114年國安情報三等-工程數學詳解

114年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

解答:$$令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow y'''=m(m-1)(m-2)x^{m-3} \\ \Rightarrow x^3y'''+2x^2y''-xy'+y=(m(m-1)(m-2)+2m(m-1)-m+1)x^m=0 \\ \Rightarrow (m-1)^2(m+1)=0 \Rightarrow m=1,-1 \Rightarrow y_h=c_1x^{-1}+c_2x+c_3x\ln x \\ y_p=Ax^{-2} \Rightarrow y_p'=-2Ax^{-3} \Rightarrow y_p''=6Ax^{-4} \Rightarrow y_p'''=-24Ax^{-5} \\ \Rightarrow x^3y_p''' +2x^2 y_p''-xy_p'+y_p = -9Ax^{-2}= 3x^{-2} \Rightarrow A=-{1\over 3} \Rightarrow y_p=-{1\over 3}x^{-2} \\ \Rightarrow y=y_h+ y_p \Rightarrow  \bbox[red, 2pt]{y=c_1x^{-1}+c_2x+c_3x\ln x-{1\over 3}x^{-2}}$$

解答:$$\cases{y_1'=y_1+4y_2+6t-t^2\\ y_2'=y_1+ y_2-1+t-t^2}, 且\cases{y_1(0)=2\\ y_2(0)=-1} \\ \Rightarrow \cases{L\{y_1'\}= L\{y_1+4y_2+6t-t^2\}\\ L\{y_2'\}= L\{y_1+ y_2-1+t-t^2\}} \Rightarrow \cases{sY_1(s)-2=Y_1(s)+4Y_2(s)+{6\over s^2}-{2\over s^3} \\ sY_2(s)+1=Y_1(s) +Y_2(s)-{1\over s}+{1\over s^2}-{2\over s^3} } \\  \Rightarrow \cases{Y_1(s)={1\over s-1}\left(4Y_2(s)+{6\over s^2}-{2\over s^3}+2 \right) \cdots(1)\\ Y_2(s) ={1\over s-1} \left( Y_1(s)  -{1\over s}+{1\over s^2}-{2\over s^3}-1\right) \cdots(2)}, 將(2)代入(1) \\\Rightarrow Y_1(s)={1\over s-1}\left({4\over s-1} \left( Y_1(s)  -{1\over s}+{1\over s^2}-{2\over s^3}-1\right)+{6\over s^2}-{2\over s^3}+2 \right) \\ \Rightarrow Y_1(s) ={-4\over s(s+1)(s-3)} +{4\over s^2(s+1)(s-3)}-{8\over s^3(s+1)(s-3)} -{4\over (s-3)(s+1)} \\\qquad +{6(s-1)\over s^2(s+1)(s-3)}-{2(s-1) \over s^3(s+1)(s-3)} +{2(s-1) \over (s-3)(s+1)} \\ \Rightarrow y_1(t) =L^{-1}\{Y_1(s)\} =t^2+2e^{-t} \Rightarrow y_1'(t) =2t-2e^{-t} \\ \Rightarrow y_2(t) ={1\over 4}\left( y_1'-y_1-6t+t^2\right) =-t-e^{-t} \Rightarrow \bbox[red, 2pt]{\cases{y_1(t)=t^2+2e^{-t} \\ y_2(t)=-t-e^{-t}}}$$

解答:$$A= \begin{bmatrix}4& 3\\ 5& 6 \end{bmatrix} =\begin{bmatrix} -1&{3\over 5}\\ 1& 1\end{bmatrix} \begin{bmatrix}1& 0\\ 0& 9 \end{bmatrix} \begin{bmatrix}-{5\over 8}& {3\over 8} \\ {5\over 8} & {5\over 8} \end{bmatrix} \Rightarrow A^{1/2} =\begin{bmatrix} -1&{3\over 5}\\ 1& 1\end{bmatrix} \begin{bmatrix} \pm 1& 0\\ 0& \pm 3 \end{bmatrix} \begin{bmatrix}-{5\over 8}& {3\over 8} \\ {5\over 8} & {5\over 8} \end{bmatrix} \\\Rightarrow A^{1/2}= \begin{bmatrix}\frac{7}{4} & \frac{3}{4} \\\frac{5}{4} & \frac{9}{4} \end{bmatrix} ,\begin{bmatrix}\frac{-7}{4} & \frac{-3}{4} \\\frac{-5}{4} & \frac{-9}{4} \end{bmatrix},\begin{bmatrix}\frac{1}{2} & \frac{3}{2} \\ \frac{5}{2} & \frac{3}{2} \end{bmatrix} ,\begin{bmatrix}\frac{-1}{2} & \frac{-3}{2} \\\frac{-5}{2} & \frac{-3}{2} \end{bmatrix} \\ \mathbf Y^2-6\mathbf Y+9\mathbf I_2 =(\mathbf Y-3\mathbf I_2)^2= A \Rightarrow \mathbf Y=A^{1/2}+3\mathbf I \\ \Rightarrow \bbox[red, 2pt]{\mathbf Y= \begin{bmatrix} \frac{19}{4} & \frac{3}{4} \\\frac{5}{4} & \frac{21}{4} \end{bmatrix} ,\begin{bmatrix}\frac{5}{4} & \frac{-3}{4} \\\frac{-5}{4} & \frac{3}{4} \end{bmatrix}, \begin{bmatrix}\frac{7}{2} & \frac{3}{2} \\ \frac{5}{2} & \frac{9}{2} \end{bmatrix} ,\begin{bmatrix}\frac{5}{2} & \frac{-3}{2} \\\frac{-5}{2} & \frac{3}{2} \end{bmatrix}}$$

