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Class 12, NCERT Chapter 1,  Exercise 1.1, Q2 Solution We have R = {(a,b): } It can be observed that ( 1 4 , 1 4 ) ∉ R Bcoz,  1 4 > ( 1 4 ) 2   R is not Reflexive. Now (1,4) ∈R as   1<16 But (4,1) ∉R as  4 ≮ 1 2 Therefore R is not symmetric. Now (3,2),(2,1.5) ∈R                                                     [As 3 < 4  and 2 < 2.25 ] But 3 > 2.25 Therefore ( 3, 1.5 ) ∉R Therefore R is not Transitive. Hence R is neither reflexive nor symmetric nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1,  Exercise 1.1, Q1  Determine whether each of the following relations are reflexive, symmetric and transitive:  (i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}  Solution Given, A={1,2,3,4,5,6,7,8,9,10,11,12,13,14} and R={(x,y): 3x-y=0} ∴ R={(1,3), (2,6), (3,9), (4,12)} R is not reflexive since (1,1), (2,2) ... (14,14) ∉  R R is not symmetric as (1,3) ∈ R, but (3,1) ∉ R   R is not transitive  as (1,3), (3,9) ∈ R, but (1,9) ∉R  Hence R is neither Reflexive, nor symmetric, nor transitive. (ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4} Solution R = {(x, y) : y = x + 5 and x < 4} = {(1,6), (2,7), (3,8)} R is not reflexive as (1,1) ∉R R is not symmetric as (1,6) ∈R, but (6,1) ∉R Since there is no pair in R such that (x,y) and (y,z) ∈R, then (x,z) cannot belong t R ∴  R is not transitive. Hence, R is neither reflex

RELATIONS AND FUNCTION Class 12, NCERT Chapter 1,  Example6  Solution Given any element a in A, both a and a must be either odd or even, so that (a, a) ∈ R.  Further, (a, b) ∈ R ⇒ both a and b must be either odd or even ⇒ (b, a) ∈ R.  Similarly, (a, b) ∈ R and (b, c) ∈ R ⇒ all elements a, b, c, must be either even or odd simultaneously ⇒ (a, c) ∈ R.  Hence, R is an equivalence relation.  Further, all the elements of {1, 3, 5, 7} are related to each other, as all the elements of this subset are odd.  Similarly, all the elements of the subset {2, 4, 6} are related to each other, as all of them, are even.  Also, no element of the subset {1, 3, 5, 7} can be related to any element of {2, 4, 6}, as elements of {1, 3, 5, 7} are odd, while elements of {2, 4, 6} are even.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1,  Example5  Solution R is reflexive, as 2 divides (a – a) for all a ∈ Z. Further, if (a, b) ∈ R, then 2 divides a – b. Therefore, 2 divides b – a. Hence, (b, a) ∈ R, which shows that R is symmetric. Similarly, if (a, b) ∈ R and (b, c) ∈ R, then a – b and b – c are divisible by 2. Now, a – c = (a – b) + (b - c) is even. So, (a-c) is divisible by 2. This shows that R is transitive. Thus, R is an equivalence relation in Z.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1,  Example4  Solution R is reflexive since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) ∈ R but (2, 1) ∉ R.  Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Theory Introduction The concept of the term ‘relation’ in mathematics has been drawn from the meaning of the relation in the English language, according to which two objects or quantities are related if there is a recognizable connection or link between the two objects or quantities If (a, b) ∈ R, we say that a is related to b under the relation R and we write as a R b. In general, (a, b) ∈ R, we do not bother whether there is a recognizable connection or link between a and b.  2. Types of Relations  In this section, we would like to study different types of relations. We know that a relation in a set A is a subset of A × A. Thus, the empty set φ and A × A are two extreme relations. For illustration, consider a relation R in the set A = {1, 2, 3, 4} given by R = {(a, b): a – b = 10}. This is the empty set, as no pair (a, b) satisfies the condition a – b = 10. Similarly, R′ = {(a, b) : | a – b | ≥ 0} is the whole se

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1,  Example 3 Solution Figure R is not reflexive, as a line L1 can not be perpendicular to itself, i.e., (L1 , L1 ) ∉ R. R is symmetric as (L1 , L2 ) ∈ R ⇒ L1 is perpendicular to L2 ⇒ L2 is perpendicular to L1 ⇒ (L2 , L1 ) ∈ R. R is not transitive. Indeed, if L1 is perpendicular to L2 and L2 is perpendicular to L3 , then L1 can never be perpendicular to L3 . In fact, L1 is parallel to L3 , i.e., (L1 , L2 ) ∈ R, (L2 , L3 ) ∈ R but (L1 , L3 ) ∉ R. Example1       Example2 Example3 ←you are here