The range of f ( x ) = 4 sin - 1 ( x 2 x 2 + 1 ) is [2023]

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Solution

Q.
The range of $f (x) = 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$ is [2023]

1 $[0, 2 π)$

2 $[0, π]$

3 $[0, 2 π]$

4 $[0, π)$

Ans.
(1)

$Given, 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1})$

$\frac{x^{2}}{x^{2} + 1} = \frac{x^{2} + 1 - 1}{x^{2} + 1} = 1 - \frac{1}{x^{2} + 1}$

$x^{2} \geq 0; x^{2} + 1 \geq 1$ $∴$ $0 < \frac{1}{x^{2} + 1} \leq 1 \Rightarrow - 1 \leq - \frac{1}{x^{2} + 1} < 0;$

$0 \leq 1 - \frac{1}{x^{2} + 1} < 1; 0 \leq 4 \sin^{- 1} (\frac{x^{2}}{x^{2} + 1}) < 2 π$

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