MA127-2019春 期中考试答案

一、判断题

1-1

False

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}{a}_n$ converges, $a_n<0$ for $n>1000$

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\left|a_n\right|=\sum _{n=1}^{1000}\left|a_n\right|-\sum _{n=1001}^{\mathrm{\infty }}{a}_n$ converges and $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$

$\therefore \underset{n\to \mathrm{\infty }}{lim}\left|a_n\right|=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n^2}{\left|a_n\right|}=0,\sum \left|a_n\right|$ converges, $\Rightarrow \sum a_n^2$ converges.

1-2

True

$\Rightarrow \sum a_n(x-1{)}^n$ converges at $x=0$

$\Rightarrow \sum a_n(x-1{)}^n$ converges absolutly for $|x-1|<1$

so $x-1=-1$, i.e., $0<x<2$

diverges at $x=2$, i.e., $x-1=1$

$\Rightarrow$ diverges for $|x-1|>1$

$\Rightarrow x>2$ or $x<0$ diverges. So converges for $0\le x<2$

since. $\sum a_n{x}^n$ converges at $x=c\ne 0$

$\Rightarrow$ converges absolutely for all $|x|<|c|$

diverges at $x=d\Rightarrow$ diverges for $|x|>|d|$

or

$\sum a_n(x-1{)}^n$ converges at $x=0$

$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}{a}_n=0\Rightarrow \mathrm{\exists }\epsilon =1$ s.t. $\left|a_n\right|<1\mathrm{\forall }n>N$

$\Rightarrow |a_n(x-1{)}^n|<|x-1{|}^n$ for $\mathrm{\forall }|x-1|<1$

and $\sum |x-1{|}^n$ converges

$\Rightarrow \sum |a_n(x-1{)}^n|$ converges for $|x-1|<1\Rightarrow 0<x<2$

$\Rightarrow$ converges at $0\le x<2$

1-3

True

Since

$f_x(x_0,y_0)=\underset{x\to x_0}{lim}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}=c$ exists

and $x-x_0\to 0\Rightarrow \underset{x\to x_0}{lim}f(x,y_0)=f(x_0,y_0)$

$f_y(x_0,y_0)=\underset{y\to y_0}{lim}\frac{f(x_0,y)-f(x_0,y_0)}{y-y_0}$ exist

and $\underset{y\to y_0}{lim}(y-y_0)=0\Rightarrow \underset{y\to y_0}{lim}f(x_0,y)=f(x_0,y_0)$

$\therefore \underset{x\to x_0}{lim}f(x,y_0)=\underset{y\to y_0}{lim}f(x_0,y)=f(x_0,y_0)$

1-4

True

$\overrightarrow{r}(t)\cdot \frac{d\overrightarrow{r}}{dt}=0\Rightarrow (\overrightarrow{r}(t)\cdot \overrightarrow{r}(t){)}^{\prime }=2\overrightarrow{r}(t)\cdot \frac{d\overrightarrow{r}}{dt}=0$

$$\Rightarrow |\overrightarrow{r}(t)|=c\Rightarrow |\overrightarrow{r}(t)|=\text{constant}$$

1-5

False

For unit circle: $x^2+y^2=1\Rightarrow$

$$\{\begin{array}{l}x=\cos t\\ y=\sin t\end{array}\Rightarrow \overrightarrow{r}(t)=(\cos t)\overrightarrow{i}+(\sin t)\overrightarrow{j}$$

$\overrightarrow{T}=\overrightarrow{v}=(-\sin t)\overrightarrow{i}+(\cos t)\overrightarrow{j}$

$\left|\frac{d\overrightarrow{T}}{dt}\right|=1,|\overrightarrow{v}(t)|=1\Rightarrow \kappa =1$

For $y=x^2\Rightarrow y^{\prime }=2x,y^{\prime \prime }=2\Rightarrow y^{\prime }(0)=0,y^{\prime \prime }(0)=2$

$$\kappa =\frac{|y^{\prime \prime }|}{(1+(y^{\prime }{)}^2{)}^{3/2}}=\frac{2}{1}=2$$

二、多选题

2-1

C.

