设函数f(x)=log1/2(1-ax/x-1)为奇函数,a是常数.求a的值?
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设函数f(x)=log1/2(1-ax/x-1)为奇函数,a是常数.求a的值?
![设函数f(x)=log1/2(1-ax/x-1)为奇函数,a是常数.求a的值?](/uploads/image/z/7594750-46-0.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dlog1%2F2%281-ax%2Fx-1%29%E4%B8%BA%E5%A5%87%E5%87%BD%E6%95%B0%2Ca%E6%98%AF%E5%B8%B8%E6%95%B0.%E6%B1%82a%E7%9A%84%E5%80%BC%3F)
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
(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
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