# Ex 1.2, 2 (iv) - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Aug. 11, 2021 by Teachoo

Last updated at Aug. 11, 2021 by Teachoo

Transcript

Ex 1.2, 2 Check the injectivity and surjectivity of the following functions: (iv) f: N → N given by f(x) = x3 f(x) = x3 Checking one-one (injective) f (x1) = (x1)3 f (x2) = (x2)3 Putting f (x1) = f (x2) ⇒ (x1)3 = (x2)3 ⇒ x1 = x2 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 ∴ It is one-one (injective) Check onto (surjective) f(x) = x3 Let f(x) = y , such that y ∈ N x3 = y x = 𝑦^(1/3) Here y is a natural number i.e. y ∈ N Let y = 2 x = 𝑦^(1/3) = 2^(1/3) So, x is not a natural number ∴ f is not onto (not surjective) Hence, function f is injective but not surjective.

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) You are here

Ex 1.2, 2 (v) Important

Ex 1.2 , 3

Ex 1.2 , 4

Ex 1.2, 5 Important

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)

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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.