1
向量a=(cosωx,根号三cosωx),b=(sinωx,cosωx)
∴f(x)=a●b-√3/2
=sinwxcoswx+√3cos²wx-√3/2
=1/2sin2wx+√3/2(1+cos2wx)-√3/2
=1/2sin2wx+√3/2cos2wx
=sin(2wx+π/3)
∵f（x+π）=f(x）
∴f(x)的周期为π
∴2π/(2w)=π ∴w=1
∴f(x)=sin(2x+π/3)
2
∵x∈[-π/12,5π/12]
∴2x∈[-π/6,5π/6]
∴2x+π/3∈[π/6,7π/6]
∴当2x+π/3=7π/6时,f(x)min=-1/2
当2x+π/3=π/2时,f(x)max=1
∴f(x)的值域为[-1/2,1]
3
3[f（x）]²+mf（x）-1=0
在[-π/12,5π/12]上有三个不相等的实数根
令t=f(x)=sin(2x+π/3)
2x+π/3∈[π/6,π/2)U(π/2,5π/6]时,
x与t的关系为2对1
2x+π/3∈（5π/6,7π/6]U{π/2}时.
x与t的关系为1对1
则3t²+mt-1=0的实数根t1,t2满足：
t1∈[-1/2,1/2)U{1} ,t2∈[1/2,1)
当t1=1时,t2=-1/3不合题意
t1∈[-1/2,1/2),t2∈[1/2,1)
考察函数g(t)=3t²+mt-1
则 {g(-1/2)=3/8-m/2-1≥0 ==>m≤-5/4
{g(1/2)=3/8+m/2-1≤0 ==> m≤5/4
{g(1)=2+t>0 ==>m>-2
取交集得 -2