Question -
Answer -
(i) Given A = {0,−1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3 x.
So, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}
Here, differentelements of the domain have different images in the co-domain.
Clearly, this is one-one.
Range of f = Range of f = B
so, f is a bijection and,
Thus, f -1 exists.
Hence, f -1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}
(ii) Given A ={1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x2
So, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}
Here, differentelements of the domain have different images in the co-domain.
Clearly, f is one-one.
But this is not onto because the element 0 in the co-domain (B) has nopre-image in the domain (A)
⇒ f is not abijection.
So, f -1does not exist.