Question

If $f\left(x_1\right)-f\left(x_2\right)=f\left(\frac{x_1-x_2}{1-x_1{x}_2}\right)$ for $x_1,x_2\in [-1,1]$ then $f\left(x\right)$ is

• A. $\log \left(\frac{1-x}{1+x}\right)$
• B. ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$
• C. $\log \left(\frac{1+x}{1-x}\right)$
• D. ${\tan }^{-1}\left(\frac{1+x}{1-x}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\log \left(\frac{1-x}{1+x}\right)$
B ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)$
C $\log \left(\frac{1+x}{1-x}\right)$

When $x_1=-1$ and $x_2=1$, then

$f(-1)-f\left(1\right)=f\left(\frac{-1-1}{1-1\left(1\right)}\right)=f(-1)\Rightarrow f\left(1\right)=0$

which is satisfied when $f\left(x\right)={\tan }^{-1}\left(\frac{1-x}{1+x}\right)$

when $x_1=x_2=0$, then

$f\left(0\right)-f\left(0\right)=f\left(\frac{0-0}{1-0}\right)=f\left(0\right)\Rightarrow f\left(0\right)=0$

when $x_1=-1$ and $x_2=0$, then

$f(-1)-f\left(0\right)=f\left(\frac{-1-0}{1-0}\right)=f(-1)\Rightarrow f\left(0\right)=0$

which is satisfied when $f\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$ and $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$