1．函数 $y=\frac{1}{\sqrt{-x^2+x+2}}$$y=\frac{1}{\sqrt{-x^2+x+2}}$y=(1)/(sqrt(-x^(2)+x+2))y=\frac{1}{\sqrt{-x^{2}+x+2}} 的定义域是 $$$$qquad\qquad
1 $$$$qquad\qquad . $y=\frac{1}{\sqrt{-x^2+x+2}}$$y=\frac{1}{\sqrt{-x^2+x+2}}$y=(1)/(sqrt(-x^(2)+x+2))y=\frac{1}{\sqrt{-x^{2}+x+2}} The domain of the function is
2．函数 $y=\sqrt{\frac{x}{x-4}}$$y=\sqrt{\frac{x}{x-4}}$y=sqrt((x)/(x-4))y=\sqrt{\frac{x}{x-4}} 定义域 $$$$qquad\qquad ．
2. Function $y=\sqrt{\frac{x}{x-4}}$$y=\sqrt{\frac{x}{x-4}}$y=sqrt((x)/(x-4))y=\sqrt{\frac{x}{x-4}} Define the field $$$$qquad\qquad .

3．函数 $y=3x^3$$y=3x^3$y=3x^(3)y=3 x^{3} 的凸区间是 $$$$qquad\qquad ．
3. $y=3x^3$$y=3x^3$y=3x^(3)y=3 x^{3} The convex interval of the function is $$$$qquad\qquad .

4．函数 $f(x)=x^3-3x^2-9x+5$$f(x)=x^3-3x^2-9x+5$f(x)=x^(3)-3x^(2)-9x+5f(x)=x^{3}-3 x^{2}-9 x+5 的最大值是 $$$$qquad\qquad －．
4. The $f(x)=x^3-3x^2-9x+5$$f(x)=x^3-3x^2-9x+5$f(x)=x^(3)-3x^(2)-9x+5f(x)=x^{3}-3 x^{2}-9 x+5 maximum value of the function is $$$$qquad\qquad -.

5．假定隐函数 $y=x+\frac{1}{2}\cos y$$y=x+\frac{1}{2}\cos y$y=x+(1)/(2)cos yy=x+\frac{1}{2} \cos y ，那么 $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad ．
5. Assuming the implicit function $y=x+\frac{1}{2}\cos y$$y=x+\frac{1}{2}\cos y$y=x+(1)/(2)cos yy=x+\frac{1}{2} \cos y , then $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad .

6．假定 $f(x)=x\cos x$$f(x)=x\cos x$f(x)=x cos xf(x)=x \cos x ，那么 $f^{\prime \prime }(x)=$$f^{\prime \prime }(x)=$f^('')(x)=f^{\prime \prime}(x)= $$$$qquad\qquad ．
6. Suppose $f(x)=x\cos x$$f(x)=x\cos x$f(x)=x cos xf(x)=x \cos x that then $$$$qquad\qquad . $f^{\prime \prime }(x)=$$f^{\prime \prime }(x)=$f^('')(x)=f^{\prime \prime}(x)=

7．若 $y=f(\cos x)$$y=f(\cos x)$y=f(cos x)y=f(\cos x) 且 $f(u)$$f(u)$f(u)f(u) 是可导的，那么 $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad ．
7. If $y=f(\cos x)$$y=f(\cos x)$y=f(cos x)y=f(\cos x) and $f(u)$$f(u)$f(u)f(u) is derivable, then $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad .

8．若 $y=f(\tan x)$$y=f(\tan x)$y=f(tan x)y=f(\tan x) 且 $f(u)$$f(u)$f(u)f(u) 是可导的，那么 $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad .
8. If $y=f(\tan x)$$y=f(\tan x)$y=f(tan x)y=f(\tan x) and $f(u)$$f(u)$f(u)f(u) is derivable, then $\frac{dy}{dx}=$$\frac{dy}{dx}=$(dy)/(dx)=\frac{d y}{d x}= $$$$qquad\qquad .

9．若 $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n^p}$sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} 收玫，则 $p$$p$pp 的取值范围 $$$$qquad\qquad ．
9. If the rose $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n^p}$sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} is harvested, then $p$$p$pp the value range $$$$qquad\qquad of .

