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2021年12月9日 星期四

103年地方特考-工程數學詳解

103年特種考試地方政府公務人員考試

等 別: 三等考試
類 科: 電力工程、電子工程
科 目: 工程數學
甲、申論題部分:( 50 分)

解答:$$A=\begin{bmatrix} 2 & 4\\ 3 & 1\end{bmatrix} =PDP^{-1}= \begin{bmatrix} 4 & -1\\ 3 & 1\end{bmatrix} \begin{bmatrix} 5 & 0\\ 0 & -2\end{bmatrix} \begin{bmatrix} 1/7 & 1/7\\ -3/7 & 4/7\end{bmatrix} \\ \Rightarrow e^A =\begin{bmatrix} 4 & -1\\ 3 & 1\end{bmatrix} \begin{bmatrix} e^5 & 0\\ 0 & e^{-2}\end{bmatrix} \begin{bmatrix} 1/7 & 1/7\\ -3/7 & 4/7\end{bmatrix} \\=\begin{bmatrix} 4e^5 & -e^{-2}\\ 3e^5 & e^{-2}\end{bmatrix} \begin{bmatrix} 1/7 & 1/7\\ -3/7 & 4/7\end{bmatrix} = \bbox[red,2pt]{\begin{bmatrix} {4e^5+3e^{-2}\over 7}& {4e^5-4e^{-2}\over 7}\\ {3e^5-3e^{-2}\over 7} & {3e^5+ 4e^{-2}\over 7} \end{bmatrix}}$$
解答:$$y''+y= \delta(t-\pi)-\delta(t-2\pi) \Rightarrow \mathcal{L}\{ y'' \}+\mathcal{L}\{ y \} =\mathcal{L}\{ \delta(t-\pi \}-\mathcal{L}\{ \delta(t-2\pi \}\\ \Rightarrow (s^2Y(s)-sy(0)-y'(0))+Y(s) =e^{-\pi s}-e^{-2\pi s} \Rightarrow (s^2+1)Y(s)= e^{-\pi s}-e^{-2\pi s} +1 \\ \Rightarrow Y(s)= {e^{-\pi s}\over s^2+1} -{e^{-2\pi s} \over s^2+1} +{1\over s^2+1}\\ \Rightarrow y(t)=\mathcal{L}^{-1}\{Y(s)\} =\mathcal{L}^{-1}\{{e^{-\pi s}\over s^2+1}\} -\mathcal{L}^{-1}\{{e^{-2\pi s} \over s^2+1}\} +\mathcal{L}^{-1}\{{1\over s^2+1}\}\\=u(t-\pi)(-\sin(t))-u(t-2\pi)\sin(t)+\sin(t)\\ \Rightarrow \bbox[red,2pt]{y(t)=\sin(t)(1-u(t-\pi)-u(t-2\pi))}$$
解答:$$f(z)={1\over (z-2)(z^2+4)} \Rightarrow \cases{Res(f,z=2)=\left.{1\over z^2+4}\right|_{z=2} ={1\over 8}\\ Res(f,z=2i)=\left. {1\over (x-2)(x+2i)}\right|_{z=2i} ={i-1\over 16}} \\ 由於\cases{2在實軸上\\ 2i在上半部} \Rightarrow \int_{-\infty}^\infty f(x)\;dx =Res(f,z=2)\times \pi i+ Res(f,z=2i)\times 2\pi i\\={\pi i\over 8} +{-\pi-\pi i\over 8} = \bbox[red,2pt]{-{ \pi \over 8}}$$
解答:$$f(z)={1\over z^2+2z+2} ={1\over (z-(-1+i))(z-(-1-i))}\\ \Rightarrow Res(f,z=-1+i)= \left. {1\over z-(-1-i)}\right|_{z=-1+i}={1\over 2i} \\ \Rightarrow \int_{-\infty}^\infty f(x)\;dx = 2\pi i\times Res(f,z=-1+i) = 2\pi i\times {1\over 2i} = \bbox[red,2pt]{\pi}$$

乙、測驗題部分:( 50 分)

