112年中山財管碩士班-微積分詳解

 國立中山大學112學年度碩士班暨碩士在職專班招生考試試題

科目名稱:微積分【財管系碩士班甲組】

解答: $$\textbf{(a) }f(x)=|x|+|x+3| =\begin{cases}2x+3 & x\gt 0\\ 3&-3\le x\le 0 \\-2x-3 & x\lt -3\end{cases} \\\Rightarrow\textbf{(1)} f(x)\text{ is differentiable on }x\in \bbox[red, 2pt]{\mathbb R-\{0,-3\}}\qquad \textbf{(2)} \bbox[red, 2pt]{f'(x)=\begin{cases} 2& x\gt 0\\ 0& -3\lt x\lt 0\\ -2& x\lt -3\end{cases}} \\\textbf{(b) }g(x)=x|x| = \begin{cases} x^2 & x\ge 0\\ -x^2 & x\lt 0\end{cases} \Rightarrow \cases{\textbf{(1) }g(x) \text{ is differentiable on }\bbox[red, 2pt]{x\in \mathbb R}\\\textbf{(2) } \bbox[red, 2pt]{g'(x)=\begin{cases} 2x & x\ge 0\\ -2x & x\lt 0\end{cases}}}$$

解答: $$\textbf{(a) }f(x)= x+{2\over x} \Rightarrow f'(x)={x^2-2\over x^2}=0 \Rightarrow x=\pm \sqrt 2  \\\qquad f'(x)\ge 0 \Rightarrow x^2-2 \ge 0 \Rightarrow x\ge \sqrt 2,x\le -\sqrt 2 \\ \qquad \Rightarrow \cases{\textbf{(1) }f(x) \text{ has relative extrema at } \bbox[red, 2pt] {x=\pm \sqrt 2}\\\textbf{(2) }f(x) \text{ is increasing for }x\in \bbox[red, 2pt]{(-\infty, -\sqrt 2] \cup [\sqrt 2, \infty)} \\ \textbf{(3) }f(x)\text{ is decreasing for }x\in \bbox[red, 2pt]{[-\sqrt 2,\sqrt 2] ,x\ne 0}} \\\textbf{(b) }g(x)= \sqrt x -2\sqrt{x+4}\Rightarrow g'(x)={\sqrt{x+4}-2\sqrt x\over 2\sqrt x \cdot \sqrt{x+4}}=0 \Rightarrow x={4\over 3}  \\\qquad g'(x)\ge 0 \Rightarrow x\le {4\over 3}   \Rightarrow \cases{\textbf{(1) }g(x) \text{ has relative extrema at }\bbox[red, 2pt]{x=4/3} \\\textbf{(2) }g(x) \text{ is increasing for }x\in \bbox[red, 2pt]{(0, 4/3] } \\ \textbf{(3) }g(x)\text{ is decreasing for }x\in \bbox[red, 2pt]{[4/3,\infty) }}$$

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

解答: $$\textbf{(a) }  {d\over dx}F(x) =(1+x^6)^{-1}\cdot 2x= \bbox[red, 2pt]{2x\over 1+x^6} \\\textbf{(b) }{d\over dx}F(x)  = \sqrt{1+x^2}- \sqrt{1+x^4}\cdot (2x) = \bbox[red, 2pt]{\sqrt{1+x^2}- 2x\sqrt{1+x^4}}$$

解答: $$\textbf{(a) }  x=1+t^2 \Rightarrow dx=2tdt \Rightarrow \int_0^1 t\sqrt{1+t^2}\, dt ={1\over 2}\int_1^2 \sqrt x\, dx={1\over 2} \left. \left[  {2\over 3}x^{3/2} \right] \right|_1^2 =\bbox[red, 2pt]{{1\over 3}(2\sqrt 2-1)} \\\textbf{(b) }x=1+t^3 \Rightarrow dx=3t^2\,dt \Rightarrow \int_0^2 t^2(1+t^3)^{-1/2}\,dt = {1\over 3}\int_1^9 x^{-1/2}\,dx ={1\over 3} \left. \left[ 2 \sqrt x \right] \right|_1^9 =\bbox[red, 2pt]{4\over 3} \\\textbf{(c) } x=1+\sqrt t \Rightarrow dx ={1\over 2\sqrt t}\,dt \Rightarrow \int_1^4 {\sqrt{1+\sqrt t} \over \sqrt t}\,dt = 2\int_2^3 \sqrt x\,dx =2 \left. \left[ {2\over 3}x^{3/2}\right] \right|_2^3 \\\qquad = \bbox[red, 2pt]{{4\over 3}(3\sqrt 3-2\sqrt 2)}$$

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解題僅供參考，碩士班歷年試題及詳解