RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS-1

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

RELATIONS Exercise 2A, 2B, 2C

Exercise 2A

1. (i) Find the values of a and b,
         when: (a + 3, b –2) = (5, 1)

Answer: Since the ordered pairs are equal,
Comparing corresponding elements,
 ∴ a + 3 = 5
⇒ a = 5 – 3 = 2 and
    b – 2 = 1
⇒ b = 1 + 2 = 3
∴ a = 2 and b = 3.

1. (ii) Find the values of a and b,
         when: (a + b, 2b – 3) = (4, –5)

Answer: Since the ordered pairs are equal.
Comparing corresponding elements,
 ∴ a + b = 4 …(i) and
2b – 3 = -5 …(ii)
From (ii) we get,
    2b – 3 = -5
⇒ 2b = -5 + 3
⇒ 2b = -2
∴ b = -1

Putting the value of b = – 1 in equation (i),
 we get
  ∴ a – 1 = 4
⇒ a – 1 = 4
⇒ a = 4 + 1 = 5
∴ a = 5 and b =- 1.

Q. 1. (iii) Find the values of a and b,
         when:  (^a/₃ + 1, b –  ¹/₃) = (⁵/₃, ²/₃)

Answer: Since the ordered pairs are equal.
Comparing corresponding elements,
 ∴ ^a/₃+ 1 = ⁵/₃⇒ ^a/₃= ⁵/₃– 1
⇒ ^a/₃ = ²/₃⇒ a = 2 and
    b –  ¹/₃  = ²/₃
⇒ b  = ²/₃ + ¹/₃
⇒ b = 1
 ∴ a = 2 and b = 1.

Q. 1. (iv) Find the values of a and b,
         when: (a – 2, 2b + 1 = (b – 1, a + 2)

Answer: Since the ordered pairs are equal,
Comparing corresponding elements,
 ∴ a – 2 = b – 1 …(i) and
 2b + 1 = a + 2 …(ii)
From equation (i) we get
    a – 2 = b – 1
⇒ a – b = (-1) + 2
⇒ a – b = 1 … (iii)
From equation (ii) we get
    2b + 1 = a + 2
⇒ 2b – a = 2 – 1
⇒ -a + 2b = 1 …(iv)

Add (iii) and (iv),
    a – b + (-a) + 2b = 1 + 1
⇒ a – b – a + 2b = 2
⇒ b = 2
  Putting b = 2 in equation (iii), we get
    a – 2 = 1
⇒ a = 1 + 2
⇒ a = 3
∴  a = 3 and b = 2.

Q. 2. If A = {0, 1} and B = {1, 2, 3},
         show that A × B ≠ B × A.

Answer: Given: A = {0, 1} and B = {1, 2, 3}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
∵ A × B = {(x, y): x ∈ A, y ∈ B} ∵ A × B = {(x, y): x ∈ A, y ∈ B}

 ∴ A × B
= {0, 1} × {1, 2, 3}
= {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}
B × A
= (1, 2, 3) × (0, 1)
= {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 1)}
Here (0, 1) ∈ A × B but (0, 1) ∉ B × A
∴ A × B ≠ B × A (Proved)

Q. 3. If P = {a, b} and Q = {x, y, z},
        show that P × Q ≠ Q × P.

Answer: Given: P = {a, b} and Q = {x, y, z}
∵ P × Q = {(p, q) : p ∈ P and q ∈ Q}
 ∴ P × Q
= (a, b) × (x, y, z)
= {(a, x), (a, y), (a, z), (b, x), (b, y), (b, z)}
    Q × P
= (x, y, z) × (a, b)
= {(x, a), (x, b), (y, a), (y, b), (z, a), (z, b)}
Here (a, x) ∈ P × Q but (a, x) ∉ Q × P
∴ P × Q ≠ Q × P (Proved)

Q. 4. (i) If A = {2, 3, 5} and B = {5, 7}, find: A × B

Answer: Given: A = {2, 3, 5} and B = {5, 7}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴ A × B
= {2, 3, 5} × {5, 7}
= {(2, 5),  (2, 7), (3, 5), (3, 7), (5, 5), (5, 7)} (Ans)

Q. 4. (ii) If A = {2, 3, 5} and B = {5, 7}, find: B × A

Answer: Given: A = {2, 3, 5} and B = {5, 7}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴ B × A
= {5, 7} × {2, 3, 5}
= {(5, 2), (5, 3), (5, 5), (7, 2), (7, 3), (7, 5)} (Ans)

