1.(1)${\overrightarrow{n}}_1=(1,2,-1).\overrightarrow{n_2}=(1,1,0)$

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

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

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

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

(选 C

(3) 造 C

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

$a=1$ 时 而 $f(0,1)=a=1.\Rightarrow \frac{l}{(x,y))(0,1)}=f(0,1)$

(4) . $0⩽a_n⩽\frac{1}{n}$ .(4A

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

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

日 ${\int }_1^{\mathrm{\infty }}\frac{1}{x(\ln (t+1+1{)}^2}dx={\int }_1^{\mathrm{\infty }}\frac{d(1+\ln x)}{(1+\ln x{)}^2}$

${=-\frac{1}{1+\ln x}]}_1^{\mathrm{\infty }}$

$\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(1+\tan n{)}^2}4<$

$=1.42$

$\Rightarrow \sum (-1{)}^n\frac{a_n}{(1+\ln n{)}^2}$ 他对收领, 故收领.

(B) .$\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n$ : 令 $a_n=\frac{(-1{)}^n+1}{7n}$ 则 $0\le a_n\le \frac{1}{n}$

$$\begin{array}{r}\text{ 且 }\begin{array}{rl}& \sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n=\sum _{n=1}^{\mathrm{\infty }}\frac{1+(-1{)}^n}{2n}\\ & =\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n}+\frac{(-1{)}^n}{2n})\\ & \Rightarrow \text{ 岁 发 }\end{array}\end{array}$$

(C) $a_n=\frac{1}{2n}.0<a_n<\frac{1}{n}$ .但 $\sum a_n$ 发.

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

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

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

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

而 $\sum _{n=2}^{\mathrm{\infty }}\frac{1}{n\ln n}{4}_2\Rightarrow \sum _{n=2}^{\mathrm{\infty }}\frac{1}{2n(\ln n+1)}{4}_2^{\prime }$

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

(5) .古 $\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n$ 在 $x=-2$ 处收(C)

则 $\Rightarrow |x|<2$ 日ま, $\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n$ 秉对收领

$\Rightarrow |x|<2$ 时, ${\left(\sum _{n=1}^{\mathrm{\infty }}{a}_n{x}^n\right)}^{\prime }\{$ 促对收做.

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

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

(2) . (1). $r=\csc \theta e^{r\cos \theta }$ $\Rightarrow r\sin \theta =e^{r\cos \theta }\Rightarrow y=e^x$

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

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

$$\begin{array}{rl}& a(t)=(e^t,t,\sin 2t)=(e^t-1,\frac{t^2}{2},-\frac{1}{2}\cos 2t+\frac{1}{2})+(1,0,0)+(1,0,0)\\ & +(1,0,0)\\ & =(e^{+},\frac{t^2}{2},\frac{1}{2}-\frac{1}{2}\cos 2t)\\ & \Rightarrow \gamma (t)={\int }_0^{t}v(x)dx+r(0)\\ & =(e^t-1,\frac{t^3}{6},\frac{t}{2}-\frac{1}{4}\sin 2t).\\ & \text{ (4). }\sin x=x-\frac{x^3}{3!}+\cdots \\ & \Rightarrow {\underset{x\to 0}{}}_{l_{x\to 0}}\frac{3\sin (2x)-2\sin (3x)}{6x^3+x^4}\\ & =\underset{x\to 0}{lim}\frac{3(2x-\frac{(2x{)}^3}{3!}+0\left(x^3\right))-2(3x-\frac{(3x{)}^3}{3!}+0\left(x^3\right))}{6x^3+x^4}\\ & =\underset{x\to 0}{\underset{―}{\lim }}\frac{(-4+9)x^3+o\left(x^3\right)}{6x^3+x^4}=\frac{5}{6}\text{. }\\ & \text{ (5). }S=\sum _{n=1}^{\mathrm{\infty }}\frac{(-\ln 2{)}^n}{n!}\end{array}$$

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

(1) .${\ell }_{n\to \mathrm{\infty }}|\frac{\ln (n+1)\cdot (x-1{)}^{n+1}}{n+1}\cdot \frac{n}{\ln n\cdot (x-1{)}^n}|$

$=l_{n\to \mathrm{\infty }}|\frac{\ln (n+1)}{\ln n}\cdot \frac{n}{n+1}|\cdot |x-1|=|x-1|<1$ 肘, $\sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}(x-1{)}^n$ 绝收

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

Shift

$$f^{\prime }(x)=\frac{1-\ln x}{x^2}<0(x⩾3)$$$$\Rightarrow n⩾3,\left\{\frac{\ln n}{n}\right\}b\text{, 且 }\frac{l_{n\to \mathrm{\infty }}}{}\frac{\ln n}{n}=0\text{. }$$

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

$x=2$ 肘 $\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}:f(x)=\frac{\ln x}{x}>0,\zeta (x⩾3)$, 连卖

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

$$\Rightarrow \sum _{n=3}^{\mathrm{\infty }}\frac{\ln n}{n}\mathrm{发}\mathrm{散}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{\ln n}{n}\mathrm{发}\mathrm{散}$$

$\Rightarrow$ 收敛域: $0\le x<2$ or $[0,2$ ).

