114年台綜大轉學考-工程數學D37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:D37

解答：$$A^{-1} =\begin{bmatrix} -18& 7\\ -5& 2\end{bmatrix} \Rightarrow \det(A^{-1}) =-36+35=-1 \Rightarrow A=(A^{-1})^{-1} ={1\over \det(A^{-1})} \begin{bmatrix} 2& -7\\ 5& -18\end{bmatrix} \\=\begin{bmatrix} -2& 7\\ -5& 18\end{bmatrix},故選\bbox[red, 2pt]{(E)}$$

解答：$$(A) \times: \begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left(\theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} \frac{1}{\cos\left(\theta\right)} & \frac{1}{\sin\left(\theta\right)} & 0 \\\frac{1}{\sin\left(\theta\right)} & \frac{1}{\cos\left(\theta\right)} & 0 \\0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 2 & \frac{2}{\sin\left(2 \theta\right)} & 0 \\\frac{2}{\sin \left(2 \theta\right)} & 2 & 0 \\0 & 0 & 1\end{bmatrix} \ne I \\(B)\times:\begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left(\theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} -\sin\left(\theta\right) & \cos \left( \theta\right) & 0 \\\cos\left(\theta\right) & -\sin\left(\theta \right) & 0 \\0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 0 & \cos\left(2 \theta\right) & 0 \\\cos\left(2 \theta\right) & 0 & 0 \\0 & 0 & 1\end{bmatrix} \ne I \\ (C) \times:\begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left(\theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} \frac{1}{\sin\left(\theta\right)} & \frac{1}{\cos\left(\theta\right)} & 0 \\\frac{1}{\cos \left(\theta\right)} & \frac{1}{\sin\left(\theta\right)} & 0 \\0 & 0 & 1\end{bmatrix} =\begin{bmatrix} \frac{2}{\sin\left(2 \theta\right)} & 2 & 0 \\2 & \frac{2}{\sin\left(2 \theta\right)} & 0 \\0 & 0 & 1\end{bmatrix} \ne I \\(D)\times:\begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left( \theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left(\theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & \sin\left(2 \theta\right) & 0 \\
\sin\left(2 \theta\right) & 1 & 0 \\0 & 0 & 1\end{bmatrix} \ne I \\(E)\times:\begin{bmatrix} \cos\left(\theta\right) & \sin\left(\theta\right) & 0 \\\sin\left(\theta\right) & \cos\left(\theta\right) & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} 1 & 0 & \cos\left(\theta\right) \\0 & 1 & \sin\left(\theta\right) \\\cos\left(\theta\right) & \sin\left( \theta\right) & 1\end{bmatrix} =\begin{bmatrix} \cos\left( \theta\right) & \sin\left( \theta\right) & 1 \\\sin\left(\theta\right) & \cos\left( \theta\right) & \sin\left(2 \theta\right) \\\cos\left(\theta\right) & \sin\left(\theta\right) & 1\end{bmatrix} \ne I \\ 故\bbox[cyan,2pt]{以上皆非}$$

解答：$$L\{f(t)\} =\int_1^\infty (3t-3)e^{-st}\,dt = \left. \left[ {3 \over s^2}e^{-st}(-st+s-1) \right] \right|_1^\infty = {3\over s^2}e^{-s},故選\bbox[red, 2pt]{(B)}$$

解答：$$\text{curl }\mathbf F =\begin{vmatrix} \mathbf i &\mathbf j &\mathbf k\\{\partial \over \partial x }&  {\partial \over \partial y }&{\partial \over \partial z } \\ xyz& 4x^2yz& 2x^3z\end{vmatrix} =8xyz \mathbf k+xy\mathbf j-xz\mathbf k-6x^2z \mathbf j-4x^2y\mathbf i\\= -4x^2y\mathbf i+(xy-6x^2z)\mathbf j+(8xyz-xz) \mathbf k \\\Rightarrow \text{div(curl }\mathbf F)={\partial \over \partial x}(-4x^2y) +{\partial \over \partial y}(xy-6x^2z) +{\partial \over \partial z}(8xyz-xz) =-8xy+x+8xy-x=0\\, 故選\bbox[red, 2pt] {(D)}$$

解答：$$\cases{P(x,y)=x^2-y+{kx\over x^2+y^2}  \Rightarrow P_x=2x+ {k(y^2-x^2) \over (x^2+y^2)^2}\\ Q(x,y)=x+y+{ky\over x^2+y^2} \Rightarrow Q_y= 1+{k(x^2-y^2) \over (x^2+y^2)^2}} \Rightarrow P_x \ne Q_y \Rightarrow \text{ dependent},故選\bbox[red, 2pt]{(A)}$$

