解： f(x) = √3sin2x + 1 + cos2x + m = 2sin(2x+π/6) + 1 + m x∈[0,π/2]时，2x+π/6∈[π/6,7π/6],sin(2x+π/6)∈[-1/2,1]∴ f(x)∈[m,3+m]∴ 3+m = 6 , m = 3 f(x)min = -2 + 1 + 3 = 2
