國立臺灣大學 114 學年度轉學生招生考試
科目：微積分 (C)
Part I: 5 points for each blank.
解答:$$\textbf{(1) } \lim_{x\to 1} \frac{\sqrt[3]{x+7}-2}{1-\sqrt{x}} = \lim_{x\to 1} \frac{\frac{d}{dx}(\sqrt[3]{x+7}-2)}{\frac{d}{dx}(1-\sqrt{x})}=\lim_{x\to 1} \frac{\frac{1}{3}(x+7)^{-2/3}}{-\frac{1}{2}x^{-1/2}} = \frac{\frac{1}{3}(8)^{-2/3}}{-\frac{1}{2}(1)^{-1/2}} = \frac{\frac{1}{3} \cdot \frac{1}{4}}{-\frac{1}{2}} = \bbox[red,2pt]{-\frac{1}{6}} \\ \textbf{(2) } \lim_{x\to \infty} x\tan\left( \frac{ \pi}{x}\right) = \lim_{t\to 0^+} \frac{\pi}{t}\tan(t) = \pi \cdot \lim_{t\to 0^+} \frac{ \tan(t)}{t} = \pi \cdot 1 = \pi \\ \lim_{x\to \infty} \tan^{-1}\left( \frac{x}{ \pi}\right) = \frac{\pi}{2} \Rightarrow \lim_{x\to \infty} x\tan \left( \frac{\pi}{x}\right)\tan^{-1}\left(\frac{x}{\pi}\right) =\pi \cdot {\pi\over 2}= \bbox[red, 2pt]{\pi^2\over 2}$$
解答:$$ f(x) = \frac{\sin x}{2 - \cos x} \Rightarrow \cases{f(0)=0\\ f(\pi)=0} \Rightarrow f'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2} \\f'(x)=0 \Rightarrow \cos x={1\over 2} \Rightarrow x={\pi\over 3} \Rightarrow f(\pi/3)={\sqrt 3\over 3} \Rightarrow \cases{\textbf{(3) } \text{absolute maximum: }\bbox[red, 2pt]{\sqrt 3}\\ \textbf{(4) }\text{absolute minimum: }\bbox[red, 2pt]0}$$
解答:
$$\textbf{(5) }V_x= \pi \int_1^{e^2} (\ln x)^2\,dx = \left. \left[ x(\ln x)^2-2x(\ln x-1) \right] \right|_1^{e^2} = \bbox[red, 2pt]{2\pi(e^2-1)} \\ \textbf{(6) } V_y = 2\pi \int_{1}^{e^2} x \ln x \, dx = 2\pi \left[ \frac{1}{2}x^2 \ln x - \frac{1}{4}x^2 \right]_1^{e^2} = \bbox[red, 2pt]{{\pi\over 2}(3e^4+1)}$$
解答:$$\textbf{(7) } \int_0^2 \int_{x^2}^4 \int_0^{x\sin(y^2)} 2025 \,dz\,dy\,dx = \int_0^2 \int_{x^2}^4 2025x\sin(y^2) \,dy\,dx = \int_0^4 \int_0^{\sqrt{y}} 2025x\sin(y^2) \,dx\,dy \\= \int_0^4 \frac{2025}{2}y\sin(y^2) \,dy = \frac{2025}{4} \left. \left[ -\cos(y^2) \right] \right|_0^4 = \bbox[red, 2pt] {\frac{2025}{4}(1 - \cos(16))} \\ \textbf{(8) }\cases{x=r\cos \theta \\y=r\sin \theta} \Rightarrow \int_{-1}^1 \int_{\vert{}y\vert{}}^{\sqrt{2-y^2}} (x^2+y^2)^{3/2} \,dx\,dy= \int_{-\pi/4}^{\pi/4} \int_0^{\sqrt{2}} r^3 \cdot r \,dr\,d\theta = \int_{-\pi/4}^{\pi/4} \int_0^{\sqrt{2}} r^4 \,dr\,d\theta \\= \int_{-\pi/4}^{\pi/4} \frac{4\sqrt{2}}{5} \,d\theta = \frac{4\sqrt{2}}{5} \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \frac{4\sqrt{2}}{5} \cdot \frac{\pi}{2} = \bbox[red, 2pt]{\frac{2\sqrt{2}\pi}{5} }$$
Part II: 15 points for each problem.
