(i) A = {0, –1, –3, 2}; B = {–9, –3, 0, 6} & f(x) = 3x

We have f : A → B and f(x) = 3x.

⇒ f = {(0, 3×0), (–1, 3×(-1)), (–3, 3×(-3)), (2, 3×2)}

∴ f = {(0, 0), (–1, –3), (–3, –9), (2, 6)}

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {0, –1, –3, 2} are mapped to distinct elements of the co-domain {0, –3, –9, 6}.

Hence, f is one-one.

Also, each element of the range {–9, –3, 0, 6} is the image of some element of {0, –1, –3, 2}.

Hence, f is also onto.

Thus, the function f has an inverse.

We have f^-1 = {(0, 0), (–3, –1), (–9, –3), (6, 2)}

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

We have f : A → B and f(x) = x².

⇒ f = {(1, 1²), (3, 3²), (5, 5²), (7, 7²), (9, 9²)}

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

Recall that a function is invertible only when it is both one-one and onto.

Here, observe that distinct elements of the domain {1, 3, 5, 7, 9} are mapped to distinct elements of the co-domain {1, 9, 25, 49, 81}.

Hence, f is one-one.

However, the element 0 of the range {0, 1, 9, 25, 49, 81} is not the image of any element of {1, 3, 5, 7, 9}.

Hence, f is not onto.

Thus, the function f does not have an inverse.