7.78 If $a>0, a \neq 1, \quad F(x)$ is an odd function, then the parity of the following function
$$
G(x)=F(x) \cdot\left[\frac{1}{a^{x}-1}+\frac{1}{2}\right]
$$

is
(A) Odd function.
(B) Even function.
(C) Neither odd nor even function.
(D) The parity depends on the specific value of $a$.
(China High School Mathematics League, 1984)
[Solution] Since $F(x)$ is an odd function, $\therefore F(-x)=-F(x)$. Therefore,
$$
\begin{aligned}
G(-x) & =F(-x)\left[\frac{1}{a^{-x}-1}+\frac{1}{2}\right]=-F(x)\left[\frac{1}{\frac{1}{a^{x}}-1}+\frac{1}{2}\right] \\
& =-F(x)\left[\frac{a^{x}}{1-a^{x}}+\frac{1}{2}\right]=-F(x)\left[\frac{2 a^{x}+1-a^{x}}{2\left(1-a^{x}\right)}\right] \\
& =-F(x)\left[\frac{a^{x}+1}{2\left(1-a^{x}\right)}\right]=F(x)\left[\frac{a^{x}-1+2}{2\left(a^{x}-1\right)}\right] \\
& =F(x)\left(\frac{1}{2}+\frac{1}{a^{x}-1}\right)=G(x) .
\end{aligned}
$$

Therefore, the correct choice is $(B)$.