若f（x）=2sinωx(0

若f（x）=2sinωx(0
若f（x）=2sinωx(0<ω<1>)在区间【0,π/3】上的最大值根号√2,则ω=?

若f（x）=2sinωx(0
0<ω<1,x∈[0,π/3]
∴ωx∈[0,ωπ/3]⊆[0,π/3]⊆[0,π/2]
∴f（x）=2sinωx(0<ω<1)在区间【0,π/3】上单调递增
∴f（x）max=f（π/3）=2sin﹙ωπ/3﹚=√2
∴sin﹙ωπ/3﹚=√2／2
∴ωπ/3=π/4
∴ω=3/4