國立臺灣海洋大學108學年度轉學生入學招生考試
考試科目：微積分
系組名稱：機械二、輪機能源二(G)、輪機動力二(G)、電機二(G)、資工二(B)、通訊二(B)、光電二(B)
解答：$$\mathbf{(a)}\lim_{x\to 0}{1-e^x\over e^{2x}-1} =\lim_{x\to 0}{-(e^x-1) \over (e^x+1)( e^{x}-1)} =\lim_{x\to 0}{-1 \over e^x+1 } =\bbox[red,2pt]{-1\over 2} \\\mathbf{(b)}\lim_{x\to 2^+} \left({8\over x^2-4}-{x\over x-2}\right) =\lim_{x\to 2^+} {8-x(x+2)\over x^2-4} =\lim_{x\to 2^+} {-(x+4)(x-2)\over (x+2)(x-2)} =\lim_{x\to 2^+} {-(x+4) \over x+2 }\\\quad =\bbox[red,2pt]{-3\over 2} \\\mathbf{(c)}\lim_{(x,y)\to (0,0)}{x-y\over \sqrt x-\sqrt y} =\lim_{(x,y)\to (0,0)}{(\sqrt x- \sqrt y)(\sqrt x+\sqrt y)\over \sqrt x-\sqrt y} =\lim_{(x,y)\to (0,0)}(\sqrt x+\sqrt y)=\bbox[red, 2pt]0$$
解答：$$\mathbf{(a)}f(t)={\sqrt{3t+1} \over t} \Rightarrow f'(t)={3\over 2}\cdot {1\over \sqrt{3t+1}} \cdot {1\over t}-{\sqrt{3t+1} \over t^2} =\bbox[red,2pt]{{3\over 2t \sqrt{3t+1}}-{\sqrt{3t+1} \over t^2}} \\\mathbf{(b)}f(x)= \ln \sqrt[3]{x-2\over 2x+1} ={1\over 3}\left( \ln (x-2)-\ln(2x+1)\right) \Rightarrow f'(x)={1\over 3}\left({1\over x-2}-{2\over 2x+1} \right)\\ =\bbox[red, 2pt]{ 5\over 3(x-2)(2x+1)}$$
解答：$$f(x,y)=y\cos(x-y) \Rightarrow \nabla f=(f_x,f_y)=(-y\sin(x-y), \cos(x-y)+y\sin(x-y))\\ \Rightarrow \nabla f_{(0,\pi/3)} =(-{\pi\over 3 }\sin(-{\pi\over 3 }),\cos(-{\pi\over 3 })+{\pi\over 3}\sin(-{\pi\over 3 })) = \bbox[red,2pt]{({\sqrt 3\pi\over 6}, {3-\sqrt 3\pi \over 6})}$$
解答：$$\vec u ={\overrightarrow{PQ} \over |\overrightarrow{PQ}|} ={(-2,-4,0)\over \sqrt{(-2)^2+(-4)^2}} =(-{1\over \sqrt 5},{-2\over \sqrt 5}, 0)\\ \Rightarrow 方向導數D_{\vec u} h= \nabla h\cdot \vec u =(e^zy, e^z x,e^z xy)\cdot (-{1\over \sqrt 5},{-2\over \sqrt 5}, 0)\\ =-{1\over \sqrt 5}e^zy-{2\over \sqrt 5}e^z x \Rightarrow D_{\vec u} h|_{(2,4,0)} =-{4\over \sqrt 5}-{4\over \sqrt 5} = \bbox[red, 2pt]{-{8\sqrt 5\over 5}}$$
解答：$$f(x,y)= 2xy-{1\over 2}(x^4+y^4)+ 12 \Rightarrow \cases{f_x= 2y-2x^3\\ f_y=2x-2y^3} \Rightarrow \cases{f_{xx} =-6x^2\\ f_{xy} =2\\ f_{yy}=-6y^2}\\ 因此\cases{f_x=0\\ f_y=0} \Rightarrow \mathbf{(a)}:臨界點(x,y)=\bbox[red,2pt]{(0,0),(1,1),(-1,-1)}\\ 又 \cases{f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^2 =-4\lt 0\\ f_{xx}(1,1)f_{yy}(1,1)-(f_{xy}(1,1))^2 =32 \gt 0\\ f_{xx}(-1,-1)f_{yy}(-1,-1)-(f_{xy}(-1,-1))^2 =32 \gt 0} \Rightarrow \mathbf{(b)}:\bbox[red,2pt]{\cases{(0,0)為鞍點\\ (1,1)及(-1,-1)為相對極值點}}$$
