115年政大國貿轉學考-微積分詳解

國立政治大學 115 學年度學士班轉學生招生考試

考試科目 微積分 
系所別 國貿系、統計系、財管、風管系、經濟系二年級

Part A Calculus

解答:$$\text{Let $a$ and $b$ be any two real numbers in $\mathbb{R}$ such that $a < b$.}\\ \text{By Mean Value Theorem, there exists at least one point $c$ in $(a,b)$ such that}\\ f'(c)={f(b)-f(a)\over b-a}\\ \text{We are given that $f'(c)=0, \forall x \in \mathbb R$. Then }f'(c)=0 \Rightarrow f(b)=f(a)\\ \text{Because $a$ and $b$ were chosen arbitrarily, $f(a) = f(b)$ for any two points in $\mathbb{R}$.}\\ \text{ This means the function outputs the exact same value for every input. }\\ \text{Therefore, $f$ is constant on $\mathbb{R}$. }\bbox[red, 2pt]{QED.}$$

解答:$$f(x)={1\over 1-x} = 1+x+x^2+ \cdots=\sum_{n=0}^\infty x^n \Rightarrow f'(x)={1 \over (1-x)^2} = \bbox[red, 2pt]{\sum_{n=1}^\infty n x^{n-1}} \\ \Rightarrow xf'(x)= {x\over (1-x)^2} = \sum_{n=1}^\infty n x^{n } \Rightarrow{1\over 3}f'({1\over 3}) ={1/3\over (2/3)^2} =\sum_{n=1}^\infty n \left({1\over 3} \right)^{n } \\ \Rightarrow \sum_{n=1}^\infty n \left({1\over 3} \right)^{n } ={1\over 3}\cdot {9\over 4} =\bbox[red, 2pt]{3\over 4}$$

解答:$$e^x \ge 1+x, \forall x \Rightarrow e^{-x} \ge 1-x \Rightarrow e^{-k/n}\ge 1-{k\over n} \Rightarrow \left( 1-{k\over n} \right)^n \le e^{-k} \\ \Rightarrow \sum_{k=0}^n \left( 1-{k\over n} \right)^n \le \sum_{k=0}^n e^{-k} ={1-(1/e)^{n+1}\over 1-1/e} ={e-e^{-n} \over e-1} \lt {e\over e-1} \\ \Rightarrow {1\over n}\sum_{k=0}^n \left( 1-{k\over n} \right)^n \le {1\over n(e-1)}, \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) }\lim_{x\to 0} {\tan x-\sin x\over x^3} =\lim_{x\to 0} {{d\over dx}(\tan x-\sin x)\over {d\over dx}x^3} = \lim_{x\to 0} {\sec^2 x-\cos x\over 3x^2}  = \lim_{x\to 0} {{d\over dx}(\sec^2 x-\cos x) \over {d\over dx}3x^2} \\\qquad = \lim_{x\to 0} { 2\sec^2 x\tan x+\sin x\over 6x}= \lim_{x\to 0} { {d\over dx}(2\sec^2 x\tan x+\sin x)\over {d\over dx}6x} = \lim_{x\to 0}{4\sec^2 x\tan^2 x+ 2\sec^4 x+\cos x\over 6} \\ \qquad ={3\over 6}= \bbox[red, 2pt]{1\over 2} \\\textbf{(b) }\lim_{x\to 0} {e^x-1-x\over x^2} =\lim_{x\to 0} {{d\over dx}(e^x-1-x)\over {d\over dx}x^2} = \lim_{x\to 0} {e^x-1\over 2x} = \lim_{x\to 0} {{d\over dx}(e^x-1)\over {d\over dx} 2x} \\\qquad = \lim_{x\to 0} {e^x\over 2} = \bbox[red, 2pt]{1\over 2}$$

解答:$$e^x \ge 1+x, \forall x \Rightarrow   1-x \le e^{-x}\Rightarrow  1-{j\over n}\le e^{-j/n} \Rightarrow \left(  1-{j\over n}\right)^n \le e^{-j}\\\Rightarrow \sum_{j=0}^{n-1} \left(  1-{j\over n}\right)^n \le \sum_{j=0}^{n-1}e^{-j} \lt \sum_{j=0}^{\infty}e^{-j}={e\over e-1} \Rightarrow 0\le {1\over n}\sum_{j=0}^{n-1} \left(  1-{j\over n}\right)^n \le {1\over n}\cdot {e\over e-1} \\ \Rightarrow \lim_{n\to \infty} {1\over n} \sum_{k=1}^n \left( {k\over n} \right)^n= \lim_{n\to \infty}\sum_{j=0}^{n-1} \left(\frac{n-j}{n}\right)^n = \lim_{n\to \infty}\sum_{j=0}^{n-1} \left(1 - \frac{j}{n}\right)^n \le \lim_{n\to \infty} {e\over n(e-1)}=0 \\ \Rightarrow 0\le \lim_{n\to \infty} {1\over n} \sum_{k=1}^n \left( {k\over n} \right)^n\le 0 \Rightarrow  \lim_{n\to \infty} {1\over n} \sum_{k=1}^n \left( {k\over n} \right)^n= \bbox[red, 2pt]0 \; \text{(by Squeeze Theorem)} $$

(B)

解答:$$x=e^t \Rightarrow dx =e^t\,dt \Rightarrow I=\int_1^\infty {(\ln x)^2 \over x^p}\,dx = \int_0^\infty t^2e^{(1-p)t}\,dt \\ \Rightarrow \cases{p\gt 1 \Rightarrow I \lt \infty \\ p=1 \Rightarrow I \to \infty\\ p\lt 1 \Rightarrow I\to \infty} \Rightarrow \text{The correct option is } \bbox[red, 2pt]{(C)}.$$

