112年專科學力鑑定-工程數學詳解

教育部 112 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: $$令\cases{M(x,y)=y\\ N(x,y)=x+xy+\cos y} \Rightarrow \cases{e^yM(x,y)=ye^y\\ e^yN(x,y)=(x+xy+\cos y) e^y} \Rightarrow \cases{\frac{\partial }{\partial y}(e^yM) =e^y+ye^y \\ \frac{\partial }{\partial x}(e^yN) =(1+y)e^y} \\ \Rightarrow \frac{\partial }{\partial y}(e^yM) =\frac{\partial }{\partial x}(e^yN) \Rightarrow e^y為積分因子，故選\bbox[red, 2pt]{(D)}$$

解答: $$y''+6y'+9y=0 \Rightarrow \lambda^2+6\lambda +9=0 \Rightarrow (\lambda+3)^2=0 \Rightarrow \lambda=-3\\ \Rightarrow y=C_1e^{-3x}+C_2 xe^{-3x}，故選\bbox[red, 2pt]{(B)}$$

解答: $$只有選項(C)的係數為常數，故選\bbox[red, 2pt]{(C)}$$

解答: $$令y=x^m \Rightarrow y'=mx^{m-1} \Rightarrow y''=m(m-1)x^{m-2} \\ \Rightarrow x^2y''+3xy'+2y=m(m-1)x^m+3mx^m+2x^m= (m^2+2m+2)x^m=0 \\ \Rightarrow m^2+2m+2 =0 \Rightarrow m=-1\pm i \Rightarrow y=c_1x^{-1+i}+c_2x^{-1-i} \\={1\over x}(c_1x^i+c_2x^{-i}) ={1\over x}(c_1e^{i\ln x}+ c_2e^{-i\ln x}) ={1\over x}(A\cos (\ln x)+B\sin (\ln x))，故選\bbox[red, 2pt]{(A)}$$

解答: $$先求齊次解:y''-2y'+2y=0 \Rightarrow \lambda^2-2\lambda+2=0 \Rightarrow \lambda=1\pm i\\ \Rightarrow y_h=e^x(A\cos x+B\sin x)\\ 令y_p=Ce^{4x} \Rightarrow y_p'=4Ce^{4x} \Rightarrow y_p''=16Ce^{4x} 代入y_p''-2y_p'+2y_p=3e^{4x} \\ \Rightarrow 16Ce^{4x}-8Ce^{4x}+2Ce^{4x} =10Ce^{4x} =3e^{4x} \Rightarrow C={3\over 10} \Rightarrow y_p={3\over 10}e^{4x} \\ \Rightarrow y=y_h+y_p =e^x(A\cos x+B\sin x)+{3\over 10}e^{4x}，故選\bbox[red, 2pt]{(C)}$$

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

解答: $$L^{-1}\left\{ {2s-1\over s^2+4s+5}\right\} =L^{-1}\left\{ {2(s+2)-5\over (s+2)^2+1}\right\} = 2L^{-1}\left\{ {s+2\over (s+2)^2+1}\right\} - 5L^{-1}\left\{ {1\over (s+2)^2+1}\right\} \\ =e^{-2t}(2\cos t-5\sin t)，故選\bbox[red, 2pt]{(D)}$$

解答: $$L\{f(t)*g(t)\} =L\{f(t)\}\cdot L\{g(t)\} =L\{te^t\}L\{\cos 2t\} =(-1)\frac{d }{ds}L\{e^t\}\cdot {s\over s^2+2^2} \\={1\over (s-1)^2}\cdot {s\over s^2+4}={s\over (s-1)^2(s^2+4)}，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=1-3\cos^2 x \Rightarrow f(x)=f(-x)\Rightarrow f(x)為偶函數\Rightarrow B_n=0,n\in \mathbb N，故選\bbox[red, 2pt]{(D)}$$

解答: $$A_0={1\over 2\pi} \int_{-\pi}^\pi f(x)\,dx ={1\over \pi }\int_0^\pi x\,dx={1\over \pi}\cdot {1\over 2}\pi^2={1\over 2}\pi \ne \pi \\ A_n={1\over \pi}\int_{-\pi}^\pi f(x)\cos nx\,dx　={2\over \pi} \int_0^\pi x\cos nx\,dx ={2\over \pi} \left. \left[ {1\over n}x\sin(nx)+{1\over n^2}\cos(nx) \right] \right|_0^\pi \\=  {2\over n^2\pi}((-1)^n-1) \Rightarrow \cases{A_1=-4/\pi \\ A_3=-4/9\pi\\ A_5=-4/25\pi}，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\vec u=(1,2,-5)\\ \vec v=(2,1,-3)} \Rightarrow \cases{\vec u\cdot \vec u=1+4+25=30\\ \vec v\cdot \vec u=2+2+15=19} \\ \Rightarrow (\vec u+2\vec v)\cdot \vec u=\vec u\cdot \vec u+2\vec v\cdot \vec u=30+2\cdot 19=68，故選\bbox[red, 2pt]{(D)}$$

