112年升官等-微積分詳解

 112年公務人員升官等考試

等 級： 薦任
類科（ 別）： 物理
科 目： 微積分
考試時間： 2 小時

解答: $$F(x)={1\over \sqrt[5]{(f(x))^4}} =(f(x))^{-4/5} \Rightarrow F'(x)=-{4\over 5}(f(x))^{-9/5}f'(x) \\ \Rightarrow F'(2)=-{4\over 5}\cdot 32^{-9/5}\cdot 20=-{4\over 5} \cdot {1\over 2^9}\cdot 20 =-{2^4\over 2^9}=\bbox[red, 2pt]{-{1\over 32}}$$

解答: $$\lim_{x\to 3} {\sqrt{x^2+16}-5\over (x-3)} =\lim_{x\to 3} {\frac{d }{dx}(\sqrt{x^2+16}- 5)\over \frac{d }{dx}(x-3)} =\lim_{x\to 3}{{x\over \sqrt{x^2+16}} \over 1} = \bbox[red, 2pt]{3\over 5}$$

解答: $$y^2+2xy=x^3+x+1 \Rightarrow 2yy'+2y+2xy'=3x^2+1 \Rightarrow y'={3x^2+1-2y\over 2x+2y} \\ \Rightarrow y'(1,1)={2\over 4} \Rightarrow 切線方程式: y={2\over 4}(x-1)+1 \Rightarrow \bbox[red, 2pt]{x-2y+1=0}$$

解答: $$y=f(x)=\sqrt{64-x^3} \Rightarrow x^3=64-y^2 \Rightarrow x= \sqrt[3]{64-y^2}\\  因此f^{-1}(x)=g(x)=\bbox[red,2pt]{\sqrt[3]{64-x^2}}\\又0\le x\le 4 \Rightarrow 0\le f(x)\le 8 \Rightarrow 0\le g(x)\le 4 \Rightarrow \bbox[red, 2pt]{\cases{定義域:[-8,8]\\ 值域:[0,4]}}$$

解答: $$f(x)=(x-6)(x+2)^3 \Rightarrow f'(x)=(x+2)^3 +3(x-6)(x+2)^2=4(x-4)(x+2)^2 \\ \Rightarrow f''(x)=4(x+2)^2+ 8(x-4)(x+2) =12(x+2)(x-2)\\ 若f'(x)=0 \Rightarrow x=4,-2 \Rightarrow \cases{f''(4)=144 \gt 0\\ f''(-2)=0} \Rightarrow \bbox[red,2pt]{\cases{f(4)=-432為相對極小值\\ 反曲點:(-2,0)}}$$

解答: $$u= x^3 \Rightarrow du=3x^2\,dx \Rightarrow \int {x^5\over \sqrt{1-2x^3}}\,dx = {1\over 3}\int {u\over \sqrt{1-2u}}\,du\\ 又令v=1-2u \Rightarrow dv=-2du \Rightarrow {1\over 3}\int {u\over \sqrt{1-2u}}\,du={1\over 3}\int {(1-v)/2\over \sqrt v}\cdot -{1\over 2}dv \\=-{1\over 12}\int {1-v\over \sqrt v}\,dv =-{1\over 12}\int {1\over \sqrt v}-\sqrt v\,dv =-{1\over 12}\left( 2\sqrt v-{2\over 3}v^{3/2} \right)+C \\=-{1\over 6}\left(  \sqrt{1-2u}-{1\over 3}(1-2u)^{3/2}\right) +C =-{1\over 6}\left(  \sqrt{1-2x^3}-{1\over 3}(1-2x^3)^{3/2}\right) +C \\= \sqrt{1-2x^3}\left(-{1\over 6}+{1\over 18}(1-2x^3) \right)+C =\bbox[red, 2pt]{-{1\over 9}(x^3+1)\sqrt{1-2x^3}+C}$$

解答: $$\Gamma(t)=\int_0^\infty e^{-x}x^{t-1}\,dx \Rightarrow \Gamma(\alpha+\beta)= \int_0^\infty e^{-x} x^{\alpha+\beta-1} \,dx \\ \Rightarrow {1\over \Gamma(\alpha)\Gamma(\beta)} \int_0^1 y^{\alpha+1}(1-y)^{\beta-1}\left( \int_0^\infty x^{\alpha+\beta-1} e^{-x}\,dx\right)\,dy ={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)} \int_0^1 y^{\alpha+1}(1-y)^{\beta-1}\,dy\\ 又\int_0^1 x^{\alpha}(1-x)^{\beta-1}\,dx ={\Gamma(\alpha+1)\Gamma(\beta) \over \Gamma(\alpha+\beta+1)}\\ \Rightarrow {\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)} \int_0^1 y^{\alpha+1}(1-y)^{\beta-1}\,dy= {\Gamma(\alpha+ \beta)\over \Gamma(\alpha) \Gamma(\beta)}\cdot {\Gamma(\alpha+2) \Gamma(\beta)\over \Gamma(\alpha+\beta+2)} \\= {\Gamma(\alpha+ \beta)\over \Gamma(\alpha) \Gamma(\beta)}\cdot {\alpha(\alpha+1)\Gamma(\alpha) \Gamma(\beta)\over (\alpha+\beta+1)(\alpha +\beta) \Gamma(\alpha+\beta)} =\bbox[red, 2pt]{\alpha(\alpha+1) \over (\alpha+\beta+1) (\alpha+ \beta)}$$

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