114年台北大學都市計劃碩士班-微積分詳解

 國立臺北大學114學年度碩士班一般入學考試

系(所)組別:都市計劃研究所
科目:微積分

解答: $$f(x)={1\over 3}x^3-ax^2+bx+c \Rightarrow f'(x)=x^2-2ax+b \\f(x)\text{ has extreme values at }x=1,2\Rightarrow \cases{f'(1)=0\\ f'(2)=0} \Rightarrow \cases{1-2a+b=0\\ 4-4a+b=0} \Rightarrow \bbox[red, 2pt]{\cases{a=3/2\\ b=2} }\\ \Rightarrow f(x)= {1\over 3}x^3-{3\over 2}x^2+ 2x+c \Rightarrow f(1)={1\over 3}-{3\over 2}+2+c=5 \Rightarrow \bbox[red, 2pt]{c={25\over 6}}$$

解答: $$f(x)={1\over 5}x^5-{1\over 2}x^4-x^3+4x^2-4x+3 \Rightarrow f'(x)=x^4-2x^3-3x^2+8x-4\\ \Rightarrow f''(x)=4x^3-6x^2-6x+8\\f'(x)=(x-1)^2(x-2)(x+2)  =0 \Rightarrow x=1,2,-2 \Rightarrow \cases{f''(1)=0\\ f''(2) =4\gt 0\\ f''(-2) =-36\lt 0} \\ \Rightarrow \cases{f(2)=7/5\\ f(-2)=103/5} \Rightarrow \bbox[red, 2pt]{\cases{\text{local minimum: }7/5\\ \text{local maximum: }103/5}}$$

解答: $$\textbf{A. }L=\left({1\over x^2} \right)^x \Rightarrow \ln L=x\ln {1\over x^2} \Rightarrow \lim_{x\to 0} \ln L= \lim_{x\to 0} {\ln 1/x^2 \over 1/x}= \lim_{x\to 0} {-2/x\over -1/x^2}= \lim_{x\to 0} x=0\\\qquad \Rightarrow \lim_{x\to 0} L= e^0= \bbox[red, 2pt]1 \\\textbf{B. }\lim_{x\to \pi/2} x\tan(x) \bbox[red, 2pt]{\text{ This limit does not exist.}} \\\textbf{C. }\lim_{t\to 1}{t-1\over \ln t} = \lim_{t\to 1}{1\over 1/t} = \lim_{t\to 1} t=\bbox[red, 2pt]1 \\\textbf{D. }\lim_{x\to 2}{x-2\over x^2-4} = \lim_{x\to 2}{x-2\over (x-2)(x+2)} = \lim_{x\to 2}{1\over x+2} = \bbox[red, 2pt]{1\over 4}$$

解答: $$\textbf{A. }u=16-x^3 \Rightarrow du=-3x^2dx \Rightarrow \int_0^{\sqrt[3]7} x^2 \sqrt{16-x^2}\,dx = \int_{16}^9-{1\over 3}\sqrt{u}\,du =-{1\over 3}\left. \left[ {2\over 3}u^{3/2}\right] \right|_{16}^9\\ \qquad =-{2\over 9}(27-64) = \bbox[red, 2pt]{74\over 9} \\\textbf{B. }\cases{u=x \\ dv=\sin x\,dx} \Rightarrow \cases{du=dx\\ v=-\cos x} \Rightarrow \int x\sin x\,dx =-x\cos x+\int \cos x\,dx \\\qquad = \bbox[red, 2pt]{-x\cos x+\sin x+C} \\\textbf{C. } u=e^x+e^{-x} \Rightarrow du=(e^x-e^{-x} )dx  \Rightarrow \int {e^x-e^{-x} \over e^x+e^{-x}}\,dx =\int {1\over u}du =\ln u+C \\\qquad = \bbox[red, 2pt]{\ln (e^x+e^{-x})+C} \\\textbf{D. } \int{x-2\over x^2+x-2}\,dx = \int {x-2\over (x+2)(x-1)}\, dx = \int \left( {4/3\over x+2} -{1/3 \over x-1}\right)\,dx\\\quad  = \bbox[red, 2pt]{{4\over 3}\ln (x+2)-{1\over 3}\ln(x-1) +C}$$

解答: $$x^3-x^2y+2xy^3=1 \Rightarrow 3x^2-2xy-x^2y'+2y^3+6xy^2y'=0 \Rightarrow y'={2xy-3x^2-2y^3\over 6xy^2-x^2} \\ \Rightarrow y'(1,1)={2-3-2\over 6-1} =-{3\over 5} \Rightarrow y=-{3\over 5}(x-1) +1\Rightarrow \bbox[red, 2pt]{3x+5y=8}$$

解答: $$\bbox[cyan,2pt]{題目有誤}, x\le y\le x-1 \Rightarrow x\le y 且y\le x-1, 矛盾$$

解答: $$\textbf{A.} \lim_{n\to \infty}\left|{x^{n+1} \over x^n} \right| \lt 1 \Rightarrow \lim_{n\to \infty}|x| \lt 1 \Rightarrow R=\bbox[red, 2pt]1 \\ \textbf{B. }\lim_{n\to \infty} \left|{x^{n+2} \over (n+3)!} \cdot {(n+2)! \over x^{n+1}} \right| =\lim_{n\to \infty}\left|{x  \over n+3} \right|=0 \Rightarrow R=\bbox[red ,2pt] \infty \\ \textbf{C. }\lim_{n\to \infty}\left|{(4x-5)^{2n+3} \over (n+1)^{1/2}} \cdot {n^{1/2} \over (4x-5)^{2n+ 1}} \right| = \lim_{n\to \infty}\left|\sqrt{n\over n+1} \cdot (4x-5)^2\right| =(4x-5)^2 \lt 1 \\\qquad \Rightarrow -1\lt 4x-5\lt 1 \Rightarrow 1\lt x\lt {3\over 2} \Rightarrow -{1\over 4}\lt x-{5\over 4}\lt {1\over 4}\Rightarrow |x-{5\over 4}| \lt {1\over 4} \Rightarrow R= \bbox[red, 2pt]{1\over 4}$$

====================== END ==========================
解題僅供參考，其他碩士班試題及詳解