QJEEM21Ma1799: JEE Main 2021 Mathematics - Class 12 JEEM21Ma1799 Solution: Let , 0 < x < 1. Then :

Question

Verified Solution • 2025–26 Session moderate

Let

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)

, 0 < x < 1. Then :

Accepted Answer

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)

{\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x

or

f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )

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

f(x) = {{1 - x} \over {1 + x}}

Now,

f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}

or

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

or

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

.

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