Question

设函数f(x)=log1/2(1-ax)/(x-1)为奇函数,a为常数。1、求a的值 2、证明f(x)在区间（1，正无穷）内单调递增

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

f(x)=log1/2(1+x)/(x-1)
(1+x)/(x-1)=(x-1+2)/(x-1)
=1+2/(x-1)
x>1时x-1递增
所以2/(x-1)递减
所以(1+x)/(x-1)是减函数
底数1/2<1
所以log1/2(x)是减函数
所以x>1时f(x)是增函数

Answer 2

令f(x)=-f(-x),
可得a=1 or a=-1
检验定义域,当a=1时定义域为空集
所以a=-1

f(x)=log1/2(1+x)/(x-1)=log2[(x-1)/(1+x)]
令1<x1<x2
f(x2)-f(x1)=log2[(x2-1)(x1+1)/(x2+1)(x1-1)]
=log2[(x1x2-x1+x2-1)/(x1x2+x1-x2-1)]
因为(x1x2-x1+x2-1)>(x1x2+x1-x2-1)
所以[(x1x2-x1+x2-1)/(x1x2+x1-x2-1)]>1
所以f(x2)-f(x1)>1
所以f(x)在区间（1，正无穷）内单调递增
证毕