Question

求解高中数学题，在线等！

函数y=3sin(kx+π/3)的最小正周期t，满足t∈(1,3)，求正整数k，并就最小的k值求出其单调区间及对称中心
（其中/是分数线，求解题过程和思路！）

Answer 1

T=2π/k
1<2π/k<3
2π/3<K<2π正整数K=3，4，5，6
K最小=3则y=3sin(3x+π/3)
增区间2kπ-π/2<=3X+π/3<=2Kπ+π/2,即【2Kπ/3-5π/18,2Kπ/3+π/18】
减区间2kπ+π/2<=3X+π/3<=2Kπ+3π/2,

Answer 2

解;1<2π/k<3, 于是得 2π/3<k<2π,∴k=3, 4, 5, 6.
当k=3时， y=3sin(3x+π/3)
单增区间:2kπ/3-5π/18≤x≤2kπ/3+π/18,
单减区间 2kπ/3+π/18≤x≤2kπ/3+7π/18