Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

2018年11月22日 星期四

107年專技高考-電子工程技師-工程數學詳解


107年專門職業及技術人員高等考試
建築師、技師、第二次食品技師考試暨普通考試不動產經紀人、記帳士考試試題
等別:高等考試
類別:電子工程技師
程目:工程數學

y=xmy=mxm1y=m(m1)xm2y1xy8x2y=0m(m1)xm2mxm28xm2=0m(m1)m8=0m22m8=0(m4)(m+2)=0m=4,2y=C1x4+C2x2,C1,C2



y3y+2y=2x+ex+cos(ex)=r(x),λ23λ+2=0(λ2)(λ1)=0λ=1,2yh=C1ex+C2e2xy1=ex,y2=e2xW=|y1y2y1y2|=|exe2xex2e2x|=2e3xe3x=e3xy=y1y2r(x)Wdx+y2y1r(x)Wdx=exe2x(2x+ex+cos(ex))e3xdx+e2xex(2x+ex+cos(ex))e3xdx=ex2x+ex+cos(ex)exdx+e2x2x+ex+cos(ex)e2xdx=ex(2xex+1+excos(ex))dx+e2x(2xe2x+ex+e2xcos(ex))dx=ex(2xex2ex+xsin(ex))+e2x(xe2x12e2xexexsin(ex)cos(ex))=ex(2xex+2exx+sin(ex))e2x(xe2x+12e2x+ex+exsin(ex)+cos(ex))=2x+2xex+exsin(ex)x12exexsin(ex)e2xcos(ex)=32+xexxexe2xcos(ex)y=yh+yp=C1ex+C2e2x+32+xxexe2xcos(ex)



f(x+25)=f(x)L{f(x)}=250estf(t)dt1e25s=1055estdt1e25s=5[1sest]|1051e25s=51e25s(1se10s+1se5s)=5e5ss(1e25s)(1e5s)


{z=a+bie2z=e2a+2bi=e2ae2bi=e2a(cos(2b)+isin(2b))3+2i=13(313+i213)=13(cosθ+isinθ){e2a=132b=θ=cos1313{a=14ln13b=12cos1313z=14ln13+i12cos1313



f(z)=2iz+sin(z)z3+z=2iz+sin(z)z(z+i)(zi)=g(z)z+iz=ilim




撲克牌有四種花色,各有13張
(一)取到同花的次數為4\times C^{13}_5,52張抽5張有C^{52}_5種抽法,因此取到同花的機率為\bbox[red,2pt]{\frac{4\times C^{13}_5}{C^{52}_5}}
(二)「順」的情況有(1-5), (2-6),..., (10,11,12,13,A) 共有10種情況,因此同花順的機率為\frac{4\times 10}{C^{52}_5}=\bbox[red,2pt]{\frac{40}{C^{52}_{5}}}



令T=\left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \\ \begin{cases} T\left( \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right]  \right) =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right]  \\ T\left( \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right]  \right) =\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right]  \end{cases}\Rightarrow \begin{cases} \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix} 2 \\ 0 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right]  \\ \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix} 0 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right]  \end{cases}\Rightarrow \begin{cases} 2a=2 \\ 2c=-4 \\ b=-1 \\ d=1 \end{cases}\\ \Rightarrow \begin{cases} a=1 \\ c=-2 \\ b=-1 \\ d=1 \end{cases}\Rightarrow \bbox[red,2pt]{T=\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] }\\ T\left( \left[ \begin{matrix} 3 \\ 6 \end{matrix} \right]  \right) =\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 6 \end{matrix} \right] =\left[ \begin{matrix} 3-6 \\ -6+6 \end{matrix} \right] =\bbox[red,2pt]{\left[ \begin{matrix} -3 \\ 0 \end{matrix} \right] }



