设 向量OA=(2sinx,cos2x),向量OB=(-cosx,1),其中x∈[0,π/2].

2个回答

  • 1.向量(OA*OB)=(-2sinx*cosx+cos2x)=-sin2x+cos2x

    =-√2*sin(2x-π/4).

    f(x)=-√2*sin(2x-π/4).

    而,x∈[0,π/2].

    0≤x≤π/2,

    -π/4≤2x-π/4≤3π/4,

    要使f(x)有最大值,则,sin(2x-π/4)=-√2/2.

    即,f(x)最大=-√2*(-√2/2)=1.

    要使f(x)最小值,则,sin(2x-π/4)=1.

    即,f(x)最小=-√2*1=-√2.

    2.向量OA⊥向量OB,则有

    向量(OA*OB)=0,

    即,-√2*sin(2x-π/4)=0,

    2X-π/4=0,

    X=π/8.

    而,向量AB=向量(OB-OA)=(-cosx-sinx,1-cos2x),

    |AB|=√[(-cosx-2sinx)^2+(1+cos2x)^2]

    =√[1+3sin^2x+4sinx*cosx+(2cos^2x)^2]

    =√[1+3/2*(1-cos2x)+4sin2x+(cos2x+1)^2].

    而,X=π/8.代入上式得,

    |AB|=√[1+3/2*(1-cos2x)+4sin2x+(cos2x+1)^2]

    =√[1+3/2*(1-√2/2)+4*√2/2+(√2/2+1)^2]

    =[√(16+9√2)]/2.