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

Last updated at Jan. 28, 2020 by Teachoo

Last updated at Jan. 28, 2020 by Teachoo

Transcript

Ex 1.2, 5 Show that the Signum Function f: R โ R, given by f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค is neither one-one nor onto. f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค For example: f(0) = 0 f(-1) = โ1 f(1) = 1 f(2) = 1 f(3) = 1 Since, different elements 1,2,3 have the same image 1 , โด f is not one-one. Check onto f: R โ R f(x) = {โ(1 for ๐ฅ >0@ 0 for ๐ฅ=0@โ1 for ๐ฅ<0)โค Value of f(x) is defined only if x is 1, 0, โ1 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

Ex 1.2

Ex 1.2, 1

Ex 1.2, 2 (i) Important

Ex 1.2, 2 (ii) Important

Ex 1.2, 2 (iii)

Ex 1.2, 2 (iv)

Ex 1.2, 2 (v) Important

Ex 1.2 , 3

Ex 1.2 , 4

Ex 1.2, 5 Important You are here

Ex 1.2 , 6 Important

Ex 1.2, 7 (i)

Ex 1.2, 7 (ii)

Ex 1.2 , 8 Important

Ex 1.2 , 9

Ex 1.2 , 10 Important

Ex 1.2 , 11 (MCQ) Important

Ex 1.2, 12 (MCQ)

Chapter 1 Class 12 Relation and Functions (Term 1)

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.