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

解答： $$yy''=(y')^2 \Rightarrow y{d^2y\over dx^2} =({dy\over dx})^2\\ 取u=y'={dy\over dx} \Rightarrow  {d^2 y\over dx^2} =y''={du\over dx}={du\over dy}{dy\over dx}= {du\over dy}u\\ \Rightarrow 原式y{du\over dy}u=u^2 \Rightarrow {1\over u}du = {1\over y}dy \Rightarrow \ln |u|=\ln |y|+C_1 \Rightarrow \ln{|u|\over |y|}=C_1 \Rightarrow  {u\over y}=C_2 \\ \Rightarrow u=C_2y \Rightarrow {dy\over dx}=C_2y \Rightarrow \int {1\over y}\,dy = \int C_2\,dx \Rightarrow \ln|y|=C_2x+C_3 \Rightarrow y=e^{C_2x+ C_3} \\ \Rightarrow \bbox[red, 2pt]{ y=Ae^{Bx},其中A與B為常數}$$

解答： $$\int_c \bar z\,dz =\int_c{1\over z}\,dz = 2\pi i\times 1=\bbox[red, 2pt]{2\pi i}$$

解答： $$\cases{平面2x-y+2z=1法向量\vec u=(2,-1,2)\\ 平面x-y=2法向量\vec v=(1,-1,0)} \Rightarrow \cos \theta={\vec u\cdot \vec v\over |\vec u||\vec v|} ={2+1\over \sqrt{9}\cdot \sqrt 2} ={1\over \sqrt 2} \Rightarrow \theta=\bbox[red,2pt]{45^\circ}$$

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

解答： $$先求齊次解：y''-y'-12y=0 \Rightarrow \lambda^2-\lambda-12=0 \Rightarrow (\lambda-4)(\lambda+3) =0\\ \Rightarrow \lambda=4,-2 \Rightarrow y_h=C_1e^{4x} +C_2e^{-3x}\\ 由於\sinh(x)={e^x-e^{-x} \over 2} \Rightarrow 2\sinh^2(x)={1\over 2}(e^{2x}-2+ e^{-2x})\Rightarrow 取y_p= ae^{2x} +be^{-2x}+C \\ \Rightarrow y_p'=2ae^{2x}-2be^{-2x} \Rightarrow y_p''=4ae^{2x}+4be^{-2x} \\ \Rightarrow y_p''-y_p'-12y_p= -10ae^{2x}-6be^{-2x}-12c={1\over 2}(e^{2x}-2+ e^{-2x}) \\ \Rightarrow \cases{a=-1/20\\ b=-1/12\\ c=1/12} \Rightarrow y=y_h+y_p =C_1e^{4x} +C_2e^{-3x} -{1\over 20}e^{2x}-{1\over 12}e^{-2x}+{1\over 12}，故選\bbox[red, 2pt]{(B)}$$

解答： $$先求3y''+12y=0 \Rightarrow y_h=C_1\cos(2x)+C_2 \sin(2x)，取\cases{y_1=\cos(2x) \\ y_2=\sin(2x)} \\ 利用參數變換法求特解，W(y_1,y_2)=\begin{vmatrix}\cos(2x)& \sin(2x)\\ -2\sin (2x)& 2\cos(2x) \end{vmatrix} =2 \\ \Rightarrow y_p =-y_1\int {y_2f(x)\over W(y_1,y_2)}\,dx + y_2 \int{y_1f(x)\over W(y_1,y_2)}\,dx \\=-\cos(2x) \int{2\sin(2x)\tan(2x)\over 2} \,dx +\sin(2x) \int{2\cos(2x)\tan(2x) \over 2}\,dx \\=-\cos(2x) \int \sin(2x)\tan(2x)\,dx+ \sin(2x) \int \sin(2x)\,dx\\ =-\cos(2x)\cdot {1\over 2}\left( \ln|\tan(2x)+\sec (2x)|-\sin(2x)\right)-{1\over 2}\sin(2x)\cos(2x) \\=-{1\over 2}\cos(2x)\cdot \ln|\tan(2x)+\sec (2x)|\\ y''+4y=2\tan(2x)的特解是-{1\over 2}\cos(2x)\cdot \ln|\tan(2x)+\sec (2x)|\\ \Rightarrow 3y''+12y=2\tan(2x)的特解是-{1\over 6}\cos(2x)\cdot \ln|\tan(2x)+\sec (2x)|，故選\bbox[red, 2pt]{(D)}$$

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

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

解答： $$A為3\times 3，且有無限多解\Rightarrow rank(C)最大為2\\ 又無限多解\Rightarrow {4\over 4}={1\over 1}={T^2-11\over -2} = {T-2\over 1} \Rightarrow \cases{T^2-11=-2 \Rightarrow T=\pm 3 \\T-2 =1 \Rightarrow T= 3} \Rightarrow T=3，故選\bbox[red, 2pt]{(D)}$$

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

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

解答： $$(A)\bigcirc: \nabla\varphi =(\varphi_x,\varphi_y, \varphi_z) =(y^3z^2,3xy^2z^2,2xy^3z) \\(B) \bigcirc: \nabla\varphi (-1,-1,2)=(-4,-12,4) \\(C)\bigcirc: \pm {(-4,-12,4) \over |(-4,-12,4)|} = \pm {(-4,-12,4) \over 4\sqrt{11}} =\pm {-1,-3,1\over \sqrt{11}} \\(D)\times: 與(B)矛盾\\，故選\bbox[red, 2pt]{(D)}$$

