115年情報三等-工程數學詳解

115年國家安全局國家安全情報人員考試

考 試 別：國家安全情報人員考試
等 別：三等考試
類科組別：電子組（選試英文）
科 目：工程數學

甲、申論題部分：（50 分）

解答:$$\det(A-\lambda I)= \lambda^2+\lambda-6= (\lambda+3)(\lambda-2)=0 \Rightarrow \lambda= -3,2\\ \lambda=-3 \Rightarrow (A+3I)v= \begin{bmatrix}1& 2\\2& 4 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix}=0 \Rightarrow x+2y=0 \Rightarrow v= \begin{bmatrix} -2\\ 1 \end{bmatrix} \Rightarrow u_1={1\over \sqrt 5} \begin{bmatrix}-2\\ 1 \end{bmatrix} \\ \lambda=2 \Rightarrow (A-2I)v=\begin{bmatrix}-4& 2\\2 &-1 \end{bmatrix} \begin{bmatrix} x\\ y  \end{bmatrix} =0 \Rightarrow 2x=y \Rightarrow v= \begin{bmatrix} 1\\2 \end{bmatrix} \Rightarrow u_2={1\over \sqrt 5} \begin{bmatrix}1\\ 2 \end{bmatrix} \\ \Rightarrow P=[u_1\; u_2] \Rightarrow \bbox[red, 2pt]{P= \begin{bmatrix}-{2\over \sqrt 4} & {1\over \sqrt 5}\\{1\over \sqrt 5} & {2\over \sqrt 5} \end{bmatrix}, D= \begin{bmatrix}-3 &0\\ 0&2\end{bmatrix}}$$

 

解答:$$y'''-2y''-y'+2y=2x-1 \Rightarrow y^3-2r^2-r+2=0 \Rightarrow (r-1)(r+1)(r-2)=0 \\ \Rightarrow r=1,-1,2 \Rightarrow y_h=c_1e^x+ c_2e^{-x}+ c_3x^{2x} \\ y_p=Ax+B\Rightarrow \cases{y_p'=A\\ y_p''=y_p'''=0} \Rightarrow y'''_p-2y''_p-y'_p+2y_p= 2Ax+(2B-A)=2x-1\\ \Rightarrow \cases{2A=2\\ 2B-A=-1} \Rightarrow \cases{A=1\\ B=0} \Rightarrow y_p=x \Rightarrow y= y_h+ y_p \Rightarrow \bbox[red, 2pt]{y= c_1e^x+ c_2e^{-x}+ c_3x^{2x}+x}$$

解答:$$f(z)={2z+1\over z^2+3iz}={2z+1\over z(z+3i)} \\分母:z(z+3i)=0 \Rightarrow z=0,-3i 皆在\gamma內\Rightarrow \text{Res}(f,z) = \left. \left[ {2z+1\over z+3i} \right] \right|_{z=0} + \left. \left[ {2z+1\over z} \right] \right|_{z=-3i}=2 \\ \Rightarrow \oint_\gamma f(z)\,dz= 2\pi i\times 2=\bbox[red, 2pt]{4\pi i}$$

解答:$$ 取\cases{u=x\\ dv=\lambda e^{-\lambda }\,dx} \Rightarrow \cases{du=dx\\ v=-e^{-\lambda x}} \Rightarrow E[X]=   \int_0^\infty x\lambda e^{-\lambda x}\,dx=\left. \left[ -xe^{-\lambda x} \right] \right|_0^\infty + \int_0^\infty e^{-\lambda x} \,dx \\=0 + \left. \left[ -{1\over \lambda} e^{-\lambda x} \right] \right|_0^\infty  ={1\over \lambda} \\ 取\cases{u=x^2\\ dv=\lambda e^{-\lambda }\,dx} \Rightarrow \cases{du =2x\,dx\\ v=-e^{-\lambda x}} \Rightarrow  E[X^2] =\int_0^\infty x^2\lambda e^{-\lambda x}\,dx = \left. \left[ -x^2e^{-\lambda x} \right] \right|_0^\infty + \int_0^\infty 2xe^{-\lambda x}\,dx \\=0+ {2\over \lambda} \int_0^\infty x (\lambda e^{-\lambda x})\,dx ={2\over \lambda }E[X] ={2\over \lambda}\times {1\over \lambda} ={2\over \lambda^2 } \Rightarrow Var(X)=E[X^2]-(E[X])^2={2\over \lambda^2}-{1\over \lambda^2} \\={1\over \lambda^2} \Rightarrow \bbox[red, 2pt]{\cases{期望值:1/\lambda\\ 變異數:1/\lambda^2}}$$

