1
向量a=(cosωx,根號(hào)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)的周期為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]
∴當(dāng)2x+π/3=7π/6時(shí),f(x)min=-1/2
當(dāng)2x+π/3=π/2時(shí),f(x)max=1
∴f(x)的值域?yàn)閇-1/2,1]
3
3[f（x）]²+mf（x）-1=0
在[-π/12,5π/12]上有三個(gè)不相等的實(shí)數(shù)根
令t=f(x)=sin(2x+π/3)
2x+π/3∈[π/6,π/2)U(π/2,5π/6]時(shí),
x與t的關(guān)系為2對(duì)1
2x+π/3∈（5π/6,7π/6]U{π/2}時(shí).
x與t的關(guān)系為1對(duì)1
則3t²+mt-1=0的實(shí)數(shù)根t1,t2滿足：
t1∈[-1/2,1/2)U{1} ,t2∈[1/2,1)
當(dāng)t1=1時(shí),t2=-1/3不合題意
t1∈[-1/2,1/2),t2∈[1/2,1)
考察函數(shù)g(t)=3t²+mt-1
則 {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