解答：$$\mathbf{(a)}\int_0^\pi \left(4x^{1/3} -2\sin x\right)\,dx =\left. \left[ 3x^{4/3}+ 2\cos x\right] \right|_0^\pi = \bbox[red, 2pt]{3\pi^{4/3} -4}\\ \mathbf{(b)}\cases{u=x \Rightarrow du=dx\\ dv= \sec^2x\,dx \Rightarrow v=\tan x} \Rightarrow \int x\sec^2 x\,dx= x\tan x-\int \tan x\,dx = \bbox[red, 2pt]{x\tan x-\ln |\sec x|+C} \\\mathbf{(c)} \int {2x^2+x+1 \over x+x^3}\,dx =\int \left({1\over x}+{x+1\over x^2+1} \right)\,dx =\int {1\over x}\,dx + \int {x\over x^2+1}\,dx +\int {1\over x^2+1}\,dx \\ \qquad = \bbox[red, 2pt]{\ln|x| +{1\over 2}\ln(x^2+1) + \tan^{-1}x +C} \\\mathbf{(d)} u=3x+1 \Rightarrow du=3dx \Rightarrow \int_0^1 x\sqrt{3x+1}\,dx =\int_1^4 {u-1\over 3}\cdot \sqrt u\cdot {1\over 3}du ={1\over 9}\int_1^4 \left( u^{3/2}-u^{1/2} \right)\,du \\\qquad ={1\over 9} \left.\left[ {2\over 5}u^{5/2} -{2\over 3}u^{3/2} \right]\right|_1^4 ={1\over 9}({2\over 5}\cdot 32-{2\over 3}\cdot 8-({2\over 5}-{2\over 3})) =\bbox[red, 2pt]{116\over 135}\\ \mathbf{(e)} u=\sqrt x \Rightarrow du ={1\over 2\sqrt x}dx \Rightarrow \int_0^1 {\ln x\over \sqrt x}dx =\int_{0}^1 {\ln (u^2)}2du =4\int_0^1 \ln u\,du =4\left. \left[ x(\ln x-1)\right]\right|_0^1\\\quad = \bbox[red,2pt]{-4} \\\mathbf{(f)} \sin(7x)\cos(5x) = \sin(5x+2x)\cos(5x) = \sin(5x)\cos(2x) \cos(5x)+ \cos^2(5x) \sin(2x) \\ \quad = {1\over 2}\sin(10x)\cos(2x)+ {1\over 2}(\cos(10x)+1)\sin(2x) ={1\over 2}\sin(12x)+{1\over 2}\sin(2x) \\\quad \Rightarrow \int \sin(7x)\cos(5x) \,dx ={1\over 2}\int \left(\sin(12x)+\sin(2x) \right)\,dx = \bbox[red, 2pt]{-{1\over 24}\cos(12x)-{1\over 4}\cos(2x)+C}$$
解答：
$$y=\sqrt{1-x^2} \Rightarrow x=\sqrt{1-y^2} \Rightarrow 欲求之體積=\int_0^{\sqrt 3/2} (1-y^2)\pi\,dy -\int_0^{\sqrt 3/2} {1\over 2}^2\pi \,dy \\ =\left({3\sqrt 3\over 8}-{\sqrt 3\over 8}\right)\pi = \bbox[red, 2pt]{{\sqrt 3\over 4}\pi}$$
解答：$$\int_0^a f(x)\,dx= \int_0^a f(a-x) \,dx\Rightarrow \int_0^{\pi/2} f(\cos x)\,dx =\int_0^{\pi/2} f(\cos(\pi/2- x))\,dx =\int_0^{\pi/2} f(\sin(x))\,dx\\ \bbox[red, 2pt]{故得證}$$
解答：$$平均速度={3-2\over \triangle t} ={\int_2^3 e^x\,dx \over 3-2} =e^3-e^2 \Rightarrow \triangle t =\bbox[red, 2pt]{1\over e^3-e^2}$$
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未公告答案，解題僅供參考
解答：$$\int_0^a f(x)\,dx= \int_0^a f(a-x) \,dx\Rightarrow \int_0^{\pi/2} f(\cos x)\,dx =\int_0^{\pi/2} f(\cos(\pi/2- x))\,dx =\int_0^{\pi/2} f(\sin(x))\,dx\\ \bbox[red, 2pt]{故得證}$$
解答：$$平均速度={3-2\over \triangle t} ={\int_2^3 e^x\,dx \over 3-2} =e^3-e^2 \Rightarrow \triangle t =\bbox[red, 2pt]{1\over e^3-e^2}$$
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