Let $f:(4,6)\to (6,8)$ be a function defined by $f(x)=x+\left[ \dfrac{x}{2} \right]$ (where $\left[ . \right]$ denotes the greatest integer function) then ${{f}^{-1}}(x)$ is equal to
A. \[x-\left[ \dfrac{x}{2} \right]\]
B. \[-x-2\]
C. \[x-2\]
D. $\dfrac{1}{x+\left[ \dfrac{x}{2} \right]}$

Hint: To find the inverse of the function we will first find the interval for $\dfrac{x}{2}$ using the domain of $x$ given then determine the value of greatest integer function $\left[ \dfrac{x}{2} \right]$ by returning the nearest integer which is less than or equal to the given number. We will then substitute the value of $\left[ \dfrac{x}{2} \right]$ in $f(x)=x+\left[ \dfrac{x}{2} \right]$.
After this we will find inverse of function by writing $y=f(x)$ and then replacing all the $x$’s with $y$’s.



Formula Used:

Complete step by step solution:We are given a function $f(x)=x+\left[ \dfrac{x}{2} \right]$ when $f:(4,6)\to (6,8)$ and we have to determine the value of ${{f}^{-1}}(x)$ where $\left[ . \right]$ denotes the greatest integer function.
We are given the domain of $x$ as $x\in (4,6)$. So the domain of $\dfrac{x}{2}$ will be,
 $\begin{align}
  & \dfrac{x}{2}\in \left( \dfrac{4}{2},\dfrac{6}{2} \right) \\
 & \dfrac{x}{2}\in (2,3) \\
\end{align}$
It means that the value of $\dfrac{x}{2}$ will be greater than $2$ and less than $3$ that is $2<\dfrac{x}{2}<3$.
We know that the greatest integer function can be defined as a function which returns the greatest integer less than or equal to the given number. So,
So the greatest integer function for $\left[ \dfrac{x}{2} \right]$will be,
$\begin{align}
  & 2<\dfrac{x}{2}<3 \\
 & \left[ \dfrac{x}{2} \right]=2 \\
\end{align}$
Now we can write the function $f(x)=x+\left[ \dfrac{x}{2} \right]$ as $f(x)=x+2$.
We know that a function has inverse when it is one to one function. The function $f(x)=x+2$ gives unique output for every input which means it is a one to one function and its inverse ${{f}^{-1}}$will exist.
To find the inverse,
We will write $y=f(x)$ and then replace all the $x$’s with $y$’s.
$\begin{align}
  & f(x)=x+2 \\
 & y=x+2 \\
 & x=y+2
\end{align}$
We will now write the equation for$y$in terms of $x$.
$y=x-2$
This $y=x-2$will be the inverse of the function $f(x)=x+2$.
We will now replace $y$ with ${{f}^{-1}}(x)=x-2$.



Option ‘D’ is correct

Note: A function is called one to one function when value of one input from the domain has exactly one output in the range.