110年地方特考-工程數學詳解

110 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：$$\cases{x_1-x_2+x_3=1\\ 2x_1+x_2+ 2x_3= 2\\ 3x_1+2x_2 -x_3=-3} \Rightarrow \begin{bmatrix} 1& -1 & 1\\ 2 & 1 & 2\\ 3 & 2& -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}= \begin{bmatrix}1 \\ 2\\ -3\end{bmatrix} \equiv Ax=b\\ \left[\begin{array}{rrr| rrr} 1& -1 & 1 & 1   & 0 & 0\\ 2 & 1 & 2 & 0 & 1 & 0\\ 3 & 2 & -1 & 0 & 0 & 1 \end{array} \right] \xrightarrow{-2r_1+r_2,-3r_1+r_3}\left[\begin{array}{rrr| rrr} 1& -1 & 1 & 1   & 0 & 0\\ 0 & 3 & 0 & -2 & 1 & 0\\ 0 & 5 & -4 & -3 & 0 & 1 \end{array} \right]\\ \xrightarrow{r_2/3,r_3/5}\left[\begin{array}{rrr| rrr} 1& -1 & 1 & 1   & 0 & 0\\ 0 & 1 & 0 & -2/3 & 1/3 & 0\\ 0 & 1 & -4/5 & -3/5 & 0 & 1/5 \end{array} \right] \xrightarrow{r_2+r_1,-r_2+r_3} \left[\begin{array}{rrr| rrr} 1& 0 & 1 & 1/3   & 1/3 & 0\\ 0 & 1 & 0 & -2/3 & 1/3 & 0\\ 0 & 0 & -4/5 & 1/15 & -1/3 & 1/5 \end{array} \right] \\ \xrightarrow{ -5/4\cdot r_3}\left[\begin{array}{rrr| rrr} 1& 0 & 1 & 1/3   & 1/3 & 0\\ 0 & 1 & 0 & -2/3 & 1/3 & 0\\ 0 & 0 & 1 & -1/12 & 5/12 & -1/4 \end{array} \right] \xrightarrow{- r_3+r_1 } \left[\begin{array}{rrr| rrr} 1& 0 & 0 & 5/12   & -1/12 & 1/4\\ 0 & 1 & 0 & -2/3 & 1/3 & 0\\ 0 & 0 & 1 & -1/12 & 5/12 & -1/4 \end{array} \right] \\ \Rightarrow A^{-1}=\begin{bmatrix} 5/12   & -1/12 & 1/4\\ -2/3 & 1/3 & 0\\ -1/12 & 5/12 & -1/4\end{bmatrix} \Rightarrow x=A^{-1}b=\begin{bmatrix} -1/2 \\ 0\\ 3/2\end{bmatrix} \Rightarrow \bbox[red,2pt]{\cases{x_1=-1/2\\ x_2=0\\ x_3=3/2}}$$

解答：$$f(z)={-3z+5\over z(z^2-3z+2)} = {-3z+5\over z(z-1)(z-2) } \Rightarrow \cases{Res(f,z=0)= \left. {-3z+5\over (z-1)(z-2) }\right|_{z=0} ={5\over 2} \\ Res(f,z=1)= \left. {-3z+5\over z (z-2) }\right|_{z=1} =-2} \\ \Rightarrow \oint_c f(z)dz =2\pi i(Res(f,z=0)+ Res(f,z=1)) =2\pi i({5\over 2}-2) = \bbox[red,2pt]{\pi i}$$

解答：$$a_0={1\over 2\pi}\int_{-\pi}^\pi x+\pi \;dx ={1\over 2\pi} \cdot 2\pi^2 = \pi\\ a_n={1\over \pi}\int_{-\pi}^\pi (x+\pi)\cos (nx) \;dx=0\\ b_n= {1\over \pi}\int_{-\pi}^\pi (x+\pi)\sin (nx) \;dx={1\over \pi }\times {-2\pi\over n}(-1)^n = -{2\over n}(-1)^n \\ \Rightarrow f(x)=x+\pi =\pi +\sum_{n=1}^\infty -{2\over n}(-1)^n \sin(nx)\\ =\pi +2(\sin(x)-{1\over 2}\sin(2x)+{1\over 3}\sin(3x)-{1\over 4}\sin(4x)+\cdots) \\ \Rightarrow {x\over 2}=\sin(x)-{1\over 2}\sin(2x)+{1\over 3}\sin(3x)-{1\over 4}\sin(4x)+\cdots\\ 取x={\pi \over 2} \Rightarrow {\pi \over 4}=1+0-{1\over 3}+0+{1\over 5}+ 0-{1\over 7}+\cdots \\ \Rightarrow {\pi \over 4}=1-{1\over 3}+{1\over 5}-{1\over 7}+\cdots,\bbox[red,2pt]{故得證}$$

