103年鐵路特考-工程數學詳解

1 0 3 年 特 種 考 試 交 通 事 業 鐵 路 人 員 考 試

等 別：高員三級鐵路人員考試
類 科：電子工程
科 目：工程數學
甲、申論題部分：（ 50 分）

解答：$$(一)(0,0)\in A且\cases{(a_1,a_2)\in A \Rightarrow a_1+a_2=0\\ (b_1,b_2)\in A \Rightarrow b_1+b_2=0} \Rightarrow a_1+b_1+a_2 +b_2=0 \Rightarrow (a_1+b_1,a_2+b_2)\in A \\ \Rightarrow (a_1,a_2)+(b_1,b_2)\in A \Rightarrow \bbox[red,2pt]{A是}R^2的子空間\\(二)\cases{(1,0)\in B\\ (0,1)\in B} \Rightarrow (1,0)+(0,1)= (1,1)\not \in B \Rightarrow \bbox[red,2pt]{B不是}R^2的子空間\\(三)(0,0)\not \in C\Rightarrow \bbox[red,2pt]{C不是}R^2的子空間$$

解答：$$\nabla \cdot (F\times G) = \nabla(f_2g_3-f_3g_2, f_3g_1-f_1g_3,f_1g_2-f_2g_1)\\={\partial \over \partial x}(f_2g_3-f_3g_2) + {\partial \over \partial y}(f_3g_1-f_1g_3) +{\partial \over \partial z}(f_1g_2-f_2g_1)\\=f_{2x}g_3+f_2g_{3x}-f_{3x}g_2-f_3g_{2x}+f_{3y}g_1+f_3g_{1y}-f_{1y}g_3-f_1g_{3y}+ f_{1z}g_2+f_1g_{2z}-f_{2z}g_1-f_2g_{1z}\\ =g_1(f_{3y}-f_{2z})+g_2(f_{1z}-f_{3x})+g_3(f_{2x}-f_{1y}) +f_1(g_{2z}-g_{3y})+f_2(g_{3x}-g_{1z})+ f_3(g_{1y}-g_{2x})\\ =G\cdot (\nabla\times F)-F\cdot (\nabla \times G) \\ \Rightarrow \nabla \cdot (F\times G) = G\cdot (\nabla\times F)-F\cdot (\nabla \times G) ，\bbox[red,2pt]{故得證}$$

解答：

(一)$$f(z)={3\over 1-iz+2z^2} ={3\over (2z+i)(z-i)} ={2i\over 2z+i}-{i\over z-i}\\ {2i\over 2z+i} ={2i\over 2(z-2i)+5i} = {2i\over 5i(1+{2\over 5i}(z-2i))} ={2\over 5}\cdot {1\over 1-{2i\over 5}(z-2i)} ={2\over 5} \sum_{n=0}^\infty \left({2i\over 5}(z-2i)\right)^n\\ {i\over z-i} ={i\over z-2i+i}={i\over i(1+{z-2i\over i} )} ={1\over 1+{z-2i\over i}} =\sum_{n=0}^\infty (-1)^n({z-2i\over i})^n\\ 因此\bbox[red,2pt]{f(z)=\sum_{n=0}^\infty\left({2\over 5}\left({2i\over 5}(z-2i)\right)^n + (-1)^n({z-2i\over i})^n\right)}$$(二)$$\cases{\left|{2i\over 5}(z-2i) \right|\lt 1 \Rightarrow  -{i\over 2}\lt z\lt {9i\over 2} \\\left|{z-2i\over i} \right|\lt 1 \Rightarrow i\lt z\lt 3i} ，兩者取交集可得收斂半徑:\bbox[red,2pt]{i\lt z\lt 3i}$$

解答：$$\mathcal{L} \{ y(t)\} =\mathcal{L} \{ 1\}+\mathcal{L} \{\int_0^t y(t-\alpha)\sin(2\alpha)\;d\alpha\}\Rightarrow Y(s)= {1\over s}+ \mathcal{L} \{  y(t)\}\mathcal{L} \{ \sin(2t)\} \\ ={1\over s}+ Y(s)\times {2 \over s^2+2^2} \Rightarrow Y(s)= {s^2+4\over s(s^2+2)} ={2\over s}-{s \over s^2+2} \\ \Rightarrow y(t)=\mathcal{L}^{-1} \{{2\over s}\}-\mathcal{L}^{-1} \{{s\over s^2+2}\} =2-\cos(\sqrt 2t) \Rightarrow \bbox[red,2pt]{y(t)=2-\cos(\sqrt 2t)}$$

