Explore topic-wise InterviewSolutions in .

Given: A and B two sets are given. 

Need to prove: (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) 

Let us consider, (x, y)∈(A × B) ∩ (B × A) 

⇒ (x, y)∈(A × B) and (x, y)∈(B × A) 

⇒ (x∈A and y∈B) and (x∈B and y∈A)

⇒ (x∈A and x∈B) and (y∈B and y∈A) 

⇒ x∈(A ×B) and y∈(B × A)

⇒ (x, y)∈(A × B) ∩ (B × A) 

From this, we can conclude that, 

⇒ (A × B) ∩ (B × A) ⊆ (A ∩ B) × (B ∩ A)..... (1) 

Let us consider again, (a, b)∈(A ∩ B) × (B ∩ A) 

⇒ a∈(A ∩ B) and b∈(B ∩ A) 

⇒ (a∈A and a∈B) and (b∈B and b∈A) 

⇒ (a∈A and b∈B) and (a∈B and b∈A) 

⇒ (a, b)∈(A × B) and (a, b)∈(B × A) 

⇒ (a, b)∈(A × B) ∩ (B × A) 

From this, we can conclude that, 

⇒ (A ∩ B) × (B ∩ A) ⊆ (A × B) ∩ (B × A) ...... (2) 

Now by the definition of set we can say that, from (1) and (2), 

(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) [Proved]

It is given that

R = {(a, b): a, b ∈ Z and (a – b) is divisible by 5}

We know that

(a, b) ∈ R where a – b is divisible by 5.

(i) Reflexive-

We know that (a, a) ∈ R as a – a = 0 divisible by 5.

Therefore, R is reflexive.

(ii) Symmetric-

We know that (a, b) ∈ R as a – b is divisible by 5.

So we get

a – b = 5k and b – a = – 5k

Hence, b – a is also divisible by 5 and (b, a) ∈ R

Therefore, R is symmetric.

(iii) Transitive-

We know that if (a, b) and (b, c) ∈ R as a – b is divisible by 5

So we get

a – b = 5m and b – c is divisible by 5 where b – c = 5n

By adding we get

a – c = 5 (m + n)

Here, a – c is divisible by 5 and (a, c) ∈ R which is transitive.

R is reflexive, symmetric and transitive and therefore is an equivalence relation.

R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3}. 

(i) Since (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive. 

(ii) (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric. 

(iii) (1, 2) ∈ R, (2, 3) ∈ R and (1, 3) ∈R ⇒ R is transitive. 

∴ Option (a) is the right answer.

An equivalence relation is one which possesses the properties of reflexivity, symmetry and transitivity.

(i) Reflexivity: A relation R on A is said to be reflexive if (a, a) є R for all a є A.

(ii) Symmetry: A relation R on A is said to be symmetrical if (a,b) є R è(b, a) є R for all (a, b) є A. 

(iii) Transitivity: A relation R on A is said to be transitive if (a, b) є R and (b, c) є R è (a, c) є R for all (a, b, c) є A.

Let S be a set of all triangles in a plane. 

(i) Since every triangle is similar to itself, it is reflexive.

(ii) If one triangle is similar to another triangle, it implies that the other triangle is also similar to the first triangle. Hence, it is symmetric. 

(iii) If one triangle is similar to a triangle and another triangle is also similar to that triangle, all the three triangles are similar. Hence, it is transitive.

It is given that

R = {(a, b): a, b ∈ Z and (a + b) is even}

Reflexive:

If a ∈ Z then a + a = 2a, which is even

So we get (a, a) ∈ R

Therefore, it is reflexive.

Symmetric:

Consider (a, b) ∈ Z then a + b is even

The same way b + a is also even.

So we know that (b, a) also belongs to R.

We get (a, b) ∈ R and (b, a) ∈ R

Therefore, R is symmetric.

