(1)设y=k(a-x)x,当x=a/2时,y=a^2,解得k=4所以y=4(a-x)x所以定义域为[0,2at/(1+2t)],t为常数且属于[0,1] (2)把y配方,则y=-4(x-a/2)^2+a^2当2at/(1+2t)>=a/2时,0.5<=t<=1,当x=a/2时,y=a^2当2at/(1+2t)当x=2at/(1+2t)时,y=8at^2/(1+2t)^2 接下来综上就可以了