108年地方特考-三等-工程數學詳解

108年特種考試地方政府公務人員考試試題

等別：三等考試
類科 ：電子工程、電力工程
科目：工程數學

解：
$$y=c_1e^x+c_2xe^x+x^2e^x \Rightarrow  \begin{cases}y_h=c_1e^x+c_2xe^x\\ y_p=x^2e^x\end{cases} \\y_h=c_1e^x+c_2xe^x \Rightarrow y_h為二次常係數微分方程 \Rightarrow  \begin{cases}y_h'= c_1e^x+c_2e^x+ c_2xe^x\\ y_h''= c_1e^x+2c_2e^x + c_2xe^x\end{cases}  \\\Rightarrow py_h''+qy_h'+ry_h=0 \Rightarrow (p+q+r)c_2xe^x+((p+q+r)c_1+ (2p+q)c_2)e^x=0\\ \Rightarrow  \begin{cases}p+q+r = 0\\ 2p+q= 0\end{cases} \Rightarrow  \begin{cases}q = -2p\\ r=p\end{cases}  \Rightarrow p(y_h''-2y_h'+y_h)=0 \Rightarrow y_h''-2y_h'+y_h=0\\ \Rightarrow 原微分方程為y''-2y'+y=p(x), 將y_p=x^2e^x 代入 \Rightarrow y_p''-2y_p'+y_p\\=2e^x+4xe^x+x^2e^x-(4xe^x+2x^2e^x)+x^2e^x=2e^x=p(x)\\  \Rightarrow 原微分方程為\bbox[red, 2pt]{y''-2y'+y=2e^x}$$

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解： $$f(t) 之週期為2T \Rightarrow L\{f(t)\}= {1 \over 1-e^{-2Ts}}\int_0^{2T}f(t)e^{-st}dt = {1 \over 1-e^{-2Ts}}\int_0^{T}\cos({\pi t\over T})e^{-st}dt \\=  {1 \over 1-e^{-2Ts}}\left. \left[-{e^{-st}\over s^2+(\pi/T)^2}(s\sin {\pi t\over T}+{\pi \over T}\cos{\pi t\over T})\right] \right|_0^T \\= {1 \over 1-e^{-2Ts}} \left[-{e^{-sT}\over s^2+(\pi/T)^2}(-{\pi \over T}) + {1\over s^2+(\pi/T)^2}({\pi \over T}) \right] = {1 \over 1-e^{-2Ts}} \left[{1+e^{-sT}\over s^2+(\pi/T)^2}({\pi \over T})  \right] \\= \bbox[red, 2pt]{{\pi/T\over (1-e^{-sT})(s^2+(\pi/T)^2)}}$$

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解：
$$\cos(z)={e^{iz}+e^{-iz} \over 2} \Rightarrow \cos(3+2i)={e^{i(3+2i)}+e^{-i(3+2i)} \over 2} = {e^{-2+3i}+e^{2-3i} \over 2} \\={1\over 2}\left( e^{-2}e^{3i} +e^2e^{-3i}\right) ={1\over 2}\left( e^{-2}(\cos 3+i\sin 3)) +e^2(\cos (-3)+i\sin(-3))\right) \\= {1\over 2}\left( e^{-2}(\cos 3+i\sin 3)) +e^2(\cos  3-i\sin3)\right) ={\cos 3(e^2+e^{-2})\over 2}+i { \sin 3(e^{-2}-e^2)\over 2}\\ =\cos 3\cosh 2-i\sin 3\sinh 2 =a+ib \Rightarrow \bbox[red, 2pt]{ \begin{cases}a = \cos 3\cosh 2 \\ b=-\sin 3\sinh 2\end{cases} }$$

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解： $$\oint_\gamma f(z)\;dz= \oint_\gamma {z \over (z+2)(z-4i)}\;dz= 2\pi i\times (\text{Res}(f,-2) +\text{Res}(f,4i) ) \\=2\pi i\left(\left. {z\over z-4i}\right|_{z=-2} + \left. {z\over z+2} \right|_{z=4i}\right) =2\pi i\left( {2\over 2+4i} +{4i\over 2+4i}\right)= \bbox[red, 2pt]{2\pi i}$$

