**NCERT
Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2**

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2
are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT
Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.2.

**Ex 2.2
Class 11 Maths Question 1.**

Let A = {1, 2, 3, …, 14}. Define a relation R from A to A by R = {(x, y): 3x –
y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

**Solution.**

We have A = (1, 2, 3, …, 14)

Given relation R = {(x, y): 3x – y = 0, where x, y ∈ A}

= {(x, y): y = 3x, where x, y ∈ A}

= {(x, 3x), where x, 3x ∈ A}

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

[∵ 1 ≤ 3x ≤ 14, ∴ 1/3 ≤ x ≤ 14/3 ⇒ x = 1, 2,
3, 4]

Domain of R = {1, 2, 3, 4}

Codomain of R = {1, 2, 3, …, 14}

Range of R = {3, 6, 9, 12}.

**Ex 2.2 Class 11
Maths Question 2.**

Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5,
x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form.
Write down the domain and the range.

**Solution.**

Given relation R = {(x, y): y = x + 5, x < 4 and x, y ∈ N}

= {(x, y): y = x + 5, x ∈ (1, 2, 3) & y ∈ N}

= {(x, x + 5): x = 1, 2, 3}

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

Domain of R = {1, 2, 3}, Range of R = {6, 7, 8}.

**Ex 2.2 Class 11
Maths Question 3.**

A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x,
y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

**Solution.**

We have, A = {1, 2, 3, 5} and B = {4, 6, 9}.

R = {(x, y): the difference
between x and y is odd; x ∈
A, y ∈ B}

= {(x, y): y – x = odd; x ∈ A, y ∈ B}

Hence, R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}.

**Ex 2.2 Class 11
Maths Question 4.**

The figure shows a relationship between the sets P and Q. Write this relation**(i)**in set-builder form

**(ii)**roster form.

What is its domain and range?

**Solution.**

**(i)** The
relation R in set builder form is written as:

R = {(x, y): x – y = 2; x

∈
P, y ∈ Q}

i.e., R = {(x, y): y = x – 2 for x = 5, 6, 7}

**(ii)** The relation R in roster form is
written as:

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

Domain of R = {5, 6, 7} = P

Range of R = {3, 4, 5} = Q

**Ex 2.2 Class 11
Maths Question 5.**

Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈ A, b is exactly divisible by a}.**(i)**Write R in roster form.

**(ii)**Find the domain of R.

**(iii)**Find the range of R.

**Solution.**

Given A = {1, 2, 3, 4, 6}

Given relation R = {(a, b): a, b ∈ A, b is exactly divisible by a}

**(i)** Roster
form of 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 = {1, 2, 3, 4, 6} = A

**(iii)** Range
of R = {1, 2, 3, 4, 6} = A

**Ex 2.2 Class 11
Maths Question 6.**

Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.

**Solution.**

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

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

∴ Domain of R = {0, 1, 2, 3, 4, 5} and

Range of R = {5, 6, 7, 8, 9, 10}.

**Ex 2.2 Class 11
Maths Question 7.**

Write the relation R = {(x, x^{3}): x is a prime number less than 10} in roster form.

**Solution.**

Given relation R = {(x, x^{3}): x is a prime number less than 10}

= {(x, x^{3}): x ∈ {2, 3, 5, 7}}

= {(2, 2^{3}), (3, 3^{3}), (5, 5^{3}), (7, 7^{3})}

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

**Ex 2.2 Class 11
Maths Question 8.**

Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.

**Solution.**

Given A = {x, y, z} and B = {1, 2}

∴ n(A) = 3 and n(B) = 2

Since n(A × B) = n(A) × n(B)

∴ n(A × B) = 3 × 2 = 6

Number of relations from A to B is equal to the number of subsets of A × B.

Since A × B contains 6 elements.

⇒ Number of subsets of A × B = 2^{6} =
64

So, there are 64 relations from A to B.

**Ex 2.2 Class 11
Maths Question 9.**

Let R be the relation on Z defined by R = {{a, b): a, b ∈ Z, a – b is an integer}. Find the domain
and range of R.

**Solution.**

Given relation R = {(a, b): a, b ∈ Z, a – b is an integer}

If a, b ∈ Z, then a – b ∈ Z ⇒ Every ordered pair of integers is contained in R.

R = {(a, b): a, b ∈ Z}

So, range of R = domain of R = Z

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