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2023年11月24日 星期五

111年地方特考-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)f=(fx,fy)=((2x2x34xy)e(x2+y2),(22x2y4y2)e(x2+y2)f(P0)=f(1,1)=(4e2,0)()u=(1,0)fu=(4e2,0)(1,0)=4e2

解答:()I(x)=e3xdx=x3()I(x)y+I(x)3xy=I(x)xx3y+3x2y=x4(x3y)=x4x3y=x4dxx3y=15x5+c1y=15x2+c1x3,y(1)=315+c1=3c1=145y=15x2+145x3


解答:()λ1=1(Aλ1I)x=0[102111102][x1x2x3]=0{x1+2x3=0x2x3=0:{x3(211)}λ2=2(Aλ2I)x=0[202101101][x1x2x3]=0x1+x3=0:{x2(010)+x3(101)}a(211),b(010),c(101),a,b,cR()vA,Av=λvA2v=λAv=λ2vA8v=λ8vvA8AA8a(211),b(010),c(101),a,b,cR


解答:()1=12π=π,,fX(x)=kxπ滿fdx=110kxπdx=1k2π=1k=2πfX(X)=2πXπ=2X ()FY(y)=P(Y<y)=P(X2<y)=P(X<y)=y02xdx=yFY(y)=yfY(y)=ddyFY(y)=1fY(y)=1

解答:()f(z)=1z2+4=1(z+2i)(z2i)f(z) has simple poles at z=±2iz=±2iC1{Res(f,2i)=1z+2i|z=2i=14iRes(f,2i)=1z2i|z=2i=14iC1f(z)dz=2πi(Res(f,2i)+Res(f,2i))=2πi×0=0(){z=2iC2z=2iC2C1f(z)dz=2πi×Res(f,2i)=π2

乙、測驗題部分:(50分)

