Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

網頁

2022年12月13日 星期二

111年地方三等-工程數學詳解

111 年特種考試地方政府公務人員考試試題

等 別: 三等考試
類 科: 電力工程、 電子工程

科 目: 工程數學

甲、 申論題部分: ( 50分)

解答()f(x,y)=(x2+2y)e(x2+y2){fx=(2x+(x2+2y)(2x))e(x2+y2)fy=(2+(x2+2y)(2y))e(x2+y2)f=(fx,fy)=((2x34xy+2x)e(x2+y2),(2x2y4y2+2)e(x2+y2))f(P0)=(4e2,0)()Duf(P0)=f(P0)(1,0)=(4e2,0)(1,0)=4e2
解答()I(x)=e(3/x)dx=e3lnx=x3()I(x)y+I(x)3yx=xI(x)x3y+3x2y=x4(x3y)=x4x3y=x4dx=15x5+Cy(1)=33=15+CC=145x3y=15x5+145y=15x2+145x3
解答λ1=2(Aλ1I)x=0x1+x3=0v1=(t,0,t)t,v2=(0,s,0)t,s,tRλ2=1(Aλ2I)x=0{x1+2x3=0x1+x2+x3=0v3=(2t,t,t)t,tRxAAx=λxA8x=Bx=λ8xxBAB[t0t],[0s0],[2rrr],r,s,tR
解答()(r)fX(X=r)2πrfX(X=x)=(2πx)kfXdx=1k=1πfX(x)=2x,0x1()
解答()f(z)=1z2+4=1(z2i)(z+2i)C1f(z)dz=2πi(Res f(2i)Res f(2i))=2πi(14i+14i)=0()C2f(z)dz=2πi×Res f(2i)=2πi×14i=π2

乙、 測驗題部分: (50分)

