Check sibling questions

Ex 1.2 , 9 - Let f(n) = {n+1/2, if n is odd n2, if n is even

Ex 1.2 , 9 - Chapter 1 Class 12 Relation and Functions - Part 2
Ex 1.2 , 9 - Chapter 1 Class 12 Relation and Functions - Part 3

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Ex 1.2, 9 Let f: N → N be defined by f (n) = {█((𝑛 + 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N. State whether the function f is bijective. Justify your answer. f (n) = {█((𝑛 + 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N. Check one-one f(1) = (1 + 1)/2 = 2/2 = 1 f(2) = 2/2 = 1 Since, f(1) = f(2) but 1 ≠ 2 "(Since 1 is odd)" "(Since 2 is even)" Both f(1) & f(2) have same image 1 ∴ f is not one-one Check onto f (n) = {█((𝑛 + 1)/2 ", if n is odd" @𝑛/2 ", if n is even" )┤ for all n ∈ N Let f(x) = y , such that y ∈ N Thus, f is onto When n is odd y = (𝑛 + 1)/2 2y = n + 1 2y – 1 = n n = 2y – 1 Hence, for y is a natural number , n = 2y – 1 is also a natural number When n is even y = 𝑛/2 2y = n n = 2y Hence for y is a natural number , n = 2y is also a natural number Thus, for every y ∈ N, there exists x ∈ N such that f(n) = y Hence, f is onto Ex 1.2, 10 Let A = R − {3} and B = R − {1}. Consider the function f: A → B defined by f (x) = ((x − 2)/(x − 3)) Is f one-one and onto? Justify your answer. f (x) = ((x − 2)/(x − 3)) Check one-one f (x1) = ((x"1 " − 2)/(x"1" − 3)) f (x2) = ((x"2 " − 2)/(x"2" − 3)) Putting f (x1) = f (x2) ((x"1 " − 2)/(x"1" − 3)) = ((x"2 " − 2)/(x"2" − 3)) Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 (x1 – 2) (x2 – 3) = (x1 – 3) (x2 – 2) x1 (x2 – 3) – 2 (x2 – 3) = x1 (x2 – 2) – 3 (x2 – 2) x1 x2 – 3x1 – 2x2 + 6 = x1 x2 – 2x1 – 3x2 + 6 – 3x1 – 2x2 = – 2x1 – 3x2 3x2 – 2x2 = – 2x1 + 3x1 x1 = x2 Hence, if f (x1) = f (x2), then x1 = x2 ∴ f is one-one Check onto f (x) = ((x − 2)/(x − 3)) Let f(x) = y such that y ∈ B i.e. y ∈ R – {1} So, y = ((x − 2)/(x − 3)) y(x – 3) = x – 2 xy – 3y = x – 2 xy – x = 3y – 2 x (y – 1) = 3y – 2 x = (3y − 2)/(y − 1) For y = 1 , x is not defined But it is given that y ∈ R – {1} Hence , x = (3y − 2)/(y − 1) ∈ R – {3} Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((3y − 2)/(y − 1)) = (((3y − 2)/(y − 1)) − 2)/(((3y − 2)/(y − 1)) −3) = ((((3y − 2) − 2(𝑦 −1))/(y − 1)))/((((3y − 2) − 3(𝑦 −1))/(y − 1)) ) = (3𝑦 − 2 − 2𝑦 + 2)/(3𝑦 − 2 − 3𝑦 + 3) = 𝑦/1 = y Thus, for every y ∈ B, there exists x ∈ A such that f(x) = y Hence, f is 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.