108年專科學力鑑定考試--工程數學-詳解

108年專科學校畢業程度自學進修學力鑑定考試

專業科目(一)：工程數學 詳解

解： $$\begin{cases}\vec{i} = <1,0,0>\\ \vec{j}=<0,1,0> \\ \vec k=<0,0,1>\end{cases}  \Rightarrow \begin{cases}\vec{u} = \vec i-4\vec j+3\vec k =<1,-4,3>\\ \vec{v}=2\vec i -3\vec k =<2,0,-3>\end{cases}  \\ \Rightarrow 2\vec u-\vec v=<2,-8,6>-<2,0,-3>  = <0,-8,9> \\\Rightarrow (2\vec u-\vec v)\cdot \vec u =<0,-8,9>\cdot <1,-4,3> =0+32+27=59 ， 故選：\bbox[red,2pt]{(C)}$$

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解： $$\cos \theta= \frac{\vec a\cdot \vec b}{|\vec a||\vec b|} = \frac{<2,3,-1>\cdot <3,-1,2>}{\sqrt{2^2+3^2+1^2}\sqrt{3^2+1^2+2^2}} =  \frac{6-3-2}{\sqrt{14}\sqrt{14}} = \frac{1}{14} ， 故選\bbox[red,2pt]{(A)}$$

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解：
$$(\vec a\times \vec b) \cdot \vec c =<3,-1,2>\times <1,-1,2>\cdot <2,1,3>=<0,-4,-2> \cdot <2,1,3> \\= -4-6=-10， 故選\bbox[red,2pt]{(C)}$$

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解： $$A= \begin{bmatrix}0 & 1 &0 \\0 & 0 &1 \\1 & 0 & 0 \end{bmatrix}  \Rightarrow A^2=\begin{bmatrix}0 & 1 &0 \\0 & 0 &1 \\1 & 0 & 0 \end{bmatrix}  \begin{bmatrix}0 & 1 &0 \\0 & 0 &1 \\1 & 0 & 0 \end{bmatrix}  =\begin{bmatrix}0 & 0 &1 \\1 & 0 &0 \\0 & 1 & 0 \end{bmatrix} ， 故選\bbox[red,2pt]{(A)}$$

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解： $$\begin{cases}A= \begin{bmatrix}1 & 2 &3 \\2 & 3 &1  \end{bmatrix} \\B= \begin{bmatrix}1 \\2 \\ 3  \end{bmatrix}\end{cases}  \Rightarrow \begin{cases}dim(A) = 2\times 3\\dim(B)=3\times 1\end{cases} \\(A)\times: \begin{cases}dim(AB)=2\times 1 \\ dim(B)=3 \times 1\end{cases} \Rightarrow AB不能與B相加\\(B)\times: \begin{cases}dim(A^T)=3\times 2 \\ dim(B)=3 \times 1\end{cases}\Rightarrow A^T不能與B相乘 \\(C)\times: \begin{cases}dim(AA^T)=2\times 2 \\ dim(B^TB)=1 \times 1\end{cases}\Rightarrow AA^T不能與B^TB相乘\\(D)\bigcirc: \begin{cases}dim(A^TA)=3\times 3 \\ dim(BB^T)=3 \times 3\end{cases} \Rightarrow dim(A^TA)=dim(BB^T) \Rightarrow兩者可相加\\， 故選\bbox[red,2pt]{(D)}$$

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解： $$rank(A)<3 \Rightarrow  det(A)=0  \Rightarrow \begin{vmatrix}t & 1 &-1 \\-1 & t &1 \\ 1&-1 &t \end{vmatrix} =0 \Rightarrow t^3-1+1+t+t+t=0 \Rightarrow t^3+3t=0\\  \Rightarrow t(t^2+3)=0 \Rightarrow t=0， 故選\bbox[red,2pt]{(B)}$$

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解： $$\begin{vmatrix}1 & 2 &3 \\2 & 3&1 \\ 3& 1 & 2 \end{vmatrix} = 6+6+6-27-8-1=-18， 故選\bbox[red,2pt]{(A)}$$

