1,f（x）=-√3sin^2x+sinxcosx=√3/2*(cos2x-1)+1/2sin2x=√3/2cos2x+1/2sin2x-√3/2=cos(2x-π/6)-√3/2,最小正周期 T=2π/2=π,2,x∈【0,π/2】,则2x-π/6∈【-π/6,5π/6】当2x-π/6=0时,f（x）取最大值1-√3/2,...帮助了你满意请采纳。