114年中央數學碩士班-微積分詳解

國立中央大學114學年度碩士班考試入學試題

系所: 數學系碩士班 應用數學組(一般生)
數學系碩士班 應用數學組(在職生)
科目: 微積分

甲(填充題):共8題,每題9分。請將答案依題號順序寫在答案卷上,不必寫演算過程。

解答:$$L=(2x)^{3/x} \Rightarrow \ln L= {3\ln (2x) \over x} \Rightarrow \lim_{x \to \infty} {3\ln (2x) \over x} =\lim_{x \to \infty}{3/x\over 1} =0 \Rightarrow \lim_{x\to \infty} L= e^0= \bbox[red, 2pt]1$$

解答:$$\sum_{n=2}^\infty {n+2\over n(n+1)2^{n+1}} =\sum_{n=2}^\infty \left[\left({2\over n}-{1\over n+1} \right) \cdot {1\over 2^{n+1}} \right]= \sum_{n=2}^\infty \left({1\over n\cdot 2^n}-{1\over (n+1)  \cdot 2^{n+1}} \right) \\ =\sum_{n=2}^\infty (a_n-a_{n+1}) =a_2-a_3+a_3-a_4+\cdots =a_2-a_\infty ={1\over 8}-0 =\bbox[red, 2pt]{1\over 8}$$

解答:$$f(x) =(x^2+1)^2\cdot \text{sech}(\ln x) \\\Rightarrow f'(x)=4x(x^2+1)\cdot \text{sech}(\ln x) +(x^2+1) \cdot (-\text{sech} (\ln x) \tanh (\ln x))\cdot {1\over x} \\ \Rightarrow f'(1)= 4\cdot 2\cdot 1+2\cdot 0\cdot 1= \bbox[red, 2pt]8$$

解答:$$\cos^2 \theta+ \sin^2\theta =1 \Rightarrow  \cos^2\theta(1+\tan^2\theta)=1 \Rightarrow \cos^2\theta ={1\over 1+\tan^2\theta} \\ \text{Let }u=\tan\theta, \text{ then }du= \sec^2 \theta\,d\theta  ={1\over \cos^2 \theta} \,d \theta =(1+u^2) \,d\theta \Rightarrow d\theta={1\over 1+u^2}du\\\Rightarrow I=\int_0^{\arctan 3} {1\over 1+2\cos^2 \theta}\,d\theta= \int_0^3 {1\over 1+{2\over 1+u^2}}\cdot {1\over 1+u^2}du =\int_0^3 {1\over u^2+3} \,du \\= \left. \left[ {1\over \sqrt 3} \arctan \left( {u\over \sqrt 3}\right) \right] \right|_0^3 = {1\over \sqrt 3} \arctan \left( {3\over \sqrt 3}\right) ={1\over \sqrt 3} \cdot {\pi\over 3} = \bbox[red, 2pt]{{\sqrt 3\over 9}\pi}$$

解答:

$$V=\int_0^1 (1-x^2)^2 \pi\,dx - \int_0^1 (1-\sqrt x)^2 \pi\,dx ={8\over 15}\pi-{1\over 6}\pi = \bbox[red, 2pt]{{11\over 30}\pi}$$

解答:$$\text{Lagrange Multiplier: } \cases{f(x,y) =x^2+y^2+4x-6y\\ g(x,y)=x^2+y^2-16} \Rightarrow \cases{f_x= \lambda g_x \\f_y= \lambda g_y\\ g=0} \Rightarrow \cases{2x+4=\lambda(2x) \\ 2y-6= \lambda(2y) \\ x^2+y^2=16} \\ \Rightarrow {2x+4\over 2y-6} ={x\over y} \Rightarrow x=-{2\over 3}y \Rightarrow {4\over 9}y^2+y^2=16 \Rightarrow y = \pm {12\over \sqrt{13}} \Rightarrow x=\mp {8\over \sqrt{13}} \\ \Rightarrow \cases{f(-8/\sqrt{13}, 12/\sqrt{13})=16-8\sqrt{13} \\f(8\sqrt{13},-12/\sqrt{13}) =16+8\sqrt{13}} \Rightarrow \text{ max of }f= \bbox[red, 2pt]{16+8\sqrt{13}}$$