Q. 4. (iii) If A = {2, 3, 5} and B = {5, 7}, find: A × A

Answer: Given: A = {2, 3, 5}∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴  A × A
= (2, 3, 5) × (2, 3, 5)
= {(2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5)} (Ans)

Q. 4. (iv) If A = {2, 3, 5} and B = {5, 7}, find: B × B

Answer: Given: B = {5, 7} ∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴ B × B
= (5, 7) × (5, 7)
= {(5, 5), (5, 7), (7, 5), (7, 7)} (Ans)

Q. 5. If A = {x ∈ N : x ≤ 3} and {x v W : x < 2}, find (A × B) and (B × A).
         Is (A × B) = (B × A)?

Answer: Given:
A = {x ∈ N: x ≤ 3}
 ∴ A = {1, 2, 3} and 
B = {x ∈ W: x < 2}
 ∴ B = {0, 1}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴ A × B = {1, 2, 3} × {0, 1}
= {(1, 0), (1, 1), (2, 0), (2, 1), (3, 0), (3, 2)} 
   B × A
= {0, 1} × {1, 2, 3}
= {(0, 1), (0, 2), (0, 3), (1, 1), (1, 2), (1, 3)}
Here (0, 1) ∈ A × B but (0, 1) ∉ B × A
∴ A × B ≠ B × A

Q. 6. (i) If A = {1, 3, 5) B = {3, 4} and C = {2, 3},
         verify that: A × (B ∪ C) = (A × B) ∪ (A × C)

Answer: Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3}
   B ∪ C = {2, 3, 4} 
∵ A × B = {(x, y): x ∈ A, y ∈ B}
L. H. S.
= A × (B ∪ C)
⇒  {1, 3, 5} ×  {2, 3, 4}
= {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}

A × B 
= {1, 3, 5} × {3, 4}
= {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} and 
A × C = {1, 3, 5} × {2, 3}
= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)} 
R. H. S.
(A × B) ⋃ (A × C)
= {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} ⋃ {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
= {(1, 2), (1, 3), (1, 4), (3, 2), (3, 3), (3, 4), (5, 2), (5, 3), (5, 4)}
⇒ L. H. S.
∴ A × (B ∪ C) = (A × B) ∪ (A × C)   (verified)

Q. 6. (ii) If A = {1, 3, 5) B = {3, 4} and C = {2, 3},
         verify that: A × (B ∩ C) = (A × B) ∩ (A × C)

Answer: Given: A = {1, 3, 5}, B = {3, 4} and C = {2, 3} 
(B ⋂ C) = {3}
L. H. S.
= A × (B ⋂ C)
⇒ {1, 3, 5} × {3}
= {(1, 3), (3, 3), (5, 3)}
Now, 
A × B = {1, 3, 5} × {3, 4}
= {(1, 3), (1, 4), (3, 3), (3, 4), (5, 3), (5, 4)} and 
A × C = {1, 3, 5} × {2, 3}
= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
R. H. S.
= (A × B) ⋂ (A × C)
⇒ {(1, 3), (3, 3), (5, 3)}
= L. H. S.
∴ A × (B ∩ C) = (A × B) ∩ (A × C) (verified)

Q. 7. (i) Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4} and C = {3, 5}.
         Verify that: A × (B ∪ C) = (A × B) ∪ (A × C)

Answer: (i) Given:
A = {x ∈ W : x < 2}
∴ A = {0, 1}
B = {x ∈ N : 1 < x ≤ 4}
∴ B = {2, 3, 4}
and C = {3, 5}
B ⋃ C = {2, 3, 4, 5}

L. H. S.
= A × (B ⋃ C)
⇒ {0, 1} × {2, 3, 4, 5}
= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}
Now, A × B
     = {0, 1} × {2, 3, 4}
     = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
and A × C
     = {0, 1} × {3, 5}
     = {(0, 3), (0, 5), (1, 3), (1, 5)}
R. H. S.
= (A × B) ⋃ (A × C)
= {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} = L. H. S.
∴ A × (B ∪ C) = (A × B) ∪ (A × C) (verified)

