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AP EAMCET · Maths · Functions

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

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

Answer & Solution

Correct Answer

(C) \((-\infty,-2), \cup[-1,2]\)

Step-by-step Solution

Detailed explanation

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 \(x^2-4 \leq 0\) From intervals Eqs. (iii) and (iv), \[ x \in(-\infty,-2) \cup[-1,2] \]