f(x)=根号3sin^2+sinxcosx=根号3（1-sin2x/2)+sin2x/2=根号3/2-根号3/2cos2x+1/2sin2x=根号3/2+sin(2x-π/3）因为x∈[π/2,π],所以2x∈[π,2π],2x-π/3∈[2π/3,5π/3],sin(2x-π/3)∈[-1,根号3/2]所以f(x)∈[根号...