不妨用换元法,因为定义域为[0,1],不妨设x=sin²t,t∈(0,π/2)
则y=sint+cost=根号2/2*sin(t+π/4),又t∈(0,π/2),则t+π/4∈(π/4,3π/4)
取t=π/4时,y取最大值 根号2/2