114年中正大學數學碩士班-微積分詳解

 國立中正大學 114學年度碩士班招生考試試題

科目名稱:微積分 
系所組別:數學系 數學系應用數學

解答:$$\sin t =t-{t^3\over 3!}+{t^5\over 5! }-\cdots \Rightarrow f(t)={\sin t\over t} =1-{t^2\over 3!}+{t^4\over 5!}- \cdots \\ \Rightarrow Si(x)= \int_0^x \left(  1-{t^2\over 3!}+{t^4\over 5!}- \cdots \right)\,dt =x-{x^3\over 18}+{x^5\over 600}-\cdots \\ \Rightarrow \lim_{x\to 0} {Si(x)-x\over x^3} =\lim_{x\to 0} \left( -{1\over 18}+{x^2\over 600}-\cdots \right) = \bbox[red, 2pt]{-{1\over 18}}$$

解答:$$f(x)=x^3-x \Rightarrow f'(x)=3x^2-1 \Rightarrow 假設切點坐標(a, f(a)) =(a,a^3-a) \\ \Rightarrow 切線斜率為f'(a)=3a^2-1 \Rightarrow 切線方程式: y=(3a^2-1)(x-a)+a^3-a \\ 切線通過(-2,2) \Rightarrow 2=(3a^2-1)(-2-a)+a^3-a \Rightarrow 2a^3+6a^2=0 \Rightarrow 2a^2(a+3) =0 \\ \Rightarrow \cases{a=0 \Rightarrow y=-x\\ a=-3 \Rightarrow y=26(x+3)} \Rightarrow 切線方程式: \bbox[red, 2pt]{\cases{y=-x\\ y=26x+54}}$$

解答:$$\textbf{a. }\sin^4 x= \left( {1-\cos 2x\over 2} \right)^2 ={1\over 4}(1-2\cos 2x+\cos^2 2x)  ={1\over 4}(1-2\cos 2x+ {1+\cos 4x\over 2}) \\={3\over 8}-{1\over 2}\cos 2x+{1\over 8}\cos 4x \Rightarrow \int_0^{\pi/2} \sin^4 x\,dx = \left. \left[ {3\over 8}x-{1\over 4}\sin 2x+{1\over 32}\sin 4x \right] \right|_0^{\pi/2}  =\bbox[red, 2pt]{{3\over 16}\pi} \\\textbf{b. }u=\sqrt{4x+1} \Rightarrow x={u^2-1\over 4} \Rightarrow dx= {u\over 2}du \Rightarrow \int_2^\infty {1\over x^2\sqrt{4x+1}}\,dx = \int_3^\infty  {(u/2)du\over ({u^2-1\over 2})^2 \cdot u} \\= 8\int_3^\infty {1\over (u^2-1)^2}\,du =2 \int_3^\infty \left( {1\over (u-1)^2} +{1\over (u+1)^2} +{1\over 2(u+1)} -{1\over 2(u-1)}\right)\,du \\= 2 \left. \left[ -{1\over u-1}-{1\over u+1}+{1\over 2} \ln|u+1|-{1\over 2}\ln|u-1| \right] \right|_3^\infty = \left. \left[ {-4u\over u^2-1}+2\ln {u+1\over u-1} \right] \right|_3^\infty \\=0- \left( -{3\over 2}+2\ln 2 \right) = \bbox[red, 2pt]{{3\over 2}-2\ln 2}$$

解答:$$\textbf{Case I }p \in (-\infty, 0] \Rightarrow a_n={\ln n\over n^p} \Rightarrow \lim_{n\to \infty}a_n  \ne 0 \Rightarrow \text{divergence} \\\textbf{Case II }p\in (0,1] \Rightarrow {\ln n\over n^p}\gt {1\over n^p} , n\ge 3. \sum {1\over n^p} \text{ diverges for all }p\le 1 \Rightarrow \sum{\ln n\over n^p} \text{ diverges} \\ \textbf{Case III }p\in (1,\infty)  \Rightarrow \int_2^\infty {\ln x\over x^p}\,dx =\lim_{t\to \infty} \left. \left[ {\ln x\over (1-p)x^{p-1}} -{1\over (1-p)^2 x^{p-1}}\right] \right|_2^t \lt \infty \\ \qquad \text{By Integral Test, }\sum {\ln n\over n^p} \text{ converges} \\ \Rightarrow \bbox[red,2pt]{\sum_{n=2}^\infty {\ln n\over n^p} \begin{cases}\text{diverges} & p\in(-\infty,0] \\ \text{diverges}& p\in(0,1]\\ \text{converges} & p\in(1,\infty)\end{cases}}$$

解答:$$\textbf{a. }f(x)={1\over \sqrt{1+x}} =(1+x)^{-1/2} \Rightarrow f'(x)=-{1\over 2} (1+x)^{-3/2} \Rightarrow f''(x)= {1\cdot 3\over 2^2}(1+x)^{-5/2} \\ \qquad \Rightarrow f'''(x)=-{1\cdot 3\cdot 5\over 2^3}(1+x)^{-7/2} \Rightarrow \bbox[red, 2pt]{f^{(k)}(x) =(-1)^k\cdot {1\cdot 3\cdot 5\cdots(2k-1) \over 2^k}(1+x)^{-(2k+1)/2} }\\\textbf{b. }f^{(n)}(0) =(-1)^n\cdot {1\cdot 3\cdot 5\cdots (2n-1) \over 2^n} \Rightarrow {f^{(n)}(0) \over n!} =(-1)^n\cdot {1\cdot 3\cdot 5\cdots (2n-1) \over n!\cdot  2^n}  \\ \qquad \Rightarrow \text{The Maclaurin series is: } \bbox[red, 2pt]{f(x) =\sum_{n=0}^\infty (-1)^n\cdot {1\cdot 3\cdot 5\cdots (2n-1) \over n!\cdot  2^n}x^n} \\\textbf{c. } a_n=(-1)^n\cdot {1\cdot 3\cdot 5\cdots (2n-1) \over n!\cdot  2^n}x^n =(-1)^2 {-1/2 \choose n}x^n \\\qquad \Rightarrow L=\lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left|{{-1/2\choose n+1} x^{n+1}\over {-1/2\choose n}x^n} \right| =\lim_{n\to \infty} \left(   |x|\cdot {2n+1\over 2n+2}\right) =|x| \lt 1 \\ \qquad \Rightarrow \text{ The radius of convergence is : }\bbox[red, 2pt]1$$

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解題僅供參考，碩士班歷年試題及詳解