5. If $a>0, a \neq 1, F(x)$ is an odd function, then $G(x)=F(x) \cdot\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right)$ is
A. Odd function
B. Even function
C. Neither odd nor even function
D. The odd or even nature depends on the specific value of $a$
Let $f(x)=\frac{1}{a^{x}-1}+\frac{1}{2} \Rightarrow f(-x)=\frac{1}{a^{-x}-1}+\frac{1}{2}=\frac{a^{x}}{1-a^{x}}+\frac{1}{2}$ $=-1+\frac{1}{1-a^{x}}+\frac{1}{2}=-\frac{1}{a^{x}-1}-\frac{1}{2}=-f(x) \Rightarrow f(x)$ is an odd function, thus $G(x)=F(x) \cdot f(x)$ is an even function, so the answer is $B$.