JEE MAIN - Mathematics (2023 - 13th April Evening Shift - No. 13)
The range of $$f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$$ is
$$[0,2 \pi]$$
$$[0,2 \pi)$$
$$[0, \pi)$$
$$[0, \pi]$$
Explanation
$$
\begin{aligned}
& \frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}<1 \\\\
\therefore & 0 \leq \frac{x^2}{1+x^2}<1 \\\\
\Rightarrow & 0 \leq \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<\frac{\pi}{2} \\\\
\Rightarrow & 0 \leq 4 \sin ^{-1}\left(\frac{x^2}{1+x^2}\right)<2 \pi
\end{aligned}
$$
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