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解： $$A= \begin{bmatrix}-1 & -2 \\1 & 4 \\  2& 2 \end{bmatrix}  \Rightarrow A^T=\begin{bmatrix}-1 &1 &2 \\-2 & 4 &2 \end{bmatrix}\\ \Rightarrow  \begin{cases}A^TA= \begin{bmatrix}-1 &1 &2 \\-2 & 4 &2 \end{bmatrix} \begin{bmatrix}-1 & -2 \\1 & 4 \\  2& 2 \end{bmatrix}=\begin{bmatrix}6 &10 \\10 & 24 \end{bmatrix} \\ A^TB= \begin{bmatrix}-1 &1 &2 \\-2 & 4 &2 \end{bmatrix} \begin{bmatrix}3 \\-2 \\  7 \end{bmatrix}= \begin{bmatrix}9 \\0 \end{bmatrix}\end{cases}  \Rightarrow A^TAX=A^TB \\ \Rightarrow \begin{bmatrix}6 &10 \\10 & 24 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}9 \\0 \end{bmatrix} \Rightarrow \begin{cases}6x_1+ 10x_2=9\\ 10x_1+24x_2=0\end{cases} \Rightarrow \begin{cases}x_1=54/11\\ x_2=-45/22\end{cases} \\ \Rightarrow \bbox[red, 2pt]{x=\begin{bmatrix} 54/11 \\ -45/22 \end{bmatrix}}使得||Ax-B||最小$$

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乙、測驗題部分：(50分)

解： $$\begin{cases}\vec u=(1,-1,-1)\\ \vec v=(-3,4,6) \\ \vec w=(-2,-4,2)\end{cases}  \Rightarrow |(\vec u\times \vec v)\cdot \vec w|=|((1,-1,-1)\times (-3,4,6))\cdot (-2,-4,2)| \\=|(-2,-3,1)\cdot (-2,-4,2)| =18，故選\bbox[red,2pt]{(C)}$$

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解： $$A= \begin{bmatrix}1 & -1 &1\\0 & 2 & -1 \\-2 & -2 & 3 \end{bmatrix}  \Rightarrow det (A-\lambda I)=0 \Rightarrow  \begin{vmatrix}1-\lambda & -1 &1\\0 & 2-\lambda & -1 \\-2 & -2 & 3-\lambda \end{vmatrix}=0\\ \Rightarrow (\lambda-3)(\lambda-2)(1-\lambda)-2+2(2-\lambda)-2(1-\lambda)=0 \Rightarrow (\lambda-3)(\lambda-2)(\lambda-1)=0\\ \Rightarrow 特徵值\lambda=1,2,3 \Rightarrow 最大的特徵值為3，\\\text{依 Rayleigh Principle } { x^TAx \over x^Tx}的最大值為最大的特徵值，即\lambda=3，故選\bbox[red,2pt]{(C)}$$

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解： $$A= \begin{bmatrix}5 & 10 &-10\\10 & 5 & -20 \\ 5 & -5 & -10 \end{bmatrix}  \Rightarrow det (A-\lambda I)=0 \Rightarrow \lambda(\lambda+5)(\lambda-5) \Rightarrow \lambda=0,15,-15\\ \lambda=0 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}5 & 10 &-10\\10 & 5 & -20 \\ 5 & -5 & -10 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=0 \Rightarrow  \begin{cases}x=2z \\y=0\end{cases}  \Rightarrow 取u_1=\begin{bmatrix}2 \\ 0 \\ 1 \end{bmatrix}\\ \lambda=15 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}-10 & 10 &-10\\10 & -10 & -20 \\ 5 & -5 & -25 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=0 \Rightarrow  \begin{cases}x=y \\z=0\end{cases}  \Rightarrow 取u_2=\begin{bmatrix}1\\ 1 \\ 0 \end{bmatrix}\\ \lambda=-15 \Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}20 & 10 &-10\\10 & 20& -20 \\ 5 & -5 & 5 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=0 \Rightarrow  \begin{cases}x=0 \\y=z\end{cases}  \Rightarrow 取u_3=\begin{bmatrix}0 \\1 \\ 1 \end{bmatrix}\\  \Rightarrow P=[u_1 u_2 u_3]=\begin{bmatrix}2 & 1 & 0\\0 &1 & 1 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix}a & b & 0\\0 &1 & 1 \\ 1 & 0 & c \end{bmatrix}\Rightarrow  \begin{cases}a=2 \\b=1 \\c=1\end{cases} \Rightarrow a-b-c=0，故選\bbox[red,2pt]{(C)}$$

