Question

Please help me to solve this...... complete steps...

11th standard

Answer 1

Hi...☺

Here is your answer...✌

GIVEN THAT,

$f(x) = log \frac{1 + x}{1 - x} \:\:\: \:\:\: .....(1)$
TO PROVE :

$f( \frac{2x}{1 + {x}^{2} } ) = 2f(x)$

PROOF :

LHS

$f( \frac{2x}{1 + {x}^{2} }) = log \: \frac{1 + \frac{2x}{1 + {x}^{2} } }{1 - \frac{2x}{1 + {x}^{2} } } \\ \\ = log \: \frac{ \frac{1 + {x}^{2} + 2x }{1 + {x}^{2} }}{ \frac{1 + {x}^{2} - 2x}{1 + {x}^{2} } } \\ \\ = log \: \frac{1 + {x}^{2} + 2x }{1 + {x}^{2} - 2x } \\ \\ = log \: \frac{{(1 + x )}^{2} }{{(1 - x)}^{2} } \\ \\ = log \: {( \frac{1 + {x}}{1 - x} ) }^{2} \\ \\ = 2 \: log( \frac{1 + x}{1 - x} ) \: \: \: \: \: \: \: \: \: [ \because log( {a})^{b} = b log(a) ] \\ \\ = 2f(x) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [from \: (1)]$

= RHS [ Proved ]