$|\overrightarrow{a}+\overrightarrow{b}|=\sqrt{|\overrightarrow{a}{|}^2+2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2}=\sqrt{|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2}$

$|\overrightarrow{a}-\overrightarrow{b}|=\sqrt{|\overrightarrow{a}{|}^2-2\overrightarrow{a}\cdot \overrightarrow{b}+|\overrightarrow{b}{|}^2}=\sqrt{|\overrightarrow{a}{|}^2+|\overrightarrow{b}{|}^2}$

$\therefore |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$

2-2

D.

$\Rightarrow {\overrightarrow{v}}_1=(1,2,-1),{\overrightarrow{v}}_2=(-2,1,1)\therefore {\overrightarrow{v}}_1\cdot {\overrightarrow{v}}_2=-2+2-1=-1,\left|{\overrightarrow{v}}_1\right|=\left|{\overrightarrow{v}}_2\right|=\sqrt{6}$

$\Rightarrow \cos \theta =-\frac{1}{6}$

But if they intersect:

$$\{\begin{array}{l}x=t=1-2t\\ y=2t=t\end{array}\Rightarrow \{\begin{array}{l}t=\frac{1}{3}\\ t=0\end{array}$$

2-3

D

$0\le a_n<\frac{1}{n}\Rightarrow a_n^2\le \frac{1}{n^2}\Rightarrow \sum a_n^2$ converges

$\Rightarrow \sum (-1{)}^n{a}_n^2$ converges

三

3-1

$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{|a_n|}=\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{(-1{)}^n(1-\frac{1}{n}{)}^{n^2}}=\underset{n\to \mathrm{\infty }}{lim}(1-\frac{1}{n}{)}^n=e^{-1}<1$

$\Rightarrow$ converges absolutely

3-2

$\Rightarrow \underset{n\to \mathrm{\infty }}{lim}{e}^{-\frac{1}{n}}=e^0=1\ne 0\Rightarrow$ diverges

3-3

$\Rightarrow$ For $\sum _{n=2019}^{\mathrm{\infty }}\frac{1}{\sqrt{n^2-2019n+1}},\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{\sqrt{n^2-2019n+1}}}{\frac{1}{n}}=1,\sum \frac{1}{n}$ diverges $\Rightarrow$ diverges

$u_n=\frac{1}{\sqrt{n^2-2019n+1}}>0,u_n>u_{n+1},u_n\to 0\Rightarrow$ converges conditionally

四

4-1

$\Rightarrow \left|\frac{u_{n+1}}{u_n}\right|=\frac{2^{n+1}\sqrt{n^2+n+1}}{2^n\sqrt{(n+1{)}^2+(n+1)+1}}|x|\to 2|x|<1\Rightarrow |x|<\frac{1}{2}\Rightarrow R=\frac{1}{2}$

4-2

when $x=\frac{1}{2}$: $\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{\sqrt{n^2+n+1}}$ converges conditionally since

$\frac{1}{\sqrt{n^2+n+1}}>0,\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n^2+n+1}}=0$

$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{\sqrt{n^2+n+1}}\cdot n=1,\sum \frac{1}{n}$ diverges

$\Rightarrow \sum \frac{1}{\sqrt{n^2+n+1}}$ diverges

when $x=-\frac{1}{2}$

$\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}\cdot (-1{)}^n}{\sqrt{n^2+n+1}}=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^{2n+1}}{\sqrt{n^2+n+1}}=\sum _{n=0}^{\mathrm{\infty }}\frac{-1}{\sqrt{n^2+n+1}}$ diverges

