109年鐵路特考-工程數學詳解

109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： $${d^2y\over dt^2} +3{dy\over dt}+2y=\delta(t-1) \Rightarrow L\{{d^2y\over dt^2}\} +3L\{{dy\over dt}\} +2L\{y\} =L\{\delta(t-1)\} \\ =s^2Y(s)-sy(0)-y'(0)+3(sY(s)-y(0))+2Y(s)=e^{-s}\\ \Rightarrow s^2Y(s)+3sY(s)+2Y(s)=e^{-s} \Rightarrow Y(s) = {1\over s^2+3s+2}e^{-s} ={1\over (s+2)(s+1)}e^{-s} \\=({1\over s+1} -{1\over s+2})e^{-s} \Rightarrow y(t)=L^{-1}\{ ({1\over s+1} -{1\over s+2})e^{-s}\} =L^{-1}\{{1\over s+1}e^{-s}\}-L^{-1}\{ {1\over s+2}e^{-s}\} \\=f(t-1)u(t-1)-g(t-1)u(t-1),其中\cases{L\{ f(t)\}={1\over s+1} \\L\{g(t)\}={1\over s+2} \\u(\cdot):\text{unit step function}}\\ =e^{-(t-1)}u(t-1)-e^{-2(t-1)}u(t-1)=\bbox[red, 2pt]{(e^{1-t}-e^{2-2t})u(t-1)}$$ 註:本題${dy\over dt}=0$應該是${dy\over dt}(0)=0$!!

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解： $$f(z)={ z\over 1+z^2} = {z\over (z+i)(z-i)} ={g_1(z)\over z-i} + {g_2(z) \over z+i}\\ \Rightarrow Res(f(z))= g_1(i)+ g_2(-i) = {i\over 2i} +{-i\over -2i}=1 \\ \Rightarrow \oint_C f(z) = 2\pi i\times Res(f(z)) =\bbox[red, 2pt]{2\pi i}$$

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解： $$a_0= {1\over 2\pi}\int_{-\pi}^\pi f(x)\;dx ={1\over 8\pi}\int_{-\pi}^\pi x^2\;dx ={1\over 8\pi}\left .\left[ {1\over 3} x^3\right]\right|_{-\pi}^\pi = {1\over 8\pi}\times {2\over 3}\pi^3 ={1\over 12 }\pi^2\\ a_n={1\over \pi}\int_{-\pi}^\pi f(x)\cos (nx)\;dx = {1\over 4\pi}\int_{-\pi}^\pi x^2\cos (nx)\;dx \\={1\over 4\pi}\left. \left[ {x^2\over n}\sin (nx) +{2x\over n^2}\cos(nx)-{2\over n^3} \sin(nx)\right] \right|_{-\pi}^\pi\\ = {1\over 4\pi} \times {4\pi \over n^2}(-1)^n ={1\over n^2}(-1)^n\\ b_n=0 (\because f(x)為偶函數)\\ 因此f(x)的傅立葉級數為a_0+\sum_{n=1}^\infty a_n\cos(nx) =\bbox[red, 2pt]{ {1\over 12 }\pi^2+ \sum_{n=1}^\infty {1\over n^2}(-1)^n\cos(nx)}$$

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解：
$$A=\left[ \matrix{1& 2\\-3 & -4}\right] \Rightarrow det(A-\lambda I)=0 \Rightarrow \left| \matrix{1-\lambda & 2\\ -3 & -4-\lambda}\right|=0 \Rightarrow (\lambda+1)(\lambda+2)=0 \Rightarrow \cases{\lambda_1=-1 \\ \lambda_2=-2}\\\lambda_1=-1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \left[ \matrix{2& 2\\-3 & -3}\right] \left[ \matrix{x_1\\x_2}\right]=0 \Rightarrow x_1+x_2=0,取u_1=\left[ \matrix{-1\\1}\right]\\ \lambda_2=-2 \Rightarrow (A-\lambda_2 I)X=0 \Rightarrow \left[ \matrix{3& 2\\-3 & -2}\right] \left[ \matrix{x_1\\x_2}\right]=0 \Rightarrow 3x_1+2x_2=0,取u_2=\left[ \matrix{2\\ -3}\right]\\ 令P=[u_1 u_2]=\left[ \matrix{-1& 2\\1 & -3}\right] \Rightarrow P^{-1}=\left[ \matrix{-3 & -2\\-1 & -1}\right] \Rightarrow A=P\left[ \matrix{\lambda_1& 0\\ 0 & \lambda_2}\right]P^{-1}\\ \Rightarrow \cases{A^7= P\left[ \matrix{\lambda_1^7& 0\\ 0 & \lambda_2^7}\right]P^{-1} = \left[ \matrix{-1& 2\\1 & -3}\right]\left[ \matrix{-1& 0\\ 0 & -128}\right]\left[ \matrix{-3 & -2\\-1 & -1}\right]= \left[ \matrix{253 & 254\\ -381 & -382}\right] \\A^2= P\left[ \matrix{\lambda_1^2 & 0\\ 0 & \lambda_2^2}\right]P^{-1} =\left[ \matrix{-1& 2\\1 & -3}\right]\left[ \matrix{1& 0\\ 0 & 4}\right]\left[ \matrix{-3 & -2\\-1 & -1}\right] = \left[ \matrix{-5 & -6\\ 9 & 10}\right] } \\ \Rightarrow f(A)=\left[ \matrix{253 & 254\\ -381 & -382}\right]-2\left[ \matrix{-5 & -6\\ 9 & 10}\right] +\left[ \matrix{1& 2\\-3 & -4}\right]-2\left[ \matrix{1& 0\\0 & 1}\right] \\ =\left[ \matrix{253 & 254\\ -381 & -382}\right]+\left[ \matrix{10 & 12\\ -18 & -20}\right] +\left[ \matrix{1& 2\\-3 & -4}\right]+\left[ \matrix{-2& 0\\0 & -2}\right]= \bbox[red,2pt]{\left[ \matrix{262 & 268\\ -402 & -408}\right]}$$

