Modulus got into Floor

We are given that $\lfloor |x| \rfloor \leq 2$ . Using the definition of floor function , we get

$|x|<3.$

And now the above absolute value inequality gives us $-3<x<3.$

Writing in the interval notation, we get $x \in (-3,3). \square$

A simple solution.

$We\quad are\quad given\\ \left\lfloor \left| x \right| \right\rfloor \quad \le \quad 2\\ First\quad we\quad break\quad it\quad up\quad from\quad the\quad inside.\\ we\quad know\quad that\quad for\quad that\quad \left| x \right| \quad \le \quad 2\\ x\quad \in \quad (-2,2)\\ Notice\quad the\quad open\quad brackets,\quad therefore\quad it\quad implies\\ that\quad -2\quad <\quad x\quad <\quad 2\quad .\\ But\quad due\quad to\quad the\quad floor\quad function,\quad the\quad value\quad will\quad \\ jump\quad down\quad to\quad the\quad lower\quad integer.\quad Therefore\quad to\\ avoid\quad that\quad we\quad go\quad one\quad integer\quad higher.\quad i.e\quad -3,3.\\ \quad \therefore \boxed { \quad x\quad \in \quad (-3,3)\quad } \quad \bigstar$