⑴f(x)=4cos(ωx-π/6)sinωx-cos(2ωx+π)
=4(√3/2cosωx+1/2sinωx)sinωx+cos2ωx
=2√3cosωxsinωx+1/2(sinωx)^2+(cosωx)^2sinωx)^2
=√3sin2ωx+1,
∵-1≤sin2ωx≤1,
∴函数y=f(x)的值域是[1-√3,1+√3]
数学教学试卷分析 shuxue.chazidian.com/jingyan
⑵∵y=sinx在每个区间[2kπ-π/2,2kπ+π/2],k∈z上为增函数,
令2kπ-π/2≤2ωx≤2kπ+π/2,又ω>0,
∴解不等式得
kπ/ω-π/4ω≤x≤kπ/ω+π/4ω,
即f(x)=√3sin2ωx+1,(ω>0)在每个闭区间[kπ/ω-π/4ω,kπ/ω+π/4ω],k∈z上是增函数
又有题设f(x)在区间[-3π/2,π/2]上为增函数
∴[-3π/2,π/2]∈[kπ/ω-π/4ω,kπ/ω+π/4ω],对某个k∈z成立,
于是有
-3π/2≥-π/4ω①
π/2≤π/4ω②
解得ω≤1、6
∴ω的最大值是1/6