解答:$$\int_0^1 \int_{\sqrt[3] z}^1 \int_0^{\ln 3}  {e^{2x} \sin(\pi y^2) \over y^2} \,dxdydz =\int_0^1 \int_{\sqrt[3] z}^1 \left. \left[ {e^{2x} \sin(\pi y^2) \over 2y^2} \right] \right|_0^{\ln 3} \,dydz \\= \int_0^1 \int_{\sqrt[3] z}^1{ 4\sin(\pi y^2) \over  y^2} \,dy dz = \int_0^1 \int_0^{y^3} { 4\sin(\pi y^2) \over  y^2} \,dzdy = \int_0^14y\sin(\pi y^2) \,dy \\= \left. \left[ -{2\over \pi}\cos(\pi y^2) \right] \right|_0^1 ={2\over \pi} (1+1) = \bbox[red, 2pt]{4\over \pi}$$

解答:$$\cases{P(x,y)=xy\\ Q(x,y)=x^2y^3} \Rightarrow \cases{P_y=x\\ Q_x=2xy^3} , \text{Green theorem,} \oint_C Pdx+Qdy =\iint (Q_x-P_y) \,dA \\=\int_0^1 \int_0^{2x} (2xy^3-x)\,dydx = \int_0^1 \left. \left[ {1\over 2}xy^4-xy\right] \right|_0^{2x}\,dx =\int_0^1 (8x^5-2x^2)\,dx = \left. \left[ {4\over 3}x^6-{2\over 3}x^3 \right] \right|_0^1\\ = \bbox[red, 2pt]{2\over 3}$$

乙(計算、證明題):共2大題,共28分。須詳細寫出計算及證明過程,否則不予計分。

解答:$$D=d(O,P) =\sqrt{x^2+y^2+z^2} =\sqrt{z^2+z^2} =\sqrt{2z^2} =\sqrt 2|z|\\ z^2=x^2+y^2 =(2z+3)^2+y^2 \Rightarrow y^2=-(3z^2+12z+9) \ge 0 \Rightarrow 3z^2+12z+9\le 0 \\ \Rightarrow 3(z+1)(z+3)\le 0 \Rightarrow -3\le z\le -1 \Rightarrow \cases{z=-3 \Rightarrow D=3\sqrt 2\\ z=-1 \Rightarrow D=\sqrt 2} \Rightarrow \cases{\bbox[red, 2pt]{(i) \max=3\sqrt 2}\\ \bbox[red, 2pt]{(ii)\min=\sqrt 2}}$$

解答:$$\textbf{(i) }a_n= {\alpha \choose n} x^n \Rightarrow \lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left|{{\alpha \choose n+1} x^{n+1} \over {\alpha \choose n} x^n} \right| =\lim_{n\to \infty} \left|{(\alpha-n) x  \over n+1} \right| =|x| \lt 1 \\\qquad \Rightarrow \text{ the radius of covergence is }1. \quad \bbox[red, 2pt]{QED.} \\\textbf{(ii) }f(x) =\sum_{n=0}^\infty {\alpha \choose n}x^n  \Rightarrow f'(x)=\sum_{n=1}^\infty n{\alpha \choose n}x^{n-1} \Rightarrow (1+x)f'(x) =\sum_{n=1}^\infty n{\alpha \choose n}x^{n-1} +\sum_{n=1}^\infty n{\alpha \choose n}x^{n} \\= \sum_{k=0}^\infty (k+1){\alpha \choose k+1}x^{k} +\sum_{k=1}^\infty k {\alpha \choose k}x^{k} = \alpha +\sum_{k=1}^\infty \left[ (k+1){\alpha \choose k+1} +k {\alpha \choose k}\right] x^k \\= \alpha +\sum_{k=1}^\infty \left[ (\alpha-k){\alpha \choose k} +k {\alpha \choose k}\right] x^k = \alpha +\sum_{k=1}^\infty \alpha {\alpha \choose k}x^k = \alpha \left[ 1+ \sum_{k=1}^\infty  {\alpha \choose k}x^k\right] \\= \alpha \sum_{k=0}^\infty  {\alpha \choose k}x^k = \alpha \sum_{n=0}^\infty  {\alpha \choose n}x^n =\alpha f(x) \Rightarrow (1+x)f'(x)= \alpha f(x)\quad \bbox[red, 2pt]{QED.} \\\textbf{(iii) } y=f(x) \Rightarrow (1+x)y'= \alpha y \Rightarrow {1\over y}y'={\alpha \over 1+x} \Rightarrow \int {1\over y}\,dy = \int{\alpha\over 1+x}\,dx \\ \quad \Rightarrow \ln y= \alpha \ln(1+x)+c_1 \Rightarrow y=c_2(1+x)^\alpha \Rightarrow y(0)=c_2=1 \Rightarrow y=(1+x)^\alpha  \\ \quad \Rightarrow (1+x)^\alpha =\sum_{n=0}^\infty {\alpha \choose n}x^n \quad \bbox[red, 2pt]{QED.}$$

============================ END ==========================

解題僅供參考，其他碩士班試題及詳解