乙、測驗題部分： （50 分）

解答：$$\nabla(f/g)=\langle{\partial\over \partial x }(f/g), {\partial\over \partial y }(f/g),{\partial\over \partial z }(f/g)\rangle =\langle {f_x\over g}-{fg_x\over g^2},{f_y\over g}-{fg_y\over g^2},{f_z\over g}-{fg_z\over g^2}\rangle \\ ={1\over g^2}\langle gf_x-fg_x,gf_y-fg_y,gf_z-fg_z\rangle ={1\over g^2}\left(g\langle f_x,f_y,f_z \rangle -f\langle g_x,g_y,g_z  \rangle\right) \\ ={1\over g^2}\left(g\nabla f-f\nabla g \right) \ne {1\over g^2}\left(f\nabla g-g\nabla f \right) ，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{x=r\cos\theta\\ y=r\sin \theta} \Rightarrow I_x=\int y^2\;dA =\int_0^{\pi/2}\int_0^1 r^2\sin^2\theta \cdot r\;drd\theta=\int_0^{\pi/2}\int_0^1 r^3\sin^2\theta\;drd\theta\\ =\int_0^{\pi/2}{1\over 4}\sin^2\theta\;d\theta ={1\over 8}\int_0^{\pi/2}(1-\cos(2\theta))\;d\theta ={1\over 8}\left. \left[ \theta-{1\over 2}\sin(2\theta) \right]\right|_0^{\pi/2} ={\pi\over 16} \ne {\pi\over 8}，故選\bbox[red,2pt]{(C)}$$

解答：$$\nabla \phi=(\phi_x,\phi_y,\phi_z) =({-x\over \sqrt{x^2+y^2}},{-y\over \sqrt{x^2+y^2}},1)，故選\bbox[red,2pt]{(A)}$$

解答：$$\nabla f=(f_x,f_y,f_z) =(2y+e^z,2x,xe^z) \Rightarrow \left.(2y+e^z,2x,xe^z)\right|_{(1,1,1)}=(2+e,2,e)，故選\bbox[red,2pt]{(C)}$$

解答：$$A為2\times 2矩陣，因此\det(cA)=c^2\det(A)，故選\bbox[red,2pt]{(C)}$$

解答：$$A=\begin{bmatrix}2 & 1 & 0\\ 1 & 3& 2\\ 1& 4 & 1 \end{bmatrix} \Rightarrow \\A_{11}= \begin{vmatrix}    3& 2\\   4 & 1 \end{vmatrix} =-5,A_{12}= \begin{vmatrix} 1 &   2\\ 1&   1 \end{vmatrix} =-1,A_{13}=\begin{vmatrix}  1 & 3 \\ 1& 4   \end{vmatrix} =1\\ A_{21}=\begin{vmatrix}  1 & 0 \\ 4& 1   \end{vmatrix} =1,A_{22}=\begin{vmatrix}  2 & 0 \\ 1& 1   \end{vmatrix} =2 ,A_{23}=\begin{vmatrix}  2 & 1 \\ 1 & 4   \end{vmatrix} =7,\\A_{31}=\begin{vmatrix}  1 & 0 \\ 3& 2   \end{vmatrix} =2,A_{32}=\begin{vmatrix}  2 & 0 \\ 1& 2   \end{vmatrix} =4, A_{33}=\begin{vmatrix}  2 & 1 \\ 1 & 3   \end{vmatrix} =5\\ \Rightarrow 輔因子矩陣=\begin{bmatrix}A_{11} & -A_{12} & A_{13}\\ -A_{21} & A_{22} & -A_{23}\\ A_{31} & -A_{32} & A_{33} \end{bmatrix} =\begin{bmatrix}-5 & 1 & 1\\ -1 & 2& -7\\ 2& -4 & 5 \end{bmatrix}  ，故選\bbox[red,2pt]{(A)}$$

解答：$$\begin{bmatrix} 1 & 2 & 1& 0\\ 2 & 5 & 5 & 1\\ -2&-3 & 0 & 3\\ 1& 4&6& 4\end{bmatrix} \xrightarrow{-r_1+r_4,r_2+r_3} \begin{bmatrix} 1 & 2 & 1& 0\\ 2 & 5 & 5 & 1\\ 0& 2 & 5 & 4\\ 0& 2 &5 & 4\end{bmatrix} \xrightarrow{-2r_1+r_2,-r3+r_4} \begin{bmatrix} 1 & 2 & 1& 0\\ 0 & 1 & 3 & 1\\ 0& 2 & 5 & 4\\ 0& 0 &0 & 0\end{bmatrix}\\ \Rightarrow \text{Rank}(A)=3，故選\bbox[red,2pt]{(C)}$$

解答：$$M=PDP^{-1} \Rightarrow M^{-1}=PD^{-1}P^{-1}，由於D為對角矩陣， D^{-1}也是對角矩陣\\，因此M與M^{-1}有相同之特徵向量，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{z_1=1-i\\ z_2=-2+4i\\ z_3=\sqrt 3-2i} \Rightarrow {z_1z_2\over z_3} ={(1-i)(-2+4i)\over \sqrt 3-2i} ={2+6i\over \sqrt 3-2i} ={(2+6i)( \sqrt 3+2i)\over (\sqrt 3-2i)(\sqrt 3+2i)}\\ = {-10+(4+6\sqrt 3)i\over 7}  \Rightarrow 虛部為{4+6\sqrt 3\over 7}，故選\bbox[red,2pt]{(A)}$$

