一(1)A

$$\begin{array}{ll}f(x-az,y-bz)=0& \Rightarrow \end{array}\begin{array}{l}u=x-az\\ v=y-bz& \\ y,z)=f(ax-az,y-bz)=0& f_u(1-a\frac{\partial z}{\partial x})+f_v(-\frac{\partial z}{\partial x})+f_v(1-b\frac{\partial z}{\partial y})=0\Rightarrow (a{f}_u+b{f}_v)\frac{\partial z}{\partial x}=f_u\\ & \end{array}$$

or $F(x,y,z)=f(ax-az,y-bz)=0$

$\Rightarrow :\frac{\partial z}{\partial x}=-\frac{F_x}{F_z},\frac{\partial z}{\partial y}=-\frac{F_y}{F_z}\Rightarrow a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=a\cdot \frac{f_u}{a{f}_u+b{f}_v}+b\cdot \frac{f_v}{a{f}_u+b{f}_v}=1$

$F_z=-a{f}_u-t{f}_v,F_x=f_u,F_y=f_v\Rightarrow$

(2) $C$

(2) $n>0,a_n>0.\underset{n\to \mathrm{\infty }}{lim}n{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{\frac{1}{n}}=1\ne 0\dots ,\sum \frac{1}{n}$ diverges $\Rightarrow \sum a_n$ diverges

(3) $B$

$2x^2+y^2=z^2$ 椭圆锥.

$$\begin{array}{rlr}f(x,y)& =\phi (x+y)+\phi (x-y)+{\int }_{x-y}^{x+y}\psi (t)dt.& \text{ let }\begin{array}{l}x+y=u\\ x-y=v\end{array}\\ \Rightarrow \frac{\partial f}{\partial x}& ={\phi }_u+{\phi }_v+\psi (x+y)-\psi (x-y)& \frac{\partial f}{\partial y^{\prime }}={\phi }_u{\phi }_v+\psi (x+y)+\psi (x-y)\\ \frac{\partial f}{\partial x^2}& ={\phi }_{uu}+{\phi }_{vu}+{\psi }_u(x+y)-{\psi }_v(x-y)\frac{{\partial }^{2}f}{\partial y^2}={\phi }_{uu}+{\phi }_u+{\psi }_u(x+y)-{\psi }_v(x-y)\\ & \Rightarrow \frac{\partial f}{\partial x^2}=\frac{{\partial }^{2}f}{\partial y^2}& \end{array}$$

$\underset{x,y)\to (0,0)}{lim}(1+xy{)}^{\frac{1}{x^2+y^2}}=\underset{x\to 0}{lim}{(1+x^2)}^{\frac{1}{x^2}}=\underset{x\to 0}{lim}{\left[{(1+x^2)}^{\frac{1}{x^2}}\right]}^{\frac{1}{2}}=e^{\frac{1}{2}}\Rightarrow$

$\Rightarrow$ dees not exist

$\underset{\begin{array}{c}(x,y)\to (0,0)\\ y=-x\end{array}}{lim}(1+xy{)}^{\frac{1}{x^2+y^2}}=\underset{x\to 0}{lim}{(1-x^2)}^{\frac{1}{2x^2}}=\underset{x\to 0}{lim}{\left[{(1-x^2)}^{\frac{1}{-x^2}}\right]}^{\frac{1}{-2}}=e^{\frac{1}{-2}}$

(I)

$$\text{begin{aligned} \& =text { (1) } frac{-frac{3}{2}}{vec{a}+vec{b}+vec{c}=overrightarrow{0}} \& Rightarrow leftlvert, begin{array}{l} |vec{a}|^2+vec{a} cdot vec{b}+vec{a} cdot vec{c}=0 vec{a} cdot vec{b}+|vec{b}|^2+vec{b} cdot vec{c}=0 vec{a} cdot vec{c}+vec{b} cdot vec{c}+|vec{c}|^2=0 end{array}right end{aligned}}$$

(1)

(2)

(1) + (2)+ (3)$\Rightarrow 2(\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{b}\cdot \overrightarrow{c})=-3$

$$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}=-\frac{3}{2}$$

(2) $2\overrightarrow{i}-4\overrightarrow{i}+6\overrightarrow{k}$

(1)

$\therefore \overrightarrow{c}+\overrightarrow{c}+\overrightarrow{c}+\overrightarrow{a}\times \overrightarrow{a}|=\overrightarrow{a}+\overrightarrow{i}+\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ 1& 2& 1\\ -1& 1& 1\end{array}|=\overrightarrow{i}-2\overrightarrow{j}+3\overrightarrow{k}$