解答:$$\textbf{(a) } \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots \\\Rightarrow \frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \dots \approx 1 - \frac{x^2}{6} + \frac{x^4}{120} \\ \textbf{(b) }f(x) =1 - \frac{x^2}{6} + \frac{x^4}{120} \Rightarrow f'(x)= -\frac{x}{3} + \frac{x^3}{30} = \frac{x(x^2 - 10)}{30} \Rightarrow f''(x)=-{1\over 3}+{x^2\over 10}\\ f'(x)=0 \Rightarrow x(x^2-10)=0 \Rightarrow x=0, \pm \sqrt{10} \Rightarrow \cases{f''(0)=-1/3\lt 0\\ f''(\pm \sqrt{10}) =2/3 \gt 0}\\ \Rightarrow \cases{\text{local max: }f(0)=1\\ \text{local min:} f(\pm \sqrt{10}) =1/6} \\ f''(x)=0 \Rightarrow -\frac{1}{3} + \frac{x^2}{10} = 0 \Rightarrow x^2 = \frac{10}{3} \Rightarrow x= \pm\sqrt{\frac{10}{3}} \Rightarrow f \left( \pm \sqrt{10\over 3} \right)= {29\over 54} \\ \Rightarrow \text{inflection points: } {\left(\sqrt{\frac{10}{3}}, \frac{29}{54}\right)} , {\left(-\sqrt{\frac{10}{3}}, \frac{29}{54}\right)} $$
解答:$$\textbf{(a) }y=f(x)=mx+b \Rightarrow {dy\over dx}=m=y-2x=(mx+b)-2x=(m-2)x+b \\ \quad \Rightarrow \bbox[red, 2pt]{\cases{m=2\\ b=2}} \\ \textbf{(b) } {dy\over dx}=y-2x \Rightarrow y'-y=-2x \Rightarrow \cases{P(x)=-1\\ Q(x)=-2x} \Rightarrow I(x)=e^{\int -1\,dx}=e^{-x} \\ \quad \Rightarrow y=e^x \int e^{-x}\cdot(-2x)\,dx = e^x (2x e^{-x} + 2 e^{-x} + C) \Rightarrow \bbox[red, 2pt]{y=2x+2+Ce^x}$$
解答:$$\textbf{(a) }\cases{F(x,y,z) =x^2+y^2+z^2\\ g(x,y,z)=x^3yz^2-2} \Rightarrow \cases{F_x=\lambda g_x\\ F_y = \lambda g_y\\ F_z= \lambda g_z \\ g=0} \Rightarrow \cases{2x= \lambda\cdot 3x^2yz^2 \cdots(1)\\ 2y=\lambda\cdot x^3z^2 \cdots(2)\\ 2z=\lambda\cdot 2x^3yz \cdots(3)\\ x^3yz^2=2} \\\quad \text{Eq.}(3) \Rightarrow \lambda= {1\over x^3y} \Rightarrow \cases{\text{Eq.(1) }\Rightarrow 2x = 3\left(\frac{1}{x^3y}\right) x^2y(2y^2) = \frac{6y^2}{x} \Rightarrow x^2 = 3y^2 \Rightarrow x = \sqrt{3}y \\ \text{Eq.(2) }\Rightarrow 2y = \left(\frac{1}{x^3y}\right) x^3z^2 = \frac{z^2}{y} \Rightarrow 2y^2 = z^2 \Rightarrow z = \sqrt{2}y } \\ \text{Eq.(4) }\Rightarrow (\sqrt{3}y)^3 (y) (\sqrt{2}y)^2 = 2 \Rightarrow y^6 = \frac{2}{6\sqrt{3}} = \frac{1}{3\sqrt{3}} = 3^{-3/2} \Rightarrow y={1\over \sqrt[4]3} \Rightarrow \cases{x=\sqrt 3y= \sqrt[4]3 \\ z=\sqrt 2y=\sqrt 2/\sqrt[4]3} \\ \Rightarrow \bbox[red, 2pt] {\left(3^{1/4}, 3^{-1/4}, \sqrt{2} \cdot 3^{-1/4}\right) } \text{ is closest to the origin} \\\textbf{(b) } f(x,y) = x^2 + y^2 + \frac{2}{x^3y}\Rightarrow \cases{f_x= 2x - \frac{6}{x^4y} = 0 \Rightarrow x^5y = 3 \\f_y=2y-{2\over x^3y^2} =0 \Rightarrow xy=1} \Rightarrow x^5\cdot {1\over x}=3 \Rightarrow x=\pm \sqrt[4]3 \\ \quad \Rightarrow y= \pm 1/\sqrt[4]3 \Rightarrow \text{critical point in the first octant: }\bbox[red, 2pt]{(\sqrt[4]3,1/\sqrt[4]3) } \\ z^2={2\over x^3y} \Rightarrow x^2+y^2+z^2 = x^2 + y^2 + \frac{2}{x^3y} =f(x,y) \\\Rightarrow \text{finding the critical points of $f(x,y)$ is equivalent to finding the points ($x,y$) which}\\ \text{minimize the distance in (a)} \\ \textbf{(c) }\cases{f_{xx}=2+{24\over x^5y} \\ f_{xy}={6\over x^4y^2} \\ f_{yy}= 2+{4\over x^3y^3}} \Rightarrow \cases{f_{xx}(\sqrt[4]3,1/\sqrt[4]3) =10 \\f_{xy}(\sqrt[4]3,1/\sqrt[4]3)=2\sqrt 3\\ f_{yy}(\sqrt[4]3,1/\sqrt[4]3)= 6} \Rightarrow D=f_{xx}f_{yy}-(f_{xy})^2= 48 \Rightarrow \bbox[red, 2pt]{\cases{f_{xx}=10\\ f_{xy}=2\sqrt 3\\ f_{yy}=6\\ D=48}}$$
解答:$$ \text{Volume}(E) = \frac{1}{8} \cdot \left( \frac{4}{3} \pi \cdot 1^3 \right) = \frac{\pi}{6}\\ \iiint_E (x + y + z) \, dV = \iiint_E x \, dV + \iiint_E y \, dV + \iiint_E z \, dV =3\iiint_E z \, dV \\= 3 \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{1} (\rho \cos\phi) (\rho^2 \sin\phi) \, d\rho \, d\phi \, d\theta = 3\left( \int_{0}^{\pi/2} 1 \, d\theta \right) \left( \int_{0}^{\pi/2} \sin\phi \cos\phi \, d\phi \right) \left( \int_{0}^{1} \rho^3 \, d\rho \right) \\= 3\cdot {\pi\over 2}\cdot {1\over 2}\cdot {1\over 4}= {3\pi\over 16} \Rightarrow \text{Average Value} = \frac{1}{\text{Volume}(E)} \iiint_E F(x,y,z) \, dV ={1\over \pi/6} \cdot {2\pi\over 16}= \bbox[red, 2pt]{9\over 8}$$
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