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

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

Edited By Vishal kumar | Updated on Nov 02, 2023 08:45 AM IST

## NCERT Solutions for Class 11 Maths Chapter 2 - Relations And Functions Exercise 2.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 2: Relations and Functions Exercise 2.1- In the previous chapter, you have learned about the sets and their applications. In the NCERT solutions for Class 11 Maths chapter 2 exercise 2.1, you will get questions related to basic definitions of relations, and cartesian products of sets. The relation is defined as the link between a pair of objects from two sets and the function is a special type of relation that qualifies for the function. In the 11th class maths exercise 2.1 answers, you will get questions related to the equality of ordered pairs. This exercise 2.1 class 11 maths is the basic exercise about ordered pairs before getting into relations and functions. The basic definition of ordered pair and some properties of ordered pair are given in the NCERT textbook before this exercise. You can go through these properties to get conceptual clarity. In the Class 11th Maths chapter 2 exercise 2.1, you will learn about the ordered triplet also. Here you can check for NCERT Solutions to get detailed solutions from Class 6 to Class 12 for Science and Maths.

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The 11th class maths exercise 2.1 answers, prepared by subject experts at Careers360, offer detailed, step-by-step explanations to help students understand the concepts effectively. Additionally, the availability of a PDF version of the class 11 maths ex 2.1 solutions makes it convenient for students to access them at their convenience, and the best part is that these resources are provided free of charge. This is a valuable resource for students looking to enhance their understanding of mathematics and excel in their studies.

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## Access Relations And Functions Class 11 Chapter 2Exercise: 2.1

It is given that
$\left ( \frac{x}{3}+1 , y - \frac{2}{3} \right ) = \left ( \frac{5}{3},\frac{1}{3} \right )$
Since the ordered pairs are equal, the corresponding elements will also be equal
Therefore,
$\frac{x}{3}+1= \frac{5}{3} \ \ \ and \ \ \ y - \frac{2}{3}= \frac{1}{3}$
$\frac{x}{3}= \frac{5}{3}-1 \ \ \ and \ \ \ y = \frac{1}{3}+ \frac{2}{3}$
$\frac{x}{3}= \frac{5-3}{3} \ \ \ and \ \ \ y = \frac{1+2}{3}$
$\frac{x}{3}= \frac{2}{3} \ \ \ and \ \ \ y = \frac{3}{3}$
$x= 2 \ \ \ and \ \ \ y = 1$

Therefore, values of x and y are 2 and 1 respectively

It is given that set A has 3 elements and the elements in set B are 3 , 4 , and 5
Therefore, the number of elements in set B is 3
Now,
Number of elements in $(A \times B)$
= ( Number of elements in set A ) $\times$ ( Number of elements in set B)
= 3 $\times$ 3
= 9
Therefore, number of elements in $(A \times B)$ is 9

It is given that
G = {7, 8} and H = {5, 4, 2}
We know that the cartesian product of two non-empty sets P and Q is defined as
P $\times$ Q = {(p,q) , where p $\epsilon$ P , q $\epsilon$ Q }
Therefore,
G $\times$ H = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
And
H $\times$ G = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

FALSE
If P = {m, n} and Q = { n, m}
Then,
$P \times Q = \left \{ (m,m),(m,n),(n,m),(n,n) \right \}$

It is a TRUE statement

$\because$ If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that $x \epsilon A$ and $y \epsilon B$

This statement is TRUE

$\because$ If A = {1, 2}, B = {3, 4}, then

$B\cap\phi=\phi$

There for

$A \times (B \cap \phi ) = \phi$

It is given that
A = {–1, 1}
A is an non-empty set
Therefore,
Lets first find $A \times A$
$A \times A = \left \{ -1,1 \right \} \times \left \{ -1,1 \right \} = \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$
Now,
$A\times A \times A=A\times (A \times A) = \left \{ -1,1 \right \} \times \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$$=\left \{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right \}$

It is given that
$A \times B$ = {(a, x),(a , y), (b, x), (b, y)}
We know that the cartesian product of two non-empty set P and Q is defined as

$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we know that A is the set of all first elements and B is the set of all second elements
Therefore,
$A= \left \{ a,b \right \} \ \ \ and \ \ \ B = \left \{ x , y \right \}$

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,
$B \cap C = \left \{ 1,2,3,4 \right \} \cap \left \{ 5,6 \right \} = \Phi$
Now,
$A \times ( B \cap C ) = A \times \phi = \phi \ \ \ \ \ \ \ \ \ \ \ \ -(i)$

$A \times B = \left \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4) \right \}$
And
$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
Now,
$(A \times B)\cap (A \times C) =\phi \ \ \ \ \ \ \ \ \ \ \ \ -(ii)$
From equation (i) and (ii) it is clear that
$L.H.S. = R.H.S.$
Hence,
$A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$

