110年鐵路特考-工程數學詳解

110年特種考試交通事業鐵路人員考試試題

考 試 別： 鐵路人員考試
等 別： 高員三級考試
類 科 別： 電力工程、 電子工程
科 目： 工程數學
甲、申論題部分：(50分)

解答： $$y''+2y'+2y=f(t) \Rightarrow \mathcal{L}\{y''\}+ 2\mathcal{L}\{y'\} +2\mathcal{L}\{y\}=\mathcal{L}\{ f(t)\} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0)+2(sY(s)-y(0))+2Y(s)=\mathcal{L}\{ f(t)\}\\ \Rightarrow Y(s)(s^2+2s+2)-1=\mathcal{L}\{ f(t)\}=\cases{2/s,\text{if }0\lt t\le \pi\\ 0,\text{ if }t\gt \pi} \\ \Rightarrow \cases{t\gt \pi \Rightarrow Y(s)={1\over (s+1)^2+ 1} \Rightarrow y(t)=e^{-t}\sin(t)\\ 0\lt t\le \pi \Rightarrow Y(s)={s+2\over s(s^2+2s+2)} ={1\over s}-{s+1\over (s+1)^2+1} \Rightarrow y(t)=1-e^{-t}\cos(t) }\\ \Rightarrow \bbox[red,2pt]{y(t)=\begin{cases}e^{-t}\sin(t), & t\gt \pi\\ 1-e^{-t}\cos(t),& 0\lt t\le \pi \end{cases}}$$

解答：

(一) $$A=\begin{bmatrix} 5 & -4 & 4\\ 12 & -11 & 12\\ 4 & -4 & 5\end{bmatrix} \Rightarrow det(A-\lambda I)=0 \Rightarrow \lambda^3+ \lambda^2-5\lambda+ 3=0 \Rightarrow (\lambda-1)^2(\lambda+3)=0\\ \Rightarrow A的3個特徵值為\bbox[red,2pt]{1,1,-3}$$ (二) $$\lambda^3+ \lambda^2-5\lambda+ 3=0 \Rightarrow  A^3+A^2-5A+3I=0 \Rightarrow \cases{A^3+A^2=5A-3I \cdots(1)\\ A^5+A^4-5A^3+3A^2=0 }\\ \Rightarrow A^5+A^4=5A^3-3A^2\cdots(2) \\因此欲求之A^5+A^4-4A^3+4A^2-4A+4I  = 5A^3-3A^2-4A^3+4A^2-4A+4I(式(2)代入)\\ =A^3+A^2-4A+4I =5A-3I-4A+4I(式(1)代入)=A+I\\ = \bbox[red,2pt]{\begin{bmatrix} 6 & -4 & 4\\ 12 & -10 & 12\\ 4 & -4 & 6\end{bmatrix}}$$

解答： $$\Gamma(t)=3e^{it} \Rightarrow \Gamma'(t) = 3ie^{it}dt \Rightarrow \int_{\Gamma}Im(z)dz = \int_0^{\pi/2} 3\cdot {e^{it}-e^{-it}\over 2i}\cdot 3ie^{it}\;dt\\ = {9\over 2}\int_0^{\pi/2}(e^{2it}-1)\;dt ={9\over 2}\left . \left[ {1\over 2i}e^{2it}-t \right] \right|_0^{\pi/2} ={9\over 2}(i-{\pi\over 2}) =\bbox[red,2pt]{{9\over 2}i-{9\over 4}\pi}$$

