Let [f(x) = ( 2 ^ - 1 ( ^ - 1 1 - x x ) )], 0

Question

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

Solutions · 1