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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

The domain of the function \(f(x)=\sin ^{-1}\left(\frac{|x|+5}{x^{2}+1}\right)\) is \((-\infty,-\mathrm{a}] \cup[\mathrm{a}, \infty) .\) Then \(a\) is equal to 

  1. A \(\frac{1+\sqrt{17}}{2}\)
  2. B \(\frac{\sqrt{17}-1}{2}\)
  3. C \(\frac{\sqrt{17}}{2}+1\)
  4. D \(\frac{\sqrt{17}}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1+\sqrt{17}}{2}\)

Step-by-step Solution

Detailed explanation

\(f(x)=\sin \left(\frac{|x|+5}{x^{2}+1}\right)\) For domain : \(-1 \leq \frac{|x|+5}{x^{2}+1} \leq 1\) since \(|x|+5 \& x^{2}+1\) is always positive So \(\frac{|x|+5}{x^{2}+1} \geq 0 \forall x \in R\) So for domain : \(\frac{|x|+5}{x^{2}+1} \leq 1\)…
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