设f(x)=log1/21−axx−1为奇函数,则a=_.
问题描述:
设f(x)=log
1 2
为奇函数,则a=______. 1−ax x−1
答
∵f(x)=log121−axx−1为奇函数,∴f(-x)=-f(x),即f(x)+f(-x)=0,则log121−axx−1+log121+ax−x−1=log12(1−axx−1•1+ax−x−1)=0,即a2x2−1x2−1=1,即a2x2-1=x2-1,即a2=1,解得a=1或a=-1,当a=1...