114年專技高考電子工程技師-工程數學詳解

114年專門職業及技術人員高等考試

等 別：高等考試
類 科：電子工程技師
科 目：工程數學（包括線性代數、微分方程、向量分析、複變函數與機率）

解答:$$\textbf{W} =a \mathbf V_1 +b \mathbf V_2+ c \mathbf V_3 \Rightarrow \begin{bmatrix}1 & 0& -1\\ 4& 3& 0\\3& 2& 1 \end{bmatrix} \begin{bmatrix}a\\ b\\c \end{bmatrix} = \begin{bmatrix}2\\1\\ 1 \end{bmatrix} \\ \Rightarrow \begin{bmatrix}a\\ b\\c \end{bmatrix} = \begin{bmatrix}1 & 0& -1\\ 4& 3& 0\\3& 2& 1 \end{bmatrix}^{-1} \begin{bmatrix}2\\1\\ 1 \end{bmatrix} = \begin{bmatrix}\frac{3}{4} & \frac{-1}{2} & \frac{3}{4} \\-1 & 1 & -1 \\\frac{-1}{4} & \frac{-1}{2} & \frac{3}{4} \end{bmatrix}  \begin{bmatrix}2\\1\\ 1 \end{bmatrix} = \begin{bmatrix}\frac{7}{4} \\-2 \\ \frac{-1}{4} \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\textbf{W} ={7\over 4} \mathbf V_1 -2 \mathbf V_2-{1\over 4} \mathbf V_3}$$

解答:$$y'''-4y''-y'+4=0 \Rightarrow \lambda^3-4\lambda^2-\lambda+4=0 \Rightarrow (\lambda-4)(\lambda-1)(\lambda+1) =0 \\ \Rightarrow \lambda=4,1,-1 \Rightarrow y_h= c_1e^{4x}+ c_2e^x +c_3e^{-x} \\ y_p=Ae^{3x} \Rightarrow y_p'=3Ae^{3x} \Rightarrow y_p''=9Ae^{3x} \Rightarrow y_p'''= 27Ae^{3x} \\ \Rightarrow y_p'''-4y_p''-y_p'+4y_p=-8Ae^{3x} =16e^{3x}\Rightarrow A=-2 \Rightarrow y_p=-2e^{3x} \\ \Rightarrow y = y_h+ y_p \Rightarrow \bbox[red, 2pt] {y=c_1e^{4x}+ c_2e^x +c_3e^{-x}-2e^{3x}}$$

解答:$$令\cases{P = \frac{-2y}{x^2+y^2}\\ Q = \frac{2x}{x^2+y^2}} \Rightarrow  P_y=  \frac{2y^2 - 2x^2}{(x^2+y^2)^2} =Q_x \\ \textbf{Case I }C不包含原點: 依格林定理可得\oint_C P \, dx + Q \, dy = \iint_D \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA= \iint_D0\,dA=0\\ \textbf{Case II }C包含原點:構造一個足夠小的圓 \mathbf{C_{\epsilon}} (半徑 \epsilon)，圓心在原點，且 C_{\epsilon} 位於 C 的內部。\\\qquad 考慮由 C 和 C_{\epsilon} 圍成的環形區域 D'。在 D' 區域內，函數 P 和 Q 沒有奇異點，\\\qquad 因此在 D' 內 \frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}  仍然成立。根據格林定理，對於 D' 的邊界 ( C 為正向， C_{\epsilon} 為負向),\\\qquad \oint_{C} P \, dx + Q \, dy - \oint_{C_{\epsilon}} P \, dx + Q \, dy = \iint_{D'} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA = \iint_{D'} 0 \, dA = 0 \\ \qquad \Rightarrow \oint_{C} P \, dx + Q \, dy = \oint_{C_{\epsilon}} P \, dx + Q \, dy = \int_{0}^{2\pi} \left[ \frac{-2(\epsilon \sin\theta)}{\epsilon^2} (-\epsilon \sin\theta \, d\theta) + \frac{2(\epsilon \cos\theta)}{\epsilon^2} (\epsilon \cos\theta \, d\theta) \right] \\\qquad = \int_{0}^{2\pi} \left[ \frac{2\epsilon^2 \sin^2\theta}{\epsilon^2} + \frac{2\epsilon^2 \cos^2\theta}{\epsilon^2} \right] d\theta= \int_{0}^{2\pi} 2 (\sin^2\theta + \cos^2\theta) \, d\theta =4\pi \\ \Rightarrow \bbox[red, 2pt]{\cases{不包含原點: 0\\ 包含原點:4\pi}}$$

解答:$$\oint_C{(1+z)^2-2z\over (z-1)^2(z+ 2i)} \,dz =\oint_C{z^2+1\over (z-1)^2(z+ 2i)} \,dz = \oint_C f(z)\,dz \\ \Rightarrow \cases{\text{Res}(f, 1) =\displaystyle \lim_{z \to 1} \displaystyle \frac{d}{dz} \left[ \frac{1+z^2}{z+2i} \right] = \frac{4i}{4i - 3} \\ \text{Res}(f, -2i) = \displaystyle \lim_{z \to -2i}   \displaystyle \frac{1+z^2}{(z-1)^2  }= \frac{-3}{4i - 3}} \Rightarrow \oint_C f(z)\,dz = 2\pi i( \text{Res}(f, 1) + \text{Res}(f, -2i)) \\=2\pi i\cdot {-3+4i\over 4i-3} =2\pi i \cdot 1=\bbox[red, 2pt]{2\pi i}$$

解答:$$\int f(x)\,dx =1 \Rightarrow \int_0^1 kx(2-x)\,dx =\int_0^1 (2kx-kx^2)\,dx = \left. \left[ kx^2-{1\over 3}kx^3 \right] \right|_0^1 ={2\over 3}k=1 \Rightarrow \bbox[red, 2pt]{k={3\over 2}} \\ F(x)=P(X\le x) = \int_{-\infty}^x f(t)\,dt = \int_0^x {3\over 2}t(2-t)\,dt ={1\over 2}x^2(3-x) \\ \Rightarrow \bbox[red, 2pt]{F(x) = \begin{cases}0, & x < 0 \\ \frac{1}{2} x^2 (3-x), & 0 \le x \le 1 \\ 1, & x > 1 \end{cases}}$$

 

====================== END ==========================
解題僅供參考，其他國家考試試題及詳解