解答：

(一) $$\mathcal{F}\{f_1(t)\} =\int_{-\infty}^\infty e^{-5|t|}\cdot e^{-j\omega t}\;dt  =\int_{-\infty}^0 e^{5t}\cdot e^{-j\omega t}\;dt + \int_{0}^\infty e^{-5t}\cdot e^{-j\omega t}\;dt \\ =\int_{-\infty}^0   e^{(5-j\omega) t}\;dt + \int_{0}^\infty   e^{-(5+j\omega) t}\;dt = \left. \left[{1\over 5-j\omega} e^{(5-j\omega) t}\right]\right|_{-\infty}^0 +\left. \left[ {1\over -(5+j\omega)} e^{-(5+j\omega) t} \right]\right|_{0}^\infty \\ ={1\over 5-j\omega}+{1\over 5+j\omega} = {5+j\omega\over 25+\omega^2}+{5-j\omega\over 25+ \omega^2} = \bbox[red,2pt]{10\over \omega^2+25}$$ (二) $$\cos(100t)= {1\over 2}(e^{j100t} +e^{-j100t}) \Rightarrow  \mathcal{F}\{f_2(t)\} =\int_{-\infty}^\infty (\cos 100t)e^{-5|t|}\cdot e^{-j\omega t}\;dt \\ \int_{-\infty}^0 (\cos 100t)e^{5t}\cdot e^{-j\omega t}\;dt + \int_{0}^\infty (\cos 100t)e^{-5t}\cdot e^{-j\omega t}\;dt \\ = {1\over 2}\int_{-\infty}^0 e^{(5-j\omega+j100)t} +e^{(5-j\omega-j100)t}\;dt +{1\over 2}\int_{0}^\infty e^{(-5-j\omega+j100)t} +e^{(-5-j\omega-j100)t}\;dt\\  = {1\over 2}\left. \left[{1\over 5-j\omega+j100} e^{(5-j\omega+j100) t} +{1\over 5-j\omega-j100} e^{(5-j\omega-j100) t}\right]\right|_{-\infty}^0 \\\qquad +{1\over 2}\left. \left[ {1\over -5-j\omega+ j100} e^{(-5-j\omega+j100) t} +{1\over -5-j\omega- j100} e^{(-5-j\omega-j100) t}\right]\right|_{0}^\infty \\ = {1\over 2}  \left[{1\over 5-j\omega+j100}   +{1\over 5-j\omega-j100}  \right] +{1\over 2}  \left[ {-1\over -5-j\omega+ j100}   +{-1\over -5-j\omega- j100}  \right] \\={1\over 2}  \left[{1\over 5-j(\omega-100)}   +{1\over 5-j(\omega+100)}  \right] +{1\over 2}  \left[ {1\over 5+j(\omega- 100)}   +{1\over 5+j(\omega+100)}  \right]\\ ={1\over 2}\left[{5+j(\omega-100) \over 25+(\omega-100)^2} + {5+j(\omega+100) \over 25+(\omega+100)^2} + {5-j(\omega-100) \over 25+(\omega-100)^2} +{5-j(\omega+100) \over 25+(\omega+100)^2} \right] \\ =\bbox[red,2pt]{{5  \over 25+(\omega-100)^2}+{5  \over 25+(\omega+100)^2}}$$

解答： $$z(x,y)=3500-2x^2-3y^2 \Rightarrow \cases{\left.{\partial z\over \partial x}\right|_{(-3,2)}= \left. -4x\right|_{(-3,2)} =12\\ \left.{\partial z\over \partial y}\right|_{(-3,2)}= \left. -6y\right|_{(-3,2)}=-12} \Rightarrow (12,-12)，故選\bbox[red,2pt]{(C)}$$

解答： $$\begin{Vmatrix} 2 & 0 & 3\\ 0 & 4 & 1\\ 5 & 6 & 0\end{Vmatrix} =|-60-12|=72，故選\bbox[red,2pt]{(B)}$$

解答： $$u_1= {u\cdot v\over |v|^2}v ={5\over 10}\begin{bmatrix} 0 \\1\\3 \end{bmatrix} =\begin{bmatrix} 0 \\0.5\\1.5 \end{bmatrix} \Rightarrow u_2=u-u_1=\begin{bmatrix} 1 \\2\\1 \end{bmatrix}-\begin{bmatrix} 0 \\0.5\\1.5 \end{bmatrix} =\begin{bmatrix} 1 \\1.5\\-0.5 \end{bmatrix}，故選\bbox[red,2pt]{(B)}$$

解答： $$A=\begin{bmatrix} 1 & 0 & 1 & 1\\ 2 & 0 & 2 & 2\\ 0 & 1 & 0 & 3\end{bmatrix} \Rightarrow A^TA=\begin{bmatrix} 1 & 0 & 1 & 1\\ 2 & 0 & 2 & 2\\ 0 & 1 & 0 & 3\end{bmatrix}  \begin{bmatrix} 1 & 2 & 0\\ 0 & 0 & 1\\ 1 & 2 & 0 \\ 1& 2 & 3\end{bmatrix} =\begin{bmatrix} 5 & 0 & 5 & 5\\ 0 & 1 &0 & 3\\ 5 & 0 & 5 & 5\\5& 3 & 5 & 14\end{bmatrix} \\ \xrightarrow{-r_1+r_3,-r_1-3r_2+r_4}\begin{bmatrix} 5 & 0 & 5 & 5\\ 0 & 1 &0 & 3\\ 0 & 0 & 0 & 0\\0& 0 & 0 & 0\end{bmatrix} \Rightarrow \text{rank }=2，故選\bbox[red,2pt]{(B)}$$

