函数f(x)=2sin^2 (x+π/4)- √3cos2x-1
=-cos(2x+π/2)- √3cos2x
=sin2x- √3cos2x
=2sin(2x-π/3)
1.当x属于R时,函数f(x)的最小正周期T=2π/2=π.
2.当x属于[π/12,13π/24]时,求f(x)的最大值和最小值.
13π/24-π/12=11π/24