111年國安及鐵路特考-工程數學詳解

111年家安全局國家安全情報人員考試及

111年特種考試交通事業鐵路人員考試

考 試 別： 鐵路人員考試、 國家安全情報人員考試
等 別： 高員三級考試、 三等考試
類科組別： 電力工程、 電子工程、 電子組（ 選試英文）
科 目： 工程數學

解答：$$y''+y = \sin 2t \Rightarrow \mathcal{L}\{y''\}+ \mathcal{L}\{ y\}= \mathcal{L}\{ \sin 2t\} \Rightarrow s^2Y(s)-sy(0)-y'(0) +Y(s)= {2\over s^2+2^2}\\ \Rightarrow Y(s)(s^2+1)-1={2\over s^2+4} \Rightarrow Y(s)= {s^2+6\over (s^2+1)(s^2+4)} ={5/3\over s^2+1}-{2/3\over s^2+4}\\ \Rightarrow y=\mathcal{L^{-1}}\{Y(s)\} ={5\over 3}\mathcal{L^{-1}}\{{1\over s^2+1}\} -{1\over 3}\mathcal{L^{-1}}\{{2\over s^2+4}\} ={5\over 3}\sin t-{1\over 3}\sin 2t\\ \Rightarrow \bbox[red,2pt]{y(t)= {5\over 3} \sin t-{1\over 3}\sin 2t}$$

解答：

(一)$$A=\begin{bmatrix} 0 & -2 \\ 1 & 3\end{bmatrix} \Rightarrow \det(A-\lambda I)= \begin{vmatrix}-\lambda & -2\\ 1 & 3-\lambda \end{vmatrix} =0 \Rightarrow \lambda^2-3\lambda +2=0 \Rightarrow (\lambda-1)(\lambda -2)=0 \\ \Rightarrow 特徵值 \lambda = 1,2\\ \lambda_1=1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} -1 & -2 \\ 1 & 2\end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix} =0 \Rightarrow x_1+2x_2=0,取v_1=\begin{bmatrix} -2 \\ 1\end{bmatrix}\\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)X=0 \Rightarrow \begin{bmatrix} -2 & -2 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix} =0 \Rightarrow x_1+x_2=0,取v_2=\begin{bmatrix} -1 \\ 1\end{bmatrix}\\ 因此特徵值為\bbox[red,2pt]{1,2}及其相對應的特徵向量為\bbox[red,2pt]{\begin{bmatrix} -2 \\ 1\end{bmatrix}, \begin{bmatrix} -1 \\ 1\end{bmatrix}}$$(二)$$令p(t)=\det(A-t I)= t^2-3t+2 \Rightarrow p(A)=0=A^2-3A+2I=0\\ \Rightarrow 欲求之\;A^5-2A^4-A^3+2A^2+3A+2I = (A^3+A^2 )(A^2-3A+2 I) +3A+ 2I \\ =3A+2I = \begin{bmatrix} 0 & -6 \\ 3 & 9\end{bmatrix}+\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} =\bbox[red,2pt]{\begin{bmatrix} 2 & -6 \\ 3 & 11\end{bmatrix}}$$(三)$$由(1)可取\cases{P=[v_1 v_2]=\begin{bmatrix} -2 & -1 \\ 1 & 1\end{bmatrix} \Rightarrow P^{-1}=\begin{bmatrix} -1 & -1 \\ 1 & 2\end{bmatrix} \\D=\begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix} =\begin{bmatrix} 1 & 0 \\ 0 & 2\end{bmatrix}}\Rightarrow A=PDP^{-1}\\ \Rightarrow A^n= PD^nP^{-1}= \bbox[red,2pt]{\begin{bmatrix} -2 & -1 \\ 1 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 2^n\end{bmatrix} \begin{bmatrix} -1 & -1 \\ 1 & 2\end{bmatrix}}$$