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解： $$m=n=r \Rightarrow 列向量獨立，有唯一解，故選\bbox[red,2pt]{(B)}$$

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解： $$A= \begin{bmatrix}1 & 3 &0\\3 & 1 & 0 \\ 0 & 0 & -2 \end{bmatrix}  \Rightarrow det (A-\lambda I)=0 \Rightarrow (\lambda-4)(\lambda+2)^2=0  \Rightarrow \lambda=4,-2\\ \lambda=4 \Rightarrow (A-\lambda I)X=0 \Rightarrow  \begin{bmatrix}-3 & 3 &0\\3 & -3 & 0 \\ 0 & 0 & -6 \end{bmatrix} \begin{bmatrix}x \\ y \\ z \end{bmatrix}=0 \Rightarrow  \begin{cases}x=y \\z=0\end{cases}  \Rightarrow 取u_1=\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix}\\ \lambda=-2 \Rightarrow (A-\lambda I)X=0 \Rightarrow  \begin{bmatrix}3 & 3 &0\\3 & 3 & 0 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z \end{bmatrix}=0 \Rightarrow  \begin{cases}x+y=0\end{cases}  \Rightarrow 取u_2=\begin{bmatrix}1\\ -1 \\ 0 \end{bmatrix},u_3=\begin{bmatrix}0\\ 0 \\ 1 \end{bmatrix}\\  \Rightarrow u_1,u_2,u_3為線性獨立的特徵向量，故選\bbox[red,2pt]{(C)}$$

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解： $$A= \begin{bmatrix}0 & -2 \\1 & 3 \end{bmatrix}  \Rightarrow A=PDP^{-1}=\begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & 2 \end{bmatrix} \begin{bmatrix}-1 & -1 \\1 & 2 \end{bmatrix}\\ e^A= \begin{bmatrix}-2 & -1 \\1 & 1 \end{bmatrix}\begin{bmatrix}e & 0 \\ 0 & e^2 \end{bmatrix} \begin{bmatrix}-1 & -1 \\1 & 2 \end{bmatrix}= \begin{bmatrix}2e-e^2 & 2e-2e^2 \\ -e+e^2 & -e+2e^2 \end{bmatrix} =\begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}\\  \Rightarrow a_{11}+a_{22} =2e-e^2-e+2e^2=e^2+e，故選\bbox[red,2pt]{(B)}$$

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解： $$(A)(-1-i)^2=2i \Rightarrow (-1-i)^4=(2i)^2=-4\ne -64\\(B)(-2-i)^2=3+4i \Rightarrow (-2-i)^4= (3+4i)^2 \ne -64\\(C)(-1-2i)^2= -3+4i \Rightarrow (-1-2i)^4=(-3+4i)^2 \ne -64\\(D)(-2-2i)^2 =8i \Rightarrow (-3+4i)^4= (8i)^2=-64\\，故選\bbox[red,2pt]{(D)}$$

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解： $$\ln(1-i\sqrt 3) = \ln(2({1\over 2}-i{\sqrt 3 \over 2})) =\ln (2(\cos ({-\pi \over 3})+i\sin ({-\pi \over 3})))= \ln(2e^{i(-\pi/3)}) \\ =\ln 2+i(-{\pi \over 3}+2n\pi)，故選\bbox[red,2pt]{(A)}$$

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解： $$z=t+it \Rightarrow dt=dt+idt=(1+i)dt \Rightarrow \int_ \varphi z^2\;dz=\int_0^2 (t+it)^2(1+i)\; dt  =(1+i)^3\int_0^2 t^2\; dt \\={8\over 3}(1+i)^3 ={8\over 3}(-2+i2) ={16\over 3}(i-1)，故選\bbox[red,2pt]{(B)}$$

