2. The functions $y=f(x)$ and $y=g(x)$ have the same domain. For any $x$ in the domain, $f(x)+f(-x)=0$, $g(x) g(-x)=1$, and when $x \neq 0$, $g(x) \neq 1$. Then $F(x)=\frac{2 f(x)}{g(x)-1}+f(x)$ is $(\quad)$.
(A) odd function
(B) even function
(C) both odd and even function
(D) neither odd nor even function
2. (B).

From the given, when $x \neq 0$,
$$
\begin{array}{l}
F(-x)=\frac{2 f(-x)}{g(-x)-1}+f(-x) \\
=\frac{-2 g(x) f(x)}{1-g(x)}-f(x) \\
=\frac{2 f(x)+2 f(x)[g(x)-1]}{g(x)-1}-f(x) \\
=\frac{2 f(x)}{g(x)-1}+f(x)=F(x) .
\end{array}
$$

Therefore, $F(x)=\frac{2 f(x)}{g(x)-1}+f(x)$ is an even function.