10．若 $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n^p}$sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} 是发散的，则 $p$$p$pp 的取值范围 $$$$qquad\qquad ．
10. If it $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^p}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n^p}$sum_(n=1)^(oo)(1)/(n^(p))\sum_{n=1}^{\infty} \frac{1}{n^{p}} is divergent, then the range $p$$p$pp $$$$qquad\qquad of values of .

11．函数 $\frac{1}{2+x}$$\frac{1}{2+x}$(1)/(2+x)\frac{1}{2+x} 在 $x=0$$x=0$x=0x=0 处可幂级数展开 则 $\frac{1}{2+x}=$$\frac{1}{2+x}=$(1)/(2+x)=\frac{1}{2+x}= $$$$qquad\qquad .
11. The function $\frac{1}{2+x}$$\frac{1}{2+x}$(1)/(2+x)\frac{1}{2+x} can be expanded $x=0$$x=0$x=0x=0 by a power series $\frac{1}{2+x}=$$\frac{1}{2+x}=$(1)/(2+x)=\frac{1}{2+x}= $$$$qquad\qquad at .

12．函数 $\frac{1}{1+x}$$\frac{1}{1+x}$(1)/(1+x)\frac{1}{1+x} 在 $x=0$$x=0$x=0x=0 处可幂级数展开．那么 $\frac{1}{1+x}=$$\frac{1}{1+x}=$(1)/(1+x)=\frac{1}{1+x}= $$$$qquad\qquad ．
12. The function $\frac{1}{1+x}$$\frac{1}{1+x}$(1)/(1+x)\frac{1}{1+x} can be expanded $x=0$$x=0$x=0x=0 by a power series at . $\frac{1}{1+x}=$$\frac{1}{1+x}=$(1)/(1+x)=\frac{1}{1+x}= Then $$$$qquad\qquad .

1．假定 $f(x)=\{\begin{array}{ll}(x-1{)}^2& -1<x\le 0\\ x+1,& 0<x\le 1\end{array}$$f(x)=\left\{\begin{array}{l}(x-1{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1<x\le 0\\ x+1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x\le 1\end{array}\right.$f(x)={[(x-1)^(2),-1 < x <= 0],[x+1",",0 < x <= 1]:}f(x)=\left\{\begin{array}{ll}(x-1)^{2} & -1<x \leq 0 \\ x+1, & 0<x \leq 1\end{array}\right. ，那么 $\underset{x\to 0}{lim}f(x)=()$$\underset{x\to 0}{lim} f(x)=()$lim_(x rarr0)f(x)=(quad)\lim _{x \rightarrow 0} f(x)=(\quad) ．
1. Suppose $f(x)=\{\begin{array}{ll}(x-1{)}^2& -1<x\le 0\\ x+1,& 0<x\le 1\end{array}$$f(x)=\left\{\begin{array}{l}(x-1{)}^2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-1<x\le 0\\ x+1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0<x\le 1\end{array}\right.$f(x)={[(x-1)^(2),-1 < x <= 0],[x+1",",0 < x <= 1]:}f(x)=\left\{\begin{array}{ll}(x-1)^{2} & -1<x \leq 0 \\ x+1, & 0<x \leq 1\end{array}\right. that then $\underset{x\to 0}{lim}f(x)=()$$\underset{x\to 0}{lim} f(x)=()$lim_(x rarr0)f(x)=(quad)\lim _{x \rightarrow 0} f(x)=(\quad) .
A． 1
B． 0
C．-1
D．不存在  D. Non-existent

2．函数 $f(x)=|x-1|()$$f(x)=|x-1|()$f(x)=|x-1|quad(quad)f(x)=|x-1| \quad(\quad) ．  2. Functions $f(x)=|x-1|()$$f(x)=|x-1|()$f(x)=|x-1|quad(quad)f(x)=|x-1| \quad(\quad) .
A．在 $x=1$$x=1$x=1x=1 处连续且可导  A. Continuous and directable $x=1$$x=1$x=1x=1 at
B．在 $x=1$$x=1$x=1x=1 处不连续  B. Discontinuity $x=1$$x=1$x=1x=1 at
C．在 $x=0$$x=0$x=0x=0 处连续且可导  C. Continuous and conductible $x=0$$x=0$x=0x=0 at
D．在 $x=0$$x=0$x=0x=0 处不连续  D. Discontinuity $x=0$$x=0$x=0x=0 at

