106年調查人員三等考試--工程數學詳解

106年法務部調查局調查人員考試

等 別：三等考試

類科組：電子科學組

科 目：工程數學

工程數學 詳解

解： $$y''+4y'+4y=f(t)\Rightarrow L\left\{y''\right\}+4L\left\{y'\right\}+4L\left\{y\right\}=L\left\{f(t)\right\}\\ \Rightarrow(s^2L\left\{y\right\}-sy(0)-y'(0))+4(sL\left\{y\right\}-y(0))+4L\left\{y\right\}=\int_0^2{e^{-st}\,dt}\\ \Rightarrow (s^2+4s+4)L\left\{y\right\}-s-6=-\frac{1}{s}e^{-2s}+\frac{1}{s}\\\Rightarrow (s+2)^2L\left\{y\right\}=\frac{1}{s}+s+6-\frac{1}{s}e^{-2s}=\frac{s^2+6s+1}{s}-\frac{1}{s}e^{-2s}\\\Rightarrow L\left\{y\right\}=\frac{s^2+6s+1}{s(s+2)^2}-\frac{1}{s(s+2)^2}e^{-2s}\\=\frac{1}{4}\cdot\frac{1}{s}+\frac{3}{4}\cdot\frac{1}{s+2}+\frac{7}{2}\cdot\frac{1}{(s+2)^2}-\left(\frac{1}{4}\cdot\frac{1}{s}-\frac{1}{4}\cdot\frac{1}{s+2}-\frac{1}{2}\cdot\frac{1}{(s+2)^2}\right)e^{-2s}\\\Rightarrow y=L^{-1}\left\{\frac{1}{4}\cdot\frac{1}{s}+\frac{3}{4}\cdot\frac{1}{s+2}+\frac{7}{2}\cdot\frac{1}{(s+2)^2}-\left(\frac{1}{4}\cdot\frac{1}{s}-\frac{1}{4}\cdot\frac{1}{s+2}-\frac{1}{2}\cdot\frac{1}{(s+2)^2}\right)e^{-2s}\right\}\\= \bbox[red,2pt]{\frac{1}{4}+\frac{3}{4}e^{-2t}+\frac{7}{2}te^{-2t}-\left(\frac{1}{4}-\frac{1}{4}e^{-2(t-2)}-\frac{1}{2}(t-2)e^{-2(t-2)}\right)u(t-2)}$$

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解：
(一) $$T\left( x_{ 1 },x_{ 2 },x_{ 3 } \right) =\left( 3x_{ 1 }+x_{ 2 },-2x_{ 1 }-4x_{ 2 }+3x_{ 3 },5x_{ 1 }+4x_{ 2 }-2x_{ 3 } \right) =\left[ \begin{matrix} 3 & 1 & 0 \\ -2 & -4 & 3 \\ 5 & 4 & -2 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \end{matrix} \right] \\ det\left( \left[ \begin{matrix} 3 & 1 & 0 \\ -2 & -4 & 3 \\ 5 & 4 & -2 \end{matrix} \right]  \right) =24+0+15-0-4-36=-1\\ \Rightarrow \left[ \begin{matrix} 3 & 1 & 0 \\ -2 & -4 & 3 \\ 5 & 4 & -2 \end{matrix} \right] ^{ -1 }=-\left[ \begin{matrix} \left| \begin{matrix} -4 & 3 \\ 4 & -2 \end{matrix} \right|  & -\left| \begin{matrix} 1 & 0 \\ 4 & -2 \end{matrix} \right|  & \left| \begin{matrix} 1 & 0 \\ -4 & 3 \end{matrix} \right|  \\ -\left| \begin{matrix} -2 & 3 \\ 5 & -2 \end{matrix} \right|  & \left| \begin{matrix} 3 & 0 \\ 5 & -2 \end{matrix} \right|  & -\left| \begin{matrix} 3 & 0 \\ -2 & 3 \end{matrix} \right|  \\ \left| \begin{matrix} -2 & -4 \\ 5 & 4 \end{matrix} \right|  & -\left| \begin{matrix} 3 & 1 \\ 5 & 4 \end{matrix} \right|  & \left| \begin{matrix} 3 & 1 \\ -2 & -4 \end{matrix} \right|  \end{matrix} \right] =\left[ \begin{matrix} 4 & -2 & -3 \\ -11 & 6 & 9 \\ -12 & 7 & 10 \end{matrix} \right] \\ \Rightarrow T^{ -1 }\left( x_{ 1 },x_{ 2 },x_{ 3 } \right) =\bbox[red,2pt] {\left( 4x_{ 1 }-2x_{ 2 }-3x_{ 3 },-11x_{ 1 }+6x_{ 2 }+9x_{ 3 },-12x_{ 1 }+7x_{ 2 }+10x_{ 3 } \right)}$$ (二) $$T^{ -1 }\left( 1,1,0 \right) =\left( 4-2,-11+6,-12+7 \right) =\bbox[red,2pt]{\left( 2,-5,-5 \right)}$$

