9. Given the function
$$
f(x)=\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right) x^{2}+b x+6,
$$

where $a$ and $b$ are constants, $a>1$, and
$$
f\left(\lg \log _{8} 1000\right)=8.
$$

Then the value of $f(\lg \lg 2)$ is ( ).
(A) 8
(B) 4
(C) -4
(D) -8
9. B.

From the given information,
$$
\begin{array}{l}
f\left(\lg \log _{8} 1000\right)=f\left(\lg \log _{2} 10\right) \\
=f(-\lg \lg 2)=8 . \\
\text { Also, } \frac{1}{a^{-x}-1}+\frac{1}{2}=\frac{a^{x}}{1-a^{x}}+\frac{1}{2} \\
=-1+\frac{1}{1-a^{x}}+\frac{1}{2} \\
=-\frac{1}{a^{x}-1}-\frac{1}{2},
\end{array}
$$

Let $F(x)=f(x)-6$. Clearly,
$$
\begin{array}{l}
F(-x)=-F(x) \\
\Rightarrow f(-x)=12-f(x) .
\end{array}
$$

Therefore, $f(\lg \lg 2)=12-f(-\lg \lg 2)=4$.