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乙、測驗題部分：(50分)

解： $$\cases{\vec u=(3,-2,-5) \\ \vec v=(1,4,-4) \\ \vec w=(0,3,2)} \Rightarrow \vec  u\cdot (\vec v\times \vec w)=(3,-2,-5)\cdot (1,4,-4)\times (0,3,2) = (3,-2,-5)\cdot(20, -2,3)\\ = 60+4-15= 49，故選\bbox[red,2pt]{(A)}$$

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解： $$(D)\;\langle\mathbf{u},\mathbf{v}\rangle = \parallel\mathbf{u}\parallel\parallel\mathbf{v} \parallel \cos \theta，故選\bbox[red,2pt]{(D)}$$

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解：
$$(A)\times: \cases{A=\begin{bmatrix}1 & 1\\ 0 & 0\end{bmatrix} \\ B=\begin{bmatrix}0 & 1\\ 0 & 1\end{bmatrix}} \Rightarrow \cases{AB =\begin{bmatrix}0 & 2\\ 0 & 0\end{bmatrix} \\BA=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix}} \Rightarrow AB\ne BA \\(B) \times: (AB)^T =B^TA^T \ne A^TB^T \\ (D)\times:  \cases{A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix} \\ B=\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}} \Rightarrow A+B=\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} \Rightarrow \cases{det(A+B)=1 \\det(A)=0\\ det(B)=0} \Rightarrow det(A+B)\ne det(A)+det(B)\\，故選\bbox[red,2pt]{(C)}$$

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解： $$只有(D)為實對稱矩陣，即A^T=A，可用正交矩陣對角化，故選\bbox[red,2pt]{(D)}。$$

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解： $$\begin{vmatrix} 4-\lambda & -2 \\ -2 & 1-\lambda\end{vmatrix} =0 \Rightarrow \lambda^2-5\lambda+6=0 \Rightarrow (\lambda-3)(\lambda-2)=0 \Rightarrow \lambda=2,3\\ 只有(D)\begin{bmatrix} -2 & 1 \\ 3 & 5\end{bmatrix}的特徵值不是2與3，故選\bbox[red,2pt]{(D)}。$$

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解：
$$\bbox[red, 2pt]{(試題有疑義)}$$

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解： $$1+i=\sqrt 2({1\over \sqrt 2}+{1\over \sqrt 2}i)= \sqrt 2(\cos({\pi\over 4}\pm 2n\pi) +i\sin({\pi\over 4}\pm 2n\pi)) =\sqrt 2e^{i({\pi\over 4}\pm 2n\pi)} \\ \Rightarrow (1+i)^{1-i} =e^{(1-i)\ln(1+i)} =e^{(1-i)\left(\ln \sqrt 2+\ln e^{i({\pi\over 4}\pm 2n\pi)}\right)} =e^{(1-i)\left(\ln \sqrt 2+i({\pi\over 4}\pm 2n\pi)\right)} \\ = e^{(\ln \sqrt 2+{\pi\over 4} \pm 2n\pi) +i({\pi\over 4} -\ln \sqrt 2\pm 2n\pi)} =\sqrt 2 e^{{\pi\over 4} \pm 2n\pi}\left( \cos ({\pi\over 4}-\ln \sqrt 2)+ i\sin ({\pi\over 4}-\ln \sqrt 2)\right)，故選\bbox[red, 2pt]{(B)}$$

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解： $$f(z)={1\over z^2-1} = {1\over (z+1)(z-1)} ={g(z)\over z-1} \Rightarrow Res_{z=1}f(z)= g(1) ={1\over 2}\\ \Rightarrow \int_C f(z)\;dz =2\pi i\times Res\;f(z) =2\pi i\times {1\over 2}=\pi i，故選\bbox[red, 2pt]{(B)}$$

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解：
$$f(z)={2iz-\cos (z) \over z^3+z} = {2iz-\cos (z) \over z(z+i)(z-i)} = {g(z)\over z+i}\\ \Rightarrow Res_{z=-i}f(z)=g(-i) = {2-\cos(-i) \over -i(-2i)} ={2-\cos(i)\over -2} =-1+{\cos (i)\over 2}，故選\bbox[red,2pt]{(C)}。$$