解答:$$x^2+4y^2=16,對x微分可得2x+8yy'=0 \Rightarrow y'=-{x\over 4y}\\ 將(2\sqrt 2,\sqrt 2)代入可得y'=-{2\sqrt 2\over 4\sqrt 2} =-{1\over 2} \Rightarrow 切線方程式: y=-{1\over 2}(x-2\sqrt 2)+\sqrt 2 \\ \Rightarrow x+2y=4\sqrt 2,故選\bbox[red,2pt]{(A)}$$
解答:$$\text{div}(u\times v)=v\cdot \text{curl } u- u\cdot \text{curl }v,故選\bbox[red,2pt]{(B)}$$
解答:$$a\times b=-b\times a \ne b\times a,故選\bbox[red,2pt]{(A)}$$
解答:$$z-\sqrt{x^2+y^2}=0 \Rightarrow \vec n=\nabla(z-\sqrt{x^2+y^2})=(-{x\over \sqrt{x^2+y^2}},-{y\over \sqrt{x^2+y^2}},1) \\ 將(1,1,\sqrt 2)代入\vec n=(-{1\over \sqrt 2}-{1\over \sqrt 2},1) \Rightarrow  切平面:-{1\over \sqrt 2}(x-1)-{1\over \sqrt 2}(y-1)+(z-\sqrt 2)=0\\ \Rightarrow x+y=\sqrt 2z,故選\bbox[red,2pt]{(B)}$$
解答:$$\det(A-\lambda I)=0 \Rightarrow (x+2)(x-1)^2=0 \Rightarrow 有重根,\\因此特徵向量不會三個都線性獨立(重根的特徵向量不獨立),故選\bbox[red,2pt]{(C)}$$
解答:$$\text{Trace}(AB)=\text{Trace}(BA) \Rightarrow \text{Trace}(B^{-1}AB) =\text{Trace}(BB^{-1}A) =\text{Trace}( A),故選\bbox[red,2pt]{(A)}$$
解答:$$\begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 1 \\ 1 & 0 & 5\end{bmatrix} \xrightarrow{-1r_1+r_3}\begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 1 \\ 0 & -2 & 2\end{bmatrix} \xrightarrow{-2r_2+r_3} \begin{bmatrix} 1 & 2 & 3\\ 0 & -1 & 1 \\ 0 & 0 & 0\end{bmatrix} \xrightarrow{2r_2+r_1} \begin{bmatrix} 1 & 0 & 5\\ 0 & -1 & 1 \\ 0 & 0 & 0\end{bmatrix} \\ 令\cases{u=(1,0,5)\\ v=(0,-1,1)} \Rightarrow \cases{(A)(1,1,4)=u-v\\ (B)(2,2,8)=2u-2v\\ (C)(2,1,9)=2u-v},只有(D)(3,4,2)無法以u,v線性組合而成\\,故選\bbox[red,2pt]{(D)}$$
解答:$$A=\begin{bmatrix} 2 & 1 & 5\\ 0 & 3 & 7 \\ 4 & 2 & 6\end{bmatrix} \Rightarrow f(x)=\det(A-x I)=-x^3+11x^2-2x-24 \\ \Rightarrow f(A)=0 \Rightarrow A^3-11X^2+2A+24I=0 \\\Rightarrow 欲求之A^3-11X^2+3A+25I =A+I =\begin{bmatrix} 3 & 1 & 5\\ 0 & 4 & 7 \\ 4 & 2 & 7\end{bmatrix},故選\bbox[red,2pt]{(A)}$$
解答:$$將\cases{f(z)=e^z \\a=i}代入泰勒級數 \sum_{n=0}^\infty {f^{[n]}(a)\over n!}(z-a)^n =\sum_{n=0}^\infty {e^i\over n!}(z-i)^n,故選\bbox[red,2pt]{(C)}$$
解答:$$e^z=1+z +{z^2\over 2!}+{z^3\over 3!}+\cdots \Rightarrow e^{2z}=1+2z +{(2z)^2\over 2!}+{(2z)^3\over 3!}+\cdots \\ \Rightarrow {e^{2z}\over z^4} ={1\over z^4}+{2\over z^3} +{2\over z^2} +\color{blue}{4\over 3}\cdot {1\over z} +\cdots\\ \Rightarrow \int_c{e^{2z}\over z^4}\;dz = \color{blue}{4\over 3}\cdot 2\pi i= {8\pi i\over 3},故選\bbox[red,2pt]{(B)}$$
解答:$$z=-{1\over 2}+i{\sqrt 3\over 2} =\cos({2\over 3}\pi) +i\sin({2\over 3}\pi) =e^{i{2\over 3}\pi} \Rightarrow z^{31}= \left( e^{i{2\over 3}\pi}\right)^{31} =e^{i{62\over 3}\pi} =e^{i{2\over 3}\pi}=z,故選\bbox[red,2pt]{(D)}$$
解答:$${2s-1\over s^2(s-1)^2} ={-1\over s^2}+ {1\over (s-1)^2} \Rightarrow \mathcal{L}^{-1}\{{2s-1\over s^2(s-1)^2}\}=-\mathcal{L}^{-1}\{{1\over s^2}\}+\mathcal{L}^{-1}\{{1\over (s-1)^2}\} =-t+te^t\\,故選\bbox[red,2pt]{(D)}$$
