朱式幸福: 112年地方特考-工程數學詳解

112 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：電力工程、電子工程
科目：工程數學

甲、申論題部分：（50分）

解答:

$先求齊次解: y''+4y=0 \Rightarrow \lambda^2+4=0 \Rightarrow \lambda =\pm 2i \Rightarrow y_h=c_1\cos(2x)+ c_2\sin(2x)\\ 令y_p = a\cos x+b\sin x+cx +d \Rightarrow y_p'=-a\sin x+b\cos x+c \Rightarrow y_p''=-a\cos x-b\sin x\\ \Rightarrow y_p''+4y_p=3a\cos x+3b\sin x+4cx+4d = 1+x+\sin x \Rightarrow \cases{a=0\\ b=1/3\\ c=1/4\\ d=1/4} \\ \Rightarrow y_p={1\over 3}\sin x+{1\over 4}x+{1\over 4} \Rightarrow y=y_h+y_p =c_1\cos(2x)+ c_2\sin(2x)+ {1\over 3}\sin x+{1\over 4}x+{1\over 4}\\ \Rightarrow y'=-2c_1\sin (2x)+2c_2\cos(2x)+{1\over 3}\cos x+{1\over 4} \Rightarrow \cases{y(0)=c_1+{1\over 4}=0\\ y'(0)=2c_2+{1\over 3}=0} \\ \Rightarrow \cases{c_1=-1/4\\ c_2=-1/3} \Rightarrow \bbox[red, 2pt]{y= -{1\over 4}\cos(2x)- {1\over 3}\sin(2x)+ {1\over 3}\sin x+{1\over 4}x+{1\over 4}}$

解答:

$\mathbf{(一)}\; A=\begin{bmatrix} 0.9& 0.1\\ 0.2& 0.8 \end{bmatrix} \Rightarrow \det(A-\lambda I)=0 \Rightarrow (\lambda-0.7)(\lambda-1)=0 \Rightarrow 特徵值為\lambda=\bbox[red, 2pt]{0.7及1} \\ \lambda_1=0.7 \Rightarrow (A-\lambda_1 I)v =0 \Rightarrow \begin{bmatrix} 0.2& 0.1\\ 0.2& 0.1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}=0 \Rightarrow v=\begin{bmatrix} -k/2\\ k \end{bmatrix},取v_1= \begin{bmatrix} -1/2\\ 1 \end{bmatrix} \\ \lambda_2=1 \Rightarrow (A-\lambda_2 I)v =0 \Rightarrow \begin{bmatrix} -0.1& 0.1\\ 0.2& -0.2 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix}=0 \Rightarrow v=\begin{bmatrix} k \\ k \end{bmatrix},取v_2= \begin{bmatrix} 1\\ 1 \end{bmatrix}\\ \Rightarrow 相對應的特徵向量為\bbox[red, 2pt]{\begin{bmatrix} -1/２\\ 1 \end{bmatrix}及\begin{bmatrix} 1\\ 1 \end{bmatrix}} \\\mathbf{(二)}\; Av=\lambda v \Rightarrow A(Av)=A(\lambda v)=\lambda Av=\lambda^2 v\\ \Rightarrow A^2的特徵值為\lambda^2,即\bbox[red, 2pt]{0.49,1}, 相對應的特徵向量與A相同,即\bbox[red, 2pt]{\begin{bmatrix} -1/２\\ 1 \end{bmatrix}及\begin{bmatrix} 1\\ 1 \end{bmatrix}}　\\\mathbf{(三)}\; Ａ=PDP^{-1}=\begin{bmatrix} -1/２& 1\\ 1 &1 \end{bmatrix} \begin{bmatrix} 0.7& 0\\0& 1 \end{bmatrix} \begin{bmatrix} -2/3 & 2/3\\ 2/3& 1/3 \end{bmatrix} \Rightarrow A^n= PD^nP^{-1} \\ \Rightarrow \lim_{n\to \infty}A^n =\begin{bmatrix} -1/２& 1\\ 1 &1 \end{bmatrix} \begin{bmatrix} 0& 0\\0& 1 \end{bmatrix} \begin{bmatrix} -2/3 & 2/3\\ 2/3& 1/3 \end{bmatrix} = \bbox[red, 2pt]{\begin{bmatrix} 2/3 & 1/3\\ 2/3& 1/3 \end{bmatrix}}$

解答:

