# Let S = {a, b, c} and T = {1, 2, 3}. Find F ^{- }^{1} of the following functions F from S to T, if it exists.

(i). F = {(a, 3), (b, 2), (c, 1)}

(ii). F = {(a, 2), (b, 1), (c, 1)}

**Solution:**

In general, a function is invertible as long as each input features a unique output. That is, every output is paired with exactly one input.

That way, when the mapping is reversed, it'll still be a function!

A monotonic function i.e. bijection function is usually invertible.

S = {a, b, c} and T = {1, 2, 3}.

(i). F : S → T is defined by

F = {(a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F (c) = 1

Therefore,

F ^{- }^{1} : T → S is given by

F ^{- 1} = {(3, a), (2, b), (1, c)}

(ii). F : S → T is defined by

F = {(a, 2), (b, 1), (c, 1)}

Since, F (b) = F (c) = 1,

F is not one-one.

Hence,

F is not invertible i.e., F ^{- 1} does not exist

NCERT Solutions for Class 12 Maths - Chapter 1 Exercise ME Question 11

## Let S = {a, b, c} and T = {1, 2, 3}. Find F ^{- }^{1} of the following functions F from S to T, if it exists. (i). F = {(a, 3), (b, 2), (c, 1)} (ii). F = {(a, 2), (b, 1), (c, 1)}

**Summary:**

For the function (i). F = {(a, 3), (b, 2), (c, 1)} , F ^{- 1} is {(3, a), (2, b), (1, c)} and for the function (ii). F = {(a, 2), (b, 1), (c, 1)}, F is not invertible i.e., F ^{- 1} does not exist

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