f(x)=√3sin(2wx)-cos(2wx)+λ
=2sin(2wx-π/6)+λ
(1)对称轴为x=π
则f(π)=±2+λ
2sin(2πw-π/6)+λ=±2+λ
sin(2πw-π/6)=±1
2πw-π/6=kπ+π/2
w=k/2+1/3
因为 w∈(1/2,1)
所以 w=5/6
(1)T=2π/(5/6)=12π/5
(2) f(x)=2sin[(5/3)x-π/6]+λ
f(π/4)=2sin(π/4)+λ=0
所以 λ=-√2
所以 f(x)=2sin[(5/3)x-π/6]-√2
值域为【-2-√2,2-√2】