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解：
$$y(t)=t+\int_0^t y(\tau)\sin(t-\tau)d\tau \Rightarrow L\{y(t)\}=L\{t\} +L\{\int_0^t y(\tau)\sin(t-\tau)d\tau\} =L\{t\} +L\{y(t)* \sin(t)\} \\= L\{t\} +L\{y(t)\} \times L\{\sin(t)\} ={1\over s^2}+L\{y(t)\} \times {1\over 1+s^2} \Rightarrow L\{y(t)\}= {1+s^2 \over s^4} = {1\over s^4}+{1\over s^2} \\ \Rightarrow y(t)=L^{-1}\{{1\over s^4}\} +L^{-1}\{{1\over s^2}\} = {1\over 6}t^3+t，故選\bbox[red,2pt]{(C)}。$$

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解： $$線性不能有y^2，故選\bbox[red,2pt]{(C)}$$

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解： $$y^{(4)}-y=0 \Rightarrow y=c_1e^x+c_2e^{-x}+c_3\sin x+c_4\cos x \\=c_1(\sinh x+\cosh x)+c_2(\cosh x-\sinh x)+c_3\sin x+c_4\cos x \\ = (c_1+c_2)\cosh x+(c_1-c_2)\sinh x+c_3\sin x+c_4\cos x，故選\bbox[red,2pt]{(D)}$$

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解： $$3y'=5x^3-{3y\over x} \Rightarrow 3xy'+3y=5x^4 \Rightarrow (3xy)'=5x^4 \Rightarrow 3xy= \int 5x^4\;dx+c=x^5+c\\ \Rightarrow y={x^4\over 3}+{c\over 3x}，故選\bbox[red, 2pt]{(A)}$$

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解： $$y(x)=a_0+a_1x+a_2x^2 +a_3x^3+\cdots \\ \Rightarrow y'(x)=a_1+2a_2x+3a_3x^2 + 4a_4x^3+\cdots \\ \Rightarrow y''(x)=2a_2+6a_3x+12a_4x^2 + 20a_5x^3+\cdots \\ \Rightarrow xy'(x)=a_1x+2a_2x^2 +3a_3x^3+\cdots \\ \Rightarrow y''+xy'-2y=(2a_2-2a_0)+ (6a_3-a_1)x+12a_4x^2 +\cdots =e^{5x} \\ =1+5x+{5^2\over 2!}x^2 + \cdots \Rightarrow 12a_4={5^2\over 2!} = {25\over 2} \Rightarrow a_4={25\over 24}，故選\bbox[red,2pt]{(C)}$$

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解： $${-1 \over (s-1)(s-2)} ={1\over s-1}- {1\over s-2} \Rightarrow f(t)=L^{-1}\{{1\over s-1}\} -L^{-1}\{{1\over s-2}\} =e^t-e^{2t}，故選\bbox[red,2pt]{(D)}$$

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解：
$$f(t)=|\sin(t)| 為一週期函數，其週期為\pi；\\因此\;L\{f(t)\} ={1\over 1-e^{-s\pi}}\times L\{f_1(t)\},其中f_1(t)=f(t)(u(t)-u(t-\pi)) =\sin(t)(u(t)-u(t-\pi)) \\ =\sin(t)u(t)- \sin(t)u(t-\pi) =\sin(t)u(t)+ \sin(t-\pi)u(t-\pi) \\ \Rightarrow L\{f_1(t)\} =L\{\sin(t)u(t)\}+L\{\sin(t-\pi)u(t-\pi)\} = {1\over 1+s^2}+ {e^{-\pi s} \over 1+s^2} \\ \Rightarrow L\{f(t)\}= {1\over 1-e^{-s\pi}}\times \left( {1\over 1+s^2}+ {e^{-\pi s} \over 1+s^2}\right) ={1+e^{-\pi s} \over 1-e^{-\pi s}}\cdot {1\over 1+s^2}，故選\bbox[red,2pt]{(A)}$$

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解： $$只有(A)正確，故選\bbox[red,2pt]{(A)}$$

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解： $$72=2^3\times 3^2 \Rightarrow 正因數有(3+1)(2+1)=12個；其中大於15的正因數有72,36,24,18，共4個\\，因此機率為4/12=1/3，故選\bbox[red,2pt]{(A)}$$

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解： $$P(X>Y) = f(3,0)+f(3,1)+f(3,2) +f(2,0)+f(2,1) +f(1,0) = (3+4+5+2+3+1)/30\\ =18/30=2/5，故選\bbox[red,2pt]{(A)}。$$

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解： $$\cases{f_X=\int_x^18xy\;dy =4x-4x^3 \\f_Y= \int_0^y8xy\;dx =4y^3} \Rightarrow \cases{EX =\int_0^1 x(4x-4x^3)\;dx =8/15\\ EY=\int _0^14y^4\;dy =4/5}，故選\bbox[red,2pt]{(A)}$$

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解題僅供參考~~

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