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【財管系碩士班甲組】
解答:$$L= x^{\tan x} \Rightarrow \ln L= \tan x \ln x ={\ln x\over \cot x} \Rightarrow \lim_{x\to 0^+} \ln L= \lim_{x\to 0^+} {(\ln x)'\over (\cot x)'} =\lim_{x\to 0^+} {1/x\over -\csc^2 x} \\ =\lim_{x\to 0^+} -{\sin^2 x\over x} =\lim_{x\to 0^+} \left( -{\sin x\over x}\cdot \sin x \right) =-1\cdot \sin(0)=0 \Rightarrow \lim_{x \to 0^+} L=e^0= \bbox[red, 2pt]1$$
解答:$$\lim_{y\to 0} \left( \sqrt{y^4+y^2+4} -\sqrt{y^4+y^2} \right) =\sqrt{0+0+4}-\sqrt{0+0} = \bbox[red, 2pt] 2$$
解答:$$\int_0^1 (\sqrt{1+t}-\sqrt t)\,dt = \left. \left[ {2\over 3}(1+t)^{3/2}-{2\over 3}t^{3/2} \right] \right|_0^1 ={2\over 3}(2^{3/2}-2) ={4\over 3}(\sqrt 2-1) \\ \Rightarrow \int_0^1 \int_{\sqrt s}^{\sqrt{1+s}} {X\over \int_0^1 (\sqrt{1+t}-\sqrt t)\,dt } \,dy ds = \int_0^1 \int_{\sqrt s}^{\sqrt{1+s}} {X\over {4\over 3}(\sqrt 2-1)} \,dy ds\\ = {3X\over 4(\sqrt 2-1)} \int_0^1(\sqrt{1+s}-\sqrt s)\, ds = {3X\over 4(\sqrt 2-1)} \cdot {4\over 3}(\sqrt 2-1) = \bbox[red, 2pt]X$$
解答:$$\int_0^1 \int_0^t e^s \,dsdt = \int_0^1 (e^t-1)\,dt = \left. \left[ e^t-t \right] \right|_0^1=e-2 \\ \Rightarrow \int_0^1 \int_0^x {Z\over \int_0^1 \int_0^t e^s\,dsdt}e^y\,dy dx =\int_0^1 \int_0^x {Z\over e-2}e^y\,dy dx = {Z\over e-2} \int_0^1 (e^x-1)\,dx ={Z\over e-2} \cdot (e-2) \\= \bbox[red, 2pt]Z$$
解答:$$\int_0^x \cos (x) \left[ \cos({x\over 2}) \right]^2 \,dx -{1\over 4}x =\int_0^x \cos (x) \left[ {1\over 2}(\cos(x)+1) \right] \,dx -{1\over 4}x\\ ={1\over 2}\int_0^x \left[ \cos^2(x)+ \cos(x) \right] \,dx -{1\over 4}x ={1\over 2}\int_0^x \left[ {1\over 2}(\cos(2x)+1)+ \cos(x) \right] \,dx -{1\over 4}x \\={1\over 2} \left( {1\over 4}\sin (2x)+{1\over 2}x+\sin (x) \right)-{1\over 4}x= \bbox[red, 2pt]{{1\over 8}\sin (2x)+{1\over 2}\sin (x)}$$
解答:$$\int {2\sin x\cos x\over \cos^2 x-\sin^2 x} \,dx = \int {\sin 2x\over \cos 2x}\,dx= \bbox[red, 2pt]{-{1\over 2}\ln |\cos 2x|+C}$$
解答:$$\textbf{(1) } \cases{f(0) =P=96 \\ D=-{1\over P}\cdot {dP\over dy} \Rightarrow f'(0)={dP\over dy}=-D\cdot P \\ C={1\over P}\cdot {d^2P\over dy^2} \Rightarrow f''(0) ={d^2P\over dy^2}=C\cdot P} \\ \textbf{Scenario A: } \Delta P \approx f'(0)x=-D\cdot P\cdot \Delta y =-4\cdot 96\cdot (-0.005)=1.92 \\\qquad \Rightarrow \text{ Estimated Price = }96+1.92=97.92 \\ \textbf{Scenario B: }\Delta P \approx -4\cdot 96\cdot 0.005 =-1.92 \Rightarrow \text{ Estimated Price = }96-1.92= 94.08 \\\textbf{(2)} P_{new} \approx P-D\cdot P\cdot \Delta y+{1\over 2}C\cdot P\cdot (\Delta y)^2 \Rightarrow {1\over 2}C\cdot P\cdot (\Delta y)^2={1\over 2}\cdot24\cdot 96\cdot (\pm 0.005)^2=0.0288 \\ \textbf{Scenario A: }P_{new} \approx 97.92+0.0288=97.9488 \\ \textbf{Scenario B: }P_{new} \approx 94.08+0.0288= 94.1088 \\ \textbf{Under a rate cut}, \text{ the positive convexity term adds to the price increase, meaning the} \\ \qquad \text{bond price rises more than the linear duration estimate predicts.} \\\textbf{Under a rate hike}, \text{the positive convexity term acts as a cushion, offsetting some of the} \\\qquad \text{price decay, meaning the bond price falls less than the linear duration estimate predicts.}$$
解答:$$u=y^{-1} \Rightarrow u'=-{y'\over y^2} \Rightarrow y'+y =xy^2 \Rightarrow {y'\over y^2}+{1\over y}=x \Rightarrow -u'+u=x \\ \Rightarrow u'-u=-x \Rightarrow e^{-x}u'-e^{-x }u=-xe^{-x} \Rightarrow (ue^{-x}) '=-xe^{-x} \Rightarrow ue^{-x}= \int -xe^{-0x}\,dx \\ \Rightarrow ue^{-x}=xe^{-x}+e^{-x}+c_1 \Rightarrow u={1\over y}=x+1+c_1e^x \Rightarrow \bbox[red, 2pt]{y={1\over x+1+ c_1e^x}}$$
解答:$$\textbf{(1) } \text{$x$ is uniformly distributed on [0,2] } \Rightarrow \text{ PDF }f(x)={1\over 2-0}= {1\over 2} \\ \Rightarrow E[\pi(x)] = \int_0^2 \pi(x)f(x)\,dx = \int_0^2 (20-4x^2)\cdot {1\over 2}\,dx = \int_0^2 (10-2x^2)\,dx = \left. \left[ 10x-{2\over 3}x^3 \right] \right|_0^2 \\ \qquad =20-{16\over 3} = \bbox[red, 2pt]{44\over 3} \\\textbf{(2) } \text{The economic meaning of this result highlights the negative impact of volatility/uncertainty}\\\text{ on profits. If there were perfectly stable market conditions and no price fluctuation ($x = 0$), the firm} \\\text{would make its maximum daily profit of $20$ million USD. However, because political instability} \\\text{introduces a random price fluctuation index between $0$ and $2$, this fluctuation acts as a penalty. }$$
解答:$$\textbf{(1) }E(R) = \int_0^1 E(R|x) f(x)\,dx = \int_0^1 (20-4x^2)(2x) \,dx = \int_0^1 (40x-8x^3) \,dx = \left. \left[ 20x^2-2x^4 \right] \right|_0^1 \\\qquad = \bbox[red, 2pt]{18} \\\textbf{(2) }\text{The unconditional expected return $E(R) = 18$ differs from $E(R|x=0)=20$} \\\qquad \text{ because market sentiment $x$ is not fixed at 0; it is a random variable distributed } \\\qquad \text{ across the interveal $[0,1].$}$$
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