1. The equation $[x]+\left[\frac{1}{x}\right]=3$ has ( ) solutions.
A. 2
B. 4
C. 8
D. infinitely many
1. D Obviously $x>0$, by symmetry, we may assume $x \geqslant 1$, then by $x \neq 1$ we get $x>1$, so $\left[\frac{1}{x}\right]=0,[x]+$ $\left[\frac{1}{x}\right]=3$, at this time $3 \leqslant x<4$. And when $0<x<1$, $\frac{1}{4}<x \leqslant \frac{1}{3}$, hence $3 \leqslant x<4, \frac{1}{4}<x \leqslant \frac{1}{3}$ are the solutions to the original equation.