解答：$$z=\cos \theta+ i\sin \theta = e^{i\theta} \Rightarrow \cases{dz = ie^{i\theta}d\theta =izd\theta \Rightarrow d\theta = {1\over iz }dz \\ \sin \theta ={e^{i\theta}-e^{-i\theta}\over 2i} ={z-1/z\over 2i}}\\ 因此\int_0^{2\pi }{d\theta \over 5+4\sin \theta } =\oint_C \cfrac{1}{5+4\cdot {z-1/z\over 2i}}\cdot \cfrac{1}{iz}dz =\oint_C \cfrac{1}{5iz-2+ 2 z^2}dz =\oint_C \cfrac{1}{(2z+i)(z+2i)}dz\\ =\cfrac{1}{2}\oint_C \cfrac{1}{(z+i/2)(z+2i)}dz ={1\over 2}\cdot 2\pi i\cdot \text{Res}_{z=-i/2} \cfrac{1}{(z+i/2)(z+2i)} =\pi i\cdot \cfrac{1}{-i/2+2i} \\= \pi i \cdot {2\over 3i} =\bbox[red, 2pt]{{2\over 3}\pi}$$

解答：$$(一)Y=3X+2 \Rightarrow E(Y) = E(3X+2) =3E(X)+2 = 3\cdot 0+2 = 2\\ F(y) = P(Y\le y) = P(3X+2\le y) = P(X\le {y-2\over 3}) =\int_{-\infty}^{(y-2)/3} {1\over \sqrt{2\pi} }e^{-x^2/2} \,dx\\ \Rightarrow f(y)= F'(y) = {1\over \sqrt{2\pi} }e^{-((y-2)/3)^2/2} \cdot {1\over 3} ={1\over 3\sqrt{2\pi}} e^{-(y-2)^2/18}\\ \Rightarrow \bbox[red,2pt]{\cases{E(Y)= 2\\ f(y)= {1\over 3\sqrt{2\pi}} e^{-(y-2)^2/18}, -\infty\lt y\lt \infty}}\\(二)Z=X^2 \Rightarrow E(Z)= E(X^2) = Var(X)+(E(X))^2 = 1+0 =1\\ F(z)= P(Z\le z) = P(  X^2\le z) \Rightarrow P(-\sqrt z\le X\le \sqrt z) =\int_{-\sqrt z}^{\sqrt z} {1\over \sqrt{2\pi}} e^{-x^2/2} =\int_{0}^{\sqrt z} {2\over \sqrt{2\pi}} e^{-x^2/2} \\ \Rightarrow f(z)=F'(z) = {2\over \sqrt{2\pi}} e^{-z /2}\cdot {1\over 2\sqrt z} ={1\over \sqrt{2\pi z}}e^{-z/2}\\ \Rightarrow \bbox[red,2pt]{\cases{E(Z)=1 \\ f(z)={1\over \sqrt{2\pi z}}e^{-z/2}, 0\le z\lt \infty}} $$

解答：$$A=\begin{bmatrix} 3 & 1\\ 2 & 4\end{bmatrix} \Rightarrow \det(A)=12-2=10 \Rightarrow A^{-1}=\begin{bmatrix} 4/10 & -1/10\\ -2/10 & 3/10\end{bmatrix} =\begin{bmatrix}0.4 & -0.1\\ -0.2 & 0.3\end{bmatrix} =\begin{bmatrix} a & b\\ c & d \end{bmatrix}\\，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{A=\begin{bmatrix} -1 & 0 & 0 & 0\\ 8 & 1 & 0 & 0\\ 4 & -8 & 1 & 0 \\ 5 & -4 & 8 & -1\end{bmatrix} \Rightarrow \det(A)= (-1)\cdot 1\cdot 1\cdot (-1)=1\\[1ex]   B=\begin{bmatrix} -1 & 1 & -1 & -3\\ 0 & 2 & 2 & -1\\ 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & 4 \end{bmatrix} \Rightarrow \det(B)= (-1)\cdot 2\cdot (-3)\cdot (4)= 24} \\\Rightarrow \det(AB)= \det(A)\det(B) = 24，故選\bbox[red,2pt]{(A)}$$

