Relation and Function

NCERT Solution

Exercise 1.1 Part 5

Question 10: Give an example of a relation, which is

1. Symmetric but neither reflexive nor transitive
2. Transitive but neither reflexive nor symmetric
3. Reflexive and symmetric but not transitive
4. Reflexive and transitive but not symmetric
5. Symmetric and transitive but not reflexive.

Solution:

(i) Let R = {(1, 2), (2, 1)}

(ii) Let R = {(1, 2), (2, 3), (1, 3)}

(iii) R = {(1, 1) (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}

(iv) R ={(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (, 2), (2, 3), (3, 2)}

(v) R = {(2, 3), (3, 2), (1, 2), (1, 3), (3, 1)}

Thus, R is symmetric.

Thus, R is transitive.

Question 11: Show that the relation R in the set A of points in a plane given by R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Solution: Let O is the origin

Given, R = {(P, Q): distance of the point P from the origin is same as the distance of the point Q from the origin}

Therefore, R = {(P, Q): OP = PQ}

Now, let OP = y

Thus, R is reflexive relation.

Let OP = OQ = y

Thus, R is symmetric.

Again, Let OP = OQ = y and OQ = OR = y

Thus, R is transitive.

Thus, all distance related to P from the origin is same as OP. As a circle is the locus of all points having same distance from a point, in the given case from O, therefore, the set of the points related to P is a circle passing through P with O as the centre, a fixed point.

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