 \(a<4,b<0\).

又

\(\left| \lambda \boldsymbol{E}-\left( \begin{array}{ccc} 1 & 2 & 0 \\ 2 & a & 0 \\ 0 & 0 & b \end{array} \right) \right|=\left( b-\lambda \right)\left[ \lambda^{2}-\left( a-1 \right)\lambda-4 \right]\).

由于 \(\lambda^{2}-\left( a-1 \right)\lambda-4=0\) 有两个异号的根，故第三个特征值必须为负，即 \(b<0\)，于是 \(a<4\)，\(b<0\).

9

下列矩阵中，可以经过若干初等变换得到矩阵 \(\left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right)\) 的是

A. \(\left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 2 & 1 & 3 \\ 2 & 3 & 1 & 4 \end{array} \right)\)

B. \(\left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 1 & 2 & 5 \\ 1 & 1 & 1 & 3 \end{array} \right)\)

C. \(\left( \begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 3 \\ 0 & 1 & 0 & 0 \end{array} \right)\)

D. \(\left( \begin{array}{cccc} 1 & 1 & 2 & 3 \\ 1 & 2 & 2 & 3 \\ 2 & 3 & 4 & 6 \end{array} \right)\)

答案：B

解析：

对 B 进行初等行变换：

\(\left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 1 & 1 & 2 & 5 \\ 1 & 1 & 1 & 3 \end{array} \right)\to \left( \begin{array}{cccc} 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 0 \end{array} \right)\).

10

设 3 阶矩阵 \(\boldsymbol{A}\)，\(\boldsymbol{B}\) 满足 \(r\left( \boldsymbol{A}\boldsymbol{B} \right)=r\left( \boldsymbol{B}\boldsymbol{A} \right)+1\)，则

A. 方程组 \(\left( \boldsymbol{A}+\boldsymbol{B} \right)\boldsymbol{x}=0\) 只有零解.

B. 方程组 \(\boldsymbol{A}\boldsymbol{x}=0\) 与方程组 \(\boldsymbol{B}\boldsymbol{x}=0\) 均只有零解.

C. 方程组 \(\boldsymbol{A}\boldsymbol{x}=0\) 与方程组 \(\boldsymbol{B}\boldsymbol{x}=0\) 没有公共非零解.

D. 方程组 \(\boldsymbol{A}\boldsymbol{B}\boldsymbol{A}\boldsymbol{x}=0\) 与方程组 \(\boldsymbol{B}\boldsymbol{A}\boldsymbol{B}\boldsymbol{x}=0\) 有公共非零解.

答案：D

解析：

取

\(\boldsymbol{A}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 0 & 0 & 0 \end{array} \right)\)，\(\boldsymbol{B}=\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array} \right)\).

则

\(\boldsymbol{A}\boldsymbol{B}=\left( \begin{array}{ccc} 3 & 3 & 3 \\ -3 & -3 & -3 \\ 0 & 0 & 0 \end{array} \right)\)，\(\boldsymbol{B}\boldsymbol{A}=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)\).

故 \(r\left( \boldsymbol{A}\boldsymbol{B} \right)=1=r\left( \boldsymbol{B}\boldsymbol{A} \right)+1\).

由此可知，A 错；\(\boldsymbol{A}\boldsymbol{x}=0\) 与 \(\boldsymbol{B}\boldsymbol{x}=0\) 均有非零解，故 B 错；并且二者有公共非零解，故 C 错；故选 D.

二、填空题

第 11～16 小题，每小题 5 分，共 30 分.

11

设 \(\int_{1}^{+\infty}\dfrac{a}{x\left( 2x+a \right)}\mathrm{d} x=\ln 2\)，则 \(a=\) __.

答案：2

解析：

\(\int_{1}^{+\infty}\dfrac{a}{x\left( 2x+a \right)}\mathrm{d} x=2\int_{1}^{+\infty}\left( \dfrac{1}{x}-\dfrac{1}{2x+a} \right)\mathrm{d} x\).

