1 Crore+ students have signed up on EduRev. Have you? Download the App 
R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3)} be a relation on A = {1, 2, 3}, then R is
A relation R on a non empty set A is said to be reflexive if fx Rx for all x ∈ R, Therefore , R is not reflexive.
A relation R on a non empty set A is said to be symmetric if fx Ry ⇔ yRx, for all x, y ∈ R Therefore, R is not symmetric.
A relation R on a non empty set A is said to be antisymmetric if fx Ry and y Rx ⇒ x = y, for all x, y ∈ R. Therefore, R is not antisymmetric.
If R is a relation from a set A to a set B and S is a relation from B to C, then the relation S ^{∘} R.
If R is a relation from A→B and S is a relation fromB→C, then S ^{∘ }R is a composite function from A to C.
Let A be the set of all students of a boys school. The relation R in A given by R = {(a, b) : a is sister of b} is
Because, the relation is defined over the set A which is the set of all students of a boys school.
y is defined If x ≥ 0, i.e.
By definition, The Signum function =
Let A = {1, 2, 3}. Which of the following is not an equivalence relation on A ?
A relation R on a non empty set A is said to be reflexive iff xRx for all x ∈ R . A relation R on a non empty set A is said to be symmetric iff xRy ⇔ yRx, for all x , y ∈ R .
A relation R on a non empty set A is said to be transitive iff xRy and yRz ⇒ xRz, for all x ∈ R. An equivalence relation satisfies all these three properties.
None of the given relations satisfies all three properties of equivalence relation.
The binary operation * defined on the set of integers as a∗b = a−b1 is:
Here * is commutative as b*a = b−a−1 = a−b−1 = a∗b.
Because ,−x = x for all x ∈ R.
Let T be the set of all triangles in a plane with R a relation in T given by R = {(T_{1}, T_{2}) : T_{1} is congruent to T_{2}}. Then R is
Let T be the set of all triangles in a plane with R a relation in T given by R = {(T_{1}, T_{2}): T_{1} is congruent to T_{2}}. (T_{1}T_{2}) ∈ R iff T_{1 }is congruent to T_{2}.
Reflexivity :T_{1}≅T_{1} ⇒ (T_{1}T_{1}) ∈ R . Symmetry :(T_{1}, T_{2})∈ R ⇒T_{1}≅T_{2} ⇒ T_{2}≅T_{1} ⇒ (T_{2}, T_{1}) ∈ R
Transitivity :(T_{1},T_{1}) ∈ R and (T_{2}, T_{3}) ∈ R ⇒ T_{1}≅T_{2} and T_{2}≅T_{3} ⇒ T_{1}≅T_{3} ⇒ (T_{1}, T_{3}) ∈ R .
Therefore, R is an equivalence relation on T.
The given function is defined as Signum function. i.e.
domain is R and Range is {1 , 0 , 1}.
For even function : f(x) = f(x) , therefore, f( x) = (− x)^{2}+ sin (− x)
= x^{2}−sin x ≠ f(x).
For an odd function : f(x) =  f(x) , therefore, f(x) = (− x)^{2}+ sin (− x)
= x^{2}− sin x
≠ − f(x).
Therefore f(x) is neither even nor odd.
If A is a finite set containing n distinct elements, then the number of relations on A is equal to
R is a relation from { 11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation R^{−1}
R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3. The relation is given by x = y + 3,from{8, 10, 12} to {11, 12, 13} ⇒ relation = {(8,11),(10,13)}.
If A = {(1, 2, 3}, then the relation R = {(2, 3)} in A is
A = {1,2,3}
B = {(2,3)} is not reflexive or symmetric on A but it is transitive
∵ if (a,b) exists but (b,c) does not exist then (a,c) does not need to exist and the relation is still transitive.
The only possible integral values of sin x are {1 ,0, 1}.
The period of the function f(x) = sin^{2}x + tan x is
We have the function f(x) = sin^{2}x + tan x, therefore, f(π+x)
= {sin^{2}(π+x)+tan(π+x)}
= {sin^{2}x+tanx} = f(x).
This implies that period of the function f(x) is π.
Let A = {1, 2, 3}, then the domain of the relation R = {(1, 1), (2, 3), (2, 1)} defined on A is
Since the domain is represented by the x co ordinate of the ordered pair (x , y).Therefore, domain of the given relation is {1, 2}.
Given the relation R = {(1, 2), (2, 3)} on eht set {1, 2, 3}, the minimum number of ordered pairs which when added to R make it an equivalence
To make the relation an equivalence relation , the following ordered pairs are required (1,1),(2,2),(3,3)(2,1)(3,2)(1,3),(3,1)
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relations on the set A = {1, 2, 3, 4}. the relation R is
Given, R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}.
(a) Since, (1, 1), (2, 2), (3, 3), (4, 4) ∉ R. So, R is not reflexive.
(b) Since, (1, 3) ∈ R and (3, 1) ∈ R but (1, 1) ∉ R. So, R is not transitive.
(c) Since, (2, 3) ∈ R but (3,2) ∉ R. So, B is not symmetric.
(d) Since, (2, 4) ∈ R and (2, 3) ∈ R. So, R is not a function.
The relation R in the set Z of integers given by R = {(a, b): 2 divides a – b}(a, b) where a, b ∈ Z is
Since [x] ≤ x, therefore , x – [x] ≥ 0. Also, x – [x] <1,∴0⩽x−[x]<1. Therefore ,Rf = [0,1).
Let A contain n distinct numbers. How many bijections from A to A can be defined?
A bijection from A to A is infact an arrangement of its n elements, taken all at a time, which can be done in ways. Hence, the number of bijections from A to A is
Number of relations that can be defined on the set A = {a, b, c, d} is
No. of elements in the set A = 4. Therefore , the no. of elements in A × A = 4 × 4 = 16. As, the no. of relations in A × A = no. of subsets of A × A = 2^{16}
A relation R in a set A is called universal relation, if
The relation R = A x A is called Universal relation.
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
Consider any a ,b , c ∈A .
1) Since both a and a must be either even or odd, so (a , a) ∈R ⇒ R is reflexive.
2) 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.
3) 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.
If f(x) = x – x^{2}, then f(a + 1) – f(a – 1) , a ∈ R is :
f(a+1)−f(a−1)
=(a+1)−(a+1)2−{(a−1)−(a−1)2} = 2−4a(a−1)2} = 2−4a
209 videos218 docs139 tests

209 videos218 docs139 tests
