f(x)=2sin²(π/4+x)-根号3cos2x=1-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x+1=2sin(2x-π/3)+1∵ x∈[π/4,π/2]∴ 2x-π/3∈[π/6,2π/3]∴ sin(2x-π/3)∈[1/2,1]∴ 2x-π/3=π/6时,f(x)有最小值22x-π/3=π/2时,f(x)有最大值3
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