Rbse Solutions for Class 11 maths Chapter 2 Exercise 2.2 | Relations

Last Updated on November 24, 2025 by Aman Singh

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 2 Exercise 2.2. Learn how to define a Relation ($R$) from $A$ to $B$ in Roster Form and Set-Builder Form. Identify the Domain, Codomain, and Range of various relations, including those on Natural Numbers and Integers. Also, calculate the number of possible relations between two sets.

This exercise covers defining a relation $R$ as a subset of the Cartesian product $A \times B$, and identifying its domain, codomain, and range.

1. Relation $R = \{(x, y) : 3x – y = 0, \text{ where } x, y \in A\}$

Given the set $A = \{1, 2, 3, \dots, 14\}$. The relation $R$ is from $A$ to $A$ (i.e., $R \subset A \times A$).

The condition is $y = 3x$. We find the pairs $(x, y)$ such that both $x$ and $y$ are in $A$.

Thus, the relation in roster form is:

$$R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$$

• Domain: The set of all first components of the ordered pairs in $R$.$$\mathbf{\text{Domain}(R) = \{1, 2, 3, 4\}}$$
• Codomain: The entire set $A$ (the set where the second elements are taken from).$$\mathbf{\text{Codomain} = \{1, 2, 3, \dots, 14\}}$$
• Range: The set of all second components of the ordered pairs in $R$.$$\mathbf{\text{Range}(R) = \{3, 6, 9, 12\}}$$

2. Relation $R = \{(x, y) : y = x + 5, x \in N, x < 4\}$

The set $N$ is the set of natural numbers $\{1, 2, 3, \dots\}$.

The condition for $x$ is $x \in N$ and $x < 4$, so $x \in \{1, 2, 3\}$.

• Roster Form:$$R = \{(1, 6), (2, 7), (3, 8)\}$$
• Domain:$$\mathbf{\text{Domain}(R) = \{1, 2, 3\}}$$
• Range:$$\mathbf{\text{Range}(R) = \{6, 7, 8\}}$$

3. Relation $R = \{(x, y) : \text{the difference between } x \text{ and } y \text{ is odd}\}$

Given: $A = \{1, 2, 3, 5\}$ and $B = \{4, 6, 9\}$.

The difference $|x – y|$ is odd. This happens when one number is odd and the other is even.

• $A$’s Odd elements: $\{1, 3, 5\}$. $A$’s Even elements: $\{2\}$.
• $B$’s Even elements: $\{4, 6\}$. $B$’s Odd elements: $\{9\}$.

We form pairs using (Odd from $A$, Even from $B$) and (Even from $A$, Odd from $B$).

1. (Odd from $A$, Even from $B$):
   • $(1, 4), (1, 6)$
   • $(3, 4), (3, 6)$
   • $(5, 4), (5, 6)$
2. (Even from $A$, Odd from $B$):
   • $(2, 9)$

• Roster Form:$$R = \{(1, 4), (1, 6), (3, 4), (3, 6), (5, 4), (5, 6), (2, 9)\}$$

4. Relation from Figure 2.7 (P to Q)

The figure shows arrows from P to Q, representing the relation $R \subset P \times Q$.

$$P = \{5, 6, 7\}, \quad Q = \{3, 4, 5\}$$

The ordered pairs shown are: $R = \{(5, 3), (6, 4), (7, 5)\}$.

(i) Set-Builder Form:

Observe the relationship between $x \in P$ and $y \in Q$: $5 = 3 + 2$, $6 = 4 + 2$, $7 = 5 + 2$.

The rule is $x = y + 2$ or $y = x – 2$.

$$\mathbf{R = \{(x, y) : y = x – 2, x \in P, y \in Q\}}$$

(ii) Roster Form:

$$\mathbf{R = \{(5, 3), (6, 4), (7, 5)\}}$$

• Domain:$$\mathbf{\text{Domain}(R) = \{5, 6, 7\}}$$
• Range:$$\mathbf{\text{Range}(R) = \{3, 4, 5\}}$$

5. Relation $R = \{(a, b): a, b \in A, b \text{ is exactly divisible by } a\}$

Given: $A = \{1, 2, 3, 4, 6\}$. $R$ is a relation on $A$ (i.e., $R \subset A \times A$).

