Question

If $f:R$ - {$\pm 1$} $\to R$ is defined by $f\left(x\right)=\log \left|\frac{1+x}{1-x}\right|$, then show that $f\left(\frac{2x}{1+x^2}\right)=2f\left(x\right)$

Answer 1

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

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

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

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

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

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

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

from (i) & (ii)

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