Question

已知函数f(x)=Asin(ωx+φ) (x∈R,ω＞0,0＜φ＜π/2)的部分图像如图所示

1 求函数f(x)的解析式
2 求函数g(x)=f(x-π/12)-f(x+π/12)的单调递增区间

Answer 1

1. T/2=π/ω=11π/12-5π/12=π/2，

   ∴ω=2.

   2*5π/12+φ=π，φ=π/6，

   Asin(π/6)=1,A=2,

   ∴f(x)=2sin(2x+π/6).

2. g(x)=f(x-π/12)-f(x+π/12)

   =2[sin2x-sin(2x+π/3)]

   =-√3cos2x,

   其增区间由2kπ<=2x<=(2k+1)π,k∈Z确定，

   各除以2，得kπ<=x<=(k+1/2)π,为所求.