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.