解答(1,1,3)×(2,0,4)=(4,10,2)=(a,b,c)a×b×c=80(B)
解答Ax=b[122436][x1x2]=[102030]x1+2x2=10(C)
解答{u=(1,3,2)v=(0,1,4)cosϕ=uv|u||v|=03+81417=5238sin2ϕ=1cos2ϕ=125238=2132380.894(A)
解答(3,1,α)=a(1,2,3)+b(1,0,1)=(a+b,2a,3ab){a+b=32a=13ab=α{a=1/2b=5/2α=3252=1(C)
解答det
解答令A=\begin{bmatrix} 2 & -3 &1 \\ 1 & -2 & 1\\ 1 & -3 & 2\end{bmatrix},則\\(A)\times: A \begin{bmatrix} 0 \\1 \\ 1 \end{bmatrix}= \begin{bmatrix} -2 \\-1 \\ -1 \end{bmatrix} \ne \lambda \begin{bmatrix} 0 \\1 \\ 1 \end{bmatrix} ,\forall \lambda \in \mathbb{R}\\(B)\times: A \begin{bmatrix} 1 \\0 \\ 1 \end{bmatrix}= \begin{bmatrix} 3 \\2 \\ 3 \end{bmatrix} \ne \lambda \begin{bmatrix} 1 \\0 \\ 1 \end{bmatrix}  ,\forall \lambda \in \mathbb{R}\\(C)\times: A \begin{bmatrix} 0 \\0 \\ 1 \end{bmatrix}= \begin{bmatrix} 1 \\1 \\ 2 \end{bmatrix} \ne \lambda \begin{bmatrix} 0 \\0 \\ 1 \end{bmatrix} ,\forall \lambda \in \mathbb{R}  \\(D)\bigcirc: A \begin{bmatrix} -1 \\0 \\ 1 \end{bmatrix}= \begin{bmatrix} -1 \\0 \\ 1 \end{bmatrix} = 1\begin{bmatrix} 1 \\0 \\ 1 \end{bmatrix}\\,故選\bbox[red,2pt]{(D)}
解答 A=\begin{bmatrix}0.8 & 0.3 \\0.2 & 0.7\end{bmatrix} =\begin{bmatrix}-1 & 3/2 \\1 & 1 \end{bmatrix}\begin{bmatrix}1/2 & 0 \\0  & 1\end{bmatrix} \begin{bmatrix}-0.4 & 0.6 \\0.4 & 0.4 \end{bmatrix} \\ \Rightarrow A^n=\begin{bmatrix}0.8 & 0.3 \\0.2 & 0.7\end{bmatrix} =\begin{bmatrix}-1 & 3/2 \\1 & 1 \end{bmatrix}\begin{bmatrix}(1/2)^n & 0 \\0  & 1^n\end{bmatrix} \begin{bmatrix}-0.4 & 0.6 \\0.4 & 0.4 \end{bmatrix} \\ \Rightarrow \lim_{n\to \infty}A^n =\begin{bmatrix}0.8 & 0.3 \\0.2 & 0.7\end{bmatrix} =\begin{bmatrix}-1 & 3/2 \\1 & 1 \end{bmatrix}\begin{bmatrix}0 & 0 \\0  & 1\end{bmatrix} \begin{bmatrix}-0.4 & 0.6 \\0.4 & 0.4 \end{bmatrix}= \begin{bmatrix}0.6 & 0.6 \\0.4  & 0.4\end{bmatrix}\\ \Rightarrow \lim_{x\to \infty} x(k)= \begin{bmatrix}0.6 & 0.6 \\0.4  & 0.4\end{bmatrix}\begin{bmatrix} 100 \\0 \end{bmatrix} =\begin{bmatrix} 60 \\40 \end{bmatrix},故選\bbox[red,2pt]{(C)}
解答\bar z={1\over z} \Rightarrow z=0不可解析,故選\bbox[red,2pt]{(D)}
解答Res f(0)=5-2i \Rightarrow \cases{a=5\\ b=-2} \Rightarrow a+b=3,故選\bbox[red,2pt]{(C)}
解答z=x+iy = t+it^2 \Rightarrow dz=(1+2it)dt \Rightarrow \int_\Gamma z^2dz = \int_0^1 (t+it)^2(1+2it)dt \\=\int_0^1 (-4t^3+2t^2 i)dt =\left.\left[ -t^4+{2\over 3}t^3i \right] \right|_0^1 =-1+{2\over 3}i \Rightarrow \cases{a=-1\\ b=2/3} \Rightarrow ab=-2/3,故選\bbox[red,2pt]{(B)}
解答令f(s)= s^2-s+2\Rightarrow f'(s)= 2s-1 \Rightarrow g(1)=\int_C{f(s) \over (s-1)^2}\,ds =2\pi i\times f'(1) = 2\pi i\\,故選\bbox[red,2pt]{(A)}
解答f(x,y)=x^2\sin(xy) \Rightarrow \nabla f=(f_x,f_y)= (2x\sin(xy)+x^2y\cos(xy),x^3\cos(xy))\\ \Rightarrow \nabla f(1,\pi)= (2\sin \pi+\pi\cos(\pi),\cos (\pi)) =(-\pi,-1),故選\bbox[red,2pt]{(C)}
解答先求齊次解,即y''+y'+y=0 \Rightarrow r^2+r+1=0 \Rightarrow r=-{1\over 2}\pm {\sqrt 3 \over 2}i \\ \Rightarrow y_h =e^{t/2}(C_1 \cos({\sqrt 3\over 2}t)+C_2\sin({\sqrt 3\over 2}t)),又y_p亦為三角函數之線性組合,故選\bbox[red,2pt]{(D)}
解答y=a_0 +a_1x+ a_2x^2 +\cdots \Rightarrow y'=a_1+ 2a_2x +3a_3x^2+4a_4x^3 +\cdots\\ \Rightarrow y'' = 2a_2 + 6a_3x+ 12a_4x^2 + \cdots\\ 將x=0代入 y''+xy'+e^xy = x^2+1\Rightarrow y''(0)+ 2=1 \Rightarrow y''(0)=-1=2a_2 \Rightarrow a_2=-{1\over 2}\\,故選\bbox[red,2pt]{(B)}
解答假設f(t)=g(t)=1, \Rightarrow \cases{L\{f\}=L\{g\}=1/s \\ L\{fg\} = 1/s} \Rightarrow L\{fg\} \ne {1\over s^2},故選\bbox[red,2pt]{(C)}
解答f(t)=\begin{cases} 1,& 0\le t\le 2\\ 0,& \text{otherwise}\end{cases} \Rightarrow F(f(t))=F(\omega)=\int_0^2 e^{-j\omega t}\,dt =-{1\over j\omega }(e^{-2j\omega}-1) \\=-{1\over j\omega }e^{-j\omega}(e^{-j\omega}-e^{j\omega}) =-{1\over j\omega }e^{-j\omega}(-2j\sin \omega) =2e^{-j\omega}{\sin \omega\over \omega},故選\bbox[red,2pt]{(A)}
解答f(x)=\begin{cases}1-x,& 0\le x\le 1\\ 1+x,&-1 \le x\le 0 \end{cases} \Rightarrow \cases{E(X)=0\\ E(X^2)=\int_{-1}^0 x^2(1+x)\,dx +\int_0^1 x^2(1-x)\,dx = 1/6}  \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 = {1\over 6},故選\bbox[red,2pt]{(C)}
解答\iint f_{XY}\,dxdy=1 \Rightarrow \int_0^1 \int_0^1 Axy^2\,dxdy = \int_0^1 {1\over 2}Ay^2\,dy ={1\over 6}A=1 \Rightarrow A=6\\ \Rightarrow f_X(x)= \int_0^1 6xy^2\,dy = 2x \Rightarrow E(X)= \int_0^1 2x^2\,dx = {2\over 3},故選\bbox[red,2pt]{(B)}
解答假設正面出現k次\Rightarrow \cases{P(k=5)=0.4^5\\ P(k=4)=C^5_1\cdot 0.6\cdot 0.4^4\\ P(k=3)=C^5_2\cdot 0.6^2\cdot 0.4^3} \Rightarrow \sum_{i=3}^5 P(k=i) =0.4^3(0.16+ 1.2+ 3.6)\\ =0.064\times 4.96 \approx 0.317,故選\bbox[red,2pt]{(C)}
解答i^{2i} =e^{2i\ln i} =e^{2i\ln e^{\pi i/2 }} = e^{2i\cdot {\pi i\over 2}} =e^{-\pi},故選\bbox[red,2pt]{(D)}

======================== END =====================================

考選部未公布申論題答案,解題僅供參考

沒有留言:

張貼留言