112年台師大經營碩士班-微積分詳解

國立臺灣師範大學112學年度碩士班招生考試試題

 科目:微積分 適用系所:全球經營與策略研究所

解答:

$$f(x)=x^4+x^3-3x^2+1 \Rightarrow f'(x)=4x^3+3x^2-6x \Rightarrow f''(x)=12x^2+6x-6=6(2x-1)(x+1) \\ f''(x)=0 \Rightarrow x=-1,{1\over 2} \Rightarrow \begin{cases}f''(x) \gt 0& x\gt 1/2\\ f''(x)\lt 0 & -1\lt x\lt 1/2\\ f''(x) \gt 0 & x\lt -1 \\f(1/2)=7/16\\ f(-1) =-2\end{cases} \\\Rightarrow \bbox[red, 2pt]{\cases{ 上凹區間: ({1\over 2}, \infty) \cup (-\infty, -1) \\ 下凹區間, x\in (-1,{1\over 2})\\ 反曲點: ({1\over 2},{7\over 16}),(-1,-2)}}$$

解答: $$\textbf{(1) }f(x)={\log|2x+1| \over 2x+1} \Rightarrow f'(x)={2(1-\log|2x+1|)\over (2x+1)^2}=0 \Rightarrow |2x+1|=e \Rightarrow x=-{1\over 2}\pm {1\over 2}e \\ \Rightarrow \begin{cases}(-\infty,-(1+e)/2):f'\lt 0\\ (-(1+e)/2, -(1-e)/2): f'\gt 0\\ (-(1-e)/2, \infty):f'\lt 0 \end{cases} \Rightarrow \bbox[red, 2pt]{\cases{f(-(1+e)/2) =-1/e 為相對極小值\\ f(-(1-e)/2) =1/e為相對極大值}} \\ \textbf{(2) }f''(x) ={8 \log|2x+1|-12\over (2x+1)^3} \Rightarrow \cases{f''(-(1+e)/2)=4/e^3 \gt 0\\ f''(-(1-e)/2)=-4/e^3\lt 0} \\\qquad \Rightarrow \bbox[red, 2pt]{\cases{f(-(1+e)/2) =-1/e 為相對極小值\\ f(-(1-e)/2) =1/e為相對極大值}}$$

解答: $$\textbf{(1) }\lim_{x\to -\infty} {3-3^x\over 5-5^x} ={3-0\over 5-0} = \bbox[red, 2pt]{3\over 5} \\ \textbf{(2) }L= (1+3x)^{1/2x} \Rightarrow \ln L= {\ln(1+3x) \over 2x} \Rightarrow \lim_{x\to 0} \ln L= \lim_{x\to 0}{{d\over dx} \ln(1+3x) \over {d\over dx}(2x)} =\lim_{x\to 0} {3\over 2(1+3x)} \\\qquad ={3\over 2} \Rightarrow \lim_{x\to 0} L= \bbox[red, 2pt]{e^{3/2}} \\\textbf{(3) } \lim_{x\to 0} \left( {1\over e^x-1}-{1\over x}\right) =\lim_{x\to 0}   {x-e^x+1\over x(e^x-1)}   = \lim_{x\to 0}   {{d\over dx}(x-e^x+1)\over {d\over dx}x(e^x-1)}   =\lim_{x \to 0}  {1-e^x\over xe^x +e^x-1} \\\qquad   =\lim_{x\to 0}  {{d\over dx}(1-e^x) \over {d\over dx}(xe^x +e^x-1)}   =\lim_{x\to 0}  {-e^x\over xe^x +2e^x} = \bbox[red, 2pt]{-{1 \over 2}} \\\textbf{(4) }\lim_{x\to \infty}{\ln(\ln x) \over \ln x} = \lim_{x\to \infty}{{d\over dx}\ln(\ln x) \over {d\over dx}\ln x} =  \lim_{x\to \infty}{1\over \ln x} = \bbox[red, 2pt]0$$

解答: $$\textbf{(1)} \cases{u=xe^x\\ dv=(1+x)^{-2}\,dx} \Rightarrow \cases{du= (e^x+xe^x)\,dx\\ v=-(1+x)^{-1}} \Rightarrow \int {xe^x \over (1+x)^2} =-{xe^x \over 1+x}+ \int {e^x+xe^x\over 1+x}\,dx \\\qquad =-{xe^x \over 1+x}+ \int e^x\,dx= -{xe^x\over 1+x}+e^x +c = \bbox[red, 2pt]{{e^x\over 1+x}+c}\\ \textbf{(2)} \cases{u=x\cdot 9^{-x} \\ dv=(1- x \ln 9)^{-2}\, dx} \Rightarrow \cases{du =9^{-x}(1-x \ln 9) \\v=(\ln 9-x\ln^2 9)^{-1}} \\\qquad \Rightarrow \int {x3^{-2x} \over (1-(2\ln 3)x)^2}\,dx =\int{ x\cdot 9^{-x}\over (1-x\ln 9)^2} ={x\cdot 9^{-x} \over \ln 9-x\ln^2 9} -\int {9^{-x}(1-x\ln 9)\over \ln 9-x\ln^2 9}\,dx \\={x\cdot 9^{-x} \over \ln 9-x\ln^2 9} -\int {9^{-x} \over \ln 9 }\,dx   =\bbox[red, 2pt]{{x\cdot 9^{-x} \over \ln 9-x\ln^2 9}+{1\over 9^x \ln^2 9}+c}$$

解答: $$\textbf{(1)} \int_{-2}^2 \int_0^{4-x^2}\,dy dx = \int_{-2}^2 (4-x^2)\,dx = \left. \left[ 4x-{1\over 3}x^3 \right] \right|_{-2}^2 =\bbox[red, 2pt]{32\over 3} \\\textbf{(2) }y=4-x^2 \Rightarrow x=\pm \sqrt{4-y} \Rightarrow \int_{-2}^2 \int_0^{4-x^2}\,dy dx =\int_0^4 \int_{-\sqrt{4-y}}^\sqrt{4-y} \,dxdy = \int_0^4 2\sqrt{4-y} \,dy \\\qquad = \left. \left[-{{4\over 3}(4-y)^{3/2}} \right] \right|_0^4 =0-\left(-{32\over 3} \right)= \bbox[red, 2pt] {32\over 3}$$

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解題僅供參考，碩士班歷年試題及詳解