114年成大企管碩士班-微積分詳解

 國立成功大學114學年度碩士班招生考試試題

系 所:企業管理學系
科目:微積分

解答: $$A.\bigcirc: f'(x)=3x^2-3 =0 \Rightarrow x= \pm 1 \Rightarrow f''(x)=6x \Rightarrow \cases{f''(1) \gt 0\\ f''(-1) \lt 0} \Rightarrow \text{minimum: }f(1)=-2\\B. \bigcirc:\text{maximum: }f(-1)=2 \\C. \bigcirc: f(1)\lt f(x)=1 \lt f(-1) \Rightarrow \text{ three solutions}\\ D. \times: f(-x)=-x^3+3x =-f(x)\\故選\bbox[red, 2pt]{(ABC)}$$

解答: $$A.\bigcirc: y=2^x = e^{\ln 2^x} =e^{x\ln 2} \\ B.\times: y=e^{x\ln 2 } \Rightarrow y'=\ln 2e^{x\ln 2} =2^x \ln 2 \\C.\times: \int 2^x\,dx = \int e^{x\ln 2}\,dx ={1\over \ln 2}2^x+C \\D.\bigcirc: \lim_{n\to \infty} \sum_{k=0}^n {1\over f(k)} =\lim_{n\to \infty} \sum_{k=0}^n {1\over 2^k} ={1\over 1-{1\over 2}} =2 \\故選\bbox[red, 2pt]{(AD)}$$

解答: $$A.\bigcirc: y=0 \Rightarrow x^2=1 \Rightarrow x=\pm 1 \Rightarrow \text{ two intersection points }\\B.\times: x=0 \Rightarrow y^3=1 \Rightarrow y=1 \Rightarrow \text{ one intersection point} \\C. \bigcirc: 2x+3y^2y'+y+xy'=0 \Rightarrow y'=-{2x+y \over x+3y^2} \Rightarrow y'(0,1)=-{1\over 3} \\\qquad\Rightarrow \text{ tangent line: }y=-{1\over 3}x+1 \Rightarrow x+3y=3 \\D. \bigcirc: y'(-1,1) ={1\over 2} \Rightarrow \text{ tangent line: }y={1\over 2}(x+1)+1 \Rightarrow 2y-x=3\\ 故選\bbox[red, 2pt]{(ACD)}$$

解答: $$A. \times: f(x)=(4-x^2)^{1/2} \Rightarrow f'(x)={1\over 2}(4-x^2)^{-1/2}(-2x) =-{x\over \sqrt{4-x^2}} \\B. \bigcirc: f'(x)=-{x\over \sqrt{4-x^2}} \Rightarrow f''(x)=-{1\over \sqrt{4-x^2}}-{x(-2x) \over 2(4-x^2)^{3/2}} =-{4\over (4-x^2)^{3/2}}\\C.:\times: \int_{-1}^1 \sqrt{4-x^2}\,dx =  2\int_0^1 \sqrt{4-x^2}\,dx =2 \left. \left[ {x\sqrt{4-x^2}\over 2} +2\sin^{-1}{x\over 2}\right] \right|_0^1 =\sqrt 3+{2\over 3}\pi \\ D.\bigcirc: f'(-x)=-f'(x) \Rightarrow f\text{ is odd} \Rightarrow \int_{-1}^1 f(x)\,dx =0\\ 故選\bbox[red, 2pt]{(BD)}$$

解答: $$f'(x)= \lim_{h\to 0} {f(x+h)-f(x)\over h} \Rightarrow f'(2) \approx{f(2.01)-f(2)\over 0.01} \Rightarrow f(2.01) \approx 0.01\times f'(2)+f(2) \\=0.01\times 4+3= \bbox[red, 2pt]{3.04}$$

解答: $$f(x)=g(x) \Rightarrow x(4-x)=x^2 \Rightarrow 2x^2-4x=0 \Rightarrow x=0,2 \Rightarrow \Omega =\int_0^2 \left(x(4-x)-x^2) \right)\,dx \\=\int_0^2 (4x-2x^2)\,dx = \bbox[red, 2pt]{8\over 3}$$

解答: $$\int_0^2 (f^2-g^2)\pi\,dx = \pi \int_0^2 (16x^2-8x^3)\,dx = \bbox[red, 2pt]{{32\over 3}\pi}$$

解答: $$f(x)=\ln(1+x^2) \Rightarrow f'(x)={2x\over 1+x^2} =2x(1-x^2+x^4-x^6+x^8-x^{10}+ \cdots) \\=2x-2x^3+2x^5-2x^7+2x^9-2x^{11}+\cdots \Rightarrow f''(x)=2-6x^2+10x^4-14x^6+18x^8-22x^{10}+\cdots \\ \Rightarrow f'''(x)=-12x+40x^3-84x^5+144x^7-220x^9+\cdots \\\Rightarrow f^{[4]}(x) =-12+120x^2-420x^4 +1008x^6-1980x^8 +\cdots \\ \Rightarrow f^{[5]}(x) =240x-1680x^3+6048x^5-15840x^7+ \cdots \Rightarrow f^{[6]}(x) =240-5040x^2+\cdots \\ \Rightarrow f(0)=0,f'(0)=0, f''(0)=2, f'''(0)=0, f^{[4]}(0)=-12, f^{[5]}(0)=0, f^{[6]}(0)=240\\ \Rightarrow f(x)=f(0)+f'(0)x+ {f''(0) \over 2!}x^2 +{f'''(0) \over 3!}x^3 +{f^{[4]}(0)\over 4!}x^4  +{f^{[5]}(0)\over 5!}x^5  +{f^{[6]}(0)\over 6!}x^6 +\cdots \\ \Rightarrow \bbox[red, 2pt]{f(x)=x^2-{1\over 2}x^4+{1\over 3}x^6-\cdots}$$

解答: $$\int {1\over x^2(x+1)}\,dx =\int \left({1\over x+1}-{1\over x}+{1\over x^2} \right)\,dx =\bbox[red, 2pt]{\ln|x+1|-\ln |x|-{1\over x}+C}$$

解答: $$\int_0^2 \int_0^x xy\,dydx= \int_0^2 \left. \left[ {1\over 2}xy^2\right] \right|_0^x\,dx =\int_0^2 {1\over 2}x^3\,dx = \left. \left[ {1\over 8}x^4 \right] \right|_0^2 =\bbox[red, 2pt]2$$

解答: $${dy\over dx}=y-xy =y(1-x) \Rightarrow \int {1\over y}\,dy =\int (1-x)\,dx \Rightarrow \ln y=x-{1\over 2}x^2+C \\ y(0)=1 \Rightarrow 0=C \Rightarrow \ln y=x-{1\over 2}x^2 \Rightarrow \bbox[red, 2pt]{y=e^{x-x^2/2}}$$

解答: $$D(x)=120-x \Rightarrow D'(x)=-1, C(q)=q^2+4q+200 \Rightarrow C'(q) =2q+4\\P(x) =x\cdot D(x)-C(D(x)) \Rightarrow P'(x)=D(x)+xD'(x)-D'(x)C'(D(x)) \\=(120-x)-x+(2(120-x)+4) =364-4x \Rightarrow P''(x)=-4\\ P'(x)=0 \Rightarrow x=91 \Rightarrow P(91) =91D(91)-C(D(91)) =91\cdot 29-C(29) =2639-1157=1482\\ \bbox[red, 2pt] {\text{optimal price: 91 },\text{max profit: }1482}$$

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