f(x)=√3/2sin2x+1/2(2cos²x-1)+1/2
=√3/2sin2x+1/2cos2x+1/2
=sin(2x+π/6)+1/2
所以最小正周期为：
T=2π/2=π
∵x∈［0,π/2］
∴2x+π/6∈[π/6,7π/6]
∴sin(x+π/6)∈[-1/2,1]
∴f(x)∈[0,3/2]
即值域为[0,3/2]