JEE MAIN - Mathematics Hindi (2021 - 26th August Morning Shift - No. 5)
यदि $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1 हो, तो :
$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$
$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$
$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$
$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$
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