f(x) = √3sin(2x-π/6)-2cos²(x-π/12) + 1
= √3sin(2x-π/6)-cos(2x-π/6)
= 2{sin(2x-π/6)cosπ/6-cos(2x-π/6)sinπ/6}
= 2sin(2x-π/6-π/6)
= 2sin(2x-π/3) ∈【-2,2】
最小值,-2,最大值2,最小正周期2π/2=π
将函数f(x)的图像向左平移 π/6 个单位得到函数g(x)
g(x) = 2sin[2(x+π/6)-π/3] = 2sin2x
2x∈（2kπ-π/2,2kπ+π/2)时g(x)单调增
故g(x)单调增区间x∈（kπ-π/4,kπ+π/4)
对称轴方程x=kπ/2±π/4