2)$\Rightarrow \overrightarrow{c}=K(1,-2\times 3)$ of $x$

$$\because \overrightarrow{c}\cdot (\overrightarrow{ı}-2\overrightarrow{j}+\overrightarrow{k})=16\Rightarrow 8k=16\Rightarrow k=2\Rightarrow \overrightarrow{c}=(2,-4,6)$$

(3) . 1

Since $\sum _{n=2}^{\mathrm{\infty }}(\tan \frac{1}{n}-k\ln (1-\frac{1}{n}))$ converges, let $x=\frac{1}{n},n\to \mathrm{\infty }\Rightarrow x\to 0$

$\Rightarrow \underset{x\to 0}{lim}\left|\frac{\tan x-k\ln (1-x)}{x^p}\right|=c⩾0,p>1$

$$\Rightarrow \underset{x\to 0}{lim}|\frac{{\sec }^{2}x}{p{x}^{p-1}\to 0}+\frac{k}{1-x}|=c⩾0,p>\mid {\omega }_{1}p{x}^{p-1}\to 0$$$$\Rightarrow 1+k\to 0\Rightarrow k=-1$$

8 (when $k=-1\underset{x\to 0}{lim}\frac{\tan x+\ln (1-x)}{x^p}=\underset{x\to 0}{lim}\frac{{\sec }^{2}x-\frac{1}{1-x}}{p{x}^{p-1}}=\underset{x\to 0}{lim}\frac{(1-x{)}^{s^2}x-1}{p{x}^{p-1}}=$

$$=\underset{x\to 0}{lim}\frac{-{\sec }^{2}x+2(1-x){\sec }^{2}x\tan x}{p(p-1)x^{p-2}}=\underset{x\to 0}{lim}\frac{-1+2(1-x)\tan x\to -1\ne 0}{p(p-1)x^{p-2}}\Rightarrow p-2=0$$

$$\begin{array}{rl}& f=y(x)=\sin x\Rightarrow f^{\prime }(x)=\cos x,f^{\prime \prime }(x)=-\sin x\\ & k=\frac{|f^{\prime \prime }(x)|}{{(1+{(f^{\prime }(x))}^2)}^{\frac{3}{2}}}=\frac{|-\sin x|}{{(1+{\cos }^{2}x)}^{\frac{3}{2}}}=\frac{|\sin x|}{{(2-{\sin }^{2}x)}^{\frac{3}{2}}}\end{array}$$

when $\mid \sin x$ max, ${\sin }^{2}x$ is max

$$\begin{array}{ll}\Rightarrow 2-{\sin }^{2}x\text{ is min. }& \Rightarrow |\sin x|\cdot \frac{1}{{(2-{\sin }^{2}x)}^{\frac{3}{2}}}\text{ is max }\\ \Rightarrow {(2-{\sin }^{2}x)}^{\frac{3}{2}}\text{ is min. }\\ \Rightarrow \frac{1}{{(2-{\sin }^{2}x)}^{\frac{3}{2}}}\text{ is max }& \Rightarrow k\text{ is max }\\ & \Rightarrow |\sin x|=1\Rightarrow k=1\end{array}$$

(5) $2-2\ln 2$

$$\begin{array}{rl}{(z^2+y)}^x=xy& \Rightarrow x\ln (z+y)=\ln xy\\ & \Rightarrow \ln (z+y)+x\cdot \frac{1}{z+y}\cdot \frac{\partial z}{\partial x}=\frac{1}{xy}\cdot y\\ & \text{ let }x=1,y=2\cdot z=0\\ & \Rightarrow \ln 2+\frac{1}{2}\cdot \frac{\partial z}{\partial x}=1\Rightarrow \frac{\partial z}{\partial x}=2-z\ln 2\end{array}$$

