Question

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. $(-\infty,-2) \cup(2, \infty)$
• D. $(-\infty,-1) \cup(3, \infty)$

Answer 1

Answer: (D)

We have,
$f(x)=\frac{1}{\sqrt{[x]^{2}-[x]-2}}$
$f(x)$ is defined when
${[x]^{2}-[x]-2 > 0} $
$([x]-2)([x]+1) > 0$
${[x]>2} $ and ${[x]<-1} $
$x \geq 3 $ and $ x < -1$
$\therefore $ Domain of $f(x)$ is $x \in(-\infty,-1) \cup(3, \infty)$