103年國安情報人員-工程數學詳解

103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
等 別：三等考試
類 科 組：電子組
科 目：工程數學
甲、申論題部分：（ 50 分）

解答：

(一) $$x^TAx= \begin{bmatrix} x_1& x_2\end{bmatrix} \begin{bmatrix} 4 & 5\\ 1 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix} =\begin{bmatrix} 4x_1+x_2 & 5x_1- 4x_2\end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix} =4x_1^2+x_1x_2 +5x_1x_2- 4x_2^2\\ =4x_1^2 +6x_1x_2- 4x_2^2 \Rightarrow 判別式=6^2-4\times 4\times (-4)\gt 0 \Rightarrow   Q為\bbox[red,2pt]{雙曲線}$$ (二) $$Q=x^TAx= 4x_1^2 +6x_1x_2- 4x_2^2= \begin{bmatrix} x_1& x_2\end{bmatrix} \begin{bmatrix} 4 & 3\\ 3 & -4 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2\end{bmatrix}  \\ 取B=\begin{bmatrix} 4 & 3\\ 3 & -4 \end{bmatrix} \Rightarrow Q=x^TBx (以B取代A, 因為B符合B=B^T)\\ 將B對角化可得B=\begin{bmatrix} 3/\sqrt{10} & 1/\sqrt{10}\\ 1/\sqrt{10} & -3/\sqrt{10} \end{bmatrix}\begin{bmatrix} 5 & 0\\ 0 & -5 \end{bmatrix} \begin{bmatrix} 3/\sqrt{10} & 1/\sqrt{10}\\ 1/\sqrt{10} & -3/\sqrt{10} \end{bmatrix} \equiv PDP^T\\ \Rightarrow 正交主軸轉換矩陣=\bbox[red,2pt]{\begin{bmatrix} 3/\sqrt{10} & 1/\sqrt{10}\\ 1/\sqrt{10} & -3/\sqrt{10} \end{bmatrix}}$$

解答： $$先求齊次解 ，即y''-2y'-8y=0 \Rightarrow 特徵方程式\lambda^2-2\lambda-8=0 \Rightarrow (\lambda-4)(\lambda+2)=0\\ \Rightarrow \lambda =4,-2 \Rightarrow y_h=c_1e^{4x} +c_2e^{-2x}\\ 令\cases{y_1=e^{4x}\\ y_2=e^{-2x}} \Rightarrow W=\begin{vmatrix} y_1 & y_2 \\ y_1' &y_2'\end{vmatrix} =\begin{vmatrix} e^{4x} & e^{-2x} \\ 4e^{4x} & -2e^{-2x}\end{vmatrix} = -6e^{2x} \\ \Rightarrow y_p=-y_1\int {y_2\cdot r(x)\over W}\;dx +y_2\int { y_1r(x)\over W}\;dx,其中r(x)=10e^{-x}+8e^{2x} \\ \Rightarrow y_p =-e^{4x}\int {10e^{-3x}+ 8\over -6e^{2x}}\;dx +e^{-2x}\int { 10e^{3x}+8 e^{6x}\over -6e^{2x}}\;dx \\=-e^{4x} \left({1\over 3}e^{-5x}+{2\over 3}e^{-2x} \right) +e^{-2x} \left(-{5\over 3}e^x-{1\over 3}e^{4x} \right) =-2e^{-x}-e^{2x} \\ \Rightarrow y=y_h+y_p = c_1e^{4x} +c_2e^{-2x}-2e^{-x}-e^{2x} \Rightarrow y'=4c_1e^{4x}-2c_2e^{-2x}+2e^{-x} -2e^{2x}\\ 再將初始值\cases{y(0)=1\\ y'(0)=4}代入，可得\cases{c_1+c_2-2-1=1\\ 4c_1-2c_2+ 2-2=4} \Rightarrow \cases{c_1=2 \\ c_2=2} \\ \Rightarrow \bbox[red,2pt]{y=2e^{4x}-e^{2x}-2e^{-x}+2e^{-2x}}$$

解答： $$x^4= 1+i= \sqrt 2({1\over \sqrt 2}+i{1\over \sqrt 2}) = \sqrt 2(\cos{\pi\over 4}+i \sin{\pi \over 4}) =\sqrt 2e^{i(\pi/4+2k\pi)},k\in \mathbb{Z} \\ \Rightarrow x_k= 2^{1/8} e^{i(\pi/16+k\pi/2)},k=0,1,2,3 \Rightarrow 四次方根為\bbox[red,2pt]{\sqrt[8]2e^{\pi i/16}, \sqrt[8]2e^{9\pi i/16}, \sqrt[8]2e^{17\pi i/16}, \sqrt[8]2e^{25 \pi i /16}}$$