The condition is $b$ is exactly divisible by $a$, or $a$ divides $b$ ($a|b$).

(i) Roster Form:

We check which pairs $(a, b)$ satisfy $a|b$ (where $a, b \in A$):

• $a=1$: $1|1, 1|2, 1|3, 1|4, 1|6$. Pairs: (1, 1), (1, 2), (1, 3), (1, 4), (1, 6).
• $a=2$: $2|2, 2|4, 2|6$. Pairs: (2, 2), (2, 4), (2, 6).
• $a=3$: $3|3, 3|6$. Pairs: (3, 3), (3, 6).
• $a=4$: $4|4$. Pair: (4, 4). ($4 \nmid 2, 4 \nmid 6$)
• $a=6$: $6|6$. Pair: (6, 6).

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

(ii) Domain of $R$: The set of all first components.

$$\mathbf{\text{Domain}(R) = \{1, 2, 3, 4, 6\}}$$

(iii) Range of $R$: The set of all second components.

$$\mathbf{\text{Range}(R) = \{1, 2, 3, 4, 6\}}$$

(In this specific case, since $R$ is on $A$ and every element divides itself, the domain and range are both equal to $A$.)

6. Domain and Range of $R = \{(x, x + 5) : x \in \{0, 1, 2, 3, 4, 5\}\}$

The relation is defined for $x \in \{0, 1, 2, 3, 4, 5\}$.

• Domain: The set of all possible first components ($x$ values).$$\mathbf{\text{Domain}(R) = \{0, 1, 2, 3, 4, 5\}}$$
• Range: The set of all possible second components ($y = x + 5$ values).
  • $x=0 \implies y=5$
  • $x=1 \implies y=6$
  • $x=2 \implies y=7$
  • $x=3 \implies y=8$
  • $x=4 \implies y=9$
  • $x=5 \implies y=10$$$\mathbf{\text{Range}(R) = \{5, 6, 7, 8, 9, 10\}}$$

7. Relation $R = \{(x, x^3) : x \text{ is a prime number less than } 10\}$ in Roster Form

The prime numbers less than 10 are $\{2, 3, 5, 7\}$.

• Roster Form:$$\mathbf{R = \{(2, 8), (3, 27), (5, 125), (7, 343)\}}$$

8. Number of Relations from $A$ to $B$

Given: $A = \{x, y, z\}$, so $n(A) = 3$. $B = \{1, 2\}$, so $n(B) = 2$.

1. Cartesian Product: The number of elements in $A \times B$ is $n(A \times B) = n(A) \cdot n(B) = 3 \times 2 = 6$.
2. Relations: A relation $R$ from $A$ to $B$ is any subset of $A \times B$.The total number of subsets of a set with $m$ elements is $2^m$.Number of relations $= 2^{n(A \times B)} = 2^6 = \mathbf{64}$.

9. Domain and Range of $R = \{(a, b): a, b \in Z, a – b \text{ is an integer}\}$

$Z$ is the set of all integers $\{\dots, -2, -1, 0, 1, 2, \dots\}$.

The condition is that the difference $a – b$ must be an integer.

Since $a$ is an integer and $b$ is an integer, the difference of two integers is always an integer.

Thus, the relation $R$ includes all possible pairs $(a, b)$ where $a, b \in Z$.

$$R = Z \times Z$$

• Domain: The set of all possible first components ($a$ values), which can be any integer.$$\mathbf{\text{Domain}(R) = Z}$$
• Range: The set of all possible second components ($b$ values), which can be any integer.$$\mathbf{\text{Range}(R) = Z}$$