已知向量a=[cos(3x/2),sin(3x/2)],已知向量b=[cos(x/2),-sin(x/2)],x属于[0

1个回答

  • (1)向量a=(cos3x/2,sin3x/2),

    向量b=(cosx/2),-sinx/2),

    向量a·b=(cos3x/2)*(cosx/2)+(sin3x/2)*(-sinx/2),

    =cos(3x/2+x/2)=cos2x,

    向量a+b=(cos3x/2+cosx/2,sin3x/2-sinx/2),

    |a+b|=√[(cos3x/2+cosx/2)^2+(sin3x/2-sinx/2)^2]

    =√{(2cos2xcosx/2)*2+(2cos2xsinx/2)]

    =2√(cos2x)^2*[cosx/2)^2+(sinx/2)^2]

    =2|cos2x|,

    F(x)=a·b/|a+b|

    =cos2x/(2|cos2x|),

    x∈[0,π/3],

    2x∈[0,2π/3],

    当2x∈[0,π/2]时,cos2x>0,

    F(x)=1/2,最大.

    (2)λa·b-(1/2)|a+b|+λ-1≤0,x∈[0,π/3],

    λcos2x+λ≤cos2x/(4|cos2x|)+1,

    λ(1+cos2x)≤cos2x/(4|cos2x|)+1,

    2x∈[0,2π/3],

    1+cos2x≥0,

    λ≤[cos2x/(4|cos2x|)+1]/(1+cos2x),

    当2x∈[0,π/2]时,

    λ≤(5/4)(1+cos2x),

    1+cos2最大为2,

    λ≤5/8,

    2x∈[π/2,2π/3]时,

    λ≤(3/4)/(1+cos2x),

    (1+cos2x)最大为1/2,(1-1/2)

    λ≤3/2,

    取其交集,

    λ≤5/8.