114年清華生科碩士班-微積分詳解

國立清華大學 114學年度碩士班考試入學試題

系所班組別:生命科學暨醫學院丙組
考試科目(代碼):微積分 (0601)

解答:$$\textbf{(A) }\lim_{x\to 1} {x^{1/5}-1\over x-1} = \lim_{x\to 1} {(x^{1/5}-1)'\over (x-1)'}= \lim_{x\to 1} {x^{-4/5} \over 5} = \bbox[red, 2pt]{1\over 5} \\\textbf{(B) }e^x =\sum_{k=0}^\infty {x^k\over k!} \Rightarrow \sum_{k=1}^\infty{x^{k-1} \over k!} ={e^x-1\over x} \Rightarrow \lim_{x\to 0}\sum_{k=1}^\infty{x^{k-1} \over k!} =\lim_{x\to 0} {e^x-1\over x}  =\lim_{x\to 0} {(e^x-1)'\over (x)'} \\ \qquad =\lim_{x\to 0} e^x =\bbox[red, 2pt]1$$

解答:$$\textbf{(A) }f(x)=x為奇函數 \Rightarrow f'(x)=1為偶函數 \Rightarrow \bbox[red, 2pt]{不正確} \\ \textbf{(B) }g(x) =|x| \Rightarrow g(x^2) =|x^2| =x^2 於x=0可微分, 但g(x)=|x|不可微分 \Rightarrow \bbox[red, 2pt]{不正確}$$

解答:$$\textbf{(A) }f(x)=e^x \sin({\pi\over 6}x) \Rightarrow f'(x)= e^x \sin({\pi\over 6}x) + {\pi\over 6}e^x \cos ({\pi\over 6}x) = \bbox[red, 2pt]{e^x \left(  \sin({\pi\over 6}x) +{\pi\over 6}\cos({\pi\over 6}x) \right)}\\ \textbf{(B) }f'(1)=e({1\over 2}+{\pi\over 6}\cdot {\sqrt 3\over 2}) =e \left( {1\over 2}+{\sqrt 3\pi\over 12} \right) \Rightarrow 切線方程式: y=e \left( {1\over 2}+{\sqrt 3\pi\over 12} \right)(x-1)+{e\over 2} \\\qquad \Rightarrow \bbox[red, 2pt]{y={e\over 12}((6+\sqrt 3\pi)x-\sqrt 3\pi)}$$

解答:$$\cases{f(x,y,z) =xyz\\ g(x,y,z)=xy+2xz+ 2yz-12} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{yz= \lambda(y+2z) \cdots(1)\\ xz= \lambda(x+2z) \cdots(2)\\ xy= \lambda(2x+2y) \cdots(3)\\ xy+2xz+2yz=12 \cdots(4)} \\ \Rightarrow \cases{{(1)\over (2)} ={y\over x}={y+2z\over x+2z} \Rightarrow x=y\\ {(2)\over (3)} ={z\over y} ={x+2z\over 2(x+y)} \Rightarrow y=2z} \Rightarrow \cases{x=2z\\ y=2z} 代入(4) \Rightarrow 4z^2+4z^2+4z^2=12z^2=12 \\ \Rightarrow \cases{z=1 \Rightarrow x=y=2\\ z=-1 \Rightarrow x=y=-2} \Rightarrow \cases{f(2,2,1) =4\\ f(-2,-2,-1) =-4} \Rightarrow 最大值\bbox[red, 2pt]4$$

解答:$$\int_{1/128}^{x^2/2} g(t)\,dt =x\tan x-{1\over 8}\tan{1\over 8} \Rightarrow {d\over dx} \int_{1/128}^{x^2/2} g(t)\,dt = {d\over dx} \left( x\tan x-{1\over 8}\tan{1\over 8} \right) \\ \Rightarrow xg(x^2/2)=\tan x+x\sec^2 x \Rightarrow g(x^2/2) ={1\over x}\tan x+ \sec^2 x \\ \Rightarrow g(\pi^2/72) = \left. {1\over x}\tan x+ \sec^2 x\right|_{\pi/6} ={6\over \pi}\cdot {1\over \sqrt 3}+{4\over 3} = \bbox[red, 2pt]{{4\over 3}+{2\sqrt 3\over \pi}}$$