It is given that
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
Now,

$A \times C = \left \{ (1,5),(1,6),(2,5),(2,6) \right \}$
And
$B \times D = \left \{ (1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),(3,8),(4,5),(4,6),(4,7),(4,8) \right \}$
We can clearly observe that all the elements of the set $A \times C$ are the elements of the set $B \times D$
Therefore, $A \times C$ is a subset of $B \times D$

It is given that
A = {1, 2} and B = {3, 4}
Then,
$A \times B = \left \{ (1,3),(1,4),(2,3),(2,4) \right \}$
$\Rightarrow n\left ( A \times B \right ) = 4$
Now, we know that if C is a set with $n(C) = m$
Then,
$n[P(C)]= 2^m$
Therefore,
The set $A \times B$ has $2^4=16$ subsets.

It is given that
n(A) = 3 and n(B) = 2 and If (x, 1), (y, 2), (z, 1) are in A × B.

By definition of Cartesian product of two non-empty Set P and Q:
$P \times Q = \left \{ (p,q) : p \ \epsilon \ P , q \ \epsilon \ Q \right \}$
Now, we can see that
P = set of all first elements.
And
Q = set of all second elements.
Now,
$\Rightarrow$ (x, y, z) are elements of A and (1,2) are elements of B
As n(A) = 3 and n(B) = 2
Therefore,
A = {x, y, z} and B = {1, 2}

It is given that Cartesian product A × A having 9 elements among which are found (–1, 0) and (0,1).
Now,
Number of elements in (A× B) = (Number of elements in set A) × (Number of elements in B)
$n(A \times A) = n(A) \times n(A)$
It is given that $n(A \times A) = 9$
Therefore,
$n(A) \times n(A) = 9$
$\Rightarrow n(A) = 3$
Now,
By definition A × A = {(a, a): a ? A}
Therefore,
-1, 0 and 1 are the elements of set A
Now, because, n(A) = 3 therefore, A = {-1, 0, 1}
Therefore,
the remaining elements of set (A × A) are
(-1,-1), (-1,1), (0,0), (0, -1), (1,1), (1, -1) and (1, 0)

## More About NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1:-

Class 11 Maths chapter 2 exercise 2.1 solutions consist of basic questions related to properties of ordered pairs and cartesian product of sets. The basic knowledge of the set, subset, and power set is required to understand the cartesian products of sets, sets of ordered pairs. There are six examples and few definitions given before the exercise 2.1 Class 11 Maths. You must go through these solved examples and definitions in order to understand the concept clearly. There are a total of 10 questions in the class 11 maths ex 2.1 that you can solve by yourself.

Also Read| Relations And Functions Class 11th Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 2 Exercise 2.1:-

• Exercise 2.1 Class 11 Maths contains the basic questions related cartesian product of sets. This concept will be needed to understand the upcoming exercises.
• NCERT syllabus Class 11 Maths chapter 2 exercise 2.1 solutions are useful when you are stuck while solving the NCERT problems.
• 11th class maths exercise 2.1 answers solutions are useful for reference while solving the NCERT problems.
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## Key Features of NCERT Solutions for Class 11 Maths Chapter 2 Exercise

1. Detailed step-by-step solutions: The class 11 maths ex 2.1 solutions provide a comprehensive, step-by-step explanation for each problem, making it easier for students to understand the concepts and problem-solving techniques.

2. Clarity and accuracy: The ex 2.1 class 11 solutions are clear and accurate, ensuring that students can confidently prepare for their exams and grasp the concepts effectively.

3. Aligned with NCERT curriculum: The solutions are specifically designed to align with the NCERT curriculum, ensuring that they cover the topics and concepts as per the official syllabus.

4. Free access: These class 11 maths chapter 2 exercise 2.1 solutions are often available for free, making them accessible to a wide range of students.

5. Format options: PDF versions of the class 11 ex 2.1 solutions is provided, allowing students to download and use them anytime and anywhere for their convenience.

## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Do the ordered pairs (2,1) and (1,2) are equal ?

No, two ordered pairs are equal, if and only if the corresponding elements of both the pairs are equal.

2. If (x + 1, 1) = (3,1), find the values of x ?

x +1 = 3

x = 3 -1 =2

3. If the P and Q are two different sets, does cartesian product P x Q and Q x P are same ?

No, cartesian product  P x Q and Q x P are not same. The number of elements in each set are the same.

4. If P = {1, 2}, than find the set P x P ?

P = { 1,2}

P x P = { (1,1), (1,2), (2,1), (2,2) }

5. If P = {1, 3}, than find the set P x P x P ?

P = { 1,3}

P x P x P=  {(1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3)}.

6. If the number of elements in the set A is 2 and in set B is 3 than find the number of elements in the A x B ?

n(A) = 2 and n(B) = 3

n(A × B) = 2x3 = 6

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