已知向量a=（cos3/2x,sin3/2x）,b=（cosx/2,-sinx/2）,且x属于[0,π/2],若f（x）=a·b-2λ|a+b|的最小值是-3/2,求实数λ的值(化简步骤稍微详细一点,谢谢)

Question

已知向量a=（cos3/2x,sin3/2x）,b=（cosx/2,-sinx/2）,且x属于[0,π/2],若f（x）=a·b-2λ|a+b|的最小值是-3/2,求实数λ的值(化简步骤稍微详细一点,谢谢)

Answer 1

a*b=cos[(3/2)x]cos(x/2)-sin[(3/2)x]sin(x/2)=cos2x,
|a+b|^2=[cos(3/2x)+cos(x/2)]^2+[sin(3/2x)-sin(x/2)]^2
=2+2cos2x=4(cosx)^2,
∴f(x)=cos2x-4λ|cosx|
=2(cosx)^2-4λ|cosx|-1
=2[|cosx|-λ]^2-2λ^2-1,
x∈[0,π/2],cosx∈[0,1],
λ∈[0,1]时f(x)|min=-2λ^2-1=-3/2,λ^2=1/4,λ=1/2;
λ1时f(x)|min=1-4λ=-3/2,4λ=5/2,λ=5/8(舍）.
综上,λ=1/2.