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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