114年政大保險碩士班-微積分詳解

 國立政治大學114學年度碩士班招生考試

考試科目:微積分
系所別:風險管理與保險學系

解答: $$\textbf{(a) }u=x+1 \Rightarrow \int_0^3 {x^2\over \sqrt{x+1}}\,dx =\int_1^4 {(u-1)^2 \over \sqrt u}\,du =\int_1^4 (u^{3/2}-2u^{1/2}+u^{-1/2})\,du \\\quad = \left. \left[ {2\over 5} u^{5/2}-{4\over 3}u^{3/2} +2 u^{1/2}\right] \right|_1^4 = \bbox[red, 2pt]{76\over 15} \\ \textbf{(b) }\int{3x^3-2x-2\over x^2(x^2+1)}\,dx = \int \left( {5x\over x^2+1} +{2\over x^2+1}-{2\over x}-{2\over x^2}\right)\,dx \\ \qquad = \bbox[red, 2pt]{{5\over 2}\ln (x^2+1)+2\tan^{-1}x-2\ln |x|+{2\over x}+C}$$

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

解答: $$\textbf{(a) }  F(x)=\int_0^{x^2} \cos(t^2+t)\,dt \Rightarrow F'(x)= \bbox[red, 2pt]{2x\cos(x^4+x^2)} \\\textbf{(b) } \lim_{x\to 0}{F(x)\over x^2} = \lim_{x\to 0}{F'(x)\over (x^2)'} = \lim_{x\to 0}{2x \cos(x^4+x^2)\over 2x} =  \lim_{x\to 0} \cos(x^4+x^2) =\bbox[red, 2pt]1$$

解答: $$\textbf{(a) }  f(x)=\tan^{-1}x \Rightarrow f'(x)={1\over 1+x^2} =1-x^2+ x^4-x^6+ \cdots   \\ \Rightarrow f''(x)= -2x+4x^3-6x^5+ \cdots \Rightarrow f'''(x)=-2+12x^2-30x^4+ \cdots \\\Rightarrow f^{[4]}(x)=  24x-120x^3 +\cdots \Rightarrow f^{[5]}(x)=24-360x^2+ \cdots\\ \Rightarrow \cases{f(0) =0\\ f'(0) =1\\ f''(0) =0\\ f'''(0)=-2 \\f^{[4]}(0)=0\\ f^{[5]}(0)=24} \Rightarrow \tan^{-1}x=x-{2x^3\over 3!}+{24x^5\over 5!} -\cdots =x-{x^3\over 3}+{x^5\over 5} -\cdots\\ \Rightarrow \bbox[red, 2pt]{\tan^{-1}x = \sum_{n=0}^\infty (-1)^{n}{1\over 2n+1}x^{2n+1}}\\ \textbf{(b) }\cases{\sin x= x-{x^3\over 3!}+{x^5\over 5!}- \cdots\\ \sin (ax) =ax-{{a^3x^3\over 3!}+{a^5x^5\over 5!}} -\cdots\\ \tan^{-1} x=x-{x^3\over 3}+{x^5\over 5}-\cdots}\\ \Rightarrow \sin(ax)-\sin x-\tan^{-1}x =(a-2)x+{3-a^3\over 3!}x^3+{a^5-25 \over 5!}x^5+\cdots \\ \Rightarrow {\sin(ax)-\sin x-\tan^{-1}x \over x^3} ={a-2\over x^2}+{3-a^3\over 3!}+ {a^5-25\over 5!}x^2+\cdots\\ \Rightarrow\bbox[red, 2pt]{ a=2} \\\textbf{(c) }\lim_{x\to 0}{\sin(ax)-\sin x-\tan^{-1}x \over x^3}={3-a^3\over 3!} ={3-2^3\over 3!} =\bbox[red, 2pt]{-{5\over 6}}$$

解答: $$\cases{f(x,y)=3x^2-2y^2\\ g(x,y)=2x^2-2xy+y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y=\lambda g_y \\ g=0} \Rightarrow \cases{6x=\lambda(4x-2y) \\ -4y= \lambda(-2x+2y) \\ 2x^2-2xy+y^2=1} \\ \Rightarrow {6x\over -4y} ={4x-2y\over -2x+2y} \Rightarrow 3x^2-7xy+2y^2=0 \Rightarrow (3x-y)(x-2y)=0 \\ \Rightarrow \cases{x=y/3\\ x=2y} \Rightarrow \cases{g(y/3,y) =5/9y^2=1 \\ g(2y,y) =5y^2=1} \Rightarrow \cases{y=3/\sqrt 5  \Rightarrow x=1/\sqrt 5\\ y=-3/\sqrt 5  \Rightarrow x=-1/\sqrt 5\\ y=1/\sqrt 5 \Rightarrow x=2/\sqrt 5 \\ y=-1/\sqrt 5 \Rightarrow x=-2/\sqrt 5} \\\Rightarrow \cases{f(\pm1/\sqrt 5,\pm 3/\sqrt 5)=-3 \\ f(\pm 2/\sqrt 5, \pm 1/\sqrt 5)=2}  \Rightarrow \bbox[red, 2pt]{\cases{\text{maximum: 2} \\\text{minimum: }-3}}$$

解答: $$\textbf{(a) }e^x = \sum_{n=0}^\infty {x^n\over n!} \Rightarrow xe^x = \bbox[red, 2pt]{\sum_{n=0}^\infty {x^{n+1}\over n!}} \\\textbf{(b) } g(x) =\int_0^x te^t\,dt = \left. \left[ te^t-e^t\right] \right|_0^x =\bbox[red, 2pt]{xe^x-e^x+1} \\\qquad \Rightarrow g(x)= \sum_{n=0}^\infty {x^{n+1}\over n!} -\sum_{n=0}^\infty {x^{n}\over n!} +1 = \sum_{n=1}^\infty {x^{n} \over (n-1)!} -\left( 1+\sum_{n=1}^\infty {x^n\over n!}\right)+1 \\\qquad= \sum_{n=1}^\infty \left( {1\over (n-1)!} -{1\over n!}\right)x^n  =\sum_{n=2}^\infty {n-1\over n!}x^n =\sum_{n=0}^\infty {n+1\over (n+2)!}x^{n+2} =\bbox[red, 2pt]{\sum_{n= 0}^\infty {1\over n!(n+2)}x^{n+2}} \\\textbf{(c) }  g(1)=e^1-e^1+1= \sum_{n= 0}^\infty {1\over n!(n+2)} \Rightarrow \sum_{n=0}^\infty {1\over n!(n+2)} =\bbox[red, 2pt]1$$

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解題僅供參考，碩士班歷年試題及詳解