若函数f(x)=sinωx(ω>0)在[0,π/3]单调递增,在[π/2,2π/3]单调递减,ω范围

题目

若函数f(x)=sinωx(ω>0)在[0,π/3]单调递增,在[π/2,2π/3]单调递减,ω范围

答案

令ωx=u
则f(x)=sinu
显然-π/2≤u≤π/2时f(x)单调递增
而当π/2≤u≤3π/2时f(x)单调递减
当0≤x≤π/3时0≤ωx≤ωπ/3
则有0≤ωx≤ωπ/3≤π/2
即0