104年專技高考_電機工程技師-工程數學詳解

104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：

$$令\mathcal{L}\left\{y(t) \right\}= Y(s) \Rightarrow \begin{cases} \mathcal{L}\left\{y'(t) \right\}= sY(s)-y(0) =sY(s)\\ \mathcal{L}\left\{y''(t) \right\}= s^2Y(s)-sy(0)-y'(0) =s^2Y(s)+7\\ \mathcal{L}\left\{ty(t) \right\}=-{d\over dt} \mathcal{L}\left\{y'(t) \right\} =-{d\over dt} sY(s) =-Y(s)-sY'(s)\end{cases}\\ \Rightarrow \mathcal{L}\left\{y'' \right\} +4\mathcal{L}\left\{ty' \right\} -4\mathcal{L}\left\{y(t) \right\}  =s^2Y(s)+7 -4Y(s)-4sY'(s)-4Y(s) =0\\ \Rightarrow  (s^2-8)Y(s)-4sY'(s)+7=0 \Rightarrow Y(s)={Ce^{s^2/8} \over s^2}-{7\over s^2}\\ 由於\lim_{s\to \infty}{sY(s)}= \lim_{t\to 0}{y(t)} =0 \Rightarrow \lim_{s\to \infty}{sY(s)} 存在 \Rightarrow \lim_{s\to \infty}{Ce^{s^2/8} \over s}-{7\over s}<\infty \Rightarrow C=0\\ Y(s)= -{7\over s^2} \Rightarrow y(t)=\mathcal{L}^{-1}\left\{-7\over s^2 \right\}=-7t \Rightarrow \bbox[red, 2pt]{y(t)=-7t}$$

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解：
(一) $$det(A-\lambda I)=0 \Rightarrow \begin{bmatrix} 1-\lambda & 0 & 0\\ 0 & 1-\lambda & 1\\ 0 & 1 & 1-\lambda \end{bmatrix}=0 \Rightarrow (1-\lambda)^3 -(1-\lambda)=0\\ \Rightarrow \lambda(\lambda-1)(\lambda-2)=0 \Rightarrow \bbox[red, 2pt]{特徵值\lambda = 0,1,2}$$ (二) $$\lambda=2 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix} -1 & 0 & 0\\ 0 & -1 & 1\\ 0 & 1 & -1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1=0\\ x_2=x_3\end{cases} \Rightarrow 取u_1= \begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix}\\ \lambda=1 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \begin{cases} x_2=0\\ x_3=0\end{cases} \Rightarrow 取u_2= \begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}\\ \lambda=0 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1=0\\ x_2+x_3=0\end{cases} \Rightarrow 取u_3= \begin{bmatrix} 0\\ 1\\ -1 \end{bmatrix}\\ \Rightarrow \bbox[red, 2pt]{ 特徵向量為\begin{bmatrix} 0\\ 1\\ 1 \end{bmatrix},\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}及\begin{bmatrix} 0\\ 1\\ -1 \end{bmatrix}}$$ (三) $$取P=[u_1 u_2 u_3]=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 1 & 0 & -1 \end{bmatrix} \Rightarrow P^{-1}=\begin{bmatrix} 0 & 1/2 & 1/2\\ 1 & 0 & 0\\ 0 & 1/2 & 1/2 \end{bmatrix} \Rightarrow P^{-1}AP= \begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}=\Lambda \\\Rightarrow \bbox[red, 2pt]{P=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 1 & 0 & -1 \end{bmatrix}, \Lambda= \begin{bmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}}$$

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解： $$\int_{-\infty}^\infty{3x+2\over x(x-4)(x^2+9)}dx =\int_c{3z+2\over z(z-4)(z+3i)(z-3i)}dz =\int_c{f(z)}dz \\ \Rightarrow \begin{cases} \text{Res}(f,0)= \left.{3z+2\over (z-4)(z^2+9)} \right|_{z=0}= -{1\over 18}\\\text{Res}(f,4)= \left.{3z+2\over z(z^2+9)} \right|_{z=4}= {7\over 50}\\ \text{Res}(f,3i)= \left.{3z+2\over z(z-4)(z+3i)} \right|_{z=3i}= {2+9i \over 72-54i} = {-19+42i \over 450}\end{cases}\\ z=0,z=4在實數軸上,且z=3i在上半圓 \Rightarrow \int_c{f(z)}dz=\pi i(\text{Res}(f,0)+\text{Res}(f,4)) +2\pi i(\text{Res}(f,3i)) \\ =\pi i(-{1\over 18}+ {7\over 50}) +2\pi i\cdot {-19+42i \over 450} = \pi i\left({19\over 225}+ {-19+42i\over 225} \right) = \bbox[red, 2pt]{-{42 \over 225}\pi}$$

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解： $$f_{X,Y}(x,y)= \begin{cases} x(y+1.5) & 0< x,y <1\\ 0 & 其它 \end{cases}\\ \Rightarrow \begin{cases} f_X(x)=\int_0^1{x(y+1.5) \; dy} =\left. \left[{1\over 2}xy^2+1.5xy \right] \right|_0^1 =2x, &0< x <1\\ f_Y(y)=\int_0^1{x(y+1.5)\; dx} =\left. \left[{1\over 2}x^2y+ {3\over 4}x^2 \right] \right|_0^1 = {1\over 2}y+ {3\over 4},& 0< y< 1\end{cases}$$ (一) $$E[X]=\int_0^1{xf_X(x)\;dx} =\int_0^1 {2x^2\;dx} =\left. \left[{2\over 3}x^3 \right] \right|_0^1=\bbox[red, 2pt]{2\over 3}$$ (二) $$E[Y]= \int_0^1{yf_Y(y)\;dy} =\int_0^1{{1\over 2}y^2 +{3\over 4}y\;dy} =\left. \left[{1\over 6}y^3 +{3\over 8}y^2 \right] \right|_0^1=\bbox[red, 2pt]{13\over 24}$$ (三) $$E[X^2] =\int_0^1 {x^2f_X(x)\;dx} =\int_0^1{2x^3\;dx} =\left. \left[{1\over 2}x^4 \right] \right|_0^1=\bbox[red, 2pt]{1\over 2}$$ (四) $$E[XY]=\int_0^1\int_0^1{xyf_{X,Y}(x,y)\;dydx} =\int_0^1\int_0^1{x^2y^2 +1.5x^2y\;dydx} =\int_0^1\left. \left[{1\over 3}x^2y^3 +{3\over 4}x^2y^2 \right] \right|_0^1\\ =\int_0^1{{13\over 12}x^2} = \left. \left[{13\over 36}x^3 \right] \right|_0^1=\bbox[red, 2pt]{13\over 36}$$

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解：
$$(1,1,1)\to (-2,1,3) \Rightarrow \begin{cases} x(t)=1-3t\\ y(t)=1 \\ z(t)=1+2t\end{cases},t:0\to 1 \Rightarrow  \begin{cases} dx=-3dt\\ dy=0 \\ dz=2dt\end{cases}\\ \Rightarrow \int_C{xyzdx-\cos{(yz)}dy+xzdz} = \int_0^1{(1-3t)(1+2t)(-3dt)-0+ (1-3t)(1+2t)(2dt)}\\ = \int_0^1{6t^2+t-1\;dt} = \left. \left[ 2t^3 +{1\over 2}t^2-t \right] \right|_0^1 = \bbox[red, 2pt]{3\over 2}$$

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