Let f ( x ) = (2 ^ - 1 ( ^ - 1 (1 - x)/(x) )), 0 < x < 1 . Then:

Put $x = \sin^{2}\theta, 0 < x < 1$

$sin\theta = \sqrt{x}$

$⇒ f\left(x\right) = \cos \left(2\tan^{- 1}\sin \left(\cot^{- 1}\sqrt{\dfrac{1 - \sin^{2}\theta}{\sin^{2}\theta}}\right)\right)$

$⇒ f\left(x\right) = \cos \left(2\tan^{- 1} ( sin\theta )\right)$

$⇒ f\left(x\right) = \cos \left(2\tan^{- 1}\sqrt{x}\right)$

$= \dfrac{1 - \tan^{2}\left(\tan^{- 1}\sqrt{x}\right)}{1 + \tan^{2}\left(\tan^{- 1}\sqrt{x}\right)}$

$⇒ f\left(x\right) = \dfrac{1 - x}{1 + x}$

$⇒ {f}^{'}\left(x\right) = \dfrac{\left(1 + x\right)\left(- 1\right) - \left(1 - x\right) \cdot 1}{{\left(1 + x\right)}^{2}}$

$⇒ {f}^{'}\left(x\right) = \dfrac{- 2}{{\left(1 + x\right)}^{2}}$

Multiply, ${\left(1 - x\right)}^{2}$ on both sides

$⇒ {\left(1 - x\right)}^{2}{f}^{'}\left(x\right) = \dfrac{- 2{\left(1 - x\right)}^{2}}{{\left(1 + x\right)}^{2}}$

Now, $2{\left(f\left(x\right)\right)}^{2} = \dfrac{2{\left(1 - x\right)}^{2}}{{\left(1 + x\right)}^{2}}$

$∴ {\left(1 - x\right)}^{2}{f}^{'}\left(x\right) + 2{\left(f\left(x\right)\right)}^{2} = 0$ option \left(1\right) satisfied