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Last updated at Jan. 28, 2020 by Teachoo

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

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Example 15 Not in Syllabus - CBSE Exams 2021

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Example 17 Not in Syllabus - CBSE Exams 2021 You are here

Example 18 Important Not in Syllabus - CBSE Exams 2021

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Chapter 1 Class 12 Relation and Functions

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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 9 years. He provides courses for Maths and Science at Teachoo.