Question

If $f\left( x \right) = \dfrac{{2x + 1}}{{3x - 2}}$ then $f\left( {f\left( 2 \right)} \right)$ is equal to
(A) $1$
(B) $3$
(C) $4$
(D) $2$

Answer 1

In this type of problem of functions firstly find the value of internal function $f\left( 2 \right)$ by putting the $x = 2$ in the given equation then put the value obtained for $f\left( 2 \right)$ in place of $x$ in the given equation of $f\left( x \right)$, It will give the value of the function $f\left( {f\left( 2 \right)} \right)$.

Complete step by step answer:
Here, The given function is $f\left( x \right) = \dfrac{{2x + 1}}{{3x - 2}}$. This is an identity function.
We have to find the value of $f\left( {f\left( 2 \right)} \right)$. If we compare this function with the function $f\left( x \right)$ we can say that to find $f\left( {f\left( 2 \right)} \right)$ we replace $x$ by value of $f\left( 2 \right)$ in the given function.
So, firstly find the value of $f\left( 2 \right)$
By, Putting the value $x = 2$ in the given equation we get the value of $f\left( 2 \right)$ as
$f\left( 2 \right) = \dfrac{{2 \times 2 + 1}}{{3 \times 2 - 2}}$
$\Rightarrow f\left( 2 \right) = \dfrac{{4 + 1}}{{6 - 2}}$
$\therefore f\left( 2 \right) = \dfrac{5}{4}$
To find the value of $f\left( {f\left( 2 \right)} \right)$ , we should replace the $x$ of given function by $f\left( 2 \right)$.
It gives $f\left( {f\left( 2 \right)} \right) = \dfrac{{2f\left( 2 \right) + 1}}{{3f\left( 2 \right) - 2}}$
Above, we get $f\left( 2 \right)$ is equal to $\dfrac{5}{4}$ , put the value of $x = f\left( 2 \right)$ in the given function $f\left( x \right)$
Then, put $f\left( 2 \right) = \dfrac{5}{4}$ in the above equation.

This implies
$\Rightarrow f\left( {\dfrac{5}{4}} \right) = \dfrac{{\dfrac{{10 + 4}}{4}}}{{\dfrac{{15 - 8}}{4}}}$
$\Rightarrow f\left( {\dfrac{5}{4}} \right) = \dfrac{{14}}{7}$
$\therefore f\left( {\dfrac{5}{4}} \right) = 2$
Thus, the required value of the function $f\left( {f\left( 2 \right)} \right) = 2$

Hence, option D is the correct option.

Note:
The given function $f\left( {f\left( x \right)} \right)$ is an identity function. We can verify it by putting $x = y$ in the given function. If a function is an identity function then its value will remain the same as that of the variable of that function.
Proof:Put in the equation $f\left( x \right) = \dfrac{{2x + 1}}{{3x - 2}}$
$x = y$, Then, we get $f\left( y \right) = \dfrac{{2y + 1}}{{3y - 2}}$
$f\left( {f\left( y \right)} \right) = \dfrac{{2f\left( y \right) + 1}}{{3f\left( y \right) - 2}}$
And then the value of
\eqalign{ & \Rightarrow f\left( {f\left( y \right)} \right) = \dfrac{{2\left( {\dfrac{{2y + 1}}{{3y - 2}}} \right) + 1}}{{3\left( {\dfrac{{2y + 1}}{{3y - 2}}} \right) - 2}} \cr & \Rightarrow f\left( {f\left( y \right)} \right) = \dfrac{{\dfrac{{4y + 2 + 3y - 2}}{{3y - 2}}}}{{\dfrac{{6y + 3 - 6y + 4}}{{3y - 2}}}} \cr & \Rightarrow f\left( {f\left( y \right)} \right) = \dfrac{{7y}}{7} \cr & \Rightarrow f\left( {f\left( y \right)} \right) = y \cr}
This derivation shows that this function is an identity function so the value of $f\left( {f\left( x \right)} \right)$ is the same as the value of $x$.