115年政治大學企管碩士-微積分詳解

 

 國立政治大學115學年度碩士班暨碩士在職專班招生考試試題

考試科目：微積分
系所別：企業管理研究所(MBA學位學程)乙組

解答:$$\textbf{(a) }  \lim_{x\to -\infty} {\sqrt{9x^2+3} \over 4x-1} =\lim_{x\to -\infty} {-x\sqrt{9+3/x^2} \over x(4-1/x)} = \bbox[red, 2pt]{-{3\over 4}} \\\textbf{(b) }\cases{\sin x=x-{x^3\over 6}+ O(x^5) \\ e^u=1+u+u^2/2+ u^3/6+ u^4/24+O(u^5)} \\ \Rightarrow e^{\sin x} =1+ \left( x-{x^3\over 6} \right) +{1\over 2}\left( x-{x^3\over 6} \right)^2 +{1\over 6} \left( x-{x^3\over 6} \right)^3+ {1\over 24} \left( x-{x^3\over 6} \right)^4+O(x^5) \\\qquad =1+x+ {x^2\over 2}-{1\over 8}x^4+O(x^5)\\ \Rightarrow \lim_{x\to 0}  \left( {1\over 2x^2}+{1\over x^3}+ {1\over x^4}-{e^{\sin x} \over x^4} \right)  \\= \lim_{x\to 0} {(1/2)x^2+x+1-(1+x+ {x^2\over 2}-{1\over 8}x^4+O(x^5))\over x^4} = \lim_{x\to 0} {(1/8)x^4-O(x^5)\over x^4} =\bbox[red, 2pt]{1\over 8}$$

解答:$$\textbf{(a) }  f(x)=\ln(x^3(4x+1)^5) = \ln x^3+ \ln(4x+1)^5= 3\ln x+5 \ln(4x+1) \\ \Rightarrow f'(x)={3\over x}+{20\over 4x+1} = \bbox[red, 2pt]{32x+3\over 4x^2+x}\\ \textbf{(b) } \arctan (x)=  \sum_{n=0}^\infty {(-1)^n\over 2n+1}x^{2n+1} \Rightarrow g(x)= \sum_{n=0}^\infty {(-1)^n\over 2n+1}x^{2n+3} \\\Rightarrow \text{Coefficient of }x^{83} ={1\over 81} \Rightarrow g^{(83)}(0) = \bbox[red, 2pt]{83!\over 81}$$

解答:$$\textbf{(a) } I_1= \int {2\over \cos^2 x} \sqrt{6+{\sin x\over \cos x}} \,dx =\int {2 \sec^2 x} \sqrt{6+ \tan x} \,dx\\ u=6+\tan x \Rightarrow du=\sec^2 x\,dx \Rightarrow I_1=\int 2\sqrt u\,du={4\over 3}u^{3/2} +C_1={4\over 3}(6+ \tan x)^{3/2}+C_1 \\ I_2= \int x\,dx ={1\over 2}x^2+C_2 \\ \Rightarrow  \int \left( {2\over \cos^2 x} \sqrt{6+{\sin x\over \cos x}} +x\right)\,dx =I_1+ I_2 =\bbox[red, 2pt]{{4\over 3}(6+ \tan x)^{3/2}+{1\over 2}x^2+C_3}\\ \textbf{(b) } I=\int_0^2 \int_{y/2}^1 e^{3x^2+2} \,dx dy= \int_0^1 \int_0^{2x} e^{3x^2+2} \,dydx = \int_0^1 2x e^{3x^2+2} \,dx \\ \quad u=3x^2+2 \Rightarrow du=6x\,dx \Rightarrow I=\int_2^5 {1\over 3}e^u\,du ={1\over 3} \left. \left[ e^u \right] \right|_2^5 = \bbox[red, 2pt]{{1\over 3}(e^5-e^2)} \\\textbf{(c) }\cases{u= \csc x\\ dv=\csc^2 x\,dx} \Rightarrow \cases{du=-\csc x\cot x\,dx \\ v=-\cot x} \Rightarrow I=  \int \csc^3 x\,dx  =   -\csc x\cot x-\int \csc x \cot^2 x\,dx \\  =- \csc x\cot x-  \int \csc x(\csc^2 x-1)\,dx  =- \csc x\cot x-\int \csc^3 x\,dx + \int\csc x\,dx \\ \Rightarrow 2I=- \csc x\cot x + \int\csc x\,dx =-\csc x\cot x+ \ln |\csc x-\cot x|+C_1 \\ \Rightarrow I=-{1\over 2 }\csc x\cot x+{1\over 2} \ln|\csc x -\cot x| +C_2 \\ \Rightarrow \int {4\over \sin^3 x}\,dx = 4I = \bbox[red, 2pt]{-2\csc x\cot x+ 2\ln|\csc x-\cot x|+C}$$

