已知函数f{x}=2sin方{x+π除以4}-根号3cos2x-1.当x属于{π除以4,π除以2},时的最大最小值

f(x)=2sin²(x+π/4)-√3cos2x-1
=2(sinxcos(π/4)+cosxsin(π/4))²-√3cos2x-1
=2(√2/2)²(sinx+cosx)²-√3cos2x-1
=sin²x+cos²x+2sinxcosx-√3cos2x-1
=sin2x-√3cos2x
=-2(cos(2π/3)sin2x+sin(2π/3)cos2x)
=-2sin(2x+2π/3)=2sin(2x-π/3)
∵x∈[π/4,π/2]
∴2x∈[π/2,π]
2x-π/3∈[π/6,2π/3]
所以f(x)∈[2sin(π/6),2sin(π/2)]=[1,2]