114年政大轉學考-微積分詳解

 國立政治大學114學年度學士班轉學生招生考試

科目:微積分
系所別:經濟系、國貿系、統計系、資管系、財管系、風管系
(A)

解答:$$f(x)={3\sin (x)\over 2+\sin (x)} =k \Rightarrow \sin (x)={2k\over 3-k} \Rightarrow -1\le {2k\over 3-k} \le 1 \Rightarrow -3\le k\le 1 \\ \textbf{a.} \lim \text{inf}_{x\to +\infty}f(x)= \bbox[red, 2pt]{-3} \\ \textbf{b. }\lim \text{sup}_{x\to +\infty}f(x) = \bbox[red, 2pt]1$$

解答:$$ f(x) = \frac{\sin(x)}{x} \Rightarrow f'(x)={x\cos (x)-\sin (x) \over x^2}\\ \text{Let }g(x)=x\cos(x)-\sin (x), \text{ then }g'(x)=-x\sin(x). \text{ Since $x\gt 0 $ and $\sin x\gt 0$} \\ \text{we have }g'(x)\lt 0, \text{ for $x\in (0,{\pi\over 6})$. That is , $g(x)$ is decreasing. And $f'(x)={g(x)\over x} \lt 0$} \\ \Rightarrow f(x) \text{ is decreasing }\Rightarrow f(x)\gt f({\pi\over 6})={3\over \pi} \Rightarrow {\sin (x)\over x}\gt {3\over \pi}, \forall x\in (0,{\pi\over 6}) \quad \bbox[red, 2pt]{QED.}$$

解答:$$\text{Case I }b(x)=0 \Rightarrow  \cases{\lim_{\epsilon \to 0^+} \frac{b(x + b(x)\sqrt{\epsilon}) - b(x)}{\sqrt{\epsilon}} =  \lim_{\epsilon \to 0^+} \frac{0}{\sqrt{\epsilon}} = 0 \\ b(x)b'(x)= 0\cdot b'(x)=0} \\ \qquad \Rightarrow  \lim_{\epsilon \to 0^+} \frac{b(x + b(x)\sqrt{\epsilon}) - b(x)}{\sqrt{\epsilon}} = b(x)b'(x) \\ \text{Case II }b(x)\ne 0\\ \qquad \text{Let }h=b(x)\sqrt \epsilon, \text{ then }\lim_{\epsilon \to 0^+} h=0 \Rightarrow  \lim_{\epsilon \to 0^+} \frac{b(x + b(x)\sqrt{\epsilon}) - b(x)}{\sqrt{\epsilon}} = \lim_{h \to 0} \frac{b(x + h) - b(x)}{\frac{h}{b(x)}} \\ \qquad  = b(x) \cdot \lim_{h \to 0} \frac{b(x + h) - b(x)}{h} =b(x)\cdot b'(x) \\ \bbox[red, 2pt]{QED.}$$

解答:$$\textbf{(a) } \frac{1}{1-t} = \sum_{n=0}^{\infty} t^n = 1 + t + t^2 + t^3 + \dots \Rightarrow  \int_{0}^{x} \frac{1}{1-t} \,dt = \int_{0}^{x} \sum_{n=0}^{\infty} t^n \,dt  \\ \Rightarrow -\ln(1-x)= \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \Rightarrow \ln(1-x)= -\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = \bbox[red, 2pt]{-\sum_{n=1}^\infty{x^n\over n}} \\\qquad a_n= -{x^n\over n} \Rightarrow  \lim_{n\to\infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n\to\infty} \left\vert{} \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right\vert{} = \vert{}x\vert{} \lim_{n\to\infty} \frac{n}{n+1} = \vert{}x\vert{} \lt 1\\ x=1 \Rightarrow -\sum_{n=1}^\infty{x^n\over n}= -\sum_{n=1}^\infty{1\over n} \text{ diverges} \\ x=-1 \Rightarrow -\sum_{n=1}^\infty{x^n\over n} =-\sum_{n=1}^{\infty} \frac{(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \text{ converges} \\ \qquad \Rightarrow \text{interval of convergence: }\bbox[red, 2pt]{[-1,1)} \\ \textbf{(b) }\ln(1-x)=-\sum_{n=1}^\infty {x^n\over n} \Rightarrow \ln(1-{1\over 2}) =-\sum_{n=1}^\infty {(1/2)^n\over n} \Rightarrow \ln {1\over 2}=- \sum_{n= 1}^\infty{1 \over n\cdot 2^n} \\ \qquad \Rightarrow \ln 2 =\sum_{n= 1}^\infty{1 \over n\cdot 2^n}\quad \bbox[red, 2pt]{QED.}$$

