112年專技高考電機工程技師-工程數學詳解

 112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：$$g(t)=\begin{cases} 1,& 5\le t\lt 20\\ 0, &t\lt 5或t\gt 20\end{cases} \Rightarrow g(t)=u(t-5)-u(t-20) \Rightarrow L\{g(t)\}={1\over s}(e^{-5s}-e^{-20s})\\ 2L\{y''\}+ L\{y'\}+2L\{y\} =L\{g(t)\}\\ \Rightarrow 2(s^2Y(s)-sy(0)-y'(0))+sY(s)-y(0)+2Y(s)={1\over s}(e^{-5s}-e^{-20s})\\ \Rightarrow (2s^2+s+2)Y(s)={1\over s}(e^{-5s}-e^{-20s}) \Rightarrow Y(s)={1\over s(2s^2+s+2)}(e^{-5s}-e^{-20s})\\ L^{-1}\left\{ {1\over s(2s^2+s+2)}\right\}= L^{-1}\left\{{1\over 2s}-{1\over 2}{s+1/2\over 2s^2+s+2}  \right\} = L^{-1}\left\{{1\over 2s}-{1\over 2}{s+1/2\over 2(s+1/4)^2+15/8}  \right\} \\ ={1\over 2}+{1 \over 2}e^{-t/4}\left({1\over 2}\cos {\sqrt{15}\over 4}t+ {1\over 2\sqrt{15}}\sin{\sqrt{15}\over 4}t \right) \\ \Rightarrow y= L^{-1}\left\{ {1\over s(2s^2+s+2)}\cdot {1\over s}(e^{-5s}-e^{-20s})\right\} \\ \Rightarrow \bbox[red, 2pt]{y={1\over 2}u(t-5)\left(1-e^{-(t-5)/4}\left(\cos {\sqrt{15}\over 4}(t-5)- {1\over \sqrt{15}}\sin{\sqrt{15}\over 4} (t-5) \right)\right)-}\\ \qquad \bbox[red, 2pt]{{1\over 2}u(t-20)\left(1-e^{-(t-20)/4}\left( \cos {\sqrt{15}\over 4}(t-20)- {1\over \sqrt{15}}\sin{\sqrt{15}\over 4} (t-20) \right)\right)}$$

解答：$$f(x)=\begin{cases}-2x,& -2\le x\lt 0\\ 2x,& 0\le x\lt \bbox[cyan,2pt]2 \end{cases}, f(x+4)=f(x) \Rightarrow f(x)為偶函數 \Rightarrow b_n=0\\ a_0={1\over 4}\int_{-2}^2 f(x)\,dx = {1\over 4 } \left(\int_{-2}^0 -2x\,dx +\int_0^2 2x\,dx \right)=2\\ a_n={1\over 2}\int_{-2}^2 f(x) \cos{n\pi x\over 2}\,dx ={1\over 2}\left(\int_{-2}^0 -2x\cos{n\pi x\over 2} \,dx + \int_0^2 2x\cos{n\pi x\over 2}\,dx\right)\\=  \int_0^2 2x\cos{n\pi x\over 2}\,dx= 2\left. \left[{2\over n\pi}x\sin {n\pi x\over 2}+{4\over n^2\pi^2} \cos{n\pi x\over 2} \right]\right|_0^2 =2\left({4\over n^2\pi^2}((-1)^n-1) \right) \\= {8\over n^2\pi^2}\left[ (-1)^n-1\right]  \Rightarrow \bbox[red, 2pt]{f(x)=2+ \sum_{n=1}^\infty{8\over n^2\pi^2}\left[ (-1)^n-1\right]\cos (nx)}\\ $$

