设f(x)=log1/2(1-a
f(x)=log[(1-ax)/(x-1)]
=log[(x-1)/(1-ax)]
则,f(-x)=log[(-x-1)/(1+ax)]
已知f(x)为奇函数,所以:f(-x)=-f(x)
即:log[(-x-1)/(1+ax)]=-log[(x-1)/(1-ax)]
=log[(1-ax)/(x-1)]
===> (-x-1)/(1+ax)=(1-ax)/(x-1)
===> (1+ax)*(1-ax)=(-x-1)*(x-1)
===> 1-a^2*x^2=1-x^2
===> (1-a^2)*x^2=0
上式对于定义域内的x均成立
所以,1-a^2=0
所以,a=±1
而,当a=1时...全部
f(x)=log[(1-ax)/(x-1)]
=log[(x-1)/(1-ax)]
则,f(-x)=log[(-x-1)/(1+ax)]
已知f(x)为奇函数,所以:f(-x)=-f(x)
即:log[(-x-1)/(1+ax)]=-log[(x-1)/(1-ax)]
=log[(1-ax)/(x-1)]
===> (-x-1)/(1+ax)=(1-ax)/(x-1)
===> (1+ax)*(1-ax)=(-x-1)*(x-1)
===> 1-a^2*x^2=1-x^2
===> (1-a^2)*x^2=0
上式对于定义域内的x均成立
所以,1-a^2=0
所以,a=±1
而,当a=1时,f(x)=log[(1-x)/(x-1)]无意义,舍去
所以,a=-1。
收起