Ex 1.1

Ex 1.1, 1 (i)

Ex 1.1, 1 (ii)

Ex 1.1, 1 (iii) Important

Ex 1.1, 1 (iv)

Ex 1.1, 1 (v)

Ex 1.1, 2

Ex 1.1, 3

Ex 1.1, 4

Ex 1.1, 5 Important

Ex 1.1, 6

Ex 1.1, 7

Ex 1.1, 8 You are here

Ex 1.1, 9 (i)

Ex 1.1, 9 (ii)

Ex 1.1, 10 (i)

Ex 1.1, 10 (ii)

Ex 1.1, 10 (iii) Important

Ex 1.1, 10 (iv)

Ex 1.1, 10 (v)

Ex 1.1, 11

Ex 1.1, 12 Important

Ex 1.1, 13

Ex 1.1, 14

Ex 1.1, 15 (MCQ) Important

Ex 1.1, 16 (MCQ)

Chapter 1 Class 12 Relation and Functions

Serial order wise

Last updated at Jan. 28, 2020 by Teachoo

Ex 1.1, 8 (Introduction) Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a,b):|a – b| is even} , is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. Modulus function |1| = 1 |2| = 2 |0| = 0 |−1| = 1 |−3| = 3 Ex 1.1, 8 Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b):|a – b| is even} , is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. R = {(a, b):|a – b| is even} where a, b ∈ A Check Reflexive Since |a – a| = |0| = 0 & 0 is always even ⇒ |a – a| is even ∴ (a, a) ∈ R, ∴ R is reflexive. Check symmetric We know that |a – b| = |b – a| Hence, if |a – b| is even, then |b – a| is also even Hence, If (a, b) ∈ R, then (b, a) ∈ R ∴ R is symmetric Check transitive If |a – b| is even , then (a – b) is even Similarly, if |b – c| is even , then (b – c) is even |a – b| = |–(b – a)| = |b – a| Now, Sum of even numbers is also even a – b + b – c is even ⇒ a – c is even Hence, |a – c| is even So, If |a – b| &|b – c| is even , then |a – c| is even i.e. If (a, b) ∈ R & (b, c) ∈ R , then (a, c) ∈ R ∴ R is transitive Since R is reflexive, symmetric and transitive, it is equivalence relation R = {(a,b):|a – b| is even} Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}. In {1, 3, 5}, All elements are odd, So, difference between any 2 numbers is always even Hence, Modulus of difference between any 2 numbers is always even Hence, element of {1, 3, 5} are related to each other In {2, 4}, All elements are even, So, difference between any 2 numbers is always even Hence, Modulus of difference between any 2 numbers is always even Hence, element of {2, 4} are related to each other In {1, 3, 5} & {2, 4}, Elements of {1, 3, 5} are odd Elements of {2, 4} are even Difference of one element from {1, 3, 5} and one element from {2, 4} is odd As Difference of even and odd number is always odd ∴ Difference is not even If difference not even, Modulus of difference also not even Hence, element of {1, 3, 5} & {2, 4} are not related to each other