Q. 7. (ii) Let A = {x ∈ W : x < 2}, B = {x ∈ N : 1 < x ≤ 4} and C = {3, 5}.
         Verify that: A × (B ∩ C) = (A × B) ∩ (A × C)

Answer: Given:
A = {x ∈ W : x < 2}
∴ A = {0, 1}
B = {x ∈ N : 1 < x ≤ 4}
∴ B = {2, 3, 4}
and C = {3, 5}
B ⋂ C = {3}
L. H. S.
= A × (B ⋂ C)
⇒ {0, 1} × {3}
= {(0, 3), (1, 3)}
Now, A × B
      = {0, 1} × {2, 3, 4}
      = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
and A × C
    = {0, 1} × {3, 5}
    = {(0, 3), (0, 5), (1, 3), (1, 5)}
R. H. S.
= (A × B) ⋂ (A × C)
= {(0, 3), (1, 3)} = L. H. S.
∴ A × (B ∩ C) = (A × B) ∩ (A × C) (verified)

Q. 8. If A × B = {(–2, 3), (–2, 4), (0, 3), (0, 4), (3, 3), (3, 4), find A and B.

Answer: Here, A × B = {(–2, 3), (–2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} 
∵ A × B = {(x, y): x ∈ A, y ∈ B}
Clearly, A is the set of all first entries in ordered pairs in A × B
∴ A = {-2, 0, 3}
and B is the set of all second entries in ordered pairs in A × B
∴ B = {3, 4}

Q. 9. Let A = {2, 3} and B = {4, 5}. Find (A × B).
         How many subsets will (A × B) have?

Answer: Given: A = {2, 3} and B = {4, 5}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
   A × B
= {2, 3} × {4, 5}
= {(2, 4), (2, 5), (3, 4), (3, 5)}
Number of elements of A× B
 = n(A×B) = 4
∴ Number of subsets of A × B
 = 2ⁿ = 2⁴ = 16
∴ A × B has 16 subsets.

Q. 10. Let A × B = {(a, b): b = 3a – 2}. if (x, –5) and (2, y) belong to A × B,
           find the values of x and y.

Answer: Given: A × B = {(a, b): b = 3a – 2} and ((x, -5), (2, y)) ∈ A × B
∵ b = 3a – 2
For (x, -5) ∈ A × B
    -5 = 3.x – 2
⇒ 3x =  -5 + 2
⇒ 3x = -3
∴ x = -1
For (2, y) ∈ A × B
   2 = a and
    y = b
⇒ y = 3a – 2
⇒ y = 3.2 – 2
∴ y = 4
∴ x = -1 and y = 4

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Q. 11. Let A and B be two sets such that n(A) = 3 and n(B) = 2. If a ≠ b ≠ c and (a, 0), (b, 1), (c, 0) is in A × B, find A and B.

Answer:  (a, 0), (b, 1), (c, 0) are the elements of A × B.
∴ a, b, c ∈ A and 0, 1 ∈ B
Given: n(A) = 3 and n(B) = 2 
∴ A = {a, b, c} and 
   B = {0, 1}

Q. 12. (i) Let A = {–2, 2} and B = {0, 3, 5}. Find: A × B
Represent each of the above (a) graphically and (b) by arrow diagram.

Answer: Given: A = {-2, 2} and B = {0, 3, 5}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
A × B
= {–2, 2} × {0, 3, 5}
= {(-2, 0), (-2, 3), (-2, 5), (2, 0), (2, 3), (2, 5)}

GRAPH:

ARROW DIAGRAM:

Q. 12. (ii) Let A = {–2, 2} and B = {0, 3, 5}. Find: B × A
Represent each of the above (a) graphically and (b) by arrow diagram.

Answer: Given: A = {-2, 2} and B = {0, 3, 5}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
B × A
= {0, 3, 5} × {–2, 2}
= {(0, -2), (0, 2), (3, -2), (3, 2), (5, -2), (5, 2)}

GRAPH:

ARROW DIAGRAM:

Q. 12. (iii) Let A = {–2, 2} and B = {0, 3, 5}. Find: A × A
Represent each of the above (a) graphically and (b) by arrow diagram.

Answer: Given: A = {-2, 2}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
A × A
= {-2, 2} × {-2, 2}
= {(-2, -2), (-2, 2), (2, -2), (2, 2)} 

GRAPH:

ARROW DIAGRAM:

Q. 12. (iv) Let A = {–2, 2} and B = {0, 3, 5}. Find: B × B
Represent each of the above (a) graphically and (b) by arrow diagram.

