115年警察大學碩士班-微積分詳解

 中央警察大學 115 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:$$\textbf{(一) }\lim_{x\to \infty}{3x^2+x+1\over 2x^2-x+2} =\lim_{x\to \infty}{3+1/x+1/x^2\over 2-1/x+2/x^2} = \bbox[red, 2pt]{3\over 2} \\ \textbf{(二) }\lim_{x\to 0}{\tan x\over x} =\lim_{x\to 0}{(\tan x)'\over (x)'} =\lim_{x\to 0}{\sec^2 x\over 1} = \bbox[red, 2pt]1 \\ \textbf{(三) } \lim_{x\to \pi/2}({\sec x-\tan x}) =\lim_{x\to \pi/2} {1-\sin x\over \cos x} =\lim_{x\to \pi/2} {(1-\sin x)'\over (\cos x)'} = \lim_{x\to \pi/2} {-\cos x\over -\sin x} = \bbox[red, 2pt] 0 \\ \textbf{(四) } \lim_{x\to 2} {x^2-4\over x-2} = \lim_{x\to 2} {(x-2)(x+2)\over x-2} =  \lim_{x\to 2}(x+2) = \bbox[red, 2pt]4 \\ \textbf{(五) }\lim_{x\to \infty} (x-\sqrt{x^2-1}) = \lim_{x\to \infty} {(x-\sqrt{x^2-1}) (x+\sqrt{x^2-1}) \over x+\sqrt{x^2-1}} = \lim_{x\to \infty} {1 \over x+\sqrt{x^2-1}} = \bbox[red, 2pt] 0$$

解答:$$假設底面在平面z=0上，圓心在原點，則底面面積=r^2\pi \\\Rightarrow 體積=\int_0^h r^2\pi\,dz= \left. \left[ r^2\pi z \right] \right|_{z=0}^h = \bbox[red, 2pt]{\pi r^2h}$$

解答:$$P(t^2/2,t) 在拋物線上，則P與Q(1,4)的距離=\sqrt{({t^2\over 2}-1)^2+(t-4)^2} \\ 取f(t)= ({t^2\over 2}-1)^2+(t-4)^2 \Rightarrow f'(t)=2t({t^2\over 2}-1)+2(t-4)=0 \Rightarrow t^3=8 \Rightarrow t=2 \\ \Rightarrow P \bbox[red, 2pt]{(2,2)}$$

解答:$$r=2(1-\cos \theta) \Rightarrow {dr\over d\theta}=2\sin \theta \Rightarrow r^2+ \left( {dr\over d\theta} \right)^2= 4(1-2\cos \theta+ \cos^2 \theta) +4\sin^2\theta =4(2-2\cos \theta) \\=8(1-\cos \theta)= 16\sin^2{ \theta\over 2} \Rightarrow ds= \sqrt{r^2+ \left( {dr\over d\theta} \right)^2} =4\sin{\theta\over 2}\,d\theta \\ 表面積=\int_\alpha^\beta 2\pi y\,ds =\int_0^{\pi} 2\pi(r\sin \theta) \cdot 4\sin{\theta\over 2}\,d\theta =16\pi\int_0^{\pi} (1-\cos \theta) \sin \theta\sin {\theta\over 2}\,d\theta \\=16\pi\int_0^{\pi} \left[ 2\sin^2 {\theta\over 2}\right] \left[ 2\sin {\theta\over 2} \cos {\theta\over 2}\right] \sin{\theta\over 2}\,d\theta =64\pi \int_0^\pi \sin^4{\theta\over 2} \cos{\theta\over 2}\,d\theta \\=64\pi \int_0^1 u^4(2\,du) =128\pi\int_0^1 u^4\,du = \bbox[red, 2pt]{128\pi\over 5} \;  \left( u=\sin{\theta\over 2} \Rightarrow du={1\over 2}\cos {\theta\over 2} d\theta \right)$$

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解題僅供參考，其他碩士班題及詳解