$f(z)={1\over (1+z^2)^2}= {1\over (z+i)^2 (z-i)^2} \text{ has two poles of order 2 at }z= i,-i\\只考慮上半部\Rightarrow Res(f,i)= {d\over dz} \left. {1\over (z+i)^2}\right|_{z=i} =\left. -{2\over (z+i)^3}\right|_{z=i} ={1\over 4i}\\ 又f(-x)=f(x)\Rightarrow f\text{ is even function } \Rightarrow \int_0^\infty {1\over 1+x^2}\,dx ={1\over 2}\int_{-\infty}^\infty {1\over 1+x^2}\,dx \\={1\over 2}\oint_c f(z)\,dz ={1\over 2}\cdot 2\pi i\times Res(f,i) =\bbox[red, 2pt]{\pi \over 4}$

解答:

$f_Z(z)=\begin{cases}\lambda^2 e^{-\lambda z} , & 0\le z\lt \infty\\ 0,& z\lt 0\end{cases} \Rightarrow \int f_z(z)\,dz =1 \Rightarrow \int_0^\infty \lambda^2 e^{-\lambda z}\,dz=1 \\\Rightarrow \left. \left[ -\lambda e^{-\lambda z}\right] \right|_0^\infty =\lambda =1 \Rightarrow f_Z(z) =\begin{cases} e^{- z} , & 0\le z\lt \infty\\ 0,& z\lt 0\end{cases} \\ \mathbf{(一)}\; E(Z)= \int_0^\infty ze^{-z}\,dz = \left. \left[ -ze^{-z}-e^{-z} \right] \right|_0^\infty =\bbox[red, 2pt]1 \\\mathbf{(一)}\; F(Z) =f(Z\lt z)= \int_0^z e^{-x}\,dx =\left. \left[ -e^{-x}\right] \right|_0^z =\bbox[red, 2pt]{1-e^{-z}}$

乙、測驗題部分：（50分）

解答:

$\phi=2x^2y-yz^2 \Rightarrow \nabla\phi =(\phi_x,\phi_y, \phi_z)= (4xy,2x^2-z^2,-2yz) \Rightarrow \nabla \phi(1,-1,2)=(-4,-2,4)\\ \bbox[red, 2pt]{無解},公布的答案是\bbox[cyan, 2pt]{(B)}$

解答:

$(2,-1,0)\cdot ((1,3,-1)\times (3,0,-1)) =(2,-1,0) \cdot (-3,-2,-9) = -4, 故選\bbox[red, 2pt]{(A)}$

解答:

$\oint_c y^2dx+ (xy+x^2)dx \Rightarrow \cases{P=y^2\\ Q=xy+x^2} \Rightarrow u=Q_x-P_y=y+2x-2y=2x-y\\ 所圍面積 = \int_0^1 \int_{x^2} ^x (2x-y)\,dydx ={1\over 10}, 故\bbox[red, 2pt]{無解},公布的答案是\bbox[cyan,2pt]{(C)}$

解答:

$A=\begin{bmatrix}1& 4& 7\\ 2& 5& 8\\ 3& 6 & 9 \end{bmatrix} \Rightarrow rref(A)= \begin{bmatrix}1& 0& -1\\ 0& 1& 2\\ 0& 0 & 0 \end{bmatrix} \Rightarrow 列空間=\{a(1,0,-1)+ b(0,1,2) \mid a,b \in \mathbb R\} \\ \cases{(B) (2,2,2)=2(1,0,-1)+ 2(0,1,2) \\(C) (6,15,24)= 6(1,0,-1)+ 15(0,1,2)\\ (D) (1,2,3)=(1,0,-1)+ 2(0,1,2)} \Rightarrow 只有(A)不在列空間中,故選\bbox[red, 2pt]{(A)}$

解答:

$由題意可知特徵值為2,-1\\ \lambda_1=2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} -6& -6\\3 & 3\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow v=\begin{bmatrix} -k\\ k\end{bmatrix}, 取v_1= \begin{bmatrix} 2\\ -2\end{bmatrix} \\ \lambda_2=-1 \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow \begin{bmatrix} -3& -6\\3 & 6\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow v=\begin{bmatrix} -2k \\ k\end{bmatrix}, 取v_2= \begin{bmatrix} -2\\ 1\end{bmatrix} \\ \Rightarrow P=[v_1\; v_2]= \begin{bmatrix} 2 &-2\\ -2 & 1 \end{bmatrix}故選\bbox[red, 2pt]{(C)}$

解答:

$\det(A)=3 \Rightarrow \cases{\det(A^{-1})=1/3\\ \det(A^T)=3} \Rightarrow \det(3A^T)=3^3\times 3=81\\ \Rightarrow \det(3A^T)+ \det(A^{-1}) = 81+{1\over 3}= {244\over 3} \Rightarrow c=244, 故選\bbox[red, 2pt]{(C)}$

解答:

