已知函数f(x)=2+1/a-1/(a²x),设0

f(x)为增函数
f(m)=m
f(n)=n
2+1/a-1/(a^2m)=m
2+1/a-1/ (a^2n)=n
1/(2+1/a-m)=a^2m
2a^2m+am-a^2m^2=1
a^2m^2+2am(a+1/2)-1=0
(am+a+1/2)^2-(a+1/2)^2-1=0
mmin=-(a+1/2)^2-1
[an-(a+1/2)]^2+(a+1/2)^2-1=0
nmax=(a+1/2)^2-1
于是n-m最大=2(a+1/2)^2