If set P has m elements and set Q has n elements n P X Q has m*n elements.
m*n=10 => if m=1 n n=10,
if m=2 n n=5,
if m=5 n n=2 and if m=10 n n=1.

{(1,5), (2,5), (3,5),……}. This function involves relation {(x, y): y=constant} which is involved in constant function. So, above function is constant function.

We know, domain of a relation is set from which relation is defined i.e. set A.
So, domain = {1,2,3,4,5}.

In each ordered pair of A X B, first element belongs to set A and second element belongs to set B.
a, b, c ∈ B so, B = {a, b, c}.

We know radical cannot be negative. So, 9-x,² ≥ 0
(3-x) (3+x) ≥ 0 => (x-3) (x+3) ≤ 0 => x∈[-3,3].

If set P has m elements and set Q has n elements n P X Q has m*n elements.
Here, m=4 and n=5 refore P X Q has 4*5=20 elements.

{(-1,1), (1,1), (-2,2), (2,2), (-3,3), (3,3), ……}. This function involves relation {(x, y): y = |x|} which is involved in modulus function. So, above function is modulus function.

Range is set of elements of codomain which have ir preimage in domain.
Relation R = {(1,1), (1, -1), (4,2), (4, -2), (9,3), (9, -3)}.
Range = {-1,1, -2,2, -3,3}.

In each ordered pair of A X B, first element belongs to set A and second element belongs to set B.
1,2 ∈ A so, A = {1, 2}.

A relation from a set A to a set B is said to be a function if every element of set A has one and one image in set B.
In {(1,2), (1,4), (2,5), (3,8)}, since element 1 has two images 2 and 4 which is not possible in a function so, it is not a function. Rest all have one and only one image so y can be called a function.