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

2018年12月6日 星期四

104年鐵路人員特考--工程數學詳解


104 年公務人員特種考試警察人員、一般警察人員考試及 104 年特種考試交通事業鐵路人員、退除役軍人轉任公務人員考試試題

等別:高員三級鐵路人員考試
類 科 :電力工程、電子工程
科 目:工程數學


()det(AλI)=0|1λ015λ|=0(λ+5)(λ+1)=0Aλ=1,5()λ=1[0014][x1x2]=0x1=4x2[x1x2]=k1[41]u1=[41]λ=5[4010][x1x2]=0x1=0[x1x2]=k2[01]u2=[01](41)(01)()P=[4011]P1=[1/401/41]A=P[1005]P1A20=P[(1)2000(5)20]P1=P[100520]P1=[4011][100520][1/401/41]=[401520][1/401/41]=[10(1520)/4520]



a0=14(10x2dx+411dx)=14(13+3)=56an=12(10x2cosnπx2dx+41cosnπx2dx)=12([(2x2nπ16n3π3)sinnπx2+8xn2π2cosnπx2]|10+[2nπsinnπx2]|41)=12((2nπ16n3π3)sinnπ2+8n2π2cosnπ22nπsinnπ2)=12(16n3π3sinnπ2+8n2π2cosnπ2)=4n2π2cosnπ28n3π3sinnπ2bn=12(10x2sinnπx2dx+41sinnπx2dx)=12([(2x2nπ+16n3π3)cosnπx2+8xn2π2sinnπx2]|10+[2nπcosnπx2]|41)=12((2nπ+16n3π3)cosnπ2+8n2π2sinnπ216n3π32nπ+2nπcosnπ2)=12(16n3π3cosnπ2+8n2π2sinnπ216n3π32nπ)=8n3π3cosnπ2+4n2π2sinnπ28n3π31nπf(x)=56+n=1[ancosnπx2+bnsinnπx2],an=4n2π2cosnπ28n3π3sinnπ2,bn=8n3π3cosnπ2+4n2π2sinnπ28n3π31nπ


f(z)=2z5(z+1)(z2+4)=2z5(z+1)(z2i)(z2i)z=1,±2i1,2iResz=1f(z)=limz12z5z2+4=251+4=75Resz=2if(z)=limz2i2z5(z+1)(z+2i)=4i5(2i+1)(4i)=710320if(z)dz=πi×Resz=1f(z)+2πi×Resz=2if(z)=πi×(75)+2πi×(710320i)=3π10


:(一)σ2x=2E[X2](E[X])2=2E[X2]=2+02=2E[W2]=E[(2X+Y)2]=E[4X2+4XY+Y2]=4E[X2]+4E[XY]+E[Y2]=4×2+4×(2)+4=4(二)E[WU]=E[(2X+Y)(X3Y)]=E[2X27XY3Y2]=2E[X2]7E[XY]3E[Y2]=2×27×(2)3×4=2(三)σ2Y=E[Y2](E[Y])2=4(1)2=3

乙、測驗題部分:(50分)

v(t)v(t)(C)


F=x2i+y2j+2z2kF=xx2+yy2+x2z2=2x+2y+4zF|(1,1,1)=2+24=0(B)




{u=zi+xj+ykv=xyi+yzj+zxku×v=|ijkzxyxyyzzx|=x2zi+yz2k+xy2jx2ykxz2jy2zi=(x2zy2z)i+(xy2xz2)j+(yz2x2y)k(u×v)=x(x2zy2z)+y(xy2xz2)+z(yz2x2y)=2xz+2xy+2yz(B)


C(t+1,π),0t1F=2(t+1)cos(2π)i2(t+1)2sin(2π)j=(2t+2)iCFdR=10(2t+2)dt=[t2+2t]|10=3(A)


(A)|1212012|=12±1(B)(1,1)(D)(23,13)(C)


det(A)=122=101=det(I)=det(AA1)=det(A)det(A1)=10det(A1)det(A1)=110(A)


