If we have a function as $f\left( x \right)=\dfrac{x+2}{3x-1}$, then \[ff\left( x \right)\] is,
A. $x$
B. $-x$
C. $\dfrac{1}{x}$
D. $-\dfrac{1}{x}$
E. 0

Hint: We will be using the concept of functions to solve the problem. We will be using the concepts of compound function to find the value of \[ff\left( x \right)\] and further simplify the solution.

Complete step-by-step solution-
Now, we have been given that $f\left( x \right)=\dfrac{x+2}{3x-1}$ and we have to find the value of \[ff\left( x \right)\].
Now, we know that $fg\left( x \right)$ means $f\left( g\left( x \right) \right)$. So, we have to find \[f\left( f\left( x \right) \right)\].
Now, for this we have to substitute f(x) for x in f(x). Therefore, we have,
\[f\left( f\left( x \right) \right)=\dfrac{f\left( x \right)+2}{3f\left( x \right)-1}\]
Now, we will put $f\left( x \right)=\dfrac{x+2}{3x-1}$ and further simplify it. So, that we have,
$=\dfrac{\dfrac{x+2}{3x-1}+2}{\dfrac{3\left( x+2 \right)}{3x-1}-1}$
Now, we will take $\left( 3x-1 \right)$ as LCM in both numerator and denominator,
$=\dfrac{x+2+2\left( 3x-1 \right)}{3\left( x+2 \right)-1\left( 3x-1 \right)}$
Now, we will expand in numerator and denominator and solve them,
$=\dfrac{x+2+6x-2}{3x+6-3x+1}$
On further simplifying we have,
$\begin{align}
  & =\dfrac{7x}{7} \\
 & =x \\
\end{align}$
So, on simplifying we have \[ff\left( x \right)=x\].
Hence, the correct option is (A).

Note: To solve these types of questions one must know the basic concepts of compound function that \[ff\left( x \right)=f\left( f\left( x \right) \right)\]. Also it is important to note that to find \[f\left( f\left( x \right) \right)\] we have put $f\left( x \right)$ in $f\left( x \right)$.