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解： $$\begin{cases}x+2y-z=-3\\ 2x+y+2z=9 \\ 3x+3y+z=t\end{cases}  \Rightarrow \left[ \begin{array}{rrr|r} 1 & 2 & -1 & -3\\ 2 & 1 & 2 & 9 \\ 3 & 3& 1 & t\end{array}\right] \xrightarrow{-2r_1+r_2,-3r_1+r_3}  \left[ \begin{array}{rrr|r} 1 & 2 & -1 & -3\\ 0 & -3 & 4 & 15 \\ 0 & -3& 4 & 9+t\end{array}\right] \\ \Rightarrow 9+t=15 \Rightarrow t=6，故選\bbox[red,2pt]{(B)}$$

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解： $$det(A)=t(t^2-2t)-(t-2) = (t^2-1)(t-2)\ne 0 \Rightarrow t\ne \pm 1,t\ne 2 \Rightarrow t=0，故選\bbox[red,2pt]{(B)}$$

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解： $$A恰有一特徵值 \Rightarrow det(A-\lambda I)=0有重根 \Rightarrow \begin{vmatrix} 1-\lambda & b\\ c & 3-\lambda\end{vmatrix}=0有重根\\ \Rightarrow \lambda^2-4\lambda +3-bc=0有重根 \Rightarrow 判別式=0 \Rightarrow 16-4(3-bc)=0 \Rightarrow bc=-1，故選\bbox[red,2pt]{(D)}$$

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解： $$(y^2+y^{2019})dx + (x^2+x^{108})dy=0 \Rightarrow {1\over x^2+x^{108}}dx=-{1\over y^2+ y^{2019}}dy \Rightarrow 可分離\\，故選\bbox[red,2pt]{(A)}$$

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解： $$(A)\times: \begin{cases} \frac{\partial}{\partial y}(2xy-\tan^{-1}y)  = 2x-{1\over 1+y^2}\\ \frac{\partial}{\partial x}(x^2+2019x)  = 2x+2019\end{cases}  \Rightarrow 兩者不相等 \\(B)\bigcirc: \begin{cases} \frac{\partial}{\partial y}(2xy-\tan^{-1}x)  = 2x\\ \frac{\partial}{\partial x}(x^2+2019y)  = 2x\end{cases}  \Rightarrow 兩者相等 \\(C)\times: \begin{cases} \frac{\partial}{\partial y}(2xy-\cot^{-1}x)  = 2x\\ \frac{\partial}{\partial x}(y^2+2019y)  = 0\end{cases}  \Rightarrow 兩者不相等 \\(D)\times: \begin{cases} \frac{\partial}{\partial y}(2xy-\cot^{-1}y)  = 2x+{1\over 1+y^2}\\ \frac{\partial}{\partial x}(y^2+2019x)  = 2019\end{cases}  \Rightarrow 兩者不相等 \\ ，故選\bbox[red,2pt]{(B)}$$

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解： $$令 \begin{cases}M(x,y) = 20y\\ N(x,y)=-19x\end{cases}   \Rightarrow 20ydx-19xdy=0 \equiv Mdx+Ndy=0 \\(A)\times: \begin{cases} \frac{\partial}{\partial y}(20x+17y)M  = \frac{\partial}{\partial y}(400xy+340y^2) =400x+680y\\ \frac{\partial}{\partial x}(20x+17y)N  =  \frac{\partial}{\partial x} (-380x^2-323xy) = -760x-323y\end{cases}  \Rightarrow 兩者不相等 \\(B)\times: \begin{cases} \frac{\partial}{\partial y}(x^{20}+y^{17})M  = \frac{\partial}{\partial y}(20x^{20}y+20y^{18}) =20x^{20}+360y^{17}\\ \frac{\partial}{\partial x}(x^{20}+y^{17})N  =  \frac{\partial}{\partial x} (-19x^{21}-19xy^{17}) = -399x^{20} -19y^{17}\end{cases}  \Rightarrow 兩者不相等 \\(C)\bigcirc: \begin{cases} \frac{\partial}{\partial y}\left({-y^{18}\over x^{21}}M \right) = \frac{\partial}{\partial y}\left({-20y^{19}\over x^{21}} \right) =-380{y^{20} \over x^{21}}\\ \frac{\partial}{\partial x}\left({-y^{18}\over x^{21}}N \right)  = \frac{\partial}{\partial x}\left(19{y^{18}\over x^{20}} \right) =-380{y^{18} \over x^{21}}\end{cases}  \Rightarrow 兩者相等 \\(D)\times: \begin{cases} \frac{\partial}{\partial y}(xyM) =\frac{\partial} {\partial y} (20xy^2)  = 40xy\\ \frac{\partial}{\partial x}(xyN) =\frac{\partial}{\partial x}(-19x^2y)  = -38xy\end{cases}  \Rightarrow 兩者不相等 \\，故選\bbox[red,2pt]{(C)}$$

