112年高考三級-工程數學詳解

112年公務人員高等考試三級考試試題

類 科：電力工程、電子工程、電信工程
科 目：工程數學

甲、申論題部分：（50分）

解答: $$令u(y)=y' \Rightarrow u'y'=y'' \Rightarrow u'u=y'' 代回原式 \Rightarrow yu'u=u^2 \Rightarrow yu'=u \\ \Rightarrow {1\over u}du ={1\over y}dy \Rightarrow \ln u=\ln y+c_1= \ln (c_2y) \Rightarrow  u=c_2y \Rightarrow y'=c_2y \Rightarrow y=c_3e^{c_2x}\\ \Rightarrow \bbox[red,2pt]{y=Ae^{Bx},A,B皆為常數}$$

解答: $$逆時針單位圓c可用z=\cos\theta+i\sin \theta =e^{i\theta}來表示 \Rightarrow dz=ie^{i\theta}d\theta \\ \Rightarrow \oint_c \bar z\,dz = \int_0^{2\pi} e^{-i\theta}\cdot ie^{i\theta}\,d\theta =\int_0^{2\pi}i\,d\theta=\bbox[red, 2pt]{2\pi i}$$

解答: $$\cases{E_1:2x-y+2z=1 \\ E_2:x-y=2} \Rightarrow \cases{\vec n_1=(2,-1,2) \\ \vec n_2 =(1,-1,0)} \Rightarrow \cos \theta ={\vec n_1\cdot \vec n_2\over \Vert \vec n_1\Vert\cdot \Vert \vec n_2\Vert} ={2+1+0\over \sqrt 9 \cdot \sqrt 2}={1\over \sqrt 2}\\ \Rightarrow \theta =\bbox[red, 2pt]{45^\circ}$$

解答: $$A=\begin{bmatrix}2& 0 & 0\\ 1& 0 & 2\\ 0& 0& 3 \end{bmatrix} \Rightarrow \det(A-\lambda I)= -\lambda^3+5\lambda^2-6\lambda =-\lambda(\lambda-2)(\lambda-3)=0\\ \Rightarrow \lambda=0,2,3\\ \lambda_1=0 \Rightarrow (A-\lambda_1 I)v= \begin{bmatrix}2& 0 & 0\\ 1& 0 & 2\\ 0& 0& 3 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=0\\ x_3=0} \Rightarrow v=\begin{bmatrix}0\\ k\\0 \end{bmatrix},k\in \mathbb R,\\\qquad取v_1=\begin{bmatrix}0\\ 1\\0 \end{bmatrix} \\\lambda_2=2 \Rightarrow (A-\lambda_2 I)v= \begin{bmatrix}0& 0 & 0\\ 1& -2 & 2\\ 0& 0& 1 \end{bmatrix}\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=2x_2\\ x_3=0} \Rightarrow v=\begin{bmatrix}2k\\ k\\0 \end{bmatrix},k\in \mathbb R,\\\qquad 取v_2=\begin{bmatrix}2\\ 1\\0 \end{bmatrix} \\ \lambda_3=3 \Rightarrow (A-\lambda_3 I)v= \begin{bmatrix}-1& 0 & 0\\ 1& -3 & 2\\ 0& 0& 0 \end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=0\\ 3x_2=2x_3} \Rightarrow v=\begin{bmatrix}0\\ 2k/3\\k \end{bmatrix},k\in \mathbb R,\\ \qquad 取 v_3=\begin{bmatrix} 0 \\ 2/3\\1 \end{bmatrix} \\ 因此特徵值為\bbox[red,2pt]{0,2,3},相對應的特徵向量為 \bbox[red, 2pt]{\begin{bmatrix} 0\\ 1\\0 \end{bmatrix}, \begin{bmatrix}2\\ 1\\0 \end{bmatrix}, \begin{bmatrix} 0 \\ 2/3\\1 \end{bmatrix}}$$

乙、測驗題部分：（50分）

解答: $$先求齊次解,y''-y'-12y=0 \Rightarrow \lambda^2-\lambda-12=0 \Rightarrow (\lambda-4)(\lambda+3)=0 \\ \Rightarrow \lambda=4,-3 \Rightarrow y_h=c_1e^{4x}+c_2 e^{-3x}\\又\sinh x={1\over 2}(e^x-e^{-x}) \Rightarrow \sinh^2 x={1\over 4}(e^{2x}-2+e^{-2x}) \Rightarrow 2\sinh^2 x= {1\over 2}(e^{2x}+e^{-2x})-1\\ 取y_p=Ae^{2x}+Be^{-2x}+C \Rightarrow y_p'=2Ae^{2x}-2Be^{-2x} \Rightarrow y_p''=4Ae^{2x} +4Be^{-2x} \\ \Rightarrow y_p''-y_p'-12y=-10Ae^{2x}-6Be^{-2x}-12C= {1\over 2}(e^{2x}+e^{-2x})-1 \\ \Rightarrow \cases{-10A=1/2\\ -6B=1/2\\ -12C=-1} \Rightarrow \cases{A=-1/20\\ B=-1/12\\ C=1/12} \Rightarrow y_p=-{1\over 20}e^{2x}-{1\over 12}e^{-2x}+{1\over 12}\\ \Rightarrow y=y_h+y_p =c_1e^{4x}+c_2 e^{-3x}-{1\over 20}e^{2x}-{1\over 12}e^{-2x}+{1\over 12}\\ ,故選\bbox[red, 2pt]{(B)}$$

