设f(x)=㏒1/2[(1-ax）/(x-1)]为奇函数,a为常数,证明：f(x)在区间（1,+∞）上为增函数

1)因为f(x)为奇函数,所以f(x)+f(-x)=0,
即(1-ax)/(x-1)*(1+ax)/(-x-1)=0.5^0=1,
所以1-a^2x^2=1-x^2,所以a^2=1.
当a=1时,(1-ax)/(x-1)=-1不合,所以a=-1；
2)f(x)=log0.5[(1+x)/(x-1)]
任取x1>x2>1
则f(x1)-f(x2)=log1/2[(x1+1)/(x1-1)]-log1/2[
(x2+1)/(x2-1)]
=log1/2[(x1+1)(x2-1)/(x2+1)(x1-1)]
=log1/2[(x1x2+x2-x1-1)/(x1x2-x2+x1-1)]
=log1/2{[x1x2-(x1-x2)-1]/[x1x2+(x1-x2)-1]}
由x1>x2>1
则x1-x2>0,x1x2>1
x1x2-1>0
则x1x2-(x1-x2)-10
则x1>x2>1时,f(x1)>f(x2)
则f(x)在区间（1,正无穷）内单调递增.

设f(x)=㏒1/2[(1-ax）/(x-1)]为奇函数，a为常数，证明：f(x)在区间（1，+∞）上为增函数