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解： $$z=0不在C內\Rightarrow \oint_C f(z)dz=0 ，故選\bbox[red,2pt]{(A)}$$

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解： $$y=k_1e^{ax}+k_2e^{bx}+e^{cx} \Rightarrow  \begin{cases}y_h=k_1e^{ax}+k_2e^{bx} \\ y_p=e^{cx}\end{cases}  \Rightarrow  \begin{cases}y_h''-6y_h'+8y_h= 0\\ y_p''-6y_p'+8y_p= 3e^x \end{cases} \\y_h''-6y_h'+8y_h= 0 \Rightarrow (a^2-6a+8)k_1e^{ax}+(b^2-6b+8)k_2e^bx=0\\ \Rightarrow  \begin{cases}a^2-6a+8= 0\\b^2-6b+8=0\end{cases} \Rightarrow  \begin{cases}(a-4)(a-2)= 0\\(b-4)(b-2)=0 \\a\ne b\end{cases}  \Rightarrow (a,b)= \begin{cases}(2,4)\\(4,2) \end{cases} \Rightarrow  a+b=6\\ 又y_p''-6y_p'+8y_p= 3e^x \Rightarrow (c^2-6c+8)e^{cx}=3e^x \Rightarrow c=1\\ \Rightarrow a+b+c=6+1=7，故選\bbox[red,2pt]{(C)}$$

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解： $$\int_cf(z)\;dz=\int_c{e^{-z}\over z-(\pi i/2)}\;dz= 2\pi i\times \text{Res}(f,\pi i/2)= 2\pi i \times e^{-\pi i/2}\\=2\pi i(\cos (-\pi/2)+i\sin(-\pi/2)) =2\pi i\times (-i)=2\pi，故選\bbox[red,2pt]{(A)}$$

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解： $$\begin{cases}x_1'=-x_2\\ x_2'=1.01x_1-0.2x_2\end{cases}  \Rightarrow x_2''=1.01x_1'-0.2x_2' = -1.01x_2-0.2x_2' \\\Rightarrow x_2''+0.2x_2'+1.01x_2=0  \Rightarrow 二階常係數齊次解x_2=e^{\alpha t}(A\cos \beta t+B\sin \beta t),其中\alpha=-{0.2\over 2}=-0.1\\ \Rightarrow x_2=e^{-0.1 t}(A\cos \beta t+B\sin \beta t) \Rightarrow \lim_{t\to \infty}x_2=0，故選\bbox[red,2pt]{(B)}$$

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解： $$F(s)={s+1\over s^2(s^2+1)} ={a\over s} +{b\over s^2}+ {cs+d \over s^2+1} ={(a+c)s^3 +(b+d)s^2+as+b \over s^2(s^2+1)}\\ \Rightarrow  \begin{cases}a  = 1\\b =1 \\a+c=0 \\ b+d=0 \end{cases} \Rightarrow  \begin{cases}a  = 1\\b =1 \\c=-1 \\ d=-1 \end{cases}  \Rightarrow F(s)= {1\over s}+ {1\over s^2}+{-s-1\over s^2+1} \\\Rightarrow L^{-1}\{F(s)\} =L^{-1}\{{1\over s}\} +L^{-1}\{{1\over s^2}\} -L^{-1}\{{s\over s^2+1}\}  -L^{-1}\{{1\over s^2+1}\} =1+t-\cos t-\sin t\\，故選\bbox[red,2pt]{(C)}$$

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解： $$題目的\begin{cases}K=a^2y\\ T=a^2x\end{cases}  應該是\begin{cases}K=\alpha^2y\\ T=\alpha^2x\end{cases},\text{直接用代的比較快:} \\ (B)u=c_1e^{\alpha^2y}\cosh 2\alpha x+c_2 e^{\alpha^2y}\sinh 2\alpha x \Rightarrow u_x=2\alpha c_1e^{\alpha^2y}\sinh 2\alpha x+2\alpha c_2e^{\alpha^2y}\cosh 2\alpha x\\ \Rightarrow u_{xx}=4\alpha^2c_1\cosh 2\alpha x+4\alpha^2 c_2e^{\alpha^2y}\sinh 2\alpha x =4u_y，故選\bbox[red,2pt]{(B)}$$

