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TS EAMCET · Maths · Functions

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 \([x]\)

  1. A \((-\infty,-2) \cup(1, \infty)\)
  2. B \((-\infty,-2) \cup(0, \infty)\)
  3. C \((-\infty,-2) \cup(2, \infty)\)
  4. D \((-\infty,-1) \cup(3, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((-\infty,-1) \cup(3, \infty)\)

Step-by-step Solution

Detailed explanation

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…