\(=\ln \left| \dfrac{2x}{2x+a} \right| \Big|_{1}^{+\infty}=\ln \left| \dfrac{2+a}{2} \right|=\ln 2\).

所以 \(\left| \dfrac{2+a}{2} \right|=2\).

解得 \(a=2\) 或 \(a=-6\).若 \(a=-6\)，则积分在 \(x=3\) 处发散，舍去.故 \(a=2\).

12

曲线 \(y=\sqrt[3]{x^{3}-3x^{2}+1}\) 的渐近线方程为 __.

答案：\(y=x-1\)

解析：

\(\lim\limits_{x\to \infty} y=\infty\)，无水平渐近线.

\(\lim\limits_{x\to \infty}\dfrac{y}{x}=1\).

又

\(\lim\limits_{x\to \infty}\left( y-x \right)=\lim\limits_{x\to \infty}\left( \sqrt[3]{x^{3}-3x^{2}+1}-x \right)\)

\(=\lim\limits_{x\to \infty}\dfrac{x^{3}-3x^{2}+1-x^{3}}{\sqrt[3]{\left( x^{3}-3x^{2}+1 \right)^{2}}+x\sqrt[3]{x^{3}-3x^{2}+1}+x^{2}}=-1\).

所以有斜渐近线 \(y=x-1\).

13

\(\lim\limits_{n\to \infty}\dfrac{1}{n^{2}}\left[ \ln \dfrac{1}{n}+2\ln \dfrac{2}{n}+\cdots+\left( n-1 \right)\ln \dfrac{n-1}{n} \right]=\) __.

答案：\(-\dfrac{1}{4}\)

解析：

\(\lim\limits_{n\to \infty}\dfrac{1}{n^{2}}\left[ \ln \dfrac{1}{n}+2\ln \dfrac{2}{n}+\cdots+\left( n-1 \right)\ln \dfrac{n-1}{n} \right]\)

\(=\lim\limits_{n\to \infty}\sum_{k=1}^{n-1}\dfrac{k}{n}\ln \dfrac{k}{n}\cdot \dfrac{1}{n}\)

\(=\int_{0}^{1} x\ln x \mathrm{~d} x=-\dfrac{1}{4}\).

14

已知函数 \(y=y\left( x \right)\) 由

\(x=\ln \left( 1+2t \right)\)，

\(2t-\int_{1}^{y+t^{2}} e^{-u^{2}} \mathrm{~d} u=0\)

确定，则 \(y^{\prime}\left( 0 \right)=\) __.

答案：\(e\)

解析：

\(t=0\) 代入方程得 \(y=1\).

对 \(2t-\int_{1}^{y+t^{2}} e^{-u^{2}} \mathrm{~d} u=0\) 关于 \(t\) 求导，得

\(2-e^{-\left( y+t^{2} \right)^{2}}\left( \dfrac{\mathrm{d} y}{\mathrm{d} t}+2t \right)=0\).

代入 \(t=0\)，\(y=1\)，得 \(\left. \dfrac{\mathrm{d} y}{\mathrm{d} t} \right|_{t=0}=2e\).

又

\(\left. \dfrac{\mathrm{d} x}{\mathrm{d} t} \right|*{t=0}=\left. \dfrac{2}{1+2t} \right|*{t=0}=2\).

所以

\(y^{\prime}\left( 0 \right)=\left. \dfrac{\mathrm{d} y / \mathrm{d} t}{\mathrm{d} x / \mathrm{d} t} \right|_{t=0}=e\).

15

微分方程 \(\left( 2y-3x \right)\mathrm{d} x+\left( 2x-5y \right)\mathrm{d} y=0\) 满足条件 \(y\left( 1 \right)=1\) 的解为 __.

答案：\(5y^{2}-4xy+3x^{2}=4\)

解析：

\(\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{2y-3x}{5y-2x}=\dfrac{2\dfrac{y}{x}-3}{5\dfrac{y}{x}-2}\).