解答：$$\mathbf{(一)}\;\cases{甲射50發皆不中靶心的機率=(1-0.75)^{50} =0.25^{50} \\ 乙射53發皆不中靶心的機率=(1-0.72)^{53} =0.28^{53} \\ 丙射60發皆不中靶心的機率=(1-0.7)^{60} =0.3^{60} } \\ \Rightarrow 三人射中靶心的機率=1-三人都沒射中靶心的機率=\bbox[red, 2pt]{1-0.25^{50}\cdot 0.28^{53}\cdot 0.3^{60}} \\ \mathbf{(二)}\; {乙至少射中一發靶心機率\over 三人至少射中一發靶心的機率}=\bbox[red, 2pt]{{1-0.28^{53} \over 1-0.25^{50}\cdot 0.28^{53}\cdot 0.3^{60}}}$$

解答：$$u=e^{-x}(x\sin y-y\cos y) \Rightarrow \cases{\frac{\partial u}{\partial x} =e^{-x}(-x\sin y+y\cos y+\sin y)\\ \frac{\partial u}{\partial y} =e^{-x}(x\cos y-\cos y+y\sin y)} \\ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} \Rightarrow v=\int e^{-x}(-x\sin y+y\cos y+\sin y)\,dy=e^{-x}(x\cos y+ y\sin y )+ \phi(x)\\ 又\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \Rightarrow v=-\int e^{-x}(x\cos y- \cos y+y \sin y)\,dx =e^{-x}(x\cos y+y\sin y)+ \varphi(y)\\ \Rightarrow \phi(x) =\varphi(y)=C 為常數 \Rightarrow \bbox[red, 2pt]{v=e^{-x}(x\cos y+y \sin y)+C}$$

解答：$$\vec F=(\cos t+t\sin t, \sin t-t\cos t,t^2) \Rightarrow \vec F'=( t\cos t,  t\sin t, 2t) \\ \Rightarrow \Vert \vec F'\Vert = \sqrt{t^2\cos^2 t+t^2\sin^2 t+4t^2}=\sqrt 5 t\\ \Rightarrow \vec T={\vec F'\over \Vert\vec F'\Vert} ={1\over \sqrt 5 t}( t\cos t,  t\sin t, 2t) = \bbox[red, 2pt]{({\cos t \over \sqrt 5}, {\sin t\over \sqrt 5}, {2\over \sqrt 5})}\\ \vec F'(t)=(t\cos t,t\sin t,2t) \Rightarrow \vec F''(t)=(\cos t-t\sin t,\sin t+t\cos t,2)\\ \Rightarrow \vec F'(t) \times \vec F''(t)=(-2t^2\cos t,-2t^2\sin t, t^2) \Rightarrow \Vert \vec F'(t) \times \vec F''(t) \Vert = \sqrt 5 t^2\\ \Rightarrow \kappa = \cfrac{\Vert \vec F'(t) \times \vec F''(t) \Vert}{ \Vert \vec F'(t) \Vert^3 } ={\sqrt 5 t^2\over (\sqrt 5 t)^3} =\bbox[red, 2pt]{1\over 5t}$$

