網頁

2019年8月17日 星期六

108年國安三等考試_電子組(選試英文)--工程數學詳解


108年國家安全局國家安全情報人員考試試題
考試別:國家安全情報人員
等別:三等考試
類 科組 :電子組
科 目:工程數學


:$$\left(e^{-2x}y' \right)'+(1+\lambda)e^{-2x}y=0 \Rightarrow e^{-2x}y''-2e^{-2x}y'+(1+\lambda)e^{-2x}y=0\\
\Rightarrow y''-2y'+(1+\lambda)y=0 \Rightarrow k^2-2k+(1+\lambda)=0 \Rightarrow k={2\pm\sqrt{4-4(1+\lambda)}\over 2}\\ \Rightarrow k=1\pm i\sqrt{\lambda} \Rightarrow y(x)=C_1e^x\cos{\sqrt{\lambda}x} +C_2e^x\sin{\sqrt{\lambda}x} \\
\begin{cases} y(0)=0 \\y(1)=0\end{cases} \Rightarrow \begin{cases} C_1=0 \\C_1e\cos{\sqrt{\lambda}} +C_2e\sin{ \sqrt{\lambda}}=0\end{cases} \Rightarrow C_2e\sin{ \sqrt{\lambda}}=0\\
\Rightarrow \sin{ \sqrt{\lambda}}=0 \Rightarrow \sqrt{\lambda}=n\pi\Rightarrow 特徵值\bbox[red, 2pt] {\lambda=n^2\pi^2},n=1,2,\dots\\ \Rightarrow 特徵函數\bbox[red, 2pt]{y=Ce^x\sin{(n\pi x)}},n=1,2,\dots$$


:$$將D'Alembert 的解代入本題,可得: y(x,t)= {1\over 2}(f(x-t)+f(x+t))+ {1\over 2}\int_{x-t}^{x+t}{g(z)\,dz}\\
其中\begin{cases} f(x)=y(x,0)=0\\ g(x)= {\partial y\over \partial t}(x,0)\end{cases} \Rightarrow y(x,t)= {1\over 2}\int_{x-t}^{x+t}{g(z)\,dz}\\
現在我們要求g(x)的解\\
令y(x,t)=X(x)T(t) \Rightarrow \begin{cases}
{\partial^2 y\over \partial t^2} =XT'' \\
{\partial^2 y\over \partial x^2} =X''T
\end{cases},
因此{\partial^2 y\over \partial t^2}={\partial^2 y\over \partial x^2} \Rightarrow XT''=X''T \Rightarrow {X''\over X}= {T''\over T}\\
由於此為波動方程式,可令 {X''\over X}= {T''\over T}=-\omega^2 \Rightarrow \begin{cases} X''+\omega^2X=0 \\ T''+\omega^2T=0 \end{cases} \Rightarrow \begin{cases} X(x)=A\cos{\omega x}+B\sin{\omega x} \\ T(t)=C\cos{\omega t}+D\sin{\omega t} \end{cases} \\
\Rightarrow g(x)={\partial y\over \partial t}(x,0)= X(x)T'(0)= (A\cos{\omega x}+B\sin{\omega x})(-\omega C\sin{0}+\omega D\cos{0}) \\= \omega D(A\cos{\omega x}+B\sin{\omega x}) =\omega(A'\cos{\omega x}+B'\sin{\omega x})\\
\Rightarrow y(x,t)={1\over 2}\int_{x-t}^{x+t}{\omega\left[ A'\cos{\omega z+ B'\sin{\omega z}} \right]\,dz}
={1\over 2} \left. \left [A'\sin{\omega z} -B'\cos{\omega z}\right] \right|_{x-t}^{x+t}\\
={1\over 2}\left [\left( A'\sin{\omega(x+t)} -B'\cos{\omega(x+t)}\right) -\left( A'\sin{\omega(x-t)} -B'\cos{\omega(x-t)} \right) \right]\\
={1\over 2}\left [A'\left( \sin{\omega(x+t)} -\sin{\omega(x-t)}\right) -B'\left(\cos{\omega(x+t)} - B'\cos{\omega(x-t)} \right) \right]\\
={1\over 2}\left [A'\left(2\cos{\omega x}\sin{\omega t}\right) -B'\left(-2\sin{\omega x} \sin{\omega t} \right) \right] = \left( A'\cos{\omega x}+B'\sin{\omega x}\right) \sin{\omega t}\\
\Rightarrow y(x,t)=\left( A'\cos{\omega x}+B'\sin{\omega x}\right) \sin{\omega t}= \sum_{n=1}^\infty {\left(A_n\cos{ \omega_n x }+B_n\sin{\omega_n x} \right)\sin{\omega_n t}}
$$ -- 待續 - -