乙、測驗題部分：（50 分）

解答:$$(C)\times: 若外積為w=[-1 \; 1\; -1]^T \Rightarrow u\cdot w=v\cdot w=0, 但v\cdot w=0-1-1=-2\ne 0，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: p(\lambda)= \det(\lambda I-A) \Rightarrow p(0)=\det(-A) =(-1)^n \det(A)\\\qquad 此題n=7 \Rightarrow p(0)=-\det(A) \\(B) \times: \text{degree of }p(\lambda) =1+1+2+2+ 1=7 \Rightarrow A為 7\times 7矩陣 \Rightarrow \text{dim}(A)=7 \\(C)\bigcirc: \det(A) =-p(0)=-((-1)\cdot 1\cdot (-2)^2\cdot 2^2\cdot (-3)) =-48 \\(D) \times:\det(A)=-48\ne 0 \Rightarrow A^{-1}存在 \Rightarrow \text{rank}(A)=7\ne 5\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\bigcirc: u_1= \left( u\cdot v\over v\cdot v \right)v={1\over 2} \begin{bmatrix}0\\1\\1 \end{bmatrix} = \begin{bmatrix}0&{1\over 2}& {1\over 2} \end{bmatrix}^T \\(B)\times: u=u_1+u_2 \Rightarrow u_2=u-u_1= \begin{bmatrix}1&0&1 \end{bmatrix}^T- \begin{bmatrix}0&{1\over 2} &{1\over 2} \end{bmatrix} = \begin{bmatrix}1&-{1\over 2}&{1\over 2} \end{bmatrix}^T \\(C)\bigcirc: ||u_2||=\sqrt{1+{1\over 4}+ {1\over 4}} ={\sqrt 6\over 2} \\(D)\bigcirc:  u_1\cdot u_2=0-{1\over 4}+{1\over 4}=0\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$T_{B\to B'} = \begin{bmatrix}a&b\\ c& d \end{bmatrix} \Rightarrow \cases{u_1=av_1+cv_2\\ u_2=bv_1+dv_2}\\ \Rightarrow \cases{ \begin{bmatrix} 0 \\ -1 \end{bmatrix} = a \begin{bmatrix} -1 \\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\ -1 \end{bmatrix} \implies -a = 0, -c = -1 \implies a = 0, c = 1\\ \begin{bmatrix} -1 \\ 0 \end{bmatrix} = b \begin{bmatrix} -1 \\ 0 \end{bmatrix} + d \begin{bmatrix} 0 \\ -1 \end{bmatrix} \implies -b = -1, -d = 0 \implies b = 1, d = 0} \\ \Rightarrow \cases{(A) \times: ab=0\ne 1 \\(B)\bigcirc: bc=1 \\(C)\times: ac=0\ne1 \\ (D)\times: cd=0\ne 1}，故選\bbox[red, 2pt]{(B)}$$

解答:$$T \left( \begin{bmatrix}x\\y \end{bmatrix} \right) = \begin{bmatrix}x+y\\ x+2y \end{bmatrix} = \begin{bmatrix}1& 1\\ 1&2  \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \\A= \begin{bmatrix}1& 1\\1& 2 \end{bmatrix} \Rightarrow \det(A)=1 \ne 0 \Rightarrow \text{rank}(A)=2 \Rightarrow \text{dim(column space of A)}=2 \ne 0，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{A= I\\ B=2I\\ C=0} \Rightarrow AC=0=BC , 但A\ne B，故選\bbox[red, 2pt]{(C)}$$

