JEE Math Practice Question
Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :
- A.$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
- B.$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
- C.$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
- D.$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
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