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
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)$$T^{ -1 }\left( 1,1,0 \right) =\left( 4-2,-11+6,-12+7 \right) =\bbox[red,2pt]{\left( 2,-5,-5 \right)} $$
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)$$z=\cos{\theta}+i\sin{\theta}=e^{i\theta}\Rightarrow \begin{cases}\cos{\theta}=\frac{e^{i\theta}+e^{-i\theta}}{2}=\frac{z+1/z}{2} \\ dz=ie^{i\theta}d\theta=izd\theta\end{cases}\Rightarrow \int_0^{2\pi}{\frac{1}{5+4\cos\theta}d\theta} \\
= \oint_{|z|=1}{\frac{1}{5+4\left((z+1/z)/2\right)}\frac{dz}{iz}}=\frac{1}{i}\oint_{|z|=1}{\frac{1}{2z^2+5z+2}dz}=\frac{1}{i}\oint_{|z|=1}{\frac{1}{(2z+1)(z+2)}dz}\\
=2\pi i\times\frac{1}{i}\times Res\,f(-1/2)=2\pi\times \lim_{z\to (-1/2)}{\frac{z+(1/2)}{(2z+1)(z+2)}}=2\pi\times\frac{1}{3}=\bbox[red,2pt]{\frac{2\pi}{3}}$$
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)$$P\left\{ X+Y\le 1 \right\} =\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1-y }{ \left( x+y \right) } dxdy } =\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 2 } x^{ 2 }+xy \right] \right| _{ 0 }^{ 1-y }dy } =\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 2 } { \left( 1-y \right) }^{ 2 }+\left( 1-y \right) y \right) dy } \\ =\int _{ 0 }^{ 1 }{ \left( -\frac { 1 }{ 2 } { y }^{ 2 }+\frac { 1 }{ 2 } \right) dy } =\left. \left[ -\frac { 1 }{ 6 } y^{ 3 }+\frac { 1 }{ 2 } y \right] \right| _{ 0 }^{ 1 }=-\frac { 1 }{ 6 } +\frac { 1 }{ 2 } =\bbox[red,2pt]{\frac { 1 }{ 3 } }$$
解:
$$由起終點可知直線方程式為L:\frac{x-1}{3}=\frac{y-1}{3}=\frac{z-1}{3}\Rightarrow \begin{cases}x=3t+1\\y=3t+1\\z=3t+1 \end{cases}\\
\Rightarrow \begin{cases}dx=3dt\\dy=3dt\\dz=3dt \end{cases}\Rightarrow \int_c{x^2dx-2yzdy+zdz}=\int_0^1{(3t+1)^23dt-2(3t+1)^23dt+(3t+1)3dt}\\
=(-9)\int_0^1{(3t^2+t)dt}=(-9)\left.\left[t^3+\frac{1}{2}t^2\right]\right|_0^1=(-9)\times\frac{3}{2}= \bbox[red,2pt]{-\frac{27}{2}} $$
考選部未公布答案,解題僅供參考
沒有留言:
張貼留言