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1. Chapter 1 Class 12 Relation and Functions
2. Serial order wise
3. Ex 1.2

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 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 –4x1 = –4x2 x1 = x2. Hence, if f(x1) = f(x2) , Then x1 = x2 ∴ function f is one-one Onto f(x) = 3 − 4x Let f(x) = y , such that y ∈ R 3 – 4x = y –4x = y – 3 x = (𝑦 − 3)/(−4) Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((𝑦 − 3)/(−4)) = 3 − 4((𝑦 − 3)/(−4)) = 3 + (y − 3) = y Thus, for every y ∈ R, there exists x ∈ R such that f(x) = y 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 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

Ex 1.2 