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

Transcript

Ex 1.2, 9 Let f: N โ†’ N be defined by f (n) = {โ–ˆ((๐‘› + 1)/2 ", if n is odd" @๐‘›/2 ", if n is even" )โ”ค for all n โˆˆ N. State whether the function f is bijective. Justify your answer. f (n) = {โ–ˆ((๐‘› + 1)/2 ", if n is odd" @๐‘›/2 ", if n is even" )โ”ค for all n โˆˆ N. Check one-one f(1) = (1 + 1)/2 = 2/2 = 1 f(2) = 2/2 = 1 Since, f(1) = f(2) but 1 โ‰  2 "(Since 1 is odd)" "(Since 2 is even)" Both f(1) & f(2) have same image 1 โˆด f is not one-one Check onto f (n) = {โ–ˆ((๐‘› + 1)/2 ", if n is odd" @๐‘›/2 ", if n is even" )โ”ค for all n โˆˆ N Let f(x) = y , such that y โˆˆ N Thus, f is onto When n is odd y = (๐‘› + 1)/2 2y = n + 1 2y โ€“ 1 = n n = 2y โ€“ 1 Hence, for y is a natural number , n = 2y โ€“ 1 is also a natural number When n is even y = ๐‘›/2 2y = n n = 2y Hence for y is a natural number , n = 2y is also a natural number Thus, for every y โˆˆ N, there exists x โˆˆ N such that f(n) = y Hence, f is onto Ex 1.2, 10 Let A = R โˆ’ {3} and B = R โˆ’ {1}. Consider the function f: A โ†’ B defined by f (x) = ((x โˆ’ 2)/(x โˆ’ 3)) Is f one-one and onto? Justify your answer. f (x) = ((x โˆ’ 2)/(x โˆ’ 3)) Check one-one f (x1) = ((x"1 " โˆ’ 2)/(x"1" โˆ’ 3)) f (x2) = ((x"2 " โˆ’ 2)/(x"2" โˆ’ 3)) Putting f (x1) = f (x2) ((x"1 " โˆ’ 2)/(x"1" โˆ’ 3)) = ((x"2 " โˆ’ 2)/(x"2" โˆ’ 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 โ€“ 2) (x2 โ€“ 3) = (x1 โ€“ 3) (x2 โ€“ 2) x1 (x2 โ€“ 3) โ€“ 2 (x2 โ€“ 3) = x1 (x2 โ€“ 2) โ€“ 3 (x2 โ€“ 2) x1 x2 โ€“ 3x1 โ€“ 2x2 + 6 = x1 x2 โ€“ 2x1 โ€“ 3x2 + 6 โ€“ 3x1 โ€“ 2x2 = โ€“ 2x1 โ€“ 3x2 3x2 โ€“ 2x2 = โ€“ 2x1 + 3x1 x1 = x2 Hence, if f (x1) = f (x2), then x1 = x2 โˆด f is one-one Check onto f (x) = ((x โˆ’ 2)/(x โˆ’ 3)) Let f(x) = y such that y โˆˆ B i.e. y โˆˆ R โ€“ {1} So, y = ((x โˆ’ 2)/(x โˆ’ 3)) y(x โ€“ 3) = x โ€“ 2 xy โ€“ 3y = x โ€“ 2 xy โ€“ x = 3y โ€“ 2 x (y โ€“ 1) = 3y โ€“ 2 x = (3y โˆ’ 2)/(y โˆ’ 1) For y = 1 , x is not defined But it is given that y โˆˆ R โ€“ {1} Hence , x = (3y โˆ’ 2)/(y โˆ’ 1) โˆˆ R โ€“ {3} Now, Checking for y = f(x) Putting value of x in f(x) f(x) = f((3y โˆ’ 2)/(y โˆ’ 1)) = (((3y โˆ’ 2)/(y โˆ’ 1)) โˆ’ 2)/(((3y โˆ’ 2)/(y โˆ’ 1)) โˆ’3) = ((((3y โˆ’ 2) โˆ’ 2(๐‘ฆ โˆ’1))/(y โˆ’ 1)))/((((3y โˆ’ 2) โˆ’ 3(๐‘ฆ โˆ’1))/(y โˆ’ 1)) ) = (3๐‘ฆ โˆ’ 2 โˆ’ 2๐‘ฆ + 2)/(3๐‘ฆ โˆ’ 2 โˆ’ 3๐‘ฆ + 3) = ๐‘ฆ/1 = y Thus, for every y โˆˆ B, there exists x โˆˆ A such that f(x) = y Hence, f is onto

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.