f(x)=√3sinxcosx+cos2x+1=(√3/2)sin2x+cos2x+1=[(√7)/2][(√3/√7)sin2x+(2/√7)cos2x]+1=[(√7)/2]sin(2x+α)+1其中,sinα=2/√7,cosα=√3/√7,α=arcsin(2/√7)可见,最小正周期为 T=π单调递减区间应满足π/2+...