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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 2 Relations – Exercise 2.3 | Set 1

### Question 1. If A = {1, 2, 3}, B = {4, 5, 6}, which of the following are relations from A to B? Give reasons in support of your answer.

(i) {(1, 6), (3, 4), (5, 2)}

(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}

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

(iv) A × B

Solution:

Given:

A = {1, 2, 3}, B = {4, 5, 6}

A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}

(i) {(1, 6), (3, 4), (5, 2)}

No, it is not a relation from A to B. The given set is not a subset of A × B

because (5, 2) is not a part of the relation from A to B.

(ii) {(1, 5), (2, 6), (3, 4), (3, 6)}

Yes, it is a relation from A to B. Hence, the given set is a subset of A × B.

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

No, it is not a relation from A to B. The given set is not a subset of A × B

because (4, 2), (4, 3), (5, 1) are not a part of the relation from A to B.

(iv) A × B

A × B is a relation from A to B:

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

### Question 2. A relation R is defined from a set A = {2, 3, 4, 5} to a set B = {3, 6, 7, 10} as follows: (x, y) R x is relatively prime to y. Express R as a set of ordered pairs and determine its domain and range.

Solution:

Given: (x, y) ∈ R = x is relatively prime to y          -(Relatively prime numbers are also known as co-prime numbers)

Here,

2 is co-prime to 3 and 7.

3 is co-prime to 7 and 10.

4 is co-prime to 3 and 7.

5 is co-prime to 3, 6 and 7.

∴ R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

So, Domain(R) = {2, 3, 4, 5} and Range(R) = {3, 6, 7, 10}

### (ii) The Range of R.

Solution:

Given: A= {1, 2, 3, 4, 5}           -(A is set of first five natural numbers)

(x, y) R x ≤ y

1 is less than 2, 3, 4 and 5.

2 is less than 3, 4 and 5.

3 is less than 4 and 5.

4 is less than 5.

5 is not less than any number A

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

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

(i) Domain(R-1) = {2, 3, 4, 5}

(ii) Range(R) = {2, 3, 4, 5}

Note: You can see that Domain of R-1 is same as Range of R. Similarly, Domain of R is same as Range of R-1

### (iii) R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

Solution:

(i) Given: R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

So the inverse relation R-1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}

(ii) Given: R= {(x, y) : x, y ∈ N; x + 2y = 8}

Here, x + 2y = 8

x = 8 – 2y

As y ∈ N, Put the values of y = 1, 2, 3,…… till x ∈ N

On putting y = 1,  x = 8 – 2(1) = 8 – 2 = 6

On putting y = 2, x = 8 – 2(2) = 8 – 4 = 4

On putting y = 3, x = 8 – 2(3) = 8 – 6 = 2

On putting y = 4, x = 8 – 2(4) = 8 – 8 = 0

Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

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

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

(iii) Given: R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3

So, x = {11, 12, 13} and y = (8, 10, 12}

y = x – 3

On putting x = 11, y = 11 – 3 = 8 ∈ (8, 10, 12}

On putting x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}

On putting x = 13, y = 13 – 3 = 10 ∈ (8, 10, 12}

Therefore, R = {(11, 8), (13, 10)}

R-1 = {(8, 11), (10, 13)}

### (i) A relation R from the set {2, 3, 4, 5, 6} to the set {1, 2, 3} defined by x = 2y.

Solution:

Let A = {2, 3, 4, 5, 6} and B = {1, 2, 3}

Given: x = 2y where x = {2, 3, 4, 5, 6} and y = {1, 2, 3}

On putting y = 1, x = 2(1) = 2 A

On putting y = 2, x = 2(2) = 4 A

On putting y = 3, x = 2(3) = 6 A

∴ R = {(2, 1), (4, 2), (6, 3)}

### (ii) A relation R on the set {1,2,3,4,5,6,7} defined by (x, y)∈ R <=>x is relatively prime to y.

Solution:

Given: (x, y) R x is relatively prime to y

Here, 2 is co-prime to 3, 5 and 7.

3 is co-prime to 2, 4, 5 and 7.

4 is co-prime to 3, 5 and 7.

5 is co-prime to 2, 3, 4, 6 and 7.

6 is co-prime to 5 and 7.

7 is co-prime to 2, 3, 4, 5 and 6.

