存在. a=b=1
Proof: f'(x)=x(2-x)=0 --> x1=0, x2=2
f(x) is maximized at x=1, with f(x)=1.
当x属于【1,1]时g(x)=f(x)且g(x)的值域为[1/1,1/1】,并a=b=1.