解答：$$\textbf{(一)}\;[A\; b]x=c \Rightarrow \begin{bmatrix}1 & 3 & 0 & 2 \\0 & 0 & 1 & 4 \\1 & 3 & 1 & 6 \end{bmatrix} \begin{bmatrix}x_1  \\x_2\\x_3\\x_4 \end{bmatrix} =\begin{bmatrix}1 \\6\\ 7 \end{bmatrix}\\ \text{Let }B=[A\; b\mid c]=\left[ \begin{array}{rrrr|r}1 & 3 & 0 & 2 & 1 \\0 & 0 & 1 & 4 &6\\1 & 3 & 1 & 6 & 7\end{array} \right] \Rightarrow rref(B)=\left[ \begin{array}{rrrr|r}1 & 3 & 0 & 2 & 1\\0 & 0 & 1 & 4 & 6\\0 & 0 & 0 & 0 & 0 \end{array} \right] \\ \Rightarrow \cases{x_1 +3x_2+2x_4=1\\ x_3+4x_4=6} \Rightarrow \mathbf x=x_2 \begin{pmatrix}-3 \\1\\ 0\\0 \end{pmatrix} +x_4 \begin{pmatrix}-2 \\0 \\ -4\\1 \end{pmatrix} +\begin{pmatrix}1 \\0\\ 6\\0 \end{pmatrix} \\ \Rightarrow \mathbf x= \bbox[red, 2pt]{\left\{s \begin{pmatrix}-3 \\1\\ 0\\0 \end{pmatrix}+t \begin{pmatrix} -2 \\0 \\ -4\\1 \end{pmatrix} +\begin{pmatrix}1 \\0\\ 6\\0 \end{pmatrix} \middle | s,t \in \mathbb R \right\}} \\\textbf{(二)}\;A=\begin{bmatrix}1 & 3 & 0  \\0 & 0 & 1  \\1 & 3 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=-\lambda(\lambda^2-2\lambda-2) =0 \Rightarrow \lambda=0,1\pm \sqrt 3\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)\mathbf x=0 \Rightarrow \begin{bmatrix}1 & 3 & 0  \\0 & 0 & 1  \\1 & 3 & 1 \end{bmatrix} \begin{bmatrix}x_1 \\x_2\\x_3 \end{bmatrix} =0 \Rightarrow \cases{x_1+3x_2=0\\ x_3=0} \\ \qquad \Rightarrow \mathbf x= k \begin{pmatrix}-3\\1\\ 0 \end{pmatrix}, k\in \mathbb R, \text{we choose }v_1= \begin{pmatrix}-3\\1\\ 0 \end{pmatrix} \\ \lambda_2=1-\sqrt 3 \Rightarrow (A-\lambda_2 I)\mathbf x=0 \Rightarrow \begin{bmatrix}\sqrt{3} & 3 & 0 \\0 & \sqrt{3}-1 & 1 \\1 & 3 & \sqrt{3} \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3 \end{bmatrix} =0 \\\qquad \Rightarrow \cases{2x_1=(3+\sqrt 3)x_3\\ 2x_2+(1+\sqrt 3)x_3=0} \Rightarrow \mathbf x=k \begin{pmatrix} 3+\sqrt 3\\-1-\sqrt 3\\ 2 \end{pmatrix}, k\in \mathbb R, \text{we choose }v_2 =\begin{pmatrix}3+\sqrt 3\\-1-\sqrt 3\\ 2 \end{pmatrix} \\ \lambda_3=1+\sqrt 3 \Rightarrow (A-\lambda_3 I)\mathbf x=0 \Rightarrow \begin{bmatrix}-\sqrt{3} & 3 & 0 \\0 & -\sqrt{3}-1 & 1 \\1 & 3 & -\sqrt{3} \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3 \end{bmatrix} =0 \\ \qquad \Rightarrow \cases{2x_1+(-3+\sqrt 3)x_3=0\\ 2x_2=(\sqrt 3-1)x_3} \Rightarrow \mathbf x=k \begin{pmatrix} 3-\sqrt 3\\-1+\sqrt 3\\ 2 \end{pmatrix}, k\in \mathbb R, \text{we choose }v_3= \begin{pmatrix} 3-\sqrt 3\\-1+\sqrt 3\\ 2 \end{pmatrix}\\ \Rightarrow \text{Eigenvalues: }\bbox[red, 2pt]{0,1-\sqrt 3,1+\sqrt 3}\text { and the associated unit eigenvectors: }{v_1\over \Vert v_1\Vert}, {v_2\over \Vert v_2\Vert}, {v_3\over \Vert v_3\Vert} = \\ \bbox[red, 2pt]{{1\over \sqrt{10}} \begin{pmatrix} -3\\1\\ 0 \end{pmatrix}, {1\over \sqrt{20+ 8\sqrt 3}} \begin{pmatrix} 3 +\sqrt 3\\-1-\sqrt 3\\ 2 \end{pmatrix},  {1\over \sqrt{20- 8\sqrt 3} } \begin{pmatrix} 3-\sqrt 3\\-1+\sqrt 3\\ 2 \end{pmatrix}}$$

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