Question

Let , a function $f:\left( 0,\infty \right)\to \left( 0,\infty \right)$ will be defined by $f\left( x \right)=\left| 1-\dfrac{1}{x} \right|$ .Then $f$ is :
(a) Injective only
(b) Not injective but it is surjective
(c)Both injective as well as surjective
(d) Neither injective nor surjective

Answer 1

Hint: Use the definition of injective and surjective to check whether the given function is injective or not and surjective or not . A function is said to be one – one if no two ordered pairs with different first coordinates have the same second coordinate. A function is said to be onto if for every element in the co domain there exists an element in the domain which maps to it. Use these definitions and choose the correct option wisely.

Complete step-by-step answer:
First, we will check if the function is one-one or injective .
For that , we will use the definition that , A function is said to be one – one if no two ordered pairs with different first coordinates have the same second coordinate.
Now, we will try to search counter examples such ${{x}_{1}}\ne {{x}_{2}}$ , but $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$ .
If we are able to find such an example , then the given function is not injective.
Now, let
${{x}_{1}}=2$ and ${{x}_{2}}=\dfrac{2}{3}$
Then,

$\begin{align}
  & f\left( {{x}_{1}} \right)=\left| 1-\dfrac{1}{2} \right| \\
 & =\left| \dfrac{1}{2} \right| \\
 & =\dfrac{1}{2}
\end{align}$

and

$\begin{align}
  & f\left( {{x}_{2}} \right)=\left| 1-\dfrac{1}{\dfrac{2}{3}} \right| \\
 & =\left| 1-\dfrac{3}{2} \right| \\
 & =\left| -\dfrac{1}{2} \right| \\
 & =\dfrac{1}{2}
\end{align}$
Hence, $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$
So , we have found an example where ${{x}_{1}}\ne {{x}_{2}}$ but $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$
Hence , the function is not injective.
Hence, option (a) and (c) are not correct .
 Now, we will check if the function is onto
The given function is surjective for every element in codomain , we have its preimage in the domain

So, the correct answer is “Option b”.

Note: We can check one – one and onto through graphs also. First plot the graph of the given function and then draw horizontal lines if no horizontal line intersects the graph of function in more than one point then function is one one and for onto draw vertical lines , all the vertical lines cut the graph at least one time , then the function is onto.