111年一般警察人員考試
考 試 別：一般警察人員考試
等 別：三等考試
類科組別：消防警察人員
科 目：微積分
解答：$$令g(x)=\frac{x(x^2+1)^2}{\sqrt{2x^3+1}}，則g'(x)=\cfrac{(x^2+1)^2 +4x^2(x^2+1)}{\sqrt{2x^3+1}} -\cfrac{3x^3(x^2+1)^2}{(\sqrt{2x^3+1})^3} \\ 因此f(x)=\ln g(x) \Rightarrow f'(x)={g'(x)\over g(x)} =\left( \cfrac{(x^2+1)^2 +4x^2(x^2+1)}{\sqrt{2x^3+1}} -\cfrac{3x^3(x^2+1)^2}{(\sqrt{2x^3+1})^3}\right)\cdot \cfrac{\sqrt{2x^3+1}}{x(x^2+1)^2}\\ = \cfrac{(x^2+1)^2 +4x^2(x^2+1)}{x(x^2+1)^2} -\cfrac{3x^3(x^2+1)^2}{( 2x^3+1 ) \cdot x(x^2+1)^2}\\={1\over x}+\cfrac{4x}{x^2+1}-\cfrac{3x^2}{2x^3+1 } \Rightarrow \bbox[red,2pt]{f'(x)= \cfrac{1}{x}+\cfrac{4x}{x^2+1}-\cfrac{3x^2}{2x^3+1 } }$$
解答：$$f'(x)=3x^2-1 \Rightarrow f(x)=x^3-x+C，又圖形通過(1,2)，即f(1)=2 \Rightarrow 1-1+C=2 \Rightarrow C=2\\ \Rightarrow \bbox[red, 2pt]{y=x^3-x+2}$$
解答：$$令f(x)=x-\sqrt{x^2+1} \Rightarrow f'(x) = 1-\cfrac{x}{\sqrt{x^2+1}}\\ 因此y=\tan^{-1}f(x) \Rightarrow y'=\cfrac{f'(x)}{1+(f(x))^2} =\cfrac{1-\cfrac{x}{\sqrt{x^2+1}}}{2(x^2+1)-2x\sqrt{x^2+1}} =\cfrac{ \cfrac{\sqrt{x^2+1}- x}{\sqrt{x^2+1}}}{2x^2+2-2x\sqrt{x^2+1}}\\ =\cfrac{\sqrt{x^2+1}-x}{(2x^2+2)\sqrt{x^2+1}-2x(x^2+1)} \Rightarrow \bbox[red, 2pt]{y'=\cfrac{\sqrt{x^2+1}-x}{(2x^2+2)\sqrt{x^2+1}-2x(x^2+1)}}$$
解答：$$先求\lim_{n\to \infty} 3n\ln(1+{3\over n}) = \lim_{n\to \infty} \cfrac{\ln(1+{3\over n})}{1\over 3n} =\lim_{n\to \infty} \cfrac{ -3\over n(n+3)}{-1\over 3n^2} = \lim_{n\to \infty} \cfrac{9n^2}{n(n+3)}=9\\ 因此=\lim_{n\to \infty} \left( 1+{3\over n}\right)^{3n} =\lim_{n\to \infty} e^{3n\ln(1+3/n)} =\bbox[red,2pt]{e^9}$$
解答：$$\int_2^\infty \cfrac{x+3}{x^3-x^2+x-1}\,dx =\int_2^\infty \cfrac{x+3}{(x^2+1)(x-1)}\,dx =\int_2^\infty \cfrac{2}{x-1 } -\cfrac{2x+1}{x^2+1}\,dx\\ =\int_2^\infty \cfrac{2}{x-1 } -\cfrac{2x }{x^2+1}-\cfrac{1}{ x^2+1}\,dx =\left.\left[2\ln(x-1)-\ln(x^2+1)-\tan^{-1} x \right] \right|_2^\infty \\ =\left.\left[ \ln\cfrac{(x-1)^2}{x^2+1}-\tan^{-1} x \right] \right|_2^\infty = (0-{\pi\over 2})-(\ln {1\over 5}-\tan^{-1} 2) = \bbox[red,2pt]{\ln 5+\tan^{-1} 2-{\pi \over 2}}$$
解答：$$\int_{-\infty }^0\cfrac{1}{x^2-3x+2}dx =\int_{-\infty }^0\cfrac{1}{(x-2)(x-1)}dx =\int_{-\infty }^0\cfrac{1}{x-2}-\cfrac{1}{x-1}dx\\ = \left. \left[ \ln(x-2)-\ln(x-1)\right] \right|_{-\infty}^0 = \left. \left[ \ln{x-2\over x-1}\right] \right|_{-\infty}^0 =\bbox[red,2pt]{\ln 2}$$
