g(x)=f(x)-x
x^2+(a-1)x+a=0
两个根都在0和1之间
则必须同时满足
(1)判别式大于0
(2)g(0)>0,g(1)>0
(3)g(x)对称轴在(0,1)内
(1)判别式大于0
(a-1)^2-4a>0
a^2-6a+1>0
a>3+2√2,a<3-2√2
(2)g(0)>0,g(1)>0
g(0)=a>0
g(1)1+a-1+a>0
a>0
(3)g(x)对称轴在(0,1)内
对称轴x=-(a-1)/2
0<-(a-1)/2<1
-2-1综上0f(0)=a,f(1)=2a+1
f(0)f(1)-f(0)=a(2a+1)-a=2a^2
又01/16=0.0625
故f(0)f(1)-f(0)<1/16