Question

已知函数f(x)=-√3sinωxcosωx+cos²ωx,x∈R,ω>0⑴求函数f(x)的值域

⑵若函数f(x)的最小正周期为π/2，则当x∈[0，π/2]时，求f(x)的单调递减区间

Answer 1

解答:
f(x)=
-√3sinωxcosωx+cos²ωx
=-(√3/2)sin(2ωx)+[1+cos(2ωx)]/2
=cos(2ωx)*cos(π/3)-sin(2ωx)*sin(π/3)+1/2
=cos(2ωx+π/3)+1/2
(1)函数的值域是[-1/2,3/2]
(2)最小正周期是T=2π/(2ω)=π/2
∴
ω=2
∴
f(x)=cos(4x+π/3)+1/2
先求所有的减区间
∴
2kπ≤4x+π/3≤2kπ+π
∴
2kπ-π/3≤4x≤2kπ+2π/3
∴
kπ/2-π/12≤x≤kπ/2+π/6
∵
x∈[0，π/2]
∴
函数的单调减区间是[0,π/6]和[5π/12,π/2]