A=[102100131321113039012]2×r1+r3,3×r2+r4[10210013130131300033](1)r2+r3,r4/3[10210013130000000011](1)r5+r1,(1)r5+r2[10201013040000000011]Ax=0{x12x3+x5=0x23x3+4x5=0x4x5=0{x1=2x3x5x2=3x3+4x5=0x3=x3x4=x5x5=x5Null(A)={s(23100)+t(14011)}dim of Null(A)=2(B)


det(AλI)=0|2λ2321λ612λ|=λ(λ1)(λ+2)+12+12+3(λ1)+4λ+12(λ+2)=λ3+λ221λ45=(λ5)(λ+3)2=0λ=5,3λ=5[723246125][x1x2x3]=0[8080816125][x1x2x3]=0{x1=x3x2=2x3[x1x2x3]=C1[121],u1=[121]λ=3[123246123][x1x2x3]=0[123000000][x1x2x3]=0x1+2x2=3x3[x1x2x3]=C2[101/3]+C3[012/3],u2=[101/3],u3=[012/3](A)[210]=2u2+u3,(B)[121]=u1,(D)[301]=3u2(C)


z=6+8i=10(35+45i)=10(cosθ+isinθ),{cosθ=35sinθ=45tanθ=43(r,θ)=(10,tan143)(D)


()


g(z)=1z=1x+iy=xiy(x+iy)(xiy)=xiyx2+y2=xx2+y2+iyx2+y2xx2+y2,yx2+y2(D)



2.5xy2.5xy若題目改為2.5xy,則解法如下:
y=xmy=mxm1y


\frac { dr }{ d\theta  } =b\left[ \frac { dr }{ d\theta  } \cos { \theta  } +r\sin { \theta  }  \right] \Rightarrow (1-b\cos { \theta  }) \frac { dr }{ d\theta  } =br\sin { \theta  } \\ \Rightarrow \frac { 1 }{ r } dr=\frac { b\sin { \theta  }  }{ 1-b\cos { \theta  }  } d\theta \Rightarrow \int { \frac { 1 }{ r } dr } =\int { \frac { b\sin { \theta  }  }{ 1-b\cos { \theta  }  } d\theta  } +C\\ \Rightarrow \ln { r } =\ln { \left( 1-b\cos { \theta  }  \right)  } +C\Rightarrow r=K\left( 1-b\cos { \theta  }  \right) \\ r\left( \frac { \pi  }{ 2 }  \right) =\pi \Rightarrow \pi =K\Rightarrow r=\pi \left( 1-b\cos { \theta  }  \right)  ,故選\bbox[red,2pt]{(A)}

y_{ 1 }=x\Rightarrow y_{ 2 }=v(x)\cdot x\Rightarrow y_{ 2 }'=v(x)+v'(x)x\Rightarrow y_{ 2 }''=v''(x)x+2v'(x)\\ \Rightarrow (x^{ 2 }-x)y''-xy'+y=(x^{ 2 }-x)\left( v''(x)x+2v'(x) \right) -x\left( v(x)+v'(x)x \right) +v(x)\cdot x\\ =x(x^{ 2 }-x)v''+\left( x^{ 2 }-2x \right) v'=0\Rightarrow (x^{ 2 }-x)v''=\left( 2-x \right) v'\Rightarrow \frac { v'' }{ v' } =\frac { 2-x }{ x^{ 2 }-x } \\ \Rightarrow \int { v''dv' } =\int { \frac { 2-x }{ x^{ 2 }-x } dx } \Rightarrow \ln { \left| v' \right|  } =\ln { \left| x-1 \right| -2\ln { \left| x \right|  }  } \Rightarrow v'=\frac { x-1 }{ x^{ 2 } } \\ \Rightarrow v=\ln { x } +\frac { 1 }{ x } \Rightarrow y_{ 2 }=\left( \ln { x } +\frac { 1 }{ x }  \right) x=x\ln { x } +1,故選\bbox[red,2pt]{(C)}