解答:|ijk113204|=4i+10j+2k=(4,10,2)=(a,b,c)a×b×c=80,(B)
解答:[111111][x1x2]=[11]x1+x2=1,(C)
解答:cosϕ=(1,3,2)(0,1,4)
解答:無法構成基底代表三向量為線性相依,即a(1,2,3)+b(1,0,-1)=(3,1,\alpha) \\ \Rightarrow \cases{a+b=3\\2a=1\\ 3a-b=\alpha} \Rightarrow \cases{a=1/2\\ b=5/2} \Rightarrow \alpha={3\over 2}-{5\over 2}=-1 ,故選\bbox[red, 2pt]{(C)}
解答: \det(A)=7 \ne 0 \Rightarrow A為可逆,故選\bbox[red, 2pt]{(A)}
解答:\begin{bmatrix} 2& -3 & 1\\ 1& -2& 1\\ 1& -3 & 2\end{bmatrix} \begin{bmatrix}-1\\ 0 \\ 1 \end{bmatrix} =\begin{bmatrix} -1\\ 0 \\ 1\end{bmatrix} 符合Av=v\Rightarrow v是特徵向量 ,故選\bbox[red, 2pt]{(D)}
解答:A=\left(\begin{matrix}\frac{4}{5} & \frac{3}{10} \\\frac{1}{5} & \frac{7}{10}\end{matrix}\right) =\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1\end{matrix}\right) \left(\begin{matrix}\frac{1}{2} & 0 \\0 & 1\end{matrix}\right)\left(\begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5} \end{matrix}\right) \Rightarrow A^\infty=\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1 \end{matrix} \right) \left(\begin{matrix}\frac{1}{2^\infty} & 0 \\0 & 1^\infty\end{matrix}\right) \left( \begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right) \\=\left(\begin{matrix}-1 & \frac{3}{2} \\1 & 1 \end{matrix} \right) \left(\begin{matrix}0 & 0 \\0 & 1 \end{matrix}\right) \left( \begin{matrix}\frac{-2}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right)=  \left( \begin{matrix}\frac{3}{5} & \frac{3}{5} \\\frac{2}{5} & \frac{2}{5}\end{matrix} \right) \Rightarrow A^\infty \begin{pmatrix} 100\\ 0\end{pmatrix}=\begin{pmatrix} 60 \\ 40\end{pmatrix} ,故選\bbox[red, 2pt]{(C)}
解答:f(z)=\bar z \Rightarrow \lim_{z\to 0}{f(z)-f(0)\over z-0} =\lim_{z\to 0}{\bar z\over z} =\lim_{(x,y)\to (0,0)}{x-iy\over x+iy}=\begin{cases} 1,& x\to 0,y=0\\ -1,& x=0, y\to 0\end{cases} \\ \Rightarrow \bar z在z=0不可微 ,故選\bbox[red, 2pt]{(D)}
解答:留數={1\over z}的係數=5-2i=a+bi \Rightarrow a+b=3 ,故選\bbox[red, 2pt]{(C)}
解答: \Gamma可以表示成r(t)=(t,t^2),0\le t\le 1,因此z=t+it^2 \Rightarrow dz=dt+2itdt =(1+2it)dt\\ \Rightarrow \int_\Gamma z^2\,dz = \int_0^1 (t+it^2)^2(1+2it)dt = \int_0^1 ((t^2-5t^4)+(4t^3-2t^5)i)dt\\ ={1\over 3}-1+(1-{1\over 3})i=-{2\over 3}+{2\over 3}i \Rightarrow a\cdot b=-{2\over 3}\times{2\over 3}=-{4\over 9}=-0.44,故選\bbox[red, 2pt]{(B)}
解答:g(z)=\int_C {s^2-s+2\over (s-z)^2} \,ds\Rightarrow g(1)=\int_C {s^2-s+2\over (s-1)^2} \,ds\\ 令f(s)=s^2-s+2 \Rightarrow f'(s)=2s-1 \Rightarrow f'(1)=1 \Rightarrow \int_C {s^2-s+2\over (s-1)^2} \,ds =2\pi i f'(1)=2\pi i\\ ,故選\bbox[red, 2pt]{(A)}
解答: f=x^2\sin(xy) \Rightarrow \nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial 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:y''+y'+y=0的解為e^{-t/2}(c_1\cos(\sqrt 3t/2)+ c_2\sin (\sqrt 3t/2)) \\ \Rightarrow y''+y'+y=\sin (\omega t)的解為週期函數,但頻率不一定是\omega ,故選\bbox[red, 2pt]{(D)}
解答:y=\sum_{n=0}^\infty a_n x^n\Rightarrow y'=\sum_{n=1}^\infty na_nx^{n-1} \Rightarrow  y''=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}  \Rightarrow y''(0)=2a_2\\ 將x=0代入原式y''(x)+xy'(x)+ e^xy(x)=x^2+1 \Rightarrow y''(0)+0+2=1 \Rightarrow 2a_2+2=1 \\ \Rightarrow a_2=-{1\over 2},故選\bbox[red, 2pt]{(B)}
解答:L與L^{-1}皆為線性轉換,因此(A)及(B)皆正確\\ 又\cases{u=f(t)\\ dv=e^{-st}dt} \Rightarrow \cases{du= f'(t)dt\\ v=-{1\over s}e^{-st}} \Rightarrow sL\{f\}= s\int_0^\infty f(t)e^{-st}\,dt \\= s \left( \left.-{1\over s}f(t)e^{-st}\right|_0^{\infty} +{1\over s}\int_0^\infty f'(t)e^{-st}\,dt\right) =f(0)+L\{f'(t)\} =L\{f'(t)\}\\ \Rightarrow sL\{f\}= L\{f'\},因此(D)正確 ,故選\bbox[red, 2pt]{(C)}
解答:若g(x)=\begin{cases} 1,& |x|\lt 1\\ 0,& 其它\end{cases} \Rightarrow G(\omega)={2\sin(\omega)\over \omega} \Rightarrow f(t)=g(t-1) \\\Rightarrow F(\omega)=G(\omega)e^{-j\omega} ={2\sin(\omega)\over \omega} e^{-j\omega} ,故選\bbox[red, 2pt]{(A)}
解答: f(x)=\begin{cases} 1-|x|,& |x|\le 1\\ 0,& |x|\gt 1\end{cases} =\begin{cases} 1+x,& -1\le x\le 0\\ 1-x,& 0\le x\le 1\\0,& 其它\end{cases} \Rightarrow E(X)=0 (圖形對稱y軸)\\ \Rightarrow E(X^2)=\int_{-1}^0 x^2(1+x)\,dx + \int_0^1 x^2(1-x)\,dx = {1\over 6} \Rightarrow Var(X)=E(X^2)-(E(X))^2 ={1\over 6}\\ ,故選\bbox[red, 2pt]{(C)}
解答: \iint f(x,y)\,dxdy=1 \Rightarrow \int_0^1 \int_0^1 Axy^2\,dx dy= \int_0^1{1\over 2}Ay^2 \,dy= {1\over 6}A=1 \Rightarrow A=6\\ f_X(x) =\int_0^1 6xy^2\,dy=2x \Rightarrow E(X) =\int_0^1 x f_X(x) \,dx = \int_0^1 2x^2\,dx = {2\over 3},故選\bbox[red, 2pt]{(B)}
解答: \cases{P(n=5)=0.4^5\\ P(n=4)=C^{5}_4\cdot 0.4^4\cdot 0.6\\ P(n=3)= C^5_3 \cdot 0.4^3\cdot 0.6^2} \Rightarrow P(n=5)+ P(n=4)+P(n=3) \\=0.4^3(0.4^2+3\cdot 0.4+10\cdot 0.6^2) =0.064\cdot 4.96 =0.31744,故選\bbox[red, 2pt]{(C)}
解答: e^{i\theta} =\cos \theta+i \sin \theta \Rightarrow e^{\pi i/2} =i \Rightarrow \ln i={\pi \over 2}i\\因此 i^{2i} =e^{2i\ln i} = e^{2i \cdot  \pi i/2} =e^{-\pi},故選\bbox[red, 2pt]{(D)}

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