113年政大碩士班-微積分詳解

 國立政治大學113學年度碩士班招生考試

考試科目:微積分
系所別:應用數學系

解答: $$\textbf{A. } u= \cos \theta \Rightarrow du=-\sin \theta \,d\theta \Rightarrow \int_0^{\pi/2} \sin^3 \theta \cos^2 \theta\,d\theta = \int_0^{\pi/2} \sin \theta (1-\cos^2 \theta)\cos^2 \theta\,d\theta \\\qquad =\int_1^{0} -(1-u^2)u^2\,du =\int_0^1 \left( u^2-u^4 \right)\,du={1\over 3}-{1\over 5} = \bbox[red, 2pt]{2\over 15} \\ \textbf{B. }\int_0^1 \int_{x^2}^1 \sqrt y\sin y\,dydx = \int_0^1 \int_0^\sqrt y \sqrt y\sin y\,dxdy = \int_0^1 y\sin y\,dy = \left. \left[ -y\cos y+\sin y \right] \right|_0^1\\\qquad = \bbox[red, 2pt]{\sin 1-\cos 1}  \\\textbf{C. }\iiint_E {1\over x^3}\,dV = \int_0^1 \int_0^{y^2} \int_1^{z+1} {1\over x^3}\, dxdzdy = \int_0^1 \int_0^{y^2} \left. \left[-{1\over 2}x^{-2} \right] \right|_1^{z+1} \,dzdy \\\qquad = \int_0^1 \int_0^{y^2} {1\over 2}\left(1-{1\over (z+1)^2} \right)\,dzdy =\int_0^1 \left. \left[{1\over 2}\left(z+{1\over 1+z} \right) \right] \right|_0^{y^2}\, dy\\\qquad = {1\over 2}\int_0^1 \left( y^2+{1\over 1+y^2} -1\right)\,dy  ={1\over 2} \left. \left[ {1\over 3}y^3+ \tan^{-1}y-y\right] \right|_0^1 ={1\over 2}\left( {1\over 3}+ {\pi\over 4} -1\right) = \bbox[red, 2pt]{{\pi\over 8}-{1\over 3}}$$

解答: $$f(x) =\int_0^x x^2 \sin(t^2)dt \Rightarrow f'(x) = \bbox[red, 2pt]{x^2 \sin(x^2) +\int_0^x 2x\sin(t^2)\,dt}$$

解答: $$h(x) =(\sin x)^x = e^{x\ln \sin x} \Rightarrow h'(x)=(\ln \sin x+{x \cos x\over \sin x}) e^{x\ln \sin x} =(\sin x)^x (\ln \sin x+ x\cot x), \bbox[red, 2pt]{QED.}$$

解答: $$\cases{u=\ln x\\ dv=dx/\sqrt x} \Rightarrow \cases{du= dx/x\\ v=2\sqrt x} \Rightarrow \int  {\ln x\over \sqrt x}\,dx = 2\sqrt x \ln x-\int {2\over \sqrt x}\,dx =2\sqrt x \ln x-4\sqrt x \\ \Rightarrow   \int_0^4{\ln x\over \sqrt x}\,dx = \left. \left[ 2\sqrt x \ln x-4\sqrt x \right] \right|_0^4 = \bbox[red, 2pt]{8(\ln 2-1)}$$

解答: $$f(x,y) =\begin{cases}{x^3y-xy^3\over x^2+y^2} & \text{if }(x,y)\ne (0,0)\\ 0& \text{if }(x,y)=(0,0) \end{cases} \Rightarrow f_x(x,y)=\begin{cases}{y(x^4+4x^2y^2-y^4)\over (x^2+y^2)^2} & \text{if }(x,y)\ne (0,0)\\ 0& \text{if }(x,y)=(0,0) \end{cases} \\ \Rightarrow f_{xy}(0,0)= \lim_{h\to 0}{f_x(0, h)-f_x(0,0) \over h}= \lim_{h\to 0} {h(-h^4)/h^4\over h}= \lim_{h\to 0} {-h\over h} =\bbox[red, 2pt]{-1}$$