解答：$$假設生產總量為N\Rightarrow \cases{A機器產量為0.25N，其中瑕疵品數量為0.25N\times 0.05\\ B機器產量為0.35N，其中瑕疵品數量為0.35N\times 0.03 \\ C機器產量為0.3N，其中瑕疵品數量為0.3N\times 0.04 } \\ \Rightarrow 瑕疵品數量為0.25N\times 0.05+ 0.35N\times 0.03+ 0.3N\times 0.04= 0.035N\\ \Rightarrow 取出一瑕疵品，\cases{此瑕疵品為A機器出產的機率為 0.25N\times 0.05/0.035N = 5/14\\此瑕疵品為B機器出產的機率為 0.35N\times 0.03/0.035N =3/10\\此瑕疵品為C機器出產的機率為 0.3N\times 0.04/0.035N =12/35}\\ \Rightarrow 取出一瑕疵品，此瑕疵品為A、B、C機器出產的機率各是\bbox[red,2pt]{{5\over 14},{3\over 10}, {12\over 35}}$$

乙、 測驗題部分： （50分）

解答：$$k_1p_1+k_2p_2 +k_3p_3=0 \Rightarrow (k_1+2k_2+k_3)x^2 +(-2k_1+ak_2+k_3)x+ (k_1-k_2+bk_3)=0 \\ \Rightarrow \cases{k_1+2k_2+k_3=0 \\-2k_1+ak_2+k_3=0 \\k_1-k_2+bk_3=0} \Rightarrow A=\begin{bmatrix} 1 & 2 & 1\\ -2& a & 1\\ 1 & -1 & b\end{bmatrix} \Rightarrow f(a,b)=\det(A)=ab-a+4b+5\\ \Rightarrow  \cases{f(2,1/2)=6 \ne 0  \\f(-1,-2)= f(1,-4/5) =f(0,-5/4)=0}，故選\bbox[red,2pt]{(A)}$$

解答：$$T=\begin{bmatrix} 1 & -\sqrt 3\\ \sqrt 3 & 1\end{bmatrix} =2\begin{bmatrix} \cos (\pi/3) & -\sin(\pi/3)\\ \sin(\pi/3) & \cos (\pi/3) \end{bmatrix} \Rightarrow \cases{r=2\\ \theta=60^\circ}，故選\bbox[red,2pt]{(C)}$$

解答：$$\det(A)=2-4=-2 \Rightarrow \det(A^5)=(-2)^5=-32，故選\bbox[red,2pt]{(D)}$$

解答：$$  v=\begin{bmatrix} 1\\1\\1\end{bmatrix} \Rightarrow Av = 1\cdot v \Rightarrow v是特徵向量，故選\bbox[red,2pt]{(A)}$$

解答：$$z=x+iy \Rightarrow \bar z=x-iy \Rightarrow (\bar z)^2 = (x^2-y^2)-2xy i \Rightarrow e^{(\bar z)^2} =e^{x^2-y^2}\cdot e^{-2xyi} \\ =e^{x^2-y^2}(\cos(-2xy)+i\sin(-2xy))   =e^{x^2-y^2}(\cos(2xy)-i\sin(2xy)) \\ \Rightarrow e^{(\bar z)^2} 的虛部為-e^{x^2-y^2}\sin(2xy)，故選\bbox[red,2pt]{(B)}$$

解答：$$\det(A-\lambda I)= \begin{vmatrix} 1-\lambda & 2 & 1\\ a & -\lambda & 2\\ 1 & 3 & b-\lambda\end{vmatrix} = -\lambda^3+(b+1)\lambda^2 +(2a-b+7)\lambda+3a-2ab-2\\ \Rightarrow \cases{b+1=-c\\ 2a-b+7=-3\\ 3a-2ab-2=2} \Rightarrow \cases{\cases{a=-1/4\\ b=19/2\\ c=-21/2} \Rightarrow 4a+b+c=-2\\\cases{a=-4\\ b=2\\ c=-3}\Rightarrow 4a+b+c=-17} ，故選\bbox[red,2pt]{(D)}$$

