99年專科學力鑑定-工程數學詳解

教育部 99 年專科學校畢業程度自學進修學力鑑定考試

科 別：冷凍空調、電機工程、電子工程、電訊工程、資訊工程
科目名稱：專業科目(一)
考 科：工程數學

解答：$$8x^2-2x為二次多項式，故選\bbox[red, 2pt]{(C)}$$

解答：$$兩向量垂直，故選\bbox[red, 2pt]{(B)}$$

解答：$$\left[\begin{matrix}1 & 2 & -1 & 3\\-3 & 4 & 0 & -1\\1 & 0 & -2 & 7\end{matrix}\right] \underrightarrow{-r_1+r_3, 3r_1+r_2} \left[\begin{matrix}1 & 2 & -1 & 3\\0 & 10 & -3 & 8\\0 & -2 & -1 & 4\end{matrix}\right] \\\underrightarrow{r_3+r_1, 5r_3+r_2} \left[\begin{matrix}1 & 0 & -2 & 7\\0 & 0 & -8 & 28\\0 & -2 & -1 & 4\end{matrix}\right] \underrightarrow{r_2\div (-4), r_3\div (-2)} \left[\begin{matrix}1 & 0 & -2 & 7\\0 & 0 & 2 & -7\\0 & 1 & \frac{1}{2} & -2\end{matrix}\right]\\ \underrightarrow{r_2+r_1,-4r_2+r_3} \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 0 & 2 & -7\\0 & 1 & 0 & - \frac{1}{4}\end{matrix}\right] \underrightarrow{r_2,r_3互換} \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & - \frac{1}{4}\\0 & 0 & 2 & -7\end{matrix}\right] \underrightarrow{r_3/2} \left[\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & - \frac{1}{4}\\0 & 0 & 1 & - \frac{7}{2}\end{matrix}\right] \\ ，故選\bbox[red, 2pt]{(C)}\\公布的答案是\bbox[blue, 2pt]{(B)}$$

解答：$$開根號為非線性，又最高微分次數為一，故選\bbox[red, 2pt]{(D)}$$

解答：$$(C)\times: (\vec A\times \vec B)'= \vec A'\times \vec B+\vec A\times \vec B'\\ (D)\times: (\vec A+\vec B)'= \vec A'+\vec B'\\，故選\bbox[red, 2pt]{(B)}$$

解答：$$\mathcal{L}(\sin t)={1\over s^2+1} \Rightarrow \mathcal{L}(t\sin t) =-{d\over dt}\left( {1\over s^2+1}\right) ={2s \over (s^2+1)^2}，故選\bbox[red, 2pt]{(D)}$$

解答：$$\mathcal{L}\{t^ne^{at}\} ={n!\over (s-a)^{n+1}} \Rightarrow \mathcal{L^{-1}}\left\{{1\over (s+2)^4}\right\} = {1\over 3!} t^3e^{-2t}，故選\bbox[red, 2pt]{(C)}$$

解答：$$令\cases{M(x,y)=y^2\\ N(x,y)=1+xy\\I(x,y)=1/y} \Rightarrow  {\partial \over \partial y}(MI)= 1={\partial \over \partial x}(NI) \Rightarrow {1\over y}是積分因子，故選\bbox[red, 2pt]{(D)}$$

解答：$$(\sin x)dy=2y(\cos x)dx \Rightarrow {1\over 2y}dy={\cos x\over \sin x}dx \Rightarrow {1\over 2}\ln y=\ln \sin x+c \Rightarrow \ln y=2\ln \sin x+ 2c\\ \Rightarrow \ln y=\ln \sin^2 x+\ln e^{2c}=\ln( e^{2c}\cdot \sin^2 x) \Rightarrow y=k\sin^2 x,k=e^{2c}為一常數，故選\bbox[red, 2pt]{(C)}$$

