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

From $\sqrt{x-2} \leqslant 0$ we get $x=2$, so $A=\left\{2 \nmid\right.$. From $10^{x^{2}-2}=10^{x}$ we get $x^{2}-x-2=0$. Therefore, $B=\{-1,2\}$. Hence $A \cap \bar{B}=\phi$.