110年專技高考-電機工程技師-工程數學詳解

 110年專門職業及技術人員高等考試

等 別： 高等考試
類 科： 電機工程技師
科 目： 工程數學（ 包括線性代數、 微分方程、 複變函數與機率）

解答： $$f(t)=2t^2+\int_0^t f(t-\tau)e^{-\tau}\;d\tau \Rightarrow \mathcal{L}\{ f(t)\} =\mathcal{L}\{ 2t^2\} +\mathcal{L}\{ \int_0^t f(t-\tau)e^{-\tau}\;d\tau\} \\ \Rightarrow F(s)=2\cdot {2\over s^3} +\mathcal{L}\{ f(t)\}\mathcal{L}\{ e^{-t}\} ={4\over s^3}+ F(s)\cdot {1\over s+1} \Rightarrow {s\over s+1}F(s)= {4\over s^3}\\ \Rightarrow F(s)={4\over s^3}+{4\over s^4} \Rightarrow \mathcal{L}^{-1}\{ F(s) \}=\mathcal{L}^{-1}\{ {4\over s^3} \}+\mathcal{L}^{-1}\{ {4\over s^4} \} \\ \Rightarrow \bbox[red,2pt]{f(t)= 2t^2+ {2\over 3}t^3}$$

解答： $$y'=y^2e^{-2x} \Rightarrow \int {1\over y^2}\;dy=\int e^{-2x}\;dx \Rightarrow -{1\over y}=-{1\over 2}e^{-2x}+C_1 \Rightarrow {1\over y}={1\over 2}e^{-2x}-C_1 \\ \Rightarrow y={1\over {1\over 2}e^{-2x} -C_1} ={2\over e^{-2x}-C_2} ={2e^{2x}\over 1-C_2e^{2x}} \Rightarrow \bbox[red,2pt]{y={2e^{2x}\over 1-Ce^{2x}},C為常數}$$

解答： $$a_0={1\over 2\pi}\int_{-\pi}^\pi f(x)\;dx ={1\over 2\pi}\int_{-\pi}^\pi x-x^2\;dx ={1\over 2\pi}\left. \left[ {1\over 2}x^2-{1\over 3}x^3 \right] \right|_{-\pi}^\pi = -{1\over 3}\pi^2\\ a_n={1\over \pi}\int_{-\pi}^\pi f(x)\cos(nx)\;dx ={1\over \pi}\int_{-\pi}^\pi (x-x^2)\cos(nx)\;dx \\=-{1\over \pi}\int_{-\pi}^\pi x^2\cos(nx)\;dx (\because x\cos(nx)為奇函數) =-{1\over n^3\pi} \left.\left[ (n^2x^2-2)\sin(nx)+2nx\cos(nx) \right] \right|_{-\pi}^\pi\\ =-{1\over n^3\pi}\times 4n\pi \cos(n\pi)= -{4\over n^2}(-1)^{n} \\ b_n={1\over \pi}\int_{-\pi}^\pi f(x)\sin(nx)\;dx ={1\over \pi}\int_{-\pi}^\pi (x-x^2)\sin(nx)\;dx \\={1\over \pi}\int_{-\pi}^\pi x\sin(nx)\;dx(\because x^2\sin(nx)為奇函數) ={1\over n^2\pi} \left.\left[ \sin(nx)-nx\cos(nx) \right] \right|_{-\pi}^\pi\\= {1\over n^2\pi} \times (-2n\pi)\cos(n\pi) =-{2\over n}(-1)^n \\ \Rightarrow f(x)= a_0+ \sum_{n=0}^\infty a_n\cos(nx)+ b_n\sin(nx)\\ = -{1\over 3}\pi^2+ \sum_{n=0}^\infty \left(  -{4\over n^2}(-1)^{n}\right)\cos(nx)+ \left(-{2\over n}(-1)^n \right)\sin(nx) \\\Rightarrow \bbox[red,2pt]{f(x)= -{1\over 3}\pi^2- \sum_{n=0}^\infty(-1)^{n} \left(  {4\over n^2} \cos(nx)+  {2\over n} \sin(nx)\right),x\in (-\pi,\pi)}$$

解答： $$\cases{x(t)=t^3\\ y(t)=-t\\ z(t)=t^2} \Rightarrow \cases{dx=3t^2\;dt\\ dy=-dt\\ dz=2t\;dt} \Rightarrow \int_c xdx-yzdy+e^zdz =\int_1^2 t^3 3t^2\;dt-(-t^3)(-dt)+e^{t^2}2t\;dt \\ =\int_1^2 3t^5-t^3+2te^{t^2}\;dt =\left. \left[ {1\over 2} t^6- {1\over 4}t^4+ e^{t^2}\right]\right|_1^2 =\bbox[red,2pt]{{111\over 4}+e^4-e}$$

