Question

已知函数f(X)=2sin2(π/4+x)-根号3cos2x

Answer 1

(1)
f(x)=2sin²(π/4+x)-√3cos2x
=1-cos(π/2+2x)-√3cos2x
=1+sin2x-√3cos2x
=2sin(2x-π/3)+1
所以最小值是-2+1=-1
最大值是2+1=3
所以值域是[-1,3]
(2)
f(x)的最小正周期是T=2π/2=π
求f(x)的单调递减区间：
令2kπ+π/2＜2x-π/3＜2kπ+3π/2,k∈Z
得kπ+5π/12＜x＜kπ+11π/12,k∈Z
所以f(x)的单调递减区间是(kπ+5π/12,kπ+11π/12),k∈Z

Answer 2

f(X)=2sin^(π/4+x)-√3cos2x
=1-cos(π/2+2x)-√3cos2x
=1+sin2x-√3cos2x
=1+2sin(2x-π/3),
(1)f(x)的值域是[-1,3].
(2)f(x)的周期是π，递减区间由（2k+1/2)π<2x-π/3<(2k+3/2)π,k∈Z确定，
各加π/3，得（2k+5/6)π<2x<(2k+11/6)π,
各除以2，得（k+5/12)π<x<(k+11/12)π,为所求。

Answer 3