解答：$$(A) 行列式值=0\\ (B)行列式值= 0\\(C) 行列式值= -54\ne 0\\ (D)行列式值=0\\，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{\vec a=(1,2,3)\\ \vec b=(-4,-5,-6)} \Rightarrow \cases{|\vec a|=\sqrt{1+4+9}=\sqrt{14}\\ |\vec b|= \sqrt{16+25 +36} =\sqrt{77} \\ \vec a\cdot \vec b=-4-10-18=-32} \Rightarrow \cos \theta ={\vec a\cdot \vec b\over |\vec a||\vec b|} =-{32\over 7\sqrt{22}} =-{16\sqrt{22}\over 77}\\，故選\bbox[red,2pt]{(B)}$$

解答：$$T\left[\left(\matrix{x\\ y} \right) \right] =\left(\matrix{ax+by\\ cx+dy} \right) \Rightarrow \cases{T\left[\left(\matrix{1\\ 1} \right) \right] =\left(\matrix{a+b\\ c+d} \right) =\left(\matrix{2\\ 3} \right)\\[1ex] T\left[\left(\matrix{1\\ -1} \right) \right] =\left(\matrix{a-b\\ c-d} \right) =\left(\matrix{4\\ -1} \right)} \Rightarrow \cases{a=3\\ b=-1\\ c=1\\ d=2}，故選\bbox[red,2pt]{(D)}$$

解答：$$A=\begin{bmatrix} -3 & 0 & 0\\ 0 & 4 & 0 \\ 0 & 0 & -1\end{bmatrix} \Rightarrow A^2=\begin{bmatrix} 9 & 0 & 0\\ 0 & 16 & 0 \\ 0 & 0 & 1\end{bmatrix} \Rightarrow A^2的特徵值=9,16,1，故選\bbox[red,2pt]{(D)}$$

解答：$$B=A^2-2A= \begin{bmatrix} -2 & -2 & -3\\ 1 & 1 & 3\\ 0 & 0 & -1\end{bmatrix} \Rightarrow \det(B-\lambda I)=0 \Rightarrow \lambda(\lambda+1)^2 =0 \Rightarrow \lambda= 0,-1\\ \lambda=0 \Rightarrow AX=0 \Rightarrow \cases{x_1+x_2=0\\ x_3=0} \cdots(1)  \\ \lambda=-1 \Rightarrow (A-\lambda I)X=0 \Rightarrow x_2+2x_2+ 3x_3=0 \cdots(2)\\ 而\begin{bmatrix} -1 \\ 1  \\  1\end{bmatrix}不符合(1)也不符合(2)，因此不是A^2-2A的特徵向量，故選\bbox[red,2pt]{(C)}$$

解答：$$(A) \det(A-\lambda I)=0 \Rightarrow 有兩相異特徵值-1,1\Rightarrow 可對角化\\(B)  \det(B-\lambda I)=0 \Rightarrow 只有一特徵值-1 \Rightarrow 無法找到線性獨立的特徵向量\\ (C)\det(C-\lambda I)=0 \Rightarrow 有兩相異特徵值-1,5\Rightarrow 可對角化 \\(D) \det(D-\lambda I)=0 \Rightarrow 有兩相異特徵值-1,3\Rightarrow 可對角化\\，故選\bbox[red,2pt]{(B)}$$

解答：$$Z=X+iY = (1-i)^{20} = (\sqrt 2(\cos(7\pi/4)+i \sin(7\pi/4))^{20} =(\sqrt 2e^{i 7\pi/4})^{20} = 2^{10}e^{i 35\pi} \\=2^{10} e^{i\pi}= -2^{10} \Rightarrow \cases{Z=-2^{10}\\ r= 2^{10} \\ \theta =\pi \ne 0\\ Y=0}，故選\bbox[red,2pt]{(C)}$$

