设函数f(x)=log1/2(1-ax/x-1)为奇函数,a是常数.求a的值?

题目

设函数f(x)=log1/2(1-ax/x-1)为奇函数,a是常数.求a的值?

答案

f(-x)=log1/2(1+ax)/(-x-1)=-f(x)=-log1/2(1-ax)/(x-1)=log1/2(x-1)/(1-ax)
(1+ax)/(-x-1)=(x-1)/(1-ax)
1-x^2=1-a^2x^2
a^2=1
a=1或-1
若a=1
则f(x)=log1/2(1-x)/(x-1)=log1/2(-1)
无意义
所以a=-1