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1.Relation and Function
medium
The range of $f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$ is
A
$[0, \pi]$
B
$[0,2 \pi)$
C
$[0, \pi)$
D
$[0,2 \pi]$
(JEE MAIN-2023)
Solution
$f(x)=4 \sin ^{-1}\left(\frac{x^2}{x^2+1}\right)$
$\frac{x^2+1-1}{x^2+1}=1-\frac{1}{x^2+1} \Rightarrow[0,1)$
Range of $f(x)=[0,2 \pi)$
Standard 12
Mathematics