Check sibling questions

Example 17 - Show that f(x) = 3x+4/5x-7, g(x) = 7x+4/5x-3

Example 17 - Chapter 1 Class 12 Relation and Functions - Part 2
Example 17 - Chapter 1 Class 12 Relation and Functions - Part 3

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Transcript

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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.