114年嘉義大學應數碩士班-微積分詳解

國立嘉義大學 114 學年度

應用數學系碩士班(甲組)招生考試試題 

科目: 微積分 (每題 10 分，共 100 分) 

解答:$$f(x)={x^2-2x+4\over x-2} =x+{4\over x-2} \Rightarrow y=x為斜漸近線\\ 分母x-2=0 \Rightarrow x=2 為垂直漸近線\\ \Rightarrow \bbox[red, 2pt]{\text{Slant asympotote: }y=x, \text{vertical asymptote: }x=2}$$

解答:$$左式: y={1-\cos x\over \sin x} \Rightarrow {dy\over dx} ={(1-\cos x)' \over \sin x} +(1-\cos x) \left( {1\over \sin x} \right)' ={\sin x\over \sin x}+(1-\cos x)\cdot {-\cos x\over \sin^2 x} \\=1+{-\cos x+\cos^2x\over \sin^2 x} ={\sin^2x+ \cos^2x -\cos x\over \sin^2 x} ={1-\cos x\over \sin^2x} \\ 右式: y=\csc x-\cot x \Rightarrow {dy\over dx } =-\csc x\cot x+\csc^2 x =\csc x(\csc x-\cot x) \\={1\over \sin x} \left( {1\over \sin x}-{\cos x\over \sin x} \right) ={1-\cos x\over \sin^2 x} \\ \Rightarrow 左式=右式\Rightarrow \bbox[red, 2pt]{{dy\over dx} ={1-\cos x\over \sin^2 x}}$$

解答:$$y= \left( {3x-1\over x^2+3} \right)^2 \Rightarrow {dy\over dx} = 2\left( {3x-1\over x^2+3} \right) \left( {3x-1\over x^2+3} \right)' = 2\left( {3x-1\over x^2+3} \right)  \left( -3x^2+2x+9\over (x^2+3)^2 \right) \\= \bbox[red, 2pt]{2(3x-1)(-3x^2+2x+9) \over (x^2+3)^2}$$

解答:$$x^2+y^2=25 \Rightarrow 2x+2yy'=0\Rightarrow y'=-{x\over y} \Rightarrow 2+2y'y'+2yy''=0 \Rightarrow y''={-2-2(y')^2\over 2y} \\={-1-(y')^2\over y} ={-1-{x^2\over y^2} \over y} ={-(x^2+y^2)/y^2 \over y}={-25/y^2\over y}\Rightarrow \bbox[red, 2pt]{y'' =-{25\over y^3}}$$

解答:$$F(x) =\int_{\pi/2}^{x^2} \cos t\,dt \Rightarrow F'(x)= \cos(x^2)\cdot {d\over dx}x^2= \bbox[red, 2pt]{2x\cos(x^2)}$$

解答:$$u=\ln x \Rightarrow du={1\over x}dx \Rightarrow \int {1\over x\sqrt{4-(\ln(x))^2}} \,dx =\int{1\over \sqrt{4-u^2}}du =\sin^{-1}{u\over 2}+C \\= \bbox[red, 2pt]{\sin^{-1}{\ln x\over 2}+C}$$

解答:$$\int {1\over x^2(1+x)}\,dx = \int \left( -{1\over x}+{1\over x^2}+{1\over 1+x} \right)\,dx = -\ln |x|-{1\over x}+\ln|x+1|+C\\ = \bbox[red, 2pt]{\ln \left|{x+1\over x} \right|-{1\over x}+C}$$

解答:$$\cases{u=(\ln x)^2 \\ dv=x^3\,dx} \Rightarrow \cases{du=2 {\ln x\over x} dx \\ v={1\over 4}x^4} \Rightarrow I= \int x^3(\ln x)^2\,dx ={1\over 4}x^4 (\ln x)^2-{1\over 2}\int x^3 \ln x\,dx \\ \cases{u= \ln x \\ dv=x^3\,dx} \Rightarrow \cases{du=dx/x\\ v={1\over 4}x^4} \Rightarrow I={1\over 4}x^4 (\ln x)^2-{1\over 2} \left( {1\over 4}x^4 \ln x-{1\over 4}\int x^3\,dx \right) \\={1\over 4}x^4 (\ln x)^2-{1\over 8}x^4 \ln x-{1\over 16}x^4 +C \\ \Rightarrow \lim_{t\to 0^+}\left. \left[ {1\over 4}x^4 (\ln x)^2-{1\over 8}x^4 \ln x-{1\over 16}x^4 \right] \right|_t^1 ={1\over 32}-0=\bbox[red, 2pt] {1\over 32}$$

解答:$$\sum_{n=1}^\infty {1\over (3n-2)(3n+1)} =\sum_{n=1}^\infty {1\over 3}  \left( {1\over 3n-2}-{1\over 3n+1} \right)  ={1\over 3} \left[ \left( {1\over 1}-{1\over 4} \right) +\left( {1\over 4}-{1\over 7} \right) + \cdots \right] \\= \bbox[red, 2pt]{1\over 3}$$

解答:$$\sum_{n=1}^\infty {(-1)^n\over\ln(n+1)} =\sum_{n=1}^\infty (-1)^n b_n \Rightarrow \cases{\ln(n+1)\gt 0 \Rightarrow b_n \gt 0, n\in \mathbb N\\ b_{n+1}\le b_n\\ \lim_{n\to \infty} b_n=0}\\由交錯級數審斂法\text{(alternating series test)} 可知該級數\bbox[red, 2pt]{收斂\text{(convergence)}}$$

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解題僅供參考，碩士班歷年試題及詳解