Check sibling questions

Ex 1.2, 5 - Show Signum Function is neither one-one nor onto

Ex 1.2, 5 - Chapter 1 Class 12 Relation and Functions - Part 2


Transcript

Ex 1.2, 5 Show that the Signum Function f: R → R, given by f(x) = {█(1 for 𝑥 >[email protected] 0 for 𝑥[email protected]−1 for 𝑥<0)┤ is neither one-one nor onto. f(x) = {█(1 for 𝑥 >[email protected] 0 for 𝑥[email protected]−1 for 𝑥<0)┤ For example: f(0) = 0 f(-1) = −1 f(1) = 1 f(2) = 1 f(3) = 1 Since, different elements 1,2,3 have the same image 1 , ∴ f is not one-one. Check onto f: R → R f(x) = {█(1 for 𝑥 >[email protected] 0 for 𝑥[email protected]−1 for 𝑥<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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.