解答:$$(A)\times: V_1 \parallel V_2 \Rightarrow V_2=kV_1 \Rightarrow {-2\over 1}={2\over 2}={0\over 1}=k \Rightarrow k值不存在\\ (B)\times: 三維空間的基底至少是三個向量，兩個向量無法不足以達成\\ (C)\bigcirc: aV_1+bV_2=[a-2b,2a+2b,b]=0 \Rightarrow a=b=0 \\(D)\times V_1\cdot V_2=-2+4+0=2\ne 0，故選\bbox[red, 2pt]{(C)}$$

解答:$$\det(A-\lambda I)= (\lambda-4) (\lambda-2)=0 \Rightarrow 特徵值為2,4，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(z)= \sec z={1\over \cos z} \Rightarrow 取\cases{p(z)=1\\ q(z)=\cos z} \Rightarrow \text{Res}(f,z_0)={p(z_0) \over q'(z_0)} ={1\over -\sin (z_0)} \\ \Rightarrow  \text{Res}(f,-\pi/2) ={1\over -\sin(-\pi/2)}=1，故選\bbox[red, 2pt]{(B)}$$

解答:$$z^2-2z+2 =0 \Rightarrow z=1\pm i \Rightarrow \cases{1+i在C內\\ 1-i不在C內} \Rightarrow \text{Res}(f,1+i) = \left. \left[ {z-1\over z-(1-i)} \right] \right|_{z=1+i} = {1\over 2} \\ \Rightarrow \oint_C f(z)\,dz= 2\pi i\times {1\over 2}=\pi i \Rightarrow \cases{a=0\\ b=\pi}，故選\bbox[red, 2pt]{(C)}$$

解答:$$|z|\gt 2 \Rightarrow \left| {2\over z}\right| \lt 1且\left| {1\over z}\right| \lt 1 \Rightarrow f(x)={1\over (z-1)(z-2)} ={1\over z-2}-{1\over z-1} ={1\over z \left( 1-{2\over z} \right)}-{1\over z \left( 1-{1\over z} \right)} \\= {1\over z} \left( 1+{2\over z} +{4\over z^2}+{8\over z^3}+ {16\over z^4}+ \cdots\right)- {1\over z} \left( 1+{1\over z}+ {1\over z^2}+ {1\over z^3}+ {1\over z^4}+\cdots \right) \\={1\over z^2} +{3\over z^3}+{7\over z^4}+{15\over z^5}+\cdots \Rightarrow \cases{a=1\\b=3\\ c=7\\ d=15}，故選\bbox[red, 2pt]{(D)}$$

解答:$$取\cases{u(x,y)=x^2+ay^2\\ v(x,y)=2xy} \Rightarrow \cases{u_x=2x\\ u_y=2ay\\v_x=2y\\ v_y=2x} \Rightarrow \cases{u_x=v_y \\ u_y=-v_x \Rightarrow 2ay=-2y \Rightarrow a=-1}，故選\bbox[red, 2pt]{(B)}$$

解答:$$y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow 4x^2(m(m-1)x^{m-2})-4x(mx^{m-1})-5x^m\\=(4m^2-4m-4m-5)x^m=(4m^2-8m-5)x^m=0 \Rightarrow 4m^2-8m-5=0 \\ \Rightarrow (2m-5)(2m+1)=0 \Rightarrow m={5\over 2},-{1\over 2} \Rightarrow y=c_1x^{5/2}+c_2x^{-1/2}，故選\bbox[red, 2pt]{(B)}$$

解答:$$(y+x^2) dx=xdy \Rightarrow x{dy\over dx}=x^2+y \Rightarrow {dy\over dx}-{y\over x}=x \Rightarrow 積分因子I(x)={1\over x} \\ \Rightarrow {1\over x}{dy\over dx}-{1\over x^2}y=1 \Rightarrow \left( {1\over x}y \right)'=1 \Rightarrow {y\over x}= \int 1\,dx =x+C \Rightarrow y=x^2+Cx\\ y(1)=2代入上式 \Rightarrow 2=1+C \Rightarrow C=1 \Rightarrow y=x^2+x \Rightarrow \cases{a=1\\b=1\\c=1}，故選\bbox[red, 2pt]{(D)}$$

