104年身障特考三等考試_電力工程--工程數學詳解

104 年公務人員特種考試關務人員考試、104 年公務人員特種考試身心障礙人員考試及104 年國軍上校以上軍官轉任公務人員考試考試

考試別：身心障礙人員考試
等別：三等考試
類 科 ：電力工程
科 目：工程數學

解： $$y''+5y'+6y=\begin{cases}-2 & 0\le t<3\\0&t\ge 3\end{cases}\Rightarrow L\{y''+5y'+6y\}=\int_0^3-2e^{-st}\,dt\\ \Rightarrow s^2L\{y\}-sy(0)-y'(0)+5\left(sL\{y\}-y(0)\right)+6L\{y\}=\left.\left[\frac{2}{s}e^{-st}\right]\right|_0^3\\ \Rightarrow s^2L\{y\}+5sL\{y\}+6L\{y\}=\frac{2}{s}e^{-3s}-\frac{2}{s}\Rightarrow L\{y\}\left(s^2+5s+6\right)=\frac{2}{s}e^{-3s}-\frac{2}{s}\\ \Rightarrow L\{y\}=\frac{2}{s(s+2)(s+3)}e^{-3s}-\frac{2}{s(s+2)(s+3)}\\ =e^{-3s}\left(\frac{1}{3}\cdot\frac{1}{s}-\frac{1}{s+2}+\frac{2}{3}\cdot\frac{1}{s+3}\right)-\left(\frac{1}{3}\cdot\frac{1}{s}-\frac{1}{s+2}+\frac{2}{3}\cdot\frac{1}{s+3}\right)\\ \Rightarrow y=L^{-1}\left\{e^{-3s}\left(\frac{1}{3}\cdot\frac{1}{s}-\frac{1}{s+2}+\frac{2}{3}\cdot\frac{1}{s+3}\right)-\left(\frac{1}{3}\cdot\frac{1}{s}-\frac{1}{s+2}+\frac{2}{3}\cdot\frac{1}{s+3}\right)\right\}\\ \Rightarrow \bbox[red,2pt]{y=u(t-3)\left(\frac{1}{3}-e^{-2(t-2)}+\frac{2}{3}e^{-3(t-2)}\right)-\left(\frac{1}{3}-e^{-2t}+\frac{2}{3}e^{-3t}\right)}$$

