105年專技高考_電機工程技師-工程數學詳解

105年專門職業及技術人員高等考試

等        別：高等考試
類        科：電機工程技師
科        目：工程數學

解：

$$y=e^{3x}\left( 2+c_1\sin{(x)}+ c_2\cos{(x)}\right)= c_1e^{3x}\sin{(x)} +c_2e^{3x}\cos{(x)}+2e^{3x}\\
\Rightarrow \begin{cases} y_h=c_1e^{3x}\sin{(x)} +c_2e^{3x}\cos{(x)}\\ y_p= 2e^{3x}\end{cases} \Rightarrow \lambda=3\pm i \Rightarrow \lambda^2-6\lambda +10=0\\ \Rightarrow y''_h-6y'_h+10y_h=0 \Rightarrow 微分方程為y''_h-6y'_h+10y_h=y_p \\ \Rightarrow \bbox[red, 2pt]{y''-6y'+10y=2e^{3x}} $$

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解：$$L\left\{ |\sin{(\omega t)}| \right\} =\int_0^\infty {|\sin{(\omega t)}|e^{-st}\;dt} =\sum_{n=0}^\infty {\int_{n\pi/\omega}^{(n+1)\pi/\omega}|\sin{(\omega t)}|e^{-st}\;dt}\\
=\sum_{n=0}^\infty {e^{-sn\pi/\omega}}\int_{0}^{\pi/\omega}\sin{(\omega t)}e^{-st}\;dt ={1\over 1-e^{s\pi/\omega}} \int_{0}^{\pi/\omega}{\sin{(\omega t)}e^{-st}\;dt}\\
={1\over 1-e^{s\pi/\omega}} \left. \left[{-s\sin{(\omega t)e^{-st}-\omega\cos{(\omega t)}e^{-st}}\over s^2+\omega^2} \right] \right|_0^{\pi/\omega} ={1\over 1-e^{s\pi/\omega}}\cdot {\omega\left(1+e^{-s\pi/\omega} \right)\over s^2+\omega^2}\\ = {\omega \over s^2+\omega^2}\cdot {1+e^{-s\pi/\omega} \over 1-e^{s\pi/\omega}}
={\omega \over s^2+\omega^2}\cdot {e^{s\pi\over 2\omega}+e^{-s\pi/2\omega} \over e^{s\pi/2\omega}-e^{s\pi/2\omega}} =\bbox[red, 2pt]{{\omega \over s^2+\omega^2} \coth{s\pi \over 2\omega}}$$

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解：$$z(t)=t^3+it \Rightarrow z'(t)=3t^2+i \Rightarrow \int_{\varphi}{Im(z)}dz = \int_0^1{Im(z(t))z'(t)}dt\\
= \int_0^1{t(3t^2+i)}dt = \int_0^1{3t^3+it}\;dt = \left . \left[ {3\over 4}t^4+{i\over 2} t^2 \right] \right|_0^1 = \bbox[red, 2pt]{{3\over 4}+ {i\over 2}}$$

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解：$$\oint_\gamma{f(z)}dz =\oint_\gamma{z^2-2z+i \over z-1+i}dz \Rightarrow \text{Res}(f,1-i)= \left. z^2-2z+i \right|_{z=1-i} =(1-i)^2-2(1-i)+i \\= -2+i \Rightarrow \oint_\gamma{f(z)}dz = 2\pi i\times \text{Res}(f,1-i)= 2\pi i\times (-2+i) = \bbox[red, 2pt]{-4\pi i-2\pi} $$

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解：
(一)$$P(1/2\le X\le 3/2) = F(3/2)-F(1/2)= (3/2)/2 - (1/2)/3= \bbox[red, 2pt]{7\over 12}$$(二)$$P(X>3/2)= 1-P(X\le 3/2)=1-F(3/2) =1-(3/2)/2=1-3/4= \bbox[red, 2pt]{1\over 4}$$

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解：
(一)$$第1顆是白球且第2顆也是白球的機率: 4/9\times 3/8 =\bbox[red, 2pt]{1/6}$$(二)$$第1顆是紅球且第2顆也是紅球的機率: 3/9\times 2/8 =1/12\\ 第1顆是不是紅球但第2顆是紅球的機率: 6/9 \times 3/8 = 1/4\\ 第二顆是紅球的機率為 1/12+1/4= \bbox[red, 2pt]{1\over 3}$$(三)$$(1)兩顆都是白球的機率:4/9 \times 4/9=\bbox[red, 2pt]{16/81};\\ (2)第2顆是紅球的機率: 3/9=\bbox[red, 2pt]{1/3}$$

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解：$$T為線性轉換 \Rightarrow aT(x)+bT(y)=T(ax+by)\\
\begin{bmatrix} 7\\6\end{bmatrix}= 7\begin{bmatrix} 1\\0\end{bmatrix}+6\begin{bmatrix} 0\\1\end{bmatrix} \Rightarrow T\left( \begin{bmatrix} 7\\6\end{bmatrix}\right)=T\left( 7\begin{bmatrix} 1\\0\end{bmatrix}+6\begin{bmatrix} 0\\1\end{bmatrix}\right) = 7T\left( \begin{bmatrix} 1\\0\end{bmatrix}\right) +6T\left(\begin{bmatrix} 0\\1\end{bmatrix} \right)\\ =7\begin{bmatrix} 2\\-3 \end{bmatrix}+ 6\begin{bmatrix} -1\\3 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} 8\\-3 \end{bmatrix}}$$

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解：$$A=\begin{bmatrix} 0& 1& 1\\1 & 0 & 1\\1 & 1 & 0\end{bmatrix} \Rightarrow det(A-\lambda I)=0 \Rightarrow (\lambda -2)(\lambda+1)^2=0 \Rightarrow \lambda=2, -1\\
\lambda =2\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix} -2& 1& 1\\1 & -2 & 1\\1 & 1 & -2\end{bmatrix} \begin{bmatrix} x_1\\x_2\\ x_3\end{bmatrix}=0\Rightarrow \begin{cases} x_2+x_2=2x_1 \\ x_1+x_3=2x_2\\ x_1+x_2=2x_3 \end{cases}\Rightarrow 取u_1=\begin{bmatrix} 1\\1\\ 1\end{bmatrix}\\
\lambda =-1\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix} 1& 1& 1\\1 & 1 & 1\\1 & 1 & 1\end{bmatrix} \begin{bmatrix} x_1\\x_2\\ x_3\end{bmatrix}=0\Rightarrow x_1+x_2+x_3=0 \\\Rightarrow 取u_2=\begin{bmatrix} 1\\0\\ -1\end{bmatrix},u_3= \begin{bmatrix} 1\\-2\\ 1\end{bmatrix}\\
令Y=[{u_1\over |u_1|}\;{u_2\over |u_2|}\;{u_3\over |u_3|}]= \bbox[red, 2pt]{\begin{bmatrix} 1/\sqrt{3}& 1/\sqrt{2}& 1/\sqrt{6}\\ 1/\sqrt{3} & 0 & -2/\sqrt{6}\\1/\sqrt{3} & -1/\sqrt{2} & 1/\sqrt{6}\end{bmatrix}} \Rightarrow Y^TAY= \begin{bmatrix} 2& 0& 0\\0 & -1 & 0\\0 & 0 & -1\end{bmatrix}$$

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