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Ex 1.2 , 7 - Chapter 1 Class 12 Relation and Functions - Part 4

Ex 1.2 , 7 - Chapter 1 Class 12 Relation and Functions - Part 5
Ex 1.2 , 7 - Chapter 1 Class 12 Relation and Functions - Part 6

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Ex 1.2, 7 In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (ii) f: R → R defined by f(x) = 1 + x2 f(x) = 1 + x2 Checking one-one f (x1) = 1 + (x1)2 f (x2) = 1 + (x2)2 Putting f (x1) = f (x2) 1 + (x1)2 = 1 + (x2)2 (x1)2 = (x2)2 ∴ x1 = x2 or 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 x1 does not have unique image, It is not one-one Eg: f(–1) = 1 + (–1)2 = 1 + 1 = 2 f(1) = 1 + (1)2 = 1 + 1 = 2 Here, f(–1) = f(1) , but –1 ≠ 1 Hence, it is not one-one Check onto f(x) = 1 + x2 Let f(x) = y , such that y ∈ R 1 + x2 = y x2 = y – 1 x = ±√(𝑦−1) Note that y is a real number, it can be negative also Putting y = −3 x = ±√((−3)−1) = ±√(−4) Which is not possible as root of negative number is not real Hence, x is not real So, f is not 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.