Q3.A relation R on a non-empty set A is said to be an equivalence relation if
Answer : Option AExplaination / Solution: A relation R on a non empty set A is said to be reflexive if xRx for all x ∈ A . . A relation R on a non empty set A is said to be symmetric if x Ry⇔y Rx, for all x , y ∈A . A relation R on a non empty set A is said to be transitive if x Ry and y Rz⇒x Rz, for all x,y,z ∈ A. An equivalence relation satisfies all these three properties.
Q5.Let L be the set of all lines in a plane and R be the relation on L defined as R = {(L1, L2): L1 is perpendicular to L2}. Then R is
Answer : Option BExplaination / Solution: The relation R is symmetric only , because if L1 is perpendicular to L2 ,then L2 is also perpendicular to L1,but no other cases that is reflexive and transitive is not possible.
Q6.The relation R on the set Z of integers given by R = {(a, b): 2 divides a – b} ,∀ a, b ∈ Z is
Answer : Option CExplaination / Solution:
Since a – a = 0 , and 0 is divisible by 2 , therefore, aRa i.e. R is reflexive.
If aRb , then a – b is divisible by 2. ⇒ - ( a- b ) is divisible by 2. ⇒(b – a ) is divisible by 2. ⇒bRa i.e. R is symmetric. .
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If aRb and bRc , then a – b is divisible by 2 and b – c is divisible by 2 ⇒ a – b = 2q and b – c = 2q’ where q and q’ are integers. ⇒( a – b ) + ( b – c ) = 2 ( q + q’) ⇒a – c =2( q + q’) ,,but (q +q’) is an integer. ⇒(a –c ) is divisible by 2. ⇒aRc i.e. R is transitive. .
Q7.Let R be the relation defined in the set A = {1, 2, 3, 4, 5, 6, 7} by R = {(a, b) : both a and b are either odd or even}. Then R is
Answer : Option AExplaination / Solution:
Consider any a ,b , c ∈A .
Since both a and a must be either even or odd, so (a , a) ∈R ⇒R is reflexive.
Let (a ,b) ∈R ⇒ both a and b must be either even or odd, ⇒ both b and a must be either even or odd, ⇒ ( b ,a) ∈R .Thus , (a ,b) ∈R ⇒ ( b ,a) ∈R ⇒R is symmetric.
Let (a ,b) ∈R and (b ,c) ∈R ⇒ both a and b must be either even or odd, also ,both b and c must be either even or odd, ⇒ all elements a, b and c must be either even or odd, ⇒ ( a ,c) ∈R . Thus , (a ,b) ∈R ⇒ ( b ,c) ∈R ⇒ (a ,c) ∈R ⇒R is transitive.
Q8.Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then R is
Answer : Option DExplaination / Solution:
The relation R is not symmetric , (1,2) ∈R , but (2,1) ∉ R , (1,3) ∈R ,but (3,1) ∉ R , (3,2) ∈R, but (2,3) ∉R.
Q10.The function f:R→R given by f(x)= cosx∀x∈R is :
Answer : Option CExplaination / Solution:
f (0) = cos 0 = 1 ,and f (2π) =cos (2π)= 1. So , different elements in R may have the same image . Hence , f is not an injective function .Also, range of f(x) is not equal to its co-domain . So, f is not surjective.
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0