12.B因为f(x)=a+e-xhna,所以f(r)=alna+e-ha=(a2-1)lna+e当a>1时,对任意的r∈[0.1],a-1≥0.ha>0,恒有f(x)>0:当00,所以f(x)在x∈[0,1是单调递增的那么对任意的x1,x2∈[0,1,不等式f(x2)-f(x1)1≤a-2恒成立,只要(x)=-f(x)m1≤a-2,f(x)m=f(1)=a+e-lna,f(x)m=f(0)=1+1=2.所以a-2a+lna-2,即a≥e