令 \(\dfrac{y}{x}=u\)，则 \(y=ux\)，代入得

\(u+x\dfrac{\mathrm{d} u}{\mathrm{d} x}=\dfrac{2u-3}{5u-2}\).

化简为

\(\dfrac{5u-2}{5u^{2}-4u+3}\mathrm{d} u=-\dfrac{1}{x}\mathrm{d} x\).

两边积分得 \(5y^{2}-4xy+3x^{2}=C\).

将 \(y\left( 1 \right)=1\) 代入得 \(C=4\).

故解为 \(5y^{2}-4xy+3x^{2}=4\).

16

设矩阵 \(\boldsymbol{A}=\left( \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} \right)\).若 \(\alpha_{1},\alpha_{2},\alpha_{3}\) 线性无关，且 \(\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}\)，则方程组 \(\boldsymbol{A}\boldsymbol{x}=\alpha_{1}+4\alpha_{4}\) 的通解 \(\boldsymbol{x}=\) __.

答案：\(\boldsymbol{x}=k\left( \begin{array}{c} 1 \ 1 \ -1 \ -1 \end{array} \right)+\left( \begin{array}{c} 1 \ 0 \ 0 \ 4 \end{array} \right)\)

解析：

由 \(\alpha_{1}+\alpha_{2}=\alpha_{3}+\alpha_{4}\)，得 \(\alpha_{4}=\alpha_{1}+\alpha_{2}-\alpha_{3}\).

故 \(r\left( \alpha_{1},\alpha_{2},\alpha_{3},\alpha_{4} \right)=r\left( \alpha_{1},\alpha_{2},\alpha_{3} \right)=3=r\left( \boldsymbol{A} \right)\).

所以齐次方程 \(\boldsymbol{A}\boldsymbol{x}=0\) 的基础解系中有 1 个线性无关解向量，

\(\left( 1,1,-1,-1 \right)^{T}\) 为 \(\boldsymbol{A}\boldsymbol{x}=0\) 的解.

而 \(\left( 1,0,0,4 \right)^{T}\) 为 \(\boldsymbol{A}\boldsymbol{x}=\alpha_{1}+4\alpha_{4}\) 的一个解.

故通解为

\(\boldsymbol{x}=k\left( \begin{array}{c} 1 \ 1 \ -1 \ -1 \end{array} \right)+\left( \begin{array}{c} 1 \ 0 \ 0 \ 4 \end{array} \right)\).

三、解答题

第 17～22 小题，共 70 分.解答应写出文字说明、证明过程或演算步骤.

17

本题满分 10 分.

计算 \(\int_{0}^{1}\dfrac{1}{\left( x+1 \right)\left( x^{2}-2x+2 \right)}\mathrm{d} x\).

答案：\(\dfrac{3}{10}\ln 2+\dfrac{\pi}{10}\)

解析：

设

\(\dfrac{1}{\left( 1+x \right)\left( x^{2}-2x+2 \right)}=\dfrac{a}{1+x}+\dfrac{bx+c}{x^{2}-2x+2}\).

则

\(\dfrac{1}{\left( 1+x \right)\left( x^{2}-2x+2 \right)}=\dfrac{\left( a+b \right)x^{2}+\left( -2a+b+c \right)x+2a+c}{\left( 1+x \right)\left( x^{2}-2x+2 \right)}\).

比较系数得

\(\left\{ \begin{array}{l} a+b=0 \\ -2a+b+c=0 \ 2a+c=1 \end{array} \right.\)，

解得

\(\left\{ \begin{array}{l} a=\dfrac{1}{5} \\ b=-\dfrac{1}{5} \\ c=\dfrac{3}{5} \end{array} \right.\).

于是

\(\int_{0}^{1}\dfrac{1}{\left( 1+x \right)\left( x^{2}-2x+2 \right)}\mathrm{d} x=\int_{0}^{1}\left[ \dfrac{1}{5\left( 1+x \right)}+\dfrac{-\dfrac{1}{5}x+\dfrac{3}{5}}{x^{2}-2x+2} \right]\mathrm{d} x\).

