國立高雄大學 115 學年度轉學招生考試試題(轉二年級)
科目： 微積分
系所：資訊工程學系(無組別)
是否使用計算機： 是
一﹑選擇題 Choose the best answer for each question (30%)
解答:$$\lim_{x\to 0}{\tan(3x) \over \tan (5x)} = \lim_{x\to 0}{{d\over dx}\tan(3x) \over {d\over dx}\tan (5x)} = \lim_{x\to 0}{3\sec^2(3x) \over 5\sec^2 (5x)} = {3\over 5}，故選\bbox[red, 2pt]{(b)}$$
解答:$$\lim_{x\to 0}{26x^2 \over \sqrt{26x^2+4}-2} = \lim_{x\to 0}{26x^2 (\sqrt{26x^2+4}+2)\over (\sqrt{26x^2+4}-2)\sqrt{26x^2+4}+2} = \lim_{x\to 0}{26x^2 (\sqrt{26x^2+4}+2)\over 26x^2} \\ = \lim_{x\to 0} (\sqrt{26x^2+4}+2) =2+2=4，故選\bbox[red, 2pt]{(b)}$$
解答:$$\lim_{x\to 0^+}{\operatorname{arccot} x \over \arctan{1/x}} =\lim_{x\to 0^+}{\operatorname{arccot} x \over \operatorname{arccot}{x}} =1，故選\bbox[red, 2pt]{(c)}$$
解答:$$ f(x) = 4\sin(\frac{x}{2}) \Rightarrow f'(x) = \frac{d}{dx} \left[ 4\sin\left(\frac{x}{2}\right) \right] = 4 \cos\left(\frac{x}{2}\right) \cdot \frac{d}{dx}\left(\frac{x}{2}\right) = 4 \cos\left(\frac{x}{2}\right) \cdot \frac{1}{2} = 2\cos\left(\frac{x}{2}\right) \\，故選\bbox[red, 2pt]{(c)}$$
解答:$$2x^2 - y^2 = 8 \Rightarrow \frac{d}{dx}(2x^2 - y^2) = \frac{d}{dx}(8) \Rightarrow 4x - 2y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{4x}{2y} = \frac{2x}{y} \\ \Rightarrow \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{2x}{y}\right) = \frac{2 \cdot y - 2x \cdot \frac{dy}{dx}}{y^2} = \frac{2y - 2x(\frac{2x}{y})}{y^2} = \frac{2y - \frac{4x^2}{y}}{y^2} = \frac{2y^2 - 4x^2}{y^3} \\={2(2x^2-8)-4x^2\over y^3}=-{16\over y^3}，故選\bbox[red, 2pt]{(a)}$$
解答:$$ \int_{0}^{2} \frac{4x^3}{(x^4-6)^2} dx = \lim_{c \to (\sqrt[4]{6})^-} \int_{0}^{c} \frac{4x^3}{(x^4-6)^2} dx + \lim_{d \to (\sqrt[4]{6})^+} \int_{d}^{2} \frac{4x^3}{(x^4-6)^2} dx \\ \lim_{c \to (\sqrt[4]{6})^-} \int_{0}^{c} \frac{4x^3}{(x^4-6)^2} dx = \lim_{c \to (\sqrt[4]{6})^-} \left[ -\frac{1}{x^4-6} \right]_{0}^{c} = \lim_{c \to (\sqrt[4]{6})^-} \left( -\frac{1}{c^4-6} - \left(-\frac{1}{-6}\right) \right) = \infty - \frac{1}{6} = \infty \\ \Rightarrow \text{divergence}，故選\bbox[red, 2pt]{(d)}$$
二﹑填充題 Please fill in the following blanks (25%)
解答:$$u=x+3 \Rightarrow du=dx \Rightarrow $\int_{1}^{6} 5x\sqrt{x+3} dx = \int_{4}^{9} 5(u-3)\sqrt{u} du = 5 \int_{4}^{9} (u^{3/2} - 3u^{1/2}) du \\ = 5 \left[ \frac{2}{5}u^{5/2} - 3 \cdot \frac{2}{3}u^{3/2} \right]_{4}^{9} = 5 \left[ \frac{2}{5}u^{5/2} - 2u^{3/2} \right]_{4}^{9} = \left( \frac{216}{5} - \left(-\frac{16}{5}\right) \right) = \bbox[red, 2pt]{232 }$$
解答:$$\cases{u=8x^2 \\ dv=\sin(2x)\,dx} \Rightarrow \cases{du=16x\,dx\\ v=-{1\over 2}\cos(2x)} \Rightarrow \int 8x^2\sin(2x) dx = -4x^2 \cos(2x) + \int8x\cos(2x)\,dx \\=-4x^2 \cos(2x)+ 8 \left( {x\over 2}\sin(2x)+{1\over 4} \cos(2x) \right)+C = -4x^2\cos(2x) + 4x\sin(2x) + 2\cos(2x)+C \\ \int_0^{\pi}8x^2\sin(2x) dx = \left. \left[ -4x^2\cos(2x) + 4x\sin(2x) + 2\cos(2x) \right] \right|_0^{\pi} = (-4\pi^2 + 2) - 2 = \bbox[red, 2pt]{-4\pi^2}$$
