Question

The domain of $f\left(x\right)=\sqrt{[x{]}^2-[x]-6}$ is

• A. $(-\infty ,-1)\cup [3,\infty )$
• B. $[-3,-2)$
• C. $(-2,3)$
• D. $(-\infty ,\infty )$

Answer 1

The correct option is A $(-\infty ,-1)\cup [3,\infty )$

$f$ is defined if term under root is positive.
$\Rightarrow [x{]}^2-[x]-6\ge 0$
$\left(\right[x]+2)\left(\right[x]-3)\ge 0$
$\Rightarrow \left[x\right]\ge 3$ or $\left[x\right]\le -2$
$\Rightarrow x\in (-\infty ,-1)\cup [3,\infty )$