解答: $$假設y=-{1\over 6}\cos(2x)\ln|\sec(2x)+\tan(2x)| \Rightarrow y'' ={1\over 3}\left( 2\cos(2x)\ln|\sec(2x)+\tan(2x)|+2\tan(2x)\right) \\ \Rightarrow 3y''+12y=2\tan(2x),故選\bbox[red, 2pt]{(D)}$$

解答: $$L\{\sin (\omega t)\} ={\omega \over s^2+\omega^2} \Rightarrow L\{e^{-2t}\sin (\omega t)\} ={\omega \over (s+2)^2+\omega^2} \\ \Rightarrow L\{te^{-2t}\sin (\omega t)\} =-\frac{d }{ds} {\omega \over (s+2)^2+\omega^2} = {2(s+2)\omega \over ((s+2)^2+\omega^2)^2} ,故選\bbox[red, 2pt]{(A)}$$

解答: $$\left[\begin{array}{rrr|rrr}1& 0 & 1& 1& 0 & 0\\ -1& 1& 1& 0 & 1& 0\\ 2& -1& 1& 0 & 0 & 1\end{array} \right] \xrightarrow{R_1+R_2\to R_2,-2R_1+R_3\to R_3} \left[\begin{array}{rrr|rrr}1& 0 & 1& 1& 0 & 0\\ 0& 1& 2& 1 & 1& 0\\ 0& -1& -1& -2 & 0 & 1\end{array} \right] \\ \xrightarrow{R_2+R_3\to R_3}\left[\begin{array}{rrr|rrr}1& 0 & 1& 1& 0 & 0\\ 0& 1& 2& 1 & 1& 0\\ 0& 0& 1& -1 & 1 & 1\end{array} \right] \xrightarrow{-R_3+R_1\to R_1,-2R_3+R_2\to R_2}\left[\begin{array}{rrr|rrr}1& 0 & 0& 2& -1 & -1\\ 0& 1& 0& 3 & -1& -2\\ 0& 0& 1& -1 & 1 & 1\end{array} \right] \\ \Rightarrow A^{-1} =\begin{bmatrix} 2& -1& -1\\ 3& -1& -2\\ -1& 1& 1\end{bmatrix},故選\bbox[red, 2pt]{(B)}$$

解答: $$a_0={1\over 2} \int_0^1 x\,dx ={1\over 4}\\ a_n=  \int_0^1 x\cos(n\pi x)\,dx ={1\over n^2 \pi^2}((-1)^n-1) \Rightarrow \cases{a_1=-2/\pi^2\\ a_2=0\\ a_3=-2/9\pi^2},故選\bbox[red, 2pt]{(C)}$$

解答: $$無限多解 \Rightarrow \begin{vmatrix}4 &1  &T^2-11  \\4 &1  &-2  \\4 & 6 &7\end{vmatrix} =20(T^2-11)+40=0 \Rightarrow T^2=9 \Rightarrow T=\pm 3\\ 又無限多解 \Rightarrow rank(C)\lt 3 \Rightarrow rank(C)最大值為2,故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{\vec H=(2,8t,t^2)\\ \vec G=(-3t,2e^t,\ln t)} \Rightarrow \vec H\times \vec G =(8t\ln t-2t^2e^t, -3t^3-2\ln t,4e^t+24t^2) \\ \Rightarrow \frac{d }{dt}[\vec H\times \vec G ] =(8\ln t+8-4te^t-2t^2e^t, -9t^2-{1\over t},4e^t+48t),故選\bbox[red, 2pt]{(A)}$$

解答: $$\vec F=(x(t),y(t),z(t))  \Rightarrow \vec F'=(x'(t),y'(t),z'(t)) =(e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t,e^t)\\ \Rightarrow 單位切線向量={\vec F'\over \Vert \vec F'\Vert} =(x'(t),y'(t),z'(t)) ={1\over \sqrt 3e^t}(e^t\cos t-e^t\sin t,e^t\sin t+e^t\cos t,e^t) \\={1\over \sqrt 3}(\cos t-\sin t,\cos t+\sin t,1)={1\over \sqrt 3}(\cos t-\sin t)\hat i +{1\over \sqrt 3}(\cos t+\sin t)\hat j +{1\over \sqrt 3}\hat k,故選\bbox[red, 2pt]{(B)}$$