解答:$$ f(x, y) = e^y \cos(x) \Rightarrow \nabla f(x,y)=(f_x,f_y) =(-e^y\sin x,e^y\cos x) \Rightarrow \nabla f(\pi, 0)= \bbox[red, 2pt]{(0,-1)}$$
解答:$$\sum_{n=1}^{\infty} \frac{1}{9n^2+3n-2} = \sum_{n=1}^{\infty} \frac{1}{3}\left( \frac{1}{3n-1} - \frac{1}{3n+2} \right) = \frac{1}{3} \left( \left(\frac{1}{2} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right) + \cdots\right) \\={1\over 3} \cdot {1\over 2} = \bbox[red, 2pt]{1\over 6}$$
解答:$$\ln(x^3 + 1) = \int_{0}^{x^2+1} f(t) dt \Rightarrow {d\over dx}\ln(x^3 + 1) ={d\over dx} \int_{0}^{x^2+1} f(t) dt \Rightarrow {3x^2 \over x^3+1} =f(x^2+1)\cdot 2x\\ x=2 \Rightarrow {3\cdot 4\over 8+1}=f(5)\cdot 4 \Rightarrow f(5)={3\over 9}=\bbox[red, 2pt]{1\over 3}$$
三﹑ 計算題 Problems (45%)
解答:$$ T(x,y) = 400 - 2x^2 - y^2 \Rightarrow \nabla T(x, y) = \left\langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right\rangle = \langle -4x, -2y \rangle \\ \text{the moving path of a particle: } \mathbf{r}(t) = \langle x(t), y(t) \rangle \Rightarrow \langle \frac{dx}{dt}, \frac{dy}{dt} \rangle \parallel \nabla T \Rightarrow \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2y}{-4x} = \frac{y}{2x} \\ \Rightarrow \frac{1}{y} dy = \frac{1}{2x} dx \Rightarrow \int \frac{1}{y} dy = \int \frac{1}{2x} dx \Rightarrow \ln y=\ln \sqrt x+C \Rightarrow y=k\sqrt{x}\\ P(8,2) \Rightarrow 2=k\sqrt 8 \Rightarrow k={1\over \sqrt 2} \Rightarrow \bbox[red, 2pt]{y=\sqrt{x\over 2}}$$
解答:$$x^2+y^2=4 \Rightarrow \cases{x=2\cos t\\ y=2\sin t} \Rightarrow (2,0)\to (1\sqrt 3) \Rightarrow t=0\to {\pi\over 3} \\ S = \int_{a}^{b} 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt = \int_{0}^{\pi/3} 2\pi (2\sin(t))\sqrt{(-2\sin t)^2+(2\cos t)^2}\, dt \\=8\pi \int_0^{\pi/3} \sin t\,dt=8\pi \left. \left[ -\cos t \right] \right|_0^{\pi/3} = 8\pi \left( -\frac{1}{2} + 1 \right) =\bbox[red, 2pt]{4\pi}$$
解答:$$ f(x) = x^7 + x^3 + x + 1 \Rightarrow \cases{f(0)=1 \gt 0\\ f(-1)=-2\lt 0} , \text{ by Intermediate Value Theorem, there exists } \\c\in (-1,0) \text{ such that }f(c)=0. \text{ That is, there is at least one real solution.}\\ \text{Suppose, for contradiction, that the equation has two distinct real solutions $a$ and $b ,\;(a\lt b),$}\\ \text{Then, }f(a)=f(b)=0. \text{Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. }\\ \text{By Roll's Theorem, there exists some $d\in (a,b)$ such that }f'(d)=0\\ \text{However, }f'(x)=7x^6+3x^2+1 \gt 0, \text{ for all }x. \text{ This contradicts $f'(d)=0$. } \\ \text{Therefore, the equation cannot have two distinct real solutions. It has exactly one real solution.}\\ \bbox[red, 2pt]{QED.}$$
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