3 下列哪个函数在［－1，1］满足罗尔定理．（ ）
3 Which of the following functions satisfies Roll's theorem at [-1,1].( )
A．$f(x)=\frac{1}{x^2}$$f(x)=\frac{1}{x^2}$f(x)=(1)/(x^(2))f(x)=\frac{1}{x^{2}}
B．$f(x)=|x|$$f(x)=|x|$f(x)=|x|f(x)=|x|
C．$f(x)=1-x^2$$f(x)=1-x^2$f(x)=1-x^(2)f(x)=1-x^{2}
D．$f(x)=x^2-2x-1$$f(x)=x^2-2x-1$f(x)=x^(2)-2x-1f(x)=x^{2}-2 x-1

4．当 $x\to \mathrm{\infty }$$x\to \mathrm{\infty }$x rarr oox \rightarrow \infty ，下列有极限的是？（）
4. When $x\to \mathrm{\infty }$$x\to \mathrm{\infty }$x rarr oox \rightarrow \infty , what are the following limits? （）
A，$x\sin \frac{1}{x}$$x\sin \frac{1}{x}$x sin((1)/(x))x \sin \frac{1}{x}
B， $\sin \frac{1}{x^2}$$\sin \frac{1}{x^2}$sin((1)/(x^(2)))\sin \frac{1}{x^{2}}
C，$\frac{x+1}{x^2-x}$$\frac{x+1}{x^2-x}$(x+1)/(x^(2)-x)\frac{x+1}{x^{2}-x}
D， $\cot x$$\cot x$cot x\cot x

5． ${\int }_0^{+\mathrm{\infty }}\frac{1}{1+x^2}dx=()$${\int }_0^{+\mathrm{\infty }} \frac{1}{1+x^2}dx=()$int_(0)^(+oo)(1)/(1+x^(2))dx=(quad)\int_{0}^{+\infty} \frac{1}{1+x^{2}} d x=(\quad) ．
A．$\frac{1}{\pi }$$\frac{1}{\pi }$(1)/(pi)\frac{1}{\pi}
B．$\frac{\pi }{2}$$\frac{\pi }{2}$(pi)/(2)\frac{\pi}{2}
C． 1
D．$\frac{4}{\pi }$$\frac{4}{\pi }$(4)/(pi)\frac{4}{\pi}

6． $\int \frac{1}{1-2x}dx=()$$\int \frac{1}{1-2x}dx=()$int(1)/(1-2x)dx=(quad)\int \frac{1}{1-2 x} d x=(\quad) ．
A．$-\frac{1}{2}\mathrm{In}|1-2x|+C$$-\frac{1}{2}\mathrm{In}|1-2x|+C$-(1)/(2)In|1-2x|+C-\frac{1}{2} \operatorname{In}|1-2 x|+C
B．In $|1-2x|+C$$|1-2x|+C$|1-2x|+C|1-2 x|+C  B． In $|1-2x|+C$$|1-2x|+C$|1-2x|+C|1-2 x|+C
C．$-\frac{1}{2}\mathrm{In}|1-x|+C$$-\frac{1}{2}\mathrm{In}|1-x|+C$-(1)/(2)In|1-x|+C-\frac{1}{2} \operatorname{In}|1-x|+C
D．$-\frac{1}{2}\mathrm{In}|1-2x|$$-\frac{1}{2}\mathrm{In}|1-2x|$-(1)/(2)In|1-2x|-\frac{1}{2} \operatorname{In}|1-2 x|

7． $\int x{e}^{-x^2}dx=()$$\int x{e}^{-x^2}dx=()$int xe^(-x^(2))dx=(quad)\int x e^{-x^{2}} d x=(\quad) ．
A．$-\frac{1}{2}{e}^{-x^2}+C$$-\frac{1}{2}{e}^{-x^2}+C$-(1)/(2)e^(-x^(2))+C-\frac{1}{2} e^{-x^{2}}+C
B．$-2e^{-x^2}+C$$-2e^{-x^2}+C$-2e^(-x^(2))+C-2 e^{-x^{2}}+C
C．$\frac{1}{2}{e}^{-x^2}+C$$\frac{1}{2}{e}^{-x^2}+C$(1)/(2)e^(-x^(2))+C\frac{1}{2} e^{-x^{2}}+C
D．$e^{-x^2}+C$$e^{-x^2}+C$e^(-x^(2))+Ce^{-x^{2}}+C

