Question

设ω＞0,若函数f(x)=sinωx/2cosωx/2在区间[-π/3,π/3]上单调递增，则ω的

设ω＞0,若函数f(x)=sinωx/2cosωx/2在区间[-π/3,π/3]上单调递增，则ω的取值范围是多少？(写出详细答案)

Answer 1

设ω＞0,若函数f(x)=sinωx/2cosωx/2在区间[-π/3,π/3]上单调递增，则ω的取值范围是多少？
解析：∵ω＞0,若函数f(x)=sinωx/2cosωx/2在区间[-π/3,π/3]上单调递增，
f(x)=sinωx/2cosωx/2=1/2sin(ωx)
单调递增区：
2kπ-π/2<=ωx<=2kπ+π/2==>2kπ/ω-π/(2ω)<=x<=2kπ/ω+π/(2ω)（k∈Z）
令-π/(2ω)<=-π/3==>ω<=3/2
∴ω的取值范围是0<ω<=3/2

追问

为什么要令-π/2ω＜=-π/3？

追答

∵函数f(x)=sinωx/2cosωx/2在区间[-π/3,π/3]上单调递增
其含义是f(x)的单调增区间至少为[-π/3,π/3]，
当单调增区间为[-π/3,π/3]时，X=-π/3为最小值点，x=π/3为最大值
当单调增区间>[-π/3,π/3]时，f(x)最小值点xπ/3
∵x=-π/(2ω)是f(x)的最小值点
∴令-π/(2ω)ω<=3/2