Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 1.1, 2 Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a b2} is neither reflexive nor symmetric nor transitive R = {(a, b) : a b2} Checking for reflexive, If the relation is reflexive, then (a, a) R i.e. a a2 Let us check Hence, a a2 is not true for all values of a. So, the given relation it is not reflexive. Checking for symmetric, To check whether symmetric or not, If (a, b) R, then (b, a) R i.e., if a b2, then b a2 Since b a2 is not true for all values of a & b. Hence, the given relation is not symmetric Checking transitive To check whether transitive or not, If (a, b) R & (b, c) R , then (a, c) R i.e., if a b2, & b c2 then a c2 Since if a b2, & b c2 then a c2 is not true for all values of a, b, c. Hence, the given relation it is not transitive Therefore, the given relation is neither reflexive, symmetric or transitive

Ex 1.1

Ex 1.1, 1 (i)

Ex 1.1, 1 (ii)

Ex 1.1, 1 (iii) Important

Ex 1.1, 1 (iv)

Ex 1.1, 1 (v)

Ex 1.1, 2 You are here

Ex 1.1, 3

Ex 1.1, 4

Ex 1.1, 5 Important

Ex 1.1, 6

Ex 1.1, 7

Ex 1.1, 8 Important

Ex 1.1, 9 (i) Important

Ex 1.1, 9 (ii)

Ex 1.1, 10 (i)

Ex 1.1, 10 (ii)

Ex 1.1, 10 (iii) Important

Ex 1.1, 10 (iv)

Ex 1.1, 10 (v)

Ex 1.1, 11

Ex 1.1, 12 Important

Ex 1.1, 13

Ex 1.1, 14

Ex 1.1, 15 (MCQ) Important

Ex 1.1, 16 (MCQ)

Chapter 1 Class 12 Relation and Functions (Term 1)

Serial order wise

About the Author

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.