NCERT Solutions for Exercise 2.1 Class 11 Maths Chapter 2

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

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

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Access Relations And Functions Class 11 Chapter 2 Exercise: 2.1

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

Answer:

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

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

Answer:

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

Question:3 If G = {7, 8} and H = {5, 4, 2}, find $G \times H \: \: and \: \: H \times G$

Answer:

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)}

Question:4 (i) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If P = {m, n} and Q = { n, m}, then $P \times Q$ = {(m, n),(n, m)}.

Answer:

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

Question:4(ii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. 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 $y \epsilon B$

Answer:

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 $y \epsilon B$

Question:4 (iii) State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly. If A = {1, 2}, B = {3, 4}, then $A \times (B \cap \phi ) = \phi$

Answer:

This statement is TRUE

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

$B\cap\phi=\phi$

There for

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

Question:5 If A = {–1, 1}, find $A \times A \times A$

Answer:

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,
$A\times A \times A=A\times (A \times A) = \left \{ -1,1 \right \} \times \left \{ (-1,-1),(-1,1),(1,-1),(1,1) \right \}$ $=\left \{ (-1,-1,-1),(-1,-1,1),(-1,1,-1),(-1,1,1),(1,-1,-1),(1,-1,1),(1,1,-1),(1,1,1) \right \}$

Question:6 If $A \times B$ = {(a, x),(a , y), (b, x), (b, y)}. Find A and B.

Answer:

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,
$A= \left \{ a,b \right \} \ \ \ and \ \ \ B = \left \{ x , y \right \}$

Question:7 (i) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$ .

Answer:

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,
$A \times ( B \cap C ) = ( A \times B ) \cap ( A \times C )$

Question:7 (ii) Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that $A \times C$ is a subset of $B \times D$

Answer:

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 $B \times D$

Question:8 Let A = {1, 2} and B = {3, 4}. Write $A \times B$ . How many subsets will $A \times B$ have?
List them.

Answer:

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.

Question: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 $A \times B$ , find A and B, where x, y and z are distinct elements.

Answer:

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}

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

Answer:

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)

Key Features of NCERT Solutions for Class 11 Maths Chapter 2 Exercise

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.
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.
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.
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.
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.

Also see-

NCERT Solutions of Class 11 Subject Wise

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Frequently Asked Question (FAQs)

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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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

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

Access Relations And Functions Class 11 Chapter 2 Exercise: 2.1

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

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