Question

The domain of the function $f\left(x\right) = \sqrt{\frac{4-x^{2}}{\left[x\right] + 2}}$ . where [.r] denotes the greatest integer not more than $x$. is

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

Answer 1

Answer: (C)

Given function $f(x)=\sqrt{\frac{4-x^{2}}{[x]+2}}$, is define if
$\frac{4-x^{2}}{[x]+2} \geq 0$
$\Rightarrow \frac{x^{2}-4}{[x]+2} \leq 0$
So, either $x^{2}-4 \leq 0$
and $[x]+2>0$ ...(i)
or $x^{2}-4 \geq 0$
and $\quad[x]+2<0$ ...(ii)
From Eq. (i),
$x \in[-2,2]$ and $x \in[-1, \infty)$
So, $x \in[-1,2]$ ...(iii)
From Eq. (ii).
$x \in(-\infty,-2] \cup[-2,-\infty]$ and $x \in(-\infty,-2)$
So, $x \in(-\infty,-2)$ ...(iv)
From intervals Eqs. (iii) and (iv),
$x \in(-\infty,-2) \cup[-1,2]$