解答：$$由泰勒級數f(x)=\sum_{n=0}^\infty {f^{[n]}(\pi)\over n!} (x-\pi)^n \Rightarrow \cos(x)=-1+{1\over 2}(x-\pi)^2 +\cdots \\y=a_0+ a_1(x-\pi) +a_2(x-\pi)^2 +a_3(x-\pi)^3+\cdots \\ \Rightarrow y'=a_1+ 2a_2(x-\pi)+ 3a_3(x-\pi)^2 + 4a_4(x-\pi)^3+\cdots\\ \Rightarrow y'+\cos(x) y= (a_1-a_0)+ (-a_1+2a_2)(x-\pi) +({1\over 2}a_0-a_2+ 3a_3)(x-\pi)^2 +\cdots=1\\ 由於y(\pi)= A=0=a_0 \Rightarrow \cases{a_1-a_0=1\\ -a_1+2a_2=0\\ {1\over 2}a_0-a_2+ 3a_3=0} \Rightarrow a_1=1 \Rightarrow a_2=1/2 \Rightarrow a_3=1/6，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)=e^{-2x} \Rightarrow \cases{f(-1)=e^2=A\\ f(0)=1=B\\ f(1)=e^{-2}=C} \Rightarrow A+B+C=e^2+e^{-2}+1，故選\bbox[red,2pt]{(D)}$$

解答：

$${梯形AFDE 面積\over \triangle ABC 面積} ={3/2\over 6/2} ={1\over 2}，故選\bbox[red,2pt]{(C)}$$

解答：$$\int \int f_{XY}(x,y) \;dxdy =1 \Rightarrow \int_0^2 \int_0^3 A(x+y) dydx =\int_0^2 A(3x+{9\over 2})dx =15A=1\Rightarrow A =1/15 \\ \Rightarrow f_Y(y)= \int f_{XY}(x,y)\;dx =\int_0^2 {1\over 15}(x+y)\;dx={1\over 15}(2+2y)，故選\bbox[red,2pt]{(B)}$$

解答：$$A=\left[\begin{matrix}1 & 2 & 2& 3\\2 & 5 & 4 & 8\\-1 & -3 & -2& -5\\ 0 & 2 & 0 &4\end{matrix}\right] \xrightarrow{-2r_1+r_2} \left[\begin{matrix}1 & 2 & 2 & 3\\0 & 1 & 0 & 2\\-1 & -3 & -2 & -5\\0 & 2 & 0 & 4\end{matrix}\right] \xrightarrow{r_1+r_3} \left[\begin{matrix}1 & 2 & 2 & 3\\0 & 1 & 0 & 2\\0 & -1 & 0 & -2\\0 & 2 & 0 & 4\end{matrix}\right] \\ \xrightarrow{r_2+r_3,-r_2+r_4}\left[\begin{matrix}1 & 2 & 2 & 3\\0 & 1 & 0 & 2\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right] \Rightarrow \text{rank}(A)=2，故選\bbox[red,2pt]{(B)}$$

解答：$$R^3的基底只需考慮(C)與(D)，其它數量不是3；\\(D)A=\left[\begin{matrix}1 & 2 & 2\\-1 & 2 & 1\\0 & 8 & 6\end{matrix}\right] \xrightarrow{r_1+r_2} \left[\begin{matrix}1 & 2 & 2\\0 & 4 & 3\\0 & 8 & 6\end{matrix}\right] \xrightarrow{-2r_2+r_3} \left[\begin{matrix}1 & 2 & 2\\0 & 4 & 3\\0 & 0 & 0\end{matrix}\right] \Rightarrow Rank(A)=2，非基底\\，故選\bbox[red,2pt]{(C)}$$

解答：$$(A)特徵值為三相異值\Rightarrow Rank(B)=3\\ (C)B與B^T有相同的特徵值且\det(B)=\det(B^T)= 0\cdot 1\cdot 2=0 \Rightarrow \det(B^TB)= 0\\(D) B的特徵值0,1,2 \Rightarrow B+I的特徵值=-1,0,1 \Rightarrow (B+I)^{-1}與B+I有相同特徵值\\只有(B)尚難確定，故選\bbox[red,2pt]{(B)}$$

