f(x)=log(1/2)^[(1-ax)/(x-1)]的图像关于原点对称,a为常数 ,有
f(-x)=-f(x)=log(1/2){[1-a(-x)]/[(-x)-1)]=log(1/2)[-(1+ax)/(x+1)].
∴f(x)=-log(1/2)[-(1+ax)/(x+1)]=log(1/2)[-(x+1)/(1+ax)].
而,f(x)=log(1/2)[(1-ax)/(x-1)],则有
-(x+1)/(1+ax)=(1-ax)/(x-1),
即(X+1)/(ax+1)=(1-ax)/(1-x).
x+1=1-ax,ax+1=1-x,
a=-1.
2).f(x)=log(1/2)[(1+x)/(x-1)].
当x大于1时,
f(x)+log(x-1)=log(1/2)[(1+x)/(x-1)]+log(1/2)(x-1)=log(1/2)(1+x)(1/2)^m,
x>(1/2)^m-1,
而,x>1,有(1/2)^m-1>1,
(1/2)^m>2,
2^(-m)>2,
即有-m>1,
m