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=[2431]=PDP−1=[4−131][500−2][1/71/7−3/74/7]⇒eA=[4−131][e500e−2][1/71/7−3/74/7]=[4e5−e−23e5e−2][1/71/7−3/74/7]=[4e5+3e−274e5−4e−273e5−3e−273e5+4e−27]解答:y″+y=δ(t−π)−δ(t−2π)⇒L{y″}+L{y}=L{δ(t−π}−L{δ(t−2π}⇒(s2Y(s)−sy(0)−y′(0))+Y(s)=e−πs−e−2πs⇒(s2+1)Y(s)=e−πs−e−2πs+1⇒Y(s)=e−πss2+1−e−2πss2+1+1s2+1⇒y(t)=L−1{Y(s)}=L−1{e−πss2+1}−L−1{e−2πss2+1}+L−1{1s2+1}=u(t−π)(−sin(t))−u(t−2π)sin(t)+sin(t)⇒y(t)=sin(t)(1−u(t−π)−u(t−2π))
解答:f(z)=1(z−2)(z2+4)⇒{Res(f,z=2)=1z2+4|z=2=18Res(f,z=2i)=1(x−2)(x+2i)|z=2i=i−116由於{2在實軸上2i在上半部⇒∫∞−∞f(x)dx=Res(f,z=2)×πi+Res(f,z=2i)×2πi=πi8+−π−πi8=−π8
解答:f(z)=1z2+2z+2=1(z−(−1+i))(z−(−1−i))⇒Res(f,z=−1+i)=1z−(−1−i)|z=−1+i=12i⇒∫∞−∞f(x)dx=2πi×Res(f,z=−1+i)=2πi×12i=π
乙、測驗題部分:( 50 分)
解答:x2+4y2=16,對x微分可得2x+8yy′=0⇒y′=−x4y將(2√2,√2)代入可得y′=−2√24√2=−12⇒切線方程式:y=−12(x−2√2)+√2⇒x+2y=4√2,故選(A)解答:div(u×v)=v⋅curl u−u⋅curl v,故選(B)
解答:a×b=−b×a≠b×a,故選(A)
解答:z−√x2+y2=0⇒→n=∇(z−√x2+y2)=(−x√x2+y2,−y√x2+y2,1)將(1,1,√2)代入→n=(−1√2−1√2,1)⇒切平面:−1√2(x−1)−1√2(y−1)+(z−√2)=0⇒x+y=√2z,故選(B)
解答:det
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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解題僅供參考,其他國考試題及詳解
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