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解：
(一) $$A=\left[\begin{array}{ccc}0&0.95&0.6\\0.8&0&0\\0&0.5&0\end{array}\right]\Rightarrow \left|\begin{array}{ccc}-\lambda&0.95&0.6\\0.8&-\lambda&0\\0&0.5&-\lambda\end{array}\right|=0\Rightarrow det(A-\lambda I)=0\Rightarrow \lambda^3-0.76\lambda-0.24=0\\ \Rightarrow (\lambda-1)(\lambda+0.4)(\lambda+0.6)=0\Rightarrow \lambda = 1,-0.4,-0.6\\ \lambda=1\Rightarrow \left[\begin{array}{ccc}-1&0.95&0.6\\0.8&-1&0\\0&0.5&-1\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0\Rightarrow \begin{cases}x_1=0.95x_2+0.6x_3\\x_2=0.8x_1\\x_3=0.5x_2\end{cases}\Rightarrow \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=k_1\left[\begin{array}{c}1\\0.8\\0.4\end{array}\right]\\ \quad\quad\Rightarrow 取u_1=\left[\begin{array}{c}1\\0.8\\0.4\end{array}\right]\\ \lambda=-0.4\Rightarrow \left[\begin{array}{ccc}0.4&0.95&0.6\\0.8&0.4&0\\0&0.5&0.4\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0\Rightarrow \begin{cases}0.4x_1+0.95x_2+0.6x_3=0\\0.8x_1+0.4x_2=0\\0.5x_2+0.4x_3=0\end{cases}\Rightarrow \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=k_2\left[\begin{array}{r}0.4\\-0.8\\1\end{array}\right]\\ \quad\quad\Rightarrow 取u_2=\left[\begin{array}{r}0.4\\-0.8\\1\end{array}\right]\\ \lambda=-0.6\Rightarrow \left[\begin{array}{ccc}0.6&0.95&0.6\\0.8&0.6&0\\0&0.5&0.6\end{array}\right]\left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=0\Rightarrow \begin{cases}0.6x_1+0.95x_2+0.6x_3=0\\0.8x_1+0.6x_2=0\\0.5x_2+0.6x_3=0\end{cases}\Rightarrow \left[\begin{array}{c}x_1\\x_2\\x_3\end{array}\right]=k_3\left[\begin{array}{r}0.9\\-1.2\\1\end{array}\right]\\ \quad\quad\Rightarrow 取u_3=\left[\begin{array}{r}0.9\\-1.2\\1\end{array}\right]\\ \Rightarrow 矩陣A的一組\text{eigenbasis}為\bbox[red,2pt]{\left\{\left(\begin{array}{c}1\\0.8\\0.4\end{array}\right),\left(\begin{array}{r}0.4\\-0.8\\1\end{array}\right),\left(\begin{array}{r}0.9\\-1.2\\1\end{array}\right)\right\}}$$ (二) $$令P=\left[\begin{array}{rrr}1&0.4&0.9\\0.8&-0.8&-1.2\\0.4&1&1\end{array}\right]\Rightarrow P^{-1}=\left[\begin{array}{rrr}25/56&125/224&15/56\\-10/7&5/7&15/7\\5/4&-15/16&-5/4\end{array}\right]\Rightarrow A=P\left[\begin{array}{rrr}1&0&0\\0&-0.4&0\\0&0&-0.6\end{array}\right]P^{-1}\\ \Rightarrow \lim_{k\to\infty}{x(k)}=\lim_{k\to\infty}{A^kx(0)}=\lim_{k\to\infty}{P\left[\begin{array}{rrr}1^k&0&0\\0&(-0.4)^k&0\\0&0&(-0.6)^k\end{array}\right]P^{-1}x(0)}\\ =P\left[\begin{array}{rrr}1&0&0\\0&0&0\\0&0&0\end{array}\right]P^{-1}x(0)=\left[\begin{array}{rrr}1&0.4&0.9\\0.8&-0.8&-1.2\\0.4&1&1\end{array}\right]\left[\begin{array}{rrr}1&0&0\\0&0&0\\0&0&0\end{array}\right]\left[\begin{array}{rrr}25/56&125/224&15/56\\-10/7&5/7&15/7\\5/4&-15/16&-5/4\end{array}\right]\left[\begin{array}{r}1250\\600\\400\end{array}\right]\\ =\left[\begin{array}{rrr}1&0&0\\0.8&0&0\\0.4&0&0\end{array}\right]\left[\begin{array}{rrr}25/56&125/224&15/56\\-10/7&5/7&15/7\\5/4&-15/16&-5/4\end{array}\right]\left[\begin{array}{r}1250\\600\\400\end{array}\right]=\left[\begin{array}{rrr}25/56&125/224&15/56\\5/14&25/56&3/14\\5/28&25/112&3/28\end{array}\right]\left[\begin{array}{r}1250\\600\\400\end{array}\right]\\ =\bbox[red,2pt]{\left[\begin{array}{r}1000\\800\\400\end{array}\right]}$$

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解： $$(一)\\ f_X(x)=\int{f_{X,Y}(x,y)\,dy}=\int_0^x{\lambda^3e^{-\lambda x}\,dy}=\bbox[red,2pt]{\lambda^3xe^{-\lambda x},x\ge 0}\\ (二)\\ f_{Y/X}\left(y/x\right)=\frac{f_{X,Y}(x,y)}{f_X(x)}=\frac{\lambda^3e^{-\lambda x}}{\lambda^3xe^{-\lambda x}}=\bbox[red,2pt]{\frac{1}{x},0\le y< x}\\ (三)\\ P(Y\le 0.1\mid X=0.5)=\int_0^{0.1}{\frac{1}{0.5}\,dy}=\int_0^{0.1}{2\,dy}=\bbox[red,2pt]{0.2}$$

