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

## 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, x3): x is a prime number less than 10} in roster form.

Solution.
Given relation R = {(x, x3): x is a prime number less than 10}
= {(x, x3): x
{2, 3, 5, 7}}
= {(2, 23), (3, 33), (5, 53), (7, 73)}
= {(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 = 26 = 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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