113年警大碩士班-微積分詳解

 中央警察大學113 學年度碩士班入學考試試題

所 別：消防科學研究所、交通管理研究所
科 目：微積分(同等學力加考)

解答: $$\textbf{(一)}\; \lim_{x \to 0} {\sin x-\tan x\over x^3} = \lim_{x \to 0} {(\sin x-\tan x)'\over (x^3)'} =  \lim_{x \to 0} {\cos x-\sec^2 x\over 3x^2} = \lim_{x \to 0} {(\cos x-\sec^2 x)' \over (3x^2)'} \\\qquad =  \lim_{x \to 0} {-\sin x-2\sec^2 x \tan x\over 6x} = \lim_{x \to 0} {(-\sin x-2\sec^2 x \tan x)'\over (6x)'} \\\qquad =  \lim_{x \to 0} {-\cos x-4\sec^4 x \tan^2 x-2\sec^4 x\over 6} ={-1+0-2\over 6}= \bbox[red, 2pt]{-{1\over 2}} \\\textbf{(二)}\; \lim_{x \to 0} {1-\cos x\over x\ln(1+x)} = \lim_{x \to 0} {(1-\cos x)'\over (x\ln(1+x))' } =\lim_{x \to 0} {\sin x\over \ln(1+x)+{x\over 1+x}} = \lim_{x \to 0} {(\sin x)' \over( \ln(1+x)+{x\over 1+x})'} \\\qquad =\lim_{x \to 0} {\cos x\over {2 \over 1+x}-{x\over (1+x)^2}} = {1\over 2-0} \bbox[red, 2pt]{1\over 2}$$

解答: $$\textbf{(一)}\; 假設\sin^{-1}x= y \Rightarrow x=\sin y \Rightarrow 1=y'\cos y \Rightarrow y'={1\over \cos y} ={1\over \sqrt{1-x^2}}\\\qquad \Rightarrow {d\over dx} \sin^{-1} x= y'={1\over \sqrt{1-x^2}},\bbox[red, 2pt]{\text{故得證}} \\ \textbf{(二)}\; y=\tan^{-1}(x-\sqrt{x^2+1}) \Rightarrow \tan y= x-\sqrt{x^2+1} \Rightarrow y'\sec^2 y=1-{x\over \sqrt{x^2+1}} \\\qquad =y'={ 1-{x\over \sqrt{x^2+1}}\over \sec^2 y} ={ 1-{x\over \sqrt{x^2+1}}\over 2(x^2-x\sqrt{x^2+1}+1)} ={\sqrt{x^2+1}-x \over 2\sqrt{x^2+1}(x^2+1-x\sqrt{x^2+1})} \\\qquad ={ 1-{x\over \sqrt{x^2+1}}\over 2(x^2-x\sqrt{x^2+1}+1)} ={\sqrt{x^2+1}-x \over 2{(x^2+1)}( \sqrt{x^2+1}-x)}  =\bbox[red,2pt]{1\over 2(x^2+1)}$$

解答: $$\textbf{(一)}\; \int_{1}^{\infty} (1-2x)e^{-2x}\,dx= \int_{1}^{\infty} (e^{-2x}- 2xe^{-2x})\,dx= \left. \left[-{1\over 2}e^{-2x}+{1\over 2}e^{-2x}(2x+1) \right] \right|_1^\infty \\\qquad ={1\over 2}e^{-2}-{3\over 2}e^{-2} =\bbox[red, 2pt]{-e^{-2}} \\\textbf{(二)}\; y={(x-2)^3\over (x^2+1)^{1/3}} \Rightarrow \frac{\text{d}y}{\text{d}x} ={3(x-2)^2\over (x^2+1)^{1/3}} -{(x-2)^3 \cdot (2x)\over 3(x^2+1)^{4/3}} =\bbox[red, 2pt]{ (x-2)^2(7x^2+4x+9)\over 3(x^2+ 1)^{4/ 3}}$$

解答: $$\textbf{(一)}\;未敘明物體溫度下降為何種曲線!\\\textbf{(二)}\; y= {1\over 8}(x^4+{2\over x^2}) \Rightarrow y'= {1\over 8}(4x^3-{4\over x^3}) ={1\over 2}(x^3-{1\over x^3}) \Rightarrow (y')^2={1\over 4}(x^6+{1\over x^6}-2)\\弧長=\int_1^4 \sqrt{1+(y')^2} \,dx =\int_1^4 \sqrt{1+{1\over 4}(x^6+{1\over x^6}-2)} \,dx  =\int_1^4 \sqrt{{1\over 4}(x^6+{1\over x^6}+2)} \,dx \\\qquad =\int_1^4\sqrt{{1\over 4}(x^3+{1\over x^3})^2}\,dx  =\int_1^4 {1\over 2}(x^3+{1\over x^3})\, dx =\left. \left[{1\over 2}({1\over 4}x^4-{1\over 2x^2}) \right] \right|_1^4 =\bbox[red, 2pt]{2055\over 64}$$

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解題僅供參考,其他歷年試題及詳解