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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Question 1 Show that if f : R – {7/5} → R – {3/5} is defined by f(x) = (3š‘„ + 4)/(5š‘„ āˆ’ 7) and g: R āˆ’ {3/5}→ R – {7/5} is defined by g(x) = (7š‘„ + 4)/(5š‘„ āˆ’ 3), then fog = IA and gof = IB, where A = R āˆ’ {3/5} , B = R – {7/5} ; IA (x) = x, āˆ€ x ∈ A, IB (x) = x, āˆ€ x ∈ B are called identity functions on sets A and B, respectively. f(x) = (3š‘„ + 4)/(5š‘„ āˆ’ 7) & g(x) = (7š‘„ + 4)/(5š‘„ āˆ’ 3) Finding gof g(x) = (7š‘„ + 4)/(5š‘„ āˆ’ 3) g(f(x)) = (7š‘“(š‘„) + 4)/(5š‘“(š‘„) āˆ’ 3) gof = (7 ((3š‘„ + 4)/(5š‘„ āˆ’ 7))" " + 4)/(5 (((3š‘„ + 4))/((5š‘„ āˆ’ 7) )) āˆ’ 3) = ((7(3š‘„ + 4) + 4(5š‘„ āˆ’ 7))/(5š‘„ āˆ’ 7))/((5(3š‘„ + 4) āˆ’ 3(5š‘„ āˆ’ 7))/(5š‘„ āˆ’ 7)) = (7(3š‘„ + 4) + 4(5š‘„ āˆ’ 7))/(5(3š‘„ + 4) āˆ’ 3(5š‘„ āˆ’ 7)) = (21š‘„ + 28 + 20š‘„ āˆ’ 28)/(15š‘„ + 20 āˆ’ 15š‘„ + 21) = 41š‘„/41 = x Thus, gof = x = IB Finding fog f(x) = (3š‘„ + 4)/(5š‘„ āˆ’ 7) f(g(x)) = (3š‘”(š‘„) + 4)/(5š‘”(š‘„) āˆ’ 7) = (3 ((7š‘„ + 4)/(5š‘„ āˆ’ 3)) + 4)/(5 (((7š‘„ + 4))/((5š‘„ āˆ’ 3) )) āˆ’ 7) = ((3(7š‘„ + 4) + 4(5š‘„ āˆ’ 3))/(5š‘„ āˆ’ 3))/((5(7š‘„ + 4) āˆ’ 7(5š‘„ āˆ’ 3))/(5š‘„ āˆ’ 3)) = (3(7š‘„ + 4) + 4(5š‘„ āˆ’ 3))/(5(7š‘„ + 4) āˆ’ 7(5š‘„ āˆ’ 3)) = (21š‘„ + 12 + 20š‘„ āˆ’ 12)/(35š‘„ + 20 āˆ’ 35š‘„ + 21) = 41š‘„/41 = x Thus, fog = x = IA Since fog = IA and gof = IB, Hence proved

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