已知函数f（x）=2sin（π/4+x）^2-（根号3）2x-1,x属于r.（1）求f（x）的最值和最小正周期.（2）若h（x）=_数学解答

1)f(x)=2sin(x+π/4)^2-√3cos2x-1=-√3cos2x-[1-2sin(x+π/4)^2]
=-√3cos2x- cos(2x+π/2)
=-√3cos2x+sin2x
=2sin(2x-π/3)
最小值为-2,最大值为2
最小正周期为T=π
2)h(x)=f(x+t)=2sin(2x+2t-π/3)关于点（-π/6,0)对称
则2sin(2×(-π/6)+2t-π/3)=0
故2×(-π/6)+2t-π/3=kπ,t属于（0,π）,
所以t= kπ/2+π/3,k∈Z.
t=π/3或5π/6.设p:x属于[π/4，π/2]，q：f（x）-m的绝对值<3，若p是q的充分条件，求实数m的取值范围。谢谢。。当x属于[π/4,π/2]时，f(x) =2sin(2x-π/3)2x-π/3属于[π/6,2π/3]sin(2x-π/3)属于[1/2,1]所以f(x) 属于[1,2]若p是q的充分条件，则有If(x)-mI<3-3