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  • Let X = {1, 2, 3} and Y = {4, 5}. Find whether the following subsets of X × Y are functions from X to Y or not. (i) f = {(1, 4), (1, 5), (2, 4), (3, 5)} (ii) g = {(1, 4), (2, 4), (3, 4)} (iii) h = {(1,4), (2, 5), (3, 5)} (iv) k = {(1,4), (2, 5)}.

Let X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} . Find whether the following subsets of X \times Y are functions from X to Y or not.

(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \}  

(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\

(iii) h = \{ (1,4), (2, 5), (3, 5) \}

(iv) k = \{ (1,4), (2, 5) \} .\\

Answers (1)

Here, X = \{ 1, 2, 3 \} and Y = \{ 4, 5 \} \\

Therefore, X \times Y = \{ (1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5) \} \\

(i) f = \{ (1, 4), (1, 5), (2, 4), (3, 5) \} \\

Here, f(1) = 4 and again f(1) = 5.

Therefore, f is not a function here.

As a result, there is no unique of pre- image ‘1’.

(ii) g = \{ (1, 4), (2, 4), (3, 4) \} \\

We can clearly see that g is a function. Here in g, each element of the given domain has a unique image at the given range

(iii) h = \{ (1,4), (2, 5), (3, 5) \} \\

It’s clear that h is a function of each pre-image that has a unique image.

Again, h(2) = h(3) = 5

Therefore, the function h is also many-one.

(iv) k = \{ (1, 4), (2, 5) \} \\

Here, ‘3’ does not have any image under the mapping. Therefore, k is not a function.

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