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解：

$${1\over s^2}\left({s-1\over s+1} \right)= {a\over s}+ {b\over s+1} + {c\over s^2} = {(a+b)s^2+(a+c)s+c \over s^2(s+1)} \Rightarrow \begin{cases}a+b=0\\ a+c=1\\ c=-1 \end{cases}\Rightarrow \begin{cases}a=2\\ b=-2\\ c=-1 \end{cases} \\ \Rightarrow \pmb{L^{-1}}\left\{{1\over s^2}\left({s-1\over s+1} \right)\right\}=\pmb{L^{-1}}\left\{2\over s \right\} +\pmb{L^{-1}}\left\{-2\over s+1 \right\} +\pmb{L^{-1}}\left\{-1\over s^2 \right\} =2-2e^{-t}-t，故選\bbox[red,2pt]{(B)}$$

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解： $$\left| X(jw)\right|=2(u(w+3)-u(w-3))= \begin{cases}0 & x \ge 3\\ 2 & -3\le x < 3 \\0 & x<-3 \end{cases}   \Rightarrow \left| X(jw)\right|= \begin{cases} 2 & -3\le x < 3 \\0 & \text{otherwise} \end{cases}\\  \Rightarrow x(t)={1\over 2\pi}\int_{-\infty}^\infty X(jw)e^{jwt}dw ={1\over 2\pi}\int_{-\infty}^\infty \left| X(jw)\right|e^{j\angle X(jw)}e^{jwt}dw  \\ ={1\over 2\pi}\int_{-3}^3 2e^{j(-3w/2+\pi)}e^{jwt}dw ={1\over \pi}e^{j\pi}\int_{-3}^3 e^{jw(-3/2+t)}dw = -{1\over \pi}\int_{-3}^3 e^{jw(-3/2+t)}dw\\ = -{1\over \pi} \left. \left[{1\over j(-3/2+t)} e^{jw(-3/2+t)} \right] \right|_{-3}^3 =-{1\over j(-3/2+t)\pi}\left( e^{j(-9/2+3t)} - e^{j(9/2-3t)}\right)\\ \Rightarrow x(t)=0 \Rightarrow  e^{j(-9/2+3t)} = e^{j(9/2-3t)} \Rightarrow -{9\over 2}+3t={9\over 2}-3t +2\pi \\ \Rightarrow 6t=9+2\pi \Rightarrow t={9\over 6}+{2\pi \over 6}= {3\over 2}+ {\pi \over 3}，故選\bbox[red,2pt]{(C)}$$

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解： $$30顆IC，劣品率為1/6 \Rightarrow 良品為30\times (1-1/6)=25 \Rightarrow 10顆IC都是良品機率: \frac{C^{25}_{10}}{C^{30}_{10}}，故選\bbox[red,2pt]{(B)}$$

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解： $$\begin{cases} E(X)=\int_0^\infty {1\over 4}xe^{-x/4}\;dx = \left. \left[-xe^{-x/4}-4e^{-x/4}  \right] \right|_0^\infty =0-(-4)= 4 \\ E(X^2)= \int_0^\infty {1\over 4}x^2e^{-x/4}\;dx = \left. \left[-x^2e^{-x/4}-8xe^{-x/4} -32e^{-x/4}  \right] \right|_0^\infty =0-(-32)= 32\end{cases} \\\\ \Rightarrow Var(X)=\sigma_X^2= E(X^2)-(E(X))^2 =32-4^2=16 \\\Rightarrow \sigma_Y^2 = Var(Y) = Var(3X-2)=3^2Var(X)= 9\times 16=144，故選\bbox[red,2pt]{(D)}$$

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解： $$E(X)=\int_{-\infty}^\infty xf(x)\;dx =\int_0^1 2x^2\;dx= {2\over 3}，故選\bbox[red,2pt]{(B)}$$

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