(2). $0<x<2$ 时, 绝对收敛

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

$\Rightarrow x=0$ 时, 条件收敛.

4.(1) $\underset{(x,y)⩾10,0)}{lim}\frac{y^2{\sin }^{2}x}{x^4+y^4}$

Sol.不存在

$$\begin{array}{rl}\text{ 令 }y=kx(k\ne 0).\text{ 上戒 }& =\frac{k_{x\to 0}}{}\frac{k^2}{(1+k^4)}\cdot \frac{x^2\cdot {\sin }^{2}x}{x^4}\\ & =l_{x\to 0}\frac{k^2}{1+k^4}\cdot \frac{{\sin }^{2}x}{x^2}=\frac{k^2}{1+k}\end{array}$$

(2)

$$\begin{array}{rl}& \left|\frac{2xy}{\sqrt{x^2+2y^2}}\right|=|\frac{x}{\sqrt{x^2+2y^2}}\cdot 2y|\\ & \le 2|y|\\ & x,y)\to (0,0)\text{ 时, }2|y|\to 0\Rightarrow \frac{l}{(x,y)\to (0,0)}\frac{2xy}{\sqrt{x^2+2y^2}}=0\end{array}$$

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

$l_z:\{\begin{array}{l}x=1+25\\ y=0\\ z=25.\end{array}$

$\Rightarrow {\tau }_1^1=(1,0,1).{\tau }_2=(2,v,2)$

$\Rightarrow {\tau }_2^{\prime }=2{\tau }_1^{\prime }\Rightarrow L_1\parallel L_2$

$\Rightarrow$ 存在平面 M 过 $L_{1}S{L}_2$.

在 4 上取点 $A(0,1,1)$, 在 $L_2$ 上取一点 $B(1,0,0)$

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

${\overrightarrow{n}}_n=\left|\begin{array}{ccc}i& j& k\\ 1& 0& 1\\ 1& -1& -1\end{array}\right|=(1,2,-1),M\notin$ 点 $B(1,0,0)$

$\Rightarrow$ 平面 $M$ 为: $(x-1)+2y-z=0$

邵 $x+2y-z=1$

6. $0<b<a<1$

(l)

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

而 $0<a<1,\sum _{n=1}^{\mathrm{\infty }}{a}^{n}4<\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{c_n}{\sqrt{(1+a^n)(1+b^n)}}4<$

$$\begin{array}{rl}& \text{ (2) }\\ & {\left(\frac{a^n}{n^2}\right)}^{\frac{1}{n}}<{(\frac{a^n}{n^2}+\frac{b^n}{n})}^{\frac{1}{n}}⩽{\left(\frac{2a^n}{n}\right)}^{\frac{1}{n}}\text{. }\\ & \begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}{\left(\frac{a^n}{n^2}\right)}^{\frac{1}{n}}& =h_{n\to \mathrm{\infty }}\frac{a}{{\left(n^{\frac{1}{n}}\right)}^2},h_{n\to \mathrm{\infty }}{\left(\frac{2a^n}{n}\right)}^{\frac{1}{n}}=l_{n\to \mathrm{\infty }}\frac{2^{\frac{1}{n}}\cdot a}{n^{\frac{1}{n}}}=a.\\ & =a\end{array}\\ & \Rightarrow l_{n\to \mathrm{\infty }}{(\frac{a^n}{n^2}+\frac{b^n}{n})}^{\frac{1}{n}}=a\text{. }\end{array}$$

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

(A)

Sol.$V(t)=(-\sqrt{2}\sin t,\cos t,\cos t)$

$$\begin{array}{rl}& |v(t)|=\sqrt{2{\sin }^{2}t+{\cos }^{2}t+{\cos }^{2}t}=\sqrt{2}\\ \Rightarrow & T(t)=\frac{v}{|v|}=\frac{1}{\sqrt{2}}(-\sqrt{2}\sin t,\cos t,\cos t)\\ \Rightarrow & \frac{d7}{dt}(t)=\frac{1}{\sqrt{2}}(-\sqrt{2}\cos t,-\sin t,-\sin t)\end{array}$$

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

$\Rightarrow N(t)=\frac{d7}{dt}=\frac{1}{\sqrt{2}}(-\sqrt{2}\cos t,-\sin t,-\sin t)$

(B) .$k(t)=\frac{1}{|v(t)|}\cdot \left|\frac{dT}{dt}\right|=\frac{1}{\sqrt{2}}$

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

$$\begin{array}{rl}& =\frac{4}{(x-1{)}^2+4}+\ln (1+(x-1))\\ & =\frac{1}{1+\frac{(x-1{)}^2}{4}}+\ln [1+(x-1)]\\ & \text{ 而 }\frac{1}{1+\frac{(x-1{)}^2}{4}}=\sum _{n=0}^{\mathrm{\infty }}{[-\frac{(x-1{)}^2}{4}]}^n=\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{4^n}\cdot (x-1{)}^{2n},\left|\frac{(x-1{)}^2}{4}\right|<1.\\ & \ln [1+(x-1)]=\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}(x-1{)}^n}{n},-1⩽x-1<1\text{ 即 }0⩽x<2\end{array}$$

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