The function\[f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\]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). \[f\left( x \right)+f\left( y \right)=f\left( \dfrac{x+y}{1+xy} \right)\]
(d). \[f\left( x+y \right)=f\left( x \right)f\left( y \right)\]

Hint: Calculate values of all the required terms of each option and verify that option is correct or not. Use the property of algorithm function, while verifying the options. Use following property:-
\[{{\log }_{m}}a-{{\log }_{m}}b={{\log }_{m}}\left( \dfrac{a}{b} \right)\]
\[{{\log }_{m}}a+{{\log }_{m}}b={{\log }_{m}}\left( ab \right)\]

Complete step-by-step answer:

We are given function \[f\left( x \right)\]as
\[f\left( x \right)=\log \left( \dfrac{1+x}{1-x} \right)\] \[\to (1)\]
As all the given options are different to each other, so, we need to calculate the values of all the given options to verify whether it is correct or not.
Option (A) \[f\left( x+2 \right)-2f\left( x+1 \right)+f\left( x \right)=0\]
So, L.H.S is given as
\[LHS=f\left( x+2 \right)-2f\left( x+1 \right)+f\left( x \right)\]
So, the value of\[f\left( n+1 \right),f\left( x+2 \right)\]can be calculated from equation (1) by replacing \[x\] by \[x+1,x+2\] respectively.
So, L.H.S is given as
\[LHS=\log \left( \dfrac{1+\left( x+2 \right)}{1-\left( x+2 \right)} \right)-2\log \left( \dfrac{1+\left( x+1 \right)}{1-\left( x+1 \right)} \right)+\log \left( \dfrac{1+x}{1-x} \right)\]
We know that property of logarithm functions are given as
\[{{\log }_{c}}a+{{\log }_{c}}b={{\log }_{c}}ab\] \[\to (2)\]
\[\log {{m}^{n}}=n\log m\] \[\to (3)\]
So, we get the value of L.H.S. as
\[LHS=\log \left( \dfrac{x+3}{-x-1} \right)-\log {{\left( \dfrac{x+2}{-x} \right)}^{2}}+\log \left( \dfrac{1+x}{1-x} \right)\]
Now, using the property (2), we get L.H.S as
\[LHS=\log \left( \dfrac{x+3}{-x-1} \right)\left( \dfrac{1+x}{1-x} \right)-\log {{\left( \dfrac{x+2}{-x} \right)}^{2}}\]
\[LHS=\log \left( \dfrac{\left( x+3 \right)\left( x+1 \right)}{\left( -1 \right)\left( x+1 \right)\left( 1-x \right)} \right)-\log {{\left( \dfrac{x+2}{-x} \right)}^{2}}\]
\[LHS=\log \left( \dfrac{\left( x+3 \right)}{-\left( 1-x \right)} \right)-\log {{\left( \dfrac{x+2}{-x} \right)}^{2}}\]
We know the property of logarithm function
\[{{\log }_{a}}b-{{\log }_{a}}c={{\log }_{a}}\dfrac{b}{c}\]
So, we get L.H.S. as
\[LHS=\log \left( \dfrac{\dfrac{\left( x+3 \right)}{\left( 1-x \right)}}{{{\left( \dfrac{\left( x+2 \right)}{-2} \right)}^{2}}} \right)\]
\[LHS=\log \left( \left( \dfrac{x+3}{1-x} \right)\times \dfrac{{{x}^{2}}}{{{\left( x+2 \right)}^{2}}} \right)\]
As, above expression is not independent of \['x'\], it means L.H.S. will never be 0.
So, option (a) is not true.

