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

Transcript

Example 17 Show that if f : R โ€“ {7/5} โ†’ R โ€“ {3/5} is defined by f(x) = (3๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 7) and g: R โˆ’ {3/5}โ†’ R โ€“ {7/5} is defined by g(x) = (7๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 3), then fog = IA and gof = IB, where A = R โˆ’ {3/5} , B = R โ€“ {7/5} ; IA (x) = x, โˆ€ x โˆˆ A, IB (x) = x, โˆ€ x โˆˆ B are called identity functions on sets A and B, respectively. f(x) = (3๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 7) & g(x) = (7๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 3) Finding gof g(x) = (7๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 3) g(f(x)) = (7๐‘“(๐‘ฅ) + 4)/(5๐‘“(๐‘ฅ) โˆ’ 3) gof = (7 ((3๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 7))" " + 4)/(5 (((3๐‘ฅ + 4))/((5๐‘ฅ โˆ’ 7) )) โˆ’ 3) = ((7(3๐‘ฅ + 4) + 4(5๐‘ฅ โˆ’ 7))/(5๐‘ฅ โˆ’ 7))/((5(3๐‘ฅ + 4) โˆ’ 3(5๐‘ฅ โˆ’ 7))/(5๐‘ฅ โˆ’ 7)) = (7(3๐‘ฅ + 4) + 4(5๐‘ฅ โˆ’ 7))/(5(3๐‘ฅ + 4) โˆ’ 3(5๐‘ฅ โˆ’ 7)) = (21๐‘ฅ + 28 + 20๐‘ฅ โˆ’ 28)/(15๐‘ฅ + 20 โˆ’ 15๐‘ฅ + 21) = 41๐‘ฅ/41 = x Thus, gof = x = IB Finding fog f(x) = (3๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 7) f(g(x)) = (3๐‘”(๐‘ฅ) + 4)/(5๐‘”(๐‘ฅ) โˆ’ 7) = (3 ((7๐‘ฅ + 4)/(5๐‘ฅ โˆ’ 3)) + 4)/(5 (((7๐‘ฅ + 4))/((5๐‘ฅ โˆ’ 3) )) โˆ’ 7) = ((3(7๐‘ฅ + 4) + 4(5๐‘ฅ โˆ’ 3))/(5๐‘ฅ โˆ’ 3))/((5(7๐‘ฅ + 4) โˆ’ 7(5๐‘ฅ โˆ’ 3))/(5๐‘ฅ โˆ’ 3)) = (3(7๐‘ฅ + 4) + 4(5๐‘ฅ โˆ’ 3))/(5(7๐‘ฅ + 4) โˆ’ 7(5๐‘ฅ โˆ’ 3)) = (21๐‘ฅ + 12 + 20๐‘ฅ โˆ’ 12)/(35๐‘ฅ + 20 โˆ’ 35๐‘ฅ + 21) = 41๐‘ฅ/41 = x Thus, fog = x = IA Since fog = IA and gof = IB, Hence proved

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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.