Q. 4

# Let A = {1, 2, 3, 4}; B = {3, 5, 7, 9}; C = {7, 23, 47, 79} and f : A → B, g : B → C be defined as f(x) = 2x + 1 and g(x) = x^{2} – 2. Express (gof)^{-1} and f^{-1}og^{-1} as the sets of ordered pairs and verify (gof)^{-1} = f^{-1}og^{-1}.

Answer :

We have f : A → B & f(x) = 2x + 1

⇒ f = {(1, 2×1 + 1), (2, 2×2 + 1), (3, 2×3 + 1), (4, 2×4 + 1)}

∴ f = {(1, 3), (2, 5), (3, 7), (4, 9)}

Function f is clearly one-one and onto.

Thus, f^{-1} exists and f^{-1} = {(3, 1), (5, 2), (7, 3), (9, 4)}

We have g : B → C & g(x) = x^{2} – 2

⇒ g = {(3, 3^{2} – 2), (5, 5^{2} – 2), (7, 7^{2} – 2), (9, 9^{2} – 2)}

∴ g = {(3, 7), (5, 23), (7, 47), (9, 79)}

Function g is clearly one-one and onto.

Thus, g^{-1} exists and g^{-1} = {(7, 3), (23, 5), (47, 5), (79, 9)}

We know (gof)(x) = g(f(x))

Thus, gof : A → C and

(gof)(1) = g(f(1)) = g(3) = 7

(gof)(2) = g(f(2)) = g(5) = 23

(gof)(3) = g(f(3)) = g(7) = 47

(gof)(4) = g(f(4)) = g(9) = 79

⇒ gof = {(1, 7), (2, 23), (3, 47), (4, 79)}

Clearly, gof is also both one-one and onto.

Thus, the function gof has an inverse.

We have (gof)^{-1} = {(7, 1), (23, 2), (47, 3), (79, 4)}

Now, let us consider f^{-1}og^{-1}.

We know (f^{-1}og^{-1})(x) = f^{-1}(g^{-1}(x))

Thus, f^{-1}og^{-1} : C → A and

(f^{-1}og^{-1})(7) = f^{-1}(g^{-1}(7)) = f^{-1}(3) = 1

(f^{-1}og^{-1})(23) = f^{-1}(g^{-1}(23)) = f^{-1}(5) = 2

(f^{-1}og^{-1})(47) = f^{-1}(g^{-1}(47)) = f^{-1}(7) = 3

(f^{-1}og^{-1})(79) = f^{-1}(g^{-1}(79)) = f^{-1}(9) = 4

⇒ f^{-1}og^{-1} = {(7, 1), (23, 2), (47, 3), (79, 4)}

Therefore, we have (gof)^{-1} = f^{-1}og^{-1}.

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