解答：

(一) $$\int_0^2 \int_{-x^2}^{x^2} 2cx \;dydx =\int_0^2 4cx^3\;dx = \left.\left[ cx^4 \right] \right|_0^2 =16c=1 \Rightarrow c=\bbox[red,2pt]{1\over 16}$$ (二) $$f_X(x)= \int_{-x^2}^{x^2} {1\over 8}x\;dy={1\over 4}x^3 \Rightarrow \bbox[red,2pt]{f_X(x)=\begin{cases} {1\over 4}x^3,& 0\le x\le 2\\ 0, &其它;\end{cases}}$$

(三) $$見上圖，\cases{上半部\int_{\sqrt y}^2{1\over 8}x \;dx=1/4-y/16 \\下半部\int_{\sqrt{-y}}^2 {1\over 8}x\;dx =1/4+y/16} \Rightarrow \bbox[red,2pt]{f_Y(y)=\begin{cases} {1\over 4}-{1\over 16}y, & 0\le y\le 4\\ {1\over 4}+{1\over 16}y, & -4\le y\le 0\\ 0,& 其它;\end{cases}}$$

乙、測驗題部分：（50 分）

解答： $$\left.({\partial \over \partial x} z, {\partial \over \partial y} z) \right|_{(4,1)} =\left.(-2x, -18y) \right|_{(4,1)} = (-8,-18)=8(-1,-2.25)，故選\bbox[red,2pt]{(A)}$$

解答： $$F\times G=-G\times F \ne G\times F，故選\bbox[red,2pt]{(A)}$$

解答： $$路徑C可表示成(x(t),y(t)),其中\cases{x(t)=\cos(t)\\ y(t)=\sin(t)} \Rightarrow \cases{x'(t)=-\sin(t)\\ y'(t)=\cos(t)},t=0-{\pi\over 2}\\ 因此\int_C F\cdot dr= \int_C (-y,-xy)\cdot (dx,dy) =\int_0^{\pi/2}(-\sin(t))(-\sin(t)dt)+(-\sin(t)\cos(t))(\cos(t)dt)\\ =\int_0^{\pi/2} \sin^2(t)-\sin(t)\cos^2(t)\;dt =\left.\left[{1\over 2}t-{1\over 4}t\sin(2t)+ {1\over 3}\cos^3(t) \right]\right|_0^{\pi/2} ={\pi\over 4}-{1\over 3}\\ \Rightarrow \cases{a=4\\ b=-3} \Rightarrow a+b=1，故選\bbox[red,2pt]{(A)}$$

解答： $$\cases{\vec F=(x,y,z)  \\ \nabla =({\partial \over \partial x}, {\partial \over \partial y}, {\partial \over \partial z})}\Rightarrow \cases{\nabla\cdot \vec F= {\partial \over \partial x}x+ {\partial \over \partial y}y+ {\partial \over \partial z}z =1+1+1=3\\ \nabla \times \vec F=({\partial \over \partial y}z-{\partial \over \partial z}y,{\partial \over \partial z}x -{\partial \over \partial x}z,{\partial \over \partial x}y-{\partial \over \partial y}x) =\vec 0}\\，故選\bbox[red,2pt]{(D)}$$

解答： $$\text{Hermitian }矩陣的特徵值皆為實數，故選\bbox[red,2pt]{(C)}$$

解答： $$tr(AB)=(4,5,7)\cdot(2,7,10)+ (3,6,8)\cdot (5,9,13)=113+183= 286，故選\bbox[red,2pt]{(D)}$$

解答： $$(B)(C)(D) 均不符合T(0,0)=(0,0)，非線性轉換，故選\bbox[red,2pt]{(A)}$$

解答： $$f(t)=\det(A-t I)= \begin{vmatrix}10-t& -8 \\ 4 & -2-t \end{vmatrix} =t^2-8t+12 \\\Rightarrow f(A)=0 \Rightarrow A^2-8A+12I=0 \Rightarrow A^3-8a^2+12A=0 \Rightarrow A^3-8a^2+15A=3A，故選\bbox[red,2pt]{(C)}$$

解答： $$e^z=1+i\sqrt 3=2({1\over 2}+i{\sqrt 3\over 2})=2(\cos(\pi/3)+ i\sin(\pi/3))=e^{\ln 2}\cdot e^{i\pi/3} \\=e^{\ln 2+i\pi/3} \Rightarrow z=\ln 2+i(\pi/3+2k\pi),k\in \mathbb{Z}，故選\bbox[red,2pt]{(D)}$$