解答:$$z=e^{i\theta} \Rightarrow \cases{\cos \theta=(z+1/z)/2\\ d\theta=dz/iz} \Rightarrow {1\over 2\pi} \int_0^{2\pi} {d\theta\over \sqrt 3-\cos \theta} ={1\over 2\pi} \oint_C {1\over \sqrt 3-{1\over 2}(z+{1\over z})}\cdot {dz\over iz}\\ ={1\over 2\pi i} \oint_C {1\over \sqrt 3z-{1\over 2}(z^2+1)}\cdot dz =-{1\over \pi i} \oint_C {1\over z^2-2\sqrt 3z+1}\,dz\\ =-{1\over \pi i} \oint_C {1\over (z-(\sqrt 3+\sqrt 2)) (z-(\sqrt 3-\sqrt 2) )}\,dz =-{1\over \pi i}  \left. \left[ {1\over z-(\sqrt 3+\sqrt 2)} \right] \right|_{z={\sqrt 3-\sqrt 2 }} (2\pi i)\\ =-{1\over \pi i} \cdot ({1\over -2\sqrt 2})\cdot (2\pi i)= {1\over \sqrt 2} = \bbox[red, 2pt]{\sqrt 2\over 2}$$

解答:$$z=x-\mu_x \Rightarrow E(X)=\int_{-\infty}^\infty x\cdot {1\over \sigma \sqrt \pi} e^{-(x-\mu_x)^2/\sigma^2} \,dx =\int_{-\infty}^\infty (z+\mu_x)\cdot {1\over \sigma \sqrt \pi} e^{-z^2/\sigma^2} \,dz \\=\int_{-\infty}^\infty z\cdot {1\over \sigma \sqrt \pi} e^{-z^2/\sigma^2} \,dz+ \int_{-\infty}^\infty \mu_x\cdot {1\over \sigma \sqrt \pi} e^{-z^2/\sigma^2} \,dz =0+ \int_{-\infty}^\infty \mu_x\cdot {1\over \sigma \sqrt \pi} e^{-z^2/\sigma^2} \,dz \\ z=\sigma y  \Rightarrow dz=\sigma dy \Rightarrow E(X)= \int_{-\infty}^\infty \mu_x\cdot {1\over  \sqrt \pi} e^{-y^2} \,dy =\mu_x\cdot{2\over \sqrt{\pi}} \int_0^\infty e^{-y^2}\,dy \\=\mu_x \cdot \lim_{t\to \infty}{2\over \sqrt \pi} \int_0^t e^{-y^2}\,dy =\mu_x \cdot \lim_{t\to \infty} erf(t) =\mu_x \Rightarrow E(X)=\mu_x \\z=x-\mu_x \Rightarrow Var(X)=\int_{-\infty}^\infty (x-\mu_x)^2\cdot {1\over \sigma \sqrt \pi} e^{-(x-\mu_x)^2/\sigma^2} \,dx =\int_{-\infty}^\infty z^2\cdot {1\over \sigma \sqrt \pi} e^{-z^2/\sigma^2} \,dz \\z=\sigma y  \Rightarrow dz=\sigma dy \Rightarrow Var(X)= {\sigma^2\over  \sqrt \pi}\int_{-\infty}^\infty   y^2e^{-y^2} \,dy \\t=y^2 \Rightarrow y=\sqrt t \Rightarrow dt=2ydy =2\sqrt t dy \Rightarrow Var(X) ={\sigma^2\over  \sqrt \pi}\int_{-\infty}^\infty t e^{-t}\cdot {1\over 2\sqrt t}\,dt \\={\sigma^2\over  2\sqrt \pi}\int_{-\infty}^\infty t^{3/2-1} e^{-t} \,dt= {\sigma^2\over  \sqrt \pi}\int_{0}^\infty t^{3/2-1} e^{-t} \,dt={\sigma^2\over  \sqrt \pi} \Gamma(3/2) ={\sigma^2\over  \sqrt \pi}\cdot {\sqrt \pi\over 2} ={\sigma^2\over 2} \\ \Rightarrow Var(X)={\sigma^2\over 2} =E(X^2)-\mu_x^2 \Rightarrow E(X^2)={\sigma^2\over 2}+\mu_x^2 \\\Rightarrow E(X^2+ 4X+2) =E(X^2)+4E(X)+2= \bbox[red, 2pt]{{\sigma^2\over 2}+\mu_x^2+4\mu_x+2}$$

