1.

Find f−1 if it exists: f: A → B, where (i) A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3x.(ii) A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Answer»

(i) Given as A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3x.

Therefore, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, the different elements of the domain have different images in the co-domain.

Thus, this is one-one.

Range of f = Range of f = B

Therefore, f is a bijection and,

Clearly, f -1 exists.  

Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given as A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2

Therefore, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, the different elements of the domain have different images in the co-domain.

Thus, f is one-one.

This is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

Therefore, f -1does not exist.


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