Question

F(x)=[1+2/(2^x-1)]*f(x)(x≠0)是偶函数,且f(x)不恒等于0,则f(x)是什么函数

Answer 1

F(-x)=[1+2/(2^-x-1)]f(-x)=F(x)=[1+2/(2^x-1)]*f(x)

1+2/(2^-x-1)上下乘2^x,且2^-x*2^x=2^0=1

所以1+2/(2^-x-1)=1+2*2^x/(1-2^x)

=(1-2^x+2*2^x)/(1-2^x)

=(2^x+1)/(1-2^x)

1+2/(2^x-1)

=(2^x-1+2)/(2^x-1)

=(2^x+1)/(2^x-1)

=-(2^x+1)/(1-2^x)

所以f(-x)=-f(x)

所以是奇函数