Question

已知函数f(x)=(sinx+cosx)^2-2倍根号3cos^2x+根号3

（1）求f(x)的单调递增区间
(2)求函数y=f(x+π/12),x属于[0,π/2]的值域

Answer 1

f(x)=(sinx+cosx)^2-2倍根号3cos^2x+根号3
=1+sin2x-√3（cos2x+1）+√3
=sin2x-√3cos2x
=2sin（2x-π/3）
令2kπ-π/2<=2x-π/3<=2kπ+π/2
得到kπ-π/12<=x<=kπ+5π/12
故函数的增区间是[kπ-π/12,kπ+5π/12] (k是整数)
（2）f（x+π/12）=2sin（2x-π/6）
当x属于[0,π/2]时，2x-π/6属于[-π/6,5π/6]
得到函数的值域是[-1,2]