题目

求函数y=2sin(2x-π/6)的值域,单调区间,对称轴,对称点,

提问时间：2020-12-21

答案

1.当2x-π/6=π/2+2kπ,即当x=π/3+kπ,(k∈Z)时,ymax=2×1=2,
当2x-π/6=3π/2+2kπ,即x=5π/6+kπ,(k∈Z)时,ymin=2×(-1)=-2,∴值域[-2,2].
2.由-π/2+2kπ≤2x-π/6≤π/2+2kπ,得-π/6+kπ≤x≤π/3+kπ,k∈Z,
单增区间是[-π/6+kπ,π/3+kπ],k∈Z；
由π/2+2kπ≤2x-π/6≤3π/2+2kπ,得π/3+kπ≤x≤5π/6+kπ,k∈Z,
单减区间是[π/3+kπ,5π/6+kπ],k∈Z.,
3.由2x-π/6=π/2+kπ,k∈Z,得对称轴为x=π/3+(1/2)kπ,k∈Z.
4.由2x-π/6=kπ,得x=π/12+(1/2)kπ,k∈Z,∴对称中心是(π/12+kπ/2,0),k∈Z.