已知函数f(x)=log a (mx-1)/(1-x) (a>0,a≠1,m≠1)是奇函数.(1)求实数m的值;(2)当a>1时,判断函数f(x)在(1,+无穷)上的单调性,并给出证明.

问题描述:

已知函数f(x)=log a (mx-1)/(1-x) (a>0,a≠1,m≠1)是奇函数.
(1)求实数m的值;
(2)当a>1时,判断函数f(x)在(1,+无穷)上的单调性,并给出证明.

(1)f(-x)=loga#[(-mx-1)/(1+x)]=-f(x)=loga#[(1-x)/(mx-1)]=loga#[(x-1)/(1-mx)],故m=-1
(2)f(x)=loga#[(x+1)/(x-1)],(x+1)/(x-1)>0,即x>1或x1时为减函数,证明如下:
当a>1时,设1=loga#[(x2+1)(x1-1)/(x1+1)(x2-1)],因[(x2+1)(x1-1)/(x1+1)(x2-1)]-1=2(x1-x2)/(x1+1)(x2-1)即(x2+1)(x1-1)/(x1+1)(x2-1)1时f(x)是减函数.