8．下列哪个级数是收敛（ ）  8. Which of the following series is convergence ( )
A．$\sum _{n=1}^{\mathrm{\infty }}{2}^{\frac{1}{n}}$$\sum _{n=1}^{\mathrm{\infty }} 2^{\frac{1}{n}}$sum_(n=1)^(oo)2^((1)/(n))\sum_{n=1}^{\infty} 2^{\frac{1}{n}}
B．$\sum _{n=1}^{\mathrm{\infty }}(\sqrt{n+1}-\sqrt{n})$$\sum _{n=1}^{\mathrm{\infty }} (\sqrt{n+1}-\sqrt{n})$sum_(n=1)^(oo)(sqrt(n+1)-sqrtn)\sum_{n=1}^{\infty}(\sqrt{n+1}-\sqrt{n})
C．$\sum _{n=1}^{\mathrm{\infty }}[{\left(\frac{3}{4}\right)}^n+\frac{1}{\sqrt{n+1}}]$$\sum _{n=1}^{\mathrm{\infty }} \left[{\left(\frac{3}{4}\right)}^n+\frac{1}{\sqrt{n+1}}\right]$sum_(n=1)^(oo)[((3)/(4))^(n)+(1)/(sqrt(n+1))]\sum_{n=1}^{\infty}\left[\left(\frac{3}{4}\right)^{n}+\frac{1}{\sqrt{n+1}}\right]
D．$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{3n^3}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{3n^3}$sum_(n=1)^(oo)(1)/(3n^(3))\sum_{n=1}^{\infty} \frac{1}{3 n^{3}}

9．下列哪个级数是发散的（ ）。  9. Which of the following series is divergent ( ).
A．$\sum _{n=1}^{\mathrm{\infty }}(1-\cos \frac{\pi }{n})$$\sum _{n=1}^{\mathrm{\infty }} \left(1-\cos \frac{\pi }{n}\right)$sum_(n=1)^(oo)(1-cos((pi )/(n)))\sum_{n=1}^{\infty}\left(1-\cos \frac{\pi}{n}\right)
B．$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n^2}$sum_(n=1)^(oo)(1)/(n^(2))\sum_{n=1}^{\infty} \frac{1}{n^{2}}
C．$\sum _{n=2}^{\mathrm{\infty }}\sin \frac{1}{\sqrt{n}}$$\sum _{n=2}^{\mathrm{\infty }} \sin \frac{1}{\sqrt{n}}$sum_(n=2)^(oo)sin((1)/(sqrtn))\sum_{n=2}^{\infty} \sin \frac{1}{\sqrt{n}}
D．$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{n!}$sum_(n=1)^(oo)(1)/(n!)\sum_{n=1}^{\infty} \frac{1}{n!} ；

10．级数 $\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{1}{n+1}$$\sum _{n=0}^{\mathrm{\infty }} (-1{)}^n\frac{1}{n+1}$sum_(n=0)^(oo)(-1)^(n)(1)/(n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{n+1} 是（ ）
10. Series $\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{1}{n+1}$$\sum _{n=0}^{\mathrm{\infty }} (-1{)}^n\frac{1}{n+1}$sum_(n=0)^(oo)(-1)^(n)(1)/(n+1)\sum_{n=0}^{\infty}(-1)^{n} \frac{1}{n+1} Yes ( )
A．绝对收玫  A. Absolute Harvest
B．条件收敛  B. Conditional convergence
C．发散  C. Divergence
D．收敛性不能确定  D. Convergence cannot be determined