解答:$$a_n= {n!\over 1\cdot 3\cdot 5\cdots(2n-1)}(x-2)^n \Rightarrow \lim_{n\to \infty} \left| {a_{n+1} \over a_n}\right| = \lim_{n\to \infty} {n+1\over 2n+1}\left| x-2\right| = {1\over 2}|x-2| \lt 1\\ \Rightarrow 0\lt x\lt 4 \Rightarrow \text{The series converges absolutely on the open interval $(0, 4)$.}\\ x=4 \Rightarrow a_n=  \frac{n! 2^n}{1 \cdot 3 \cdot 5 \cdots (2n-1)} ={2\cdot 4\cdot 6\cdots(2n) \over 1 \cdot 3 \cdot 5 \cdots (2n-1)} = \prod_{k=1}^n {2k\over 2k-1} \\\qquad {2k\over 2k-1}\gt 1\Rightarrow \{a_n \}\text{ is strictly increasing and }a_n \gt 1 \Rightarrow \text{divergent} \\x=0 \Rightarrow a_n=(-1)^n\cdot \prod_{k=1}^n {2k\over 2k-1} \Rightarrow |a_n| \gt 1 \Rightarrow \text{divergent} \\ \Rightarrow \text{The correct option is }\bbox[red, 2pt]{(C)}.$$

解答:$$r=2+\cos \theta \Rightarrow \cases{x =r\cos \theta=(2+\cos \theta)\cos \theta= 2\cos\theta+ \cos^2\theta\\ y=r\sin\theta = (2+\cos \theta)\sin \theta=2\sin \theta+ {1\over 2}\sin 2\theta} \\ \Rightarrow \cases{dx/d\theta =-2\sin \theta-2\cos \theta\sin \theta \\ dy/d\theta=2\cos \theta+ \cos 2\theta} \Rightarrow \cases{ \left. \displaystyle {dx\over d\theta} \right|_{\theta=\pi/3} =-3\sqrt 3/2 \\  \left. \displaystyle {dy\over d\theta} \right|_{\theta=\pi/3} =1/2}\\ \Rightarrow {dy\over dx} ={1/2\over -3\sqrt 3/2}=-{1\over 3\sqrt 3}  = \bbox[red, 2pt]{-{\sqrt 3\over 9}}$$

解答:$$\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \cases{\text{The surface: }z=x^2+y^2 \Rightarrow z=r^2\\ x^2+y^2=1 \Rightarrow r=1\\ x^2+y^2=2x \Rightarrow r^2=2r\cos \theta \Rightarrow r=2\cos \theta} \\ \text{The region $R$ is bounded radially from $r = 1$ to $r = 2\cos\theta$.}\\1=2\cos \theta \Rightarrow \cos \theta={1\over 2} \Rightarrow \theta=-{\pi\over 3}, {\pi \over 3} \\ \Rightarrow V= \iint_R (x^2+y^2)\, dA = \int_{-\pi/3}^{\pi/3} \int_1^{2\cos \theta}r^2 \cdot r\,dr\,d\theta=  2 \int_{0}^{\pi/3} \int_{1}^{2\cos\theta} r^3 \, dr \, d\theta  \\  = 2 \int_{0}^{\pi/3} \left( 4\cos^4\theta - \frac{1}{4} \right) d\theta = \int_{0}^{\pi/3} \left( 8\cos^4\theta - \frac{1}{2} \right) d\theta   = \int_{0}^{\pi/3} \left( {5\over 2} + 4\cos(2\theta) + \cos(4\theta)  \right) d\theta  \\= \left. \left[ {5\over 2}\theta +2 \sin (2\theta)+{1\over 4}\cos(4\theta) \right] \right|_0^{\pi/3}= \bbox[red, 2pt]{  \frac{5\pi}{6} + \frac{7\sqrt{3}}{8} }$$

解答:$$\textbf{(a) }\lim_{x\to \infty} {x^n\over e^x} =\lim_{x\to \infty} {nx^{n-1}\over e^x} = \lim_{x\to \infty} {n(n-1)x^{n-2}\over e^x} =\cdots=\lim_{x\to \infty} {n!\over e^x} = \bbox[red, 2pt]0 \\\textbf{(b) } \lim_{x \to \infty} \frac{P(x)}{G(x)} = \lim_{x \to \infty} \frac{a_mx^m + a_{m-1}x^{m-1} + \dots + a_1x + a_0}{e^x} \\= \lim_{x \to \infty} \left( a_m\frac{x^m}{e^x} + a_{m-1}\frac{x^{m-1}}{e^x} + \dots + a_1\frac{x}{e^x} + a_0\frac{1}{e^x} \right) \\=  a_m \left( \lim_{x \to \infty} \frac{x^m}{e^x} \right) + a_{m-1} \left( \lim_{x \to \infty} \frac{x^{m-1}}{e^x} \right) + \dots + a_1 \left( \lim_{x \to \infty} \frac{x}{e^x} \right) + a_0 \left( \lim_{x \to \infty} \frac{1}{e^x} \right) \\=a_m\cdot 0+ a_{m-1}\cdot 0+ \cdots+a_1\cdot 0+ a_0 \cdot 0 =0 \\ \Rightarrow \bbox[red, 2pt]{G(x)} \text{ grows faster than }P(x) \text{ as }x\to \infty.$$

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