解答: $$(A)\bigcirc: \nabla \varphi =(\frac{\partial }{\partial x}\varphi,\frac{\partial }{\partial y}\varphi,\frac{\partial }{\partial z}\varphi) =(y^3z^2,3xy^2z^2,2xy^3z) \\ (B)\bigcirc: \nabla \varphi(-1,-1,2)=(-4,-12,4) \\ (C)\bigcirc: {(-4,-12,4)\over \Vert (-4,-12,4)\Vert}= \pm {(-4,-12,4)\over 4\sqrt{11}}= \pm{1\over \sqrt{11}}(-1,-3,1)\\(D)\times: 應該是k(-1,-3,1) \ne (8,24,-24)\\,故選\bbox[red, 2pt]{(D)}$$

解答: $$計算的結果有差,再想想,故選\bbox[red, 2pt]{(B)}$$

解答: $$顯然f(z)=|z|在z=0不可微,故選\bbox[red, 2pt]{(A)}$$

解答: $$\det(AB)=\det(A)\det(B)=24\times 24=576,故選\bbox[red, 2pt]{(B)}$$

解答: $$(1+i)^2=2i \Rightarrow (1+i)^{12} =(2i)^6=-2^6=-64,故選\bbox[red, 2pt]{(A)}$$

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

解答: $$\det(A)=-30 \Rightarrow \det(B)=\det(A^{-1})=-{1\over 30 } \\\Rightarrow \det(B^2)=\det(B)\times \det(B)=1/(-30)^2=1/900,故選\bbox[red, 2pt]{(A)}$$

解答: $$特徵向量v需滿足Av=\lambda v,其中\lambda 為特徵值\\ \begin{bmatrix} 5&4 \\1& 2\end{bmatrix} \begin{bmatrix} 4\\ 1\end{bmatrix} =\begin{bmatrix} 24\\ 6\end{bmatrix} =6\begin{bmatrix} 4\\ 1\end{bmatrix} \Rightarrow 特徵值\lambda_1=6\\ \begin{bmatrix} 5&4 \\1& 2\end{bmatrix} \begin{bmatrix} 1\\ -1\end{bmatrix} =\begin{bmatrix} 1\\ -1\end{bmatrix} =1\begin{bmatrix} 1\\ -1\end{bmatrix} \Rightarrow 特徵值\lambda_2=1\\ \Rightarrow 對角化矩陣=\begin{bmatrix} \lambda_1 &0 \\0& \lambda_2\end{bmatrix} =\begin{bmatrix} 6&0 \\0& 1\end{bmatrix},故選\bbox[red, 2pt]{(A)}$$

解答: $$e^x y'=2(x+1)y^2 \Rightarrow {1\over y^2}dy={2(x+1)\over e^x}\,dx \Rightarrow -{1\over y}=-{2x\over e^x}-{4\over e^x}+c_1 =-{2x+4+c_2e^x\over e^x}\\ \Rightarrow y={e^x\over 2x+4+c_2e^x} \Rightarrow y(0)={1\over 4+c_2}={1\over 6} \Rightarrow c_2=2 \Rightarrow y={e^x\over 2x+4+ 2e^x} \\ \Rightarrow y={1\over (2x+4)e^{-x}+2},故選\bbox[red, 2pt]{(A)}$$

解答: $$L^{-1}\left\{{s-1\over s^2(s+1)} \right\} =L^{-1}\left\{{2\over s } -{1\over s^2}-{2\over s+1}\right\} =2-t-2e^{-t},故選\bbox[red, 2pt]{(B)}$$

解答: $$L\left\{\cosh(at)\cos(at) \right\} =L\left\{{e^{at}+e^{-at}\over 2}\cdot \cos(at) \right\} = {1\over 2}L\left\{e^{at} \cos(at) \right\}+ {1\over 2}L\left\{e^{-at} \cos(at) \right\} \\={1\over 2}{s-a\over (s-a)^2+a^2} +{1\over 2}{s+a\over (s+a)^2+a^2} ={1\over 2}\cdot {(s-a)((s+a)^2+a^2)+ (s+a)((s-a)^2+a^2) \over (s-a)^2+a^2)((s+a)^2+a^2)}\\={s^3\over s^4+4a^4},故選\bbox[red, 2pt]{(D)}$$

解答: $$f(t)={\sin(8t)\over t}為偶函數 \Rightarrow \int_{-\infty}^\infty f(t)e^{-j5t}\,dt=\int_{-\infty}^\infty f(t)(\cos(5t)-i\sin(5t))\,dt=\int_{-\infty}^\infty f(t)\cos(5t)\,dt\\=\int_{-\infty}^\infty {\sin(8t)\over t}\cos(5t)\,dt　={1\over 2}\int_{-\infty}^\infty {1\over t}(\sin(13t)+\sin(3t))\,dt =\int_0^\infty ({\sin(13t)\over t}+{\sin(3t)\over t})dt \\={\pi \over 2}+{\pi \over 2}=\pi,故選\bbox[red, 2pt]{(D)}$$