Option (b):- \[f\left( x \right)+f\left( x+1 \right)=f\left\{ x\left( x+1 \right) \right\}\]
So, L.H.S. of this equation can be given as
\[f\left( x \right)+f\left( x+1 \right)=f\left( \dfrac{1+x}{1-x} \right)+\log \left( \dfrac{1+x+1}{1-\left( 1+x \right)} \right)\]
\[L.H.S.=\log \left( \dfrac{1+x}{\left( 1-x \right)} \right)+\log \left( \dfrac{2+x}{-x} \right)\]
Appling the property given in equation (2), we get
\[\begin{align}
  & L.H.S.=\log \left( \dfrac{\left( 1+x \right)\left( 2+x \right)}{\left( 1-x \right)\left( -x \right)} \right) \\
 & \\
 & L.H.S.=\log \left( \dfrac{{{x}^{2}}+3x+2}{{{x}^{2}}-x} \right) \\
\end{align}\]
R.H.S can be calculated using equation (1) as
\[\begin{align}
  & R.H.S.=\log \left( \dfrac{1+x\left( x-1 \right)}{1-x\left( x-1 \right)} \right) \\
 & \\
 & R.H.S.=\log \left( \dfrac{{{x}^{2}}-x+1}{-{{x}^{2}}+x+1} \right) \\
\end{align}\]
As,\[L.H.S.\ne R.H.S.\]
So, option (b) is not correct.

Option (d):- \[f\left( x+y \right)=f\left( x \right)f\left( y \right)\]
So, we get value of L.H.S. that is \[f\left( x+y \right)\]from the equation (1) as
\[\begin{align}
  & f\left( x+y \right)=\log \left( \dfrac{1+\left( x+y \right)}{1-\left( x+y \right)} \right) \\
 & \\
 & f\left( x+y \right)=\log \left( \dfrac{x+y+1}{-x-y+1} \right) \\
\end{align}\]
R.H.S. of the expression can be calculated as
\[R.H.S.=f\left( x \right)f\left( y \right)=\log \left( \dfrac{1+x}{1-x} \right)+\log \left( \dfrac{1+y}{1-y} \right)\]
Use the property given in equation (2), we get
\[\begin{align}
  & R.H.S.=\log \left( \dfrac{\left( 1+x \right)\left( 1+y \right)}{\left( 1-x \right)\left( 1-y \right)} \right) \\
 & \\
 & R.H.S.=\log \left( \dfrac{x+y+xy+1}{xy+1-x-y} \right) \\
\end{align}\]
Hence, \[L.H.S.\ne R.H.S.\]
So, this option is also wrong.

Option (c):- \[f\left( x \right)+f\left( y \right)=f\left( \dfrac{x+y}{1+xy} \right)\]
L.H.S. of the given expression is given as
\[f\left( x \right)f\left( y \right)=\log \left( \dfrac{1+x}{1-x} \right)+\log \left( \dfrac{1+y}{1-y} \right)\]
Now, use property (2) and hence, we get
\[\begin{align}
  & f\left( x \right)f\left( y \right)=\log \left( \dfrac{\left( 1+x \right)\left( 1+y \right)}{\left( 1-x \right)\left( 1-y \right)} \right) \\
 & \\
 & f\left( x \right)f\left( y \right)=\log \left( \dfrac{x+y+xy+1}{xy+1-x-y} \right) \\
\end{align}\]
R.H.S. of the expression can be calculated as
\[\begin{align}
  & R.H.S.=f\left( \dfrac{x+y}{1+xy} \right)=\log \left( \dfrac{1+\dfrac{x+y}{1+xy}}{1-\dfrac{x+y}{1+xy}} \right) \\
 & \\
 & R.H.S.=\log \left( \dfrac{1+xy+x+y}{1+xy-x-y} \right) \\
\end{align}\]
Hence,\[L.H.S.=R.H.S.\], So, the given option is the correct answer to the problem.

Note: Calculation is an important side of this question, as we need to verify all four options, given to the problem. Don’t confuse the property of logarithms. One may use identity of \[{{\log }_{c}}\left( \dfrac{a}{b} \right)\] as \[{{\log }_{c}}a+{{\log }_{c}}b\], which is wrong. Correct property of logarithm functions are given as
 \[\begin{align}
  & {{\log }_{c}}\left( \dfrac{a}{b} \right)={{\log }_{c}}a-{{\log }_{c}}b \\
 & \\
 & {{\log }_{c}}ab={{\log }_{c}}a+{{\log }_{c}}b \\
 & \\
 & {{\log }_{c}}{{m}^{n}}=n{{\log }_{c}}m\text{ } \\
\end{align}\]