解答： $$\begin{array} {}X & Y & X^2 & Y^2\\\hline 88& 84 &7744& 7056\\ 77 & 60& 5929& 3600\\ 40& 74 & 1600& 5476\\ 58& 50& 3364 & 2500\\ 72& 95 & 5184& 9025\\ 92 & & 8464\\\hdashline 427 & 363 & 32285 & 27657\end{array} \Rightarrow \cases{\cases{E(X)=427/6\\ E(X^2)=32285/6} \Rightarrow \sigma_x=\sqrt{{32285\over 6}-\left( {427\over 6}\right)^2} \approx 17.78\\ \cases{E(Y)=363/5\\ E(Y^2)=27657/5} \Rightarrow \sigma_y= \sqrt{{27657\over 5}-\left( {363 \over 5}\right)^2} \approx 16.14}\\，故選\bbox[red, 2pt]{(B)}$$

解答： $$z=x+iy \Rightarrow |z|= \sqrt{x^2+y^2}+0i \Rightarrow \cases{u(x,y)=\sqrt{x^2+y^2}\\ v(x,y)=0} \Rightarrow u_x\ne 0=v_y \\ \Rightarrow |z|非可解析函數，故選\bbox[red, 2pt]{(Ａ)}$$

解答： $$A,B均為三角矩陣，其行列式=對角線相乘 \Rightarrow \cases{\det(A)=1\cdot 2\cdot 3\cdot 1\cdot 4=24\\ \det(B)=1 \cdot 2\cdot 3\cdot 4\cdot 1=24} \\ \Rightarrow \det(AB) =\det(A) \cdot \det(B)=24^2=576，故選\bbox[red, 2pt]{(B)}$$

解答： $$(1+i)^2=2i \Rightarrow (1+i)^4=(2i)^2=-4 \Rightarrow (1+i)^{12}=(-4)^3 =-64，故選\bbox[red, 2pt]{(A)}$$

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

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

解答： $$特徵向量v滿足Av=\lambda v,因此A\begin{bmatrix}4\\1 \end{bmatrix} =\begin{bmatrix}24 \\6 \end{bmatrix} =6 \begin{bmatrix}4\\1 \end{bmatrix} \Rightarrow \lambda=6，故選\bbox[red, 2pt]{(A)}$$

解答： $$e^xy'=2(x+1)y^2 \Rightarrow \int {1\over y^2}\,dy = \int {2(x+1)\over e^x}\,dx \Rightarrow -{1\over y}=-2({x+2\over e^x})+C\\ \Rightarrow y={e^x \over 2(x+2)+Ce^x},將y(0)=1/6代入左式 \Rightarrow {1\over 6}={1\over 4+C} \Rightarrow C=2 \\ \Rightarrow y={e^x \over 2(x+2)+2e^x}={1\over 2(x+2)e^{-x}+ 2}，故選\bbox[red, 2pt]{(A)}$$

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

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

解答： $$e^{-j5t} =\cos(5t)-j\sin(5t) \Rightarrow \int_{-\infty}^\infty f(t)e^{-j5t}\,dt = \int_{-\infty}^\infty \frac{\sin(8t)}{t}(\cos(5t)-j\sin(5t))\,dt \\=\int_{-\infty}^\infty\frac{\sin(8t)\cos(5t)}{t}\,dt -j\int_{-\infty}^\infty \frac{\sin(8t)\sin(5t)}{t}\,dt =2 \int_0^{\infty}\frac{\sin(8t)\cos(5t)}{t}\,dt-0(奇函數＝0) \\=\int_0^\infty{ \sin(13t)\over t}\,dt +\int_0^\infty {\sin(3t)\over t}\,dt ;用\text{Laplace transform }來計算積分值\\令I_1(s)= \int_0^\infty{ \sin(13t)\over t} e^{-st}\,dt \Rightarrow I_1(0)= \int_0^\infty{ \sin(13t)\over t}\,dt且\lim_{s\to \infty} I_1(s)=0 \\ 因此I_1'(s)=- \int_0^\infty \sin(13t)e^{-st}\,dt =-L\{ \sin(13t)\} =-{13\over s^2+13^2} \Rightarrow I_1(s)= -\int  {13\over s^2+13^2}\,ds\\=-\tan^{-1}{s\over 13}+C  \Rightarrow \lim_{s\to \infty}I_1(s)=-{\pi\over 2}+C=0 \Rightarrow C={\pi\over 2} \Rightarrow I_1(0)=C= {\pi\over 2} \\同理,\int_0^\infty {\sin(3t)\over t}\,dt={\pi \over 2}; 因此\int_{-\infty}^\infty f(t)e^{-j5t}\,dt ={\pi \over 2}+{\pi \over 2}=\pi，故選\bbox[red, 2pt]{(D)}$$
 

＝＝＝＝＝＝＝＝＝＝＝＝＝　ＥＮＤ　＝＝＝＝＝＝＝＝＝＝＝＝

解題僅供參考，其他歷年高普考試題及詳解