解答：$$只有(D)符合y(0)=2，故選\bbox[red, 2pt]{(D)}$$

解答：$$只有(D)符合y'=4x^3+1，故選\bbox[red, 2pt]{(D)}$$

解答：$$n=1，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{3x_1- 2x_2+ 2x_3=10\\ x_1+2x_2-3x_3 =-1\\ 4x_1+ x_2 +2x_3=3} \Rightarrow Ax=b，其中A=\left[\begin{matrix}3 & -2 & 2\\1 & 2 & -3\\4 & 1 & 2\end{matrix}\right],x=\left[\begin{matrix}x_1 \\x_2 \\x_3 \end{matrix}\right] ,b=\left[ \begin{matrix} 10\\-1\\3\end{matrix}\right] \\ \Rightarrow \cases{x_1= {\left|\begin{matrix}10 & -2 & 2\\-1 & 2 & -3\\3 & 1 & 2 \end{matrix} \right| \over \det(A)} ={70 \over 35}=2 \\ x_2= {\left| \begin{matrix}3 & 10 & 2\\1 & -1 & -3\\4 & 3 & 2\end{matrix}\right| \over \det(A)} = {-105\over 35}=-3 \\x_3= {\left|\begin{matrix}3 & -2 & 10\\1 & 2 & -1\\4 & 1 & 3 \end{matrix} \right|\over \det(A)} ={-35\over 35}=-1} \Rightarrow \cases{x_1=2\\ x_2=-3\\ x_3=-1}，故選\bbox[red, 2pt]{(B)}$$

解答：$$各函數在區間[-\pi,pi]積分皆為0(除了1之外)，因此任相異函數相乘積分仍為0，故選\bbox[red, 2pt]{(B)}$$

解答：$$(A) \times: 偶函數對稱y軸\\(B) \bigcirc: 例:\sin(x)+\cos(x)既非偶函數，亦非奇函數\\(C)\times: f(-x)=-f(x)\Rightarrow f(x)為奇函數\\ (D)\times: f(x)為奇函數\Rightarrow \int_{-\ell}^\ell f(x)\,dx=0\\，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{\vec A=(1,-3, 1) \\ \vec B=(2,-1,1) \\ \vec C=(3,1,-1)} \Rightarrow \vec A\cdot (\vec B\times \vec C)= \vec A\cdot (0,5,5)= -10，故選\bbox[red, 2pt]{(A)}$$

解答：$$A^T=A \Rightarrow \cases{\alpha=5\\ \beta=4}，故選\bbox[red, 2pt]{(D)}$$

解答：$$A=\begin{bmatrix}1 & 3\\ 2& 7\end{bmatrix} \Rightarrow A^T= \begin{bmatrix}1 & 2\\ 3& 7\end{bmatrix} \Rightarrow \det(A^T)=7-6=1 \Rightarrow (A^T)^{-1}={1\over \det(A^T)}\begin{bmatrix}7 & -2\\ -3& 1\end{bmatrix} \\ =\begin{bmatrix}7 & -2\\ -3& 1\end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答：$$(A)\times: \vec A+(\vec B+\vec C)= (\vec A+\vec B)+\vec C\\ (B)\times: \vec A-\vec B=-(\vec B-\vec A)\\(C)\times:\cases{\vec A=(a,b)\\ \vec B=(c,d)} \Rightarrow \cases{m\vec A\cdot \vec B=(ma,mb)\cdot (c,d)= mac+mbd\\ \vec A\cdot m\vec B=(a,b)\cdot (mc,md)=mac+mbd} \Rightarrow m\vec A\cdot \vec B=\vec A\cdot m\vec B \\(D)\bigcirc::\cases{\vec A=(a,b)\\ \vec B=(c,d)} \Rightarrow \cases{m(n\vec A)= m(na,nb)= (mna,mnb) \\(mn)\vec A=(mna,mnb)} \Rightarrow m(n\vec A)= (mn)\vec A\\，故選\bbox[red, 2pt]{(D)}，公布的答案是\bbox[blue,2pt]{(C)}$$

解答：$$ \begin{vmatrix}-\sin n\theta & \cos n\theta\\ \cos n\theta & \sin n\theta\end{vmatrix}  =-\sin^2 n\theta-\cos^2 n\theta =-1，故選\bbox[red, 2pt]{(A)}$$

========================= END =============================

解題僅供參考，其他歷屆試題及詳解