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2019年8月26日 星期一

105年國安三等考試_電子組(選試英文)-工程數學詳解


105年國家安全局國家安全情報人員考試試題
考試別:國家安全情報人員
等別:三等考試
類 科組 :電子組
科 目:工程數學


v=ω×r×v=×(ω×r)=ω(r)r(ω)+(r)ω(ω)r



(一)A=[5/21/21/25/2]det(AλI)=0|5/2λ1/21/25/2λ|=0(λ5/2)2(1/2)2=0(λ2)(λ3)=0λ=3,2λ=3(AλI)X=0[1/21/21/21/2][x1x2]=0x1=x2u1=[11]λ=2(AλI)X=0[1/21/21/21/2][x1x2]=0x1+x2=0u2=[11][11][11](二)P=[u1u2]=[1111]P1=[1/21/21/21/2]PAP1=[3002]


,y

解:f_X(x)= \int{f_{X,Y}\;dy}= \int_0^1{{6(x+y^2)\over 5}\;dy}= \int_0^1{{6x\over 5}+ {6y^2\over 5}\;dy}= \left. \left[ {6xy\over 5} +{2y^3\over 5} \right] \right|_0^1 = {1\over 5}(6x+2)\\ \Rightarrow \bbox[red, 2pt]{f_X(x)={2\over 5}(3x+1),0\le x\le 1};\\ f_Y(y)= \int{f_{X,Y}\;dx}= \int_0^1{{6(x+y^2)\over 5}\;dx}= \int_0^1{{6x\over 5}+ {6y^2\over 5}\;dx}= \left. \left[ {3x^2\over 5} +{6xy^2\over 5} \right] \right|_0^1 ={1\over 5}(3+6y^2)\\ \Rightarrow \bbox[red, 2pt]{f_Y(y)={3\over 5}(2y^2+1),0\le y\le 1};

乙、測驗題部分:(50分)

det(e^A)= e^{tr(A)} \Rightarrow e^{At}= e^{tr(At)} = e^{tr\begin{bmatrix} 0 & t\\ t& 0\end{bmatrix}} =e^0=1,故選\bbox[red,2pt]{(B)}


\sqrt{|\vec{u}|^2|\vec{v}|^2-(\vec{u}\cdot\vec{v})^2} =\sqrt{(3^2+4^2+1^2) (0^2+2^2+6^2)-(0-8+6)^2} =\sqrt{1040-4} =\sqrt{1036},\\故選\bbox[red,2pt]{(A)}




\left|{2+12-18+1 \over \sqrt{2^2+3^2+6^2}} \right|= \left|{-3 \over \sqrt{49}} \right|={3\over 7},故選\bbox[red,2pt]{(A)}


det(A-\lambda I)=0 \Rightarrow \begin{bmatrix}3-\lambda & 2 & 0 & 0\\1 & 2-\lambda & 0 & 0\\ 0 & 0 &1-\lambda & 1\\ 0 & 0 & -2 &4-\lambda \end{bmatrix}=0 \Rightarrow (\lambda-4)(\lambda-3)(\lambda-2)(\lambda-1)=0\\ \Rightarrow \lambda=1,2,3,4\\ \lambda=4\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}-1 & 2 & 0 & 0\\1 & -2 & 0 & 0\\ 0 & 0 & -3 & 1\\ 0 & 0 & -2 & 0 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\\ x_3\\ x_4 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1=2x_2 \\x_4=3x_3\\ x_3=0\end{cases} \Rightarrow 取u_1= \begin{bmatrix}2 \\1\\ 0\\ 0 \end{bmatrix}\\ \lambda=3\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}0 & 2 & 0 & 0\\1 & -1 & 0 & 0\\ 0 & 0 & -2 & 1\\ 0 & 0 & -2 & 1 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\\ x_3\\ x_4 \end{bmatrix}=0 \Rightarrow \begin{cases} x_2=0 \\x_1=x_2\\ 2x_3=x_4\end{cases} \Rightarrow 取u_2= \begin{bmatrix}0 \\0\\ 1\\ 2 \end{bmatrix} \\ \lambda=2\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}1 & 2 & 0 & 0\\1 & 0 & 0 & 0\\ 0 & 0 & -1 & 1\\ 0 & 0 & -2 & 2 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\\ x_3\\ x_4 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1+2x_2=0 \\x_1=0\\ x_3=x_4\end{cases} \Rightarrow 取u_3= \begin{bmatrix}0 \\0\\ 1\\ 1 \end{bmatrix}\\ \lambda=1\Rightarrow (A-\lambda I)X=0 \Rightarrow \begin{bmatrix}2 & 2 & 0 & 0\\1 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -2 & 3 \end{bmatrix}\begin{bmatrix}x_1 \\x_2\\ x_3\\ x_4 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1+x_2=0 \\x_3=0\\ 2x_3=3x_4\end{cases} \Rightarrow 取u_4= \begin{bmatrix}1 \\-1\\ 0\\ 0 \end{bmatrix}\\ ,故選\bbox[red,2pt]{(B)}


