MCQ
The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is
  • A
    $[0, \pi]$
  • $[0,2 \pi)$
  • C
    $[0, \pi)$
  • D
    $[0,2 \pi]$

Answer

Correct option: B.
$[0,2 \pi)$
b
$f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$

$\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1} \Rightarrow[0,1)$

Range of $f(x)=[0,2 \pi)$

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