1．计算 $\underset{x\to 1}{lim}\frac{x^2-1}{x^2+2x-3}$$\underset{x\to 1}{lim} \frac{x^2-1}{x^2+2x-3}$lim_(x rarr1)(x^(2)-1)/(x^(2)+2x-3)\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+2 x-3} ．  1. Calculation $\underset{x\to 1}{lim}\frac{x^2-1}{x^2+2x-3}$$\underset{x\to 1}{lim} \frac{x^2-1}{x^2+2x-3}$lim_(x rarr1)(x^(2)-1)/(x^(2)+2x-3)\lim _{x \rightarrow 1} \frac{x^{2}-1}{x^{2}+2 x-3} .
2．计算 $\underset{x\to \mathrm{\infty }}{lim}\frac{2x^3+3x^2+5}{7x^3+4x^2-1}$$\underset{x\to \mathrm{\infty }}{lim} \frac{2x^3+3x^2+5}{7x^3+4x^2-1}$lim_(x rarr oo)(2x^(3)+3x^(2)+5)/(7x^(3)+4x^(2)-1)\lim _{x \rightarrow \infty} \frac{2 x^{3}+3 x^{2}+5}{7 x^{3}+4 x^{2}-1} ．  2. Calculation $\underset{x\to \mathrm{\infty }}{lim}\frac{2x^3+3x^2+5}{7x^3+4x^2-1}$$\underset{x\to \mathrm{\infty }}{lim} \frac{2x^3+3x^2+5}{7x^3+4x^2-1}$lim_(x rarr oo)(2x^(3)+3x^(2)+5)/(7x^(3)+4x^(2)-1)\lim _{x \rightarrow \infty} \frac{2 x^{3}+3 x^{2}+5}{7 x^{3}+4 x^{2}-1} .
3．若 $y=e^x\cos x+\frac{x}{1-x}$$y=e^x\cos x+\frac{x}{1-x}$y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} ，则 $y^{\prime }$$y^{\prime }$y^(')y^{\prime} ．
3. If $y=e^x\cos x+\frac{x}{1-x}$$y=e^x\cos x+\frac{x}{1-x}$y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} , then $y^{\prime }$$y^{\prime }$y^(')y^{\prime} .
4．若 $y=e^x\cos x+\frac{x}{1-x}$$y=e^x\cos x+\frac{x}{1-x}$y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} ，则 $y^{\prime }$$y^{\prime }$y^(')y^{\prime} ．
4. If $y=e^x\cos x+\frac{x}{1-x}$$y=e^x\cos x+\frac{x}{1-x}$y=e^(x)cos x+(x)/(1-x)y=e^{x} \cos x+\frac{x}{1-x} , then $y^{\prime }$$y^{\prime }$y^(')y^{\prime} .
5．计算 ${\int }_0^1(\sqrt{x}-x^2)dx$${\int }_0^1 \left(\sqrt{x}-x^2\right)dx$int_(0)^(1)(sqrtx-x^(2))dx\int_{0}^{1}\left(\sqrt{x}-x^{2}\right) d x  5 ${\int }_0^1(\sqrt{x}-x^2)dx$${\int }_0^1 \left(\sqrt{x}-x^2\right)dx$int_(0)^(1)(sqrtx-x^(2))dx\int_{0}^{1}\left(\sqrt{x}-x^{2}\right) d x . Calculations
6．计算 ${\int }_0^{\pi }(\sin x+x)dx$${\int }_0^{\pi } (\sin x+x)dx$int_(0)^(pi)(sin x+x)dx\int_{0}^{\pi}(\sin x+x) d x  6 ${\int }_0^{\pi }(\sin x+x)dx$${\int }_0^{\pi } (\sin x+x)dx$int_(0)^(pi)(sin x+x)dx\int_{0}^{\pi}(\sin x+x) d x . Calculations
7．求 ${\int }_0^{2}x\sqrt{4-x^2}dx$${\int }_0^2 x\sqrt{4-x^2}dx$int_(0)^(2)xsqrt(4-x^(2))dx\int_{0}^{2} x \sqrt{4-x^{2}} d x ．  7. Seek ${\int }_0^{2}x\sqrt{4-x^2}dx$${\int }_0^2 x\sqrt{4-x^2}dx$int_(0)^(2)xsqrt(4-x^(2))dx\int_{0}^{2} x \sqrt{4-x^{2}} d x .
8．求 ${\int }_0^{\pi }x\cos xdx$${\int }_0^{\pi } x\cos xdx$int_(0)^(pi)x cos xdx\int_{0}^{\pi} x \cos x d x ．  8. Seek ${\int }_0^{\pi }x\cos xdx$${\int }_0^{\pi } x\cos xdx$int_(0)^(pi)x cos xdx\int_{0}^{\pi} x \cos x d x .
