解：f(x)=√3sin2x+cos2x+a+1 =2[sin2x*(√3/2)+cos2x*(1/2)]+a+1 =2[sin2x*cos(π/6)+cos2x*sin(π/6)]+a+1 =2sin(2x+π/6)+a+1正弦函数的值域是[-1,1]所以 f(x)的最大值为 2+a+1=3+a=2所以 a=-1
