Relations and Functions - Online Test
Q1. A relation R in a set A is called universal relation, if
Answer : Option B
Explaination / Solution:
The relation R = A x A is called Universal relation.
Q2. A relation R in a set A is called reflexive,
Answer : Option B
Explaination / Solution:
A relation R on a non empty set A is said to be reflexive if x Rx for all x ∈ R , Therefore , R is reflexive
Q3. A relation R on a non-empty set A is said to be an equivalence relation if
Answer : Option A
Explaination / 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 Ryy 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 Rzx Rz, for all x,y,z A.
An equivalence relation satisfies all these three properties.
A relation R on a non empty set A is said to be symmetric if x Ryy 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 Rzx Rz, for all x,y,z A.
An equivalence relation satisfies all these three properties.
Q4. Let A be the set of all students of a boys school. The relation R on A is given by R = {(a, b) : a is sister of b} is
Answer : Option A
Explaination / Solution:
Because , the relation is defined over the set A be the set of all students of a boys school.
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 B
Explaination / 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 C
Explaination / 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. .
- .
- 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 A
Explaination / 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 D
Explaination / 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.
Q9. Let A be a set containing n distinct elements. The number of functions that can be defined from A to A is =
Answer : Option B
Explaination / Solution:
The number of functions from a finite set A to a finite set B = .
Q10. The function f:R→R given by f(x)= cosx∀x∈R is :
Answer : Option C
Explaination / 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.