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

解答:$$y'+y=x \Rightarrow y'e^x+ye^x= xe^x \Rightarrow (ye^x)'=xe^x \Rightarrow ye^x =\int xe^x\,dx =xe^x-e^x+C \\ \Rightarrow y=x-1+Ce^{-x} \Rightarrow y(0)=-1+C=7 \Rightarrow C=8 \Rightarrow y=x-1+8e^{-x}，故選\bbox[red, 2pt]{(B)}$$

解答:$$(\cos x-2xy)+(e^y-x^2){dy\over dx}=0 \Rightarrow (\cos x-2xy)dx+ (e^y-x^2)dy=0 \\ 取\cases{P(x,y)=\cos x-2xy\\ Q(x,y)=e^y-x^2} \Rightarrow P_y=-2x=Q_x \Rightarrow Exact \\ \Rightarrow \Psi(x,y)=\int P(x,y) \,dx=\int Q(x,y)\,dy \Rightarrow \int(\cos x-2xy)\,dx = \int(e^y-x^2)\,dy \\ \Rightarrow \sin x-x^2y+ \rho(y) =e^y-x^2y+ \phi(x) \Rightarrow \Psi(x,y)=\sin x-x^2y+e^y=C \\ 將y(0)=1代入可得C=e \Rightarrow \sin x-x^2y+e^y=e，故選\bbox[red, 2pt]{(A)}$$

解答:$$(A)\times: \int_0^1 \int_x^1 kx^2y\,dydx=1 \Rightarrow \int_0^1 {1\over 2}kx^2(1-x^2)\,dx =1 \Rightarrow \int_0^1 {1\over 2}\left( kx^2-kx^4 \right)\,dx=1\\ \Rightarrow \left. {k\over 2}\left[ {1\over 3}x^3-{1\over 5}x^5 \right] \right|_0^1= {k\over 15}=1 \Rightarrow k=15\ne 10 \\(B) \times: f_Y(y)= \int_0^y 15x^2y\,dx =5y^4 \ne 5y^3 \\(C)\bigcirc: E[X] = \int_0^1 \int_x^1 x(15x^2y)\,dy dx= {5\over 8} \\(D) \times E[Y]= \int_0^1 \int_x^1 y(15x^2y)\, dxdy={5\over 6} \ne{3\over 8}\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$X\sim Geo(p=1/4) \Rightarrow E[X]={1\over p} =4，故選\bbox[red, 2pt]{(D)}$$

解答:$$P_{Y|X}(y|x) ={P(X=x,Y=y) \over P_X(x)} \Rightarrow P(X=x,Y=y)=P_{Y|X}(y|x)\times P_X(x) \\ \Rightarrow \cases{X=0 \Rightarrow P_X(0)=0.4 \Rightarrow \cases{P(X=0,Y=0)=P_{Y|X}(0|0)\times P_X(0)=0.8\times 0.4=0.32\\ P(X=0,Y=1)=P_{Y|X}(1|0)\times P_X(0)=0.2\times 0.4=0.08} \\X=2 \Rightarrow P_X(2)=0.6 \Rightarrow \cases{P(X=2,Y=0)=P_{Y|X}(0|2)\times P_X(2)=0.5\times 0.6=0.3\\ P(X=2,Y=1)=P_{Y|X}(1|2)\times P_X (2)=0.5\times 0.6= 0.3}} \\ \Rightarrow \quad\begin{array}{r|rr}& Y=0 & Y=1\\ \hline X=0& 0.32& 0.08\\\hline X=2& 0.3& 0.3 \end{array} \Rightarrow P_{X|Y}(0|0) ={P(X=0, Y=0) \over P_Y(Y=0)} ={0.32\over 0.32+0.3} ={0.32\over 0.62} ={16\over 31}\\，故選\bbox[red, 2pt]{(C)}$$

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