解答： $$f(x,y,z)=x^2y-xy^2+xz^2 \Rightarrow \cases{f_x=2xy-y^2+z^2\\ f_y=x^2-2xy\\ f_z=2xz} \Rightarrow \nabla f=(f_x,f_y,f_z)\\ 令\cases{P(1,-1,1)\\ Q(1,-2,1)} \Rightarrow \vec u=\overrightarrow{PQ}=(0,-1,0)為一單位向量\\ 因此\nabla f(P)\cdot u =(-2,3,2)\cdot (0,-1,0)= \bbox[red,2pt]{-3}$$

解答：

(一) $$\det(A)= \begin{vmatrix} 3 & 0 & -2\\ 0 & 2 & 0 \\ -2 & 0 & 0\end{vmatrix} =0-8= \bbox[red,2pt]{-8}$$ (二) $$\det(A-\lambda I)=0 \Rightarrow  \begin{vmatrix} 3-\lambda & 0 & -2\\ 0 & 2-\lambda & 0 \\ -2 & 0 & -\lambda\end{vmatrix} =0 \Rightarrow -\lambda(\lambda-2)(\lambda-3)-4(2-\lambda)=0\\ \Rightarrow (\lambda-2)(-\lambda^2+3\lambda+4)=0 \Rightarrow (\lambda-2)(\lambda-4)(\lambda+1)=0 \Rightarrow \lambda= -1,2,4\\ \lambda_1=-1 \Rightarrow (A-\lambda_1 I)X=0 \Rightarrow \begin{bmatrix} 4 & 0 & -2\\ 0 & 3 & 0 \\ -2 & 0 & 1\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} =0 \Rightarrow \cases{2x_1=x_3\\ x_2=0}，取v_1= \begin{bmatrix} 1\\ 0 \\ 2\end{bmatrix} \\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)X=0 \Rightarrow \begin{bmatrix} 1 & 0 & -2\\ 0 & 0 & 0 \\ -2 & 0 & -2\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1=0\\  x_3=0}，取v_2= \begin{bmatrix} 0\\ 1 \\ 0 \end{bmatrix} \\\lambda_3=4 \Rightarrow (A-\lambda_3 I)X=0 \Rightarrow \begin{bmatrix} -1 & 0 & -2\\ 0 & -2 & 0 \\ -2 & 0 & -4\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+2x_3=0\\  x_2=0}，取v_3= \begin{bmatrix} 2\\ 0 \\ -1 \end{bmatrix} \\因此\bbox[red,2pt]{特徵值為\cases{\lambda_1=-1\\ \lambda_2=2\\ \lambda_3=4}，及相對應的特徵向量為\cases{v_1=(1,0,2)\\ v_2=(0,1,0)\\ v_3=(2,0,-1)}}$$ (三) $$取P=[v_1 v_2 v_3]=\bbox[red,2pt]{\begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & 0 \\ 2 & 0 & -1\end{bmatrix} }\Rightarrow P^{-1}=\begin{bmatrix} 1/5 & 0 & 2/5\\ 0 & 1 & 0 \\ 2/5 & 0 & -1/5\end{bmatrix} \\\Rightarrow P\begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}P^{-1}=A \Rightarrow P^{-1}AP= \begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3\end{bmatrix}為一對角化矩陣$$

解答： $$\int_{-\infty}^\infty p(x)\;dx =\int_0^2 a(x+1)\;dx =1 \Rightarrow \left.\left[ a({1\over 2}x^2+x) \right] \right|_0^2 =4a=1 \Rightarrow a={1\over 4}\\ 期望值E(X)=\int_0^2 xp(x)\;dx = {1\over 4}\int_0^2 x^2+x\;dx = {1\over 4}\left. \left[ {1\over 3}x^3+ {1\over 2}x^2\right] \right|_0^2 = {7\over 6}\\ E(X^2)=\int_0^2 x^2p(x)\;dx = {1\over 4}\int_0^2 x^3+x^2 \;dx = {1\over 4}\left. \left[ {1\over 4}x^4+ {1\over 3}x^3\right] \right|_0^2 ={5\over 3} \\ 因此變異數Var(X)= E(X^2)-(E(X))^2 = {5\over 3}-({7\over 6})^2 = {11\over 36}\\ 即\bbox[red,2pt]{期望值={7\over 6}，變異數={11\over 36}}$$

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