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Chapter 1 Class 12 Relation and Functions
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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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