\(=\dfrac{1}{5}\ln \left( 1+x \right)\Big|*{0}^{1}-\dfrac{1}{5}\int*{0}^{1}\dfrac{x-3}{x^{2}-2x+2}\mathrm{d} x\).

\(=\dfrac{1}{5}\ln 2-\dfrac{1}{5}\left[ \dfrac{1}{2}\int_{0}^{1}\dfrac{1}{x^{2}-2x+2}\mathrm{d}\left( x^{2}-2x+2 \right)-\int_{0}^{1}\dfrac{2}{x^{2}-2x+2}\mathrm{d} x \right]\).

\(=\dfrac{1}{5}\ln 2-\dfrac{1}{10}\ln \left( x^{2}-2x+2 \right)\Big|*{0}^{1}+\dfrac{2}{5}\int*{0}^{1}\dfrac{1}{\left( x-1 \right)^{2}+1}\mathrm{d} x\).

\(=\dfrac{1}{5}\ln 2-\dfrac{1}{10}\left( 0-\ln 2 \right)+\dfrac{2}{5}\arctan \left( x-1 \right)\Big|_{0}^{1}\).

\(=\dfrac{1}{5}\ln 2+\dfrac{1}{10}\ln 2+\dfrac{2}{5}\left( 0+\dfrac{\pi}{4} \right)=\dfrac{3}{10}\ln 2+\dfrac{\pi}{10}\).

18

本题满分 12 分.

设函数 \(f\left( x \right)\) 在 \(x=0\) 处连续，且

\(\lim\limits_{x\to 0}\dfrac{xf\left( x \right)-e^{2\sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}=-3\)，

证明 \(f\left( x \right)\) 在 \(x=0\) 处可导，并求 \(f^{\prime}\left( 0 \right)\).

答案：\(f^{\prime}\left( 0 \right)=5\)

解析：

由题设，

\(-3=\lim\limits_{x\to 0}\dfrac{xf\left( x \right)-e^{2\sin x}+1}{\ln \left( 1+x \right)+\ln \left( 1-x \right)}\)

\(=\lim\limits_{x\to 0}\dfrac{xf\left( x \right)-e^{2\sin x}+1}{\ln \left( 1-x^{2} \right)}\)

\(=\lim\limits_{x\to 0}\dfrac{xf\left( x \right)-e^{2\sin x}+1}{-x^{2}}\).①

又

\(=\lim\limits_{x\to 0}\dfrac{f\left( x \right)+\dfrac{1-e^{2\sin x}}{x}}{-x}\).

因为

\(\lim\limits_{x\to 0}\dfrac{1-e^{2\sin x}}{x}=-2\).

由 ① 知

\(-3=\lim\limits_{x\to 0}\dfrac{x\left[ f\left( x \right)-f\left( 0 \right) \right]+2x-e^{2\sin x}+1}{-x^{2}}\)

\(=-\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}+\lim\limits_{x\to 0}\dfrac{2x-e^{2\sin x}+1}{-x^{2}}\)

\(=-\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}+\lim\limits_{x\to 0}\dfrac{2-2\cos x , e^{2\sin x}}{-2x}\)

\(=-\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}+\lim\limits_{x\to 0}\dfrac{1-\cos x+\cos x\left( 1-e^{2\sin x} \right)}{-x}\)

\(=-\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}+2\).

所以

\(\lim\limits_{x\to 0}\dfrac{f\left( x \right)-f\left( 0 \right)}{x}=5\)，

即 \(f\left( x \right)\) 在 \(x=0\) 处可导，且 \(f^{\prime}\left( 0 \right)=5\).

19

本题满分 12 分.

设函数 \(f\left( x,y \right)\) 可微且满足 \(\mathrm{d} f\left( x,y \right)=-2xe^{-y}\mathrm{d} x+e^{-y}\left( x^{2}-y-1 \right)\mathrm{d} y\)，\(f\left( 0,0 \right)=2\)，求 \(f\left( x,y \right)\)，并求 \(f\left( x,y \right)\) 的极值.

