g(x)=f(x)-x
x^2+(a-1)x+a=0
兩個(gè)根都在0和1之間
則必須同時(shí)滿足
(1)判別式大于0
(2)g(0)>0,g(1)>0
(3)g(x)對(duì)稱軸在(0,1)內(nèi)
(1)判別式大于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)對(duì)稱軸在(0,1)內(nèi)
對(duì)稱軸x=-(a-1)/2
0<-(a-1)/2<1
-2-1綜上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