Answer: Given: B = {0, 3, 5} ∵ A × B = {(x, y): x ∈ A, y ∈ B}
B × B
= {0, 3, 5} × {0, 3, 5}
= {(0, 0), (0, 3), (0, 5), (3, 0), (3, 3), (3, 5), (5, 0), (5, 3), (5, 5)}

GRAPH:

ARROW DIAGRAM:

Q. 13. (i) If A = {5, 7), find: A × A

Answer: Given: A = {5, 7}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
∴ A × A
=  {5, 7}× {5, 7}
= {(5, 5), (5, 7), (7, 5), (7. 7)}

Q. 13. (ii) If A = {5, 7), find: A × A × A.

Answer: Given: A = {5, 7}
∵ A × B × C = {(x, y, z): x ∈ A, y ∈ B, z ∈ C}
∴ A × A × A
=  {5, 7}× {5, 7}× {5, 7}
={(5, 5, 5), (5, 5, 7), (5, 7, 5), (5, 7, 7), (7, 5, 5), (7, 5, 7), (7, 7, 5), (7, 7, 7)}

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

Q. 14. (i) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: A × B

Answer: Given: A = {-3, -1} and B = {1, 3}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
 A × B
= {-3, -1} × {1, 3}
= {(-3, 1), (-3, 3), (-1, 1), (-1, 3)}

Q. 14. (ii) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: (A × B) × C

Answer: Given: A = {-3, -1} and B = {1, 3} and C = {3, 5}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
(A × B) × C
= [{-3, -1} × {1, 3}] × {3, 5}
⇒ {(-3, 1), (-3, 3), (-1, 1), (-1, 3)} × (3, 5)
= (-3, 1, 3), (-3, 1 , 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}

Q. 14. (iii) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: B × C

Answer: Given: B = {1, 3} and C = {3, 5}
∵ A × B
= {(x, y): x ∈ A, y ∈ B}
B × C
= (1, 3) × (3, 5)
= {(1, 3), (1, 5), (3, 3), (3, 5)}

Q. 14. (iv) Let A = {–3, –1}, B = {1, 3) and C = {3, 5). Find: A × (B × C)

Answer: Given: A = {-3, -1}, B = {1, 3} and C = {3, 5}
∵ A × B = {(x, y): x ∈ A, y ∈ B}
 B × C
=  {1, 3} × {3, 5}
= {(1, 3), (1, 5), (3, 3), (3, 5)} 
A × (B × C)
= {-3, -1} × {(1, 3), (1, 5), (3, 3), (3, 5)}
= (-3, 1, 3), (-3, 1, 5), (-3, 3, 3), (-3, 3, 5), (-1, 1, 3), (-1, 1, 5), (-1, 3, 3), (-1, 3, 5)}

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

Exercise 2B

Q. 1 (i) For any sets A, B and C prove that:
A × (B ∪ C) = (A × B) ∪ (A × C)

Answer: Let us consider, (x, y) ∈ A × (B ∪ C)
⇒ x ∈ A and y ∈ (B ∪ C)
⇒ x ∈ A and (y ∈ B or y ∈ C)
= (x ∈ A and y ∈ B) or (x ∈ A and y ∈ C)
⇒ (x, y) ∈ (A × B) or (x, y) ∈ (A × C)
⇒ (x, y) ∈ (A × B) ∪ (A × C)
∴ A × (B ∪ C) ⊆ (A × B) ∪ (A × C) —- (i)

Let us consider again (a, b) ∈ (A × B) ∪ (A × C)
⇒ (a, b) ∈ (A × B) or (a, b) ∈ (A × C)
⇒ (a ∈ A and b ∈ B) or (a ∈ A and b ∈ C)
= a ∈ A and (b ∈ B or b ∈ C)
⇒ a ∈ A and b ∈ (B ∪ C)
⇒ (a, b) ∈ A × (B ∪ C)
∴ (A × B) ∪ (A × C) ⊆ A × (B ∪ C) —- (ii)
From (i) and (ii) we can conclude that,
A × (B ∪ C) = (A × B) ∪ (A × C) (Proved)