解答: $$Z軸向量=\vec b=(0,0,1) \Rightarrow \cos \gamma={\vec a\cdot \vec b\over |\vec a||\vec b|}= {(2,3,-2)\cdot (0,0,1)\over \sqrt{2^2+3^2+(-2)^2}\cdot \sqrt 1}={-2\over \sqrt{17}}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{\vec v=(-3,4,4)\\ \vec w=(0,1,2)} \Rightarrow \vec v\times \vec w =(4,6,-3) \Rightarrow \vec u\cdot (\vec v\times \vec w)=(2,3,a)\cdot (4,6,-3)=26-3a=29\\ \Rightarrow a=-1，故選\bbox[red, 2pt]{(B)}$$

解答: $$(\vec u\times \vec v)\cdot \vec v=0 \Rightarrow (1,7,-5)\cdot (a,3,5)=a-4=0 \Rightarrow a=4，故選\bbox[red, 2pt]{(D)}$$

解答: $$2A+3B=2\begin{bmatrix}1 & -2 \\-3 &4\end{bmatrix}+3\begin{bmatrix}1 & 3 \\ 5 &a\end{bmatrix} =\begin{bmatrix}5 & 5 \\9 &8+3a\end{bmatrix}\\ \Rightarrow (2A+3B)C= \begin{bmatrix}5 & 5 \\9 &8+3a\end{bmatrix}\begin{bmatrix}b & 3 \\3 &5 \end{bmatrix}= \begin{bmatrix}25 & 40 \\ 51 & 82 \end{bmatrix} \Rightarrow \cases{5b+15=25\\ 67+15a=82} \\ \Rightarrow \cases{b=2\\ a=1} \Rightarrow a-b=-1，故選\bbox[red, 2pt]{(B)}$$

解答: $$\begin{vmatrix}1 & 3 & 1 \\2 &a  &3  \\4 &7  &1\end{vmatrix} =-3a+23=8 \Rightarrow a=5，故選\bbox[red, 2pt]{(D)}$$

解答: $$A=\begin{bmatrix} 6& 7\\ 3 & 4\end{bmatrix} \Rightarrow A^{-1}={1\over \det(A)} \begin{bmatrix} 4& -7\\ -3 & 6\end{bmatrix}={1\over3} \begin{bmatrix} 4& -7\\ -3 & 6\end{bmatrix}=\begin{bmatrix} 4/3& -7/3\\ -1 & 2\end{bmatrix} \\ \Rightarrow \cases{c=-1\\ d=2} \Rightarrow c+d=1，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{vmatrix} 8 & 4\\ 4& 2\end{vmatrix}=16-16=0,其餘行列式皆不為0，故選\bbox[red, 2pt]{(A)}$$

解答: $$\begin{bmatrix} -4 & 1 & 11\\ 1& 2& 4\\ 2& 1& -1\end{bmatrix} \xrightarrow{4R_2+R_1 \to R_1,-2R_2+R_3\to R_3}\begin{bmatrix} 0 & 9 & 27\\ 1& 2& 4\\ 0& -3& -9\end{bmatrix} \xrightarrow{3R_3+R_1\to R_1} \begin{bmatrix} 0 & 0 & 0\\ 1& 2& 4\\ 0& -3& -9\end{bmatrix} \\ \xrightarrow{R_3/(-3) \to R_3} \begin{bmatrix} 0 & 0 & 0\\ 1& 2& 4\\ 0& 1& 3\end{bmatrix} \xrightarrow{(R_1,R_2)交換,(R_2,R_3)交換)}\begin{bmatrix}1& 2& 4\\ 0& 1& 3\\0 & 0 & 0\end{bmatrix} \Rightarrow 秩=2，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{x+2y-z=3 \cdots(1)\\ 4x+3y+5z=4 \cdots(2)\\ 2x-y+7z=t\cdots(3)} ,由(1)得z=x+2y-3代入(3)及(4) \Rightarrow \cases{9x+13y=19\\ 9x+13y=t+21}\\ 有無限多解\Rightarrow 19=t+21 \Rightarrow t=-2，故選\bbox[red, 2pt]{(A)}$$

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