1
f(x)=a·b+2λ|a+b|
a·b=(cos(3x/2),sin(3x/2))·(cos(x/2),-sin(x/2))
=cos(2x)
|a+b|^2=|a|^2+|b|^2+2a·b
=2+2cos(2x)=4cosx^2,x∈[0,π/2]
故：|a+b|=2cosx
即：f(x)=a·b+2λ|a+b|=cos(2x)+2λcosx
2
f(x)=2cosx^2+2λcosx-1
=2(cosx+λ/2)^2-(λ^2/2+1)
cosx∈[0,1]
当：0≤-λ/2≤1,即：-2≤λ≤0时
f(x)的最小值是：-(λ^2/2+1)
当：-λ/2>1,即：λ