# Relation and Function

## NCERT Exemplar Problem

### Long Answer Type Part 3

Question 22: Each of the following defines a relation on N: Determine which of the above relations are reflexive, symmetric and transitive.

Solution: Hence, the given relation is only transitive. Therefore, R is not reflexive. Therefore, R is symmetric. Therefore, R is not transitive.

Thus, R is only symmetric. Therefore, R is reflexive. Therefore, it is clear that R is symmetric. Therefore, R is transitive. Therefore, R is not reflexive. Therefore, R is not symmetric.

Since, there is no element which begins with Therefore, R is a transitive.

Question 23: Let A = {1, 2, 3, ………, 9} and R be the relation in A x A defined by (a, b) R (c, b) if a + d = b + c for (a, b), (c, d) in A × A. Prove that R is an equivalence relation and also obtain the equivalent class [(2, 5)].

Solution: Therefore, R is reflexive.

Let (a, b) R (c, d) Therefore, R is symmetric.

Let (a, b) R (c, d) and (c, d) R (e, f) Therefore, R is transitive.

Thus, R is reflexive, symmetric and transitive.

Therefore, R is an equivalence relation.

Equivalence class containing {(2, 5)} is {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.

Question 24: Using the definition, prove that the function is invertible if and only if f is both one-one and onto.

Solution: By the definition of an invertible function:

A function is defined to be and invertible function, if there exists a function The function g is called the inverse of f and is denoted by f – 1. has to one-one and onto.

Therefore, f(x) should be both one-one and onto.