8.33 It is known that there is only one value of $x$ that satisfies the equation
$$
\left(1-\lg ^{2} a\right) x^{2}+(3+\lg a) x+2=0,
$$

then the real number $a$ is equal to
(A) $\frac{1}{\sqrt[3]{10}}$.
(B) 10 .
(C) $\frac{1}{10}$.
(D) None of the above.
(Shanghai Junior High School Mathematics Competition, 1985)
［Solution］From $\Delta=(3+\lg a)^{2}-8\left(1-\lg ^{2} a\right)=9 \lg ^{2} a+6 \lg a+1=0$, solving yields $\lg a=-\frac{1}{3}$, i.e., $a=10^{-\frac{1}{3}}$. At this point, $1-\lg ^{2} a \neq 0$ indicates that the equation has two equal real roots.
When $1-\lg ^{2} a=0$, i.e., $\lg a= \pm 1$ or $a=10$ and $\frac{1}{10}$, the original equation becomes $4 x+2=0$ or $2 x+2=0$, at which point the original equation has only one real root in each case.
Therefore, the answer is $(D)$.