f(x)=m[√3sinxcosx+(cosx)^2]+1
=m[(√3/2)sin(2x)+(1/2)(1+cos(2x)] +1
=m[(√3/2)sin(2x)+(1/2)cos(2x)]+m/2 +1
=msin(2x+π/6) +m/2 +1
最小正周期Tmin=2π/2=π
π/6≤x≤π/3
π/2≤2x+π/6≤5π/6
1/2≤sin(2x+π/6)≤1
m>0时,当sin(2x+π/6)=1时,f(x)有最大值m+ m/2 +1=4,解得m=2
此时,当sin(2x+π/6)=1/2时,f(x)有最小值m/2+m/2+1=m+1=2+1=3
m0,舍去)
综上,得f(x)在此区间的最小值为3.