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解： $$(一)C:\frac{x-1}{2-1}=\frac{y-2}{-1-2}=\frac{z-3}{4-3}=t\Rightarrow C:(x,y,z)=\bbox[red,2pt]{(t+1,-3t+2,t+3),0\le t\le 1}\\ (二)\int_C{\vec{F}\cdot d\vec{r}}=\int_C{(3y,3x,2z)\cdot(dx,dy,dz)}=\int_C{(3ydx+3xdy+2zdz)}\\ \quad\quad =\int_0^1{(3(-3t+2)dt+3(t+1)(-3dt)+2(t+3)dt)}=\int_0^1{(-16t+3)\,dt}=-8+3=\bbox[red,2pt]{-5}\\ (三)\vec{F}=3y\vec{i}+3x\vec{j}+2z\vec{k}\Rightarrow \begin{cases}\frac{\partial f}{\partial x}=3y\\\frac{\partial f}{\partial y}=3x\\ \frac{\partial f}{\partial z}=2z\end{cases}\Rightarrow \begin{cases}f(x,y,z)=3xy+g_1(y,z)\\f(x,y,z)=3xy+g_2(x,z)\\ f(x,y,z)=z^2+g_3(x,y)\end{cases} \\ \Rightarrow \bbox[red,2pt]{f(x,y,z)=3xy+z^2+C,C為常數}$$

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乙、測驗題部分：(50分)

解： $$F\times G=-G\times F ，故選\bbox[red,2pt]{(C)}$$

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解： $$F=xy\vec { i } +\left( zx-\sin { y }  \right) \vec { j } +yz\vec { k } \Rightarrow \nabla \cdot F=\frac { \partial  }{ \partial x } xy+\frac { \partial  }{ \partial y } \left( zx-\sin { y }  \right) +\frac { \partial  }{ \partial x } yz\\ =y-\cos { y } \Rightarrow \left. \nabla \cdot F \right| _{ \left( -1,0,1 \right)  }=0-\cos { 0 } =-1，故選\bbox[red,2pt]{(B)}$$

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解： $$\nabla \times fA=f\nabla \times A，故選\bbox[red,2pt]{(D)}$$

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解： $$依定義，故選\bbox[red,2pt]{(C)}$$

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解： $$A=\left[ \begin{matrix} -3 & 2 \\ -10 & 6 \end{matrix} \right] \Rightarrow 特徵值為1,2\Rightarrow A=P\left[ \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right] P^{ -1 }\Rightarrow f\left( A \right) =A^{ 3 }-2A^{ 2 }+A+I\\ =P\left[ \begin{matrix} 1 & 0 \\ 0 & 8 \end{matrix} \right] P^{ -1 }+P\left[ \begin{matrix} -2 & 0 \\ 0 & -8 \end{matrix} \right] P^{ -1 }+P\left[ \begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix} \right] P^{ -1 }+P\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] P^{ -1 }\\ =P\left[ \begin{matrix} 1 & 0 \\ 0 & 3 \end{matrix} \right] P^{ -1 }\Rightarrow \left| f\left( A \right)  \right| =\left| \begin{matrix} 1 & 0 \\ 0 & 3 \end{matrix} \right| =3，故選\bbox[red,2pt]{(B)}$$

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解： $$A=P\left[ \begin{matrix} \lambda _{ 1 } & 0 & 0 \\ 0 & \lambda _{ 2 } & 0 \\ 0 & 0 & \lambda _{ 3 } \end{matrix} \right] P^{ -1 }\Rightarrow A^{ n }=P\left[ \begin{matrix} \lambda _{ 1 }^{ 2 } & 0 & 0 \\ 0 & \lambda _{ 2 }^{ 2 } & 0 \\ 0 & 0 & \lambda _{ 3 }^{ 2 } \end{matrix} \right] P^{ -1 }\\\Rightarrow A^{ n }=\begin{cases} P\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] P^{ -1 } & n是偶數 \\ P\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] P^{ -1 }=A & n是奇數 \end{cases}\\ \left( 2A+I \right) A\left( A+2I \right) =2A^{ 3 }+5A^{ 2 }+2A=2A+5A^{ 2 }+2A=4A+5A^{ 2 }\\ =P\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & -4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] P^{ -1 }+P\left[ \begin{matrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{matrix} \right] P^{ -1 }=P\left[ \begin{matrix} 9 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 9 \end{matrix} \right] P^{ -1 }\Rightarrow 9,1,9為其特徵值，故選\bbox[red,2pt]{(B)}$$

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解： $$A=\left[ \begin{matrix} 1 & 3 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & -2 \end{matrix} \right] \Rightarrow det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 1-\lambda  & 3 & 0 \\ 3 & 1-\lambda  & 0 \\ 0 & 0 & -2-\lambda  \end{matrix} \right| =0\Rightarrow (\lambda -1)^{ 2 }(-2-\lambda )+9(\lambda +2)=0\\ \Rightarrow \left( \lambda +2 \right) \left( 3^{ 2 }-(\lambda -1)^{ 2 } \right) =\left( \lambda +2 \right) \left( \lambda +2 \right) \left( -\lambda +4 \right) \Rightarrow \lambda =-2,-2,4，故選\bbox[red,2pt]{(D)}$$

