f(x)=sin²x+cos²x+2sinxcosx+2cos²x
=1+sin2x+2cos²x-1+1
=sin2x+cos2x+2
=(√2)(sin2xcosπ/4+cos2xsinπ/4)+2
=(√2)sin(2x+π/4)+2
x∈[0,π/2],所以2x+π/4∈[π/4,5π/4]
所以-√2/2≤sin(2x+π/4)≤1
f(x)最大值为2+√2
f(x)最小值为1