若函数f(x)=√3*sin2x+2cos²x+m在区间[0,π/2]上的最大值为6,求常数m的值及此函数当x

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