Transitive:

Consider (a, b) and (b, c) ∈ R then a + b = 2k and b + c = 2r is even

By adding them

a + 2b + c = 2k + r

We get

a + c = 2 (k + r – b)

So a + c is even (a, c) ∈ R

Therefore, R is transitive.

We know that R is reflexive, symmetric and transitive.

Therefore, R is equivalence.

R = {(0, 0), (1, – 1), (2, – 2), (3, – 3) ...} = {(x, y) : y = – x, x ∈ W} 

Domain = {0, 1, 2, 3, ....} = W, Range = {...,– 3, – 2, – 1, 0}

Reflectivity:

Consider ∆ as an arbitrary element on A

We know that

∆ – ∆ => (∆, ∆) ∈ R Ɐ ∆ ∈ R

Hence, R is reflective.

Symmetric:

Consider ∆₁ and ∆₂ ∈ A where (∆₁, ∆₂) ∈ R

We get (∆₁, ∆₂) ∈ R => ∆₁ ~ ∆₂

So ∆₁ ~ ∆₂ ∈ R => (∆₁, ∆₂) ∈ R

Hence, R is symmetric.

Transitivity:

Consider ∆₁, ∆₂ and ∆₃ ∈ A where (∆₁, ∆₂) and (∆₂, ∆₃) ∈ R

We get

(∆₁, ∆₂) ∈ R => ∆₁ ~ ∆₂

(∆₂, ∆₃) ∈ R => ∆₂ ~ ∆₃

It can be written as

(∆₁, ∆₃) ∈ R => ∆₁ ~ ∆₃

Hence, R is transitive.

Given, 

R = {(x, x³) : x is a prime number less than 10} 

Prime numbers less than 10 are 2, 3, 5 and 7 

∴ R = {(2,2³), (3,3³), (5,5³), (7,7³)} 

⇒ R = {(2, 8), (3, 27), (5, 125), (7, 343)} 

So, 

Domain of relation R = {2, 3, 5, 7} 

Range of relation R = {8, 27, 125, 343}

Given, 

R= {a, b) : a N, a < 5, b = 4} 

Natural numbers less than 5 are 1, 2, 3 and 4 

Therefore,

a {1, 2, 3, 4} and b {4} 

⇒ R = {(1, 4), (2, 4), (3, 4), (4, 4)} 

So, 

Domain of relation R = {1, 2, 3, 4} 

Range of relation R = {4}

The given relation stated in words is 

R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

A relation R in a set A is a subset of A × A.

Thus, R is a relation in a set A R ⊆ A × A.

If (a, b) ∈ R then we say that a is related to b and write, a R b.

If (a, b) ∉ R then we say that a is not related to b and write, a Ꞧ b.

Consider R be a relation in a set A. Then, the set of all first coordinates of elements of R is called the domain of R, written as dom (R) and the set of all second coordinates of R is called the range of R, written as range (R).

For example:

R = {(-1, 1), (1, 1), (-2, 4), (2, 4)}

dom (R) = {-1, 1, -2, 2}

range (R) = {1, 4}

Given as A = {2, 3}, B = {4, 5} and C = {5, 6}

Let us find the A x (B ∪ C) and (A x B) ∪ (A x C)

(B ∪ C) = {4, 5, 6}

A × (B ∪ C) = {2, 3} × {4, 5, 6}

= {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(A × B) = {2, 3} × {4, 5}

= {(2, 4), (2, 5), (3, 4), (3, 5)}

(A × C) = {2, 3} × {5, 6}

= {(2, 5), (2, 6), (3, 5), (3, 6)}

Thus, (A × B) ∪ (A × C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

A × (B ∪ C) = {(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

Given as

A = {1, 2, 3}, B = {4} and C = {5}

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

Let us consider the LHS  (B ∪ C)

(B ∪ C) = {4, 5}
A × (B ∪ C) = {1, 2, 3} × {4, 5}

= {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}

Then, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) ∪ (A × C) = {(1, 4), (2, 4), (3, 4), (1, 5), (2, 5), (3, 5)}

