109年專科學力鑑定-工程數學詳解

教育部109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：$$(A)\times:非齊次，若改為y''-4y=0就是齊次且線性\\(B) \times: 有y^3項，非線性\\ (C)\times: 有y^2項，非線性\\ (D)\bigcirc:齊次且線性，故選\bbox[red,2pt]{(D)}$$

解：$$f(t)={1\over 2}\cos(2t)-3e^{-2t} \Rightarrow L\{f(t)\} = {1\over 2}L\{\cos(2t)\}-3L\{e^{-2t}\} ={1\over 2}\cdot {s\over s^2+2^2}-3 \cdot {1\over s+2} \\ ={s\over 2s^2+8} -{3\over s+2}，故選\bbox[red,2pt]{(A)}$$

解：$$\begin{vmatrix} 2 & 3 & 1\\ a & 0 & -1\\ 5 & 1 & 3\end{vmatrix}=3 \Rightarrow a-15-9a+2=3 \Rightarrow 8a=-16 \Rightarrow a=-2，故選\bbox[red,2pt]{(A)}$$

解：$$\cases{3x+2y+z=6\\ 2x+3y+z=7\\ x+2y+3z=2} \Rightarrow \cases{9x+6y+3z=18\cdots (1)\\ 6x+9y+3z=21 \cdots (2)\\ x+2y+3z=2\cdots (3)} \Rightarrow \cases{(1)-(2)\\ (2)-(3)} \Rightarrow \cases{3x-3y=-3\\ 5x+7y=19} \\ \Rightarrow \cases{x=y-1\cdots(4)\\ 5x+7y=19 \cdots(5)},(4)代入(5) \Rightarrow 5y-5+7y=19 \Rightarrow 12y=24 \Rightarrow y=2 代入(4)\\ \Rightarrow x=2-1=1 \Rightarrow x+y=1+2=3，故選\bbox[red,2pt]{(C)}$$

解：$$y'=e^x \Rightarrow y=e^x+C;\\y(0)=1 \Rightarrow e^0+C=1 \Rightarrow C=0 \Rightarrow y=e^x \Rightarrow y(1)=e，故選\bbox[red,2pt]{(B)}$$

解：$$\cases{\vec u=(a,2,1)\\ \vec v=(-1,b,-2)\\ \vec w=(-2,1,c)} \Rightarrow \vec u+2\vec v-\vec w=(a,2,1)+(-2,2b,-4)+(2,-1,-c) \\ =(a,2b+1,-3-c)=(1,5,-2) \Rightarrow \cases{a=1\\ 2b+1=5\\ -3-c=-2} \Rightarrow \cases{a=1\\ b=2 \\c=-1} \\ \Rightarrow a+2b =1+4=5，故選\bbox[red,2pt]{(D)}$$

解：$$\cases{\vec s=(2,3,-1)\\ \vec t=(a,2,5)} \Rightarrow \vec s \cdot \vec t =2a+6-5=5 \Rightarrow a=2，故選\bbox[red,2pt]{(C)}$$

解：$$\cases{\vec u=(2,a,5)\\ \vec v=(1,0,-2)} \Rightarrow \vec u \times \vec v =(\begin{vmatrix} a& 5\\ 0& -2\end{vmatrix}, \begin{vmatrix}  5& 2\\  -2& 1\end{vmatrix},\begin{vmatrix} 2& a\\ 1& 0\end{vmatrix}) =(-2a,9,-a) =(-6,9,-3)\\ \Rightarrow a=3，故選\bbox[red,2pt]{(C)}$$

解：$$(mx^3y+3xy^2)dx +(x^4+nx^2y)dy=0 \Rightarrow \cases{M(x,y)=mx^3y+3xy^2\\ N(x,y)=x^4+nx^2y} \\ \Rightarrow M_y=N_x \Rightarrow mx^3+6xy = 4x^3+2nxy \Rightarrow \cases{m=4\\ 6=2n\Rightarrow n=3} \\ \Rightarrow 3m+2n=12+6=18，故選\bbox[red,2pt]{(??)}$$公告的答案為(B)=>最後送分

解：$$A=\begin{bmatrix}2 & 3 \\ 5 & 4 \end{bmatrix} \Rightarrow det(A-\lambda I)=0 \Rightarrow \begin{vmatrix}2-\lambda & 3 \\ 5 & 4-\lambda \end{vmatrix}=0 \Rightarrow \lambda^2-6\lambda+8-15=0\\ \Rightarrow (\lambda-7)(\lambda+1)=0 \Rightarrow \lambda=7,-1 \Rightarrow 7-(-1)=8，故選\bbox[red,2pt]{(D)}$$

解：$$只有(B)的行列式不為0，故選\bbox[red,2pt]{(B)}$$

解：$$(A)\times: f_1(-1)=0 \ne -f(1)=-2\\(B)\times: f_2(\pi/2)=\sin(\pi/2)=1 \ne -\sin(-\pi/2)=0\\(C) \times: f_3(\pi/2)=-\pi^2/4 \ne f_3(-\pi/2)= \pi^2/4\\(D)\bigcirc: f_4(x)=f_4(-x)=x^2+1\\，故選\bbox[red,2pt]{(D)}$$