解答：$$(A) \times:\cases{u(x,y)=x^2 \Rightarrow u_x=2x\\ v(x,y)=-y^2 \Rightarrow v_y= -2y} \Rightarrow u_x\ne v_y\\ (B) \times:\cases{u(x,y)=x \Rightarrow u_x=1\\ v(x,y)=-y  \Rightarrow v_y= -1}\Rightarrow u_x\ne v_y \\(C)\times: \cases{u(x,y)=x^2 +y^2-2y \Rightarrow u_y=2y-2\\ v(x,y)=2xy-2x \Rightarrow v_x=  2y-2}\Rightarrow u_y\ne -v_x \\(D)\bigcirc: \cases{u(x,y)=x^2 -y^2-2y \Rightarrow \cases{u_x=2x\\ u_y=-2y-2}\\ v(x,y)=2xy+2x \Rightarrow \cases{v_y = 2x\\ v_x=  2y+2} } \Rightarrow u_x=v_y 且u_y=-v_x\\，故選\bbox[red,2pt]{(D)}$$

解答：$$f(z)= e^{iz} \Rightarrow f''(z)=-e^{iz} \Rightarrow \oint_C {e^{iz}\over z^3} ={2 \pi i\over 2!} f''(0) =-\pi i，故選\bbox[red,2pt]{(A)}$$

解答：$$f(z) = \sum_{k=1}^\infty {1\over z^k} +\sum_{k=0}^\infty ({z\over 3})^k \Rightarrow 收斂區域為\left|{1\over z} \right|\lt 1 且\left|{z\over 3} \right|\lt 1 \Rightarrow 1\lt |z| 且|z|\lt 3 \Rightarrow 1\lt |z|\lt 3\\，故選\bbox[red,2pt]{(A)}$$

解答：$$y''+4y'+3 y=0 \Rightarrow \lambda^2+4\lambda +3=(\lambda+3)(\lambda+1)=0 \Rightarrow \lambda=-1,-3\\ \Rightarrow y=C_1e^{-x} +C_2e^{-3x} \Rightarrow y'=-C_1e^{-x}-3C_2e^{-3x};\\ 將初始值\cases{y(0)=3\\ y'(0)=-5}代入上式\Rightarrow \cases{C_1+C_2 = 3\\ -C_1-3C_2= -5} \Rightarrow \cases{C_1=2\\ C_2=1} \Rightarrow y=2e^{-x}+ e^{-3x}\\，故選\bbox[red,2pt]{(D)}$$

解答：$$取y=(x-2)^m \Rightarrow y'=m(x-2)^{m-1} \Rightarrow y''=m(m-1)x^{m-2}\\ \Rightarrow (x-2)^2y''-5(x-2)y'+8y = m(m-1)x^m-5m(x-2)^m +8(x-2)^m =0\\ \Rightarrow m(m-1)-5m+8 = m^2-6m+8=0 \Rightarrow (m-2)(m-4)=0\\ \Rightarrow m=2,4 \Rightarrow y= c_1(x-2)^2 + c_2(x-2)^4，故選\bbox[red,2pt]{(D)}$$

解答：$$y''+5y'+6y = 2\delta(t-1) \Rightarrow \mathcal{L}\{y'' \} +5\mathcal{L}\{ y'\}+6 \mathcal{L}\{ y\}=2 \mathcal{L}\{\delta(t-1) \} \\ \Rightarrow s^2Y(s)-sy(0)-y'(0) +5(sY(s)-y(0))+6Y(s)= 2e^{-s} \\\Rightarrow s^2Y(s)  +5sY(s) +6Y(s)= 2e^{-s} \Rightarrow Y(s)= \cfrac{2e^{-s}}{s^2+5s+6} =2e^{-s}\left({1\over s+2} -{1\over s+3}\right) \\ \Rightarrow y(t) = \mathcal{L^{-1}}\{Y(s) \} =2\mathcal{L^{-1}}\{{e^{-s}\over s+2} \} -2\mathcal{L^{-1}}\{{e^{-s}\over s+3} \} =2e^{-2(t-1)}-2 e^{-3(t-1)}，故選\bbox[red,2pt]{(B)}$$