解答:$$A=\begin{bmatrix} 1& 2& 3& 1\\ 2& 5& 8& 2\\ 3& 8& 14& 3\end{bmatrix} \Rightarrow rref(A) = \begin{bmatrix} 1& 0& 0& 1\\ 0& 1& 0& 0\\ 0& 0& 1& 0 \end{bmatrix} \Rightarrow \mathbf x=\begin{bmatrix} 1 \\ 0\\ 0\end{bmatrix} 為唯一解，故選\bbox[red, 2pt]{(A)}$$

解答:$$B= \left[ \begin{matrix}1 & 1 & -1 & -1\\1 & 2 & -3 & -2\\1 & 0 & 1 & 0\end{matrix} \right] \Rightarrow rref(B)= \left[ \begin{matrix}1 & 0 & 1 & 0\\0 & 1 & -2 & -1\\0 & 0 & 0 & 0\end{matrix} \right] \Rightarrow rank(B)=2，故選\bbox[red, 2pt]{(B)}$$

解答:$$(A)\times: rref(A) =\left[\begin{matrix}1 & 1\\0 & 0\end{matrix}\right] \Rightarrow rank(A)=1\\ (B) \bigcirc: \det(A-\lambda I) = \lambda(\lambda-2)=0 \Rightarrow 特徵值\lambda=0,2 \\(C)\times: \det(A)=0 \Rightarrow A \text{ is NOT invertible} \\(D)\times: A \text{ is  positive semi-definite}\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\det(A-\lambda I) =\begin{vmatrix}-\lambda & 1& 0\\ 0& -\lambda & 1\\ -1&-3& -3-\lambda \end{vmatrix}= -\lambda^3-3\lambda^2-3\lambda-1 =-(\lambda^3+3\lambda^2+3\lambda+1)，故選\bbox[red, 2pt]{(B)}$$

解答:$$rref(A) = \left[\begin{matrix}1 & 0 & 1\\0 & 1 & 4\\0 & 0 & 0\end{matrix} \right] \Rightarrow \cases{x_1+x_3=0\\ x_2+4x_3=0} \Rightarrow v= x_3\begin{bmatrix} -1\\ -4\\ 1\end{bmatrix} \\ 取x_3=-1 可得v=[1\; 4 \;-1]^T，故選\bbox[red, 2pt]{(D)}$$

解答:$$Q(x_1,x_2) =4x_1^2+12x_1x_2 +9x_2^2 =[x_1 \;x_2] \begin{bmatrix}4&6\\6& 9 \end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} \\ A=\begin{bmatrix}4&6\\6& 9 \end{bmatrix} \Rightarrow \det(A-\lambda I) = \lambda(\lambda-13) =0 \Rightarrow 特徵值\lambda=0,13 \Rightarrow \text{positive semi-definite}\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$\det(A^TA) =\det(A^T) \det(A)=\det(A) \det(A) =(\det(A))^2=k=196 \Rightarrow k=14，故選\bbox[red, 2pt]{(B)}$$

解答:$$A= \left[\begin{matrix}0 & 3\\1 & 0\end{matrix}\right] \Rightarrow A^2= \left[\begin{matrix}3 & 0\\0 & 3\end{matrix} \right] \Rightarrow A^{20} =(A^2)^{10} = \left[ \begin{matrix}3^{10} & 0\\0 & 3^{10}\end{matrix} \right]，故選\bbox[red, 2pt]{(D)}$$

解答:$$B=\int_0^1 e^{At}\,dt \Rightarrow \text{trace}(B) =\int_0^1 \left(e^{2t} + e^{3t} +e^{4t} \right)\,dt ={1\over 2}(e^2-1)+{1\over 3}(e^3-1) +{1\over 4}(e^4-1)\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$(A)\times: {\partial \over \partial y} y \ne -{\partial \over \partial x}x \\ (B)\times: {\partial \over \partial y} (x^2+y^2)=2y \ne -{\partial \over \partial x}(2xy)=-2y \\(C)\bigcirc: \cases{{\partial \over \partial x} e^x \cos y=e^x \cos y = {\partial \over \partial y}e^x \sin y \\{\partial \over \partial y} e^x \cos y=-e^x\sin y \ne -{\partial \over \partial x}e^x \sin y } \Rightarrow \text{analytic} \\(D)\times: {\partial \over \partial x}e^y\cos x=-e^y\sin x \ne {\partial \over \partial y}e^y\sin x=e^y\sin x\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(z)={1\over z^2(1-z)} ={1\over z^2}(1+z+ z^2+\cdots) ={1\over z^2}+ {1\over z}+1+\cdots，故選\bbox[red, 2pt]{(A)}$$

