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**CBSE Mathematics Guide
- Class XI NCERT Mathematics Solutions **

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**Chapter
2, relations and functions**

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__Cbse
Ncert Solutions Class 11 Maths Exercise 2.1__

__Cbse Ncert Solutions Class 11 Maths Exercise 2.1__

*(*

__Cbse Class 11 Ncert Math Textbook Page 33, 34__)**Question 1:**

__Solution__**:**

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

__Solution__**:**It is given that set A has 3 elements while the elements of set B are 3, 4, and 5.

i.e., Number of elements in set B = 3

Number of elements in (A × B)

= (Number of elements in A) × (Number of elements in B)

= 3 × 3 = 9

Hence, the number of elements in (A × B)
is 9.

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

__Solution__**:**

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

We know that the Cartesian product P × Q of two non-empty
sets P and Q is defined as -

P × Q = {(

*p*,*q*):*p*∈ P,*q*∈ Q}
∴ G ×
H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}, and

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

**Question 4:**State whether each of the following statement 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 ∩ Î¦) = Î¦.

__Solution__**:**

(i) False.

Since,
P = {

*m*,*n*} and Q = {*n*,*m*} then
P
× Q = {(

*m*,*m*), (*m*,*n*), (*n, m*), (*n*,*n*)}
(ii)
True. (iii) True.

**Question 5:**If A = {–1, 1}, find A × A × A.

__Solution__**:**

For any non-empty set A,

A × A × A = {(

*a*,*b*,*c*):*a*,*b*,*c*∈ A}
It is given that A = {–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)}

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**CBSE Class 11 NCERT Mathematics
Solutions ****Chapter 2, Relations and Functions Exercise 2.1**

**Question 6:**If A × B = {(

*a*,

*x*), (

*a*,

*y*), (

*b*,

*x*), (

*b*,

*y*)}. Find A and B.

**Solution****:**Given that A × B = {(

*a*,

*x*), (

*a, y*), (

*b*,

*x*), (

*b*,

*y*)}

For two non-empty sets P and Q is
defined as P × Q = {(

*p*,*q*):*p*∈ P,*q*∈ Q}
Since, A is the set of all
first elements and B is the set of all second elements.

Thus, A = {

*a*,*b*} and B = {*x*,*y*}

**Question 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****:**

(i)

__To verify__: A × (B ∩ C) = (A × B) ∩ (A × C)
From
the given elements we have, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Î¦

L.H.S
= A × (B ∩ C) = A × Î¦ = Î¦

From
the given elements we have,

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)}

R.H.S
= (A × B) ∩ (A × C) = Î¦

Thus,
L.H.S = R.H.S verified.

(ii)

__To verify__: A × C is a subset of B × D
From
the given elements we can write,

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

A
× 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)}

We can observe that all the elements of set A
× C are also present in the elements of set B × D.

Therefore, A × C is a subset of B × D,
verified.

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

(Hint: If C is a set with

*n*(C) =*m*, then*n*[P(C)] = 2^{m}. Try yourself)**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 × B, find A and B, where

*x*,

*y*and

*z*are distinct elements.

**Solution****:**Given that

*n*(A) = 3 and

*n*(B) = 2; and (

*x*, 1), (

*y*, 2), (

*z*, 1) are in A × B.

A and B are
Sets of the first and second elements respectively of the ordered pair elements
of A × B

As

*x*,*y*, and*z*are the elements of A; and 1 and 2 are the elements of B, so*n*(A) = 3 and

*n*(B) = 2.

Therefore, it is clear that A = {

*x*,*y*,*z*} and B = {1, 2}.**Question 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.

**We know that if**

__Solution__:*n*(A) =

*p*and

*n*(B) =

*q,*then

*n*(A × B) =

*pq*.

So,

*n*(A × A) =*n*(A) ×*n*(A)
Given that the Cartesian
product

*n*(A × A) has 9 elements so,*n*(A × A) = 9

Or,

*n*(A) ×*n*(A) = 9
Or,

*n*(A) = 3
The ordered pairs (–1, 0)
and (0, 1) are two of the nine elements of A × A.

We know that A × A = {(

*a, a*):*a*∈ A}. Hence –1, 0, and 1 are elements of A.
Since

*n*(A) = 3, then A = {–1, 0, 1} and the remaining elements of set A × A are
(–1, –1), (–1, 1), (0, –1),
(0, 0), (1, –1), (1, 0), and (1, 1).

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