State with reason whether following functions have inverse (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)}

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(i) f: {1, 2, 3, 4} → {10}defined as: f = {(1, 10), (2, 10), (3, 10), (4, 10)} From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10 ∴f is not one-one. Hence, function f does not have an inverse. (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4}...
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(i) f: {1, 2, 3, 4} → {10}defined as: f = {(1, 10), (2, 10), (3, 10), (4, 10)} From the given definition of f, we can see that f is a many one function as: f(1) = f(2) = f(3) = f(4) = 10 ∴f is not one-one. Hence, function f does not have an inverse. (ii) g: {5, 6, 7, 8} → {1, 2, 3, 4} defined as: g = {(5, 4), (6, 3), (7, 4), (8, 2)} From the given definition of g, it is seen that g is a many one function as: g(5) = g(7) = 4. ∴g is not one-one, Hence, function g does not have an inverse. (iii) h: {2, 3, 4, 5} → {7, 9, 11, 13} defined as: h = {(2, 7), (3, 9), (4, 11), (5, 13)} It is seen that all distinct elements of the set {2, 3, 4, 5} have distinct images under h. ∴Function h is one-one. Also, h is onto since for every element y of the set {7, 9, 11, 13}, there exists an element x in the set {2, 3, 4, 5}such that h(x) = y. Thus, h is a one-one and onto function. Hence, h has an inverse. read less
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