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2018年9月13日 星期四

106年調查人員三等考試--工程數學詳解


106年法務部調查局調查人員考試
等 別:三等考試
類科組:電子科學組
科 目:工程數學
工程數學 詳解

y+4y+4y=f(t)L{y}+4L{y}+4L{y}=L{f(t)}(s2L{y}sy(0)y(0))+4(sL{y}y(0))+4L{y}=20estdt(s2+4s+4)L{y}s6=1se2s+1s(s+2)2L{y}=1s+s+61se2s=s2+6s+1s1se2sL{y}=s2+6s+1s(s+2)21s(s+2)2e2s=141s+341s+2+721(s+2)2(141s141s+2121(s+2)2)e2sy=L1{141s+341s+2+721(s+2)2(141s141s+2121(s+2)2)e2s}=14+34e2t+72te2t(1414e2(t2)12(t2)e2(t2))u(t2)



(一)T(x1,x2,x3)=(3x1+x2,2x14x2+3x3,5x1+4x22x3)=[310243542][x1x2x3]det([310243542])=24+0+150436=1[310243542]1=[|4342||1042||1043||2352||3052||3023||2454||3154||3124|]=[423116912710]T1(x1,x2,x3)=(4x12x23x3,11x1+6x2+9x3,12x1+7x2+10x3)(二)T1(1,1,0)=(42,11+6,12+7)=(2,5,5)


f(x)=15+4cosxf(π+x)=f(πx)2π0f(x)dx=2π0f(x)dxu=tan(x/2)cosx=1u21+u2dx=2du1+u22π015+4cosxdx=2π015+4cosxdx=2π015+41tan2(x/2)1+tan2(x/2)dx=2π01+tan2(x/2)9+tan2(x/2)dx=2π0(189+tan2(x/2))dx=2π16π0(19+tan2(x/2))dx19+tan2(x/2)dx=2(9+u2)(1+u2)du=14(11+u219+u2)du=14arctanu112arctan(u/3)+C=14arctantan(x/2)112arctan(tan(x/2)/3)+C=2π16π0(19+tan2(x/2))dx=2π[4arctantan(x/2)43arctan(tan(x/2)/3)]|π0=2π[4π243π2]=2π(2π2π3)=2π3另解:利用柯西-留數定理(Cauchy's Residue Theorem)z=cosθ+isinθ=eiθ{cosθ=eiθ+eiθ2=z+1/z2dz=ieiθdθ=izdθ2π015+4cosθdθ=|z|=115+4((z+1/z)/2)dziz=1i|z|=112z2+5z+2dz=1i|z|=11(2z+1)(z+2)dz=2πi×1i×Resf(1/2)=2π×lim



(一)\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ A\left( x+y \right)  } dxdy } =1\Rightarrow A\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 2 } x^{ 2 }+xy \right]  \right| _{ 0 }^{ 1 }dy } =A\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 2 } +y \right) dy } =A\left. \left[ \frac { 1 }{ 2 } y+\frac { 1 }{ 2 } y^{ 2 } \right]  \right| _{ 0 }^{ 1 }\\ =A\left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 }  \right) =1\Rightarrow \bbox[red,2pt]{A=1} (二)P\left\{ X+Y\le 1 \right\} =\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1-y }{ \left( x+y \right)  } dxdy } =\int _{ 0 }^{ 1 }{ \left. \left[ \frac { 1 }{ 2 } x^{ 2 }+xy \right]  \right| _{ 0 }^{ 1-y }dy } =\int _{ 0 }^{ 1 }{ \left( \frac { 1 }{ 2 } { \left( 1-y \right)  }^{ 2 }+\left( 1-y \right) y \right) dy } \\ =\int _{ 0 }^{ 1 }{ \left( -\frac { 1 }{ 2 } { y }^{ 2 }+\frac { 1 }{ 2 }  \right) dy } =\left. \left[ -\frac { 1 }{ 6 } y^{ 3 }+\frac { 1 }{ 2 } y \right]  \right| _{ 0 }^{ 1 }=-\frac { 1 }{ 6 } +\frac { 1 }{ 2 } =\bbox[red,2pt]{\frac { 1 }{ 3 } }



由起終點可知直線方程式為L:\frac{x-1}{3}=\frac{y-1}{3}=\frac{z-1}{3}\Rightarrow \begin{cases}x=3t+1\\y=3t+1\\z=3t+1 \end{cases}\\ \Rightarrow \begin{cases}dx=3dt\\dy=3dt\\dz=3dt \end{cases}\Rightarrow \int_c{x^2dx-2yzdy+zdz}=\int_0^1{(3t+1)^23dt-2(3t+1)^23dt+(3t+1)3dt}\\ =(-9)\int_0^1{(3t^2+t)dt}=(-9)\left.\left[t^3+\frac{1}{2}t^2\right]\right|_0^1=(-9)\times\frac{3}{2}= \bbox[red,2pt]{-\frac{27}{2}}


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