Let $f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$, 0 < x < 1. Then :

Question

${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$

${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$

${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$

${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$

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