114年政大企管碩士班-微積分詳解

 國立政治大學114學年度碩士班暨碩士在職專班招生考試試題

考試科目 微積分 系所別 企業管理研究所(MBA 學位學程)乙組 

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

解答:$$\cases{f(x) =ax^2+b(e^x-1)+ \sin(cx) \\ g(x)=2x^2+3x^5} \Rightarrow \cases{f'(x)=2ax+be^x+c\cos(cx) \\ g'(x)=4x+15x^4} \\\Rightarrow \cases{f''(x) =2a +be^x -c^2\sin(cx) \\ g''(x)=4+60x^3} \\\lim_{x\to 0} {f(x)\over g(x)} = 1 \Rightarrow \cases{f'(0)=g'(0) =0 \Rightarrow b+c=0 \cdots(1)\\ f''(0)/g''(0)=1 \Rightarrow 2a+b=4 \cdots(2)} \Rightarrow (2)\times 2-(1)=4a+b-c= \bbox[red, 2pt]8$$

解答:$$g(u) = (\sin u)^{2025} \Rightarrow g(-u)=-(\sin u)^{2025}=-g(u) \Rightarrow g(u) \text{ is an odd function } \\ \Rightarrow  \int_{-nx^2}^{nx^2} (\sin u)^{2025} du=0 \Rightarrow  f(x)={\sec x\over 1+\tan x} + \sum_{n=1}^\infty \int_{-nx^2}^{nx^2} (\sin u)^{2025} du = {\sec x\over 1+\tan x} \\={1/\cos x \over 1+\sin x/\cos x} = {1\over \sin x+\cos x} \Rightarrow  f'(x)= -{\cos x-\sin x\over (\sin x+ \cos x)^2} = \bbox[red, 2pt]{\sin x-\cos x\over 1+ \sin 2x }$$

解答:$$g(x) = \sqrt{x+\sqrt{x+\sqrt x}} \Rightarrow g'(x)={ (x+\sqrt{x+\sqrt x})' \over 2\sqrt{x+ \sqrt{x+ \sqrt x}}} = {  1+{(x+\sqrt x)'\over 2\sqrt{x+\sqrt x}} \over 2\sqrt{x+ \sqrt{x+ \sqrt x}}} \\={  1+{1+{1\over 2\sqrt x}\over 2\sqrt{x+\sqrt x}} \over 2\sqrt{x+ \sqrt{x+ \sqrt x}}} ={  1+{1+2\sqrt x\over 4\sqrt x\cdot \sqrt{x+\sqrt x}} \over 2\sqrt{x+ \sqrt{x+ \sqrt x}}} = \bbox[red, 2pt]{4\sqrt x \sqrt{x+\sqrt x} + 2\sqrt x+1 \over 8\sqrt x \sqrt{x+\sqrt x} \sqrt{x+\sqrt x+\sqrt x}}$$

解答:$$u=\cos x \Rightarrow du =-\sin x\,dx \Rightarrow \int_0^{\pi/2} {\cos x\sin x\over 3+ \cos^2 x}dx = \int_1^0 {-u\over 3+u^2}du =\int_0^1 {u\over 3+u^2}du  \\ \left. \left[ {1\over 2}\ln(3+u^2) \right] \right|_0^1= {1\over 2}(\ln 4-\ln 3) = \bbox[red, 2pt]{{1\over 2}\ln{4\over 3}}$$

解答:$$\cases{u=\tan^{-1}x\\ dv=x^2dx} \Rightarrow \cases{du=dx/(1+x^2) \\ v=x^3/3} \Rightarrow \int_0^1 x^2\tan^{-1} \,dx = \left. \left[ {1\over 3}x^3\tan^{-1}x \right] \right|_0^1-{1\over 3} \int_0^1{x^3\over 1+x^2}\,dx \\={\pi\over 12}-{1\over 3} \int_0^1 \left( x-{x\over 1+x^2} \right)\,dx ={\pi\over 12}-{1\over 3} \left. \left[ {1\over 2}x^2-{1\over 2}\ln(1+x^2) \right] \right|_0^1 = \bbox[red, 2pt]{{\pi\over 12}-{1\over 6}(1-\ln 2) }$$

