Q41 / 160 · TS EAMCET 2019 (04 May Shift 2) · Mathematics
The domain of the function $f(x)=\frac{1}{\sqrt{[x]^2-[x]-2}}$ is
Here $[x]$ denotes the greatest integer not exceeding the value of $[x]$
- A$(-\infty,-2) \cup(1, \infty)$
- B$(-\infty,-2) \cup(0, \infty)$
- C$(-\infty,-2) \cup(2, \infty)$
- Correct answerD$(-\infty,-1) \cup(3, \infty)$
Solution
We have,
$$
f(x)=\frac{1}{\sqrt{[x]^2-[x]-2}}
$$
$f(x)$ is defined when
$$
\begin{aligned}
& {[x]^2-[x]-2>0} \\
& ([x]-2)([x]+1)>0 \\
& {[x]>2 \text { and }[x] < -1} \\
& x \geq 3 \text { and } x < -1
\end{aligned}
$$
$\therefore$ Domain of $f(x)$ is $x \in(-\infty,-1) \cup(3, \infty)$
Hence, the correct answer is option: 'D'.