答案：\(f\left( x,y \right)=-x^{2}e^{-y}+\left( y+2 \right)e^{-y}\)；\(f\left( 0,-1 \right)\) 为极大值点.

解析：

由题意知

\(f_{x}^{\prime}\left( x,y \right)=-2xe^{-y}\)，\(f_{y}^{\prime}\left( x,y \right)=e^{-y}\left( x^{2}-y-1 \right)\).

故

\(f\left( x,y \right)=\int -2xe^{-y}\mathrm{d} x=-x^{2}e^{-y}+C\left( y \right)\).

于是

\(f_{y}^{\prime}\left( x,y \right)=x^{2}e^{-y}+C^{\prime}\left( y \right)=e^{-y}\left( x^{2}-y-1 \right)\).

所以

\(C^{\prime}\left( y \right)=-\left( y+1 \right)e^{-y}\)，

从而

\(C\left( y \right)=\left( y+2 \right)e^{-y}+C\).

故

\(f\left( x,y \right)=-x^{2}e^{-y}+\left( y+2 \right)e^{-y}+C\).

由 \(f\left( 0,0 \right)=2\)，得 \(C=0\).

因此

\(f\left( x,y \right)=-x^{2}e^{-y}+\left( y+2 \right)e^{-y}\).

令

\(\left\{ \begin{array}{l} f_{x}^{\prime}\left( x,y \right)=-2xe^{-y}=0 \\ f_{y}^{\prime}\left( x,y \right)=e^{-y}\left( x^{2}-y-1 \right)=0 \end{array} \right.\)，

解得

\(\left\{ \begin{array}{l} x=0 \\ y=-1 \end{array} \right.\).

再求二阶偏导数：

\(f_{xx}^{\prime\prime}\left( x,y \right)=-2e^{-y}\)，

\(f_{xy}^{\prime\prime}\left( x,y \right)=2xe^{-y}\)，

\(f_{yy}^{\prime\prime}\left( x,y \right)=-e^{-y}\left( x^{2}-y-1 \right)-e^{-y}\).

在 \(\left( 0,-1 \right)\) 处，

\(A=-2e\)，\(B=0\)，\(C=-e\)，

故

\(AC-B^{2}>0\)，且 \(A<0\).

所以 \(\left( 0,-1 \right)\) 为极大值点，极大值为

\(f\left( 0,-1 \right)=e\).

20

本题满分 12 分.

已知平面有界区域 \(D=\left\{ \left( x,y \right)\mid x^{2}+y^{2}\le 4x,\\ x^{2}+y^{2}\le 4y \right\}\)，计算二重积分 \(\iint_{D}\left( x-y \right)^{2}\mathrm{d} x \mathrm{d} y\).

答案：\(12\pi-\dfrac{112}{3}\)

解析：

由对称性，

\(\iint_{D}\left( x-y \right)^{2}\mathrm{d} x \mathrm{d} y=2\iint_{D_{1}}\left( x^{2}+y^{2}-2xy \right)\mathrm{d} x \mathrm{d} y\).

化为极坐标，取

\(x=r\cos \theta\)，\(y=r\sin \theta\)，

则

\(=2\int_{0}^{\frac{\pi}{4}} \mathrm{d} \theta \int_{0}^{4\sin \theta}\left( r^{2}-2r^{2}\cos \theta \sin \theta \right)r\mathrm{d} r\).

\(=128\int_{0}^{\frac{\pi}{4}}\left( \sin^{4}\theta-2\cos \theta \sin^{5}\theta \right)\mathrm{d} \theta\)

\(=128\left[ \int_{0}^{\frac{\pi}{4}}\left( \dfrac{1-\cos 2\theta}{2} \right)^{2}\mathrm{d} \theta-\dfrac{1}{3}\sin^{6}\theta \Big|_{0}^{\frac{\pi}{4}} \right]\)

\(=128\left[ \dfrac{1}{4}\int_{0}^{\frac{\pi}{4}}\left( 1-2\cos 2\theta+\cos^{2}2\theta \right)\mathrm{d} \theta-\dfrac{1}{24} \right]\)

\(=128\left[ \dfrac{1}{4}\left( \dfrac{\pi}{4}-\sin 2\theta \right)\Big|*{0}^{\frac{\pi}{4}}+\dfrac{1}{2}\int*{0}^{\frac{\pi}{2}}\cos^{2} t \mathrm{d} t-\dfrac{1}{24} \right]\)

\(=128\left( \dfrac{3\pi}{32}-\dfrac{7}{24} \right)=12\pi-\dfrac{112}{3}\).