\begin{bmatrix}1 & 1 & 2 \\0 & 1 & 1 \\ 1 & 3 & 4 \end{bmatrix} \xrightarrow{-r_1+r_3} \begin{bmatrix}1 & 1 & 2 \\0 & 1 & 1 \\ 0 & 2 & 2 \end{bmatrix} \xrightarrow{-2r_2+r_3} \begin{bmatrix}1 & 1 & 2 \\0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix} \xrightarrow{-r_2+r_1} \begin{bmatrix}1 & 0 & 1 \\0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\\ \begin{bmatrix}1 & 0 & 1 \\0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}x_1 \\x_2 \\ x_3 \end{bmatrix}=0 \Rightarrow \begin{cases} x_1+x_3=0\\ x_2+x_3=0\end{cases} \Rightarrow \begin{bmatrix}1 \\1 \\ -1 \end{bmatrix}為其一解,故選\bbox[red,2pt]{(A)}


L=1不一定收斂,故選\bbox[red,2pt]{(D)}


det(A-\lambda I)=0 \Rightarrow \lambda=0,0,1 \Rightarrow A=P\begin{bmatrix} 0& 0 &0\\ 0 & 0 & 0\\ 0 & 0 & 1\end{bmatrix} P^{-1}(不用把P求出)\\ \Rightarrow A^{99} =P\begin{bmatrix} 0& 0 &0\\ 0 & 0 & 0\\ 0 & 0 & 1^{99}\end{bmatrix}P^{-1} =A,故選\bbox[red,2pt]{(A)}


e^{-z+i}=1-i = \sqrt{2}\left({1\over \sqrt{2}} -{1\over \sqrt{2}}i \right) =\sqrt{2}\left(\cos{-\pi\over 4} +i\sin{-\pi\over 4} \right) =e^{\ln{\sqrt{2}}}\cdot e^{-{\pi \over 4}i} = e^{\ln{\sqrt{2}}-{\pi\over 4}i} \\\Rightarrow -z+i= \ln{\sqrt{2}}-{\pi\over 4}i \Rightarrow z=-\ln{\sqrt{2}}+ \left( {\pi\over 4} +1\right)i,故選\bbox[red,2pt]{(A)}


若z,\omega皆為實數,則\overline{z\omega}=z\omega \ne -z\omega,故選\bbox[red,2pt]{(B)}


\oint_c{f(z)\;dz}= \oint_c{{1 \over z^2+4}\;dz} \Rightarrow \text{Res}(f,2i)=\left . {1\over z+2i}\right|_{2i} = {1\over 4i} \\\Rightarrow \oint_c{f(z)\;dz}= 2\pi i\times \text{Res}(f,2i)= 2\pi i\times {1\over 4i}= {\pi \over 2},故選\bbox[red,2pt]{(C)}


y'''+y''-4y'-4y=0 \Rightarrow \lambda^3+\lambda^2 -4\lambda -4=0 \Rightarrow (\lambda^2-4)(\lambda+1)=0\\ \Rightarrow \lambda=-1,-2,2 \Rightarrow y=c_1e^{-x}+c_2e^{-2x}+c_3e^{2x},故選\bbox[red,2pt]{(D)}


L\left\{e^{at}\cos{\omega t} \right\} ={ s-a\over (s-a)^2+\omega^2},故選\bbox[red,2pt]{(A)}