9．求 $\int x^3\ln xdx$$\int x^3\ln xdx$intx^(3)ln xdx\int x^{3} \ln x d x ．  9. Seek $\int x^3\ln xdx$$\int x^3\ln xdx$intx^(3)ln xdx\int x^{3} \ln x d x .
10．求 $\int x\cos x^{2}dx$$\int x\cos x^{2}dx$int x cos x^(2)dx\int x \cos x^{2} d x ．  10. Seek $\int x\cos x^{2}dx$$\int x\cos x^{2}dx$int x cos x^(2)dx\int x \cos x^{2} d x .
11 求出下列级数的收玫区间 $\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{2n+1}$$\sum _{n=0}^{\mathrm{\infty }} \frac{x^{2n+1}}{2n+1}$sum_(n=0)^(oo)(x^(2n+1))/(2n+1)\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1} ．
11 Find the harvest interval for the following series $\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{2n+1}$$\sum _{n=0}^{\mathrm{\infty }} \frac{x^{2n+1}}{2n+1}$sum_(n=0)^(oo)(x^(2n+1))/(2n+1)\sum_{n=0}^{\infty} \frac{x^{2 n+1}}{2 n+1} .
12．求出级数的和 $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}+\frac{1}{3^n}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{2^n}+\frac{1}{3^n}$sum_(n=1)^(oo)(1)/(2^(n))+(1)/(3^(n))\sum_{n=1}^{\infty} \frac{1}{2^{n}}+\frac{1}{3^{n}} ．  12. Find the sum of the series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}+\frac{1}{3^n}$$\sum _{n=1}^{\mathrm{\infty }} \frac{1}{2^n}+\frac{1}{3^n}$sum_(n=1)^(oo)(1)/(2^(n))+(1)/(3^(n))\sum_{n=1}^{\infty} \frac{1}{2^{n}}+\frac{1}{3^{n}} .
四．应用题  Four. Vocabulary questions
1．一人在家制作手工艺品，用以在网上销售，每周制作 $x$$x$xx 件的总成本 $C_{\text{是制作}}$$C_{\text{是制作 }}$C_("是制作 ")C_{\text {是制作 }}量的函数 $C(x)=10x^2+60x-100$$C(x)=10x^2+60x-100$C(x)=10x^(2)+60 x-100C(x)=10 x^{2}+60 x-100 元，如果每件工艺品售价 300 元，且所制作的工艺品可以全部售出，求使利润最大的周制作量及最大利润
1. A person makes handicrafts at home for online sales, and the total cost of the $x$$x$xx pieces produced per week is $C(x)=10x^2+60x-100$$C(x)=10x^2+60x-100$C(x)=10x^(2)+60 x-100C(x)=10 x^{2}+60 x-100 a function $C_{\text{是制作}}$$C_{\text{是制作 }}$C_("是制作 ")C_{\text {是制作 }} of yuan, if each handicraft is sold for 300 yuan, and the handicrafts made can be sold in full, so as to make the largest weekly production volume and the maximum profit

2．已知某产品的总收益函数为 $R=100+33x-4x^2$$R=100+33x-4x^2$R=100+33 x-4x^(2)R=100+33 x-4 x^{2}（万元），总成本函数为 $C=x^3-7x^2+24x+6$$C=x^3-7x^2+24x+6$C=x^(3)-7x^(2)+24 x+6C=x^{3}-7 x^{2}+24 x+6（万元），其中 $\mathit{x}$$\mathit{x}$x\boldsymbol{x} 表示产品的产量（单位千件），求利润函数以及该产品获得最大利润时的产量和最大利润。
2. It is known that the total revenue function of a product is $R=100+33x-4x^2$$R=100+33x-4x^2$R=100+33 x-4x^(2)R=100+33 x-4 x^{2} (10,000 yuan) and the total cost function is $C=x^3-7x^2+24x+6$$C=x^3-7x^2+24x+6$C=x^(3)-7x^(2)+24 x+6C=x^{3}-7 x^{2}+24 x+6 (10,000 yuan), where $\mathit{x}$$\mathit{x}$x\boldsymbol{x} represents the output of the product (in 1,000 pieces), finds the profit function and the output and maximum profit when the product obtains the maximum profit.