21

本题满分 12 分.

设函数 \(f\left( x \right)\) 其在区间 \(\left( a,b \right)\) 内可导.证明：导函数 \(f^{\prime}\left( x \right)\) 在 \(\left( a,b \right)\) 内严格单调增加的充分必要条件是：

对 \(\left( a,b \right)\) 内任意的 \(x_{1},x_{2},x_{3}\)，当 \(x_{1}<x_{2}<x_{3}\) 时，有

\(\dfrac{f\left( x_{2} \right)-f\left( x_{1} \right)}{x_{2}-x_{1}}<\dfrac{f\left( x_{3} \right)-f\left( x_{2} \right)}{x_{3}-x_{2}}\).

答案：略

解析：

证明：

（1）必要性

由拉格朗日中值定理，

\(\dfrac{f\left( x_{2} \right)-f\left( x_{1} \right)}{x_{2}-x_{1}}=f^{\prime}\left( \xi_{1} \right)\)，其中 \(x_{1}<\xi_{1}<x_{2}\)；

\(\dfrac{f\left( x_{3} \right)-f\left( x_{2} \right)}{x_{3}-x_{2}}=f^{\prime}\left( \xi_{2} \right)\)，其中 \(x_{2}<\xi_{2}<x_{3}\).

由于 \(f^{\prime}\left( x \right)\) 在 \(\left( a,b \right)\) 上单增，且 \(\xi_{1}<\xi_{2}\)，故

\(f^{\prime}\left( \xi_{1} \right)<f^{\prime}\left( \xi_{2} \right)\)，

即

\(\dfrac{f\left( x_{2} \right)-f\left( x_{1} \right)}{x_{2}-x_{1}}<\dfrac{f\left( x_{3} \right)-f\left( x_{2} \right)}{x_{3}-x_{2}}\).

（2）充分性

对任意 \(c_{1}<c_{2}\in \left( a,b \right)\)，有

\(f^{\prime}\left( c_{1} \right)=\lim\limits_{x\to c_{1}}\dfrac{f\left( x \right)-f\left( c_{1} \right)}{x-c_{1}}\)，

\(f^{\prime}\left( c_{2} \right)=\lim\limits_{x\to c_{2}}\dfrac{f\left( x \right)-f\left( c_{2} \right)}{x-c_{2}}\).

当 \(a<c_{1}<x<c_{2}<b\) 时，由已知条件可知

\(\dfrac{f\left( c_{1} \right)-f\left( a \right)}{c_{1}-a}<\dfrac{f\left( x \right)-f\left( c_{1} \right)}{x-c_{1}}<\dfrac{f\left( c_{2} \right)-f\left( x \right)}{c_{2}-x}\).

两边同时取极限，并利用极限保号性，可得

\(\lim\limits_{x\to c_{1}^{+}}\dfrac{f\left( x \right)-f\left( c_{1} \right)}{x-c_{1}}\le \lim\limits_{x\to c_{1}^{+}}\dfrac{f\left( c_{2} \right)-f\left( x \right)}{c_{2}-x}=\dfrac{f\left( c_{2} \right)-f\left( c_{1} \right)}{c_{2}-c_{1}}\)，

\(\lim\limits_{x\to c_{2}^{-}}\dfrac{f\left( c_{2} \right)-f\left( x \right)}{c_{2}-x}\ge \lim\limits_{x\to c_{2}^{-}}\dfrac{f\left( x \right)-f\left( c_{1} \right)}{x-c_{1}}=\dfrac{f\left( c_{2} \right)-f\left( c_{1} \right)}{c_{2}-c_{1}}\).

