111年暨大碩士班-微積分詳解

國立暨南國際大學111學年度碩士班入學考試

科目: 微積分
適用: 財金系

一、填空題(共80分，每空格10分，不需列出計算過程)

解答: $$\lim_{x\to 9}{\sqrt x(\sqrt x-3) \over x-9} =\lim_{x\to 9}{\sqrt x(\sqrt x-3) \over (\sqrt x+3)(\sqrt x-3)} =\lim_{x\to 9}{\sqrt x  \over \sqrt x+3} ={3\over 6}=\bbox[red, 2pt]{1\over 2}$$

解答: $$\lim_{n\to \infty}e^{-n}\ln n =\lim_{n\to \infty}{\ln n\over e^n} = =\lim_{n\to \infty}{{d\over dn}\ln n\over {d\over dn}e^n} =\lim_{n\to \infty}{1/ n\over e^n}= \lim_{n\to \infty}{1\over ne^n} =\bbox[red, 2pt]0$$

解答: $$f(x)=\sqrt[3]{x+|x|} =\begin{cases} \sqrt[3]{2x}& x\ge 0\\ 0 & \text{otherwise}\end{cases} \Rightarrow f'(x)=\begin{cases} {2\over 3}(2x)^{-2/3}& x\ge 0\\ 0 & \text{otherwise}\end{cases} \\ \Rightarrow f'(4)={2\over 3}\cdot 8^{-2/3} ={2\over 3}\cdot {1\over 4} =\bbox[red, 2pt]{1\over 6}$$

解答: $$f(x)=\ln(1-xe^{-x}) \Rightarrow f'(x)={-e^{-x}+ xe^{-x}\over 1-xe^{-x}} \Rightarrow f'(0)=\bbox[red, 2pt]{-1}$$

解答: $$u=4-3x \Rightarrow du=-3dx \Rightarrow \int_1^2 (4-3x)^8\,dx = -{1\over 3}\int_1^{-2} u^8\,du =-{1\over 3}\left. \left[ {1\over 9}u^9\right] \right|_1^{-2} ={513\over 27}=\bbox[red, 2pt]{19}$$

解答: $$u=2x^2+5 \Rightarrow du=4xdx \Rightarrow \int_{\sqrt 2}^{\sqrt{10}} {3x\over \sqrt{2x^2+5}}dx =\int_{9}^{25} {3\over 4}\cdot{1\over \sqrt u}\,du \\ =\left. \left[ {3\over 2}\sqrt u \right] \right|_9^{25} ={3\over 2}(5-3) =\bbox[red, 2pt]3$$

解答: $$\int_0^1 \ln(1+x)\,dx = \left. \left[ (1+x)\ln(1+x)-x \right] \right|_0^1 =\bbox[red, 2pt]{2\ln 2-1}$$

解答: $$\sum_{n=1}^\infty {2^{n+1} \over 3^{n-1}} =\sum_{n=1}^\infty 6\cdot ({2 \over 3})^n =6\cdot {2/3\over 1-2/3} =\bbox[red, 2pt]{12}$$

二、計算題(共20分，沒有列出詳細計算過程者不予計分)

解答: $$f(x)=x^2e^{-x} \Rightarrow f'(x)=2xe^{-x}-x^2e^{-x} \Rightarrow f''(x)=2e^{-x}-4xe^{-x}+ x^2e^{-x},\\因此若f'(x)=0 \Rightarrow xe^{-x}(2-x)=0 \Rightarrow \cases{x=0\\ x=2} \Rightarrow \cases{f''(0)=2 \gt 0\\ f''(2)=-2e^{-2}\lt 0} \\ \Rightarrow \bbox[red, 2pt]{\cases{f(0)=0為相對極小值\\ f(2)= 4/e^4為相對極大值}}$$

解答: $$f''(x)=0 \Rightarrow    e^{-x}(x^2-4x+2)=0 \Rightarrow x^2-4x+2=0 \Rightarrow x=2\pm \sqrt 2\\ 由於\begin{cases}f''(x)\gt 0& x\in(-\infty,2-\sqrt 2)\\ f''(x)\lt 0& x\in (2-\sqrt 2,2+\sqrt 2)  \\ f''(x)\gt & x\in (2+\sqrt 2,\infty)\end{cases},\\因此反曲點為\bbox[red, 2pt]{(2-\sqrt 2,f(2-\sqrt 2)),(2+\sqrt 2,f(2+\sqrt 2))}$$

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