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

NCERT Solutions Class 11 Maths 

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise 2.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter 2 (Relations and Functions) 

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 – Given, A = {1, 2, 3,…,14}. 
Here, the relation R from A to A is given by, R = {(x, y): 3x – y = 0, where x, y ∈ A} 
So, relation R = {(1,3), (2,6), (3,9), (4,12)} 

Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation. 
So, Domain of R = {1, 2, 3, 4} 

Now, Here the complete set A is the Codomain of relation R. 
So, Co-Domain of R = {1, 2, 3, 4,….,14} 

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation. 
So, Range of R = {3, 6, 9, 12}

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 – Here, the relation R is given by, R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈N} 
Now, As we know that the natural numbers less than 4 are 1, 2 and 3. 
So, relation R = {(1,6), (2,7), (3,8)} 

Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation. 
So, Domain of R = {1, 2, 3} 

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation. 
So, Range of R = {6, 7, 8}

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 – Given, A = {1, 2, 3, 5} and B = {4, 6, 9}
Here, the relation from A to B is given by, R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
So, relation R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}

4. The figure shows a relationship between the sets P and Q. Write this relation
(i) in set-builder form
(ii) in roster form
What is its domain and range?

Solution – From the given figure, we can see that – 
P = {5, 6, 7} and Q = {3, 4, 5} 
Now, The relation between sets P and Q – 

(i) In set-builder form
R = {(x, y): y = x – 2; x ∈ P} ‘or’ R = {(x, y): y = x – 2 for x = 5, 6, 7} 

(ii) In roster form
R = {(5,3), (6,4), (7,5)} 
Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation. 
So, Domain of R = {5, 6, 7} = P. 

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation. 
So, Range of R = {3, 4, 5} = Q.

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

(i) The relation R in roster form will be – 
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) We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation. 
So, Domain of R = {1, 2, 3, 4, 6} 

(iii) We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation. 
So, Range of R = {1, 2, 3, 4, 6}

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}}
Thus,
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
So,
Domain of R = {0, 1, 2, 3, 4, 5} and,
Range of R = {5, 6, 7, 8, 9, 10}

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}
The prime numbers less than 10 are 2, 3, 5, and 7.
Therefore,
R = {(2, 8), (3, 27), (5, 125), (7, 343)}

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}
Now,
A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
As n(A × B) = 6, the number of subsets of A × B will be 2⁶.
Thus, the number of relations from A to B is 2⁶.

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}
We know that the difference between any two integers is always an integer.
Therefore,
Domain of R = Z and Range of R = Z

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