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解： $$R=\lim _{ n\rightarrow \infty  }{ \left| \frac { a_{ n } }{ a_{ n+1 } }  \right|  } =\lim _{ n\rightarrow \infty  }{ \left( \frac { \left( 2n \right) ! }{ { \left( n! \right)  }^{ 2 } } \cdot \frac { { \left( \left( n+1 \right) ! \right)  }^{ 2 } }{ \left( 2n+2 \right) ! }  \right)  } =\lim _{ n\rightarrow \infty  }{ \left( \frac { \left( n+1 \right) \left( n+1 \right)  }{ \left( 2n+2 \right) \left( 2n+1 \right)  }  \right)  } \\ =\lim _{ n\rightarrow \infty  }{ \frac { n^{ 2 }+2n+1 }{ 4n^{ 2 }+4n+1 }  } =\frac { 1 }{ 4 } ，故選\bbox[red,2pt]{(A)}$$

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解： $$f\left( z \right) =\frac { 1 }{ \left( z+4 \right) z^{ 3 } } \Rightarrow z=-4,0皆在C之內\Rightarrow \begin{cases} \underset { z=-4 }{ Res } { f\left( z \right) =\frac { 1 }{ { \left( -4 \right)  }^{ 3 } } =-\frac { 1 }{ 64 }  } \\ \underset { z=0 }{ Res } { f\left( z \right) =\frac { 1 }{ 2 } \times \frac { 2 }{ 4^{ 3 } } =\frac { 1 }{ 64 }  } \end{cases}\\ \Rightarrow \int _{ C }{ f\left( z \right) dz } =2\pi i\left( -\frac { 1 }{ 64 } +\frac { 1 }{ 64 }  \right) =0，故選\bbox[red,2pt]{(B)}$$

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解： $$\lim _{ n\rightarrow \infty  }{ \frac { 1 }{ { 2 }^{ n } }  } =\lim _{ n\rightarrow \infty  }{ \frac { 1 }{ { 3 }^{ n } }  } =0,但\left\{ \frac { 1 }{ { 2 }^{ n } }  \right\} \neq \left\{ \frac { 1 }{ { 3 }^{ n } }  \right\} ，故選\bbox[red,2pt]{(D)}$$

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解： $$y=x^{ m }\Rightarrow y'=mx^{ m-1 }\Rightarrow y''=m(m-1)x^{ m-2 }\\ x^{ 2 }y''+2xy'-6y=0\Rightarrow m(m-1)x^{ m }+2mx^{ m }-6x^{ m }=0\\ \Rightarrow m(m-1)+2m-6=0\Rightarrow m^2+m-6=0\Rightarrow (m+3)(m-2)=0\\ \Rightarrow m=2,-3\Rightarrow y=C_1x^2+C_2x^{-3}\Rightarrow m+n=2-3=-1，故選\bbox[red,2pt]{(C)}$$

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解： $$y=a_{ 0 }+a_{ 1 }x+a_{ 2 }x^{ 2 }+\cdots \Rightarrow y'=a_{ 1 }+2a_{ 2 }x+3a_{ 3 }x^{ 2 }+\cdots \\ y(0)=0\Rightarrow a_{ 0 }=0\\ \sin { x } =x-\frac { x^{ 3 } }{ 3! } +\cdots 的常數項為0\\ e^{ y }=1+y+\frac { y^{ 2 } }{ 2! } +\cdots =1+\left( a_{ 0 }+a_{ 1 }x+\cdots  \right) +\frac { 1 }{ 2 } \left( a_{ 0 }+a_{ 1 }x+\cdots  \right) ^{ 2 }+\cdots 的常數項為1\\ \Rightarrow \sin { x } +e^{ y }的常數項為0+1=1=a_{ 1 }\\ \sin { x } =x-\frac { x^{ 3 } }{ 3! } +\cdots 的x係數為1\\ e^{ y }=1+\left( a_{ 0 }+a_{ 1 }x+\cdots  \right) +\frac { 1 }{ 2 } \left( a_{ 0 }+a_{ 1 }x+\cdots  \right) ^{ 2 }+\cdots 的x係數為a_{ 1 }+0+0\cdots =a_{ 1 }=1\\ \Rightarrow \sin { x } +e^{ y }的x係數為1+1=2=2a_{ 2 }\Rightarrow a_{ 2 }=1\\ \sum _{ n=0 }^{ 2 }{ a_n } =a_0+a_1+a_2=0+1+1=2，故選\bbox[red,2pt]{(C)}$$

