113年成功大學環境醫學碩士班-微積分詳解

國立成功大學113學年度碩士班招生考試試題

系所:環境醫學研究所
考試科目:微積分

解答:$$\textbf{(a) }\lim_{x\to 0}{(\cos^2 x-1) \sin x\over x^2} =\lim_{x\to 0}{((\cos^2 x-1) \sin x)' \over (x^2)' } =\lim_{x\to 0}{-\sin(2x)\sin x+(\cos^2 x-1) \cos x\over 2x} \\\quad = \lim_{x\to 0}{(-\sin(2x)\sin x+(\cos^2 x-1) \cos x)' \over (2x)'}\\ =\lim_{x\to 0}{-2\cos(2x)\sin x-\sin(2x)\cos x-(\cos^2 x-1)\sin x-\sin(2x)\sin x\cos x \over 2}={0\over 2}=\bbox[red, 2pt]0\\ \textbf{(b) }x=1+a \Rightarrow \lim_{x\to 1^+}{[x]^2-[x^2] \over x^2-1}=\lim_{a\to 0^+}{[1+a]^2-[(1+a)^2] \over (1+a)^2-1} =\lim_{a\to 0^+}{1-[1+2a+a^2] \over a^2+ 2a}\\\qquad  =\lim_{a\to 0^+}{1-1 \over a^2+2a}= \bbox[red, 2pt]0\\ \textbf{(c) }[x]=x-\{x\} \Rightarrow \lim_{x \to -\infty} {1+[x] \over 5x-6} =\lim_{x\to -\infty}{x+1-\{x\} \over 5x-6} =\bbox[red, 2pt]{1\over 5}\\ \textbf{(d) }\lim_{n\to +\infty} \left( {1\over \sqrt{n^2 +2n}} + {1\over \sqrt{n^2+4n}} + \cdots + {1\over \sqrt{n^2+ 2n^2}}\right) =\lim_{n\to +\infty} \sum_{k=1}^n {1\over \sqrt{n^2+ 2kn}} \\ \quad =\lim_{n\to +\infty} \sum_{k=1}^n  {1 \over n\sqrt{1+2({k\over n})}} =\int_0^1 {1\over \sqrt{1+2x}}\,dx =\left. \left[ \sqrt{1+2x} \right] \right|_0^1 =\bbox[red, 2pt]{\sqrt 3-1}$$

解答:$$\text{By comparison test, }a_k \gt \ln(1+a_k), \forall a_k\gt 0 \Rightarrow \sum_{k=1}^\infty a_k \text{ converges, then }\sum_{k=1}^\infty \ln(1+a_k) \text{ converges}\\ \quad \bbox[red, 2pt]{QED}$$

解答:$$z=x-u \Rightarrow \int_{-\infty}^\infty {1\over \sqrt{2\pi}} e^{-(x-\mu)^2/2} \,dx = \int_{-\infty}^\infty {1\over \sqrt{2\pi} }e^{-z^2/2}\,dz = 1 \\ \Rightarrow I=\int_{-\infty}^\infty {x^2\over \sqrt{2\pi}} e^{-(x-\mu)^2/2}\,dx = \int_{-\infty}^\infty {(z+\mu)^2\over \sqrt{2\pi}}e^{-z^2/2}\, dz \\\qquad=\int_{-\infty}^\infty {z^2\over \sqrt{2\pi}} e^{-z^2/2}\,dz + \int_{-\infty}^\infty {2\mu z\over \sqrt{2\pi}} +\int_{-\infty}^\infty {\mu^2\over \sqrt{2\pi}}e^{-z^2/2} \,dz =\int_{-\infty}^\infty {z^2\over \sqrt{2\pi}} e^{-z^2/2}\,dz +0+\mu^2\\ \text{let }\cases{u=z\\ dv={z\over \sqrt{2\pi}}e^{-z^2/2}dz}, \text{ then } \cases{du=dz \\ v=-{1\over \sqrt{2\pi}}e^{-z^2/2}} \\ \Rightarrow \int_{-\infty}^\infty {z^2\over \sqrt{2\pi}} e^{-z^2/2}\,dz = \left. \left[ -{z\over \sqrt{2\pi}}e^{-z^2/2}\right] \right|_{-\infty}^\infty + \int_{-\infty}^\infty {1\over \sqrt{2\pi}}e^{-z^2/2}\,dz =0+1 \\ \Rightarrow I= 1+0+\mu^2 =\bbox[red, 2pt]{1+\mu^2}$$

解答:$$g(x,y)=\int_0^{x^2+ \ln y} \cos t^3\,dt \Rightarrow \cases{\frac{\partial g}{\partial x} =\cos (x^2+\ln y)^3\cdot \frac{\partial }{\partial x}(x^2+\ln y) =2x \cos(x^2+\ln y)^3 \\ \frac{\partial g}{\partial y} =\cos (x^2+\ln y)^3\cdot \frac{\partial }{\partial y}(x^2+\ln y) ={1\over y} \cos(x^2+ \ln y)^2} \\ \Rightarrow \bbox[red, 2pt]{\cases{\frac{\partial g}{\partial x} = 2x \cos(x^2+\ln y)^3 \\ \frac{\partial g}{\partial y}  ={1\over y} \cos(x^2+ \ln y)^2}}$$

解答:$$\lim_{x\to 0} {f(ax)-f(bx)\over cx} =\lim_{x\to 0} {(f(ax)-f(bx))'\over (cx)'} =\lim_{x\to 0} {af'(ax)-bf'(bx)\over c} = \bbox[red, 2pt]{\left({a-b\over c} \right)f'(0)}$$

解答:$$I= \int_0^1 \int_x^1 \tan^{-1}y\,dy dx = \int_0^1 \int_0^y \tan^{-1}y\,dxdy =\int_0^1y\tan^{-1}y\, dy\\ \cases{u=\tan^{-1}y \\ dv=ydy} \Rightarrow \cases{du={1\over y^2+1}\,dy\\ v={1\over 2}y^2} \Rightarrow I= \left.{1\over 2}y^2 \tan^{-1}y \right|_0^1-{1\over 2}\int_0^1 {y^2\over y^2+1}\,dy \\={\pi\over 8}-{1\over 2}\int_0^1 \left( 1-{1\over y^2+1}\right)\,dy ={\pi\over 8}-{1\over 2} \left. \left[ y-\tan^{-1}y \right] \right|_0^1 ={\pi\over 8}-{1\over 2}\left( 1-{\pi\over 4}\right)= \bbox[red, 2pt]{{\pi\over 4}-{1\over 2}}$$

解答:$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \iint_K \cos(x^2+y^2)\,dxdy = \int_0^{\pi/2} \int_0^1 r\cos r^2\,drd\theta= \int_0^{\pi/2} \left.\left[ {1\over 2}\sin r^2\right] \right|_0^1\,d\theta \\ = \int_0^{\pi/2} {1\over 2} \sin 1\,d\theta =\bbox[red, 2pt]{\pi \sin 1 \over 4}$$

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解題僅供參考，碩士班歷年試題及詳解