114年成大環境醫學碩士班-微積分詳解

 國立成功大學114學年度碩士班招生考試試題

系 所:環境醫學研究所
科目:微積分

解答: $$\textbf{(a) }L=(1+ \sin(2x))^{1/3x} =e^{\ln(1+\sin(2x))/3x} \Rightarrow \ln L={\ln(1+\sin(2x)) \over 3x} \\ \qquad \Rightarrow \lim_{x\to 0} \ln L =\lim_{x\to 0} {\ln(1+\sin(2x)) \over 3x} =\lim_{x\to 0} {2\cos(2x)/(1+\sin(2x)) \over 3} ={2\over 3} \Rightarrow \lim_{x\to 0} L= \bbox[red, 2pt]{e^{2/3}} \\\textbf{(b) } \lim_{x\to 0}{\tan^{-1}(ax) \over \tan^{-1}(bx)} =\lim_{x\to 0}{a/(1+a^2x^2) \over b/(1+b^2x^2)} =\bbox[red, 2pt]{a\over b} \\ \textbf{(c) }\ln(1+x) =x-{x^2\over 2}+{x^3\over 3} + \cdots \Rightarrow \lim_{x\to 0^+} {\sqrt{x-\ln(1+x)} \over x} = \lim_{x\to 0^+} { \sqrt{x^2/2-x^3/3+ \cdots} \over x} \\ \qquad \Rightarrow \lim_{x \to0^+} \sqrt{1/2-x/3+\cdots}= \sqrt{1\over 2} =\bbox[red, 2pt]{\sqrt 2\over 2} \\ \textbf{(d) }\lim_{x\to 0^+} {\int_0^{x^2} e^t\sqrt t \sin\sqrt{t}\,dt +x\cos x-x \over x^3} =\lim_{x\to 0^+} {  e^{x^2}( 2x^2)   \sin x\ +\cos x-x\sin x-1 \over 3x^2} \\\qquad =\lim_{x\to 0^+} {  e^{x^2}(4x^3 \sin x+ 4x\sin x+2x^2 \cos x)-2 \sin x\  -x\cos x \over 6x} \\\qquad = \lim_{x\to 0^+} {  e^{x^2}(8x^4 \sin x+ 18x^2\sin x+ 8x^3 \cos x+4\sin x+8x\cos x)-3 \cos x\  +x\sin x\over 6} \\\qquad = \bbox[red, 2pt]{-{1\over 2}}$$

解答: $$g(x)= \int_1^{\tan x} \sqrt{1+t^3}\,dt \Rightarrow g'(x)= \sec^2 x\sqrt{1+\tan^3 x} \Rightarrow g'(\pi/4) = 2\sqrt{2}\\ f(x)=e^{g(x)} \Rightarrow f'(x)=g'(x)e^{g(x)} \Rightarrow f'(\pi/4) =g'(\pi/4) e^{g(\pi/4)} =2\sqrt 2 e^0 =\bbox[red, 2pt]{2\sqrt 2}$$

解答: $$G(x) =\int_0^{x^3} f(u)(x-\sqrt[3] u)\,du = x\int_0^{x^3}f(u)\,du -\int_0^{x^3} \sqrt[3]u f(u)\,du \\ \Rightarrow G'(x)= \int_0^{x^3}f(u)\,du+ 3x^3 f(x^3) -3x^3f(x^3) = \int_0^{x^3}f(u)\,du \\ \Rightarrow G(x)=C+ \int_0^x G'(u)\,du =C+ \int_0^x \left( \int_0^{u^3} f(t)\,dt \right)\,du\\ G(0) =\int_0^0 f(u)(x-\sqrt[3]u)du =0 \Rightarrow C=0 \Rightarrow G(x)=F(x). \quad \bbox[red, 2pt]{QED.}$$

解答: $$\text{Assume }a \text{ and }b \text{ are two distinct fixed points, then }\cases{f(a)=a\\ f(b)=b};\\ \text{By Mean Value Theorem, there exists }c\in (a,b) \text{ such that }f'(c)={f(b)-f(a)\over b-a} ={b-a\over b-a}=1,\\ \text{  contradicting the hypothesis. That is, }f\text{ has at most one fixed point.} \bbox[red, 2pt]{QED.}$$

解答: $$I= \int_0^1 \int_y^1 \tan(x^2)\,dx dy = \int_0^1 \int_0^x \tan(x^2)\,dy dx = \int_0^1  x\tan(x^2)\,dx \\ u=x^2 \Rightarrow du=2x\,dx \Rightarrow I=\int_0^1 {1\over 2} \tan u\,du= \left. \left[ -{1\over 2}\ln \cos u \right] \right|_0^1= \bbox[red, 2pt]{-{1\over 2} \ln \cos 1}$$

解答: $$\cases{x=uv\\ y=v-u} \Rightarrow \cases{\frac{\partial x}{\partial u} =v \\\frac{\partial x}{\partial v} =u \\ \frac{\partial y}{\partial u} =-1\\ \frac{\partial y}{\partial v} =1 } \Rightarrow \frac{\partial z}{\partial u} =\frac{\partial z}{\partial x}\frac{\partial x}{\partial u} +\frac{\partial z}{\partial y} \frac{\partial y}{\partial u} =f_xv+f_y(-1) =f_xv-f_y \\ \Rightarrow \frac{\partial^2 z}{\partial u \partial v} =\frac{\partial }{\partial v} (f_xv-f_y) =\frac{\partial }{\partial v}(f_xv)-\frac{\partial }{\partial v}f_y =(f_{xx}u + f_{xy})v+ f_x-f_{yx}u-f_{yy} \\=f_{xx}uv+ f_{xy}(v-u)+f_x-f_{yy}=  \bbox[red, 2pt]{xf_{xx}+yf_{xy}+f_x-f_{yy}}$$

解答: $$x^{2/3} +y^{2/3}+y=6 \Rightarrow {2\over 3}x^{-1/3} +{2\over 3}y^{-1/3}y'+y'=0 \Rightarrow y'={-{2\over 3}x^{-1/3} \over {2\over 3}y^{-1/3} +1} \\ \Rightarrow y'(8,1)= {-1/3 \over 5/3} =-{1\over 5} \Rightarrow \text{ tangent line: }y=-{1\over 5}(x-8)+1 \Rightarrow \bbox[red, 2pt]{x+5y= 13}$$

解答: $$\textbf{(a) }g(x)=\int_0^x f(u)du = -2+x^2+x\sin(2x)+ c\cos (2x) \Rightarrow g(0)=\int_0^0 f(u)du = 0= -2+ c  \\\qquad \Rightarrow \bbox[red, 2pt]c=2 \\\textbf{(b) }g'(x)=f(x) =2x+ \sin(2x)+ 2x\cos(2x)-4\sin(2x) \Rightarrow f'(x) =2+ 4\cos(2x) -4x\sin(2x)-8\cos(2x) \\\qquad \Rightarrow f'(\pi/4) =2+0-\pi-0= \bbox[red, 2pt]{2-\pi}$$

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