已知向量a=(cos3x/2,sin3x/2),向量b=(cosx/2,-sinx/2),且x∈[0,π/2].

1.ab=cos(3x/2)*cos(x/2)-sin(3x/2)*sin(x/2)
=cos[(3x+x)/2]
=cos(2x).

a+b=(cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)),
|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]
=√[2(1+cos2x)]
=2*|cosx|,
因为,x∈[-π/3,π/4]。则有,cosx>0,
即,
|a+b|=2*|cosx|=2cosx.
2.f（x）=a*b-2λ|a+b|=2cos^2x-1+4λcosx=2(cosx+λ)^2-2λ^2-1
0<=cosx=<1
λ>0时 x>0f(x)递增
所以f(x)min=2(0+λ)^2-2λ^2-1=-1≠-3/2
1<=λ=<0时 cosx=-λ f(x)取的最小值
f(x)min=-2λ^2-1=-3/2
λ=-1/2
λ>1时 cos x=1取得最小值
f(x)min=2-1+4λ=-3/2
λ=-5/8不符合条件舍去
综上 λ=-1/2