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
NCERT Solutions for Maths Class 9