113年高雄大學統計所碩士班-微積分詳解

國立高雄大學 113 學年度研究所碩士班招生考試試題

科目：微積分
系所：統計學研究所(無組別)

解答:$$\textbf{(1) } \sin^{-1} x= \int_0^x {1\over \sqrt{1-t^2}}\,dt = \int_0^x \sum_{n=0}^\infty (-1)^n{-1/2 \choose n} t^{2n}\,dx  = \int_0^x \sum_{n=0}^\infty {(2n)! \over 2^{2n}(n!)^2} t^{2n}\,dx \\\quad  =  \sum_{n=0}^\infty \int_0^x{(2n)! \over 2^{2n}(n!)^2} t^{2n}\,dx=\bbox[red, 2pt]{ \sum_{n=0}^\infty {(2n)! \over 2^{2n}(n!)^2 (2n+1)} x^{2n+1}} \\\textbf{(2) } \cases{x(t)= \cos t\\ y(t)=\sin t} \Rightarrow \cases{x'(t)=-\sin t\\ y'(t) =\cos t}, 0\le t\le {\pi\over 2} \\ \quad \Rightarrow \int_C xy\,ds = \int_0^{\pi/2} \cos t\sin t \sqrt{(-\sin t)^2 + \cos^2 t}\,dt ={1\over 2}\int_0^{\pi/2} \sin 2t\,dt= \left. \left[ -{1\over 4} \cos 2t\right] \right|_0^{\pi/2} = \bbox[red, 2pt]{1\over 2}$$
 

解答:$$\textbf{(1) } \lim_{x\to -\infty} (\sqrt{x^2+2x}-x) = \lim_{x\to \infty} (\sqrt{x^2-2x}+x) = \bbox[red, 2pt]\infty \\\textbf{(2) } x^2+y^2=1 \Rightarrow 2x+2yy'=0 \Rightarrow y'=-{x\over y} \Rightarrow y'({3\over 5},{4\over 5}) =-{3\over 4} \\\quad \Rightarrow 切線方程式: y=-{3\over 4}(x-{3\over 5})+{4\over 5} \Rightarrow \bbox[red, 2pt]{3x+4y=5} \\\textbf{(3) } \frac{d }{dx}\int_0^{x^2} t^3 \sin t\,dt = x^6 \sin x^2 \cdot (2x) = \bbox[red,2pt]{2x^7 \sin x^2} \\\textbf{(4) } u=e^{2x}  \Rightarrow du=2e^{2x} \,dx \Rightarrow \int{e^{2x} \over 1+e^{4x}}\,dx = \int {1/2 \over 1+u^2}\,du ={1\over 2}\tan^{-1} u+c_1= \bbox[red, 2pt]{{1\over 2}\tan^{-1} e^{2x}+c_1} \\\textbf{(5) }  L= \left({n-1\over n+1} \right)^n =e^{n\ln {n-1\over n+1}} \Rightarrow \ln L= n\ln {n-1\over n+1} \\\quad \Rightarrow \lim_{n\to \infty} \ln L= \lim_{n\to \infty}{\ln{n-1\over n+1} \over {1\over n}} =\lim_{n\to \infty}{(\ln{n-1\over n+1})' \over ({1\over n})'} =\lim_{n\to \infty}{ {2\over n^2-1} \over -{1\over n^2}}  =\lim_{n\to \infty} \left(-{2n^2 \over n^2-1} \right)=-2 \\\quad \Rightarrow \lim_{n\to \infty} L= \bbox[red, 2pt] {e^{-2}} \\\textbf{(6) }   F={x^2\over a^2} +{y^2\over b^2}+{z^2\over c^2}-1=0 \Rightarrow \cases{\frac{\partial }{\partial x} F={2x\over a^2} \\\frac{\partial }{\partial z}F ={2z\over c^2}} \Rightarrow \frac{\partial z}{\partial x} =-{\frac{\partial }{\partial x} F \over \frac{\partial }{\partial z}F} =\bbox[red, 2pt]{-{c^2x\over a^2 z}} \\\textbf{(7) } a_n={x^n\over n} \Rightarrow \lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left| {x^{n+1} \over n+1}  \cdot {n\over x^n}\right| =|x|\lt 1 \Rightarrow 收斂半徑=\bbox[red, 2pt]1$$

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解題僅供參考，碩士班歷年試題及詳解