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解： $$f(x)=\frac{1}{5+4\cos{x}}\Rightarrow f(\pi+x)=f(\pi-x)\Rightarrow \int_0^{2\pi}f(x)\,dx=2\int_0^{\pi}f(x)\,dx\\ 令u=\tan{(x/2)}\Rightarrow \cos{x}=\frac{1-u^2}{1+u^2}\;且\;dx=\frac{2du}{1+u^2}\\ \Rightarrow \int_0^{2\pi}{\frac{1}{5+4\cos{x}}dx}=2\int_0^{\pi}{\frac{1}{5+4\cos{x}}dx}=2\int_0^{\pi}{\frac{1}{5+4\frac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}}dx}\\ =2\int_0^{\pi}{\frac{1+\tan^2{(x/2)}}{9+\tan^2{(x/2)}}dx}=2\int_0^{\pi}{\left(1-\frac{8}{9+\tan^2{(x/2)}}\right)dx}=2\pi-16\int_0^{\pi}{\left(\frac{1}{9+\tan^2{(x/2)}}\right)dx}\\ 又\int{\frac{1}{9+\tan^2{(x/2)}}dx}=\int{\frac{2}{(9+u^2)(1+u^2)}du}=\frac{1}{4}\int{\left(\frac{1}{1+u^2}-\frac{1}{9+u^2}\right)\,du}\\ =\frac{1}{4}\arctan{u}-\frac{1}{12}\arctan{(u/3)}+C=\frac{1}{4}\arctan{\tan{(x/2)}}-\frac{1}{12}\arctan{(\tan{(x/2)}/3)}+C\\ 因此原式=2\pi-16\int_0^{\pi}{\left(\frac{1}{9+\tan^2{(x/2)}}\right)dx}=2\pi-\left.\left[4\arctan{\tan{(x/2)}}-\frac{4}{3}\arctan{(\tan{(x/2)}/3)}\right]\right|_0^{\pi}\\ =2\pi-\left[4\cdot\frac{\pi}{2}-\frac{4}{3}\cdot\frac{\pi}{2}\right]=2\pi-(2\pi-\frac{2\pi}{3})=\bbox[red,2pt]{\frac{2\pi}{3}}$$ 另解：利用柯西-留數定理(Cauchy's Residue Theorem) $$z=\cos{\theta}+i\sin{\theta}=e^{i\theta}\Rightarrow \begin{cases}\cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+1/z}{2} \\ dz=ie^{i\theta}d\theta=izd\theta\end{cases}\Rightarrow \int_0^{2\pi}{\frac{1}{5+4\cos\theta}d\theta} \\ = \oint_{|z|=1}{\frac{1}{5+4\left((z+1/z)/2\right)}\frac{dz}{iz}}=\frac{1}{i}\oint_{|z|=1}{\frac{1}{2z^2+5z+2}dz}=\frac{1}{i}\oint_{|z|=1}{\frac{1}{(2z+1)(z+2)}dz}\\ =2\pi i\times\frac{1}{i}\times Res\,f(-1/2)=2\pi\times \lim_{z\to (-1/2)}{\frac{z+(1/2)}{(2z+1)(z+2)}}=2\pi\times\frac{1}{3}=\bbox[red,2pt]{\frac{2\pi}{3}}$$

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解：
(一) $$\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ A\left( x+y \right)  } dxdy } =1\Rightarrow A\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 2 } x^{ 2 }+xy \right]  \right| _{ 0 }^{ 1 }dy } =A\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 2 } +y \right) dy } =A\left. \left[ \frac { 1 }{ 2 } y+\frac { 1 }{ 2 } y^{ 2 } \right]  \right| _{ 0 }^{ 1 }\\ =A\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 }  \right) =1\Rightarrow \bbox[red,2pt]{A=1}$$ (二) $$P\left\{ X+Y\le 1 \right\} =\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1-y }{ \left( x+y \right)  } dxdy } =\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 2 } x^{ 2 }+xy \right]  \right| _{ 0 }^{ 1-y }dy } =\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 2 } { \left( 1-y \right)  }^{ 2 }+\left( 1-y \right) y \right) dy } \\ =\int _{ 0 }^{ 1 }{ \left( -\frac { 1 }{ 2 } { y }^{ 2 }+\frac { 1 }{ 2 }  \right) dy } =\left. \left[ -\frac { 1 }{ 6 } y^{ 3 }+\frac { 1 }{ 2 } y \right]  \right| _{ 0 }^{ 1 }=-\frac { 1 }{ 6 } +\frac { 1 }{ 2 } =\bbox[red,2pt]{\frac { 1 }{ 3 } }$$

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解：

$$由起終點可知直線方程式為L:\frac{x-1}{3}=\frac{y-1}{3}=\frac{z-1}{3}\Rightarrow \begin{cases}x=3t+1\\y=3t+1\\z=3t+1 \end{cases}\\ \Rightarrow \begin{cases}dx=3dt\\dy=3dt\\dz=3dt \end{cases}\Rightarrow \int_c{x^2dx-2yzdy+zdz}=\int_0^1{(3t+1)^23dt-2(3t+1)^23dt+(3t+1)3dt}\\ =(-9)\int_0^1{(3t^2+t)dt}=(-9)\left.\left[t^3+\frac{1}{2}t^2\right]\right|_0^1=(-9)\times\frac{3}{2}= \bbox[red,2pt]{-\frac{27}{2}}$$

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