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

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