f(x)=√3sin2x+2cos²x+m
= √3sin2x+cos2x+1+m
=2[sin2xcos(π/6)+cos2xsin(π/6)]+1+m
=2sin(2x+π/6)+1+m
∵ 0 ≤x≤π/2
∴ π/3≤2x+π/6≤7π/6
所以 当2x+π/6=π/2时,有最大值
2+1+m=6
所以 m=3
f(x)=2sin(2x+π/6)+4