解答:$$ f(x) = 2\sqrt{x} - \sqrt{x+1} - \sqrt{x-1} \Rightarrow f'(x) = \frac{1}{\sqrt{x}} - \frac{1}{2\sqrt{x+1}} - \frac{1}{2\sqrt{x-1}} \\=   \frac{1}{2} \left( \frac{2}{\sqrt{x}} - \left( \frac{1}{\sqrt{x+1}} + \frac{1}{\sqrt{x-1}} \right) \right) \lt 0 \Rightarrow f(x) \text{ is strictly decreasing for }x\in [1,\infty] \\ \Rightarrow f(1)=2-\sqrt 2-0= \bbox[red, 2pt]{2-\sqrt 2} \text{ is the maximum value}$$

(B)

解答:$$I= \int_{0}^{\pi/6} \int_{\sin x}^{1/2} xy^2 \,dy\,dx = \int_{0}^{\pi/6} \left. x\left[ {1\over 3}y^3 \right] \right|_{\sin x}^{1/2}\,dx =  \frac{1}{3} \int_{0}^{\pi/6} \left( \frac{x}{8} - x \sin^3 x \right) \,dx \\ I_1=  \int_{0}^{\pi/6} \frac{x}{8} \,dx = \frac{1}{8} \left[ \frac{x^2}{2} \right]_{0}^{\pi/6} = \frac{1}{16} \left( \frac{\pi}{6} \right)^2 = \frac{\pi^2}{576} \\ I_2=  \int_{0}^{\pi/6} x \sin^3 x \,dx = \int_{0}^{\pi/6} x \left( \frac{3\sin x - \sin(3x)}{4} \right) \,dx = \frac{3}{4} \int_{0}^{\pi/6} x \sin x \,dx - \frac{1}{4} \int_{0}^{\pi/6} x \sin(3x) \,dx  \\= {3\over 4} \left. \left[ -x\cos x+ \sin x \right] \right|_0^{\pi/6}-{1\over 4} \left. \left[ -{1\over 3}x\cos(3x)+{1\over 9}\sin(3x) \right] \right|_0^{\pi/6} ={3\over 4} \left( {1\over 2}-{\sqrt 3\pi \over 12} \right)-{1\over 4}\cdot {1\over 9} \\={25\over 72}-{\sqrt 3\pi\over 16} \\ \Rightarrow I={1\over 3}(I_1-I_2) = \frac{1}{3} \left( \frac{\pi^2}{576} - \left( \frac{25}{72} - \frac{\sqrt{3}\pi}{16} \right) \right) = \bbox[red, 2pt]{\frac{\pi^2}{1728} - \frac{25}{216} + \frac{\sqrt{3}\pi}{48} }$$

解答:$$e^x =\sum_{n=0}^\infty {x^n\over n!} \Rightarrow e^{3x} =\sum_{n=0}^\infty {(3x)^n\over n!}  =\sum_{n=0}^\infty {3^nx^n\over n!} \Rightarrow f(x)=e^{3x+1}=  e \cdot \sum_{n=0}^{\infty} \frac{3^n x^n}{n!} = \bbox[red, 2pt]{\sum_{n=0}^{\infty} \frac{e \cdot 3^n}{n!} x^n } \\a_n = \frac{e \cdot 3^n x^n}{n!} \Rightarrow  \lim_{n\to\infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \lim_{n\to\infty} \left\vert{} \frac{\frac{e \cdot 3^{n+1} x^{n+1}}{(n+1)!}}{\frac{e \cdot 3^n x^n}{n!}} \right\vert{}  = \lim_{n\to\infty} \left\vert{} \frac{3x}{n+1} \right\vert{} =0 \Rightarrow \text{ radius of convergence:} \bbox[red, 2pt]{\infty}$$

