# Problem 5 Solution

This page contains the NCERT mathematics class 12 chapter Relations and Functions Exercise 1.2 Problem 5 Solution. Solutions for other problems are available at Exercise 1.2 Solutions
Exercise 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.