已知向量a=(5√3cosx,cosx),b=(sinx,2cosx),
则a*b=5√3sinxcosx+2(cosx)^2=(5√3/2)sin2x+cos2x+1
IbI^2=(sinx)^2+4(cosx)^2=1+3(cos2x)^2=(3/2)cos2x+5/2
所以f(x)=(5√3/2)sin2x+(5/2)cos2x+7/2
=5sin(2x+π/6)+7/2
当π/12≤x≤π/3时,π/3≤2x+π/6≤5π/6
f(x)=a+7/2=5sin(2x+π/6)+7/2
sin(2x+π/6)=a/5有两个不相等的实数根
则π/3≤2x+π/6≤π-π/3=2π/3,且2x+π/6≠π/2
即√3/2≤a/5