114年台綜大轉學考-工程數學D36詳解

 臺灣綜合大學系統114學年度學士班轉學生聯合招生考試

科目名稱:工程數學
類組代碼:D36

解答:$$y''+9=0 \Rightarrow \lambda^2+9=0 \Rightarrow \lambda=\pm 3i \Rightarrow y=c_1 \cos 3x+c_2\sin 3x，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(t)= t^2 \Rightarrow L\{f(t)\} =\int_0^\infty t^2e^{-st}\,dt =\left. \left[-{s^2t^2+2st+2\over s^3} e^{-st} \right] \right|_0^\infty ={2\over s^3},故選\bbox[red, 2pt]{(B)}$$

解答:$$b_n={2\over L}\int_0^L f(x)\sin({n\pi\over L}x)\,dx \Rightarrow b_n=0  \text{ if }f(x) \text{ is even}. 故選\bbox[red, 2pt]{(C)}$$

解答:$$A=\begin{bmatrix} 1& 2& 3\\ 4& 5& 6\\ 9& 8&7\end{bmatrix} \xrightarrow{R_2-4R_1\to R_2, R_3-9R_1\to R_3} \begin{bmatrix} 1& 2& 3\\ 0& -3& -6\\ 0& -10&-20\end{bmatrix} \xrightarrow{R_3-(10/3)R_2\to R_3} \begin{bmatrix} 1& 2& 3\\ 0& -3& -6\\ 0& 0& 0\end{bmatrix} \\ \Rightarrow \det(A) =\begin{vmatrix} 1& 2& 3\\ 0& -3& -6\\ 0& 0& 0\end{vmatrix} =0,故選\bbox[red, 2pt]{(D)}$$

解答:$$\nabla\times(\nabla\phi)= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \dfrac{\partial \phi}{\partial x}& \dfrac{\partial \phi}{\partial y} & \dfrac{\partial \phi}{\partial z} \end{vmatrix}=\\ \left(\dfrac{\partial^2 \phi}{\partial y\partial z}-\dfrac{\partial^2 \phi}{\partial y\partial z}\right)\mathbf{i}-\left(\dfrac{\partial^2 \phi}{\partial x\partial z}-\dfrac{\partial^2 \phi}{\partial x\partial z}\right) \mathbf{j}+\left(\dfrac{\partial^2 \phi}{\partial x\partial y}-\dfrac{\partial^2 \phi}{\partial x\partial y}\right)\mathbf{k}=\mathbf{0},故選\bbox[red, 2pt]{(E)}$$

解答:$$\vec F= yz\vec i+ xz\vec j+xy\vec k \Rightarrow \text{div }\vec F= {\partial \over \partial x}F_1 +{\partial \over \partial y}F_2 +{\partial \over \partial z}F_3 =0+0+0=0 \\ \text{curl }\vec F= \begin{vmatrix} \vec i& \vec j&\vec k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\F_1& F_2& F_3\end{vmatrix} = \begin{vmatrix} \vec i& \vec j&\vec k\\ {\partial \over \partial x} &{\partial \over \partial y} &{\partial \over \partial z} \\yz& xz& xy\end{vmatrix} =x\vec i+z\vec k+y\vec j-z\vec k-y\vec j-x\vec i =\vec 0 \\ \Rightarrow \bbox[red, 2pt]{\text{div }\vec F=0, \text{curl }\vec F=\vec 0}$$

解答:$$A= \begin{bmatrix} 4&2\\1& 3\end{bmatrix}  \Rightarrow \det(A-\lambda I) =(\lambda-2)(\lambda -5) =0 \Rightarrow \bbox[red, 2pt]{\text{eigenvalues: }2,5} \\ \lambda_1=2 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix} 2 & 2 \\1 & 1\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1+x_2=0 \Rightarrow v= x_2\begin{bmatrix}-1\\ 1 \end{bmatrix}, \text{choose }v_1=\begin{bmatrix}-1\\ 1 \end{bmatrix} \\ \lambda_2=5 \Rightarrow (A-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}-1 & 2 \\1 & -2\end{bmatrix} \begin{bmatrix} x_1\\ x_2\end{bmatrix} =0 \Rightarrow x_1=2x_2 \Rightarrow v= x_2\begin{bmatrix}2\\ 1 \end{bmatrix}, \text{choose }v_2 =\begin{bmatrix} 2\\ 1 \end{bmatrix} \\ \Rightarrow \bbox[red, 2pt]{\text{eigenvectors: }\begin{bmatrix}-1\\ 1 \end{bmatrix}, \begin{bmatrix} 2\\ 1 \end{bmatrix}}$$

解答:$$a_0 ={1\over \pi} \int_0^\pi x^2\,dx ={1\over \pi} \left. \left[ {1\over 3}x^3 \right] \right|_0^\pi ={1\over 3}\pi^2 \\ a_n={2\over \pi} \int_0^{\pi} x^2 \cos(2nx)\,dx={2\over \pi} \left. \left[ {1\over 4n^3} ((2n^2x^2-1)\sin(2nx)+2nx \cos(2nx))\right] \right|_0^\pi ={2\over \pi}\cdot (2n\pi) =4n\\ f(x) =x^2 \text{ is an even function }\Rightarrow b_n=0 \\ \Rightarrow f(x) \sim \bbox[red, 2pt]{{1\over 3}\pi^2+ \sum_{n=1}^\infty 4n \cos(2nx)}$$

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解題僅供參考，轉學考歷年試題及詳解