107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：$$\begin{cases} \vec { s } =\left( 1,0,-2 \right) \\ \vec { t } =\left( a,1,1 \right) \\ \vec { u } =\left( 2,1,3 \right) \\ \vec { v } =\left( 3,1,0 \right) \end{cases}\Rightarrow \begin{cases} \vec { s } \cdot \vec { t } =a-2 \\ \vec { u } \cdot \vec { v } =6+1=7 \end{cases}, a-2=7\Rightarrow a=9\Rightarrow 故選\bbox[red,2pt]{(C)}$$
解：
$$\begin{cases} \vec { u } =\left( 2,3,5 \right) \\ \vec { v } =\left( 1,0,-2 \right) \end{cases}\Rightarrow \vec { u } \times \vec { v } =\left( \begin{vmatrix} 3 & 5 \\ 0 & -2 \end{vmatrix},\begin{vmatrix} 5 & 2 \\ -2 & 1 \end{vmatrix},\begin{vmatrix} 2 & 3 \\ 1 & 0 \end{vmatrix} \right) =\left( -6,9,-3 \right) , 故選\bbox[red,2pt]{(A)}$$
解：$$\int _{ -\pi }^{ \pi }{ \cos { 2x } \cdot \cos { 3x } dx } =\frac { 1 }{ 2 } \int _{ -\pi }^{ \pi }{ \cos { 5x } +\cos { x } dx } =\frac { 1 }{ 2 } \left. \left[ \frac { 1 }{ 5 } \sin { 5x } +\sin { x } \right] \right| _{-\pi}^\pi=0, 故選\bbox[red,2pt]{(A)}$$本題公告的答案為(D), 應該是錯的！
解：$$f\left( x \right) =x\sin { x } \Rightarrow f\left( -x \right) =\left( -x \right) \sin { \left( -x \right) } =x\sin { x } =f\left( x \right) \Rightarrow f\left( x \right) 為偶函數\\ 故選\bbox[red,2pt]{(A)}$$
解：$$\begin{cases} a+b=5 \\ \begin{vmatrix} 3 & 2 \\ a & 4 \end{vmatrix}=6 \end{cases}\Rightarrow \begin{cases} a+b=5 \\ 12-2a=6 \end{cases}\Rightarrow \begin{cases} a+b=5 \\ a=3 \end{cases}\Rightarrow \begin{cases} b=2 \\ a=3 \end{cases}\\ \Rightarrow 3a-2b=9-4=5, 故選\bbox[red,2pt]{(C)}$$
解：$$\begin{cases} A=\begin{bmatrix} -1 & 2 \\ 1 & 4 \end{bmatrix} \\ B=\begin{bmatrix} 0 & 1 \\ 5 & 2 \end{bmatrix} \\ C=\begin{bmatrix} 2 & 3 \\ 3 & 1 \end{bmatrix} \end{cases}\Rightarrow \left( 2A+3B \right) C=\left( \begin{bmatrix} -2 & 4 \\ 2 & 8 \end{bmatrix}+\begin{bmatrix} 0 & 3 \\ 15 & 6 \end{bmatrix} \right) \begin{bmatrix} 2 & 3 \\ 3 & 1 \end{bmatrix}=\begin{bmatrix} -2 & 7 \\ 17 & 14 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 3 & 1 \end{bmatrix}\\=\begin{bmatrix} 17 & 1 \\ 76 & 65 \end{bmatrix}, 故選\bbox[red,2pt]{(A)}$$
解：$$\begin{cases} 2x-3y=7 \\ 3x+2y=4 \end{cases}\Rightarrow \begin{cases} 6x-9y=21 \\ 6x+4y=8 \end{cases}\Rightarrow -13y=13\Rightarrow y=-1\Rightarrow 2x+3=7\Rightarrow x=2\\\Rightarrow x+y=2-1=1，故選\bbox[red,2pt]{(B)} $$
解：$$y'=\frac { x }{ y } \Rightarrow yy'=x\Rightarrow ydy=xdx\Rightarrow \frac { 1 }{ 2 } y^{ 2 }=\frac { 1 }{ 2 } x^{ 2 }+C\Rightarrow y^{ 2 }=x^{ 2 }+C\\ y\left( 0 \right) =1\Rightarrow 1=C\Rightarrow y^{ 2 }=x^{ 2 }+1\Rightarrow { \left( y\left( 1 \right) \right) }^{ 2 }=2，故選\bbox[red,2pt]{(A)} $$
解：$$\begin{cases} \vec { u } =\left( 1,3,5 \right) \\ \vec { v } =\left( 2,0,-1 \right) \\ \vec { w } =\left( 3,-2,3 \right) \end{cases}\Rightarrow \vec { u } \cdot \left( \vec { v } \times \vec { w } \right) =\left( 1,3,5 \right) \cdot \left( \left( 2,0,-1 \right) \times \left( 3,-2,3 \right) \right) \\ =\left( 1,3,5 \right) \cdot \left( -2,-9,-4 \right) =-2-27-20=-49，故選\bbox[red,2pt]{(A)}$$
