Question

设f(x)=log1/2((1-ax)/(x-1))为奇函数,a为常数,(1)求a的值(2)证明：函数f(x)在区间(1

Answer 1

所以

log(1/2)[(1+ax)/(-x-1)]+log(1/2)[(1-ax)/(x-1)]=0

即log(1/2)[(1+ax)/(-x-1)*(1-ax)/(x-1)]=0

所以(1+ax)/(-x-1)*(1-ax)/(x-1)=1

1-a^2x^2=1-x^2