111年專科學力鑑定-工程數學詳解

教育部 111 年自學進修專科學校學力鑑定考試

專業科目(一)：工程數學

解答：$$y''-3y'+2y=0 \Rightarrow \lambda^2-3\lambda +2=0 \Rightarrow (\lambda-2)(\lambda-1)=0 \Rightarrow \lambda_1=1, \lambda_2= 2\\ \Rightarrow y=C_1e^{\lambda_1 x} +C_2e^{\lambda_2 x}=C_1e^{ x} +C_2e^{2 x}，故選\bbox[red, 2pt]{(A)}$$

解答：$$只有（B）符合係數為常數，故選\bbox[red, 2pt]{(B)}$$

解答：$$（A＋B）C＝ \left(\begin{bmatrix} a & 2\\ 4 & 3\end{bmatrix} +\begin{bmatrix} 2 & 0\\ b & 4\end{bmatrix}\right) \begin{bmatrix} -2 & 1\\ 4 & 3\end{bmatrix} =\begin{bmatrix} a+2 & 2\\ b+4 & 7\end{bmatrix} \begin{bmatrix} -2 & 1\\ 4 & 3\end{bmatrix} \\=\begin{bmatrix} -2a+4 & a+8\\ -2b+20 & b+25\end{bmatrix}  =\begin{bmatrix} 0 & 10 \\ 18 & 26\end{bmatrix} \Rightarrow \cases{a=2\\b=1} \Rightarrow a-b=1，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{\vec s=(1,2,4)\\ \vec t=(3,a,6)} \Rightarrow \vec s\cdot \vec t=1\times 3+ 2a+4\times 6 =2a+27 =15 \Rightarrow 2a=-12 \Rightarrow a=-6\\，故選\bbox[red, 2pt]{(A)}$$

解答：$$\cases{\vec u=(a,1,2)\\ \vec v=(-3,b,4)\\ \vec w=(-5,7,c)} \Rightarrow 2\vec u+\vec v-\vec w =(2a,2,4)+(-3,b,4)+ (5,-7,-c) \\=(2a+2,b-5,8-c) =(4,5,-3) \Rightarrow \cases{a=1\\ b=10} \Rightarrow 2a+b = 2+10=12，故選\bbox[red, 2pt]{(A)}$$

解答：$$\det(A-\lambda I)= \begin{vmatrix}2-\lambda & 0 & 0 \\ 0 & 1-\lambda & 0\\ 2 & 1 & 3-\lambda \end{vmatrix} =0 \Rightarrow (\lambda-1)(\lambda-2)(\lambda-3)=0 \Rightarrow \lambda=1,2,3\\，故選\bbox[red, 2pt]{(A)}$$

解答：$$令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \Rightarrow 原式為m(m-1)x^m +7mx^m +5x^m=0\\ \Rightarrow (m^2+6m +5)x^m=0 \Rightarrow m^2+6m+5 =0 \Rightarrow (m+1)(m+5)=0 \Rightarrow m=-1,-5\\ \Rightarrow y=C_1x^{-1}+ C_2x^{-5}，故選\bbox[red, 2pt]{(D)}$$

解答：$$先求齊次解，即y''-6y'+9y=0 \Rightarrow \lambda^2-6\lambda+9 =0 \Rightarrow (\lambda-3)^2=0 \Rightarrow \lambda=3(重根）\\ \Rightarrow y_h= C_1e^{3x}+ C_2xe^{3x}，只有（C）符合此條件，不用再求特解，故選\bbox[red, 2pt]{(C)}$$

解答：$$（A）\times: \cases{M= 2xy+y\\ N=x^2-2} \Rightarrow \cases{M_y= 2x+1\\ N_x=2x} \Rightarrow M_y\ne N_x \\（B）\times: \cases{M= e^{2x}+3y\\ N=\sin x-\sin y} \Rightarrow \cases{M_y= 3\\ N_x= \cos x} \Rightarrow M_y\ne N_x \\（C）\times: \cases{M= 2xy^3 +2x+y\\ N=3x^2y^2-x+y} \Rightarrow \cases{M_y= 6xy^2+1\\ N_x=6xy^2-1} \Rightarrow M_y\ne N_x \\ （D）\bigcirc: \cases{M= 1+e^{-x}\cos y\\ N= e^{-x}\sin y} \Rightarrow \cases{M_y= -e^{-x}\sin y\\ N_x=-e^{-x}\sin y} \Rightarrow M_y = N_x \\，故選\bbox[red, 2pt]{(D)}$$

