I think this function equation is quite difficult to play in senior one mathematics.

(1)∫ function y = (a2x-1-a)/(2x-1) is odd function.

∴f(-x)=-f(x)

∴[a2^(-x)-(a+ 1)]/[2^(-x)- 1]=-(a2^x-a- 1)/(2^x- 1)

The left formula is multiplied by 2x: [a-(a+1) 2x]/(1-2x) = (a2x-a-1)/(1-2x).

The corresponding coefficients are equal:-(a+1) = a.

∴a=- 1/2

(2)f(x)=[- 1/2×2^x- 1/2]/(2^x- 1)

f(x)=( 1+2^x)/[2( 1-2^x)],2^x- 1≠0,2^x≠ 1，x≠ 0

The domain of ∴ function is (-∞, 0)∞(0,+∞).

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

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

Let y = (-1/2)-[1/(2x-1)].

2 x = (2y+1)/(2y-1) > 0, then the solution is:

Y 1/2。

Therefore, the range is (-infinity,-1/2)U( 1/2,+infinity).