111年調查三等_電子科學組-工程數學詳解

 111年法務部調查局調查人員考試

考 試 別：調查人員
等 別：三等考試
類 科 組：電子科學組
科 目：工程數學

解答： $$(一)\nabla\cdot \vec F ={\partial \over \partial x}x^2 +{\partial \over \partial y}xy +{\partial \over \partial z}(y+z^2) = 2x+x+2z= \bbox[red, 2pt]{3x+2z}\\(二)\nabla \times\vec F =\left( {\partial \over \partial y}(y+2z)-{\partial \over \partial z}xy, {\partial \over \partial z}x^2-{\partial \over \partial x}(y+z^2), {\partial \over \partial x}xy- {\partial \over \partial y}x^2\right) =\bbox[red, 2pt]{(1,0,y)} \\(三) \nabla \cdot (\nabla \times \vec F)= \nabla \cdot (1,0,y) =\bbox[red, 2pt]{0} \\(四)\nabla \times (\nabla \times \vec F)= \nabla \times (1,0,y) =\bbox[red, 2pt]{ (1,0,0)}$$

解答： $$(一)只要能寫成y''+p(x)y'+ q(x)y=f(x)，就是線性二階微分方程,因此\bbox[red, 2pt]{是}\\(二) y''+6y'-7y=0 \Rightarrow 特徵方程式r^2+6r-7=0 \Rightarrow (r+7)(r-1)=0 \Rightarrow r=1,-7\\ \Rightarrow 齊次解:\bbox[red, 2pt]{y_h=C_1e^x +C_2e^{-7x},C_1及C_2為常數}\\ (三)特定解y_p= a\cos x+b\sin x \Rightarrow y_p' = -a\sin x+b\cos x \Rightarrow y_p''=-a\cos x-b\sin x\\ \Rightarrow y_p''+6y_p'-7y_p= -a\cos x-b\sin x -6a\sin x+6b\cos x -7a\cos x-7b\sin x\\ =(6b-8a)\cos x-(6a+8b)\sin x=\cos x \Rightarrow \cases{6b-8a=1\\ 6a+8b=0} \Rightarrow \cases{a=-2/25\\b= 3/50}\\ \Rightarrow 特定解\bbox[red, 2pt]{y_p=-{2\over 25}\cos x+{3\over 50}\sin x}$$

解答： $$(一)A=\begin{bmatrix} 3 & -1 \\ 1 & 1\end{bmatrix} \Rightarrow \det(A)= 4 \Rightarrow A^{-1}={1\over \det(A)}\begin{bmatrix} 1 & 1 \\ -1 & 3\end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} 1/4 & 1/4 \\ -1/4 & 3/4\end{bmatrix}} \\(二)\det(A)=\begin{vmatrix} 3 & -1 \\ 1 & 1\end{vmatrix}=3+1=\bbox[red, 2pt]{4}\\(三)\det(A-\lambda I)=0 \Rightarrow \lambda^2-4\lambda+4=0 \Rightarrow (\lambda-2)^2=0 \Rightarrow 特徵值\lambda=2\\ \quad 又(A-2I)X=0 \Rightarrow \begin{bmatrix} 1 & -1 \\ 1 & -1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix}=0 \Rightarrow x_1=x_2\\ \Rightarrow 只能找到一組線性獨立的特徵向量(k,k),k\in \mathbb{R}  \Rightarrow A無法對角化，\bbox[red, 2pt]{故得證}$$

解答： $$(一)\int_{-\infty}^\infty f_X(x)\,dx =1 \Rightarrow \int_{-\infty}^\infty Ce^{-\lambda|x|}\,dx =\int_{-\infty}^0 Ce^{\lambda x}\,dx + \int_0^{\infty} Ce^{-\lambda x}\,dx= 1\\ \quad\Rightarrow \left. \left[{C\over \lambda }e^{\lambda x} \right] \right|_{-\infty}^0 + \left. \left[ -{C\over \lambda }e^{-\lambda x} \right]\right|_{0}^\infty ={2C\over \lambda } =1 \Rightarrow C= \bbox[red, 2pt]{\lambda \over 2} \\(二)E(X) = \int_{-\infty}^\infty xf_X(x)\,dx =\int_{-\infty}^0 Cxe^{\lambda x}\,dx + \int_0^{\infty} Cxe^{-\lambda x}\,dx\\ \quad =C\left. \left[({x\over \lambda }-{1\over \lambda^2})e^{\lambda x} \right] \right|_{-\infty}^0 + C\left. \left[ (-{x\over \lambda }-{1\over \lambda^2})e^{-\lambda x} \right]\right|_{0}^\infty =\bbox[red, 2pt]{0} \\(三)x\gt x^2 \Rightarrow x\in (0,1) \Rightarrow P(X\gt X^2)=\int_0^1 f_X(x)\,dx = \int_0^1 Ce^{-\lambda x}\,dx = \left.\left[ -{C\over \lambda}e^{-\lambda x} \right] \right|_0^1 \\ =-{C\over \lambda}e^{-\lambda}+{C\over \lambda} =\bbox[red, 2pt]{{1\over 2}(1-e^{-\lambda})} \\(四)F_Y(y) = P(Y\le y)= P(|X|\le y) = P(-y\le X\le y) = F_X(y)-F_X(-y)\\ \Rightarrow f_Y(y)=\begin{cases}f_X(y)+f_X(-y) & y\ge 0\\ 0 & \text{otherwise} \end{cases} =\begin{cases}2f_X(y)   & y\ge 0\\ 0 & \text{otherwise}\end{cases} = \bbox[red, 2pt]{\begin{cases} \lambda e^{-\lambda y}   & y\ge 0\\ 0 & \text{otherwise}\end{cases}}$$

解答： $$(一)\bbox[red,2pt]{z=0} \\(二) \cos z= {1\over 2}(e^{iz}+ e^{-iz}) \Rightarrow \cos(i)= {1\over 2}(e+{1\over e}) \Rightarrow f(i)=\cos(i)/i^5 ={1\over 2}(e+{1\over e})/i \\ \quad=- {1\over 2}(e+{1\over e})i =\alpha+ i\beta \Rightarrow \bbox[red, 2pt]{\cases{\alpha=0\\ \beta =- {1\over 2}(e+{1\over e})}} \\(三)由羅倫特展開式可知Res(f(z))={1\over 24} \Rightarrow \int_C f(z)\,dz = {1\over 24}\times 2\pi i=\bbox[red, 2pt]{\pi i\over 12} \\(四)z=0不在K內，因此\int_K f(z)\,dz = \bbox[red, 2pt]0$$

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