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• NCERT Solutions for Class 11 Maths

# Class 11 NCERT Solutions- Chapter 2 Relation And Functions – Exercise 2.2

### Problem 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}

### Problem 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}

### Problem 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)}

### Problem 4: Fig.2.7 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:

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.

### Problem 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}
Here, the relation R on A is given by, R = {(a, b): a , b ∈ A, b is exactly divisible by a}

(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}

### Problem 6: Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.

Solution:

Here, the relation R is given by, R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
So, relation R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}

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 = {0, 1, 2, 3, 4, 5}

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 = {5, 6, 7, 8, 9, 10}

### Problem 7: Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.

Solution:

Here, the relation R is given by, R = {(x, x3) : x is a prime number less than 10}
Now, As we know that the prime numbers less than 10 are 2, 3, 5 and 7.
So, relation R = {(2,8), (3,27), (5,125), (7,343)}

### Problem 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, number of elements in set A, n(A) = 3
and number of elements in set B, n(B) = 2
So, n(A × B) = n(A) × n(B) = 6.
We know that, the number of relations from A to B = 2n(A × B) = 26 = 64.

‘OR’

Given, A = {x, y, z} and B = {1, 2}.
Now, A × B = {(x,1), (x,2), (y,1), (y,2), (z,1), (z,2)}
Here, number of elements in A × B, n(A × B) = 6
So, the number of subsets of A × B = 26 = 64
Thus, the number of relations from A to B are 64.

### Problem 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:

Here, the relation R is given by, R = {(a, b): a, b ∈ Z, a – b is an integer}
As we know that the difference between any two integers is always an integer.
So, Domain of R = Z and Range of R = Z.

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