$\Rightarrow$ converges absolutely: $(-\frac{1}{2},\frac{1}{2})$, converges: $(-\frac{1}{2},\frac{1}{2}]$, converges conditionally: $x=\frac{1}{2}$

五

5-1

$\frac{dx}{dt}=-3{\cos }^{2}t\sin t,\frac{dy}{dt}=3{\sin }^{2}t\cos t\Rightarrow (\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2=9{\sin }^{2}t{\cos }^{2}t$

$$\begin{array}{rl}L& ={\int }_0^{\frac{\pi }{2}}\sqrt{(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2}dt={\int }_0^{\frac{\pi }{2}}\sqrt{9{\sin }^{2}t{\cos }^{2}t}dt={\int }_0^{\frac{\pi }{2}}3\sin t\cos tdt\\ & =\frac{3}{2}{\sin }^{2}t{|}_0^{\frac{\pi }{2}}=\frac{3}{2}\end{array}$$

5-2

$S={\int }_0^{\frac{\pi }{2}}2\pi y\sqrt{(\frac{dx}{dt}{)}^2+(\frac{dy}{dt}{)}^2}dt={\int }_0^{\frac{\pi }{2}}2\pi {\sin }^{3}t\cdot 3\sin t\cos tdt={\int }_0^{\frac{\pi }{2}}6\pi {\sin }^{4}td\sin t$

=$\frac{6}{5}si{n}^{5}t{|}_0^{\frac{\pi }{2}}=\frac{6\pi }{5}$

六

$\Rightarrow$ let $P(2,-1,-1),Q(1,0,-1)\Rightarrow \overrightarrow{PQ}=(-1,1,0),\overrightarrow{n_1}=(2,3,-5)$

$$\begin{array}{r}\Rightarrow \overrightarrow{n}=\overrightarrow{PQ}\times \overrightarrow{n_1}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ -1& 1& 0\\ 2& 3& -5\end{array}\right|=-5\overrightarrow{i}-5\overrightarrow{j}-5\overrightarrow{k}\end{array}$$

$\Rightarrow$ the plane: $-5(x-2)-5(y+1)-5(z+1)=0$

$\Rightarrow x+y+z=0$

七

$\Rightarrow$ let $P(1,0,-2),Q(2,-1,3)\Rightarrow \overrightarrow{PQ}=(1,-1,5),\overrightarrow{r_0}=\overrightarrow{i}-2\overrightarrow{k}$

the direction of $\overrightarrow{v}(0)$ is $\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{1}{\sqrt{27}}(1,-1,5)=\frac{1}{3\sqrt{3}}(1,-1,5)$

So $\overrightarrow{v}(0)=|\overrightarrow{v}(0)|\cdot \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=3\cdot \frac{1}{3\sqrt{3}}(1,-1,5)=\frac{\sqrt{3}}{3}(1,-1,5)$

