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k值求出其单调区间及对称中心.

Question

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k值求出其单调区间及对称中心.

Answer 1

周期 = 2π/k∈（1,3）
所以 2π/3 < k < 2π/1
2.09 < k < 6.28
k = 3,4,或5
最小k为3
y = 3sin(3x+π/3)
T=2π/3,单调增区间为(-T/4+nT/2,T/4+nT/2) 即 (-π/6+nπ/3,π/6+nπ/3) (n为整数)
单调减区间为(T/4+nT/2,3T/4+nT/2) 即 (π/6+nπ/3,π/2+nπ/3) (n为整数)
对称中心 x = T/4 + nT/2 = π/6 + nπ/3 (n为整数)