Question

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

Answer 1

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