111年高雄大學轉學考-微積分詳解

國立高雄大學111學年度第2學期寒假轉學招生考試試題(轉二年級) 

科目：微積分
考試時間：80 分鐘
系所：電機工程學系(無組別)
本科原始成績：100 分
是否使用計算機：是 

解答：$$\lim_{x\to 2}{x^3-3x-2\over x-2} =\lim_{x\to 2}{(x-2)(x^2+2x+1)\over x-2} =\lim_{x\to 2}( x^2+2x+1) = \bbox[red, 2pt]{9}$$

解答：$${d\over dx} \sec x=\sec x\tan x \Rightarrow {d\over dx} \sec(5x) =\bbox[red, 2pt]{5\sec(5x)\tan (5x)}$$

解答：$${d\over dx}[\int_3^{\tan x} {t\over 1-t^5}\,dt] ={\tan x\over 1-\tan^5 x}\cdot {d\over dx}\tan x =\bbox[red, 2pt]{\tan x \sec^2 x\over 1-\tan^5 x}$$

解答：$$u=\sin x+1 \Rightarrow du =\cos x\,dx \Rightarrow \int \cos x(\sin x+1)^{100}\,dx = \int u^{100}\,du ={1\over 101}u^{101}+C \\=\bbox[red, 2pt]{{1\over 101}(\sin x+1)^{101}+C}$$

解答：$$u=5^x \Rightarrow du =\ln 5\cdot 5^x\,dx \Rightarrow \int 5^x \sec(5^x)\,dx = \int {1\over \ln 5}\cdot \sec u\,du \\ ={1\over \ln 5}\ln |\sec u+\tan u|+C =\bbox[red, 2pt]{{1\over \ln 5}\ln|\sec 5^x+ \tan 5^x|+C}$$

解答：$${d\over dx}[\sin^{-1}(x^2+2)] ={1\over \sqrt{1-(x^2+2)^2}}\cdot {d\over dx}(x^2+2) = \bbox[red, 2pt]{2x\over \sqrt{1-(x^2+2)^2}}$$

解答：$$f(x)={x \over (x-1)^2} \Rightarrow f(1)不存在 \Rightarrow \int_{-1}^2 f(x)\,dx \bbox[red, 2pt]{不存在}$$

解答：$$f=2x^3-6x \Rightarrow f'=6x^2-6 \Rightarrow f''=12x\\ \Rightarrow \cases{f'(1)=0\\ f''(1)=12} \Rightarrow \text{curvature} ={|f''(0)| \over (1+f'^2(0))^{3/2}}= \bbox[red, 2pt]0$$

解答：$$f(x,y)=x\tanh(y) \Rightarrow f_x(x,y)=\tanh(y) \Rightarrow f_{xy}(x,y)= \bbox[red, 2pt]{\sech^2(y)}$$

解答：$$\vec F(x,y,z) =(2xz, 3xy, 5yz) \Rightarrow \nabla\cdot \vec F= {d\over dx}(2xz) +{d\over dy}(3xy)+ {d\over dz}(5yz) \\=2z+3x+5y =\bbox[red,2pt]{3x+5y+2z}$$

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解題僅供參考,其他歷年試題及詳解