113年嘉義大學應數碩士班-微積分詳解

 國立嘉義大學113學年度應用數學系碩士班(甲組)招生考試

科目: 微積分 (每題 10 分，共 100 分) 

解答:$$\int {x^2+1\over (x-1)(x-2)^2}\,dx = \int \left( {2\over x-1}-{1\over x-2}+{5\over (x-2)^2}\right)\,dx \\ =\bbox[red, 2pt]{2\ln(x-1)-\ln(x-2)-{5\over x-2}+c_1}$$

解答:$$L=\log_2 \sqrt{(x^2+1)(x^3+3)\over x^4+5} ={{1\over 2} \ln {(x^2+1)(x^3+3)\over x^4+5}\over \ln 2} ={1\over 2\ln 2}\left( \ln(x^2+1)+ \ln(x^3+3)-\ln(x^4+5)\right) \\ \Rightarrow {d\over dx}L= \bbox[red, 2pt]{{1\over 2\ln 2}\left( {2x\over x^2+1}+ {3x^2 \over x^3+3}-{4x^3\over x^4+5}\right) }$$

解答:$$  \lim_{x\to 0 }  {\sqrt{1+2x}-\sqrt[3]{1-3x} \over x+x^2} =\lim_{x\to 0 }  {(\sqrt{1+2x}-\sqrt[3]{1-3x})' \over (x+x^2)'} =\lim_{x\to 0 }  {{1\over \sqrt{1+2x}}+{1\over (1-3x)^{2/3}} \over 1+2x} =\bbox[red, 2pt]2$$

解答:$$  \lim_{y\to 0 } {\cos y-1\over y} =\lim_{y\to 0 } {(\cos y-1)'\over (y)'} = \lim_{y\to 0 } {-\sin y \over 1} =\bbox[red, 2pt]0$$

解答:$$\lfloor x^2\rfloor =\begin{cases}0& 0\le x\lt 1\\ 1& 1\le x\lt \sqrt 2\\ 2 &\sqrt 2\le x\lt \sqrt 3 \\ 3& \sqrt 3\le x\lt 2\end{cases} \\\Rightarrow \int_0^2 \lfloor x^2\rfloor\,dx =1\cdot (\sqrt 2-1)+2(\sqrt 3-\sqrt 2)+3(2-\sqrt 3)=5-\sqrt 2-\sqrt 3 \\ \lfloor x \rfloor = \begin{cases}0 & 0\le x\lt 1\\ 1& 1\le x\lt 2\end{cases} \Rightarrow \int_0^2 \lfloor x \rfloor^2\,dx =1\cdot(2-1)=1\\  \int_0^2 \left( \lfloor x^2 \rfloor -\lfloor x \rfloor^2\right)\,dx =5-\sqrt 2-\sqrt 3-1=\bbox[red, 2pt]{4-\sqrt 2-\sqrt 3}$$

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

解答:$$f(x)=\tan^{-1}x = \int {1\over 1+x^2}\,dx  = \int  \sum_{n=0}^\infty (-x^2)^n\,dx = \int  \sum_{n=0}^\infty (-1)^n x^{2n}\,dx \\\qquad =  \bbox[red, 2pt]{\sum_{n=0}^\infty {(-1)^n\over 2n+1} x^{2n+1}}$$

解答:$$r=1+\cos \theta  \Rightarrow r'=-\sin \theta \Rightarrow \text{ arc length }=\int_0^{2\pi} \sqrt{(1+\cos\theta)^2+ (-\sin \theta)^2}\,d \theta \\= \int_0^{2\pi} \sqrt{2+2\cos \theta} \,d\theta =\int_0^{2\pi} \sqrt{2+ 2(2 \cos^2(\theta/2)-1)}\,d\theta =\int_0^{2\pi} \sqrt{4\cos^2(\theta/2)}\,d\theta \\=\int_0^{2\pi} \left|2\cos{\theta\over 2}\right|\,d\theta=4 \int_0^\pi \cos{ \theta\over 2}\,d\theta= 4\left. \left[ 2\sin{\theta\over 2} \right] \right|_0^{\pi}= \bbox[red, 2pt]8$$

解答:$$\lim_{h\to 0^+}{f(1+h)-f(1)\over h} =\lim_{h\to 0^+}{e^h-1\over h} =1\\\lim_{h\to 0^-}{f(1+h)-f(1)\over h} =\lim_{h\to 0^+}{e^{-h}-1\over h} =-1 \\ \Rightarrow \bbox[red, 2pt]{f \text{ is NOT differentiable at }x=1}$$

解答:$$y=f(x)=\sqrt{x+\sqrt{x+\sqrt x}} \Rightarrow y^2-x=\sqrt{x+\sqrt x} \Rightarrow (y^2-x)'=(\sqrt{x+\sqrt x})'\\ \Rightarrow 2yy'-1={1+{1\over 2\sqrt x}\over 2\sqrt{x+\sqrt x}} \Rightarrow f'(x)=y'= \bbox[red, 2pt]{{1+{1\over 2\sqrt x}\over 2\sqrt{x+\sqrt x}}+1 \over 2\sqrt{x+\sqrt{x+\sqrt x}}}$$

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解題僅供參考，碩士班歷年試題及詳解