$33 \cdot 32$ is less than or equal to $x$'s maximum integer and the minimum integer greater than or equal to $x$ sum to 5, then the solution set of $x$ is
(A) $\{5 / 2\}$.
(B) $\{x \mid 2 \leqslant x \leqslant 3\}$.
(C) $\{x \mid 2 \leqslant x<3\}$.
(D) $\{x \mid 2<x \leqslant 3\}$.
(E) $\{x \mid 2<x<3\}$.
(37th American High School Mathematics Examination, 1986)
[Solution] Let $[x]$ be the greatest integer not exceeding $x$, and $\{x\}$ be the smallest integer not less than $x$.
Obviously, $\quad[x] \leqslant x<[x]+1$,

and $\quad\{x\}-1<x \leqslant\{x\}$,
Adding the above two inequalities, we get $[x]+\{x\}-1<2 x<[x]+\{x\}+1$,
Given that $[x]+\{x\}=5$, then
$$
4<2 x<6 \text {, or } 2<x<3 \text {. }
$$

Therefore, the answer is $(E)$.