解：$$\cases{A=\begin{bmatrix}a & 1\\2 & 4 \end{bmatrix} \\ B=\begin{bmatrix}2 & 1\\0 & 3 \end{bmatrix} \\C=\begin{bmatrix}-1 & 3\\5 & 2 \end{bmatrix}} \Rightarrow (A+2B)C =(\begin{bmatrix}a & 1\\2 & 4 \end{bmatrix}+ \begin{bmatrix}4 & 2\\0 & 6 \end{bmatrix})\begin{bmatrix}-1 & 3\\5 & 2 \end{bmatrix} =\begin{bmatrix}a+4 & 3\\2 & 10 \end{bmatrix} \begin{bmatrix}-1 & 3\\5 & 2 \end{bmatrix} \\ =\begin{bmatrix}11-a & 3a+18\\48 & 26 \end{bmatrix} =\begin{bmatrix}8 & 27\\48 & 26 \end{bmatrix} \Rightarrow a=3，故選\bbox[red,2pt]{(A)}$$

解：$$f(x)=x \Rightarrow f(x)是奇偶數 \Rightarrow a_n=0\\ b_n={1\over \pi}\int_{-\pi}^{\pi} x\sin (nx)\;dx ={1\over \pi} \left.\left[ -{1\over n}x\cos (nx)+{1\over n^2}\sin (nx)\right] \right|_{-\pi}^{\pi} ={1\over \pi}(-{2\pi\over n}\cos(n\pi))= {2\over n}(-1)^{n+1} \\ \Rightarrow f(x)=\sum_{n=1}^\infty b_n\sin(nx) =\sum_{n=1}^\infty {2\over n}(-1)^{n+1}\sin (nx)，故選\bbox[red,2pt]{(A)}$$

解：$${2s-5 \over (s+2)(s-1)}={a\over s+2} +{b\over s-1} ={a(s-1)+b(s+2) \over (s+2)(s-1)} ={(a+b)s-a+2b \over (s+2)(s-1)} \\\Rightarrow \cases{a+b=2 \\ -a+2b=-5} \Rightarrow \cases{a=3\\ b=-1}\\ \Rightarrow L^{-1}\left\{ {2s-5 \over (s+2)(s-1)}\right\} = L^{-1}\left\{ {3 \over s+2}-{1\over s-1}\right\} =3L^{-1}\left\{ {1 \over s+2} \right\}-L^{-1}\left\{  {1\over s-1}\right\}\\=3e^{-2t}-e^t，故選\bbox[red,2pt]{(B)}$$

解：$$  L^{-1}\left\{ {-2s+6 \over s^2+9}\right\} = L^{-1}\left\{ (-2) \cdot{s \over s^2+3^2}+2\cdot {3\over s^2+ 3^2}\right\} =-2L^{-1}\left\{ {s \over s^2+3^2} \right\}+2L^{-1}\left\{  {3\over s^2+3^2} \right\}\\ =-2\cos 3t +2 \sin 3t，故選\bbox[red,2pt]{(A)}$$

解：$$先求齊次解: y''-2y'+5y=0 \Rightarrow \lambda^2-2\lambda+5 = 0 \Rightarrow \lambda = 1\pm 2i \Rightarrow y_h=e^x(A\cos(2x) +B\sin(2x)) \\ 再由待定係數求特解 y_p=Cx+D\Rightarrow y通解y=y_h+y_p = Ae^x\cos(2x) +Be^x\sin(2x)+Cx+D\\，故選\bbox[red,2pt]{(B)}$$

解：$$先求齊次解: y''+4y=0 \Rightarrow y_h=A\cos(2x) +B\sin(2x) \\ \Rightarrow 通解y=y_h+y_p = A\cos(2x) +B\sin(2x)+{1\over 4}x，故選\bbox[red,2pt]{(C)}$$

解：$$H(t)=\int_0^t e^{20\tau}(t-\tau)^{19}\;d\tau \Rightarrow L^{-1}\left\{ H(t)\right\} =L^{-1}\left\{e^{20t} \right\} L^{-1}\left\{ t^{19}\right\} = {1\over s-20}\times {19!\over s^{20}}，故選\bbox[red,2pt]{(C)}$$

解：$$y=x^r \Rightarrow y'=rx^{r-1} \Rightarrow y''=r(r-1)x^{r-2} \Rightarrow x^2y''-2xy'+2y=0 \Rightarrow r(r-1)x^r-2rx^r+2x^r=0 \\ \Rightarrow x^r(r^2-r-2r+2)=0 \Rightarrow r^2-3r+2=0 \Rightarrow (r-2)(r-1)=0 \\ \Rightarrow y=C_1x +C_2x^2，故選\bbox[red,2pt]{(D)}$$

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