Test: Relations & Functions- 2

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→B and S is a relation fromB→C, then S ^∘ R is a composite function from A to C.

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 =  

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.

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 = {(T₁, T₂): T₁ is congruent to T₂}. (T₁T₂) ∈ R iff T₁ is congruent to T₂. 
Reflexivity :T₁≅T₁ ⇒ (T₁T₁) ∈ R . Symmetry :(T₁, T₂)∈ R ⇒T₁≅T₂ ⇒ T₂≅T₁ ⇒ (T₂, T₁) ∈ R
Transitivity :(T₁,T₁) ∈ R and (T₂, T₃) ∈ R ⇒ T₁≅T₂ and T₂≅T₃ ⇒ T₁≅T₃ ⇒ (T₁, T₃) ∈ 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)
= x²−sin x ≠ f(x).
For an odd function : f(-x) = - f(x) , therefore, f(-x) = (− x)²+ sin (− x)
= x²− sin x
 ≠ − f(x).
Therefore f(x) is neither even nor odd.

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)}.

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}.

We have the function f(x) = sin²x + tan x, therefore, f(π+x)
= {sin²(π+x)+tan(π+x)}
= {sin²x+tanx} = f(x).
This implies that period of the function f(x) is π.

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}.

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)

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.

Since [x] ≤ x, therefore , x – [x] ≥ 0. Also, x – [x] <1,∴0⩽x−[x]<1. Therefore ,Rf = [0,1).

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 

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 = 2¹⁶

The relation R = A x A is called Universal relation.

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.

f(a+1)−f(a−1)
=(a+1)−(a+1)2−{(a−1)−(a−1)2} = 2−4a(a−1)2} = 2−4a