$9 \cdot 11$ If $\lg ^{2} x \lg 10 x<0$, then the value of $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x^{2}}$ is
(A) $\pm 1$.
(B) 1 .
(C) -1 .
(D) An answer different from (A)-(C).
(China Junior High School Mathematics Correspondence Competition;
China Guangzhou, Wuhan, Fuzhou and other five cities Junior High School Mathematics Competition, 1984)
[Solution] From $\lg ^{2} x \lg 10 x<0$, we know $\lg 10 x<0$, which means
$$
1+\lg x<0 \text {. }
$$

And $\lg ^{2} x+\lg 10 x^{2}=\lg ^{2} x+2 \lg x+1=(\lg x+1)^{2}$,
from $1+\lg x<0$, then $\sqrt{\lg ^{2} x+\lg 10 x^{2}}=-(1+\lg x)$.
Also, $\lg 10 x=1+\lg x$, so $\frac{1}{\lg 10 x} \sqrt{\lg ^{2} x+\lg 10 x}=-1$. Therefore, the answer is $(C)$.