解答：$$P為投影矩陣\Rightarrow P^T=P=P^2 =P^3=\cdots =P^{123}，故選\bbox[red,2pt]{(B)}$$

解答：$$|z+i| \lt \sqrt[4]{1\over 16}={1\over 2}，故選\bbox[red,2pt]{(A)}$$

解答：$$y''+3y'+2y = f(t) \Rightarrow \mathcal{L}\{ y'' \} +3\mathcal{L}\{ y' \}+2\mathcal{L}\{ y \}   =\mathcal{L}\{ f(t) \} \\ \Rightarrow \mathcal{L}\{ f(t) \}= s^2Y(s)-sy(0)-y'(0)+3(sY(s)-y(0))+2Y(s) =(s^2+3s+2)Y(s) \\ =(s^2+3s+2)\cdot {1\over (x^2+3s+2) (s-2)^2} ={1\over (s-2)^2}  \Rightarrow f(t)=\mathcal{L}^{-1}\{{1\over (s-2)^2}\} =te^{2t}，故選\bbox[red,2pt]{(C)}$$

解答：$$y(t)+\int_0^t (t-\tau)y(\tau)\;d\tau=1 \Rightarrow \mathcal{L}\{ y(t) \} +\mathcal{L}\{ \int_0^t (t-\tau)y(\tau)\;d\tau \} =\mathcal{L}\{ 1 \} \\ \Rightarrow Y(s)+ \mathcal{L}\{ t \} \mathcal{L}\{ y(t) \}={1\over s} \Rightarrow Y(s) +{1\over s^2}Y(s) ={1\over s} \Rightarrow Y(s)={s\over s^2+1} \\ \Rightarrow y(t)=  \mathcal{L}^{-1}\{ {s\over s^2+1} \} =\cos(t)，故選\bbox[red,2pt]{(A)}$$

解答：$$令\cases{M(x,y)=-y\\ N(x,y)= x \\ 積分因子I(x,y)} \\ (A)I=1/x^2 \Rightarrow \cases{IM=-y/x^2 \\ IN=1/x} \Rightarrow \cases{{\partial \over \partial y}IM =-1/x^2\\ {\partial \over \partial x}IN =-1/x^2} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\ (B)I=1/xy  \Rightarrow \cases{IM=-1/x  \\ IN=1/y} \Rightarrow \cases{{\partial \over \partial y}IM =0\\ {\partial \over \partial x}IN =0} \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(C)I=1/(x^2+y^2) \Rightarrow \cases{IM=-y/(x^2+y^2) \\ IN=x/(x^2+y^2)} \Rightarrow \cases{{\partial \over \partial y}IM = (y^2-x^2)/(x^2+y^2)^2 \\ {\partial \over \partial x}IN =(y^2-x^2)/(x^2+y^2)^2} \\\qquad \Rightarrow {\partial \over \partial y}IM ={\partial \over \partial x}IN \\(D) I=xy \Rightarrow \cases{IM=-xy^2 \\ IN=x^2y} \Rightarrow \cases{{\partial \over \partial y}IM =-2xy \\ {\partial \over \partial x}IN =2xy} \Rightarrow {\partial \over \partial y}IM \color{blue}{\ne} {\partial \over \partial x}IN \\，故選\bbox[red,2pt]{(D)}$$

解答：$$X:黑桃牌的次數 \Rightarrow \cases{P(X=0)=({3\over 4})^3 \\P(X=1)=C^3_1({3\over 4})^2({1\over 4}) } \Rightarrow P(X\ge 2) =1-P(X=0)-P(X=1)\\ = 1-({3\over 4})^3 -C^3_1({3\over 4})^2({1\over 4}) =1-{27\over 64}-{ 27 \over 64} ={10\over 64} ={5\over 32}，故選\bbox[red,2pt]{(D)}$$

解答：$$X\sim U[a,b] \Rightarrow \cases{E(X)={a+b\over 2}=0 \\\sigma(X)= {b-a\over \sqrt {12}} =\sqrt {12}} \Rightarrow \cases{a+b= 0\\ b-a=12} \Rightarrow \cases{a=-6\\ b=6}\\ \Rightarrow 區間為[14-a,14+b]=[14:00-0:06,14:00+0:06] =[13:54,14:06]，故選\bbox[red,2pt]{(A)}$$

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