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Ex 1.2 , 2 - Chapter 1 Class 12 Relation and Functions - Part 10

Ex 1.2 , 2 - Chapter 1 Class 12 Relation and Functions - Part 11
Ex 1.2 , 2 - Chapter 1 Class 12 Relation and Functions - Part 12

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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.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.