1. Let the universal set be the set of real numbers. If $A=\{x \mid \sqrt{x-2} \leqslant 0\}$, $B=\left\{x \mid 10^{x^{2}-2}=10^{x}\right\}$, then $A \cap \bar{B}$ is ( ).
(A) $\{2\}$
(B) $\{-1\}$
(C) $\{x \mid x \leqslant 2\}$
(D) $\varnothing$
$-、 1 .(\mathrm{D})$
From $\sqrt{x-2} \leqslant 0$ we get $x=2$, so $A=\{2\}$. From $10^{x^{2}-2}=$ $10^{x}$ we get $x^{2}-x-2=0$. Therefore, $B=\{-1,2\}$. Then $A \cap \bar{B}=\varnothing$.