Question

已知函数f（x）=2cos2ωx+2sinωxcosωx+1（x∈R，ω＞0）的最小值正周期是π/2，当x∈[﹣π／8，π／8]

求f（x）值域

Answer 1

f(x)=2cos2wx+sin2wx+1=√5sin(2wx+t)+1, t=arctan2
T=2π/(2w)=π/2, 得：w=2
f(x)=√5sin(4x+t）+1
-π/8<x<π/8
-π/2+t<4x+t<π/2+t
-1/√5=-cost=sin(-π/2+t)=<sin(4x+t)<=1
故0=<f(x)<=√5+1