数理答疑团为您解答,希望对你有所帮助. 函数f(x)=√2sin(ωx-π/4)(ω＞0)的最小正周期为π,则ωπ=2π,ω=2；2x-π/4=2kπ-π/2,x=kπ-π/8;2x-π/4=2kπ+π/2,x=kπ+3π/8;函数单调递增区间[kπ-π/8,kπ+...