Question

The function $\log \left(\frac{1+x}{1-x}\right)$ will satisfy the equation

• A. $f\left(x+2\right)-2f\left(x+1\right)+f\left(x\right)=0$
• B. $f\left(x\right)+f\left(x+1\right)=f\left(x\left(x+1\right)\right)$
• C. $f\left(x_1\right)f\left(x_2\right)=f\left(x_1+x_2\right)$
• D. $f\left(x_1\right)+f\left(x_2\right)=f\left(\frac{x_1+x_2}{1+x_1{x}_2}\right)$

Answer 1

The correct option is D

$f\left(x_1\right)+f\left(x_2\right)=f\left(\frac{x_1+x_2}{1+x_1{x}_2}\right)$

Explanation for the correct option

Option(D): Given function is, $\log \left(\frac{1+x}{1-x}\right)$, let it be equal to $f\left(x\right)$, i.e.,
$f\left(x\right)=\log \left(\frac{1+x}{1-x}\right)$

$f\left(x_1\right)=\log \left(\frac{1+x_1}{1-x_1}\right)$
Similarly,
$f\left(x_2\right)=\log \left(\frac{1+x_2}{1-x_2}\right)$

Adding the above two equations,

$\begin{array}{rcl}f\left(x_1\right)+f\left(x_2\right)& =& \log \left(\frac{1+x_1}{1-x_1}\right)+\log \left(\frac{1+x_2}{1-x_2}\right)\\ & =& \log \left[\left(\frac{1+x_1}{1-x_1}\right)\left(\frac{1+x_2}{1-x_2}\right)\right]\left[\because \log \left(m\right)+\log \left(n\right)=\log \left(m·n\right)\right]\\ & =& \log \left(\frac{1+x_2+x_1+x_1{x}_2}{1-x_2-x_1+x_1{x}_2}\right)\dots \left(1\right)\end{array}$

Also,
$\begin{array}{rcl}f\left(\frac{x_1+x_2}{1+x_1{x}_2}\right)& =& \log \left(\frac{1+{\displaystyle \frac{x_1+x_2}{1+x_1{x}_2}}}{1-{\displaystyle \frac{x_1+x_2}{1+x_1{x}_2}}}\right)\\ & =& \log \left(\frac{{\displaystyle \frac{1+x_1{x}_2+x_1+x_2}{{\cancel{1+x_{1}x}}_2}}}{{\displaystyle \frac{1+x_1{x}_2-x_1-x_2}{\cancel{1+x_1{x}_2}}}}\right)\\ & =& \log \left(\frac{1+x_2+x_1+x_1{x}_2}{1-x_2-x_1+x_1{x}_2}\right)\dots \left(2\right)\end{array}$

From $\left(1\right)$ and $\left(2\right)$,
$f\left(x_1\right)+f\left(x_2\right)=f\left(\frac{x_1+x_2}{1+x_1{x}_2}\right)$

So, option(D) is correct.

Explanation for the incorrect options

Option(A): $f\left(x+2\right)-2f\left(x+1\right)+f\left(x\right)=0$

Consider L.H.S.,

$\begin{array}{c}f\left(x+2\right)-2f\left(x+1\right)+f\left(x\right)\\ =\log \left[\frac{1+\left(x+2\right)}{1-\left(x+2\right)}\right]-2·\log \left[\frac{1+\left(x+1\right)}{1-\left(x+1\right)}\right]+\log \left(\frac{1+x}{1-x}\right)\\ =\log \left(x+3\right)-\log \left(-x-1\right)-2\log \left(x+2\right)-2\log \left(-x\right)+\log \left(1+x\right)-\log \left(1-x\right)\\ =\log \left(x+3\right)-\log \left[\left(-1\right)\left(x+1\right)\right]-2\log \left(x+2\right)-2\log \left[\left(-1\right)\left(x\right)\right]+\log \left(1+x\right)-\log \left(1-x\right)\\ =\log \left(x+3\right)-\log \left(-1\right)-\cancel{\log \left(x+1\right)}-2\log \left(x+2\right)-2\log \left(-1\right)-2\log \left(x\right)+\cancel{\log \left(1+x\right)}-\log \left(1-x\right)\\ =\log \left(x+3\right)-3\log \left(-1\right)-2\log \left(x+2\right)-2\log \left(x\right)-\log \left(1-x\right)\end{array}$

R.H.S. is $0$

$\Rightarrow$ L.H.S. $\ne$R.H.S.

So, option(A) is incorrect.

Option(B): $f\left(x\right)+f\left(x+1\right)=f\left(x\left(x+1\right)\right)$

Consider L.H.S.,

$\begin{array}{c}f\left(x\right)+f\left(x+1\right)\\ =\log \left[\frac{1+x}{1-x}\right]+\log \left[\frac{1+\left(x+1\right)}{1-\left(x+1\right)}\right]\\ =\log \left(1+x\right)-\log \left(1-x\right)+\log \left(x+2\right)-\log \left(-x\right)\end{array}$

Consider R.H.S.,

$\begin{array}{c}f\left[x\left(x+1\right)\right]\\ =\log \left[\frac{1+x\left(x+1\right)}{1-x\left(x+1\right)}\right]\\ =\log \left(\frac{x^2+x+1}{1-x^2-x}\right)\\ =\log \left(x^2+x+1\right)-\log \left(1-x^2-x\right)\end{array}$

$\Rightarrow$ L.H.S. $\ne$R.H.S.

So, option(B) is also incorrect

Option(C): $f\left(x_1\right)f\left(x_2\right)=f\left(x_1+x_2\right)$

Consider L.H.S.,

$\begin{array}{c}f\left(x_1\right)f\left(x_2\right)\\ =\log \left[\frac{1+x_1}{1-x_1}\right]\log \left[\frac{1+x_2}{1-x_2}\right]\\ =\left[\log \left(1+x_1\right)-\log \left(1-x_1\right)\right]\left[\log \left(1+x_2\right)-\log \left(1-x_2\right)\right]\\ =\log \left(1+x_1\right)\log \left(1+x_2\right)-\log \left(1+x_1\right)\log \left(1-x_2\right)-\log \left(1-x_1\right)\log \left(1+x_2\right)+\log \left(1-x_1\right)\log \left(1-x_2\right)\end{array}$

Consider R.H.S.,

$\begin{array}{c}f\left(x_1+x_2\right)\\ =\log \left[\frac{1+\left(x_1+x_2\right)}{1-\left(x_1+x_2\right)}\right]\\ =\log \left(1+x_1+x_2\right)-\log \left(1-x_1-x_2\right)\end{array}$

$\Rightarrow$ L.H.S. $\ne$R.H.S.

So, option(C) is also incorrect.

Hence option(D) is correct.