f(x)=sinx^2+√3*sinxcosx
=(1-cos2x)/2+(√3/2)sin2x
=1/2-[(1/2)*cos2x-(√3/2)sin2x]
=1/2-[sinπ/6*cos2x-cosπ/6sin2x]
=1/2-sin(π/6-2x)
=1/2+sin(2x-π/6)
x∈[π/4,π/2]→2x∈[π/2,π]→2x-π/6∈[π/3,5π/6]→
当2x-π/6=π/2,有最大值(1/2)+1=3/2
当2x-π/6=5π/6,有最小值(1/2)+(1/2)=1