解答： $$X\begin{bmatrix} 1 & 0 & 0 & 0\\ 2 & -1 & -1 & 1\\ 0 & 0 & -9 & 9 \\ 3 & 0 &-5 & 5\end{bmatrix} \Rightarrow det(X-\lambda I)=0 \Rightarrow \lambda^4+4 \lambda^3-\lambda^2-4\lambda =\lambda(\lambda^3+4 \lambda^2-\lambda-4)\\= \lambda(\lambda-1) (\lambda^2+5\lambda+4) = \lambda(\lambda-1) (\lambda+1) (\lambda+4)=0\\ \Rightarrow \lambda=0,1,-1,-4，故選\bbox[red,2pt]{(B)}$$

解答： $$(A)\times: \sin(x-y)非線性轉換\\(B)\times: \cases{T(1,1,1,1,1)=(-1,2,1,0,1)\\ T(2,2,2,2,2)=(-2,4,2,0,1)} \Rightarrow T(2,2,2,2,2) \ne 2T(1,1,1,1,1)\\ (C)\times: xy非線性轉換\\(D)\bigcirc: 均為線性運算\\，故選\bbox[red,2pt]{(D)}$$

解答： $$A=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} \Rightarrow e^A =\begin{bmatrix} e & 0 & 0\\ 0 & e & 0 \\ 0 & 0 & 1\end{bmatrix} \Rightarrow \text{rank}(e^A)=3，故選\bbox[red,2pt]{(C)}$$

解答： $$y=\sum_{n=0}^\infty a_n(x-1)^n = a_0+a_1(x-1) +a_2(x-1)^2+ a_3(x-1)^3\cdots\\ \Rightarrow \cases{x^2y = a_0x^2 + a_1x^2(x-1)+ a_2x^2(x-1)^2 + a_3x^2(x-1)^3+ \cdots\\y'=a_1+ 2a_2(x-1)+ 3a_3(x-1)^2+ \cdots} \\ 初始值y(1)=1 \Rightarrow a_0=1；\\又\left.(y'+x^2y) \right|_{x=1} =\cos(\pi)=-1 \Rightarrow a_0+a_1=-1 \Rightarrow a_1=-2\\又\left.(y'+x^2y) \right|_{x=0} =y'(0)=\cos(0)=1 \Rightarrow a_1-2a_2+3a_3-4a_4+\cdots=1\\ \left.(y'+x^2y) \right|_{x=2} =y'(2)+4y=\cos(2\pi)=1 \Rightarrow \sum_{n=0}^\infty na_n +\sum_{n=0}^\infty 4a_n+，故選\bbox[red,2pt]{(C)}$$

解答： $$令非齊次解y_1=C,C為常數\Rightarrow {dy_1\over dx} ={1\over 3x}y_1^2 +{1\over x}y_1-{6\over x} \Rightarrow 0={C^2\over 3x}+{C\over x}-{6\over x}\\ \Rightarrow C^2+3C-18=0 \Rightarrow (C+6)(C-3)=0 取C=3 \Rightarrow y=y_1+{1\over u} =3+{1\over u}，故選\bbox[red,2pt]{(B)}$$

解答： $$y(x)= (a+bx)e^{2x}+ ce^{-2x}+d \Rightarrow \cases{y'(x)=  (2a+b+ 2bx)e^{2x}-2ce^{-2x} \\ y''(x)= (4a+4b+ 4bx)e^{2x} +4ce^{-2x}} \\ \Rightarrow y''-2y'=2be^{2x} +8ce^{-2x} =6e^{2x} -4e^{-2x} \Rightarrow \cases{b=3\\ c=-1/2}\\ 又初始值\cases{y(0)=-1 \\ y'(0)=6} \Rightarrow \cases{a+c+d=-1 \Rightarrow d=-3/2\\ 2a+b-2c=6 \Rightarrow a=1} \Rightarrow \cases{a=1\\ b=3\\ c=-1/2\\ d=-3/2}\\，故選\bbox[red,2pt]{(A)}$$

解答： $$\int_{-\infty}^\infty f(t)\delta(t-T)\;dt = f(T) \Rightarrow \mathscr{L} \{f(t)\} = \int_0^\infty \cos(\pi t) \delta(t-1)e^{-st}\;dt =\cos(\pi)e^{-s}= -e^{-s}\\，故選\bbox[red,2pt]{(A)}$$

解答： $$\mathscr{L}\{f(t)\} =\mathscr{L}\{Ae^{-t}\cos(2t) +Be^{-t}\sin(2t)\} =A\mathscr{L}\{e^{-t}\cos(2t)\} +B\mathscr{L}\{e^{-t}\sin(2t)\} \\ =A\cdot {s+1\over (s+1)^2+4}+B\cdot {2\over (s+1)^2+4} ={As+A+2B\over s^2+2s+ 5}= {2s+5\over s^2+as+b} \\\Rightarrow \cases{a=2\\ b=5} \Rightarrow a/b=2/5=0.4，故選\bbox[red,2pt]{(D)}$$

