Question

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

• A. $2f(x)$
• B. $2f(x^2)$
• C. $(f(x))^2$
• D. $-2 f(x)$

Answer 1

Answer: (A)

$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+2x^{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)$