解答： $$|z|\lt |3+4i|=5收斂，故選\bbox[red,2pt]{(C)}$$

解答： $$令y=x+2 \Rightarrow {x-2\over x^2+4x+5} ={x-2\over (x+2)^2+1}={y-4\over y^2+1} =(y-4)(1-y^2+y^4-y^6+\cdots) \\ =(y-y^3+y^5-y^7+\cdots )-(4-4y^2+4y^4-4y^6+ \cdots)\\ \Rightarrow y^2\lt 1則上式收斂，即(x+2)^2 \lt 1 \Rightarrow  |x+2|\lt 1，收斂半徑為1，故選\bbox[red,2pt]{(B)}$$

解答： $$x^2y'+y^2=xy \Rightarrow y'-{1\over x}y=-{1\over x^2}y^2 \text{ (Bernoulli equation)}\\ 取u={1\over y} \Rightarrow u'= -{1\over y^2}y'代回原式\Rightarrow u'+{1\over x}u={1\over x^2} \Rightarrow xu'+u={1\over x} \Rightarrow (xu)'={1\over x}\\ \Rightarrow xu=\ln x+c \Rightarrow u={\ln x+C\over x} \Rightarrow y={x\over \ln x+C}，故選\bbox[red,2pt]{(D)}$$

解答： $${dy\over dx} =6e^{3x}y^2 \Rightarrow \int {1\over y^2}\;dy = \int 6e^{3x}\;dx \Rightarrow -{1\over y}=2e^{3x}+C \Rightarrow y=-{1\over 2e^{3x}+C} \\y(0)=1 \Rightarrow -{1\over 2+C}=1 \Rightarrow C=-3 \Rightarrow y=-{1\over 2e^{3x}-3} ={1 \over 3-2e^{3x}}，故選\bbox[red,2pt]{(B)}$$