解答:$$利用剝殼法(\text{shell method}), V= 2\pi \int_0^{\pi/6} x\sin^2 x\,dx = \pi \int_0^{\pi/6} (x-x\cos(2x))\,dx \\= \pi  \left(  \underbrace{\int_0^{\pi/2} x\,dx}_{I_1} - \underbrace{\int_0^{\pi/6} x\cos(2x)\,dx}_{I_2} \right)\\ I_1= \left. \left[ {1\over 2}x^2 \right] \right|_0^{\pi/6} ={\pi^2\over 72} \\ \cases{u=x\\ dv=\cos(2x)\,dx} \Rightarrow \cases{du =dx \\ v={1\over 2}\sin(2x)} \Rightarrow I_2= \left.{1\over 2}x\sin(2x) \right|_0^{\pi/6}-{1\over 2}\int_0^{\pi/6} \sin(2x)\,dx \\= \left. \left[ {1\over 2}x\sin(2x)+{1\over 4 }\cos(2x) \right] \right|_0^{\pi/6} = {\sqrt 3\pi\over 24}-{1\over 8} \\ \Rightarrow V=\pi(I_1+ I_2) = \bbox[red, 2pt]{{\pi\over 72} \left( \pi^2 -3\sqrt 3\pi +9\right)}$$

解答:$$\int e^x(f(x)+f'(x)) \,dx = e^x f(x)+C \\ \Rightarrow \int e^x \left( \ln x+{2\over x^3} \right)\,dx = \int e^x \left( \ln x+{1\over x}-({1\over x}-{1\over x^2}) -({1\over x^2}-{2\over x^3}) \right) \\ =\int e^x (\ln x+{1\over x})\,dx - \int e^x({1\over x}-{1\over x^2})\,dx -\int e^x({1\over x^2}-{2\over x^3})\,dx = \bbox[red, 2pt] {e^x(\ln x-{1\over x}-{1\over x^2})+C}$$

解答:$$f(x)={1\over 3+2x} ={1\over 3}\cdot {1\over 1-(-2x/3)} ={1\over 3} \sum_{n=0}^\infty (-{2\over 3}x)^n = \bbox[red, 2pt]{\sum_{n=0}^\infty (-1)^n {2^nx^n\over 3^{n+1}}} \\ \left|-{2\over 3}x \right| \lt 1 \Rightarrow {2\over 3}|x|\lt 1 \Rightarrow |x|\lt {3\over 2} \Rightarrow 收斂半徑R=\bbox[red, 2pt]{3\over 2} \\ f(x) ={1\over 3+2x}=\sum_{n=0}^\infty (-1)^n {2^nx^n\over 3^{n+1}}  \Rightarrow f'(x)=-{2\over (3+2x)^2} =\sum_{n=1}^\infty (-1)^n {2^n\cdot n\over 3^{n+1}}x^{n-1} \\ \Rightarrow xf'(x)=-{2x\over (3+2x)^2} =\sum_{n=1}^\infty (-1)^n {2^n\cdot n\over 3^{n+1}}x^{n } \Rightarrow {1\over 4}f'(1/4)=-{2\over 49} =\sum_{n=1}^\infty (-1)^n{2^n\cdot n\over 3^{n+1}} \cdot {1\over 4^n} \\={1\over 3} \sum_{n=1}^\infty (-1)^n{n\over 6^n} \Rightarrow \sum_{n=1}^\infty (-1)^n{n\over 6^n} =\sum_{k=1}^\infty {(-1)^k k\over 6^k} =\bbox[red, 2pt]{-{6\over 49}}$$

解答:$$\iint_R y\sin({\pi\over 2}x^3)\,dA= \int_1^2 \int_0^x y\sin({\pi\over 2}x^3)\,dy\,dx =\int_1^2  \left. \left[ {1\over 2}y^2\sin({\pi\over 2}x^3) \right] \right|_0^x \,dx \\=\int_1^2 {1\over 2}x^2\sin({\pi\over 2}x^3)\,dx =\int_1^8 {1\over 6} \sin({\pi\over 2}u)\,du = \left. \left[ -{1\over 3\pi} \cos ({\pi\over 2}u) \right] \right|_1^8 = \bbox[red, 2pt]{-{1\over 3\pi}}$$

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解題僅供參考，碩士班歷年試題及詳解