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1.Relation and Function
hard
જો $f\left( x \right) = {\log _e}\,\left( {\frac{{1 - x}}{{1 + x}}} \right)$, $\left| x \right| < 1$, તો $f\left( {\frac{{2x}}{{1 + {x^2}}}} \right)$ મેળવો.
A
$2f\left( x \right)$
B
${\left( {f\left( x \right)} \right)^2}$
C
$2f\left( x^2 \right)$
D
$ - 2f\left( x \right)$
(JEE MAIN-2019)
Solution
$f\left( x \right) = {\log _e}\left( {\frac{{1 – x}}{{1 + x}}} \right),\left| x \right| < 1$
$f\left( {\frac{{2x}}{{1 + {x^2}}}} \right) = \ell n\left( {\frac{{1 – \frac{{2x}}{{1 + 2{x^2}}}}}{{1 + \frac{{2x}}{{1 + {x^2}}}}}} \right)$
$ = \ell n\left( {\frac{{{{\left( {x – 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}}} \right) = 2\ell n\left| {\frac{{1 – x}}{{1 + x}}} \right| = 2f\left( x \right)$
Standard 12
Mathematics
