112年海洋大學碩士班-微積分詳解

 國立臺灣海洋大學112學年度碩士班考試入學招生考試

考試科目：微積分
學系組名稱：運輸科學系碩士班不分組

解答:$$y={x\over x-2}=1+{2\over x-2} \Rightarrow y-1={2 \over x-2} \Rightarrow x={2 \over y-1}+2 \Rightarrow y=g(x)= \bbox[red, 2pt]{{2\over x-1}+2}$$

解答:$$y=\sqrt{4-x^2}為一半圓,其半徑為2,面積為{1\over 2}\cdot 4\pi =2\pi \Rightarrow 平均值={2\pi \over 2-(-2)} =\bbox[red, 2pt]{\pi\over 2}$$

解答:$$f={2y\over y+\cos x} \Rightarrow \cases{f_x={2y\sin x\over (y+\cos x)^2} \\ f_y= {2\over y+\cos x}-{2y\over (y+\cos x)^2}= {2\cos x\over (y+\cos x)^2}} \Rightarrow \bbox[red,2pt]{\cases{f_x={2y\sin x \over (y+ \cos x)^2} \\f_y={2\cos x\over (y+\cos x)^2}}}$$

解答:$$ \int_0^3 {dx\over (x-1)^{2/3}} =\int_0^1 {dx\over (x-1)^{2/3}}+ \int_1^3{dx\over (x-1)^{2/3}}= \bbox[red, 2pt]{3(\sqrt[3]2+1)}$$

解答:$$\cases{u=x \\ dv=e^{-x}dx} \Rightarrow \cases{du=dx\\ v=-e^{-x}} \Rightarrow \int xe^{-x}\,dx=-xe^{-x} +\int e^{-x}\,dx = -xe^{-x}-e^{-x}+C \\ \Rightarrow \int_0^4 xe^{-x}\,dx = \left. \left[ -xe^{-x}-e^{-x} \right] \right|_0^4 = {1-4e^{-4}-e^{-4}} =\bbox[red, 2pt]{1-{5\over e^4}}$$

解答:$$xe^y+ ye^x =x \Rightarrow e^y+xy'e^y+ y'e^x+ye^x=1 \Rightarrow (xe^y+e^x)y'=1-e^y-ye^x \\ \Rightarrow y'={dy\over dx} =\bbox[red, 2pt]{1-e^y-ye^y\over xe^y+ e^x}$$

解答:$$\mathbf{(1)}\; L=(1+{1\over n})^n \Rightarrow \ln L=n\ln(1+{1\over n}) ={\ln {n+1\over n} \over {1\over n}} \\ \Rightarrow \lim_{n\to \infty} \ln L=\lim_{n\to \infty}{\ln {n+1\over n} \over {1\over n}} L=\lim_{n\to \infty}{(\ln {n+1\over n})' \over ({1\over n})'}=\lim_{n\to \infty}{n\over n+1}=1\\ \Rightarrow \lim_{n\to \infty} L=e^1= \bbox[red, 2pt]e \\\mathbf{(2)}\;\cases{\lim_{x\to -2^+}(x+3){|x+2| \over x+2}=1 \\\lim_{x\to -2^-}(x+3){|x+2| \over x+2}=-1} \Rightarrow \lim_{x\to -2}(x+3){|x+2| \over x+2}\bbox[red,2pt]{不存在}$$

解答:$$\mathbf{(a)}\;f(x)={x^2+4\over 2x} ={x\over 2}+{2\over x}\Rightarrow \bbox[red, 2pt]{\cases{\text{domain of }f=\{x\in \mathbb R,x\ne 0\}\\ 漸近線：\cases{y=x/2\\ x=0}}} \\ \mathbf{(b)}\; f'(x)={(x-2)(x+2)\over 2x^2} =0 \Rightarrow x=\pm 2 \Rightarrow \cases{f(2)=2\\ f(-2)=-2} \Rightarrow \text{critical points: }\bbox[red, 2pt]{(2,2),(-2,-2)}\\ f''(x)={4\over x^3} \ne 0 \Rightarrow \text{inflection points:}\bbox[red,2pt]{none}\\ \mathbf{(c)}\; f'(x)={(x-2)(x+2)\over 2x^2} \Rightarrow y=f(x) \text{ is}\bbox[red, 2pt]{\cases{\text{increasing, }x\in [2, \infty)\cup (-\infty, -2]\\\text{decreasing, }, x\in [-2,0)\cup (0,2]}} \\\mathbf{(d)}\; \cases{f''(2)= 1/2 \gt 0\\ f''(-2)=-1/2 \lt 0} \Rightarrow \bbox[red, 2pt]{\cases{f(2)=2為極小值\\ f(-2)=-2為極大值}} \\ \mathbf{(e)}\; f''(x)={4\over x^3} \Rightarrow \begin{cases} f''(x)\gt 0 & x\gt 0\\ f''(x)\lt 0 & x\lt 0\end{cases} \Rightarrow \bbox[red, 2pt]{\cases{f\text{ is concave up, if }x\gt 0\\f \text{ is concave down, if }x\lt 0}}\\ \mathbf{(f)}圖形如下$$

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