解答：
$$第一象限心臟線面積=\int_0^{\pi/2}{1\over 2}\cdot (2(1+\cos \theta))^2\;d\theta =\int_0^{\pi/2} 2+ 4\cos \theta+ 2\cos^2\theta \;d\theta \\ =\int_0^{\pi/2} 3+ 4\cos \theta+ \cos 2\theta \,d\theta = \left.\left[ 3\theta + 4\sin \theta +{1\over 2}\sin 2\theta\right]\right|_0^{\pi/2} =4+{3\over 2}\pi\\ 第一象限圓面積={1\over 4}\cdot 2^2\pi = \pi\\ 欲求之面積=2(4+{3\over 2}\pi-\pi)= \bbox[red,2pt]{8+\pi}$$
解答：$$質心坐標(\bar x,\bar y)，其中\bar x=\cfrac{\int_0^{\pi/4} x(\cos x-\sin x)\,dx}{\int_0^{\pi/4} \cos x-\sin x\,dx},\bar y=\cfrac{\int_0^{\pi/4} (\cos x-\sin x)^2\,dx}{\int_0^{\pi/4} \cos x-\sin x\,dx}\\ 而\cases{\int_0^{\pi/4} \cos x-\sin x\,dx =\left. \left[ \sin x+\cos x\right] \right|_0^{\pi/4} =\sqrt 2-1\\[1ex] \int_0^{\pi/4} x(\cos x-\sin x)\,dx= \left. \left[ (x-1)\sin x+ (x+1)\cos x\right] \right|_0^{\pi/4} ={\sqrt 2\over 4}\pi -1\\[1ex] \int_0^{\pi/4} (\cos x-\sin x)^2\,dx =\int_0^{\pi/4} 1-\sin(2x)\,dx= \left. \left[ x+{1\over 2}\cos(2x)\right] \right|_0^{\pi/4} ={\pi\over 4}-{1\over 2}} \\ \Rightarrow \cases{\bar x={\sqrt 2\pi/4-1 \over \sqrt 2-1}={2+\sqrt 2\over 4}\pi -(\sqrt 2+1)\\[1ex] \bar y= {\pi/4-1/2\over \sqrt 2-1} = {(\sqrt 2+1)(\pi-2)\over 4}} \Rightarrow 質心坐標=\bbox[red,2pt]{\left({2+\sqrt 2\over 4}\pi -(\sqrt 2+1), {(\sqrt 2+1)(\pi-2)\over 4}\right)}$$================ END ===============
解答：$$質心坐標(\bar x,\bar y)，其中\bar x=\cfrac{\int_0^{\pi/4} x(\cos x-\sin x)\,dx}{\int_0^{\pi/4} \cos x-\sin x\,dx},\bar y=\cfrac{\int_0^{\pi/4} (\cos x-\sin x)^2\,dx}{\int_0^{\pi/4} \cos x-\sin x\,dx}\\ 而\cases{\int_0^{\pi/4} \cos x-\sin x\,dx =\left. \left[ \sin x+\cos x\right] \right|_0^{\pi/4} =\sqrt 2-1\\[1ex] \int_0^{\pi/4} x(\cos x-\sin x)\,dx= \left. \left[ (x-1)\sin x+ (x+1)\cos x\right] \right|_0^{\pi/4} ={\sqrt 2\over 4}\pi -1\\[1ex] \int_0^{\pi/4} (\cos x-\sin x)^2\,dx =\int_0^{\pi/4} 1-\sin(2x)\,dx= \left. \left[ x+{1\over 2}\cos(2x)\right] \right|_0^{\pi/4} ={\pi\over 4}-{1\over 2}} \\ \Rightarrow \cases{\bar x={\sqrt 2\pi/4-1 \over \sqrt 2-1}={2+\sqrt 2\over 4}\pi -(\sqrt 2+1)\\[1ex] \bar y= {\pi/4-1/2\over \sqrt 2-1} = {(\sqrt 2+1)(\pi-2)\over 4}} \Rightarrow 質心坐標=\bbox[red,2pt]{\left({2+\sqrt 2\over 4}\pi -(\sqrt 2+1), {(\sqrt 2+1)(\pi-2)\over 4}\right)}$$================ END ===============
沒有留言:
張貼留言