Functions Ques 47
If $f(x)=\log _e (\frac{1-x}{1+x}),|x|<1$, then $f (\frac{2 x}{1+x^{2}})$ is equal to
(a) $2 f(x)$
(c) $(f(x))^{2}$
(b) $2 f\left(x^{2}\right)$
(d) $-2 f(x)$
(2019 Main, 8 April I)
Show Answer
Answer:
Correct Answer: 47.(a)
Solution:
- Given, $f(x)=\log _e (\frac{1-x}{1+x}),|x|<1$, then
$ f (\frac{2 x}{1+x^{2}})=\log _e (\frac{1-\frac{2 x}{1+x^{2}}}{1+\frac{2 x}{1+x^{2}}}) \quad[\because |\frac{2 x}{1+x^{2}}|<1 ]$
$= \log _e (\frac{\frac{1+x^{2}-2 x}{1+x^{2}}}{\frac{1+x^{2}+2 x}{1+x^{2}}})=\log _e( \frac{(1-x)^{2}}{(1+x)^{2}})=\log _e (\frac{1-x^{2}}{1+x}) $
$= 2 \log _e (\frac{1-x}{1+x}) $ $\quad {\left[\because \log _e|A|^{m}=m \log _e|A|\right]} $
$=2 f(x) $ $ \quad [\because f(x)=\log _e (\frac{1-x}{1+x})]$
BETA
