112年高雄大學碩士班－微積分詳解

國立高雄大學 112 學年度研究所碩士班招生考試試題

科目：微積分

考試時間：100 分鐘

系所：統計學研究所(無組別)

本科原始成績：100 分

解答：$$A:\lim_{x \rightarrow 0}{x^2+3x\over x} =\lim_{x \rightarrow 0}{x(x +3)\over x} =\lim_{x \rightarrow 0}x+3=3\\ B:\lim_{x \rightarrow 1}|x+3|=|4|=4\\ D:\lim_{x\to 3}{x-3\over x^2+4}=0\\ 故選\bbox[red, 2pt]{(C)}$$

解答：$$A:\times:\lim_{x\to 0^-}f(x)= \lim_{x\to 0^-}{x\over x+x^2}=\lim_{x\to 0^-}{1\over 1+x}=1\ne 0 \\ B:\bigcirc: \lim_{x\to 0^+}f(x)= \lim_{x\to 0^+}(x^2-1)=-1\\ C:\bigcirc: \lim_{x\to 0^+}f(x) \ne \lim_{x\to 0^-}f(x) \\ D:\bigcirc: f(0)= 0-1=-1\\故選\bbox[red, 2pt]{(A)}$$

解答：$$A:\bigcirc: f(0)=1-|1|=0 \\ B:\bigcirc: \lim_{x\to 1^+}f'(x)\ne \lim_{x\to 1^-}f'(x) \\C:\bigcirc: f(2)=1-|-1|=0\\ D:\times: f(1)是臨界點\\故選\bbox[red, 2pt]{(D)}$$

解答：$$y^2-x^2=-8 \Rightarrow 2yy'-2x=0 \Rightarrow y'={x\over y} \Rightarrow y'(3,-1)={3\over -1}=-3，故選\bbox[red, 2pt]{(A)}$$

解答：$$u=1+x^2 \Rightarrow \cases{{1\over 2}u=xdx\\ u-1=x^2} \Rightarrow \int {x^3\over \sqrt{1+x^2}}\,dx = {1\over 2}\int {u-1\over \sqrt u}\,du ={1\over 2}\int \sqrt u-{1\over \sqrt u}\,du \\={1\over 3}u^{3/2}-u^{1/2}+C =\bbox[red, 2pt]{{1\over 3}(1+x^2)^{3/2}-\sqrt{1+x^2}+C,C為常數}$$

解答：$$f(x)=e^{-x} \Rightarrow f^{[n]}(x)=\begin{cases}-e^{-x} & n是奇數\\e^{-x} & n是偶數\end{cases} \\ \Rightarrow f(x)=f(2)+f'(2)(x-2) + {f''(2)\over 2!}(x-2)^2+ {f'''(2)\over 3!}(x-2)^3+\cdots\\ =\bbox[red,2pt]{e^{-2}-e^{-2}(x-2)+{e^{-2}\over 2!}(x-2)^2-{e^{-2}\over 3!}(x-2)^3+{e^{-2}\over 4!}(x-2)^4 }+\cdots  $$

解答：$$\int_{-1}^{2} \int_{1}^{4} x^2-2xy \,dydx =\int_{-1}^{2} \left. \left[ x^2y-xy^2\right]\right|_1^4 dx = \int_{-1}^2 3x^2-15x\,dx = \left.\left[ x^3-{15\over 2}x^2 \right]\right|_{-1}^2 \\ =(8-30)-(-1-{15\over 2}) =\bbox[red,2pt]{-{27\over 2}}$$

解答：$$f(x)=x^3+4x^2+2 \Rightarrow f'(x)=3x^2+8x =x(3x+8)\\ 因此\cases{f'(x)\ge 0 \Rightarrow x\ge 0 或x\le -8/3\\ f'(x)\le 0\Rightarrow -8/3x\le 0} \Rightarrow \bbox[red,2pt]{\begin{cases}f(x)遞增 & x\ge 0 或x\le -8/3 \\f(x)遞減 & x\in [-8/3,0]\end{cases}}$$

解答：$$f(x,y)=\sqrt{x^2+2y^2} \Rightarrow f_x={1\over 2}\cdot {2x\over \sqrt{x^2+2y^2}} ={x\over \sqrt{x^2+2y^2}} \\ \Rightarrow f_{xy}=-{1\over 2}\cdot {x\cdot 4y\over (x^2+2y^2)^{3/2}} =-{2xy\over (x^2+2y^2)^{3/2}} \\ \Rightarrow\bbox[red, 2pt]{ \cases{f_x={x\over \sqrt{x^2+2y^2}}  \\ f_{xy}=-{2xy\over (x^2+2y^2)^{3/2}}}}$$

解答：$$\int_e^{e^2} \ln x\,dx=\left. \left[ x\ln x-x \right]\right|_e^{e^2}=\bbox[red, 2pt]{e^2}$$

解答：$${2k+5\over \sqrt{k^6+3k^3}} \le {2k+5\over \sqrt{k^6}}={2k+5\over k^3}\\ 依比較審斂法，\sum {2k+5\over k^3}收斂\Rightarrow \sum {2k+5\over \sqrt{k^6+3k^3}}\bbox[red, 2pt]{收斂}$$