# NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions (Ex 2.1)

One of the most important and difficult subjects for students in Class 11 is mathematics. In addition to solid conceptual knowledge, Mathematics also necessitates a lot of problem-solving practice. Practising lots of questions is the key to getting good grades in Mathematics in the annual examinations. The NCERT mathematics textbook covers Relations and Functions in Chapter 11. As questions from this chapter are frequently asked in the exams, it is an important chapter for students to study thoroughly. Compared to the other chapters in the book, this one is a little bit shorter. It contains three exercises in total. However, the concepts it contains are extremely crucial. This chapter teaches students the concept of functions, which will be very helpful to them in the future. For a solid understanding of the topics, it is crucial that students complete the problems in this chapter. Class 11 Maths Chapter 2 Exercise 2.1, deals with the Cartesian Products of Sets. Students must carry out a series of actions in order to find the Cartesian products and address the problems in the exercises. Extramarks offers NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1, to assist students in this regard.

Since they are written in straightforward language that all students can comprehend, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1 are helpful for students while they are preparing for their annual exams. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1, have been produced in a step-by-step manner to help students secure full marks in their exams. In order to prevent students from being compelled to look elsewhere for the topics covered by the curriculum, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1 have been created in accordance with the most recent syllabus. Every CBSE guideline is followed in the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1. All of the problems in the chapter’s exercises are addressed in the NCERT Solutions Class 11 Maths Chapter 2 Exercise 2.1, which is also fully error-free. The NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1, can help students prepare for exams by helping them answer questions like these fast and correctly. Since they are prepared in an easy-to-understand manner, the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1 offered by Extramarks will suffice to help students tackle NCERT questions. Using the NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1 can help students apply the concepts from Maths Class 11 Chapter 2 Exercise 2.1 more accurately and easily.

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## NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions (Ex 2.1) Exercise 2.1

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

Extramarks provide a complete response to a student’s educational needs. Despite the fact that NCERT textbooks are a great source of study material, students still struggle to locate reliable NCERT Solutions For Class 11 Maths Chapter 2 Exercise 2.1. Although the NCERT textbook contains the answers to the questions, it does not contain their in-depth explanations. Students must comprehend the steps taken to arrive at the answer while answering the questions. Writing every step in the exam is extremely crucial if students want to get full marks.

NCERT Solutions like the NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 6, NCERT Solutions Class 7, NCERT Solutions Class 8, NCERT Solutions Class 9 and NCERT Solutions Class 10 are also available at Extramarks.

For students in primary classes in CBSE-affiliated schools, Extramarks provides NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 and NCERT Solutions Class 1.

Q.1

$\text{If }\left(\frac{\text{x}}{\text{3}}\text{+1, y-}\frac{\text{2}}{\text{3}}\right)\text{=}\left(\frac{\text{5}}{\text{3}}\text{,}\frac{\text{1}}{\text{3}}\right)\text{, find the values of x and y.}$

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \text{\hspace{0.17em}}\left(\frac{\mathrm{x}}{3}+1,\mathrm{y}-\frac{2}{3}\right)=\left(\frac{5}{3},\frac{1}{3}\right)\\ \mathrm{Comparing}\text{both sides, we get}\\ \frac{\mathrm{x}}{3}+1=\frac{5}{3}⇒\frac{\mathrm{x}}{3}=\frac{5}{3}-1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{3}=\frac{2}{3}⇒\mathrm{x}=2\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}-\frac{2}{3}=\frac{1}{3}⇒\mathrm{y}=\frac{1}{3}+\frac{2}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}=\frac{3}{3}=1\\ \mathrm{Thus},\text{x}=2\text{and y}=\text{1.}\end{array}$

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

Ans.

$\begin{array}{l}\mathrm{Number}\text{of elements in set A}=3\\ \text{i}.\text{e}.,\text{ n}\left(\text{A}\right)=3\\ \mathrm{Elements}\text{in set B}=\left\{3,4,5\right\}\\ ⇒\text{ n}\left(\text{B}\right)=3\\ \mathrm{So},\text{}\\ \text{Number of elements in A}×\text{B}=\text{n}\left(\text{A}\right)×\text{n}\left(\text{B}\right)\\ \text{ n}\left(\text{A}×\text{B}\right)=3×3\\ \text{ }=9\\ \mathrm{Thus},\text{ number of elements in A}×\text{B is 9.}\end{array}$

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

Ans.

$\begin{array}{l}\text{Given: G}=\left\{\text{7},\text{8}\right\}\text{and H}=\left\{\text{5},\text{4},\text{2}\right\},\text{}\\ \text{ G}×\text{H}=\left\{\left(7,5\right),\left(7,4\right),\left(7,2\right),\left(8,5\right),\left(8,4\right),\left(8,2\right)\right\}\\ \text{ H}×\text{G}=\left\{\left(5,7\right),\left(5,8\right),\left(4,7\right),\left(4,8\right),\left(2,7\right),\left(2,8\right)\right\}\end{array}$

Q.4 State whether each of the following statements are 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 ∩ ∅) = ∅ .

Ans.
(i) False,
Since, P = {m, n} and Q = {n, m}
P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) True.

(iii) True.

Q.5 If A = {–1, 1}, find A × A × A.

Ans.
Since, A = {–1, 1}
A × A = {(–1, –1), (–1, 1,), (1,–1), (1, 1)}
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, –1), (1,–1, 1,), (1, 1,–1), (1, 1, 1)}

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

Ans.
Given, A × B = {(a, x),(a , y), (b, x), (b, y)}
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B}.
Then, A = {a, b}, B = {x, y}.

Q.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.

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

(i) B ∩ C = {}
= ∅
A × (B ∩ C)
= A × ∅
= ∅
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)}
(A × B) ∩ (A × C)
= ∅
L.H.S.= R.H.S. Hence proved.

(ii) Here,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
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)}
Since,
(A × C) ∩ (B × D) = {(1, 5), (1, 6), (2, 5), (2, 6)}
= (A × C)
So, (A × C) is a subset of (B × D).

Q.8 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.

Ans.
Number of elements in set A = 3
Number of elements in set B = 2,
Some elements of A × B are (x, 1), (y, 2), (z, 1).
Since, Cartesian product A × B of two sets A and B is given by A × B = {(a, b): a ∈ A, b ∈ B} So, A = {x, y, z} and B = {1, 2}.

Q.9 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.

Ans.
Number of elements in the cartesian product A × A = 9
Some given elements of A × A are (–1, 0) and (0, 1).
Let number of elements in set A, n(A) = 3
and number of elements in set A × A = 9
i.e., n(A × A) = n(A) × n(A)
9 = { n(A)}2
n(A) = 3
So, A = {–1, 0, 1}
Then, A × A = {(–1, –1), (–1, 0), (–1, 1), (0, –1,), (0, 0), (0,1), (1, –1,), (1, 0), (1, 1)}
Remaining elements of A × A are: (–1, –1), (–1, 1), (0, –1,), (0, 0), (1, –1,), (1, 0), (1, 1).