解∵f(x)=2x^2+2tx+2t∴f'(x)=4x+2t令f'(x)=0,得x=-t/2∴g(t)=2(-t/2)^2+2t(-t/2)+2t =t^2/2-t^2+2t =-t^2/2+2t =-(t-2)^2/2+2当t=2时g(t)取得最大值2∴选C