\begin{cases} L\left\{ u\left( t-1 \right) f\left( t-1 \right)  \right\} ={ e }^{ -s }F\left( s \right)  \\ L\left\{ t\cos { \left( 2t \right)  }  \right\} =\frac { { s }^{ 2 }-4 }{ { \left( { s }^{ 2 }+4 \right)  }^{ 2 } }  \end{cases}\\ \Rightarrow L^{ -1 }\left\{ \frac { \left( { s }^{ 2 }-4 \right) { e }^{ -s } }{ { \left( { s }^{ 2 }+4 \right)  }^{ 2 } }  \right\} =u\left( t-1 \right) \left( t-1 \right) \cos { \left( 2\left( t-1 \right)  \right)  } ,故選\bbox[red,2pt]{(A)}


y''(t)+4y'(t)+3y(t)=6,先求y''(t)+4y'(t)+3y(t)=0的解\\ \lambda ^{ 2 }+4\lambda +3=0\Rightarrow (\lambda +3)(\lambda +1)=0\Rightarrow \lambda =-1,-3\Rightarrow y_{ h }=C_{ 1 }e^{ -t }+C_{ 2 }e^{ -2t }\\ 令y_{ p }=k\Rightarrow y''_{ p }+4y'_{ p }+3y_{ p }=3k=6\Rightarrow k=2\Rightarrow y=y_{ h }+y_{ p }=C_{ 1 }e^{ -t }+C_{ 2 }e^{ -2t }+2\\ \lim _{ t\rightarrow \infty  }{ y\left( t \right)  } =\lim _{ t\rightarrow \infty  }{ \left( C_{ 1 }e^{ -t }+C_{ 2 }e^{ -2t }+2 \right)  } =2,故選\bbox[red,2pt]{(C)}


f\left( x \right) =\begin{cases} -1 & -1<x<0 \\ 1 & 0<x<1 \\ 0 & otherwise \end{cases}\Rightarrow F\left( \omega \right) =\frac { 1 }{ \sqrt { 2\pi } } \left( \int _{ -1 }^{ 0 }{ - } { e }^{ -i\omega x }dx+\int _{ 0 }^{ 1 }{ { e }^{ -i\omega x }dx } \right) \\ =\frac { 1 }{ \sqrt { 2\pi } } \left( \left. \left[ \frac { 1 }{ i\omega } { e }^{ -i\omega x } \right] \right| _{ -1 }^{ 0 }-\left. \left[ \frac { 1 }{ i\omega } { e }^{ -i\omega x } \right] \right| _{ 0 }^{ 1 } \right) =\frac { 1 }{ \sqrt { 2\pi } } \left( \frac { 2 }{ i\omega } -\frac { 1 }{ i\omega } \left( { e }^{ i\omega }+{ e }^{ -i\omega } \right) \right) \\ =\frac { 1 }{ \sqrt { 2\pi } } \left( \frac { 2 }{ i\omega } -\frac { 2 }{ i\omega } \cos { \omega } \right) =\frac { i }{ \sqrt { 2\pi } } \left( -\frac { 2 }{ \omega } +\frac { 2 }{ \omega } \cos { \omega } \right) =\frac { i\sqrt { 2 } }{ \sqrt { \pi } } \frac { \cos { \omega } -1 }{ \omega },故選\bbox[red,2pt]{(D)}


C^5_3\times\frac{1}{2^5}=10\times\frac{1}{32}=\frac{5}{16},故選\bbox[red,2pt]{(B)}


\sum _{ y=1 }^{ 3 }{ \sum _{ x=1 }^{ 2 }{ P\left( x,y \right)  }  } =1\Rightarrow A\left( 5+8+11+7+10+13 \right) =54A=1\Rightarrow A=\frac { 1 }{ 54 } ,故選\bbox[red,2pt]{(A)}


P\left( x=k \right) =\frac { { e }^{ -\lambda  }{ \lambda  }^{ k } }{ k! } \Rightarrow E\left[ X \right] =\lambda =4\\ P\left( x\le 1 \right) =1-P\left( x=0 \right) -P\left( x=1 \right) =1-{ e }^{ -4 }-4{ e }^{ -4 }=1-5{ e }^{ -4 },故選\bbox[red,2pt]{(D)}


考選部未公布答案,解題僅供參考

沒有留言:

張貼留言