9. Given the function $f(x)=\left(\frac{1}{a^{x}-1}+\frac{1}{2}\right) x^{2}+b x+6(a, b$ are constants, $a>1)$, and $f\left(\lg \log _{8} 1000\right)=8$, then the value of $f(\lg \lg 2)$ is $(\quad)$.
(A) 8
(B) 4
(C) -4
(D) -8
(9) $\mathrm{B}$ Hint: From the given information, we have
$$
f\left(\lg \log _{8} 1000\right)=f\left(\lg \frac{3}{3 \lg 2}\right)=f(-\lg \lg 2)=8 .
$$

Also,
$$
\frac{1}{a^{-x}-1}+\frac{1}{2}=\frac{a^{x}}{1-a^{x}}+\frac{1}{2}=-1+\frac{1}{1-a^{x}}+\frac{1}{2}=-\frac{1}{a^{x}-1}-\frac{1}{2} \text {. }
$$

Let $F(x)=f(x)-6$, then $F(-x)=-F(x)$. Therefore,
$$
f(-\lg \lg 2)=F(-\lg \lg 2)+6=-F(\lg \lg 2)+6=8 .
$$

This implies
$$
F(\lg \lg 2)=-2, f(\lg \lg 2)=F(\lg \lg 2)+6=4 .
$$