∴ LHS = RHS

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

Let us consider the LHS: (B ∩ C)

(B ∩ C) = ∅ (Here, no common element)

A × (B ∩ C) = {1, 2, 3} × ∅

= ∅

Then, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) ∩ (A × C) = ∅

∴ LHS = RHS

(iii) A × (B − C) = (A × B) − (A × C)

Let us consider the LHS: (B − C)
(B − C) = ∅

A × (B − C) = {1, 2, 3} × ∅

= ∅

Then, RHS

(A × B) = {1, 2, 3} × {4}

= {(1, 4), (2, 4), (3, 4)}

(A × C) = {1, 2, 3} × {5}

= {(1, 5), (2, 5), (3, 5)}

(A × B) − (A × C) = ∅

Thus, LHS = RHS

(i) Domain = {–3, –1, 1, 3}, Range = {0, 1} 

(ii) Domain = {x : x is a multiple of 3} = {3n : n ∈ Z} 

Range = {y : y is a multiple of 5} = {5n : n ∈ Z} 

(iii) Relation = {(x, x²) : x is a prime number less than 15} 

= {(2, 4), (3, 9), (5, 25), (7, 49), (11, 121), (13, 169)} 

Domain = {2, 3, 5, 7, 11, 13}, Range = {4, 9, 25, 49, 121, 169}

(i) Given: 

A = {x ϵ W : x < 2} 

Here, W denotes the set of whole numbers (non – negative integers). 

∴ A = {0, 1} 

[∵ It is given that x < 2 and the whole numbers which are less than 2 are 0 & 1] 

B = {x ϵ N : 1 < x ≤ 4} 

Here, N denotes the set of natural numbers. 

∴ B = {2, 3, 4} 

[∵ It is given that the value of x is greater than 1 and less than or equal to 4] and C = {3, 5} 

L. H. S = A × (B ⋃ C) 

By the definition of the union of two sets, (B ⋃ C) = {2, 3, 4, 5} 

= {0, 1} × {2, 3, 4, 5} 

Now, by the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q}

= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} 

R. H. S = (A × B) ⋃ (A × C) 

Now, A × B = {0, 1} × {2, 3, 4} 

= {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)} 

and A × C = {0, 1} × {3, 5} 

= {(0, 3), (0, 5), (1, 3), (1, 5)} 

Now, we have to find (A × B) ⋃ (A × C) 

So, by the definition of the union of two sets, 

(A × B) ⋃ (A × C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} 

= L. H. S 

∴ L. H. S = R. H. S is verified

(ii) Given: 

A = {x ϵ W : x < 2} 

Here, W denotes the set of whole numbers (non – negative integers). 

∴ A = {0, 1} 

[∵ It is given that x < 2 and the whole numbers which are less than 2 are 0, 1] 

B = {x ϵ N : 1 < x ≤ 4} 

Here, N denotes the set of natural numbers. 

∴ B = {2, 3, 4} 

[∵ It is given that the value of x is greater than 1 and less than or equal to 4] and C = {3, 5} 

L. H. S = A × (B ⋂ C) 

By the definition of the intersection of two sets, (B ⋂ C) 

= {3} = {0, 1} × {3} 

Now, by the definition of the Cartesian product, 

Given two non – empty sets P and Q. The Cartesian product P × Q is the set of all ordered pairs of elements from P and Q, .i.e. 

