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JEE Mains · Maths · STD 12 - 1. relation and function

\(f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)\) નો વિસ્તાર \(......\)

  1. A \([0, \pi]\)
  2. B \([0,2 \pi)\)
  3. C \([0, \pi)\)
  4. D \([0,2 \pi]\)
Verified Solution

Answer & Solution

Correct Answer

(B) \([0,2 \pi)\)

Step-by-step Solution

Detailed explanation

\(f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)\) \(\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1} \Rightarrow[0,1)\) Range of \(f(x)=[0,2 \pi)\)
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