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
$$ \begin{array}{rr} & f \frac{2 x}{1+x^{2}}=\log _e \frac{1-\frac{2 x}{1+x^{2}}}{1+\frac{2 x}{1+x^{2}}} \quad \because \frac{2 x}{1+x^{2}}<1 \\ = & \log _e \frac{\frac{1+x^{2}-2 x}{1+x^{2}}}{\frac{1+x^{2}+2 x}{1+x^{2}}}=\log _e \frac{(1-x)^{2}}{(1+x)^{2}}=\log _e \frac{1-x^{2}}{1+x} \\ = & \log _e \frac{1-x}{1+x} \\ =2 f(x) & {\left[\because \log _e|A|^{m}=m \log _e|A|\right]} \\ \because f(x)=\log _e \frac{1-x}{1+x} \end{array} $$