P × Q = {(p, q) : p Є P, q Є Q}

= {(0, 3), (1, 3)} 

R. H. S = (A × B) ⋂ (A × C) 

Now, A × B = {0, 1} × {2, 3, 4} 

= {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)} 

and A × C = {0, 1} × {3, 5} 

= {(0, 3), (0, 5), (1, 3), (1, 5)} 

Now, we have to find (A × B) ⋂ (A × C) 

So, by the definition of the intersection of two sets, 

(A × B) ⋂ (A × C) = {(0, 3), (1, 3)} 

= L. H. S 

∴ L. H. S = R. H. S is verified

Given as

A = {3, 5} and B = {7, 11}

R = {(a, b): a ∈ A, b ∈ B, a-b is odd}

Putting a = 3 and b = 7,

a – b = 3 – 7 = -4 which is not odd

Putting a = 3 and b = 11,

 a – b = 3 – 11 = -8 which is not odd

Putting a = 5 and b = 7:

a – b = 5 – 7 = -2 which is not odd

Putting a = 5 and b = 11:

a – b = 5 – 11 = -6 which is not odd

∴ R = { } = Φ

R is an empty relation from A into B.

Thus proved.

By using roster form

R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(i) Non reflexive

We know that

(1, 1) ∈ R where 1 ∈ A

Hence, R is non reflexive.

(ii) Non symmetric

We know that

(1, 2) ∈ R but (2, 1) ∉ R

Hence, R is non symmetric.

(iii) Non transitive

We know that

(1, 2) ∈ R and (2, 3) ∈ R

In the same way (1, 3) ∉ R

Hence, R is non transitive.

Given: A = {1, 2} and B = {2, 3} 

Need to write: All possible subsets of A × B 

A = {1, 2} and B = {2, 3} 

So, all the possible subsets of A × B are: 

(A × B) = {(x, y): x∈A and y∈B} 

= {(1, 2), (1, 3), (2,2), (2,3)}

Given : n (A) = 3 

n (B) = 2 

To prove : The sets A x B and B x A have an element in common if the sets A and B be two sets such that n (A) = 3 and n (B) = 2 

Proof : 

Case 1 : No elements are common 

Assuming: 

A = (a, b, c) and B = (e, f) 

So, we have : 

A × B = {(a, e), (a, f), (b, e), (b, f), (c, e), (c, f)} 

B × A = {(e, a), (e, b), (e, c), (f, a), (f, b), (f, c)} 

There are no common ordered pair in A × B and B × A. 

Case 2 : One element is common 

Assuming : 

A = (a, b, c) and B = (a, f) 

So, we have : 

A × B = {(a, a), (a, f), (b, a), (b, f), (c, a), (c, f)} 

B × A = {(a, a), (a, b), (a, c), (f, a), (f, b), (f, c)} 

Here, 

A × B and B × A have one ordered pair in common. 

Therefore, 

We can say that A × B and B × A will have elements in common if and only if sets A and B have an element in common.

Given as

A = {1, 2}, B = {3, 4}

n (A) = 2 (The number of elements in set A).

n (B) = 2 (The number of elements in set B).

As we know,

n (A × B) = n (A) × n (B)

= 2 × 2

= 4 [Since, n(x) = a, n(y) = b. The total number of relations = 2^ab]

Thus, number of relations from A to B are 2⁴ = 16.

(i) Reflexivity

Consider a as an arbitrary element on the set S

So we get a ≤ a where (a, a) ∈ R

Hence, R is reflective.

(ii) Transitivity

Consider a, b and c ∈ S where (a, b) and (b, c) ∈ S

We get

(a, b) ∈ R => a ≤ b and (b, c) ∈ R => b ≤ c

Based on the above equation we get

(a, c) ∈ R => a ≤ c

Hence, R is transitive.

(iii) Non symmetry

We know that

(5, 6) ∈ R => 5 ≤ 6

In the same way

(6, 5) ∈ R => 6 ≰ 5

Hence, R is non symmetric.

Consider O as the origin of the plane

So R = {(P, Q): OP = OQ)

By considering properties of relation R

Symmetric:

Consider P and Q as the two points in set A where (P, Q) ∈ R

We can write it as

OP = OQ where (Q, P) ∈ R

So we get (P, Q) ∈ R and (Q, P) ∈ R for P, Q ∈ A

Hence, R is symmetric.

