115年中央財金碩士班-微積分詳解

國立中央大學115學年度碩士班考試入學試題

系所:財務金融 科目:微積分

解答:$$ \lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{x - \sin x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - e^{-x} - 2x)}{\frac{d}{dx}(x - \sin x)} = \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x + e^{-x} - 2)}{\frac{d}{dx}(1 - \cos x)} \\= \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - e^{-x})}{\frac{d}{dx}(\sin x)} = \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} ={1+1\over 1}= \bbox[red, 2pt]2$$

解答:$$f(x)=g(x) \Rightarrow  x^3 - 3x + 3 = x + 3 \Rightarrow x(x-2)(x+2)=0 \Rightarrow x=-2,0,2 \\\begin{cases} f(g)\ge g(x), &x\in[-2,0] \\ g(x)\ge f(x),& x\in[0,2]\end{cases} \Rightarrow \text{area } A = \int_{-2}^{0} [f(x) - g(x)] \,dx + \int_{0}^{2} [g(x) - f(x)] \,dx \\  = \int_{-2}^{0} (x^3 - 4x) \,dx + \int_{0}^{2} -(x^3 - 4x) \,dx= \left. \left[ {1\over 4}x^4-2x^2 \right] \right|_{-2}^0+ \left. \left[ -{1\over 4}x^4+2x^2 \right] \right|_0^2=4+4=\bbox[red, 2pt]8$$

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

解答:$$y= x^x \Rightarrow \ln y=x\ln x\Rightarrow {y'\over y}=\ln x+1 \Rightarrow y'=(\ln x+1)y = \bbox[red, 2pt]{x^x(\ln x+1)}$$

解答:$$\cases{dx=1.01-1=0.01\\ dy=1.98-2=-0.02}, z=2x^2y+y^3\Rightarrow \cases{{\partial z\over \partial x}=4xy\\ {\partial z\over \partial y}=2x^2+3y^2} \Rightarrow  \cases{{\partial z(1,2)\over \partial x}= 8\\ {\partial z(1,2)\over \partial y} =14} \\ \Rightarrow dz= {\partial z\over \partial x}dx+ {\partial z\over \partial y}dy =8\cdot 0.01+14\cdot(-0.02)=\bbox[red, 2pt]{-0.2}$$

解答:$$\cases{f(x,y,z)= 2xy+6yz+ 8xz\\ g(x,y,z) =xyz-96000} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ f_z= \lambda g_z\\ g=0} \Rightarrow \cases{2y+8z= \lambda yz\\ 2x+6z= \lambda xz\\ 6y+8x = \lambda xy\\ xyz=96000} \Rightarrow \cases{2xy+8xz=\lambda xyz\\ 2xy+6yz=\lambda xyz\\ 6yz+8xz=\lambda xyz} \\ \Rightarrow 2xy+8xz=2xy+6yz= 6yz+8xz \Rightarrow \cases{y=4x/3\\ z=x/3} \Rightarrow xyz=x\cdot {4x\over 3} \cdot {x\over 3}={4\over 9}x^3=96000 \\ \Rightarrow x=60 \Rightarrow \cases{y=80\\ z=20} \Rightarrow f(60,80,20) =9600+ 9600+9600=\bbox[red, 2pt]{28800}$$

解答:

$$r=1=1-\cos \theta \Rightarrow \cos \theta=0 \Rightarrow \theta=\pi/2, -\pi/2 \\ \Rightarrow A={1\over 2} \int_{-\pi/2}^{\pi/2} \left( 1^2-(1-\cos\theta)^2 \right) \, d\theta =  \int_{0}^{\pi/2} \left( 1-(1-2\cos\theta+ \cos^2\theta) \right) \, d\theta\\ =  \int_{0}^{\pi/2} (2\cos \theta-\cos^2\theta)\,d\theta=   \left[ 2\sin(\theta) - \left( \frac{1}{2}\theta + \frac{1}{4}\sin(2\theta) \right) \right]_{0}^{\frac{\pi}{2}} =\bbox[red, 2pt]{2-{\pi\over 4}}$$

解答:$$ \int_{0}^{2/3} \int_{y}^{2-2y} (x+2y)e^{(y-x)} dx dy=  \int_{0}^{2/3} \left. \left[ (-x-1-2y)e^{y-x} \right] \right|_y^{2-2y} dy =  \int_{0}^{2/3} (3y+1-3e^{3y-2})\,dy \\= \left. \left[ {3\over 2} y^2+y-e^{3y-2} \right] \right|_0^{2/3}= \bbox[red, 2pt]{{1\over 3}+e^{-2}}$$

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解題僅供參考，碩士班歷年試題及詳解