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) $\varnothing$
(D)
1. 【Analysis and Solution】From $\sqrt{x-2} \leqslant 0$ we get $x=2$. Note that $A$ contains only one element, so substituting $x=2$ into $B$, the equation holds true, hence $A \cap \bar{B}=\varnothing$. (Note: This approach makes the calculation simple even if $B$ is more complex).