Reflexivity:

Consider P as any point in set A where OP = OP

We know that (P, P) ∈ R for all P ∈ A

Hence, R is reflexive.

Transitivity:

Consider P, Q and S as three points in a set A where (P, Q) ∈ R and (Q, S) ∈ R

We know that OP = OQ and OQ = OS

So we get OP = OS where (P, S) ∈ R

Hence, R is transitive

Therefore, R is an equivalence relation.

Consider P as a fixed point in set A and let Q be a point in set A where (P, Q) ∈ R

We know that OP = OQ where Q moves in the plane that its distance from O.

So we get O (0, 0) = OP

So the locus of Q is a circle having centre at O and OP as the radius

Therefore, the set of all points which is related to P passes through the point P having O as centre.

By substituting values we get

R = {(1, 6), (2, 7), (3, 8)}

So we get

dom (R) = {1, 2, 3}

range (R) = {6, 7, 8}

(i) Given as

R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}

∴ R = {(0, 0 + 5), (1, 1 + 5), (2, 2 + 5), (3, 3 + 5), (4, 4 + 5), (5, 5 + 5)}

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

Therefore,

The domain of relation R = {0, 1, 2, 3, 4, 5}

The range of relation R = {5, 6, 7, 8, 9, 10}

(ii) Given as

R = {(x, x³): x is a prime number less than 10}

Since, the prime numbers less than 10 are 2, 3, 5 and 7

∴ R = {(2, 2³), (3, 3³), (5, 5³), (7, 7³)}

R = {(2, 8), (3, 27), (5, 125), (7, 343)}

Therefore,

The domain of relation R = {2, 3, 5, 7}

The range of relation R = {8, 27, 125, 343}

Here, the total number of relations that can be defined from the set A to the set B is the number of possible subsets of A × B. When n (A) = p and n (B) = q, then n (A × B) = pq.

There, the total number of relations is 2^pq.

Then,

A × A = {(a, a), (a, b), (b, a), (b, b)}

The total number of relations are all possible subsets of A × A:

[{(a, a), (a, b), (b, a), (b, b)}, {(a, a), (a, b)}, {(a, a), (b, a)},{(a, a), (b, b)}, {(a, b), (b, a)}, {(a, b), (b, b)}, {(b, a), (b, b)}, {(a, a), (a, b), (b, a)}, {(a, b), (b, a), (b, b)}, {(a, a), (b, a), (b, b)}, {(a, a), (a, b), (b, b)}, {(a, a), (a, b), (b, a), (b, b)}]

n (A) = 2 ⇒ n (A × A) = 2 × 2 = 4

Thus, the total number of relations = 2⁴ = 16

Given : A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8} 

(i) To prove : A × C ⊂ B × D 

LHS : A × C = {(1, 5), (1, 6), (2, 5), (2, 6)} 

RHS : B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)} 

Since, all elements of A × C is in B × D. 

∴We can A × C ⊂ B × D 

(ii) To prove : A × (B ∩ C) = (A × B) ∩ (A × C) 

LHS : (B ∩ C) = ∅ 

A × (B ∩ C) = ∅ 

RHS : (A × B) = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)} 

(A × C) = {(1, 5), (1, 6), (2, 5), (2, 6)} 

Since, there is no common element between A × B and A × C 

⇒ (A × B) ∩ (A × C) = ∅ 

∴ LHS = RHS

Given : A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6} 

To find : (A × B) ∩ (B × C) 

(A × B) = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)} 

(B × C) = {(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)} 

∴ (A × B) ∩ (B × C) = {(3, 4)}

Option : (A)

Inverse of a relation is given by interchanging the element’s position in each pair. 

Ex : Inverse of relation P = {(x, y)} is given by P⁻¹ = {(y, x)}. 

Therefore, 

R⁻¹ = {(3, 1), (5, 2), (3, 3)}.

