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
解:
(一)T(x1,x2,x3)=(3x1+x2,−2x1−4x2+3x3,5x1+4x2−2x3)=[310−2−4354−2][x1x2x3]det([310−2−4354−2])=24+0+15−0−4−36=−1⇒[310−2−4354−2]−1=−[|−434−2|−|104−2||10−43|−|−235−2||305−2|−|30−23||−2−454|−|3154||31−2−4|]=[4−2−3−1169−12710]⇒T−1(x1,x2,x3)=(4x1−2x2−3x3,−11x1+6x2+9x3,−12x1+7x2+10x3)(二)T−1(1,1,0)=(4−2,−11+6,−12+7)=(2,−5,−5)
解:f(x)=15+4cosx⇒f(π+x)=f(π−x)⇒∫2π0f(x)dx=2∫π0f(x)dx令u=tan(x/2)⇒cosx=1−u21+u2且dx=2du1+u2⇒∫2π015+4cosxdx=2∫π015+4cosxdx=2∫π015+41−tan2(x/2)1+tan2(x/2)dx=2∫π01+tan2(x/2)9+tan2(x/2)dx=2∫π0(1−89+tan2(x/2))dx=2π−16∫π0(19+tan2(x/2))dx又∫19+tan2(x/2)dx=∫2(9+u2)(1+u2)du=14∫(11+u2−19+u2)du=14arctanu−112arctan(u/3)+C=14arctantan(x/2)−112arctan(tan(x/2)/3)+C因此原式=2π−16∫π0(19+tan2(x/2))dx=2π−[4arctantan(x/2)−43arctan(tan(x/2)/3)]|π0=2π−[4⋅π2−43⋅π2]=2π−(2π−2π3)=2π3另解:利用柯西-留數定理(Cauchy's Residue Theorem)z=cosθ+isinθ=eiθ⇒{cosθ=eiθ+e−iθ2=z+1/z2dz=ieiθdθ=izdθ⇒∫2π015+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
解:
(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 } }
解:
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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