Question

The function $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$ satisfies the equation

• A. $f(x+2)-2f(x+1)+f\left(x\right)=0$
• B. $f\left(x\right)+f(x+1)=f\left\{x(x+1)\right\}$
• C. $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1+xy}\right)$
• D. $f(x+y)=f\left(x\right)f\left(y\right)$

Answer 1

The correct option is C $f\left(x\right)+f\left(y\right)=f\left(\frac{x+y}{1+xy}\right)$

Since, $f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$
$\therefore f\left(\frac{x+y}{1+xy}\right)=\log \left(\frac{1+\frac{x+y}{1+xy}}{1-\frac{x+y}{1+xy}}\right)$
$=\log \left(\frac{1+xy+x+y}{1+xy-x-y}\right)$
Now, $f\left(x\right)+f\left(y\right)=\log \left(\frac{1+x}{1-x}\right)+\log \left(\frac{1+y}{1-y}\right)$
$=\log \left(\frac{(1+x)(1+y)}{(1-x)(1-y)}\right)$
$=\log \left(\frac{1+x+y+xy}{1+xy-x-y}\right)$
$\therefore f\left(\frac{x+y}{1+xy}\right)=f\left(x\right)+f\left(y\right)$