:$$A=\left[ \begin{array}{rrr}
-2 & 4 & 2 \\
 6 & 3 &-3 \\
 2 & 8 &-5
\end{array} \right]
\Rightarrow det(A)=30+96-24-12+120-48=162\\
\Rightarrow A^{-1}={1\over det(A)}\left[ \begin{array}{rrr}
\left| \begin{matrix} 3& -3\\8&-5 \end{matrix} \right| & -\left| \begin{matrix} 4& 2\\8&-5 \end{matrix} \right| & \left| \begin{matrix} 4& 2\\3 &-3 \end{matrix} \right| \\
-\left| \begin{matrix} 6& -3\\2&-5 \end{matrix} \right| & \left| \begin{matrix} -2& 2\\2&-5 \end{matrix} \right| & -\left| \begin{matrix} -2& 2\\6 &-3 \end{matrix} \right|
  \\
 \left| \begin{matrix} 6& 3\\2& 8 \end{matrix} \right| & -\left| \begin{matrix} -2& 4\\2 & 8 \end{matrix} \right| & \left| \begin{matrix} -2& 4\\6 & 3 \end{matrix} \right|
\end{array} \right]\\
={1\over 162}\left[ \begin{array}{rrr}
9 & 36 & -18 \\
 24 & 6 & 6 \\
 42 & 24 &-30
\end{array} \right] =\bbox[red, 2pt]{\left[ \begin{array}{rrr}
1/18 & 2/9 & -1/9 \\
4/27 & 1/27 & 1/27 \\
7/27 & 4/27 &-5/27
\end{array} \right] }$$


乙、測驗題部分:(50分)

:$$\mathbf{u}\bot(\mathbf{v}\times\mathbf{u}) \Rightarrow \mathbf{u}\cdot(\mathbf{v}\times\mathbf{u})=0,故選\bbox[red,2pt]{(A)}$$


:$$\begin{cases} (m,n,m-2n)\in W\\ (p,q,p-2q)\in W \end{cases} \Rightarrow a(m,n,m-2n)+b(p,q,p-2q)=(am,an,am-2an)+ (bp,bq,bp-2bq)\\ =(am+bp, an+bq, am-2an+bp-2bq)= (am+bp, an+bq, am+bp-2(an+bq))\in W\\
即\mathbf{u},\mathbf{v}\in W \Rightarrow a\mathbf{u}+b\mathbf{v}\in W\Rightarrow W是R^3的子空間,故選\bbox[red,2pt]{(D)}$$




:$$\left[ \begin{matrix}
a_1 & a_2 &a_3\\
b_1 & b_2 &b_3\\
c_1 & c_2 &c_3
\end{matrix}\right] \xrightarrow{-r_2+r_1}
\left[ \begin{matrix}
a_1-b_1 & a_2-b_2 &a_3-b_3\\
b_1 & b_2 &b_3\\
c_1 & c_2 &c_3
\end{matrix}\right] \xrightarrow{\bbox[red,2pt]{3}r_2}
\left[ \begin{matrix}
a_1-b_1 & a_2-b_2 &a_3-b_3\\
3b_1 & 3b_2 &3b_3\\
c_1 & c_2 &c_3
\end{matrix}\right]\\ \xrightarrow{-r_2+r_3}
\left[ \begin{matrix}
a_1-b_1 & a_2-b_2 &a_3-b_3\\
3b_1 & 3b_2 &3b_3\\
c_1-3b_1 & c_2-3b_2 &c_3-3b_3
\end{matrix}\right] \Rightarrow det\left[ \begin{matrix}
a_1 & a_2 &a_3\\
b_1 & b_2 &b_3\\
c_1 & c_2 &c_3
\end{matrix}\right]=3\times det(C)=15,故選\bbox[red,2pt]{(A)}$$


:$$\left[ \begin{matrix}
1 & 0 & 1 & 1\\
1 & 1 &-1 & 2\\
2 & 0 & 1 & 0\\
3 & 2 &-1 & 5
\end{matrix}\right] \xrightarrow{-r_1+r_2,\;-2r_1+r_3,\;-3r_1+r_4}
\left[ \begin{matrix}
1 & 0 & 1 & 1\\
0 & 1 &-2 & 1\\
0 & 0 &-1 &-2\\
0 & 2 &-4 & 2
\end{matrix}\right] \\\xrightarrow{-2r_2+r_4}
\left[ \begin{matrix}
1 & 0 & 1 & 1\\
0 & 1 &-2 & 1\\
0 & 0 &-1 &-2\\
0 & 0 &0  & 0
\end{matrix}\right] \xrightarrow{-2r_3+r_2\;,r_3+r_1}
\left[ \begin{matrix}
1 & 0 & 0 & -1\\
0 & 1 & 0 & 5\\
0 & 0 &-1 &-2\\
0 & 0 &0  & 0
\end{matrix}\right] \\ \Rightarrow Rank(A)=3\Rightarrow A之零空間維度為4-3=1
,故選\bbox[red,2pt]{(B)}$$