$$\begin{array}{rl}\overrightarrow{a}(t)& =\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k}\\ \Rightarrow \overrightarrow{v}(t)& =\int (\overrightarrow{i}+2\overrightarrow{j}+\overrightarrow{k})dt=t\overrightarrow{i}+2t\overrightarrow{j}+t\overrightarrow{k}+\overrightarrow{c}\\ \overrightarrow{v}(0)& =\frac{\sqrt{3}}{3}\overrightarrow{i}-\frac{\sqrt{3}}{3}\overrightarrow{j}+\frac{5\sqrt{3}}{3}\overrightarrow{k}\\ \Rightarrow \overrightarrow{c}& =\frac{\sqrt{3}}{3}\overrightarrow{i}-\frac{\sqrt{3}}{3}\overrightarrow{j}+\frac{5\sqrt{3}}{3}\overrightarrow{k}\\ \Rightarrow \overrightarrow{v}(t)& =(t+\frac{\sqrt{3}}{3})\overrightarrow{i}+(2t-\frac{\sqrt{3}}{3})\overrightarrow{j}+(t+\frac{5\sqrt{3}}{3})\overrightarrow{k}\\ \overrightarrow{r}(t)& =\int [(t+\frac{\sqrt{3}}{3})\overrightarrow{i}+(2t-\frac{\sqrt{3}}{3})\overrightarrow{j}+(t+\frac{5\sqrt{3}}{3})\overrightarrow{k}]dt\\ & =(\frac{1}{2}{t}^2+\frac{\sqrt{3}}{3}t)\overrightarrow{i}+(t^2-\frac{\sqrt{3}}{3}t)\overrightarrow{j}+(\frac{1}{2}{t}^2+\frac{5\sqrt{3}}{3}t)\overrightarrow{k}+\overrightarrow{c}\\ \overrightarrow{r}(0)& =\overrightarrow{i}-2\overrightarrow{k}\Rightarrow \overrightarrow{c}=\overrightarrow{r}(0)=\overrightarrow{i}-2\overrightarrow{k}\\ \overrightarrow{r}(t)& =(\frac{1}{2}{t}^2+\frac{\sqrt{3}}{3}t+1)\overrightarrow{i}+(t^2-\frac{\sqrt{3}}{3}t)\overrightarrow{j}+(\frac{1}{2}{t}^2+\frac{5\sqrt{3}}{3}t-2)\overrightarrow{k}\end{array}$$

八

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}f(x,y)=\underset{(x,y)\to (0,0)}{lim}\frac{\sin (x^3+y^2)}{x^2+y^2}=\underset{(x,y)\to (0,0)}{lim}\frac{\sin (x^3+y^2)}{x^3+y^2}\cdot \frac{x^3+y^2}{x^2+y^2}=\underset{(x,y)\to (0,0)}{lim}\frac{x^3+y^2}{x^2+y^2}$

Using polar coordinates: $x=r\cos \theta ,y=r\sin \theta$:

$=\underset{r\to 0}{lim}\frac{r^3{\cos }^3\theta +r^2{\sin }^2\theta }{r^2}=\underset{r\to 0}{lim}r({\cos }^3\theta +{\sin }^3\theta )=0$

or

$0\le \left|\frac{x^3+y^2}{x^2+y^2}\right|=|\frac{x\cdot x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}|=|x\times \frac{x^2}{x^2+y^2}+y\times \frac{y^2}{x^2+y^2}|\le |x|+|y|$

$\underset{(x,y)\to (0,0)}{lim}(|x|+|y|)=0$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}(|\frac{x^3+y^3}{x^2+y^2}|)=0$

$\Rightarrow \underset{(x,y)\to (0,0)}{lim}(\frac{x^3+y^3}{x^2+y^2})=0$

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

$f(x,y)$ is continuous at $(0,0)$

九

$\Rightarrow \frac{\partial z}{\partial x}=f^{\prime }(e^{x}y)\cdot e^{x}y,\frac{\partial z}{\partial y}=f^{\prime }(e^{x}y)e^x$

$\Rightarrow \frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=2f^{\prime }(e^{x}y)e^{x}y=1$

Let $u=e^{x}y$

$\Rightarrow 2f^{\prime }(u)u=1\Rightarrow \frac{df}{du}=\frac{1}{2u}\Rightarrow f(u)=\frac{1}{2}\ln (u)+c$

$f(1)=0\Rightarrow \frac{1}{2}\ln 1+c=0\Rightarrow c=0$

$\Rightarrow f(u)=\frac{1}{2}\ln |u|$

十

$$\begin{array}{rl}\Rightarrow f^{\prime }(x)& =\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+1}}=(1+x^2{)}^{-\frac{1}{2}}\\ & =1+\sum _{k=1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{-\frac{1}{2}}{k})x^{2k},-1<x<1\\ \Rightarrow f(x)& =x+\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2k+1}(\genfrac{}{}{0ex}{}{-\frac{1}{2}}{k})x^{2k+1},-1<x<1\end{array}$$