MA127-2023春 期中考试答案

一、选择题答案

1-1

${\overrightarrow{n}}_1=(1,2,-1),{\overrightarrow{n}}_2=(1,1,0)$

$\Rightarrow \cos \theta =\frac{{\overrightarrow{n}}_1\cdot {\overrightarrow{n}}_2}{|{\overrightarrow{n}}_1|\cdot |{\overrightarrow{n}}_2|}=\frac{3}{\sqrt{6}\cdot \sqrt{2}}=\frac{\sqrt{3}}{2}$

$\Rightarrow \theta =\frac{\pi }{6}$

答案 D

1-2

$V(t)=(\sqrt{t},-\sqrt{2-t},1),|V(t)|=\sqrt{3}$

$\Rightarrow L={\int }_{\frac{1}{4}}^{\frac{1}{2}}|v(t)|dt=\frac{\sqrt{3}}{4}$

答案 C

1-3

$\underset{(x,y)\to (0,1)}{lim}\frac{\ln (1+xy)}{x}=\underset{(x,y)\to (0,1)}{lim}\frac{\ln (1+xy)}{xy}\cdot y\Rightarrow \underset{y\to 1}{lim}y=1$

$a=1$ 时，$f(0,1)=a=1$。$\Rightarrow \underset{(x,y)\to (0,1)}{lim}f(x,y)=f(0,1)$

答案 C

1-4

$0\le a_n\le \frac{1}{n}$

(A) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{a_n}{(1+\ln n{)}^2}$：由于 $0\le |(-1{)}^n\frac{a_n}{(1+\ln n{)}^2}|\le \frac{1}{n(1+\ln n{)}^2}$

而 $f(x)=\frac{1}{x(1+\ln x{)}^2}>0$，连续 $(x\ge 1)$

且 ${\int }_1^{\mathrm{\infty }}\frac{1}{x(1+\ln x{)}^2}dx={\int }_1^{\mathrm{\infty }}\frac{d(1+\ln x)}{(1+\ln x{)}^2}=-\frac{1}{1+\ln x}{]}_1^{\mathrm{\infty }}=1$ 收敛

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(1+\ln n{)}^2}$ 收敛

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{a_n}{(1+\ln n{)}^2}$ 绝对收敛，故收敛。

(B) $\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n$：令 $a_n=\frac{(-1{)}^n+1}{2n}$ 则 $0\le a_n\le \frac{1}{n}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n& =\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\cdot \frac{(-1{)}^n+1}{2n}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{1+(-1{)}^n}{2n}\end{array}$$

发散

(C) $a_n=\frac{1}{2n}$，$0<a_n<\frac{1}{n}$，但 $\sum a_n$ 发散。

(D) $a_n=\frac{(-1{)}^n+1}{2n}$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{a_n}{1+\ln n}& =\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n+1}{2n(\ln n+1)}\\ & =\sum _{n=1}^{\mathrm{\infty }}(\frac{(-1{)}^n}{2n(\ln n+1)}+\frac{1}{2n(\ln n+1)})\end{array}$$

对于 $\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{2n(\ln n+1)}$：$\frac{1}{2n(\ln n+1)}\to 0$，$\Rightarrow$ 收敛

对于 $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n(\ln n+1)}$：$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{2n(\ln n+1)}}{\frac{1}{n\ln n}}=\frac{1}{2}$

而 $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln n}$ 发散 $\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{1}{2n(\ln n+1)}$ 发散

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n(\ln n+1)}$ 发散

答案 A

1-5

若 $\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n$ 在 $x=-2$ 处收敛

则 $\Rightarrow |x|<2$ 时，$\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n$ 绝对收敛

$\Rightarrow |x|<2$ 时，$(\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n{)}^{\prime }$ 绝对收敛，即

$$\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$$

$\Rightarrow x=1$ 时，$\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}=\sum _{n=1}^{\mathrm{\infty }}n{a}_n$ 收敛

答案 C

二、填空题答案

2-1

$r=\csc \theta e^{r\cos \theta }$ $\Rightarrow r\sin \theta =e^{r\cos \theta }\Rightarrow y=e^x$

2-2

$d=\frac{|1+2+3+4|}{\sqrt{1^2+2^2+3^2}}=\frac{10}{\sqrt{14}}$

2-3

$v(t)={\int }_0^{t}a(x)dx+v(0)=(e^x{]}_0^t,\frac{x^2}{2}{]}_0^t,-\frac{1}{2}\cos 2x{]}_0^t)+(1,0,0)$

