114年中山大學企管碩士班-微積分詳解

國立中山大學 114學年度碩士班考試入學招生考試

科目名稱:微積分【企管系企管甲班碩士班甲組選考、乙組選考、丙組選考】

解答:$$f(x,y)=x^2-e^{y^2} \Rightarrow \cases{f_x=2x\\ f_y=-2ye^{y^2}} \Rightarrow \cases{f_{xy} =0\\ f_{xx}=2\\ f_{yy} =-2e^{y^2}-4y^2e^{y^2}} \\\Rightarrow D(x,y)= f_{xx}f_{yy}-(f_{xy})^2  = -4e^{y^2}-8y^2e^{y^2} \\ 若\cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(0,0) \Rightarrow D(0,0)=-4\lt 0 \Rightarrow (0,0) 為鞍點\\ \Rightarrow \bbox[red, 2pt]{\cases{唯一臨界點(0,0)\\ (0,0)為一鞍點\\ f(x,y)無相對極值}}$$

解答:$$y'=-{4xy\over x^2+1} \Rightarrow  \int {1\over y}\,dy =\int {-4x\over x^2+1}\,dx \Rightarrow \ln y= -2\ln(x^2+1)+c_1 \\ \quad \Rightarrow y={c_2 \over (x^2+1)^2 }\Rightarrow 1={c_2 \over (1+1)^2} \Rightarrow c_2= 4 \Rightarrow \bbox[red, 2pt]{y=f(x)={4\over (x^2+1)^2}}$$

 

解答:$$x^2=x^3 \Rightarrow x^2(x-1)=0 \Rightarrow x=0,1 \Rightarrow R=\{(x,y) \mid 0\le x\le 1, x^3\le y\le x^2\} \\ \Rightarrow 所圍體積V=\iint_R 2x^2y\,dA =\int_0^1 \int_{x^3}^{x^2} 2x^2y\,dy\,dx =\int_0^1 (x^6-x^8)\,dx ={1\over 7}-{1\over 9} = \bbox[red, 2pt]{2\over 63}$$

解答:$$u=x^{-2}+1 \Rightarrow \cases{du=-2x^{-3}dx \\ x^{-2}=u-1} \Rightarrow \int x^{-5}(x^{-2}+1)^{3/2}\,dx = \int (u-1)u^{3/2}(-{du\over 2}) \\ =-{1\over 2}\int \left( u^{5/2}-u^{3/2}\right)\,du =  \left( -{1\over 7}u^{7/2} +{1\over 5}u^{5/2}\right) +C= \bbox[red, 2pt]{-{1\over 7}(x^{-2}+1)^{7/2}+{1\over 5}(x^{-2}+1)^{5/2}+C}$$

解答:$$\cases{u=x/3\\ dv=e^{3x}dx} \Rightarrow \cases{du=dx/3\\ v=e^{3x}/3} \Rightarrow \int {x\over 3}e^{3x}\,dx ={1\over 9}xe^{3x}-{1\over 9}\int e^{3x}\,dx ={1\over 9}xe^{3x}-{1\over 27}e^{3x}+C \\ \Rightarrow \int_0^{\ln x} {x\over 3}e^{3x}\,dx = \left. \left[{1\over 9}xe^{3x}-{1\over 27}e^{3x} \right] \right|_0^{\ln 2} ={1\over 9}\ln 2\cdot 8-{8\over 21}+{1\over 21} = \bbox[red, 2pt]{{8\over 9}\ln 2-{7\over 21}}$$

解答:

$$f(x)={x^2-9\over x^2-4} \Rightarrow f'(x)={10x\over (x^2-4)^2} =0 \Rightarrow f''(x)={ -30x^2-40\over (x^2-4)^3}\\ f'(x)=0\Rightarrow x=0  \Rightarrow f''(0)\gt 0 \Rightarrow 極小值f(0) \\ f(x)={x^2-9\over (x+2)(x-2)} \Rightarrow x=\pm 2為垂直漸近線, \lim_{x\to \pm \infty}f(x)=1 \Rightarrow y=1為水平漸近線$$

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

解答:$$\textbf{(i) }f(x)={1\over 2x-1} =-{1\over 1-2x} =- \left( 1+2x +(2x)^2+\cdots+ (2x)^n+ \cdots\right) \\=-1-2x-4x^2-\cdots-2^nx^n-\cdots \Rightarrow \bbox[red, 2pt]{f(x)=\sum_{n=0}^\infty -2^nx^n} \\ \textbf{(ii) } \lim_{n\to \infty} \left|{a_{n+1} \over a_n} \right| =\lim_{n\to \infty} \left|{-2^{n+1} x^{n+1} \over -2^nx^n} \right| =\lim_{n\to \infty}|2x| \Rightarrow |2x|\lt 1 \Rightarrow -{1\over 2}\lt x\lt {1\over 2} \\\quad \Rightarrow f(-1/2)= \sum_{n=0}^\infty -2^n(-1/2)^n = \sum_{n=0}^\infty (-1)^{n+1}\text{ divergent series} \\\text{ and }x\ne {1\over 2} \Rightarrow \text{ convergent interval}: \bbox[red, 2pt]{\left(-{1\over 2}, {1\over 2} \right)}$$

解答:$$a=1.424242\dots \Rightarrow 100a=142.424242\dots \Rightarrow 100a-a=99a=141 \Rightarrow a= \bbox[red, 2pt]{141\over 99}$$

解答:$$y'={dy\over dx} =x^2(1-y) \Rightarrow \int {1\over 1-y}dy = \int x^2\,dx \Rightarrow -\ln(1-y) ={1\over 3}x^3+c_1 \\ \Rightarrow {1\over 1-y} =e^{x^3/3+c_1} =c_2e^{x^3/3} \Rightarrow 1-y=c_3e^{-x^3/3} \Rightarrow y=1-c_3e^{-x^3/3} \\ \Rightarrow y(0)=1-c_3=-2 \Rightarrow c_3=3 \Rightarrow \bbox[red, 2pt]{y=1-3e^{-x^3/3}}$$

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