Question

已知函数f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)

且函数y=f(x)的图像相邻两条对称轴之间的距离为π/2，（1）求f(π/6）的值 （2）若函数f(kx+π/12）（k>0)在[-π/6，π/3]上单调递增，求k的取值范围

Answer 1

解：f(x)=sin^2ωx+√3cosωxcos(π/2-ωx)(ω>0)
=(1-cos2ωx)/2+(√3/2)sin2ωx
=sin(2ωx-π/6)+1/2
∵函数y=f(x)的图像相邻两条对称轴之间的距离为π/2即半周期=π/2
∴T=2π/2ω=π，∴ω=1，
∴f(x)=sin(2x-π/6)+1/2
（1）f(π/6)=sin(π/3-π/6)+1/2=sinπ/6+1/2=1/2+1/2=1
(2)函数f(kx+π/12)=sin(2kx+π/6-π/6)+1/2=sin2kx+1/2，其在[-π/2，π/2]上单调递增，
∴-π/2<2kx<π/2即-π/4k<x<π/4k，
又∵其在[-π/6，π/3]上单调递增
∴-π/4k<-π/6<x<π/3<π/4k，
∴0<k<3/4