解答：$$\begin{bmatrix} 1 & 3 & 1 \\1 & 1 & -1 \\1 & 2 & 1\end{bmatrix}: R_2-{\color{blue}1}R_1\to R_2 \Rightarrow \begin{bmatrix} 1 & 3 & 1 \\0 & -2 & -2 \\1 & 2 & 1\end{bmatrix} :R_3-{\color{blue}1}R_1 \to R_3 \Rightarrow \begin{bmatrix} 1 & 3 & 1 \\0 & -2 & -2 \\0 & -1 & 0\end{bmatrix} \\ R_3-({\color{blue}1/2})R_2 \to R_3 \Rightarrow \begin{bmatrix}1 & 3 & 1 \\0 & -2 & -2 \\0 & 0 & 1 \end{bmatrix}  \\ \Rightarrow L=\begin{bmatrix}1 & 0 & 0 \\ 1 & 1 & 0 \\  1 &  \frac{1}{2} & 1\end{bmatrix}, U=\begin{bmatrix}1 & 3 & 1 \\0 & -2 & -2 \\0 & 0 & 1\end{bmatrix} \Rightarrow 6. 故選\bbox[red, 2pt]{(C)}, 7.故選\bbox[red, 2pt]{(A)}$$

解答：$$\vec r(t)=\cos t\vec i+\cos t\vec j+\sqrt 2\sin t\vec k \Rightarrow \vec v(t) =-\sin t\vec i-\sin t\vec j+\sqrt 2\cos t\vec k \\ \Rightarrow \text{speed of the particle }|\vec v|=\sqrt{(-\sin t)^2+(-\sin t)^2+(\sqrt 2\cos t)^2} =\sqrt 2,故選\bbox[red, 2pt]{(B)}$$

解答：$$\text{By Green theorem,} \oint_C \vec F\cdot d\vec r = \iint_D\left( {\partial \over \partial x}6x-{\partial \over \partial y}4y\right)\,dxdy =\iint_D 2\,dxdy = 2\cdot \pi=2\pi,故選\bbox[red, 2pt]{(A)}$$

解答：$$r(t)=\sin t\mathbf i+\cos t\mathbf j+3t\mathbf k \Rightarrow r'(t)=\cos t\mathbf i-\sin t\mathbf j+3\mathbf k \Rightarrow r''(t)=-\sin t\mathbf i-\cos t\mathbf j+0\mathbf k \\ \Rightarrow r'(t) \times r''(t) =3\cos t\mathbf i-3\sin t\mathbf j-\mathbf k \\\Rightarrow \text{curvature of }r(t)={||r'(t) \times r''(t)|| \over ||r'(t)||^3} ={\sqrt{10} \over (\sqrt{10})^3} ={1\over 10},故選\bbox[red, 2pt]{(C)}$$

解答：$$\cases{x_1-2x_2+3x_3=4\\ 2x_1-4x_2+6x_3=8\\ 3x_1+x_2+6x_3=5} \Rightarrow \begin{bmatrix} 1 & -2 & 3 \\2 & -4 & 6  \\3 & 1 & 6  \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=\begin{bmatrix} 4\\ 8\\5\end{bmatrix} \\ \Rightarrow rref\left( \left[\begin{array} {rrr|r} 1 & -2 & 3 & 4\\2 & -4 & 6 & 8\\3 & 1 & 6 & 5\end{array} \right]\right) = \left[\begin{array} {rrr|r} 1 & 0 & \frac{15}{7} & 2\\0 & 1 & - \frac{3}{7} & -1\\0 & 0 & 0 & 0\end{array} \right] \Rightarrow \text{rank =2 and infinity solutions}\\ ,故選\bbox[red, 2pt]{(A)}$$

解答：$$A=\begin{bmatrix}1 & 4 \\-1 & 1 \end{bmatrix} \Rightarrow \det(A-\lambda I)=\lambda^2-2\lambda+5 =0 \Rightarrow \lambda=1\pm 2i,故選\bbox[red, 2pt]{(D)}$$

解答：$$\cases{u(x,y)=y-x\\ v(x,y)=y+x} \Rightarrow {\partial(u,v) \over \partial (x,y)} =\begin{vmatrix} -1& 1\\ 1& 1\end{vmatrix} =-2,故選\bbox[red, 2pt]{(E)}\\ \bbox[cyan,2pt]{試題有疑義}:題目未敍明u,v為何?$$

解答：$$由13題可知{\partial(x,y) \over \partial (u,v)}=1/{\partial(u,v) \over \partial (x,y)}=-{1\over 2} \Rightarrow \int_0^1 \int_0^{1-x}e^{(y-x)/(y+x)} \,dydx = \int_0^1 \int_{-v}^v e^{u/v} \left| {\partial(x,y) \over \partial (u,v)}\right|\,du \,dv\\ ={1\over 2}\int_0^1 \int_{-v}^v e^{u/v}\,dudv ={1\over 2}\int_0^1 v(e-e^{-1}) \,dv ={1\over 4}(e-e^{-1}),故選\bbox[red, 2pt]{(B)} \\ \bbox[cyan,2pt]{試題有疑義}:\int_0^1 \int_0^{1-x}e^{(y-x)/(y+x)} \,{\color{blue}dxdy} 應該是\int_0^1 \int_0^{1- x}e^{(y- x)/(y +x)} \,{\color{blue} dydx}$$

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解題僅供參考，轉學考歷年試題及詳解