已知函数f(x)= |lg(x-1)| 若a≠b,f(a)= f(b) ,则a+2b的取值范围是

函数f(x)= |lg(x-1)| 若a≠b,f(a)= f(b) ∴ |lg(a-1)|=|lg(b-1)|∴ lg(a-1)=lg(b-1)或lg(a-1)=-lg(b-1)即 a=b（舍）或lg(a-1)+lg(b-1)=0∴ lg[(a-1)(b-1)]=lg1即 (a-1)(b-1)=1∴ ab-a-b=0即 b=a/(a-1)由已知,则a>1∴...