Question

Let the function $f\left(x\right)=\log \left(\frac{1-x}{1+x}\right)$, the value of $a$ and $b$ for which $f\left(a\right)+f\left(b\right)=f\left(\frac{a+b}{1+ab}\right)$ lies in

• A. $(0,1)$
• B. $(-1,1)$
• C. $(-1,0)$
• D. None of these

Answer 1

The correct option is B $(-1,1)$

$f\left(x\right)=\log (1-x)-\log (1+x)$

$|1-x|>0$

$1-x>0$

$x<1$ ...(i)

$1+x>0$

$x>-1$ ...(ii)

Hence the domain of $f\left(x\right)$ is $(-1,1)$

Hence $a$ and $b$ will also belong in the same range, that is $(-1,1)$.