:$$\left[ \begin{matrix} 8 \\-5 \end{matrix}\right] =a\left[ \begin{matrix} 1 \\-1 \end{matrix}\right]+b \left[ \begin{matrix} 2 \\-1 \end{matrix}\right] \Rightarrow \begin{cases} a+2b=8 \\ -a-b=-5 \end{cases} \Rightarrow \begin{cases} a=2 \\ b=3 \end{cases}\\
\Rightarrow L\left(\left[ \begin{matrix} 8 \\5 \end{matrix}\right]\right) =L\left(a\left[ \begin{matrix} 1 \\-1 \end{matrix}\right]\right)+ L\left(b \left[ \begin{matrix} 2 \\-1 \end{matrix}\right]\right)\\
=aL\left(\left[ \begin{matrix} 1 \\-1 \end{matrix}\right]\right)+ bL\left(\left[ \begin{matrix} 2 \\-1 \end{matrix}\right]\right) = a\left[ \begin{matrix} 1 \\2 \\-1 \end{matrix}\right]+ b\left[ \begin{matrix} 0 \\1 \\2 \end{matrix}\right] = 2\left[ \begin{matrix} 1 \\2 \\-1 \end{matrix}\right]+ 3\left[ \begin{matrix} 0 \\1 \\2 \end{matrix}\right] \\=\left[ \begin{matrix} 2 \\4 \\-2 \end{matrix}\right]+ \left[ \begin{matrix} 0 \\3 \\6 \end{matrix}\right] =\left[ \begin{matrix} 2 \\7 \\4 \end{matrix}\right],故選\bbox[red,2pt]{(B)} $$


:$$det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 5-\lambda  & 1 & 3 \\ 0 & 2-\lambda  & -12 \\ 0 & 3 & -13-\lambda  \end{matrix} \right| =0\Rightarrow \lambda =5,-1,-10\\ A\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\lambda \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \Rightarrow \begin{cases} \lambda =5 \\ \lambda =-1 \\ \lambda =-10 \end{cases}\Rightarrow \begin{cases} \left[ \begin{matrix} 5x+y+3z \\ 2y-12z \\ 3y-13z \end{matrix} \right] =\left[ \begin{matrix} 5x \\ 5y \\ 5z \end{matrix} \right]  \\ \left[ \begin{matrix} 5x+y+3z \\ 2y-12z \\ 3y-13z \end{matrix} \right] =\left[ \begin{matrix} -x \\ -y \\ -z \end{matrix} \right]  \\ \left[ \begin{matrix} 5x+y+3z \\ 2y-12z \\ 3y-13z \end{matrix} \right] =\left[ \begin{matrix} -10x \\ -10y \\ -10z \end{matrix} \right]  \end{cases}\Rightarrow \begin{cases} \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} x \\ 0 \\ 0 \end{matrix} \right]  \\ \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} -7z/6 \\ 4z \\ z \end{matrix} \right]  \\ \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} -4z/15 \\ z \\ z \end{matrix} \right]  \end{cases}\\ \Rightarrow 取特徵向量為\begin{cases} \left[ \begin{matrix} x \\ 0 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right]  \\ \left[ \begin{matrix} -7z/6 \\ 4z \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -24 \\ -6 \end{matrix} \right]  \\ \left[ \begin{matrix} -4z/15 \\ z \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ -15 \\ -15 \end{matrix} \right]  \end{cases},故選\bbox[red,2pt]{(D)}$$


:$$\lim _{ z\to 0 }{ \frac { z }{ \left| z \right|  }  } =\pm 1\rightarrow 不存在\\ \lim _{ z\to 0 }{ \frac { { \left| z \right|  }^{ 2 } }{ z }  } =0\rightarrow 存在,故選\bbox[red,2pt]{(C)}$$


