Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 2 Relations and Functions Ex 2.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths Chapter 2 Relations and Functions Ex 2.1
Question 1.
If (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = (\(\frac{5}{3}\), \(\frac{1}{3}\))
find the values of x and y.
Answer.
It is given that (\(\frac{x}{3}\) + 1, y – \(\frac{2}{3}\)) = (\(\frac{5}{3}\), \(\frac{1}{3}\))
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 = \(\frac{1}{3}\) + \(\frac{2}{3}\)
⇒ \(\frac{x}{3}\) = \(\frac{5}{3}\) and y = \(\frac{3}{3}\)
⇒ x = 2 and y= 1
∴ x = 2 and 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).
Answer.
It is 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
Thus, 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.
Answer.
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)}
H × G = {(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.
Answer.
(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.
Answer.
It is known that for any non-empty set A, A × A × A is defined as
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)}
Question 6.
If A × B = {(o, x), (a, y), (b, x), (b, y)}. Find A and B.
Answer.
It is given that A × B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two non-empty sets P and Q is defined as P × Q = {(p, q): p ∈ P, q ∈ Q}
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
Answer.
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
We have B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
L.H.S. = A × (B ∩ C) = A × Φ = Φ
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) = Φ
L.H.S. = R.H.S
Hence, A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) To verify: A × C is a subset of B × D
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 x B have? List them.
Answer.
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 andz are distinct elements.
Answer.
It is given that n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
We know that 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.
Answer.
We know that if n(A) =p and n(B) =q, then n(A × B) = pq.
∴ n(A × A) = n(A) × n(A)
It is 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 x 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}.
The remaining elements of set A × A are (- 1, – 1), (- 1, 1), (0, – 1), (0, 0), (1, – 1), (1, 0), and (1, 1).
ਰੈਮ ਦੀ ਸ਼੍ਰੇਣੀ ਵੰਡ – ਰੈਮ ਨੂੰ ਹੇਠ ਲਿਖੀਆਂ ਸ਼੍ਰੇਣੀਆਂ ਵਿਚ ਵੰਡਿਆ ਜਾ ਸਕਦਾ ਹੈ-
ਬਟਨ ਤੇ ਕਲਿੱਕ ਕਰੋ ਜਾਂ Ctrl + ਬਟਨ ਤੇ ਕਲਿੱਕ ਕਰੋ ।
ਆਪਸ਼ਨ ‘ਤੇ ਕਲਿੱਕ ਕਰੋ ਅਤੇ ਜੇ ਤੁਸੀਂ ਕੋਈ ਨੰਬਰ ਪਾਉਣਾ ਚਾਹੁੰਦੇ ਹੋ ਤਾਂ ਨੰਬਰ 
ਬਟਨ ’ਤੇ ਕਲਿੱਕ ਕਰੋ ।
ਬਟਨਾਂ ’ਤੇ ਕਲਿੱਕ ਕਰੋ ।
ਬਟਨ ‘ਤੇ ਕਲਿੱਕ ਕਰੋ ਅਤੇ ਡਰਾਪਡਾਊਨ ਮੀਨੂੰ ਤੋਂ ਲੋੜੀਂਦੇ ਵਿਕਲਪ ’ਤੇ ਕਲਿੱਕ ਕਰੋ ।
ਇਹ ਵਿਕਲਪ ਪੈਰਾਗ੍ਰਾਫ ਜਾਂ ਚੁਣੇ ਟੈਕਸਟ ਦੀ ਬੈਕਗ੍ਰਾਊਂਡ ਨੂੰ ਰੰਗੀਨ ਬਣਾਉਣ ਲਈ ਕੰਮ ਕਰਦੀ ਹੈ ।
” ਇਹ ਵਿਕਲਪ ਚੁਣੇ ਸੈੱਲ ਜਾਂ ਟੈਕਸਟ ਦੀ ਬਾਰਡਰ ਨੂੰ
g