$=(e^t-1,\frac{t^2}{2},-\frac{1}{2}\cos 2t+\frac{1}{2})+(1,0,0)$

$=(e^t,\frac{t^2}{2},\frac{1}{2}-\frac{1}{2}\cos 2t)$

$\Rightarrow r(t)={\int }_0^{t}v(x)dx+r(0)$

$=(e^t-1,\frac{t^3}{6},\frac{t}{2}-\frac{1}{4}\sin 2t)$

2-4

$\sin x=x-\frac{x^3}{3!}+\cdots$

$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{3\sin (2x)-2\sin (3x)}{6x^3+x^4}& =\underset{x\to 0}{lim}\frac{3(2x-\frac{(2x{)}^3}{3!}+O(x^5))-2(3x-\frac{(3x{)}^3}{3!}+O(x^5))}{6x^3+x^4}\\ & =\underset{x\to 0}{lim}\frac{6x-\frac{8x^3}{6}-6x+\frac{54x^3}{6}+O(x^5)}{6x^3+x^4}\\ & =\underset{x\to 0}{lim}\frac{(-4+9)x^3+O(x^5)}{6x^3+x^4}=\frac{5}{6}\end{array}$$

2-5

$S=\sum _{n=0}^{\mathrm{\infty }}\frac{(-\ln 2{)}^n}{n!}=e^{-\ln 2}=\frac{1}{2}$

原文为 -1/2, 我觉得不是

三

$\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}(x-1{)}^n$

3-1

$\underset{n\to \mathrm{\infty }}{lim}|\frac{\ln (n+1)\cdot (x-1{)}^{n+1}}{n+1}\cdot \frac{n}{\ln n\cdot (x-1{)}^n}|$

$=\underset{n\to \mathrm{\infty }}{lim}|\frac{\ln (n+1)}{\ln n}\cdot \frac{n}{n+1}|\cdot |x-1|=|x-1|<1$ 时，$\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}(x-1{)}^n$ 绝对收敛

$x=0$ 时 $\Rightarrow \sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{\ln n}{n}$：令 $f(x)=\frac{\ln x}{x}$

$$f^{\prime }(x)=\frac{1-\ln x}{x^2}<0(x\ge 3)$$

当 $n\ge 3$ 时，$\{\frac{\ln n}{n}\}$ 单调递减，且 $\underset{n\to \mathrm{\infty }}{lim}\frac{\ln n}{n}=0$

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{\ln n}{n}$ 收敛

$x=2$ 时 $\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}$：$f(x)=\frac{\ln x}{x}>0$ $(x\ge 3)$，连续

且 ${\int }_3^{\mathrm{\infty }}\frac{\ln x}{x}dx=\frac{1}{2}(\ln x{)}^2{]}_3^{\mathrm{\infty }}=\mathrm{\infty }$，发散

因此：

$\Rightarrow \sum _{n=3}^{\mathrm{\infty }}\frac{\ln n}{n}$ 发散

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}$ 发散

$\Rightarrow$ 收敛域：$[0,2)$

3-2

$0<x<2$ 时，绝对收敛

而对于 $x=0$ 时，$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{\ln n}{n}$：$\sum _{n=1}^{\mathrm{\infty }}|\frac{(-1{)}^n\ln n}{n}|=\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}$ 发散

$\Rightarrow x=0$ 时，条件收敛

四

极限计算

4-1

$\underset{(x,y)\to (0,0)}{lim}\frac{y^2{\sin }^{2}x}{x^4+y^4}$

解：不存在

令 $y=kx$ $(k\ne 0)$，上式 $=\underset{x\to 0}{lim}\frac{k^2{x}^2{\sin }^{2}x}{x^4+k^4{x}^4}=\underset{x\to 0}{lim}\frac{k^2}{1+k^4}\cdot \frac{{\sin }^{2}x}{x^2}=\frac{k^2}{1+k^4}$

由于结果依赖于$k$的值，所以极限不存在。

4-2

$\underset{(x,y)\to (0,0)}{lim}\frac{2xy}{\sqrt{x^2+2y^2}}$

$$|\frac{2xy}{\sqrt{x^2+2y^2}}|=|\frac{x}{\sqrt{x^2+2y^2}}\cdot 2y|\le 2|y|$$

当$(x,y)\to (0,0)$ 时，$2|y|\to 0$ $\Rightarrow \underset{(x,y)\to (0,0)}{lim}\frac{2xy}{\sqrt{x^2+2y^2}}=0$