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解：
$$y=e^{mx}適用於常係數微分方程，故選\bbox[red,2pt]{(D)}$$

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解：
$$y=x^m適用於柯西-歐拉方程式，即n階導函數的係數為a_nx^n(a_n為常數)，故選\bbox[red,2pt]{(A)}$$

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解： $$y=(Ax^2+Bx)\cos (5x)+(Cx^2+Dx)\sin(5x) \\\Rightarrow y'=(2Ax+B)\cos(5x)-5(Ax^2+Bx)\sin (5x) +(2Cx+D)\sin (5x)+5(Cx^2+Dx)\cos(5x)\\ =(5Cx^2+(5D+2A)x+B)\cos(5x) +(-5Ax^2 +(-5B+2C)x+D)\sin(5x)\\ \Rightarrow y''=(10Cx+5D+2A)\cos(5x)+(-25Cx^2+(-25D-10A)x-5B)\sin (5x)\\ \qquad +(-10Ax-5B+2C)\sin (5x)+(-25Ax^2+(-25B+10C)x+5D)\cos(5x)\\ =(-25Ax^2+(-25B+20C)x +10D+2A) \cos(5x) +(-25Cx^2+(-20A-25D)x-10B+2C)\sin(5x)\\ \Rightarrow y''+25y = (20Cx+10D+2A)\cos(5x) +(-20Ax-10B+2C)\sin(5x)\\ = (A'x+B')\cos(5x)+ (C'x+D')\sin(5x)，無x^2項，故選\bbox[red,2pt]{(B)}$$

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解： $$反拉普拉斯轉換為線性轉換\Rightarrow L^{-1}(F+G)= L^{-1}(F)+L^{-1}(G) \Rightarrow (A)與(B)錯誤\\ 又L\{e^{at}f(t)\}= \int e^{at}f(t)e^{-st}\;dt = \int f(t)e^{-(s-a)t}\;dt =F(s-a)\\ \Rightarrow L^{-1}(F(s-a))=e^{at}L^{-1}(F(s))，故選\bbox[red,2pt]{(C)}$$

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解： $$L(e^{2019t}\cos t\sin t) =L\left({1\over 2}e^{2019t}\sin (2t)\right) ={1\over 2}L(e^{2019t}\sin (2t)) = {1\over 2}\cdot {2 \over (s-2019)^2+2^2}\\ ={1 \over (s-2019)^2+4}，故選\bbox[red,2pt]{(D)}$$

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解： $$該傅立葉級數無項次\sin{2k\pi x\over P}\Rightarrow f(x)為偶函數，即f(-x)=f(x)，僅有(A)符合此條件\\，故選\bbox[red,2pt]{(A)}$$

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解： $$(A)及(C)的週期均為3，不符條件；又a_0={1\over 2}\int_{-1}^1f(x)\;dx ={1\over 2}\int_{2018}^{2020}f(x)\;dx ={1\over 2}\times 2=1\\，故選\bbox[red,2pt]{(B)}$$

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解題僅供參考