1=f(π/2)=m·1+0=m
∴所求的解析式为y=sinx+cosx
∵ sinx+cosx=√2[(√2/2)sinx+(√2/2)cosx]=√2[sinxcos(π/4)+cosxsin(π/4)]=√2sin(x+π/4)
所以y=f(x)的最小正周期为2π