解答:$$[-1,2] \Rightarrow \cases{a=-1\\ b=2} \Rightarrow \Delta x={b-a\over n}={3\over n} \Rightarrow x_i^* =a+i\Delta x =-1+{3i\over n} \\ \Rightarrow f(x_i^*) = 3 - 2\left(-1 + \frac{3i}{n}\right) = 3 + 2 - \frac{6i}{n} = 5 - \frac{6i}{n} \\\Rightarrow A = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x  = \lim_{n \to \infty} \sum_{i=1}^{n} \left( 5 - \frac{6i}{n} \right) \left( \frac{3}{n} \right)    = \lim_{n \to \infty} \left( \frac{15}{n} \sum_{i=1}^{n} 1 - \frac{18}{n^2} \sum_{i=1}^{n} i \right) \\  = \lim_{n \to \infty} \left( 15 - \frac{9(n+1)}{n} \right)  = \lim_{n \to \infty} \left( 15 - 9 \left( 1 + \frac{1}{n} \right) \right) =15-9= \bbox[red, 2pt]6$$

解答:$$\textbf{(a) }a_n = \frac{(-1)^n(x - 1)^{2n+1}}{2n + 1}  \Rightarrow  L = \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{}   = \lim_{n \to \infty} \left\vert{} \frac{\frac{(-1)^{n+1}(x - 1)^{2(n+1)+1}}{2(n+1) + 1}}{\frac{(-1)^n(x - 1)^{2n+1}}{2n + 1}} \right\vert{} \\  = \lim_{n \to \infty} \left\vert{} (x - 1)^2 \cdot \frac{2n + 1}{2n + 3} \right\vert{} = \vert{}x - 1\vert{}^2 \lim_{n \to \infty} \frac{2n + 1}{2n + 3} = \vert{}x - 1\vert{}^2 \lt 1 \Rightarrow |x-1|\lt 1 \\ \Rightarrow -1\lt x-1\lt 1 \Rightarrow 0\lt x\lt 2\\ x=0 \Rightarrow \sum_{n=0}^\infty a_n = \sum_{n=0}^{\infty} \frac{(-1)^n(0 - 1)^{2n+1}}{2n + 1} = \sum_{n=0}^{\infty} \frac{(-1)^n(-1)^{2n+ 1}}{2n + 1} =  \sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{2n + 1} \text{ converges} \\ x=2 \Rightarrow \sum_{n=0}^\infty a_n =  \sum_{n=0}^{\infty} \frac{(-1)^n(2 - 1)^{2n+1}}{2n + 1} = \sum_{n=0}^{\infty} \frac{(-1)^n(1)^{2n+1}}{2n + 1} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n + 1} \text{ converges} \\ \qquad \Rightarrow \bbox[red, 2pt]{\cases{\text{radius of convergence: 1} \\ \text{interval of convergence: [0,2]}}} \\\textbf{(b) }\bbox[red, 2pt]{(0,2)} \quad \textbf{(c) } \bbox[red, 2pt]{\{0,2\}}$$

解答:$$a\sin 2\theta =0 \Rightarrow \theta=0,{\pi\over 2}, \pi, \dots \\A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta =  \frac{1}{2} \int_{0}^{\pi/2} (a \sin 2\theta)^2 \, d\theta =  \frac{a^2}{2} \int_{0}^{\pi/2} \sin^2 2\theta \, d\theta   = \frac{a^2}{4} \int_{0}^{\pi/2} (1 - \cos 4\theta) \, d\theta \\  = \frac{a^2}{4} \left[ \theta - \frac{\sin 4\theta}{4} \right]_{0}^{\pi/2}   = \frac{a^2}{4} \left( \frac{\pi}{2} - 0 \right) = \bbox[red, 2pt]{a^2 \pi \over 8}$$

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