∴ R = {(2,3), (2,5), (2,7), (3,2), (3,4), (3,5), (3,7), (4,3), (4.5), (4,7), (5,2), (5,3), (5,4), (5,6), (5,7), (6,5), (6,7), (7,2), (7,3), (7,4), (7,5), (7,6)}

### (iii) A relation R on the set {0,1,2,…,10} defined by 2x + 3y = 12.

Solution:

Given: (x, y) R 2x + 3y = 12

Where x and y {0,1,2,…,10}

2x + 3y = 12

⇒ 2x = 12 – 3y

⇒ x = (12-3y)/2

On putting y=0, we get

⇒ x = ( 12 – 3(0) )/2 = 6

On putting y=2,

⇒ x = ( 12 – 3(2) )/2 = 3

On putting y=4, we get

⇒ x= ( 12 – 3(4) )/2 = 0

∴ R = {(0, 4), (3, 2), (6, 0)}

### (iv) A relation R form a set A = {5, 6, 7, 8} to the set B = {10, 12, 15, 16, 18} defined by (x, y) R x divides y.

Solution:

Given: (x, y) R x divides y

Where x = {5, 6, 7, 8} and y = {10, 12, 15, 16, 18}

Here,

5 divides 10 and 15.

6 divides 12 and 18.

7 divides none of the value of set B.

8 divides 16.

∴ R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}

### Question 6. Let R be a relation in N defined by (x, y) R x + 2y = 8. Express R and R-1 as sets of ordered pairs.

Solution:

Given: (x, y) R x + 2y = 8 where x N and y N

x = 8 – 2y

As y N, Put the values of y = 1, 2, 3,…… till x N

On putting y = 1, x = 8 – 2(1) = 8 – 2 = 6

On putting y = 2, x = 8 – 2(2) = 8 – 4 = 4

On putting y = 3, x = 8 – 2(3) = 8 – 6 = 2

On putting y = 4, x = 8 – 2(4) = 8 – 8 = 0

Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.

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

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

### Question 7: Let A = {3, 5} and B = {7, 11}. Let R = {(a, b): a ∈ A, b ∈ B, a-b is odd}. Show that R is an empty relation from into B.

Solution:

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

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

When a = 3 and b = 7

a – b = 3 – 7 = -4 which is not odd

When a = 3 and b = 11

⇒ a – b = 3 – 11 = -8 which is not odd

When a = 5 and b = 7

⇒ a – b = 5 – 7 = -2 which is not odd

When a = 5 and b = 11

⇒ a – b = 5 – 11 = -6 which is not odd

∴ R = { } = Φ

⇒ R is an empty relation from into B

### Question 8: Let A = {1, 2} and B={3, 4}. Find the total number of relations from A into B.

Solution:

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

n(A) = 2         -(Number of elements in set A).

n(B) = 2         -(Number of elements in set B).

We know,

n(A × B) = n(A) × n(B) = 2 × 2 = 4

Therefore, the number of relations from A to B are 24 = 16

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

Solution:

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

Therefore, 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)}

So,

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

Range(R) = {5, 6, 7, 8, 9, 10}

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

Solution:

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

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

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

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

So,

Domain(R) = {2, 3, 5, 7}

Range(R) = {8, 27, 125, 343}

### (i) R= {a, b): a ∈ N, a < 5, b = 4}

Solution:

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

Range(R) = {4}

### (ii) S= {a, b): b = |a-1|, a ∈ Z and |a| ≤ 3}

Solution:

Given: S= {a, b): b = |a-1|, a Z and |a| ≤ 3}

Z denotes integer which can be positive as well as negative

Now, |a| ≤ 3 and b = |a-1|

∴ a {-3, -2, -1, 0, 1, 2, 3}

S = {a, b): b = |a-1|, a Z and |a| ≤ 3}

S = {a, |a-1|): b = |a-1|, a Z and |a| ≤ 3}

S = {(-3, |-3 – 1|), (-2, |-2 – 1|), (-1, |-1 – 1|), (0, |0 – 1|), (1, |1 – 1|), (2, |2 – 1|), (3, |3 – 1|)}

S = {(-3, |-4|), (-2, |-3|), (-1, |-2|), (0, |-1|), (1, |0|), (2, |1|), (3, |2|)}

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

So,

Domain(S) = {-3, -2, -1, 0, 1, 2, 3}

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

### Question 11. Let A = {a, b}. List all relations on A and find their number.

Solution:

As we know that the total number of relations that can be defined from a set A to a set B is the number of possible subsets of A × B. If n(A) = p and n(B) = q, then n(A × B) = pq. So, the total number of relations is 2pq.

Now,

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

Total number of relations are all possible subsets of A × A:

{Φ, {(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

Hence, the total number of relations = 24 = 16

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