解答：$${8s-1\over s^2+s-6} ={8s-1\over (s+3)(s-2)} ={5\over s+3}+{3\over s-2 }\\ \Rightarrow \mathcal L^{-1}\left\{{8s-1\over s^2+s-6} \right\} = 5 \mathcal L^{-1}\left\{{ 1\over s+3} \right\}+3 \mathcal L^{-1}\left\{{ 1\over  s-2} \right\} =5e^{-3t}+3 e^{2t}，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{\sin^2 x =(1-\cos 2x)/2\\ \cos^3 x = (3\cos x+ \cos 3x)/4} \Rightarrow f(x)= 3\sin^2(2x)-4\cos^3 x\\ = {3\over 2}(1-\cos (4x))-(3\cos(x)+ \cos(3x)）={3\over 2}-3\cos(x)-\cos(3x)-{3\over 2}\cos(4x) \\ \Rightarrow  A_0= 3/2 \ne -3/2，故選\bbox[red, 2pt]{(A)}$$

解答：$$\sin^2 x =(1-\cos 2x)/2 \Rightarrow f(x)=4＋3\sin^2 x = 4+{3\over 2}(1-\cos 2x) ={11\over 2}-{3\over 2}\cos 2x \\ \Rightarrow A_2=0 \ne -{3\over 2}，故選\bbox[red, 2pt]{(C)}$$

解答：$$\begin{Vmatrix}1 & 2 & 3\\ 3 & -2 & 2\\ 4 & 0 & 3 \end{Vmatrix} =|-6+16+24-18|=16，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{\vec u=(2,4,a)\\ \vec v=(3,2,1)} \Rightarrow \vec u\times \vec v=(4-2a,3a-2,-8) =(-2,7,-8) \Rightarrow a=3，故選\bbox[red, 2pt]{(C)}$$

解答：$$AB^t =\begin{bmatrix} 3 & 5 & 7\\ 2 & 6 & 3\end{bmatrix} \begin{bmatrix} 4 & 3\\ 1& 0 \\ 2& 8\end{bmatrix} =\begin{bmatrix} 31 & 65\\ 20 & 30\end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答：$$AB =\begin{bmatrix} 8 & 3\\5 & 2 \end{bmatrix} \begin{bmatrix} a & -3\\ -5& b \end{bmatrix} =\begin{bmatrix} 8a-15 & -24+3b\\ 5a-10 & -15+2b\end{bmatrix} =\begin{bmatrix} 1 & 0\\0 & 1 \end{bmatrix} \Rightarrow \cases{a=2\\ b=8}\\ \Rightarrow a+b= 10，故選\bbox[red, 2pt]{(C)}$$

解答：$$只有(B)的行列式為1\ne 0，其他矩陣的行列式均為0，故選\bbox[red, 2pt]{(B)}$$

解答：$$\cases{M=x^3+y^2+\sin x\\ N=xy} \Rightarrow \cases{xM= x^4+xy^2+ x\sin x \\xN= x^2y } \Rightarrow \cases{{\partial \over \partial y}(xM)= 2xy\\ {\partial \over \partial x}(xN) =2xy} \\ \Rightarrow {\partial \over \partial y}(xM)= {\partial \over \partial x}(xN) \Rightarrow x為積分因子，故選\bbox[red, 2pt]{(C)}$$

解答：$$\mathcal L\{f*g\} = \mathcal L\{f\} \times \mathcal L\{ g\} = \mathcal L\{1\} \times \mathcal L\{ te^{-3t}\} ={1\over s}\times {1\over (s+3)^2} ={1\over s(s+3)^2}，故選\bbox[red, 2pt]{(B)}$$

解答：$$A\mathbf x= \lambda \mathbf x \Rightarrow \begin{bmatrix} -3 & 2\\ a & 6\end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} =1 \begin{bmatrix} x_1\\ x_2 \end{bmatrix} \Rightarrow \cases{-3x_1+2x_2= x_1\\ ax_1+ 6x_2= x_2} \Rightarrow \cases{x_2=2x_1\\ x_2= -ax_1/5} \\ \Rightarrow 2=-{a\over 5} \Rightarrow a=-10，故選\bbox[red, 2pt]{(D)}$$
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解題僅供參考，其他歷年試題及詳解