﻿ Class 11 NCERT Math Solution
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TOPICS
Exercise - 2.1

Question-1 :-  If (x/3 + 1, y – 2/3) = (5/3, 1/3), find the values of x and y.

Solution :-
```   In the Ordered Pairs, the corresponding values are equal.
So, x/3 + 1 = 5/3
x/3 = 5/3 - 1
x/3 = (5 - 3)/3
3x = 2/3
x = 2/3 x 3
x = 2
y - 2/3 = 1/3
y = 1/3 + 2/3
y = (1 + 2)/3
y = 3/3
y = 1
```

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 :-
```   Given that set A has 3 elements and the elements of set B are 3, 4, and 5.
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
```

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

Solution :-
```   Given that G = {7, 8} and H = {5, 4, 2}.
G x H = {7, 8} x {5, 4, 2}
= {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 3)}
H x G = {5, 4, 2} x {7, 8}
= {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
```

Question-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 ∩ φ) = φ.

Solution :-
```(i) False, If 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 :-
```   Given that A = {–1, 1}.
A x A = {–1, 1} x {–1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
A x A x A = {(-1, -1), (-1, 1), (1, -1), (1, 1)} x {-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)}
```

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)}.
A = {a, b}
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 :-
```    Given that A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}.
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
L.H.S
= B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
= A × (B ∩ C) = A × Φ = Φ
R.H.S
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) = Φ
L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
```
```(ii) A × C is a subset of B × D
L.H.S
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)}
We can observe that all the elements of set A × C are the elements of set B × D.
Therefore, A × C is a subset of B × D.
```

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

Solution :-
```   Given that A = {1, 2} and B = {3, 4}.
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
n(A × B) = 4
We know that if C is a set with n(C) = m, then n[P(C)] = 2ᵐ.
Therefore, the set A × B has 2⁴ = 16 subsets.
These are {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, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)},
{(1, 3), (1, 4), (2, 3), (2, 4)} }
```

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. If (x, 1), (y, 2), (z, 1) are in A × B.
A = Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
x, y, and z are the elements of A; and 1 and 2 are the elements of B.
Since n(A) = 3 and n(B) = 2, 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.

Solution :-
```   We know that if n(A) = p and n(B) = q, then n(A × B) = pq. ∴ n(A × A) = n(A) × n(A)
Given that n(A × A) = 9
n(A) × n(A) = 9 = 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}.
Therefore, –1, 0, and 1 are elements of A.
Since n(A) = 3, it is clear that A = {–1, 0, 1}.
A × A = {(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), (1, 1)}.
```
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