解答:$$e^x = \sum_{n=0}^\infty {x^n\over n!} \Rightarrow e^{-x^2} = \sum_{n=0}^\infty (-1)^n{x^{2n}\over n!}  \Rightarrow f(x)= xe^{-x^2}= \sum_{n= 0}^\infty (-1)^n{x^{2n+1}\over n!}  =\sum_{k=0}^\infty {f^{(k)}(0) \over k!}x^k \\ k=2025 \Rightarrow n=1012 \Rightarrow {(-1)^{1012} \over 1012!} ={f^{(2025)}(0)\over 2025!} \Rightarrow f^{(2025)}(0)= \bbox[red, 2pt]{2025!\over 1012!}$$

解答:$$\textbf{(a) } \cases{\text{price }p=700, \text{demand }x=1000 \Rightarrow A(x,p) =(1000,700) \\\text{price }p=600, \text{demand }x=1200 \Rightarrow B(x,p) =(1200,600)}\\\qquad \Rightarrow \text{slope of }\overline{AB}:m ={600-700\over 1200-1000} =-{1\over 2} \Rightarrow \text{equation of }\overline{AB}: p-70= -{1\over 2}(x-1000) \\\qquad \Rightarrow p=-0.5x+1200 \Rightarrow \text{The weekly demand equation is: } \bbox[red, 2pt]{p(x)=-0.5x+1200} \\\textbf{(b) }\text{The revenue }R(x)=x\cdot p(x) \Rightarrow \bbox[red, 2pt]{R(x)=-0.5x^2+1200x} \\\textbf{(c) }R'(x)=-x+1200=0 \Rightarrow x=1200 \Rightarrow p(1200)=-600+1200=600 \\ \qquad \Rightarrow \text{The price that should be charged to maximize revenue is }\bbox[red, 2pt]{$600}$$

解答:$$\textbf{(a) }\text{The total number of students, }N_{total} =\int_0^T rx\,dx ={rT^2\over 2} \\\qquad \Rightarrow N_{\lt T/2} =\int_0^{T/2} rx\,dx ={rT^2\over 8} \Rightarrow \text{ Fraction }={N_{\lt T/2} \over N_{total}} ={rT^2/8 \over rT^2/2} = \bbox[red ,2pt]{1\over 4} \\\textbf{(b) }\text{Total Score }=\int_0^T G(x)\cdot n(x)\,dx =\int_0^T \left( 100{x\over T} \right) (rx)\,dx ={100rT^2\over 3} \\ \qquad \Rightarrow \text{ Average Score }={\text{Total Score} \over N_{total}} ={100rT^2/3 \over rT^2/2} = \bbox[red, 2pt]{200\over 3}$$

解答:$$f(x)={1\over 1+x^3}, n=4 \Rightarrow \cases{x_0=0 \Rightarrow f(x_0)=1\\ x_1=1 \Rightarrow f(x_1)=1/2\\ x_2= 2 \Rightarrow f(x_2)=1/9\\ x_3=3 \Rightarrow f(x_3) =1/28\\ x_4= 4 \Rightarrow f(x_4)=1/65} \\\textbf{(a) Trapezoidal Rule } \Rightarrow T_4={\Delta x\over 2} (f(x_0) +2f(x_1)+ 2 f(x_2)+ 2 f(x_3)+  f(x_4))\\\qquad ={1\over 2} \left( 1+1+{2\over 9}+{1\over 14}+{1\over 65} \right) \approx \bbox[red, 2pt]{1.1545} \\\textbf{(b) Simpson's Rule }\Rightarrow S_4={\Delta x\over 3}(f(x_0) +4 f(x_1)+ 2 f(x_2)+ 4 f(x_3)+  f(x_4)) \\\qquad ={1\over 3} \left( 1+2+{2\over 9}+{1\over 7}+{1\over 65} \right) \approx \bbox[red, 2pt]{1.1268}$$

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解題僅供參考，碩士班歷年試題及詳解