解答:$$\lim_{z\to {\pi/2}} {(x-\pi/2)\sin z \over \cos z} =\lim_{z\to {\pi/2}} {\sin z+(x-\pi/2)\cos z \over -\sin z} = -1，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{P(x,y)=x^2+y^2 \\ Q(x,y)= 2xy+\cos x} \Rightarrow P_x=2x =Q_y \Rightarrow \text{ exact} \\ \Rightarrow \Phi(x,y)=\int (x^2+ y^2)\,d y= \int (2xy+\cos x)\,dx  \Rightarrow \Phi(x,y)=x^2y+{1\over 3}y^3+ \rho(x) =x^2y+\sin x+ \phi(y) \\ \Rightarrow \Phi(x,y)=x^2y+{1\over 3}y^3+\sin x =C，故選\bbox[red, 2pt]{(D)}$$

解答:$$特徵方程式為\lambda^4+ \lambda^3-\lambda^2+1=0 \Rightarrow \lambda=-1為其中一根，故選\bbox[red, 2pt]{(C)}$$

解答:$$特徵方程式為\lambda^2-4 \lambda+5=0 \Rightarrow \lambda= 2\pm i \Rightarrow y_h= e^{2x}(c_1 \cos x+ c_2\sin x) \\(A)\bigcirc: 取c_1=1, c_2=0 \Rightarrow y_h=e^{2x} \cos x\\ (B) \bigcirc: 取c_1=0,c_2=1 \Rightarrow  y_h=e^{2x}\sin x\\ (C) \bigcirc: 取c_1=1, c_2=2 \Rightarrow y_h=e^{2x}(\cos x+2\sin x)\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a_0= \int_0^1 f(x)\,dx ={7\over 2} \\ a_n=2 \int_0^1 f(x) \cos(2n\pi x)\,dx =0 \\ b_n= 2\int_0^1 f(x) \sin(2n\pi x)dx =-{3\over n\pi} } \Rightarrow f_1= {7\over 2}-\sum_{n=1}{3\over n\pi} \sin(2n\pi x) \\ \cases{a_0= \int_0^1 f(x)\,dx ={7\over 2} \\a_n=2 \int_0^1 f(x) \cos(n\pi x)\,dx = {6\over n^2\pi^2}((-1)^n-1)} \Rightarrow f_3={7\over 2} +\sum_{n=1}^\infty {6\over n^2\pi^2}((-1)^n-1) \cos(n\pi x)\\ \Rightarrow f_1(0.5) ={7\over 2}=f_3(0.5)，故選\bbox[red, 2pt]{(D)}$$

解答:$$邊界條件一: 端點值, 如:u(0,t)=0,u(1,t)=0,...\\邊界條件二:端點梯度, 如:u_x(0,t)=0, u_x(1,t)=0,...\\ 條件一及條件二的混合, 如u(0,t)+u_x(0,t)=0, \\不可以有兩個條件二的混合, 如:u_x(0,t)+u_x(1,t)=0，故選\bbox[red, 2pt]{(B)}$$

解答:$$\mathcal{F}\{e^{iat}\} =F(\omega-a) \Rightarrow \mathcal{F}\{f(t)2i\sin 2t\} = \mathcal{F}\{f(t)(e^{i2t}-e^{-2it})\} =F(\omega-2)-F(\omega+2)，故選\bbox[red, 2pt]{(B)}$$

解答:$${正常品判定為正常 \over 正常品判定為正常+瑕疵品判定為正常} ={60\% \times 90\% \over 60\% \times 90\% +40\% \times 20\%} ={27\over 31}，故選\bbox[red, 2pt]{(A)}$$

解答:$$Var(Y) =Var({X_1+ X_2\over 2}) ={1\over 4}Var(X_1+X_2) ={1\over 4}\cdot 2\sigma_x^2 ={1\over 2}\sigma_x^2 \Rightarrow \sigma_y= {\sigma_x\over \sqrt 2}\\ 同理, \sigma_z={\sigma_x\over \sqrt 3}, 因此\sigma_x \gt \sigma_y\gt \sigma_z，故選\bbox[red, 2pt]{(C)}$$3

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