or let $F(x,y,z)=x\ln (z+y)-\ln xy$

$$\begin{array}{rl}\Rightarrow F_x& =\ln (z+y)\\ F_z& =\frac{x}{z+y}=\frac{x}{2}=\ln 2-1\\ \Rightarrow \frac{\partial z}{\partial x}& =-\frac{F_x}{Fz}=2-2\ln 2\end{array}$$$$\begin{array}{rl}A& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}[a^2(1+\cos \theta {)}^2-a^2]d\theta \text{ or }A=2{\int }_0^{\frac{\pi }{2}}\frac{1}{2}[a^2(1+\cos \theta {)}^2-a^2]d\theta \\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}[2a^2\cos \theta +a^2{\cos }^2\theta ]d\theta \\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}[2a^2\cos \theta +a^2\cdot \frac{1+\cos 2\theta }{2}]d\theta .\\ & {=\frac{1}{2}(2a^2\sin \theta +\frac{1}{2}{a}^2\theta +\frac{a^2}{4}\sin 2\theta )]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ & =2a^2\end{array}$$$$\begin{array}{rl}{A}_1)& =2{\int }_0^{\pi }\frac{1}{2}[a^2(1+\cos \theta {)}^2]\\ & ={\int }_0^{\pi }(a^2+2a^2\cos \theta +a^2{\cos }^2\theta )d\theta .\\ & ={\int }_0^{\pi }(a^2+2a^2\cos \theta +a^2\cdot \frac{(+\cos \theta }{2})d\theta .\\ & {=a^2\theta +2a^2\sin \theta +\frac{a^2}{2}\theta +\frac{a^2}{4}\sin 2\theta ]}_0^{\pi }\\ & =\frac{3}{2}{a}^2\pi \\ \Rightarrow & A_2=A_1-A=\frac{3}{2}{a}^2\pi -(2+\frac{\pi }{4})a^2=\frac{5}{4}\pi a^2-2a^2=(\frac{5}{4}\pi -2)a^2\end{array}$$

4. $\overrightarrow{r}(t)=f(t)\overrightarrow{i}+g(t)\overrightarrow{j}+h(t)\overrightarrow{k}$., $f(t)={\int }_0^t\cos \left(x^2\right)dx,g(t)=-t\cos t$

$$\begin{array}{rl}h(t)& =\sum _{n=1}^{\mathrm{\infty }}{t}^n\\ \Rightarrow h^{\prime }(t)& =\sum _{i=1}^{\mathrm{\infty }}{t}^{n-1}=\sum _{n=0}^{\mathrm{\infty }}{t}^n=\frac{1}{1-t}\\ f^{\prime }(t)& =\cos \left(t^2\right)\\ g^{\prime }(t)& =-\cos t+t\sin t\end{array}$$

$\Rightarrow f^{\prime }(0)=1,g^{\prime }(0)=-1,h^{\prime }(0)=1$

$\Rightarrow f^{\prime }(0)=(1,-1,1)$

5. (1) $f(x,y)=\{\begin{array}{cc}y\sin \frac{1}{x^2+y^2}& ,(x,y)\ne (0,0)\\ 0& ,(x,y)=(0,0)\end{array}$

$$\begin{array}{r}\underset{(x,y)\to (0,0)}{lim}f(x,y)=\underset{(x,y)\to (0,0)}{lim}y\sin \frac{1}{x^2+y^2}=0\text{ since }|\sin \frac{1}{x^2+y^2}|⩽1\\ 0⩽|f(x,y)|⩽|y|\to 0\end{array}$$

(2) $f_x(0,0)=\underset{h\to 0}{lim}\frac{f(h,0)-f(0,0)}{h}=\underset{h\to 0}{lim}\frac{0-0}{h}=0$

$$f_y(0,0)=\underset{h\to 0}{lim}\frac{f(0,h)-f(0,0)}{h}=\underset{h\to 0}{lim}\frac{h\sin h{h}^{-0}}{h}=\underset{h\to 0}{lim}\sin \frac{1}{h^2}\text{ does nt exist. }$$

6.(1) $\underset{(x,y)\to (0,0)}{lim}\frac{xy}{\sqrt{x^2+y^2}}=\underset{r\to 0}{lim}\frac{r^2\cos \theta \sin \theta }{|r|}=\underset{r\to 0}{lim}|r|\cos \theta \sin \theta =0$. since |cos $\theta \sin \theta \mid ⩽1$. or $0⩽\left|\frac{x^y}{\sqrt{x^2+y^2}}\right|⩽|y|\to 0$

(2) $\underset{(x,y)\to (0,0)}{lim}\frac{x{y}^3+2x^2{y}^4}{x^2+y^6}=\underset{(x,y)\to (0,0)}{lim}\frac{k{y}^6+2k^2{y}^7}{k^2{y}^6+y^6}=\underset{\begin{array}{c}\\ y\to 0\end{array}}{lim}\frac{k+2k^{2}y}{k^2+1}=\frac{k}{k^2+1}$ $x=k{y}^{\prime };$ $\Rightarrow$ the limit does not exist

