113年中正大學經濟碩士班-微積分詳解

 國立中正大學113學年度碩士班招生考試

科目名稱:微積分
系所組別:經濟學系國際經濟學

解答: $$\textbf{1. (1)}L=\left(1+{1\over x} \right)^x =\left({x+1\over x} \right)^x \Rightarrow \ln L={\ln(x+1)-\ln x\over 1/x} \\\quad \Rightarrow \lim_{x\to \infty} \ln L= \lim_{x\to \infty} {(\ln(x+1)-\ln x)'\over (1/x)'} = \lim_{x\to \infty}{{1\over x+1}-{1\over x}\over -{1\over x^2}} =\lim_{x\to \infty} \left(-{x^2\over x+1}+x \right)= \lim_{x\to \infty} {x\over x+1}\\\qquad =1 \Rightarrow \lim_{x\to \infty} L=e^1=\bbox[red, 2pt]e \\ \quad \textbf{(2) } \lim_{x\to 0}{2\sin x\over x} =\lim_{x\to 0}{(2\sin x)' \over (x)'} = \lim_{x\to 0}{2\cos x\over 1} =\bbox[red, 2pt]2 \\ \quad \textbf{(3)}L=(3^x+4^x)^{1/x} \Rightarrow \ln L={\ln(3^x+4^x)\over x} \Rightarrow \lim_{x\to \infty} \ln L=\lim_{x\to \infty} {(\ln(3^x+4^x))'\over (x)'} \\\qquad =\lim_{x\to \infty} {\ln 3\cdot 3^x +\ln 4\cdot 4^x\over 3^x +4^x}=\lim_{x\to \infty} {\ln 3\cdot (3/4)^x +\ln 4 \over (3/4)^x +1} =\ln 4 \Rightarrow \lim_{x\to \infty} L=e^{\ln 4} =\bbox[red, 2pt]4$$

$$\textbf{2. (4) }f(x)={1\over 1+{1\over x}} ={x\over x+1} \Rightarrow f'(x)={1\over x+1}-{x\over (x+1)^2} = \bbox[red, 2pt]{1\over (x+1)^2} \\\quad \textbf{(5) }g(x)={e^x\over 1+e^x} =1-{1\over 1+e^x} \Rightarrow g'(x) = \bbox[red, 2pt]{e^x\over (1+e^x)^2}$$

$$\textbf{3. (6) } u=\ln x \Rightarrow du ={dx\over x} \Rightarrow \int{1\over x\ln x}\,dx =\int {1\over u}\,du =\ln u+c_1 =\bbox[red, 2pt]{\ln(\ln u)+c_1} \\\quad \textbf{(7) }\cases{u=e^x\\ dv=\cos x\,dx} \Rightarrow \cases{du=e^x\,dx \\ v=\sin x} \Rightarrow I=\int e^x\cos x\,dx =e^x\sin x-\int e^x\sin \,dx\\\qquad \cases{u =e^x\\ dv= \sin xdx} \Rightarrow \cases{du=e^x\,dx \\v=-\cos x} \Rightarrow \int e^x\sin x\,dx = -e^x\cos x+ \int e^x\cos x\,dx \\ \qquad \Rightarrow I=e^x\sin x+e^x\cos x-I \Rightarrow I=\bbox[red, 2pt]{{1\over 2}e^{x}(\sin x+\cos x)+c_1} \\\quad \textbf{(8) } u={x\over 2} \Rightarrow du={1\over 2}dx \Rightarrow  \int e^{-x^2/4}\,dx =2\int e^{-u^2}\,du  =2\cdot {\sqrt \pi\over 2}erf(u) +c_1=\bbox[red, 2pt]{\sqrt{\pi} \text{erf}(x/2)+c_1}$$

$$\textbf{4. (9) } a_n={n!\over 3^n} \Rightarrow \lim_{n\to \infty}\left|{a_{n+1} \over a_n} \right| = \lim_{n\to \infty}\left|{(n+1)! \over 3^{n+1}} \cdot {3^n\over n!}\right| =  \lim_{n\to \infty}\left|{n+1 \over 3} \right| =\infty \Rightarrow \bbox[red, 2pt]{發散} \\\quad \textbf{(10) }\cases{\lim_{n\to \infty}{1\over n}=0 \\ {1\over n}\gt {1\over n+1}} \Rightarrow  \bbox[red, 2pt]{收斂}  \text{ (alternating series test) }$$

解答: $$\textbf{5. (11) } x+3y=x^2y+y^2 \Rightarrow 1+3y'=2xy+x^2y'+2yy' \Rightarrow 1-2xy=(x^2+2y-3)y'\\ \Rightarrow y'={dy\over dx} =\bbox[red, 2pt]{1-2xy\over x^2+2y-3} \\\textbf{6. (12) }f(x)= e^{-x^2} \Rightarrow f'(x)=-2xe^{-x^2} \Rightarrow f''(x)= (4x^2-2)e^{-x^2} \Rightarrow \cases{f(0)=1\\ f'(0)=0\\ f''(0)=-2} \\\qquad \Rightarrow \sum_{n=0}^2 {f^{[n]}(0)x^n \over n!}=1+0-{2\over 2!}x^2=\bbox[red, 2pt]{1-x^2} \\\quad \textbf{(13) }g(x)=\ln(1+x^2) \Rightarrow g'(x)={2x\over 1+x^2} \Rightarrow g''(x)= {-2x^2+2\over (1+x^2)^2} \Rightarrow \cases{g(0)=g'(0)=0\\ g''(0)=2} \\\qquad \Rightarrow \sum_{n=0}^2 {g^{[n]}(0)x^n \over n!}=0+0+{2\over 2!}x^2 = \bbox[red, 2pt]{x^2}$$

$$\textbf{7. (14) } R_1=2\int_1^2 \sqrt{x-1}\,dx =2\left. \left[{2\over 3} (x-1)^{3/2}\right] \right|_1^2 ={4\over 3} \\ \qquad R_2= \int_2^5 \left( \sqrt{x-1}-(x-3)\right)\,dx =\left. \left[ {2\over 3}(x-1)^{3/2}-{1\over 2}x^2+ 3x \right] \right|_2^5={19\over 6} \\ \qquad \Rightarrow R_1+R_2= \bbox[red, 2pt]{ 9\over 2}\\ \textbf{8. (15) } f(x,y)=\sqrt x+\sqrt y \Rightarrow \cases{f_x={1\over 2\sqrt x} \ne0\\ f_y={1\over 2\sqrt y} \ne 0} \\ 2x+2y=5 \Rightarrow y={5\over 2}-x \Rightarrow g(x)=f(x,{5\over 2}-x) =\sqrt{x}+ \sqrt{{5\over 2}-x} \\ \qquad \Rightarrow g'(x)={1\over 2\sqrt x}-{1\over 2\sqrt{5/2-x}} =0 \Rightarrow \sqrt x=\sqrt{{5\over 2}-x} \Rightarrow x={5\over 4} \Rightarrow y={5\over 2}-x={5\over 4} \\ \Rightarrow 最大值=f(5/4,5/4) =\sqrt{5\over 4} +\sqrt{5\over 4}= \bbox[red, 2pt]{\sqrt 5}$$

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解題僅供參考，碩士班歷年試題及詳解