5. If $a>0, a \neq 1, F(x)$ is an odd function, then $G(x)=F(x)\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right)$ is
(A) an odd function
(B) an even function
(C) neither an odd function nor an even function
(D) the parity depends on the specific value of $a$
(B)
5.【Analysis and Solution】 $\because F(x)$ is an odd function, $\therefore F(-x)=-F(x)$.
Let $g(x)=\frac{1}{a^{x}-1}+\frac{1}{2}=\frac{a^{x}+1}{2\left(a^{x}-1\right)}$,
then $g(-x)=\frac{a^{-x}+1}{2\left(a^{-x}-1\right)}=\frac{\frac{1}{a^{x}}+1}{2\left(\frac{1}{a^{x}}-1\right)}=\frac{a^{x}+1}{2\left(1-a^{x}\right)}=-g(x)$, i.e., $g(x)$ is also an odd function.
When $a>0, a \neq 1$, $G(x)=F(x) g(x)$ is defined on $\mathbf{R}$.
And $G(-x)=F(-x) g(-x)=[-F(x)][-g(x)]=F(x) g(x)$,
$\therefore G(x)$ is an even function.