(i) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

Function:

(i) all elements of the first set are associated with the elements of the second set.

(ii) An element of the first set has a unique image in the second set.

f = {(–1, 2), (-1, 3), (0, 4), (1, 5)}

Here, -1 is coming twice.

Hence, it does not have a unique (one) image.

∴ f is not a function

(ii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

Function:

(i) all elements of first set is associated with the elements of second set.

(ii) An element of first set has a unique image in second set.

g = {(0, 2), (1, 3), (2, 4)}

Here, -1 is not associated with any element of set B

Hence, it does not satisfy the condition of the function

∴ g is not a function.

(iii) Given: A = {-1, 0, 1, 2} and B = {2, 3, 4, 5}

Function:

(i) all elements of first set is associated with the elements of second set.

(ii) An element of first set has a unique image in second set.

h = {(-1, 2), (0, 3), (1, 4), (2, 5)}

Here, (i) all elements of set A are associated to element in set B.

(ii) an element of set A is associated to a unique element in set B.

∴ h is a function.