解答：$$f(x)=\begin{cases}-1, & -\pi\lt x\lt 0\\ 1, &0\lt x\lt \pi \end{cases} \Rightarrow f(x)為奇函數\Rightarrow f(x)=\sum_{n=1}^\infty b_n\sin (nx)\\ \Rightarrow b_n = {1\over \pi}\int_{-\pi}^\pi f(x)\sin(nx)\,dx ={1\over \pi}\left(\int_{-\pi}^0 -\sin(nx)\,dx + \int_0^\pi \sin(nx)\,dx \right)\\ ={1\over \pi}\left(\left.\left[{1\over n}\cos(nx) \right] \right|_{-\pi}^0 + \left.\left[-{1\over n}\cos (nx) \right] \right|_0^\pi  \right) =  {2\over n\pi} \left(1-(-1)^n \right) \Rightarrow \cases{b_1=4/\pi\\ b_2=0 \\ b_3=4/3\pi\\ b_4=0} \\ \Rightarrow b_1+b_2 +b_3+b_4={16\over 3\pi } \Rightarrow a+b+c+d = {16\over 3\pi}\times {3\pi \over 4} =4，故選\bbox[red,2pt]{(D)}$$

解答：$$f(x)=\cases{1,|x|\lt a\\ 0,其它} \Rightarrow F(\omega)= 2a {\sin(a\omega)\over a\omega}；因此當a=2時，F(\omega)=4\cdot {\sin(2\omega)\over 2\omega} ={2\sin(2\omega)\over \omega}\\，故選\bbox[red,2pt]{(D)}\\如果公式忘記，可直接計算:G(\omega)=\int_{-\infty}^\infty g(x)e^{-j\omega x}dx =\int_{-2}^2 e^{-j\omega x}dx ={1\over -j\omega}\left.\left[ e^{-j\omega x}\right]\right|_{-2}^2 \\={1\over -j\omega}(e^{-j2\omega}- e^{j2\omega}) ={1\over -j\omega}(-2j\sin(2\omega)) =\cfrac{2\sin(2\omega)}{\omega}$$

解答：$${真陽  \over 真陽+偽陽} ={10\% \times 90\%\over 10\% \times 90\% +90\%\times 5\%} ={0.09\over 0.09+ 0.045}={2\over 3}，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{E(X) = 1\cdot {1\over 2}+ 2\cdot {1 \over 4}  =1\\[1ex] E(X^2)=1^2\cdot {1\over 2}+ 2^2\cdot {1 \over 4}  ={3\over 2}} \Rightarrow \cases{a=E(X)=1 \\b= Var(X)= E(X^2)-(E(X))^2 = {3\over 2}-1={1\over 2}}，故選\bbox[red,2pt]{(B)}$$

解答：$$f_{X,Y}(x,y)= \begin{cases} 2, & 0\le y\le x\le 1\\ 0,& 其他\end{cases} \Rightarrow \cases{f_X(x)= \int_0^x 2\,dy = 2x\\ f_Y(y)= \int_y^1 2\,dx = 2(1-y)}\\ \Rightarrow \cases{f_{Y|X}(y\mid x)= {f_{X,Y}(x,y)\over f_X(x)} ={2\over 2x} ={1\over x} \\[1ex] f_{X|Y}(x\mid y)= {f_{X,Y}(x,y)\over f_Y(y)} ={2\over 2(1-y)}= {1\over 1-y}\ne {1\over y}}，故選\bbox[red,2pt]{(D)}$$
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