Functions 2 Question 3
4.
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)
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Answer:
Correct Answer: 4. (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}$
