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:(一)積分因子I(x)=e∫(3/x)dx=e3lnx=x3(二)I(x)y′+I(x)3yx=xI(x)⇒x3y′+3x2y=x4⇒(x3y)′=x4⇒x3y=∫x4dx=15x5+C將y(1)=3代入上式⇒3=15+C⇒C=145⇒x3y=15x5+145⇒y=15x2+145x−3
解答:λ1=2⇒(A−λ1I)x=0⇒x1+x3=0⇒特徵向量v1=(t,0,−t)t,v2=(0,s,0)t,s,t∈Rλ2=1⇒(A−λ2I)x=0⇒{x1+2x3=0x1+x2+x3=0⇒特徵向量v3=(−2t,t,t)t,t∈R若x為A之特徵向量⇒Ax=λx⇒A8x=Bx=λ8x⇒x為B之特徵向量⇒A與B有相同的特徵向量,其一般式為[t0−t],[0s0],[−2rrr],r,s,t∈R
解答:(一)與靶心相同距離(r)有相同的機率,且機率與面積成正比,因此fX(X=r)與周長2πr成正比因此假設fX(X=x)=(2πx)⋅k,再由∫fXdx=1⇒k=1π⇒fX(x)=2x,0≤x≤1(二)
解答:(一)f(z)=1z2+4=1(z−2i)(z+2i)⇒∮C1f(z)dz=2πi(Res f(2i)+Res f(−2i))=2πi(14i+1−4i)=0(二)∮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,故選(B)解答:Ax=b≡[122436][x1x2]=[102030]⇒x1+2x2=10有無窮多解,故選(C)
解答:{→u=(1,3,2)→v=(0,−1,4)⇒cosϕ=→u⋅→v|→u||→v|=0−3+8√14⋅√17=5√238⇒sin2ϕ=1−cos2ϕ=1−25238=213238≈0.894,故選(A)
解答:(3,1,α)=a(1,2,3)+b(1,0,−1)=(a+b,2a,3a−b)⇒{a+b=32a=13a−b=α⇒{a=1/2b=5/2⇒α=32−52=−1,故選(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)}
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考選部未公布申論題答案,解題僅供參考
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