
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. $\dfrac{1}{x+\left[ \dfrac{x}{2} \right]}$
Answer
161.1k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
