1. (2000 National High School Competition Question) Let the universal set be the set of real numbers. If $A=\{x \mid \sqrt{x-2} \leqslant 0\}, B=\{x \mid$ $\left.10^{x^{2}-2}=10^{x}\right\}$, then $A \cap \bar{B}$ is ( ).
A. $\{2\}$
B. $\{-1\}$
C. $\{x \mid x \leqslant 2\}$
D. $\varnothing$
1. D. Since $\sqrt{x-2} \leqslant 0$, and $\sqrt{x-2} \geqslant 0$, it can only be that $\sqrt{x-2}=0$, i.e., $x=2$. Thus, we get $A=\{2\}$. From $10^{x^{2}-2}=$ $10^{x}$, we get $x^{2}-2=x$, solving this, we get $x=-1, x=2$. $\bar{B}=\{x \mid x \in \mathbf{R}, x \neq-1, x \neq 2\}$, so $A \cap \bar{B}=\varnothing$, hence the answer is D.