$A=\begin{bmatrix} 1& 0 & -1\\ 0 & 1& 2\\ -1& 2& 5\end{bmatrix} =PDP^T ,其中P={1\over \sqrt{30}} \begin{bmatrix} \sqrt 5& 2\sqrt 6 & -1\\-2 \sqrt 5 & \sqrt 6& 2\\ \sqrt 5& 0& 5\end{bmatrix},D= \begin{bmatrix} 0& 0 & 0\\ 0 & 1& 0\\ 0& 0& 6\end{bmatrix} \\ \Rightarrow A可正交對角化,但\det(A)=0 ,故選\bbox[red, 2pt]{(D)}$

解答:

$(B) \times: (0,1,1) \in S 但2(0,1,1)=(0,2,2) \not \in S\\ (C)\times: (2,1,0)\in S,但2(2,1,0)=(4,2,0)\not \in S\\ (D)\times: (0,0,0)\not \in S\\ 故選\bbox[red, 2pt]{(A)}$

解答:

$(D) \cases{u(x,y)=e^x\cos y\\ v(x,y)=e^x\sin y} \Rightarrow \cases{u_x=e^x\cos y=v_y\\ u_y=-e^x\sin y=-v_x} \Rightarrow f 可解析\\　故選\bbox[red, 2pt]{(D)}$

解答:

$z^2-z+1+i=0 \Rightarrow z={1\pm \sqrt{-3-4i}\over 2} ={1\pm (-1+2i)\over 2}=i,1-i\\ |z_1| \gt |z_2| \Rightarrow \cases{z_1=1-i\\ z_2=i} \Rightarrow \cases{(A)\times:z_1+z_2=1 \ne 1+2i\\ (B)\bigcirc: z_1-z_2=1-2i \\ (C)\times: |z_1|=\sqrt 2\ne 1\\ (D)\times:|z_2|=1 \ne \sqrt 2}，故選\bbox[red, 2pt]{(B)}$

解答:

$f(z)={z(z-5)(z-i)\over (z+i)^3 (z+1)(z-4)^2} \Rightarrow \cases{零點: z=0,5,i\\ 極點:z=-i,-i,-i,-1,4,4} \Rightarrow \cases{N=2(z=0,i)\\ P=4(z=-i,-i,-i,-1)} \\ \Rightarrow \oint_c {f'(z)\over f(z)} =2\pi i \times(N-P) =2\pi i \times (2-4)=-4\pi i,故選 \bbox[red, 2pt]{(B)}$

解答:

$f(z)={1\over (1+x^2)^3} ={1\over (z+i)^3 (z-i)^3} \Rightarrow Res(f,i)={1\over 2}\cdot \left.{d^2\over dz^2} {1\over (z+i)^3} \right|_{z=i} \\=\left. {6\over (z+i)^5}\right|_{z=i} ={6\over 32i} \Rightarrow \int_{-\infty}^\infty f(x)\,dx = 2\pi i\times Res(f,i)=2\pi i\times {6\over 32 i}={3\over 8}\pi,故選 \bbox[red, 2pt]{(D)}$

解答:

$\cases{P=Ax^2y^2+ By^3\\ Q=2x^3y+ 12xy^2+5} \Rightarrow \cases{P_y=2Ax^2y +3By^2\\ Q_x=6x^2y+12y^2}\\ 正合\Rightarrow P_y=Q_x \Rightarrow \cases{2A=6\\ 3B=12} \Rightarrow \cases{A=3\\ B=4}, 故選\bbox[red, 2pt]{(C)}$

解答:

$(4-y^2)y'=x^2 \ge 0 \Rightarrow y^2\gt 4可能不會有唯一解,即y\lt -2, y\gt 2, 故選\bbox[red, 2pt]{(B)}$

解答:

$a_n= {1\over 3}\int_{-3}^3 f(x)\cos{n\pi x\over 3}\,dx ={2\over 3}\int_{0}^3 x\cos{n\pi x\over 3}\,dx ={6\over n^2\pi^2}((-1)^n-1)\\ \Rightarrow \cases{a_1=-{12\over \pi^2} \\ a_3=-{12\over 9\pi^2} \\ a_5=-{12\over 25\pi^2} \\ a_7=-{12\over 49\pi^2}},故選\bbox[red, 2pt]{(D)}$

解答:

${1\over \sqrt{2\pi}}\int_{-1}^1 e^{-i\omega t}\,dt ={1\over \sqrt{2\pi}}\cdot {1\over -i\omega}(e^{-i\omega}-e^{i\omega}) = {1\over \sqrt{2\pi}}\cdot {1\over -i\omega}\cdot (-2i)\cdot {\sin \omega } \\ ={2\over \sqrt{2\pi}} {\sin \omega \over \omega} =\sqrt{2\over \pi}{\sin \omega \over \omega},故選 \bbox[red, 2pt]{(B)}(題目應該是筆誤)$