Question

The domain of the function, $f\left(x\right)=\frac{1}{\sqrt{[x{]}^2-[x]-6}}$ is

• A. $(-\infty ,-3]\cup [4,\infty )$
• B. $(\infty ,2)\cup [4,\infty )$
• C. $(\infty ,2)\cup (4,\infty )$
• D. none of these

Answer 1

The correct option is A $(-\infty ,-3]\cup [4,\infty )$

Given : $f\left(x\right)=\frac{1}{\sqrt{[x{]}^2-[x]-6}}$

$f\left(x\right)$ is real if $\sqrt{[x{]}^2-[x]-6}\in R$

i.e.

$[x{]}^2-[x]-6>0$

$⟹[x{]}^2-3[x]+2[x]-6>0$

$⟹\left[x\right]\left(\right[x]-3)+2\left(\right[x]-3)>0$

$⟹\left(\right[x]-3)\left(\right[x]+2)>0$
$⟹\left[x\right]<-2$ and $\left[x\right]>3$

Therefore, $x\in (-\infty ,-3]\cup [4,\infty )$