从而

\(f^{\prime}\left( c_{1} \right)=\lim\limits_{x\to c_{1}^{+}}\dfrac{f\left( x \right)-f\left( c_{1} \right)}{x-c_{1}}\le \lim\limits_{x\to c_{2}^{-}}\dfrac{f\left( x \right)-f\left( c_{2} \right)}{x-c_{2}}=f^{\prime}\left( c_{2} \right)\).

故 \(f^{\prime}\left( x \right)\) 在 \(\left( a,b \right)\) 上单增.

22

本题满分 12 分.

已知矩阵

\(\boldsymbol{A}=\left( \begin{array}{ccc} 4 & 1 & -2 \\ 1 & 1 & 1 \\ -2 & 1 & a \end{array} \right)\)，

\(\boldsymbol{B}=\left( \begin{array}{ccc} k & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0 \end{array} \right)\).

（1）求 \(a\) 的值及 \(k\) 的取值范围；

（2）若存在正交矩阵 \(\boldsymbol{Q}\) 使得 \(\boldsymbol{Q}^{T}\boldsymbol{A}\boldsymbol{Q}=\boldsymbol{B}\)，求 \(k\) 及 \(\boldsymbol{Q}\).

答案：

（1）\(a=4\)，\(k>0\)；

（2）\(k=3\)，\(\boldsymbol{Q}=\left( \begin{array}{ccc} \dfrac{1}{\sqrt{3}} & -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} \ \dfrac{1}{\sqrt{3}} & 0 & -\dfrac{2}{\sqrt{6}} \ \dfrac{1}{\sqrt{3}} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} \end{array} \right)\).

解析：

（1）因为 \(\boldsymbol{A}\) 与 \(\boldsymbol{B}\) 合同，所以 \(\lambda=0\) 为 \(\boldsymbol{A}\) 的一个特征值.

于是 \(\left| \boldsymbol{A} \right|=0\)，得 \(a=4\).

再由 \(\left| \lambda \boldsymbol{E}-\boldsymbol{A} \right|=0\)，得 \(\lambda_{1}=3\)，\(\lambda_{2}=6\)，\(\lambda_{3}=0\)，故 \(k>0\).

（2）因为存在正交矩阵 \(\boldsymbol{Q}\)，使得 \(\boldsymbol{Q}^{T}\boldsymbol{A}\boldsymbol{Q}=\boldsymbol{B}\)，所以 \(k=3\).

当 \(\lambda_{1}=3\) 时，由 \(\left( 3\boldsymbol{E}-\boldsymbol{A} \right)\boldsymbol{x}=0\)，可取特征向量

\(\boldsymbol{\xi}_{1}=\left( \begin{array}{c} 1 \ 1 \ 1 \end{array} \right)\).

当 \(\lambda_{2}=6\) 时，由 \(\left( 6\boldsymbol{E}-\boldsymbol{A} \right)\boldsymbol{x}=0\)，可取特征向量

\(\boldsymbol{\xi}_{2}=\left( \begin{array}{c} -1 \ 0 \ 1 \end{array} \right)\).

当 \(\lambda_{3}=0\) 时，由 \(\left( 0\boldsymbol{E}-\boldsymbol{A} \right)\boldsymbol{x}=0\)，可取特征向量

\(\boldsymbol{\xi}_{3}=\left( \begin{array}{c} 1 \ -2 \ 1 \end{array} \right)\).

将 \(\boldsymbol{\xi}*{1}\)，\(\boldsymbol{\xi}*{2}\)，\(\boldsymbol{\xi}_{3}\) 单位化，得

\(\boldsymbol{Q}=\left( \begin{array}{ccc} \dfrac{1}{\sqrt{3}} & -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} \ \dfrac{1}{\sqrt{3}} & 0 & -\dfrac{2}{\sqrt{6}} \ \dfrac{1}{\sqrt{3}} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{6}} \end{array} \right)\).