解答： $$\\f(t)=\begin{cases} 0 & \text{for }0\le t\lt 4\\ 3 & \text{for }t\ge 4 \end{cases} \Rightarrow f(t)=3H(t-4) \Rightarrow y''+4y=3H(t-4) \\ \mathscr{L}\{y''\}= s^2Y(s)-sy(0)-y'(0) = s^2Y(s)-s \Rightarrow \mathscr{L}\{y''+4y\} = \mathscr{L}\{f(t)\} \\ \Rightarrow  (s^2+4)Y(s) -s=3\cdot {1\over s}e^{-4s } \Rightarrow Y(s)={s\over s^2+4} +{3\over s(s^2+4)}e^{-4s} = {s\over s^2+4} +{3\over 4}\left({1\over s}-{s\over s^2+4}\right)e^{-4s}\\ \Rightarrow y(t)=\cos(2t)+ {3\over 4}\left(1-\cos(2(t-4)) \right)H(t-4)  =  \cos(at) +b(c-\cos(d(t-4)))H(t-4) \\\Rightarrow \cases{a=2\\ b=3/4 \ne 3/2\\ c=1\\ d=2}，故選\bbox[red,2pt]{(B)}$$

解答： $$f(t)=5\left[H(t-3)-H(t-11) \right] = x(t-7),x(t)=5\cdot \text{rect}(t/8)\\ \mathcal{F}\{x(t)\}={5\over \omega}\cdot 2\cdot \sin ({\omega\cdot 8\over 2}) ={10\over \omega}\sin(4\omega) \Rightarrow \mathcal{F}\{x(t-7)\} ={10\over \omega}e^{-j 7\omega }\sin(4\omega)\\ \Rightarrow \cases{a=10\\ b=1\\ c=7\ne -7\\ d=4}，故選\bbox[red,2pt]{(C)}$$

解答： $$Res(f,0) = \lim_{z\to 0}zf(z)= \lim_{z\to 0}{z(z+i)\over \sin (z)}= \lim_{z\to 0}{2z+i\over \cos (z)} ={i\over 1}=i，故選\bbox[red,2pt]{(D)}$$

解答： $$\cases{f(z)=\sin(z)/(z^2(z^2+1))\\ C:|z-0.5i|=1} \Rightarrow z=0,z=i 皆在C內部\\ \Rightarrow \cases{Res(f,0)=\lim_{z\to 0}{d\over dz}z^2f(z)  =\lim_{z\to 0}{d\over dz} \sin(z)/(z^2+1)=1\\ Res(f,i)= \lim_{z\to i}(z-i)f(z) =\lim_{z\to i} \sin(z)/(z^2(z+i)) =i\sin(i)/2} \\ \Rightarrow \oint_C f(z)\;dz=2\pi i(Res(f,0)+Res(f,i)) =2\pi i(1+i\sin(i)/2)= 2\pi i-\pi \sin(i)，故選\bbox[red,2pt]{(A)}$$

解答： $$C_n=\left( {2i\over 5+i}\right)^n \Rightarrow {1\over R}=\left| {C_{n+1}\over C_n}\right| =\left| {2i\over 5+i}\right| =\left| {2+10i\over 26}\right| = \sqrt{\left({1\over 13} \right)^2 +\left({5\over 13} \right)^2} ={\sqrt{26}\over 13}\\ \Rightarrow 收斂半徑R={13\over \sqrt{26}} ={\sqrt{26}\over 2}，故選\bbox[red,2pt]{(B)}$$

解答： $$\cases{E(X)= \int_{-a}^a {x\over 2a} \;dx = 0\\ E(X^2) =\int_{-a}^a {x^2\over 2a} \;dx={a^2\over 3}} \Rightarrow Var(X)= E(X^2)-(E(X))^2 = {a^2\over 3}=4 \Rightarrow a=2\sqrt 3，故選\bbox[red,2pt]{(A)}$$

解答： $$Z={X-\mu\over \sigma} ={X-1\over 2}，故選\bbox[red,2pt]{(C)}$$

解答： $$X\sim B(n=4,p=1/2)\\(A)\bigcirc: P(X=3)=C^4_3\cdot {1\over 16}=0.25\\ (B)\bigcirc: P(X=2)=C^4_2\cdot {1\over 16}={3\over 8}=0.375\\ (C)\bigcirc: E(X)=np=4\cdot {1\over 2}=2\\ (D)\times: Var(X)= np(1-p)= 4\cdot {1\over 2} \cdot {1\over 2}=1 \ne 1.5\\，故選\bbox[red,2pt]{(D)}$$

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