A=\left[ \begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & -3 & 0 \\ -1 & -2 & 0 & -3 \end{matrix} \right] \Rightarrow det\left( A \right) =\left| \begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & -3 & 0 \\ -1 & -2 & 0 & -3 \end{matrix} \right| =2\times \left| \begin{matrix} 2 & 0 & 0 \\ 0 & -3 & 0 \\ -2 & 0 & -3 \end{matrix} \right| \\=2\times 2\times \left( -3 \right) \times \left( -3 \right) =\bbox[red,2pt]{36}\\ det\left( A-\lambda I \right) =0\Rightarrow \left| \begin{matrix} 2-\lambda  & 0 & 0 & 0 \\ 0 & 2-\lambda  & 0 & 0 \\ 1 & 0 & -3-\lambda  & 0 \\ -1 & -2 & 0 & -3-\lambda  \end{matrix} \right| =0\Rightarrow { \left( \lambda -2 \right)  }^{ 2 }{ \left( \lambda +3 \right)  }^{ 2 }=0\\\Rightarrow 特徵值\lambda =\bbox[red,2pt]{2,-3}\\ \lambda =2\Rightarrow \left[ \begin{matrix} 2-\lambda  & 0 & 0 & 0 \\ 0 & 2-\lambda  & 0 & 0 \\ 1 & 0 & -3-\lambda  & 0 \\ -1 & -2 & 0 & -3-\lambda  \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\Rightarrow \left[ \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & -5 & 0 \\ -1 & -2 & 0 & -5 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\\\Rightarrow \begin{cases} x_{ 1 }=5x_{ 3 } \\ x_{ 1 }+2x_{ 2 }+5x_{ 4 }=0 \end{cases} \Rightarrow 取特徵向量u_{ 1 }=\bbox[red,2pt]{\left[ \begin{matrix} 5 \\ 0 \\ 1 \\ -1 \end{matrix} \right] }及u_{ 2 }= \bbox[red,2pt]{ \left[ \begin{matrix} 0 \\ -5 \\ 0 \\ 2 \end{matrix} \right] }\\ \lambda =-3\Rightarrow \left[ \begin{matrix} 2-\lambda  & 0 & 0 & 0 \\ 0 & 2-\lambda  & 0 & 0 \\ 1 & 0 & -3-\lambda  & 0 \\ -1 & -2 & 0 & -3-\lambda  \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\Rightarrow \left[ \begin{matrix} 5 & 0 & 0 & 0 \\ 0 & 5 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ -1 & -2 & 0 & 0 \end{matrix} \right] \left[ \begin{matrix} x_{ 1 } \\ x_{ 2 } \\ x_{ 3 } \\ x_{ 4 } \end{matrix} \right] =0\\\Rightarrow \begin{cases} x_{ 1 }=0 \\ x_{ 2 }=0 \\ x_{ 1 }+2x_{ 1 }=0 \end{cases} \Rightarrow 取特徵向量u_{ 3 }=\bbox[red,2pt]{\left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix} \right]} 及u_{ 4 }=\bbox[red,2pt]{\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 1 \end{matrix} \right] }




(一)A=\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\2&-1&0&0\\1&-1&-1&0\end{array}\right]\xrightarrow{(-2)\times r_1+r_3,\,(-1)\times r_1+r_4}\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\0&-1&-2&0\\0&-1&-2&0\end{array}\right] \\\xrightarrow{r_2+r_3,\, r_2+r_4}\left[\begin{array}{rrr|c}1&0&1&0\\0&1&2&0\\0&0&0&0\\0&0&0&0\end{array}\right]\Rightarrow Ax=0\Rightarrow x=\alpha\left[\begin{array}{r}1\\2\\-1\end{array}\right]\Rightarrow \bbox[red,2pt]{\left\{(1,2,-1)\right\}}可為Ax=0之x一組基底(二)A=\left[\begin{array}{rrr|c}1&0&1&b_1\\0&1&2&b_2\\2&-1&0&b_3\\1&-1&-1&b_4\end{array}\right]\xrightarrow{(-2)\times r_1+r_3,\,(-1)\times r_1+r_4}\left[\begin{array}{rrr|l}1&0&1&b_1\\0&1&2&b_2\\0&-1&-2&b_3-2b_1\\0&-1&-2&b_4-b_1\end{array}\right] \\\xrightarrow{r_2+r_3,\, r_2+r_4}\left[\begin{array}{rrr|l}1&0&1&b_1\\0&1&2&b_2\\0&0&0&b_2+b_3-2b_1\\0&0&0&b_2+b_4-b_1\end{array}\right]\Rightarrow Ax=b\Rightarrow \begin{cases}b_2+b_3-2b_1=0\\b_2+b_4-b_1=0\end{cases} \\\Rightarrow b=\alpha\left[\begin{array}{}1\\0\\2\\1\end{array}\right]+\beta\left[\begin{array}{}1\\1\\1\\0\end{array}\right] \Rightarrow \bbox[red,2pt]{\left\{(1,0,2,1),(1,1,1,0)\right\}}可作為b之一組基底

解題僅供參考

沒有留言:

張貼留言