if f(x) = loge

Question:

If $f(x)=\log _{e}\left(\frac{1-x}{1+x}\right),|x|<1$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ is equal to:

 

  1. (1) $2 f(x)$

  2. (2) $2 f\left(x^{2}\right)$

  3. (3) $(f(x))^{2}$

  4. (4) $-2 f(x)$


Correct Option: 1

Solution:

$f(x)=\log \left(\frac{1-x}{1+x}\right),|x|<1$

$f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1-\frac{2 x}{1+x^{2}}}{1+\frac{2 x}{1+x^{2}}}\right)$

$=\log \left(\frac{1+x^{2}-2 x}{1+x^{2}+2 x}\right)$

$=\log \left(\frac{1-x}{1+x}\right)^{2}$

$=2 \log \left(\frac{1-x}{1+x}\right)$

$=2 f(x)$

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