解(1)y=2x^2-2ax+3=2(x-a/2)^2+3-a^2/2对称轴为x=a/2讨论：当0<=a/2<=1即0<=a<=2时,g(a)=ymin=f(a/2)=3-a^2/2当a<0时,g(a)=f(0)=3当a>2时g(a)=f(1)=5-2a所以g(a)=3,a<03-a^2/2,0<=a<25-2a,a>2(2)容易知道当a<=0时g(a)max=3