T. (1) $f(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{n+2}{n(n+1)}{x}^n$

$$\left|\frac{u_{n+1}}{u_n}\right|=\left|\frac{(n+3)}{(n+2)(n+1),(n+2)}x\right|\to |x|<1\Rightarrow -1<x<1$$

when $x=1\sum _{n=1}^{\mathrm{\infty }}\frac{n+2}{n(n+1)}$ diverges since $\frac{\frac{n+2}{n(n+1)}}{\frac{1}{n}}\to 1,\sum \frac{1}{n}$ diverges

when $x=-1,\sum _{n=1}^{\mathrm{\infty }}(-1{)}^n\frac{n+2}{n(n+1)}$ converges conditionally. Sine $\sum \frac{n+2}{n(n+1)}$ diverges

$$\Rightarrow [-1,1]$$$$a_n=\frac{n+2}{n(n+1)}>0\to 0$$

(2) $f(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{n+2}{n(n+1)}{x}^n=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n(n+1)}{x}^n=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}+2\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-2\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=2\sum _{i=1}^{\mathrm{\infty }}{x}^n\frac{\sum _n^n}{n}\frac{x^n{x}^n}{}$

$$g(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=\frac{1}{x}\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{n+1},{\left(\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{n+1}\right)}^{\prime }=\sum _{n=1}^{\mathrm{\infty }}{x}^n=\frac{x}{1-x}$$

if $x\ne 0$

$$\begin{array}{rl}\Rightarrow \sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{n+1}& =\int \frac{x}{1-x}dx=\int \frac{1-(1-x)}{1-x}dx=\int \frac{dx}{1-x}-x\\ & =-\ln (1-x)-x+c\\ \text{ let }x=0.& \Rightarrow c=0\end{array}$$

$\Rightarrow g(x)=-\frac{\ln (1-x)}{x}-1$, if $x\ne 0$

$$\begin{array}{rl}& h(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}{}^n\Rightarrow h^{\prime }(x)=\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1}=\frac{1}{1-x}\Rightarrow h(x)=-\ln (1-x)\\ & \Rightarrow \\ & f(x)=\{\begin{array}{ll}-2\ln (1-x)+\frac{\ln (1-x)}{x}+1=1+(\frac{1}{x}-2)\ln (1-x),& x\ne 0.\\ 0.& x=0.\end{array}\end{array}$$

8. $\sum _{n=2}^{\mathrm{\infty }}\frac{(-1{)}^{p+1}}{n^p(\ln n{)}^2},p>0$. Let $u_n=\frac{1}{n^p(\ln \eta {)}^2}\{\begin{array}{l}\text{ converges absolutely if }p⩾1\\ \text{ converges conditionally if }0<p<1\end{array}$

(2) if ${p=1\cdot \sum _{n-2}^{\mathrm{\infty }}\frac{1}{n(\ln n{)}^2},f(x)=\frac{1}{x(\ln x{)}^2},{\int }_2^{\mathrm{\infty }}\frac{dx}{x(\ln x{)}^2}=-\frac{1}{\mid nx}]}_2^{\mathrm{\infty }}=\frac{1}{\ln 2}$ converges $\Rightarrow \sum u_n$ ca query

$\to \mathrm{\Sigma }{\mathrm{K}}_n$ converge concrify

9. $\underset{n\to \mathrm{\infty }}{lim}((n^2-n)e^{\frac{1}{n}}-\sqrt{n^4+1})$

lef $f(x)=(x^2-x)e^{\frac{1}{x}}-\sqrt{x^4+1}$

$$\begin{array}{rl}& =(x^2-x)(1+\frac{1}{x}+\frac{1}{2!x^2}+\cdots +\frac{1}{k!x^k}+\cdots )\\ & =x^2-\frac{1}{2}-\frac{1}{3x}+\cdots \end{array}$$$$\begin{array}{rl}\sqrt{x^4+1}=x^2\sqrt{1+\frac{1}{x^4}}& =x^2(1+\frac{1}{2x^4}-\frac{1}{8x^8}+\cdots )\\ & =x^2+\frac{1}{2x^2}-\frac{1}{8x^6}+\cdots \end{array}$$$$\begin{array}{rl}& \Rightarrow f(x)=-\frac{1}{2}-\frac{1}{3x}+\cdots \\ & \Rightarrow \underset{x\to \mathrm{\infty }}{lim}f(x)=-\frac{1}{2}\end{array}$$