# Ex 1.2, 5 - Chapter 1 Class 12 Relation and Functions

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 1.2, 5 Show that the Signum Function f: R R, given by f(x) = 1 for >0 0 for =0 1 for <0 is neither one-one nor onto. f(x) = 1 for >0 0 for =0 1 for <0 For example: f(0) = 0 f(-1) = -1 f(1) = 1 f(2) = 1 f(3) = 1 Check onto f: R R f(x) = 1 for >0 0 for =0 1 for <0 Value of f(x) is defined only if x is 1, 0, 1 For other real numbers(eg: y = 2, y = 100) there is no corresponding element x Hence f is not onto Thus, f is neither one-one nor onto

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.