
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
163.2k+ 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.
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