114年中山大學財管碩士班-微積分詳解

 國立中山大學114學年碩士班考試入學招生考試

科目名稱：微積分【財管系碩士班甲組】

解答:$$L= x^{\sin x} =e^{\sin x\ln x} \Rightarrow \ln L=\sin x\ln x ={\ln x\over {1\over \sin x}} \\ \Rightarrow \lim_{x\to 0^+} \ln L=\lim_{x\to 0^+}{\ln x\over {1\over \sin x}} = \lim_{x\to 0^+}{(\ln x) '\over ({1\over \sin x})'} =\lim_{x\to 0^+} {1/x\over -{\cos x\over \sin^2 x}} = \lim_{x\to 0^+}{-\sin^2 x\over x\cos x} = \lim_{x\to 0^+}{(-\sin^2 x)'\over (x\cos x)'} \\= \lim_{x\to 0^+}{-\sin 2 x\over \cos x-x\sin x} = 0 \Rightarrow  \lim_{x\to 0^+}L= e^0= \bbox[red, 2pt]1$$

解答:$$\lim_{y\to 2}\left(\sqrt{y^4+y^3-y+1} -\sqrt{y^4-y^2+y}\right) =\sqrt{16+8-2+1}-\sqrt{16-4+2} = \bbox[red, 2pt]{\sqrt{23}-\sqrt{14}}$$

解答:

$$u=y^2 \Rightarrow du=2y dy \Rightarrow  \int_0^1 \int_{\sqrt x}^{\sqrt{x+1}} y^3e^{y^2}\,dy \,dx= \int_0^1 \int_{ x}^{x+1} {1\over 2}u e^{u} \, du\,dx \\= {1\over 2}\int_0^1 \left. \left[ ue^u-e^u\right] \right|_{x}^{x+1} \,dx = {1\over 2}\int_0^1 \left(xe^{x+1}-xe^x +e^x\right)dx ={1\over 2} \left. \left[ xe^{x+1}-e^{x+1}-xe^x +2e^x\right] \right|_0^1 \\= \bbox[red, 2pt]{e-1}$$

解答:$$\int_0^1 \int_0^x e^{x+y}\,dy\,dx = \int_0^1 \left. \left[ e^{x+y}\right] \right|_0^x \,dx =\int_0^1(e^{2x} -e^x)\,dx =\left. \left[ {1\over 2}e^{2x}-e^x\right] \right|_0^1 ={1\over 2}e^2-e +{1\over 2} \\ =\bbox[red, 2pt]{{1\over 2}(e-1)^2}$$

解答:$$u=\cos(5x) \Rightarrow du= -5\sin (5x)\,dx \Rightarrow \int \sin^3(5x) \cos^2(5x)\,dx =  \int -{1\over 5} (1-u^2)u^2du \\ ={1\over 5} \int (u^4-u^2)\,du ={1\over 5}\left( {1\over 5}u^5 -{1\over 3}u^3\right) +C={1\over 25}u^5 -{1\over 15}u^3+C \\=\bbox[red, 2pt]{{1\over 25} \cos^5(5x) -{1\over 15}\cos^3(5x)+C}$$

解答:$$y'+3y=e^x \cos(x) \Rightarrow e^{3x}y'+3ye^{3x} = e^{4x} \cos(x) \Rightarrow (e^{3x}y)'=  e^{4x} \cos(x) \\ \Rightarrow e^{3x}y= \int e^{4x}\cos (x)\,dx ={1\over 17} e^{4x}(\sin x+ 4\cos x)+c_1 \Rightarrow \bbox[red, 2pt]{y= {1\over 17} e^{x}(\sin x+ 4\cos x)+c_1e^{-3x}}$$

解答:$$\int_1^3 {-x^2+6x \over 46}\,dx = \bbox[red, 2pt]{1\over 3}$$

解答:$$\textbf{(1) }P(r+dr) \approx P(r)+{dP\over dr}(dr)+ {1\over 2}\cdot {d^2 P\over dr^2}(dr)^2 + \cdots \\ \Rightarrow dP=P(r+dr)-P(r) \Rightarrow \bbox[red, 2pt]{dP \approx {dP\over dr}(dr)+{1\over 2}\cdot {d^2P\over dr^2}(dr)^2+\cdots}\\\textbf{(2) } \cases{ \text{initial price }P=100\\ \text{modified duration }D=5\\ \text{convexity }C=20 \\ \text{change in rate }dr=-0.5\%=-0.005 \text{ (1/2 percentage point)}} \\\qquad \Rightarrow dP \approx -P\cdot D\cdot dr +{1\over 2}P\cdot C\cdot (dr)^2=-100\times 5\times (-0.005)+ {1\over 2} \times 100\times 20\times (-0.005)^2 \\ \qquad =2.5+0.025=2.525 \Rightarrow \text{new Treasury bill price }P_{new} =100+2.525 = \bbox[red, 2pt]{102.525}$$

解答:$$\text{Let }A \text{ and }B \text{ be the quantities produced for type 1 and type 2 customers, respectively.}\\ \Rightarrow \cases{ R_1= (100-10A)A= 100A-10A^2\\ R_2=(80-4B)B= 80B-4B^2} \Rightarrow \text{total cost (TC)= }100+10(A+B) \\ \Rightarrow \text{ profit function }f(A,B)=(R_1+R_2)-TC =-10A^2+90A-4B^2+70B-100 \\ \Rightarrow \cases{f_A=-20A+90=0 \Rightarrow \bbox[red, 2pt]{A=4.5}\\ f_B=-8B+70=0 \Rightarrow \bbox[red, 2pt]{B=8.75}} \Rightarrow f(4.5,8.75)= \bbox[red, 2pt]{\text{max profits}= 408.75}$$

解答:$$\textbf{(1) }\text{weight vector }w= \begin{bmatrix}GS\\ AXP \end{bmatrix} = \begin{bmatrix}0.4\\ 0.6 \end{bmatrix}, \text{covariance matrix }\Sigma= \begin{bmatrix}0.06& 0.03\\ 0.03& 0.04 \end{bmatrix} \\ \Rightarrow \text{portfolio variance }\sigma_p^2 = w^T \Sigma w= [0.4 \;0.6] \begin{bmatrix}0.06& 0.03\\ 0.03& 0.04 \end{bmatrix} \begin{bmatrix}0.4\\ 0.6 \end{bmatrix} =\bbox[red, 2pt]{0.0384} \\\textbf{(2) } \text{The portfolio's standard deviation is }\sqrt{0.0384} \approx 19.6\% \lt 25\% \\\Rightarrow \bbox[red, 2pt]{\text{Yes}},\text{ the weights are adequate for your needs.}$$

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解題僅供參考，碩士班歷年試題及詳解