:$$z^{ 3 }=i=0+i=\begin{cases} \cos { \frac { \pi  }{ 2 }  } +i\sin { \frac { \pi  }{ 2 }  }  \\ \cos { \frac { 5\pi  }{ 2 }  } +i\sin { \frac { 5\pi  }{ 2 }  }  \\ \cos { \frac { 9\pi  }{ 2 }  } +i\sin { \frac { 9\pi  }{ 2 }  }  \end{cases}\Rightarrow z=\begin{cases} \cos { \frac { \pi  }{ 6 }  } +i\sin { \frac { \pi  }{ 6 }  } =\frac { \sqrt { 3 }  }{ 2 } +\frac { 1 }{ 2 } i \\ \cos { \frac { 5\pi  }{ 6 }  } +i\sin { \frac { 5\pi  }{ 6 }  } =-\frac { \sqrt { 3 }  }{ 2 } +\frac { 1 }{ 2 } i \\ \cos { \frac { 9\pi  }{ 6 }  } +i\sin { \frac { 9\pi  }{ 6 }  } =0-i=-i \end{cases}\\,故選\bbox[red,2pt]{(C)}$$


:$$\int_{|z|=2}{\frac{f(z)}{z-1}dz}=2\pi i\times f(1)=2\pi i\times(1+2-5)=-4\pi i,故選\bbox[red,2pt]{(D)}$$


:$$y'''+3y''-y''-y'-3y=0\Rightarrow \lambda^3+3\lambda^2-\lambda-3=0 \Rightarrow (\lambda-1)(\lambda+1)(\lambda+3)=0\\ \Rightarrow \lambda=-3,-1,1 \Rightarrow y=k_1e^{-3t}+k_2e^{-t}+k_3e^t,故選\bbox[red,2pt]{(A)}$$


:$$\left| z-3+7i \right| =4\Rightarrow \left| z-(3-7i) \right| =4\Rightarrow z至(3-7i)的距離為4 \Rightarrow 其解為一圓,故選\bbox[red,2pt]{(C)}$$


:$$先求y_h,即y''+0.5y'-0.5y=0 \Rightarrow \lambda^2+0.5\lambda-0.5=0 \Rightarrow (2\lambda-1)(\lambda+1)=0\\ \Rightarrow \lambda={1\over 2},-1 \Rightarrow y_h=C_1e^{x/2}+C_2e^{-x}\\
令y_p=ae^x+b\sin{x}+c\cos{x} \Rightarrow y_p'=ae^x+b\cos{x}-c\sin{x} \Rightarrow y_p''=ae^x-b\sin{x}-c\cos{x}\\ 代回原式可得 ae^x+ \left( -{3\over 2}b-{1\over 2}c \right)\sin{x}+\left( -{3\over 2}c+{1\over 2}b \right)\cos{x}=e^x+\sin{x}+3\cos{x} \\ \Rightarrow \begin{cases} a=1\\ -{3\over 2}b -{1\over 2}c=1 \\ -{3\over 2}c+{1\over 2}b=3\end{cases} \Rightarrow \begin{cases} a=1\\ b=0 \\ c=-2\end{cases} \Rightarrow y=y_h+y_p= C_1e^{x/2}+C_2e^{-x}+e^x-2\cos{x}\\ \Rightarrow y'={1\over 2}C_1e^{x/2}-C_2e^{-x}+e^x+2\sin{x}\\
\begin{cases} y(0)=0\\ y'(0)=1.5 \end{cases} \Rightarrow \begin{cases} C_1+C_2+1-2=0\\ {1\over 2}C_1-C_2+1=1.5 \end{cases} \Rightarrow  \begin{cases} C_1+C_2=1\\ C_1-2C_2=1 \end{cases} \Rightarrow \begin{cases} C_1=1\\ C_2=0 \end{cases}\\ \Rightarrow y=e^{x/2}+e^x-2\cos{x},故選\bbox[red,2pt]{(A)}$$


:$$ L\{f(t)\}=\int_0^\infty{e^{-st}f(t)\,dt}= \int_0^{2\pi}{e^{-st}\sin{t}\,dt} + \int_{2\pi}^\infty{e^{-st}(2+\cos{t})\,dt} \\=\int_0^{2\pi}{e^{-st}\sin{t}\,dt} + 2\int_{2\pi}^\infty{e^{-st}\,dt}+ \int_{2\pi}^\infty{e^{-st}\cos{t}\,dt}\\
=\left. \left[ e^{-st}\left(-{\cos{t}\over s^2+1} -{s\sin{t}\over s^2+1}\right) \right] \right|_0^{2\pi} +2 \left. \left[ -{1\over s}e^{-st}\right] \right|_{2\pi}^\infty +\left. \left[ e^{-st}\left({\sin{t}\over s^2+1} -{s\cos{t}\over s^2+1}\right) \right] \right|_{2\pi}^\infty\\
={1\over s^2+1}\left(1-e^{-2\pi s} \right)+{2\over s}e^{-2\pi s}+e^{-2\pi s}\left(  {s\over s^2+1}\right) ={1\over s^2+1} + e^{-2\pi s}\left( {2\over s}+{s-1\over s^2+1}\right),故選\bbox[red,2pt]{(C)}$$