解答:$$\mathcal{F}(\omega) = \int_{-\infty}^\infty f(t)e^{-i\omega t}\;dt =\int_{-1}^1 e^{-i\omega t}\;dt =\left. \left[ -{1\over i\omega }e^{i\omega t} \right] \right|_{-1}^1 ={1\over i\omega}(e^{i\omega} -e^{-i\omega})\\ ={2\over \omega}\cdot {1\over 2i }(e^{i\omega} -e^{-i\omega})={2\over \omega}\sin(\omega)   \Rightarrow \left|\mathcal{F}(\omega) \right|^2 ={4\over \omega^2}\sin^2(\omega)\\ 令g(t)=\begin{cases}a-|t|,&|t|\lt a\\ 0,&|t|\ge a\end{cases} \Rightarrow \mathcal{F}'=  \int_{-\infty}^\infty g(t)e^{-i\omega t}\;dt =\int_{-a}^a (a-|t|)e^{-i\omega t}\;dt\\= \int_{-a}^0(a+t) e^{-i\omega t}\;dt +\int_{0}^a(a- t)e^{-i\omega t} \\=\left( -{a\over i\omega} -{1\over (i\omega)^2}+{1\over (i\omega)^2}e^{iwa}\right) +\left( {a\over i\omega} -{1\over (i\omega)^2}+{1\over (i\omega)^2}e^{-iwa}\right)\\ ={2\over \omega^2}-{1\over \omega^2}(e^{i\omega a}+e^{-i\omega a}) ={2\over \omega^2}-{2\over \omega^2}\cos(\omega a)  ={2\over \omega^2}-{2\over \omega^2}(2\cos^2(\omega a/2)-1)\\={4\over \omega^2}-{4\over \omega^2}\cos^2(\omega a/2) ={4\over \omega^2 }\sin^2(\omega a/2)=\left|\mathcal{F}(\omega) \right|^2 ,\text{if }a=2\\ 也就是g(t)=\begin{cases}2-|t|,&|t|\lt 2\\ 0,&|t|\ge 2\end{cases} ,故選\bbox[red,2pt]{(B)}$$
解答:$$y''+2y'+y=x^2e^{-x} \Rightarrow 齊次解y_h=c_1e^{-x}+c_2xe^{-x} \Rightarrow y'_h=(c_2-c_1)e^{-x} -c_2xe^{-x};\\將\cases{y(0)=1\\ y'(0)=0} 代入y_n \Rightarrow c_1=c_2=1 \Rightarrow y_h=e^{-x}+xe^{-x} \Rightarrow \cases{y_1=e^{-x} \\ y_2=xe^{-x}} \Rightarrow W(y_1,y_2)=e^{-2x}\\ \Rightarrow y_p=-e^{-x}\int {xe^{-x}\cdot x^2e^{-x}\over e^{-2x}}\;dx +xe^{-x} \int { e^{-x}\cdot x^2e^{-x}\over e^{-2x}}\;dx = {1\over 12}x^4e^{-x}\\ \Rightarrow y=y_h+y_p = e^{-x}+xe^{-x}+  {1\over 12}x^4e^{-x} \Rightarrow y(1)=2e^{-1}+{1\over 12}e^{-1} ={25\over 12}e^{-1},故選\bbox[red,2pt]{(A)}$$
解答:$${d\over dx}\left( \ln{x\over y}\right) ={1/y-xy'/y^2\over x/y} = {1\over x}-{y'\over y} ={y-xy'\over xy} ={y-x(dy/dx)\over xy} \\ \Rightarrow d\left( \ln{x\over y}\right) = \left( {y-x(dy/dx)\over xy}\right)dx ={ydx-xdy\over xy},故選\bbox[red,2pt]{(D)}$$
解答:$$\left| {c_{n+1}\over c_n}\right| =\left| {(-1)^{n+1}/9^{n+1}\over (-1)^n/9^n}\right| =\left|-{1\over 9}\right| ={1\over 9} \Rightarrow x^2的收斂半徑=9 \Rightarrow x的收斂半徑=\sqrt{9} =3\\,故選\bbox[red,2pt]{(B)}$$
解答:$$y=e^{2x}(c_1\sin(\sqrt 3x)+c_2\cos(\sqrt 3x)) \Rightarrow 特徵多項式的解為2\pm \sqrt 3 i\\ \Rightarrow 特徵方程式為(r-2-\sqrt 3i)(r-2+\sqrt 3i)= r^2-4r+7 \Rightarrow   y''-4y'+7y=0,故選\bbox[red,2pt]{(C)}$$
解答:$$X為指數分配,因此變異數=6^2=36,故選\bbox[red,2pt]{(D)}$$
解答:$$P(X+Y\le 3) =P(X=1,Y=1)+ P(X=1,Y=2) +P(X=2,Y=1)={1\over 36}(2+3+3)\\ ={2\over 9},故選\bbox[red,2pt]{(B)}$$
解答:$$除了8號球不符要求外,其他號碼均符合要求,因此機率為9/10,故選\bbox[red,2pt]{(D)}$$

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