(i) f : {1, 2, 3, 4} → {10} with
f= {(1, 10), (2, 10), (3, 10), (4, 10)}.
(ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}.
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}.
Solution:
(i) f : {1, 2, 3, 4} → {10} with
f= {(1, 10), (2, 10), (3, 10), (4, 10)}.
f is not one-one since 1, 2, 3, 4 have same page 10.
⇒ f has no inverse.
(ii) g: {5, 6, 7, 8} → {1,2, 3,4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}.
Here also 5 and 7 have the same image 4.
∴ g is not one-one.
Therefore, g is not invertible, i.e., It has no inverse.
(iii) h : {2,3,4,5} → {7,9,11,13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}.
Hence, f(x1) = f(x2) ⇒ x1 = x2.
i. e., each element of X = {2,3,4,5} has a unique image in Y = {7,9,11,13}.
Similarly, each member of Y has a unique pre-image in X.
⇒ f is one-one and onto.
∴ f is invertible, i.e., It has an inverse.
