Q. 4

# Let A = {1, 2, 3,

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) = x2 – 2

g = {(3, 32 – 2), (5, 52 – 2), (7, 72 – 2), (9, 92 – 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-1og-1.

We know (f-1og-1)(x) = f-1(g-1(x))

Thus, f-1og-1 : C A and

(f-1og-1)(7) = f-1(g-1(7)) = f-1(3) = 1

(f-1og-1)(23) = f-1(g-1(23)) = f-1(5) = 2

(f-1og-1)(47) = f-1(g-1(47)) = f-1(7) = 3

(f-1og-1)(79) = f-1(g-1(79)) = f-1(9) = 4

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

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

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