:$$\int_1^2{f(x)\,dx}=\int_1^2{(x-ix^2)\,dx}=\left. \left[\frac{1}{2}x^2-\frac{1}{3}ix^3 \right] \right|_1^2=(2-\frac{8}{3}i)-(\frac{1}{2}-\frac{1}{3}i)=\frac{3}{2}-\frac{7}{3}i\\,故選\bbox[red,2pt]{(A)}$$


:$$只看常數項\\y=a-x^2-{x^4\over 3}-\cdots
\Rightarrow y'=-2x-{4x^3\over 3}+\cdots
\Rightarrow y''=-2-4x^2+\cdots\\
\Rightarrow x^2y''=-2x^2-4x^4+ \cdots
\Rightarrow (1-x^2)y''=-2 -2x^2-2x^4 + \cdots \\
\Rightarrow -2xy'=4x^2+ \dots
\Rightarrow (1-x^2)y''-2xy'+2y=(-2+2a)+0\cdot x^2+\cdots =0\\
\Rightarrow -2+2a=0 \Rightarrow a=1,故選\bbox[red,2pt]{(B)}$$


:$$\int_0^t{f(\tau)d\tau} =\int_0^t{\sin(2\tau)d\tau} = {1\over 2}\left( 1-\cos{2t}\right)\\
\Rightarrow L\{\int_0^t{f(\tau)d\tau} \}= L\{{1\over 2}\left( 1-\cos{2t}\right)\}={1\over 2}\left( {1\over s}-{s\over s^2+4}  \right) ={1\over 2}\left( {s^2+4-s^2\over s(s^2+4)}  \right)\\ ={2\over s(s^2+4)},故選\bbox[red,2pt]{(D)}$$


:$$f(x)=x^2={\pi^2\over 3}+4\sum_{n=1}^\infty{(-1)^n\over n^2}\cos{nx}\\ \Rightarrow
\begin{cases} f(0)=0= {\pi^2\over 3}+4\left(-1+{1\over 2^2} -{1\over 3^2} +{1\over 4^2}-\cdots\right)\cdots(1)\\
f(\pi)=\pi^2={\pi^2\over 3}+4\left(1+{1\over 2^2} +{1\over 3^2} +{1\over 4^2}+\cdots\right)\cdots(2)
\end{cases}
\\
(2)-(1)\Rightarrow \pi^2=4\left(2+{2\over 3^2} +{2\over 5^2}+{2\over 7^2}\cdots\right)=8 \left(1+{1\over 3^2} +{1\over 5^2}+{1\over 7^2}\cdots\right)\\ \Rightarrow 1+{1\over 3^2} +{1\over 5^2}+{1\over 7^2}\cdots=\pi^2/8,故選\bbox[red,2pt]{(A)}$$


:$$\begin{cases} E(X)=\int _{ 0 }^{ 1 }{ 2x^{ 2 }\, dx } =\frac { 2 }{ 3 }  \\ E(X^{ 2 })=\int _{ 0 }^{ 1 }{ 2x^{ 3 }\, dx } =\frac { 1 }{ 2 }  \end{cases}\Rightarrow Var(X)=E(X^{ 2 })-(E(X))^{ 2 }=\frac { 1 }{ 2 } -\frac { 4 }{ 9 } =\frac { 1 }{ 18 } ,故選\bbox[red,2pt]{(A)} $$


:$$f(x)=2(1-x) \Rightarrow F(x)=\int{f(x)dx}=2x-x^2 \Rightarrow F(y)=2\sqrt{y}-(\sqrt{y})^2=2\sqrt{y}-y\\ \Rightarrow f(y)=F'(y)={1\over \sqrt{y}}-1,故選\bbox[red,2pt]{(A)} $$


:$$E[XY]=\int\int{xyf(x,y)dxdy} =\int_0^1\int_0^1{x^2y(y+1.5)dxdy} =\int_0^1{\left. \left[ {1\over 3}x^3y(y+1.5) \right] \right|_0^1\,dy} \\ =\int_0^1{{1\over 3}y(y+1.5)\,dy} =\int_0^1{({1\over 3}y^2+{1\over 2}y)\,dy} = \left. \left[ {1\over 9}y^3+ {1\over 4}y^2 \right]\right|_0^1= {1\over 9}+{1\over 4}={13\over 36},故選\bbox[red,2pt]{(C)}$$


考選部未公布答案,解題僅供參考

沒有留言:

張貼留言