Ex 1.2, 7 - State whether one-one, onto or bijective - Class 12 - To prove injective/ surjective/ bijective (one-one & onto)

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  1. Chapter 1 Class 12 Relation and Functions
  2. Serial order wise

Transcript

Ex 1.2 , 7 In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. • f: R → R defined by f(x) = 3 − 4x f(x) = 3 – 4x Checking one-one f (x1) = 3 – 4x1 f (x2) = 3 – 4x2 Putting f(x1) = f(x2) 3 – 4x1 = 3 – 4x2 –4x1 = –4x2 x1 = x2. Hence, if f(x1) = f(x2) , x1 = x2 ∴ function f is one-one Onto f(x) = 3x Let f(x) = y , such that y ∈ R 3 – 4x = y –4x = y – 3 x = 𝑦 − 3﷮−4﷯ Since y is a real number, Hence 𝑦 − 3﷮−4﷯ will also be a real number So, x will also be a real number, i.e., x ∈ R Hence, f is onto 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 f (x1) = f (x2) ⇒ 1 + (x1)2 = 1 + (x2)2 ⇒ (x1)2 = (x2)2 ⇒ x1 = x2 or 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.