1.Relation and Function
easy

यदि $f(x) = \log \left[ {\frac{{1 + x}}{{1 - x}}} \right]$, तब $f\left[ {\frac{{2x}}{{1 + {x^2}}}} \right]$ बराबर है

A

${[f(x)]^2}$

B

${[f(x)]^3}$

C

$2f(x)$

D

$3f(x)$

Solution

(c) $f(x) = \log \frac{{(1 + x)}}{{(1 – x)}}$

$\therefore \,\,\,f\left( {\frac{{2x}}{{1 + {x^2}}}} \right) = \log \,\left[ {\frac{{1 + \frac{{2x}}{{1 + {x^2}}}}}{{1 – \frac{{2x}}{{1 + {x^2}}}}}} \right] $

$= \log \,\left[ {\frac{{{x^2} + 1 + 2x}}{{{x^2} + 1 – 2x}}} \right]$

$ = \log \,{\left[ {\frac{{1 + x}}{{1 – x}}} \right]^2} = 2\,\log \,\left[ {\frac{{1 + x}}{{1 – x}}} \right] = 2\,f(x)$.

Standard 12
Mathematics

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