解答:$$\textbf{(a) }F(x)= \int_0^x \sqrt t\sin t\,dt \Rightarrow F'(x)= \sqrt x\sin x\\ \quad F'(x)=0 \Rightarrow \cases{\sqrt x=0\\ \sin x =0} \Rightarrow \cases{x=0\\ x=n\pi,n\in \mathbb Z} \Rightarrow \text{ Since the domain is restricted to }x\ge 0,\\ \text{ the cirtical points are: }\bbox[red, 2pt]{x=n\pi, n=0,1,2,3,\dots} \\\textbf{(b) }F''(x)={\sin x\over 2\sqrt x}+ \sqrt x\cos x \Rightarrow F''(\pi)=-\sqrt{\pi}\lt 0 \Rightarrow \bbox[red, 2pt]{F(x) \text{ has a local maximum at }x=\pi} \\\textbf{(c) }G(x)= \int_0^x t^2 \sin(t^2)\,dt = aF(bx^2+cx+d) \\ \Rightarrow G'(x)=x^2\sin(x^2) =aF'(bx^2+cx+d)(2bx+c) =a \sqrt{bx^2+cx+d} \sin(bx^2+cx+d) (2bx+c) \\ \Rightarrow bx^2+cx+d=x^2 \Rightarrow \cases{b=1\\ c=d=0} \Rightarrow x^2\sin(x^2)=a\cdot x\cdot  \sin(x^2)\cdot (2x) \Rightarrow 1=2a \Rightarrow a={1\over 2} \\ \Rightarrow \bbox[red, 2pt]{\cases{a=1/2\\ b=1\\ c=0\\d=0}}$$

解答:$$ {250\over q+20}=q+5  \Rightarrow q^2+25q-150=0 \Rightarrow (q+30)(q-5)=0 \Rightarrow q_0=5 \\ \Rightarrow p_0 = S(q_0=5) =5+5=10 \Rightarrow \bbox[red, 2pt]{\text{The equilibrium point is (5,10)}} \\ \text{Consumer's surplus: }CS= \int_0^{q_0} D(q)\,dq -p_0q_0 =\int_0^5 {250\over q+20} dq-10\cdot 5=250 \left. \left[ \ln|q+20| \right] \right|_0^5-50 \\= 250 \ln{25\over 20}-50 \Rightarrow \bbox[red, 2pt]{\text{Consumer's surplus is }250\ln{5\over 4}-50} \\\text{Producer's Surplus: }PS=p_0q_0-\int_0^{q_0} S(q)\,dq =10\cdot 5-\int_0^5 (q+5)\,dq ={25\over 2} \\ \Rightarrow \bbox[red, 2pt]{\text{Producer's Surplus is }{25\over 2}}$$

解答:$$4x+2y=80 \Rightarrow y=40-2x \Rightarrow P(x) =x^3+ 2(40-2x)^2 = x^3+8x^2-320x+3200 \\ \Rightarrow P'(x)=3x^2+16x-320=0 \Rightarrow (3x+40) (x-8)=0 \Rightarrow x=8 \\ \Rightarrow \cases{P(0)=3200\\ P(8) =1664\\ P(20) =8000} \Rightarrow \text{Therefore, to maximize the company's profit, they should produce}\\\textbf{ 20 units of Product A and 0 units of Product B.}$$

========================== END =========================

解題僅供參考，碩士班歷年試題及詳解