115年中山大學海科碩士班-微積分詳解

 國立中山大學115學年度碩士班考試入學招生考試試題

科目名稱：微積分【應數系碩士班乙組、海科系碩士班乙組選考】

計算題：共7題，子題分數平均分配。答題時，每題都必須寫下題號及詳細步驟

解答:$$\textbf{(a) }L=(4^n+5^n)^{1/n} \Rightarrow \ln L ={1\over n} \ln(4^n+5^n) \Rightarrow \lim_{n\to \infty} \ln L = \lim_{n\to \infty} {\ln(4^n+5^n) \over n} \\\qquad =\lim_{n\to \infty}  {{d\over dn}\ln(4^n+5^n) \over {d\over dn}n} =\lim_{n\to \infty}{4^n \ln 4+5^n \ln 5\over 4^n+5^n} =\lim_{n\to \infty}{(4/5)^n \ln 4+1^n \ln 5\over (4/5)^n+1^n} =\ln 5 \\ \qquad \Rightarrow \lim_{n\to \infty} \ln L =\ln 5 \Rightarrow \lim_{n\to \infty}  L = e^{\ln 5} =\bbox[red, 2pt]5 \\\textbf{(b) } \lim_{n\to \infty} \sum_{k=1}^n {5\over \sqrt{3n^2-k^2}} = \lim_{n\to \infty} \sum_{k=1}^n {5\over n\sqrt{3-(k/n)^2}} = I=\int_0^1 {5\over \sqrt{3-x^2}} \,dx \\\qquad x=\sqrt 3 \sin \theta \Rightarrow dx=\sqrt 3 \cos \theta\,d\theta \Rightarrow I=\int_0^{\sin^{-1}(1/\sqrt 3)} {5\cdot \sqrt 3\cos \theta\over \sqrt{3-3\sin^2 \theta}}\,d\theta =\int_0^{\sin^{-1}(1/\sqrt 3)} 5\,d\theta \\ \qquad =\bbox[red, 2pt]{5 \sin^{-1}{1\over \sqrt 3}}$$

解答:$$\textbf{(a) }  \cases{u=3-x\\ dv=e^{-2x}\,dx} \Rightarrow \cases{du=-dx\\ v=-{1\over 2}e^{-2x}}\Rightarrow \int_0^\infty (3-x)e^{-2x}\,dx =\left. {1\over 2}(x-3)e^{-2x} \right|_0^\infty -\int_0^\infty {1\over 2}e^{-2x}\,dx \\\qquad ={3\over 2} + \left. \left[ {1\over 4}e^{-2x} \right] \right|_0^\infty ={3\over 2}-{1\over 4}= \bbox[red, 2pt]{5\over 4} \\\textbf{(b) }  f(x)=x^2-2x=x(x-2) \Rightarrow \begin{cases}f(x)\ge 0,& x\ge 2, x\le 0\\ f(x)\le 0,& 0\le x\le 2 \end{cases} \\\qquad \Rightarrow \int_{-1}^2|f(x)|\,dx = \int_{-1}^0 f(x)\,dx- \int_{0}^2 f(x)\,dx = \int_{-1}^0 (x^2-2x)\,dx- \int_{0}^2 (x^2-2x)\,dx \\\qquad = \left. \left[ {1\over 3}x^3-x^2 \right] \right|_{-1}^0- \left. \left[ {1\over 3}x^3-x^2 \right] \right|_{0}^2={4\over 3}+{4\over 3}= \bbox[red, 2pt]{8\over 3}$$

解答:$$f(x)=e^{\sin x}+{1\over \sqrt{x^2+4}}+\int_0^x \sqrt{1+t^4}\,dt \Rightarrow \bbox[red, 2pt]{f'(x)= \cos xe^{\sin x}-{x\over (x^2+4)^{3/2}} +\sqrt{1+x^4} }\\ \Rightarrow f''(x)=-\sin xe^{\sin x} +\cos^2 xe^{\sin x}-{1\over (x^2+4)^{3/2}}+{3x^2\over (x^2+4)^{5/2}} +{2x^3\over \sqrt{1+x^4}} \\ \Rightarrow \bbox[red, 2pt]{f''(x)=e^{\sin x} (\cos^2x-\sin x)+{2x^2-4\over (x^2+4)^{5/2}} +{2x^3 \over \sqrt{1+x^4}}}$$

解答:$$y'=-y+2x \Rightarrow y'+y=2x \Rightarrow e^xy'+e^x y=2xe^{x} \Rightarrow \left( e^x y \right)'=2xe^x\\ \Rightarrow e^x y= \int 2xe^x\,dx =2e^x(x-1)+C \Rightarrow y=2(x-1)+Ce^{-x} \Rightarrow y(1)=Ce^{-1}=1 \Rightarrow C=e \\ \Rightarrow \bbox[red, 2pt]{y=2x-2+e^{1-x}}$$

解答:$$y={1\over 4}x^2-{1\over 2}\ln x \Rightarrow y'={1\over 2}x-{1\over 2x}  \Rightarrow (y')^2={1\over 4}x^2-{1\over 2} +{1\over 4x^2}\\\Rightarrow \text{arc length }=\int_1^e \sqrt{1+(y')^2} \,dx =\int_1^e  \sqrt{{1\over 4}x^2+{1\over 2}+{1\over 4x^2}} \,dx=\int_1^e   \sqrt{\left( {1\over 2}x+{1\over 2x} \right)^2} \,dx \\=\int_1^e \left( {1\over 2}x+{1\over 2x} \right)\,dx = \left. \left[ {1\over 4}x^2+{1\over 2}\ln x \right] \right|_1^e ={1\over 4}e^2+{1\over 2}-{1\over 4} = \bbox[red, 2pt]{{1\over 4}(e^2+1) }$$

解答:$$\textbf{(a) } \cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow I= \int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} {2\over 2+x^2+y^2} \,dx\,dy = \bbox[red, 2pt]{\int_0^{2\pi} \int_0^1 {2r\over 2+r^2}\,dr\,d\theta} \\\textbf{(b) } I= \int_0^{2\pi} \left. \left[ \ln(2+r^2) \right] \right|_0^1\,d\theta = \int_0^{2\pi} (\ln 3-\ln 2)\,d\theta =\bbox[red, 2pt]{2\pi\ln{3\over 2}}$$

解答:$$f(x,y) =x^3+y^3-3x-12y \Rightarrow \cases{f_x=3x^2-3\\ f_y=3y^2-12} \Rightarrow \cases{f_{xx}=6x\\ f_{xy} =0\\ f_{yy} =6y} \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2=36xy \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{x=\pm 1\\ y=\pm 2} \Rightarrow \text{ critical points: }(\pm 1,\pm 2) \\\Rightarrow \cases{D(1,2) =72\gt 0 \Rightarrow f_{xx}(1,2)=6\gt 0 \Rightarrow \text{local min}\\ D(1,-2)=-72\lt 0 \Rightarrow (1,-2) \text{ saddle point} \\D(-1,2)=-72\lt 0 \Rightarrow (-1,2)\text{ saddle point} \\D(-1,-2)=72\gt 0\Rightarrow f_{xx}(-1,-2)\lt 0 \Rightarrow \text{local max =}f(-1,-2)=18} \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{Local maximum value: 18} \\ \text{Saddle points:}(1,-2),(-1,2)}}$$

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