已知函数f(x)=2sin^(x+兀/4)-(根号3)cos2x,x属于[兀/4,兀/2].(1)求f(x)的最大值和最小值;(2)若不等式|f(x)-m|<2在x属于[兀/4,兀/2]上恒成立,求实数m的取值范围

已知函数f(x)=2sin^(x+π/4)-√3cos2x,x属于[π/4,π/2].(1)求f(x)的最大值和最小值;(2)若不等式|f(x)-m|<2在x属于[π/4,π/2]上恒成立,求实数m的取值范围
解:f(x)=-[1-2sin^(x+π/4)]+1-√3cos2x
=-cos[2(x+π/4)]+1-√3cos2x
=-cos[2x+(π/2)]+1-√3cos2x
=sin2x+1-√3cos2x
=1+sin2x-√3cos2x
=1+2[((1/2)*sin2x-(√3/2)*cos2x]
=1+2sin(2x+π/3)
(1).f(x)的最大值1+2=3,最小值1-2=-1
(2).|f(x)-m|<2→
-2<f(x)-m<2→
-2<1+2sin(2x+π/3)-m<2→
-3+m<2sin(2x+π/3)<2+m

x属于[π/4,π/2]→2x属于[π/2,π]→2x+π/3属于[7π/12,4π/3]→
2sin(4π/3)<2sin(2x+π/3)<2sin(7π/12)→
-√3<2sin(2x+π/3)<(√6+√2)/2→
-3+m<-√3且2+m>(√6+√2)/2→
m<3-√3且m>(√6+√2)/2-2=(-4+√6+√2)/2
∴实数m的取值范围(-4+√6+√2)/2<m<3-√3