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**NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 5 Solution**. Solutions for other problems are available at Exercise 1.2 SolutionsExercise 1.2 Problem 5 Solution

5. Show that the Signum Function f : R → R, given by {f = \begin{cases}1, & x \gt 0\\0, & x = 0\\-1, & x \lt 0\end{cases}} is neither one-one nor onto.

To check whether f is one-one:

In the given function f, all the positive real numbers in the domain have the same image 1. Similarly all the negative numbers in the domain have the same image -1 in the co-domain.

⇒ Multiples elements in the domain have the same image in the domain.

For instance,

f(1) = 1

f(9) = 1

f(55) = 1

So, the distinct elements 1, 10, 55 in the domain have the same image 1 in the co-domain.

∴ f is not one-one.

To check whether f is onto:

Elements other than -1, 0 and 1 in the co-domain are not images of any element in the domain.

For example, 10 is not image of any element.

∴ f is not onto.