解答： $$由\hat f(w)可知:f(t)為矩形函數(\text{rect function})，故選\bbox[red,2pt]{(A)}$$

解答： $$\cases{x= r\cos \theta \Rightarrow  {\partial x\over \partial r}= \cos\theta,{\partial x\over \partial \theta} =-r\sin \theta\\ y=r\sin \theta \Rightarrow  {\partial y\over \partial r}= \sin\theta, {\partial y\over \partial \theta} = r\cos \theta} ;\\因此{\partial u\over \partial r} ={\partial u\over \partial x} {\partial x\over \partial r} +{\partial u\over \partial y} {\partial y\over \partial r} ={\partial u\over \partial x} \cos\theta+{\partial u\over \partial y} \sin\theta \\ \Rightarrow {\partial^2 u\over \partial r^2} = {\partial u\over \partial r} \left({\partial u\over \partial x} \cos\theta+{\partial u\over \partial y} \sin\theta \right) ={\partial u\over \partial x}\left({\partial u\over \partial x} \cos\theta+{\partial u\over \partial y} \sin\theta \right){\partial x\over \partial r} \\\qquad \qquad + {\partial u\over \partial y} \left({\partial u\over \partial x} \cos\theta+{\partial u\over \partial y} \sin\theta \right) {\partial y\over \partial r}\\ = \left({\partial^2 u\over \partial x^2} \cos\theta+ {\partial^2 u\over \partial x \partial y} \sin\theta \right) \cos\theta +  \left({\partial^2 u\over \partial x \partial y} \cos\theta+{\partial^2 u\over \partial y^2} \sin\theta \right) \sin \theta\\ ={\partial^2 u\over \partial x^2} \cos^2\theta+ {\partial^2 u\over \partial x \partial y} \sin\theta \cos\theta + {\partial^2 u\over \partial x \partial y} \cos\theta \sin \theta+{\partial^2 u\over \partial y^2} \sin^2 \theta \\ \Rightarrow {\partial^2 u\over \partial r^2} ={\partial^2 u\over \partial x^2} \cos^2\theta  + 2{\partial^2 u\over \partial x \partial y} \cos\theta \sin \theta+{\partial^2 u\over \partial y^2} \sin^2\theta\cdots(1)\\同理{\partial u\over \partial \theta} ={\partial u\over \partial x}{\partial x\over \partial \theta} +{\partial u\over \partial y}{\partial y\over \partial \theta} ={\partial u\over \partial x}(-r\sin \theta)  +{\partial u\over \partial y}r\cos \theta \\ \Rightarrow {\partial^2 u\over \partial \theta^2} = {\partial u\over \partial x}(-r\cos \theta) +(-r\sin\theta)\left( {\partial \over \partial x}{\partial u\over \partial x} {\partial x\over \partial \theta} + {\partial \over \partial y} {\partial u\over \partial x} {\partial y\over \partial \theta}\right) +{\partial u\over \partial y}(-r\sin \theta ) \\\qquad \qquad +r\cos\theta \left( {\partial \over \partial x}{\partial u\over \partial y} {\partial x\over \partial \theta} + {\partial \over \partial y} {\partial u\over \partial y} {\partial y\over \partial \theta}\right) \\= {\partial u\over \partial x}(-r\cos \theta) +(-r\sin\theta)\left(  {\partial^2 u\over \partial x^2} (-r\sin\theta) +  {\partial^2 u\over \partial x \partial y} r\cos \theta\right) +{\partial u\over \partial y}(-r\sin \theta )\\ \qquad\qquad +r\cos\theta \left(  {\partial^2 u\over \partial x\partial y} (-r\sin \theta) +  {\partial^2 u\over \partial y^2} r\cos\theta\right) \\ ={\partial u\over \partial x}(-r\cos \theta) +  {\partial^2 u\over \partial x^2} (r^2\sin^2 \theta) -  2{\partial^2 u\over \partial x \partial y} r^2\sin \theta\cos \theta  +{\partial u\over \partial y}(-r\sin \theta ) +  {\partial^2 u\over \partial y^2} r^2\cos^2 \theta \\=-r\left( {\partial u\over \partial x}\cos \theta +{\partial u\over \partial y}\sin \theta\right) +r^2\left( {\partial^2 u\over \partial x^2} \sin^2 \theta -  2{\partial^2 u\over \partial x \partial y}  \sin \theta\cos \theta +{\partial^2 u\over \partial y^2}  \cos^2 \theta\right) \\ =-r{\partial u\over \partial r} +r^2\left( {\partial^2 u\over \partial x^2} \sin^2 \theta -  2{\partial^2 u\over \partial x \partial y}  \sin \theta\cos \theta +{\partial^2 u\over \partial y^2}  \cos^2 \theta\right) \\ \Rightarrow {1\over r^2}{\partial^2 u\over \partial \theta^2} =-{1\over r}{\partial u\over \partial r} +  {\partial^2 u\over \partial x^2} \sin^2 \theta -  2{\partial^2 u\over \partial x \partial y}  \sin \theta\cos \theta +{\partial^2 u\over \partial y^2}  \cos^2 \theta \cdots(2)\\ (1)+(2) \Rightarrow {\partial^2 u\over \partial r^2}+ {1\over r^2}{\partial^2 u\over \partial \theta^2} =-{1\over r}{\partial u\over \partial r} +{\partial^2 u\over \partial x^2} +{\partial^2 u\over \partial y^2} \\ \Rightarrow {\partial^2 u\over \partial x^2} +{\partial^2 u\over \partial y^2}={\partial^2 u\over \partial r^2} +{1\over r}{\partial u\over \partial r}+ {1\over r^2}{\partial^2 u\over \partial \theta^2}，故選\bbox[red,2pt]{(A)}$$

解答： $$y=e^{2x} \Rightarrow y'=2e^{2x} \Rightarrow y''=4e^{2x} \Rightarrow x^2y''+Axy'+By = 4x^2e^{2x}+2Axe^{2x}+Be^{2x}  \\ =e^{2x}(4x^2+2Ax+B)  \Rightarrow 無法找出固定的A與B使得4x^2+2Ax+B=0,\forall x\in \mathbb{R}，故選\bbox[red,2pt]{(D)}$$

解答： $$\int_0^{\infty}{1\over t}e^{-st}\;dt 不存在，故選\bbox[red,2pt]{(B)}$$

解答： $$F_Y(y)=f(Y\le y) = f(8X^3\le y) = f(X\le {1\over 2}y^{1/3}) = \int_0^{y^{1/3}/2} 2x\;dx ={1\over 4}y^{2/3} \\ \Rightarrow f_Y(y) = {d\over dy}F_Y(y) ={1\over 6}y^{-1/3}，故選\bbox[red,2pt]{(B)}$$

解答： $${國小女性+中學女性人數\over 女性人數}={95\over 112}，故選\bbox[red,2pt]{(C)}$$

解答： $$$P(X\ge 0.5) =\int_{0.5}^1 \int_0^{1-x} 24xy\;dydx =\int_{0.5}^1 12x(1-x)^2\; dx =\int_{0.5}^1 12x-24x^2+12^3\; dx \\=\left. \left[ 6x^2-8x^3-3x^4 \right] \right|_{0.5}^1 =1-{11\over 16}={5\over 16}，故選\bbox[red,2pt]{(A)}$$

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