1
a·b=√3sinxcosx+cosx^2=√3sin(2x)/2+(1+cos(2x))/2
=sin(2x+π/6)+1/2
故：f(x)=a·b-1/2=sin(2x+π/6)
最小正周期：T=2π/2=π
2
f(α)=sin(2α+π/6)=4/5
π/6≤α≤5π/12,即：π/2≤2α+π/6≤π
故：cos(2α+π/6)=-3/5
sin(2α)=sin(2α+π/6-π/6)
=sin(2α+π/6)cos(π/6)-cos(2α+π/6)sin(π/6)
=(4/5)(√3/2)-(-3/5)(1/2)=(3+4√3)/10
3
f(x)图像向右平移π/6个单位,得到：sin(2x-π/6)
即：g(x)=sin(2x-π/6)
x∈[0,π/2],即：2x-π/6∈[-π/6,5π/6]
g(x)=k在x∈[0,π/2]上有一个实根
即：sin(2x-π/6)=k在x∈[0,π/2]上有一个实根
sin(2x-π/6)∈[-1/2,1]
k=1时,只有一个实根
-1/2≤k