Option : (C)

We have,

x + 2y = 8 

y = \(\frac{8-x}{2}\)

Since,

x and y are Natural numbers, 

So .. x must be an even number.. 

if x = 2, y = 3; 

if x = 4, y = 2; 

if x = 6, y = 1. 

So, 

Relation R = {(2, 3), (4, 2), (6, 1)} 

Now, 

Domain of R is {2, 4, 6}.

D. i ϕ 1

In complex numbers the y-axis is taken as the imaginary axis,

Thus i ϕ 1

Given relation is R = {(2, 3)}, where A = {1, 2, 3}

∵ 2 ∈ A but (2, 2) ∉ R

∴ R is not a reflexive relation.

Also (2, 3) ∈ R but (3, 2) ∉ R

∴ R is not a symmetric relation.

∵ (2, 3) ∈ R but a ∈ A such that (3, a) ∈ R

Therefore, no requirement that (2, a) ∈ R is exists for R to be a transitive relation.

∴ There is no condition to prove that R is not a transitive relation.

∴ R is a transitive relation.

Option : (A)

Since, 

y = x – 3;

Therefore, 

for x = 11, y = 8. 

For x = 12, y = 9.

[But the value y = 9 does not exist in the given set.] 

For x = 13, y =10. 

So, we have 

R = {(11, 8), (13, 10)} 

Now, 

R⁻¹ = {(8, 11), (10, 13)}.

Option : (D)

We have,

xφy |x| = y 

By checking the options,

 A. (2 + 3i) φ 13 

x = 2 + 3i;

|x| = \(\sqrt{2^2+3^2}\) 

= √13 

Therefore, 

|x|≠ y. 

So, 

Option A is incorrect.

B. 3 φ (−3)

x = 3;

|x| = \(\sqrt{3^2}\) 

= 3 

3 ≠(-3) 

Therefore, 

Option B is incorrect.

C. (1 + i) φ 2

|x| = \(\sqrt{1^2+1^2}\) 

= √2 

√2 ≠ 2 

Therefore, 

Option C is also incorrect.

D. iφ 1 

x = i;

|x| = \(\sqrt{1^2}\) 

= 1 

1 = 1 

|x| = y. 

Therefore, 

Option D is correct.

C. {2, 4, 6}

{2, 4, 6}

∵ R={(2,3),(4,2),(6,1)}

Hence domain(R)={2,4,6}

D. none of these

A. {(8, 11), (10, 13)}

{(8, 11), (10, 13)}

According to the question R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x – 3:

⇒R={(11,8),(13,10)}

⇒R^-1={(8, 11), (10, 13)}

∵ An important property of equivalence is that it divides the set into pairwise disjoint subsets called equivalent classes whose collection is called a partition of the set. Note that the union of all equivalence classes gives the complete set.

B. reflexive

∵ R = {(a, a), (b, b), (c, c), (a, b)} is satisfuing only the property of reflexive relation.

i.e Its reflexive i.e (a,a) ∈ R ∀ a ∈ A

∵ For A = {1, 2, 3, 4, 5, 6, 7, 8, 9} the satisfying complete relation is:

R={(1, 3), (2, 6), (3, 9)}

C. transitive

∵ Understood property

Here,

R = {(2,1),(4,7),(1,-2),…} 

It is seen that all the elements of R have a rule that (x,y)ϵ R 

⇒ y + 5 = 3x 

i.e y = 3x - 5.

Option : (C)

Domain of R is a set constituting all values of x. 

Here, 

Possible values for x by equation x² + y² ≤  4 will be 0,1,-1,2,-2. 

So, 

Domain of R is : {-2, -1, 0, 1, 2}.

Option : (D)

Here, 

y = 3x; If x = 1; then y = 3. 

If x = 2; then y = 6. 

If x = 3; then y = 9. 

Therefore the required relation will be R = {(1, 3), (2, 6), (3, 9)}.