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Exercise 2.1 NCERT Solutions Relations & Functions 11 Math Chapter 2 Free PDF Download

 

NCERT Solutions for Class 11 – Chapter 2 Relations and Functions – Exercise 2.1

This section provides detailed solutions for Exercise 2.1 from Chapter 2 of the Class 11 NCERT Mathematics textbook. The exercise includes problems in identifying and representing relations and functions, determining the domain and range, and working with function notation. The solutions are presented step-by-step to help students understand the methods for solving these problems and applying the concepts of relations and functions effectively.

Exercise 2.1

1. If (x/3 + 1, y – 2/3) = (5/3, 1/3) find the values of x and y.

Solution:

Given, (x/3 + 1, y – 2/3) = (5/3, 1/3)

As the ordered pairs are equal, the corresponding elements should also be equal. Thus,

x/3 + 1 = 5/3 and y – 2/3 = 1/3

Solving, we get

x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M. and adding]

x = 2 and 3y = 3

Therefore, x = 2 and y = 1

2. If set A has 3 elements and set B = {3, 4, 5}, then find the number of elements in (A × B).

Solution:

Given, set A has 3 elements, and the elements of set B are {3, 4, and 5}.

So, the number of elements in set B = 3

Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B) = 3 × 3 = 9

Therefore, the number of elements in (A × B) will be 9.

3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.

Solution:

Given, G = {7, 8} and H = {5, 4, 2}

We know that,

The Cartesian product of two non-empty sets P and Q is given as

P × Q = {(p, q): p ∈ P, q ∈ Q}

So,

G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}

H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

4. State whether each of the following statements is true or false. If the statement is false, rewrite the given statement correctly.

(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}

(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.

(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ

Solution:

(i) The statement is false. The correct statement is

If P = {m, n} and Q = {n, m}, then

P × Q = {(m, m), (m, n), (n, m), (n, n)}

(ii) True

(iii) True

5. If A = {–1, 1}, find A × A × A.

Solution:

The A × A × A for a non-empty set A is given by

A × A × A = {(a, b, c): a, b, c ∈ A}

Here, it is given A = {–1, 1}

So,

A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}

6. If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.

Solution:

Given,

A × B = {(a, x), (a, y), (b, x), (b, y)}

We know that the Cartesian product of two non-empty sets, P and Q is given by:

P × Q = {(p, q): p ∈ P, q ∈ Q}

Hence, A is the set of all first elements, and B is the set of all second elements.

Therefore, A = {a, b} and B = {x, y}

7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that

(i) A × (B ∩ C) = (A × B) ∩ (A × C)

(ii) A × C is a subset of B × D

Solution:

Given,

A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}

(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)

Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Î¦

Thus,

L.H.S. = A × (B ∩ C) = A נΦ = Î¦

Next,

A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

Thus,

R.H.S. = (A × B) ∩ (A × C) = Î¦

Therefore, L.H.S. = R.H.S.

Hence verified

(ii) To verify: A × C is a subset of B × D

First,

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

And,

B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}

Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.

Thus, A × C is a subset of B × D.

Hence verified

8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.

Solution:

Given,

A = {1, 2} and B = {3, 4}

So,

A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}

Number of elements in A × B is n(A × B) = 4

We know that,

If C is a set with n(C) = m, then n[P(C)] = 2m.

Thus, the set A × B has 24 = 16 subsets.

And these subsets are as given below:

Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}

9. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

Solution:

Given,

n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.

We know that,

A = Set of first elements of the ordered pair elements of A × B

B = Set of second elements of the ordered pair elements of A × B

So, clearly, x, y, and z are the elements of A; and

1 and 2 are the elements of B.

As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}

10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.

Solution:

We know that,

If n(A) = p and n(B) = q, then n(A × B) = pq.

Also, n(A × A) = n(A) × n(A)

Given,

n(A × A) = 9

So, n(A) × n(A) = 9

Thus, n(A) = 3

Also, given that the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.

And, we know in A × A = {(a, a): a ∈ A}

Thus, –1, 0, and 1 have to be the elements of A.

As n(A) = 3, clearly A = {–1, 0, 1}

Hence, the remaining elements of set A × A are as follows:

(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)

FAQs on Relation And Functions

What is a relation in mathematics?

A relation in mathematics is a set of ordered pairs where each element of one set is associated with elements of another set. For example, if A={1,2,3}and B={a,b} a relation from A to B could be {(1,a),(2,b),(3,a)}

How do you represent a function graphically?

A function can be represented graphically using a coordinate plane where the x-axis represents the input values (domain) and the y-axis represents the output values (range). The function’s graph shows the relationship between the inputs and outputs.

How do you find the domain and range of a function?

The domain of a function is the set of all possible input values (x-values) for which the function is defined. The range is the set of all possible output values (y-values) that the function can produce. To find these, identify all permissible values for the input and determine the corresponding output values.

What is the difference between a relation and a function?

A relation can have multiple outputs for a single input, while a function has exactly one output for each input. In other words, functions are a subset of relations where each input has a unique output.

Relations and Functions NCERT Solutions for Class 11 Maths Chapter 2 Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions are solved in detail in the PDF given below. All the solutions to the problems in the exercises are created in such a way that it enables the students to prepare for the exam and ace it. The NCERT Solutions are prepared by the most experienced teachers in the education space, making the explanation of each solution simple, understandable, and according to the latest CBSE Syllabus. The solution helps Class 11 students to master the concept of Relations and Functions.

The solutions provide a good understanding of the fundamental concepts before they solve the equations. Through regular practice, students will know the difference between relations and functions, which are included under the syllabus, and become well-versed in its concepts. Numerous examples are present in the textbook before the exercise questions to help them understand the methodologies to be followed while solving the problems. Referring to the NCERT Class 11 Solutions PDF, students can get a glimpse of the important concepts before facing their final exams.

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