What is the number of linear functions f satisfying \[f\left( {x + f\left( x \right)} \right) = x + f\left( x \right)\forall x \in R\]?
A. 0
B. 1
C. 2
D. 3

Hint: First we assume the general linear function that represents \[f\left( x \right)\]. Then substitute the value of \[f\left( x \right)\] in the given equation. Then compare the coefficients and solve the equations.

Complete step by step solution: Given that \[f\left( x \right)\] is a linear function. The general equation of the linear function is \[ax + b\] where \[a \ne 0\].
Assume that \[f\left( x \right) = ax + b\]
Now substitute \[f\left( x \right) = ax + b\] in \[f\left( {x + f\left( x \right)} \right) = x + f\left( x \right)\forall x \in R\]
\[f\left( {x + ax + b} \right) = x + ax + b\]
\[ \Rightarrow f\left( {\left( {a + 1} \right)x + b} \right) = x + ax + b\]
Again putting \[x = \left( {a + 1} \right)x + b\] in \[f\left( x \right) = ax + b\]
\[ \Rightarrow a\left\{ {\left( {a + 1} \right)x + b} \right\} + b = x + ax + b\]
\[ \Rightarrow a\left( {a + 1} \right)x + ab + b = x + ax + b\] ….(i)
Now comparing the coefficients of x:
\[ \Rightarrow a\left( {a + 1} \right) = a + 1\]
\[ \Rightarrow {a^2} + a = a + 1\]
Subtract a from both sides
\[ \Rightarrow {a^2} = 1\]
Taking square root on both sides:
\[ \Rightarrow a = \pm 1\]
The possible value of a is 2.
Now comparing the constant term of equation (i):
\[ab + b = b\]
\[ \Rightarrow ab = 0\]
\[ \Rightarrow b = 0\] since \[a \ne 0\].
Thus the possible number of functions is 2.

 Option ‘C’ is correct

Note: Student often make mistake to solve the equation \[a\left( {a + 1} \right) = a + 1\]. They cancel out \[\left( {a + 1} \right)\] from both sides. For this reason they get only one solution. The equation is a quadratic equation thus we have to it as a quadratic equation.