112年專技高考電子工程技師-工程數學詳解

 112年專門職業及技術人員高等考試

等 別：高等考試
類 科：電子工程技師
科 目：工程數學（包括線性代數、微分方程、向量分析、複變函數與機率）

解答：$$T=\begin{bmatrix}a& b& c\\ d& e& f\\ g& h& i \end{bmatrix} \Rightarrow \cases{T\left[ \begin{pmatrix}1\\ 1\\ 1 \end{pmatrix}\right]=\begin{pmatrix}3\\ 0\\ 2 \end{pmatrix} \Rightarrow \cases{a+b+c=3\\ d+e+f=0\\ g+h+i=2} \\[1ex]T\left[ \begin{pmatrix}1\\ 2\\ 1 \end{pmatrix} \right]=\begin{pmatrix}4\\ 0\\ 2 \end{pmatrix} \Rightarrow \cases{a+2 b+c=4\\ d+2e+f=0\\ g+2 h+i=2} \\[1ex]T\left[ \begin{pmatrix}1\\ 2\\ 3 \end{pmatrix} \right]=\begin{pmatrix}4\\ 0\\ 4 \end{pmatrix} \Rightarrow \cases{a+2b+3c=4\\ d+2e+3f=0\\ g+2 h+ 3i=4}} \\ \Rightarrow \cases{\cases{a+b+c=3\\ a+2b+c=4\\ a+2b+3c=4} \Rightarrow \cases{a=2\\b=1\\c=0} \\ \cases{d+e+f=0\\ d+2e+f=0\\ d+2e+ 3f=0} \Rightarrow \cases{d=0\\e=0\\f=0} \\ \cases{g+h+i=2\\ g+2h+i=2\\ g+2h+3i=4}\Rightarrow \cases{g=1\\h =0 \\ i=1}} \Rightarrow T=\begin{bmatrix}2& 1& 0\\ 0& 0& 0\\ 1& 0& 1 \end{bmatrix} \\ \Rightarrow  T\left[ \begin{pmatrix}x\\ y\\ z \end{pmatrix}\right]= \begin{bmatrix}2& 1& 0\\ 0& 0& 0\\ 1& 0& 1 \end{bmatrix} \begin{bmatrix}x\\ y\\ z \end{bmatrix} =\bbox[red, 2pt]{\begin{pmatrix} 2x+y\\ 0\\ x+z \end{pmatrix}}$$

解答：$$L\{y''\} +4L\{y'\}+3L\{y\}=3L\{\delta(t-2)\} \\\Rightarrow s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+3Y(s)=3e^{-2s}\\ \Rightarrow (s^2+4s+3)Y(s)=3e^{-2s} \Rightarrow Y(s)={3e^{-2s} \over s^2+4s+3}={3\over 2}e^{-2s}\left( {1\over s+1}-{1\over s+3}\right)\\ \Rightarrow y(t)=L^{-1}(Y(s))\Rightarrow \bbox[red, 2pt]{y(t)={3\over 2}u(t-2)\left(e^{-(t-2)}-e^{-3(t-2)}\right)}$$

解答：$$半球底面為一圓：x^2+y^2=1,z=0 \Rightarrow \vec r(t)=(\cos t,\sin t,0),0\le t\le 2\pi\\ \Rightarrow \vec r'(t)=(-\sin t,\cos t,0)dt,又\vec F=(-y,x,-xyz)=(-\sin t, \cos t, 0)\\ 因此\oint_C\vec F\cdot d\vec r =\int_0^{2\pi} (-\sin t, \cos t,0)\cdot (-\sin t,\cos t,0)dt =\int_0^{2\pi} (\sin^2 t+\cos^2 t)\,dt \\= \int_0^{2\pi} 1\,dt =\bbox[red, 2pt]{2\pi}$$

解答：$$f_1(z)=z^2 無奇異點\Rightarrow Res(f_1)=0\\ f_2(z)={1\over z^2-16}={1\over (z-4)(z+4)} \Rightarrow z=\pm 4皆不在圓內 \Rightarrow Res(f_2)=0\\ f_3(z)={e^{\pi z}\over z^2+4} ={e^{\pi z}\over (z+2i)(z-2i)} \Rightarrow z=\pm 2i皆在圓內 \Rightarrow Res(f)=Res(f,2i)+Res(f,-2i)\\ =\lim_{z\to 2i}{e^{\pi z}\over z+2i} +\lim_{z\to -2i}{e^{\pi z}\over z-2i} ={e^{2\pi i}\over 4i}+{e^{-2\pi i}\over -4i} ={1\over 4i}-{1\over 4i}=0 \\f_4(z)=z^2e^{1/z}=z^2(1+{1\over z}+{1\over 2z^2}+{1\over 6z^3}+{1\over 24z^4}+\cdots) =(z^2+z+{1\over 2}+{1\over 6z}+{1\over 24z^2}+\cdots) \\ \Rightarrow Res(f_4,0)={1\over 6}\\ 因此\oint_C \left( z^2 +{1\over z^2-16} +{e^{\pi z}\over z^2+4}+ z^2e^{1/z}\right)\,dz =\oint_C z^2e^{1/z}\,dz =2\pi i \cdot Res(f_4,0)=2\pi i \cdot {1\over 6} =\bbox[red, 2pt]{\pi i \over 3}$$

解答：$$f_Y(y)=\int_y^2f(x,y)\,dx = \int_y^2 {1\over 2}\,dx ={1\over 2}(2-y) \Rightarrow f_{X|Y}(x|y) ={f(x,y)\over f_Y(y)} ={1/2\over (2-y)/2}={1\over 2-y} \\ \Rightarrow E[X\mid Y=y] =\int_y^2 xf_{X|Y}(x,y)\,dx = \int_y^2 {x\over 2-y}\,dx ={4-y^2\over 2(2-y)} =\bbox[red, 2pt]{2+y\over 2}$$

=========================== END ==============================

解題僅供參考