L\left\{(t^2+1)u(t-2) \right\} = e^{-2s}L\left\{(t+2)^2+1) \right\} = e^{-2s}L\left\{t^2+4t+5) \right\} = e^{-2s}\left( {2\over s^3} +{4\over s^2} + {5\over s}\right)\\ = e^{-2s}\left( {2+4s+5s^2\over s^3} \right),故選\bbox[red,2pt]{(C)}



y'={y+x \over y-x} \Rightarrow -(x+y)+(y-x)y'=0 \equiv M(x,y)+N(x,y)y'=0 \Rightarrow \begin{cases}M=-x-y \\N=y-x \end{cases} \\\Rightarrow \begin{cases}{\partial\over \partial y}M=-1 \\{\partial\over \partial x}N=-1 \end{cases} \Rightarrow {\partial\over \partial y}M={\partial\over \partial x}N \Rightarrow \begin{cases}\Psi(x,y)=\int Mdx =-{1\over 2}x^2 -xy+p(y) \\\Psi(x,y)=\int Ndy={1\over 2}y^2-xy +q(x) \end{cases} \\\Rightarrow \Psi(x,y)=-{1\over 2}x^2-xy+{1\over 2}y^2+C=0 \Rightarrow \Psi(0,-2)=2+C=0 \Rightarrow C=-2\\ \Rightarrow -{1\over 2}x^2-xy+{1\over 2}y^2-2=0 \Rightarrow y^2-2xy-x^2=4,故選\bbox[red,2pt]{(A)}


\sum_{n=2}^\infty{1\over n(n-1)} =\sum_{n=2}^\infty{{1\over n-1}-{1\over n}}=(1-{1\over 2})+ ({1\over 2}-{1\over 3})+\cdots =1,故選\bbox[red,2pt]{(B)}


y=ax^m+bx^n \Rightarrow y'=amx^{m-1}+bnx^{n-1} \Rightarrow y''= am(m-1)x^{m-2} +bn(n-1)x^{n-2}\\ \Rightarrow x^2y''+4xy'-4y = am(m-1)x^{m} +bn(n-1)x^{n} +4amx^m+4bnx^n -4ax^m-4bx^n\\ =(am^2-am+4am-4a)x^m +(bn^2-bn+4bn-4b)x^n =0\\ \Rightarrow \begin{cases}am^2-am+4am-4a=0\\ bn^2-bn+4bn-4b=0 \end{cases} \Rightarrow \begin{cases}am(m-1)+4a(m-1)=0\\ bn(n-1)+4b(n-1)=0 \end{cases} \\\Rightarrow \begin{cases}a(m+4)(m-1)=0\\ b(n+4)(n-1)=0 \end{cases} \Rightarrow (m,n)=(-4,1), (1,-4) \Rightarrow m+n=-3,故選\bbox[red,2pt]{(B)}


\iiint{f_{X,Y,Z}(x,y,z)dzdydx}=1 \Rightarrow \int_0^\infty\int_0^\infty\int_0^\infty{c\cdot x^2e^{-x(2+y+z)}\;dzdydx}\\ = \int_0^\infty\int_0^\infty \left. \left[ -c\cdot xe^{-x(2+y+z)}\right] \right|_0^\infty\;dydx = \int_0^\infty\int_0^\infty  c\cdot xe^{-x(2+y)}\;dydx = \int_0^\infty\left. \left[ -c\cdot e^{-x(2+y)}\right] \right|_0^\infty\;dx \\ = \int_0^\infty ce^{-2x}\;dx = \left. \left[ -{1\over 2}ce^{-2x} \right]\right|_0^\infty ={1\over 2}c=1 \Rightarrow c=2,故選\bbox[red,2pt]{(B)}


0.3\times 0.02+ 0.45\times 0.03 +0.25\times 0.02 = 0.006+0.0135 +0.05=0.0245,故選\bbox[red,2pt]{(B)}


連續型的機率密度函數只能計算區間的機率,不能計算單點的機率,故選\bbox[red,2pt]{(D)}


y'={N\over M} \Rightarrow My'-N=0 \Rightarrow \mu My'-\mu N=0 \Rightarrow {\partial \over \partial x}\mu M= {\partial \over \partial y}(-\mu N)\\ \Rightarrow {\partial \over \partial x}\mu M+ {\partial \over \partial y}\mu N=0,故選\bbox[red,2pt]{(B)}


考選部未公布答案,解題僅供參考

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