# Example 12 - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Jan. 28, 2020 by Teachoo

Last updated at Jan. 28, 2020 by Teachoo

Transcript

Example 12 Show that f : N โ N, given by f(x) = {โ(๐ฅ+1 , ๐๐ ๐ฅ ๐๐ ๐๐๐@๐ฅโ1, ๐๐ ๐ฅ ๐๐ ๐๐ฃ๐๐)โค is both one-one and onto. Check one-one There can be 3 cases x1 & x2 both are odd x1 & x2 both are even x1 is odd & x2 is even If x1 & x2 are both odd f(x1) = x1 + 1 f(x2) = x2 + 1 Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Putting f(x1) = f(x2) x1 + 1 = x2 + 1 x1 = x2 If x1 & x2 are both are even f(x1) = x1 โ 1 f(x2) = x2 โ 1 If f(x1) = f(x2) x1 โ 1 = x2 โ 1 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 If x1 is odd and x2 is even f(x1) = x1 + 1 f(x2) = x2 โ 1 If f(x1) = f(x2) x1 + 1 = x2 โ 1 x2 โ x1 = 2 which is impossible as difference between even and odd number can never be even Hence, if f(x1) = f(x2) , Then x1 = x2 โด function f is one-one Rough One-one Steps: 1. Calculate f(x1) 2. Calculate f(x2) 3. Putting f(x1) = f(x2) we have to prove x1 = x2 Check onto f(x) = {โ(๐ฅ+1 , ๐๐ ๐ฅ ๐๐ ๐๐๐@๐ฅโ1, ๐๐ ๐ฅ ๐๐ ๐๐ฃ๐๐)โค Let f(x) = y , such that y โ N x = {โ(๐ฆโ1 , ๐๐ ๐ฆ ๐๐ ๐๐ฃ๐๐@๐ฆ+1, ๐๐ ๐ฆ ๐๐ ๐๐๐)โค If x is odd f(x) = x + 1 y = x + 1 y โ 1 = x x = y โ 1 If x is odd, y is even If x is even f(x) = x โ 1 y = x โ 1 y + 1 = x x = y + 1 If x is even, y is odd Hence, if y is a natural number, x will also be a natural number i.e. x โ N Thus, f is onto.

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Chapter 1 Class 12 Relation and Functions (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.