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