五

直线和平面

$L_1:\{\begin{array}{l}x=t\\ y=1\\ z=1+t\end{array}$

$L_2:\{\begin{array}{l}x=1+2s\\ y=0\\ z=2s\end{array}$

$\Rightarrow \overrightarrow{v_1}=(1,0,1)$，$\overrightarrow{v_2}=(2,0,2)$

$\Rightarrow \overrightarrow{v_2}=2\overrightarrow{v_1}\Rightarrow L_1\parallel L_2$

$\Rightarrow$ 存在平面$M$过$L_1$和$L_2$

在$L_1$上取点$A(0,1,1)$，在$L_2$上取点$B(1,0,0)$

$\Rightarrow \overrightarrow{AB}=(1,-1,-1)$

$\overrightarrow{n}=\overrightarrow{v_1}\times \overrightarrow{AB}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 0& 1\\ 1& -1& -1\end{array}\right|=(1,2,-1)$，过点$B(1,0,0)$

$\Rightarrow$ 平面$M$为：$(x-1)+2y-z=0$

即 $x+2y-z=1$

六

$0<b<a<1$

6-1

$$0<\frac{C_n}{\sqrt{(1+a^n)(1+b^n)}}\le a^n$$

而 $0<a<1$，$\sum _{n=1}^{\mathrm{\infty }}{a}^n$ 收敛 $\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{C_n}{\sqrt{(1+a^n)(1+b^n)}}$ 收敛

6-2

$$(\frac{a^n}{n^2}{)}^{\frac{1}{n}}\le (\frac{a^n}{n^2}+\frac{b^n}{n}{)}^{\frac{1}{n}}\le (\frac{2a^n}{n}{)}^{\frac{1}{n}}$$$$\underset{n\to \mathrm{\infty }}{lim}(\frac{a^n}{n^2}{)}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{a}{(n^{\frac{1}{n}}{)}^2}=a$$$$\underset{n\to \mathrm{\infty }}{lim}(\frac{2a^n}{n}{)}^{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{2^{\frac{1}{n}}\cdot a}{n^{\frac{1}{n}}}=a$$

$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}(\frac{a^n}{n^2}+\frac{b^n}{n}{)}^{\frac{1}{n}}=a$

七

$r(t)=(\sqrt{2}\cos t,\sin t,\sin t)$

7-1

$v(t)=(-\sqrt{2}\sin t,\cos t,\cos t)$

$$|v(t)|=\sqrt{2{\sin }^{2}t+{\cos }^{2}t+{\cos }^{2}t}=\sqrt{2}$$$$T(t)=\frac{v(t)}{|v(t)|}=\frac{1}{\sqrt{2}}(-\sqrt{2}\sin t,\cos t,\cos t)$$$$\frac{dT}{dt}=\frac{1}{\sqrt{2}}(-\sqrt{2}\cos t,-\sin t,-\sin t)$$

而 $|\frac{dT}{dt}|=\frac{1}{\sqrt{2}}\sqrt{2{\cos }^{2}t+{\sin }^{2}t+{\sin }^{2}t}=1$

$$N(t)=\frac{dT/dt}{|dT/dt|}=\frac{1}{\sqrt{2}}(-\sqrt{2}\cos t,-\sin t,-\sin t)$$

7-2

$\kappa (t)=\frac{1}{|v(t)|}\cdot |\frac{dT}{dt}|=\frac{1}{\sqrt{2}}$

八

$f(x)=\frac{4}{x^2-2x+5}+\ln x$

$$\begin{array}{rl}f(x)& =\frac{4}{(x-1{)}^2+4}+\ln (1+(x-1))\\ & =\frac{1}{1+\frac{(x-1{)}^2}{4}}+\ln [1+(x-1)]\end{array}$$

而 $\frac{1}{1+\frac{(x-1{)}^2}{4}}=\sum _{n=0}^{\mathrm{\infty }}(-\frac{(x-1{)}^2}{4}{)}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{4^n}\cdot (x-1{)}^{2n}$，$|\frac{(x-1{)}^2}{4}|<1$

$\ln [1+(x-1)]=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}(x-1{)}^n}{n}$，$-1<x-1<1$ 即 $0<x<2$

$\Rightarrow x\in (0,2)$ 时，$f(x)=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{4^n}(x-1{)}^{2n}+\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}(x-1{)}^n}{n}$