Exercise 2.1 NCERT Class 11
Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10

Exercise 2.1 NCERT Class 11 from chapter Relations and Functions contains 10 questions.
In this exercise, the questions are based on the topic "Cartesian Product of Sets".

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Exercise 2.1 NCERT Class 11 Mathematics Solutions

The first question of Exercise 2.1 NCERT Class 11 is related to ordered pairs.

1. If (x3+1,y−23)=(53,13), Find the values of x and y.

∵ (x3+1,y−23) = (53,13)
(When ordered pairs are equal, then their corresponding first and second elements are also equal.)
So,
∴x3+1=53 ...(1)
y−23=13 ...(2)
Solving equation (1), we get
∴x3+1=53
⇒x3=53-1
⇒x3=23
⇒x=2
Solving equation (2), we get,
y−23=13
⇒y=13+23
⇒y=33
⇒y=1
Hence, the values of x and y are 2 and 1 respectively.

The second question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of two sets.

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

Here, the set A has 3 elements.
⇒ n(A) = 3
and the set B also has 3 elements.
⇒ n(B) = 3
∵ n(A×B) = n(A) × n(B) = 3 × 3 = 9
Hence, the number of elements in (A×B) is 9.

The third question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of two sets.

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

Given that G = {7, 8} and H = {5, 4, 2}
∵ G×H contains all the ordered pairs that can be formed taking first element from G and the second element from H.
So, G×H = {(7, 5), (7, 4), (7, 2), (8, 5), (8,4), (8,2)}
And H×G contains all the ordered pairs that can be formed taking the first element from H and the second element from G.
So, H×G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2,8)}

The fourth question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of two sets and ordered pairs.

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)}

The given statement is false, as the ordered pairs (m, m) and (n, n) are missing.
So, the correct statement will be
If P = {m, n} and Q = {n, m},
then P×Q = {(m, n), (m, m), (n, 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.

The given statement is true, as A×B is a non-empty set containing ordered pairs (x, y) such that x ∈ A and y ∈ B.

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

The given statement is true, as B ∩ ϕ = &#x3d5.
⇒ A×(B ∩ ϕ) = A×ϕ = ϕ

The fifth question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of three sets.

5. If A = {-1, 1}, find A × A × A.

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), (1, 1, −1), (1, 1, 1)}

The sixth question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of two sets.

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

∵ A×B = {(a,x),(a,y),(b,x),(b,y)}
We can find the set A from the set A×B by collecting the first element of each ordered pair and the set B by collecting the second elements of each ordered pair.
∴ A = {a,b} and B = {x,y}

The seventh question of Exercise 2.1 NCERT Class 11 is related to the cartesian product of sets and union and intersection of sets.

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)

Given sets are:
A = {1, 2}
B = {1, 2, 3, 4}
C = {5, 6}
D = {5, 6, 7, 8}
To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
Taking LHS,
  A×(B ∩ C)
= A×ϕ   [∵ B ∩ C = ϕ]
= ϕ
Taking RHS,
  (A×B) ∩ (A×C)
Here,
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)}
We can see that there is no common element in (A×B) and (A×C).
∴ (A×B) ∩ (A×C) = ϕ
⇒ LHS = RHS
Hence, verified.

(ii) A × C is a subset of B × D

The given sets are:
A = {1, 2}
B = {1, 2, 3, 4}
C = {5, 6}
D = {5, 6, 7, 8}
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)}
∵ All elements of A×C lie in B×D.
∴ A×C is a subset of B×D.

The eighth question of Exercise 2.1 NCERT Class 11 is related to subsets of sets.

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

We have,
A = {1, 2}
B = {3, 4}
A×B = {(1,3), (1,4), (2,3), (2,4)}
∵ Number of elements in A×B = n = 4.
∴ Number of subsets = 2n=24=16.
Subsets of A×B are listed 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, 4), (2, 3), (2, 4)},
{(2, 3), (2, 4), (1,3)}, {(2, 4), (1, 3), (1, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)}

The ninth question of Exercise 2.1 NCERT Class 11 is related to cartesian product of sets.

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.

Given: n(A) = 3 and n(B) = 2
(x,1),(y,2),(z,1)∈ A×B
∵ (x,1),(y,2),(z,1) are in A×B.
∴ the first elements of the ordered pairs belong to A and the second elements of the ordered pairs belong to B
⇒x∈A,y∈A,z∈A and n(A) = 3
⇒A={x,y,z}
Also, 1∈A,2∈A and n(B) = 2.
⇒B={1,2}
Hence, A = {x,y,z} and B = {1,2}

The tenth question of Exercise 2.1 NCERT Class 11 is related to cartesian product of sets.

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

Given that n(A×A) = 9,
(−1, 0) ∈ (A×A),
(0, 1) ∈ (A×A)
∵ n(A×A) = 9
⇒ n(A) = 3
i.e. the number of elements in A are 3.
∵ (−1, 0), (0,1) ∈ A×A
⇒ −1, 0, 1 ∈ A
And it is given that n(A) = 3.
So, it becomes clear that A = {−1, 0, 1}
∴ A×A = {(−1, −1), (−1, 0), (−1, 1), (0, −1), (0,0), (0,1), (1, −1), (1, 0), (1, 1)}
Hence, the remaining elements of A×A are
(−1, −1), (−1, 1), (0, −1), (0, 0), (1, −1), (1, 0), (1, 1)