Question

If R denotes the set of all real number, then the function f: $R \to R$ defined by $f\left( x \right) = |x|\,\,is\,$
A. only one-one
B. only onto
C. both one-one and onto
D. neither one-one nor onto

Answer 1

Hint: In this question to check if a function is one-one or many one. We have to check if we are getting one value of a function for two different inputs and if we are getting one value for more than one input, it is many-one otherwise it is one-one. Also, if the range is equal to codomain then it is onto, otherwise it is into.

Complete step by step answer:
In the above question, $f\left( x \right) = |x|$. As we know a function is one-one if and only if for every unique input there must be a unique output. But in the case of many-one functions there can be many inputs for a single output. Now,
Let, $x = 1$
$f\left( 1 \right) = |1|$
$ \Rightarrow 1$
Now let, $x = - 1$
$f\left( { - 1} \right) = | - 1|$
$ \Rightarrow 1$
Therefore, for two different values $1\,\,and\, - 1$ we have a particular output that is $1$.So, the given function is not one-one.

Now, in the above question it is given that the function is defined from $R \to R$, where R is a real number and R before ‘$ \to $’ shows domain and R after ‘$ \to $’ shows codomain.
But, according to our function $f\left( x \right) = |x|$, we never get a negative value because the modulus of any real number is a positive number. As we know, a number is onto only when the $range = codomain$.
But, here
$range \in \left( {0,\infty } \right)$
$codomain \in \left( { - \infty ,\infty } \right)$
Therefore, function is not onto.

Hence, the function is neither one-one nor onto and the correct option is $\left( D \right)$.

Note: One-one functions are special functions that return a unique range for each element in their domain. They are also known as injective functions. Also, onto functions are those functions in which range is equal to the codomain of the function. They are also known as surjective functions.