# Example 17 - Chapter 1 Class 12 Relation and Functions (Term 1)

Last updated at Jan. 28, 2020 by

Last updated at Jan. 28, 2020 by

Transcript

Example 17 Show that if f : R β {7/5} β R β {3/5} is defined by f(x) = (3π₯ + 4)/(5π₯ β 7) and g: R β {3/5}β R β {7/5} is defined by g(x) = (7π₯ + 4)/(5π₯ β 3), then fog = IA and gof = IB, where A = R β {3/5} , B = R β {7/5} ; IA (x) = x, β x β A, IB (x) = x, β x β B are called identity functions on sets A and B, respectively. f(x) = (3π₯ + 4)/(5π₯ β 7) & g(x) = (7π₯ + 4)/(5π₯ β 3) Finding gof g(x) = (7π₯ + 4)/(5π₯ β 3) g(f(x)) = (7π(π₯) + 4)/(5π(π₯) β 3) gof = (7 ((3π₯ + 4)/(5π₯ β 7))" " + 4)/(5 (((3π₯ + 4))/((5π₯ β 7) )) β 3) = ((7(3π₯ + 4) + 4(5π₯ β 7))/(5π₯ β 7))/((5(3π₯ + 4) β 3(5π₯ β 7))/(5π₯ β 7)) = (7(3π₯ + 4) + 4(5π₯ β 7))/(5(3π₯ + 4) β 3(5π₯ β 7)) = (21π₯ + 28 + 20π₯ β 28)/(15π₯ + 20 β 15π₯ + 21) = 41π₯/41 = x Thus, gof = x = IB Finding fog f(x) = (3π₯ + 4)/(5π₯ β 7) f(g(x)) = (3π(π₯) + 4)/(5π(π₯) β 7) = (3 ((7π₯ + 4)/(5π₯ β 3)) + 4)/(5 (((7π₯ + 4))/((5π₯ β 3) )) β 7) = ((3(7π₯ + 4) + 4(5π₯ β 3))/(5π₯ β 3))/((5(7π₯ + 4) β 7(5π₯ β 3))/(5π₯ β 3)) = (3(7π₯ + 4) + 4(5π₯ β 3))/(5(7π₯ + 4) β 7(5π₯ β 3)) = (21π₯ + 12 + 20π₯ β 12)/(35π₯ + 20 β 35π₯ + 21) = 41π₯/41 = x Thus, fog = x = IA Since fog = IA and gof = IB, Hence proved

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Chapter 1 Class 12 Relation and Functions (Term 1)

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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.