f(x)=lnx+(1-x)/ax
=lnx+1/ax-1/a求导
f'(x)=1/x-1/(ax^2),当f'(x)=0,即x=1/a时,函数f(x)有极值
所以当1≤1/a≤e时,即1/e≤a≤1时,minf(x)=f(1/a)=1-1/a-lna
当a＜1/e时,minf(x)=f(e)=(ae-e+1)/ae
当a>1时,minf(x)=f(1)=0