|m|>f(x)-2
|m|>2sin(2x-π/3)-1
2x-π/3∈[π/6,2π/3]
2x-π/3=π/2时,2sin(2x-π/3)-1取得最大值2-1=1
所以|m|>1
m1
已知函数f(x)=2sin(2x-π/3)+1,x属于[π/4,π/2],
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