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解： $$Bessel\quad function\quad x^{ 2 }y''+xy'+\left( x^{ 2 }-v^{ 2 } \right) y=0\Rightarrow y=AJ_{ v }\left( x \right) +BY_{ v }\left( x \right) \\ 因此x^{ 2 }y''+xy'+\left( x^{ 2 }-\left( \frac { 1 }{ 3 }  \right) ^{ 2 } \right) y=0\Rightarrow y=AJ_{ 1/3 }\left( x \right) +BY_{ 1/3 }\left( x \right) ，故選\bbox[red,2pt]{(B)}$$

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解： $$f\left( t \right) =3t^{ 2 }-e^{ -t }-\int _{ 0 }^{ t }{ f\left( \alpha  \right) { e }^{ t-\alpha  }\, d\alpha  } \Rightarrow f\left( 0 \right) =0-1-0=-1\\ \Rightarrow f'\left( t \right) =6t+e^{ -t }-\left( \int _{ 0 }^{ t }{ \frac { d }{ d\alpha  } f\left( \alpha  \right) { e }^{ t-\alpha  }\, d\alpha  } +f\left( t \right) { e }^{ t-t }\cdot \frac { d }{ dt } t-f\left( 0 \right) { e }^{ t }\cdot \frac { d }{ dt } 0 \right) \\ =6t+e^{ -t }-\int _{ 0 }^{ t }{ f\left( \alpha  \right) { e }^{ t-\alpha  }\, d\alpha  } -f\left( t \right) \Rightarrow f'\left( 0 \right) =1-f\left( 0 \right) =2\\ \Rightarrow \begin{cases} f\left( 0 \right) =-1 \\ f'\left( 0 \right) =2 \end{cases}，僅有(C)符合此條件，故選\bbox[red,2pt]{(C)}$$

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解： $$無論F(s), \lim_{t\to\infty}{f(t)}皆為0，故選\bbox[red,2pt]{(A)}$$

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解： $$(A)e^y非線性 (C) PDE 2次非線性  (D) (y')^3 非線性，故選\bbox[red,2pt]{(B)}$$

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解：
在3個小孩的家庭中，有2女1男的機率為$C^3_2\times\frac{1}{2^3}=\frac{3}{8}$
至少有3個家庭: 有3個家庭或4個家庭，機率為$C^4_3\times\frac{3^3}{8^3}\times\frac{5}{8} + \frac{3^4}{8^4} = \frac{23\times 3^3}{8^4}=\frac{621}{4096}$，故選$\bbox[red,2pt]{(B)}$

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解：
選到第1枚錢幣的機率為1/2，投擲二次皆正面的機率為$\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}$，因此機率為$\frac{1}{2}\times\frac{1}{9}=\frac{1}{18}$；
同理，選到第1枚錢幣的機率為1/2，投擲二次皆正面的機率為$\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}$，因此機率為$\frac{1}{2}\times\frac{1}{25}=\frac{1}{50}$；
上述兩種機率的和為$\frac{1}{18}+\frac{1}{50}=\frac{34}{450}=\frac{17}{225}$，故選$\bbox[red,2pt]{(C)}$

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解：

號碼為奇數或3的倍數: 0,1,3,5,6,7,9，共七個，因此機率為7/10，選項皆不正確!!
主辦單位公布的答案是$\bbox[red,2pt]{(C)}$

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解： $$\left( A \right) P\left( 0<x\le 2 \right) =1-P\left( x=0 \right) =1-2/7=5/7\\ \left( B \right) P\left( x=1 \right) =6/7-2/7=4/7\\ \left( C \right) P\left( x\le 0 \right) =P\left( x=0 \right) =2/7\\ \left( D \right) P\left( x\le 1 \right) =F\left( x=1 \right) =6/7，故選\bbox[red,2pt]{(D)}$$

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考選部未公布答案，解題僅供參考