解答: $$\cases{g(t)=f(tx,ty) \Rightarrow g'(t)=xf_x(tx,ty)+ yf_y(tx,ty) \\g(t)=t^nf(x,y) \Rightarrow g'(t)=nt^{n-1} f(x,y)} \\ \Rightarrow g'(1)= xf_x(x,y) +yf_y(x,y)=nf(x,y). \bbox[red, 2pt]{QED.}$$

解答: $$\lim_{x\to a}g(x)=c \Rightarrow \exists \delta_1: 0 < |x-a| < \delta_1 \implies |g(x)-c| < \frac{c}{2} \Rightarrow g(x)\gt {c\over 2}\\ \lim_{x\to a}f(x)=\infty \Rightarrow \exists \delta_2 > 0: 0 < |x-a| < \delta_2 \implies f(x) > \frac{2M}{c} \in \mathbb R\\ \text{Now we choose }\delta=\min\{\delta_1, \delta_2\} \Rightarrow 0\lt |x-a|\lt \delta \Rightarrow f(x)g(x) \gt {c\over 2}\cdot {2M\over c}=M \in \mathbb R \\ \Rightarrow \lim_{x\to a} f(x)g(x) =\infty, \bbox[red, 2pt]{QED}$$

解答: $$f(x)={x\over 1+x^2}-\tan^{-1}x \Rightarrow f'(x)=-{2x^2 \over (1+x^2)^2} \lt 0, \text{ for }x\gt 0 \\ \Rightarrow f(x) \text{ is strictly decreasing} \Rightarrow {x\over 1+x^2}-\tan^{-1}x \lt 0 \Rightarrow {x\over 1+x^2}\lt \tan^{-1}x, \text{ for }x\gt 0 \\ g(x)=x- \tan^{-1}x \Rightarrow g'(x)=1-{1\over 1+x^2} ={x^2\over 1+x^2} \gt 0, \text{ for }x\gt 0 \\ \Rightarrow g(x) \text{ is strictly increasing }\Rightarrow x-\tan^{-1}x \gt 0 \Rightarrow \tan^{-1}x \lt x, \text{ for }x\gt 0\\ \text{At last, we have }\cases{{x\over 1+x^2}\lt \tan^{-1}x\\ \tan^{-1}x \lt x} \Rightarrow {x\over 1+x^2}\lt \tan^{-1}x\lt x, \text{ for }x\gt 0. \bbox[red, 2pt]{QED.}$$

解答: $$\cos \theta =\sqrt{{1\over 2}(\cos 2\theta+1)} \Rightarrow \cos {\pi\over 4}={1\over 2}\sqrt 2 \Rightarrow \cos{\pi\over 8}=\sqrt{{1\over 2}\left( {1\over 2}\sqrt 2+1\right)}={1\over 2}\sqrt{2+\sqrt 2} \\ \Rightarrow \cos {\pi\over 16} =\sqrt{{1\over 2}\left(\sqrt{{1\over 2}\left( {1\over 2}\sqrt 2+1\right)} +1\right)} ={1\over 2}\sqrt{2+\sqrt{2+\sqrt 2}} \Rightarrow \cdots\\ \Rightarrow \cos{\pi \over 2^{n+1}} = \underbrace{\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+{\ldots}}}}}_\text{n times}=\frac{1}{2}a_{n} \Rightarrow a_n= 2\cos{\pi\over 2^{n+1}} \\ \Rightarrow \lim_{n\to \infty} a_n=2 \Rightarrow \lim_{n\to \infty} a_n\text{ exists and }\lim_{n\to \infty} a_n=2. \bbox[red, 2pt]{QED}$$

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解題僅供參考，碩士班歷年試題及詳解