解：$$L\left\{ f\left( t \right) \right\} =L\left\{ e^{ 2t }+\sin { 3t } \right\} =L\left\{ e^{ 2t } \right\} +L\left\{ \sin { 3t } \right\} =\frac { 1 }{ s-2 } +\frac { 3}{ s^2+3^2 } ，故選\bbox[red,2pt]{(D)}$$
解：$$L^{ -1 }\left\{ \frac { s+3 }{ s^{ 2 }+6s+10 } \right\} =L^{ -1 }\left\{ \frac { s+3 }{ \left( s+3 \right) ^{ 2 }+1^{ 2 } } \right\} =e^{ -3t }\cos { t } ，故選\bbox[red,2pt]{(B)}$$
解：$$b_{ n }=\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ f\left( x \right) \sin { \left( nx \right) } dx } =\frac { 1 }{ \pi } \int _{ -\pi }^{ \pi }{ f\left( x \right) \sin { \left( nx \right) } dx } =\frac { 1 }{ \pi } \left( \int _{ -\pi }^{ 0 }{ -2\sin { \left( nx \right) } dx } +\int _{ 0 }^{ \pi }{ 2\sin { \left( nx \right) } dx } \right) \\ =\frac { 1 }{ \pi } \left. \left[ \frac { 2 }{ n } \cos { \left( nx \right) } \right] \right| _{ -\pi }^{ 0 }+\frac { 1 }{ \pi } \left. \left[ -\frac { 2 }{ n } \cos { \left( nx \right) } \right] \right| _{ 0 }^{ \pi }=\frac { 2 }{ n\pi } \left( 1-\cos { \left( -n\pi \right) } \right) -\frac { 2 }{ n\pi } \left( \cos { \left( n\pi \right) } -1 \right) \\ \Rightarrow b_{ 1 }=\frac { 2 }{ \pi } \left( 1-\cos { \left( \pi \right) } \right) -\frac { 2 }{ \pi } \left( \cos { \left( \pi \right) } -1 \right) =\frac { 4 }{ \pi } +\frac { 4 }{ \pi } =\frac { 8 }{ \pi } ，故選\bbox[red,2pt]{(B)}$$
解：$$A=\left[ \begin{matrix} 3 & 2 \\ 5 & 4 \end{matrix} \right] \Rightarrow A^{ -1 }=\frac { 1 }{ det\left( A \right) } \left[ \begin{matrix} 4 & -2 \\ -5 & 3 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} 4 & -2 \\ -5 & 3 \end{matrix} \right] =\left[ \begin{matrix} 2 & -1 \\ -5/2 & 3/2 \end{matrix} \right] =\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \\ \Rightarrow a-b=2-(-1)=3，故選\bbox[red,2pt]{(C)}$$
解：$$\frac { \partial }{ \partial y } mx^{ n }y=\frac { \partial }{ \partial x } x^{ 3 }\Rightarrow mx^{ n }=3x^{ 2 }\Rightarrow \begin{cases} m=3 \\ n=2 \end{cases}\Rightarrow m+2n=3+4=7，故選\bbox[red,2pt]{(D)}$$
解：$$y=m\cos { nx } \Rightarrow y'=-mn\sin { nx } \Rightarrow y''=-mn^{ 2 }\cos { nx } \\ \Rightarrow y''-2y'-y=-mn^{ 2 }\cos { nx } +2mn\sin { nx } -m\cos { nx } \\ =2mn\sin { nx } -\left( mn^{ 2 }+m \right) \cos { nx } =4\sin { 2x } -5\cos { 2x } \\ \Rightarrow \begin{cases} 2mn=4 \\ m\left( n^{ 2 }+1 \right) =5 \end{cases}\Rightarrow \frac { 2n }{ n^{ 2 }+1 } =\frac { 4 }{ 5 } \Rightarrow 2n^{ 2 }-5n+2=0\Rightarrow \left( 2n-1 \right) \left( n-2 \right) =0\\ \Rightarrow n=2(n=1/2不合)\Rightarrow m=1\Rightarrow m+n=2+1=3，故選\bbox[red,2pt]{(C)}$$
解：$$y=x^m\Rightarrow y'=mx^{m-1}\Rightarrow y''=m(m-1)x^{m-2}\Rightarrow x^2y''+2xy'-6y\\=x^2m(m-1)x^{m-2}+2xmx^{m-1}-6x^m =\left(m^2+m-6\right)x^m=0 \Rightarrow (m+3)(m-2)=0\\\Rightarrow m=2,-3\Rightarrow y=C_1x+C_2x^{-3}，故選\bbox[red,2pt]{(D)} $$
解：$$f\left( t \right) \ast g\left( t \right) =\int _{ 0 }^{ t }{ f\left( u \right) g\left( t-u \right) du } =\int _{ 0 }^{ t }{ e^{ au }e^{ a(t-u) }du } =\int _{ 0 }^{ t }{ e^{ at }du } =te^{ at }，故選\bbox[red,2pt]{(D)}$$
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