Solution:

Given set S = {(x, y)|x y = 5} where x, y  R

To check whether the given set is a function or not, we need to check the following conditions:

1. Each element in the domain must be mapped to exactly one element in the range.
2. If an element in the domain is mapped to more than one element in the range, then it is not a function.

Let's take some examples to check whether the given set S satisfies the above conditions or not.

Examples:

1. Let (x1, y1) and (x2, y2) be two elements in S such that x1 ≠ x2. Then we have:

x1 y1 = 5 and x2 y2 = 5
⇒ y1 = 5/x1 and y2 = 5/x2

Since x1 ≠ x2, we have y1 ≠ y2. Hence, each element in the domain is mapped to exactly one element in the range. Therefore, S is a function.

2. Let (x1, y1) and (x2, y2) be two elements in S such that x1 = x2. Then we have:

x1 y1 = 5 and x2 y2 = 5
⇒ y1 = 5/x1 and y2 = 5/x2

Since x1 = x2, we have y1 = y2. Hence, each element in the domain is mapped to exactly one element in the range. Therefore, S is a function.

Conclusion:

Hence, we can conclude that the given set S = {(x, y)|x y = 5} where x, y  R is a function.

Option (a) is incorrect as the given set is a function.

Option (b) is incorrect as the given set is not a composite function.

Option (c) is incorrect as the given set is not a one-one mapping.

Option (d) is incorrect as the given set is a function.

Assume X= 1,2,3
condition: x+y=5
therefore y =2,1,0
By mapping this ,
we can conclude that all Elements in y has a unique pre-image in X .so one -one mapping