解答：$$\lim_{n\to \infty} \left\{(1-{1\over n}) +i(1+{1\over n}) \right\} =1+i \Rightarrow \left\{(1-{1\over n}) +i(1+{1\over n}) \right\}收斂，故選\bbox[red,2pt]{(D)}$$

解答：$$\sin(\pi)=0且分母4次方為0，故選\bbox[red,2pt]{(D)}$$

解答：$$y'+2y=1 \Rightarrow \cases{y_h=c_1e^{-2t} \\ y_p=c_2=1/2} \Rightarrow y=c_1e^{-2t}+{1\over 2} \Rightarrow \lim_{t\to \infty}y(t) ={1\over 2}，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{a_1={1\over \pi}\int_{-\pi}^\pi (x+\pi)\cos(x)\;dx = {1\over \pi}\left.\left[ x\sin(x)+\cos(x)+\pi \sin(x)\right] \right|_{-\pi}^\pi =0\\ b_1={1\over \pi}\int_{-\pi}^\pi (x+\pi)\sin(x)\;dx = {1\over \pi}\left.\left[ \sin(x)-x\cos(x)-\pi \cos(x) \right] \right|_{-\pi}^\pi = 2}\\ \Rightarrow a_1+b_1= 2，故選\bbox[red,2pt]{(C)}$$

解答：$$\mathcal{L}\{ f(t)\} =\mathcal{L}\{e^{-3t}\sin(5t)u(t)\} =\mathcal{L}\{e^{-3t}\sin(5t) \} ={5\over (s+3)^2+5^2} ={5\over s^2+6s+34}，故選\bbox[red,2pt]{(B)}$$

解答：$$f(t)=\mathcal{L}^{-1}\{ {3s+2\over s^2+4s+5}\} =\mathcal{L}^{-1}\{ {3(s+2)-4\over (s+2)^2+1}\}=3\mathcal{L}^{-1}\{ {s+2 \over (s+2)^2+1}\}-4\mathcal{L}^{-1}\{{ 1\over (s+2)^2+1}\} \\ =3e^{-2t}\cos(t)-4e^{-2t}\sin(t) \Rightarrow f(0)=3，故選\bbox[red,2pt]{(D)}$$

解答：$$y'=3x^2-{y\over x} \Rightarrow xy'+y=3x^3 \Rightarrow \cases{y_h=c_1x^{-1}\\ y_p={3\over 4}x^3} \Rightarrow y=c_1x^{-1}+{3\over 4}x^3\\ y(1)=1 \Rightarrow c_1+{3\over 4}=1 \Rightarrow c_1= {1\over 4} \Rightarrow y={1\over 4x} +{3\over 4}x^3，故選\bbox[red,2pt]{(D)}$$

解答：$$\cos(\omega_0x)={1\over 2}(e^{i\omega_0x}+e^{-i\omega_0x}) \Rightarrow   {1\over \sqrt{2\pi}} \int f'(x)\cos(\omega_0 x)e^{-i\omega x}dx \\=  {1\over 2 } \left({1\over \sqrt{2\pi}}\int f'(x)(e^{ix(\omega_0-\omega)}+ {1\over \sqrt{2\pi}}\int f'(x)e^{-ix(\omega_0+\omega )})dx \right) \\={1\over 2}(i(\omega-\omega_0)F(\omega-\omega_0)+ i(\omega+\omega_0)F(\omega+\omega_0))\\ ={i\over 2}((\omega-\omega_0)F(\omega-\omega_0)+ (\omega+\omega_0)F(\omega+\omega_0))，故選\bbox[red,2pt]{(C)}$$

解答：$$X\sim B(n=100,p=0.2) \Rightarrow E(X)=np=100\times 0.2=20 \Rightarrow E(50-2X)=50-2E(X)\\ =50-2\times 20=10，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{從第1袋取出白球(機率=4/7)放入第2袋，再從第2袋取出黑球的機率為5/9 \\從第1袋取出黑球(機率=3/7)放入第2袋，再從第2袋取出黑球的機率為6/9}；\\\Rightarrow 第2袋取出黑球的機率={4\over 7}\times {5\over 9}+ {3\over 7}\times {6\over 9}={38\over 63} ，故選\bbox[red,2pt]{(D)}$$

解答：$$\int f(x)\;dx =1 \Rightarrow \int_0^\infty Ae^{-4x}\;dx = \left.\left[ -{A\over 4}e^{-4x}\right]\right|_0^\infty ={A\over 4}=1 \Rightarrow A=4，故選\bbox[red,2pt]{(B)}$$

==================== END =======================
 解題僅供參考，其他國考試題及詳解