Q. 1 (ii) For any sets A, B and C prove that:
A × (B ∩ C) = (A × B) ∩ (A × C)

Answer: Let us consider, (x, y) ∈ A × (B ∩ C)
⇒ x ∈ A and y ∈ (B ∩ C)
⇒ x ∈ A and (y ∈ B and y ∈ C)
= (x ∈ A and y ∈ B) and (x ∈ A and y ∈ C)
⇒ (x, y) ∈ (A × B) and (x, y) ∈ (A × C)
⇒ (x, y) ∈ (A × B) ∩ (A × C)
∴ A × (B ∩ C) ⊆ (A × B) ∩ (A × C) —- (i)

Let us consider again,
(a, b)∈ (A × B) ∩ (A × C)
⇒ (a, b) ∈ (A × B) and (a, b) ∈ (A × C)
⇒ (a ∈ A and b ∈ B) and (a ∈ A and b ∈ C)
= a ∈ A and (b ∈ B and b ∈ C)
⇒ a ∈ A and b ∈ (B ∩ C)
⇒ (a, b) ∈ A × (B ∩ C)
∴ (A × B) ∩ (A × C) ⊆ A × (B ∩ C) —- (ii)
From (i) and (ii) we can conclude that,
A × (B ∩ C) = (A × B) ∩ (A × C) (Proved)

Q. 1 (iii) For any sets A, B and C prove that:
A × (B – C) = (A × B) – (A × C)

Answer: Let us consider, (x, y) ∈ A × (B – C)
⇒ x ∈ A and y ∈ (B – C )
⇒ x ∈ A and (y ∈ B and y ∉ C)
= (x ∈ A and y ∈ B) and (x ∈ A and y ∉ C)
⇒ (x, y) ∈ (A × B) and (x, y) ∉ (A × C)
⇒ (x, y) ∈ [(A × B) – (A × C)]
∴  A × (B – C) ⊆ (A × B) – (A × C) —- (i)

Let us consider again, (a, b) ∈ [(A × B) – (A × C)]
⇒ (a, b) ∈ (A × B) and (a, b) ∉ (A × C)
⇒ (a ∈ A and b ∈ B) and (a ∈ A and b ∉ C)
= a ∈ A and (b ∈ B and b ∉ C)
⇒ a ∈ A and b ∈ (B – C)
⇒ (a, b) ∈ A × (B ∪ C)
∴ (A × B) – (A × C) ⊆ A × (B – C) —- (ii)
From (i) and (ii) we can conclude that,
A × (B – C) = (A × B) – (A × C) (Proved)

Q. 2. For any sets A and B, prove that
(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A).

Answer: Let us consider, (x, y) ∈ (A × B) ∩ (B × A)
⇒ (x, y) ∈ (A × B) and (x, y) ∈ (B × A)
⇒ (x ∈ A and y ∈ B) and (x ∈ B and y ∈ A)
= (x ∈ A and x ∈ B) and (y ∈ B and y ∈ A)
⇒ x ∈ (A × B) and y ∈ (B × A)
⇒ (x, y) ∈ (A × B) ∩ (B × A) 
∴ (A × B) ∩ (B × A) ⊆ (A ∩ B) × (B ∩ A) —- (i)

Let us consider again, (a, b) ∈ (A ∩ B) × (B ∩ A)
⇒ a ∈ (A ∩ B) and b ∈ (B ∩ A)
⇒ (a ∈ A and a ∈ B) and (b ∈ B and b ∈ A)
= (a ∈ A and b ∈ B) and (a ∈ B and b ∈ A)
⇒ (a, b) ∈ (A × B) and (a, b) ∈  (B × A)
⇒ (a, b) ∈ (A × B) ∩ (B × A)
∴ (A ∩ B) × (B ∩ A) ⊆ (A × B) ∩ (B × A) —- (ii)
From (i) and (ii) we can conclude that,
(A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A) (Proved)

Q. 3. If A and B are nonempty sets, prove that
A × B = B × A ⇔ A = B

Answer: Given: A = B, where A and B are nonempty sets.
Let us consider, (x, y) ∈ (A × B)
⇒ x ∈ A and y ∈ B
⇒ x ∈ B and y ∈ A….. [A = B]
= (x, y) ∈ (B × A)
∴ (A × B) ⊆ (B × A) —- (i)
Let us consider again,
(x, y) ∈ (B × A)
⇒ x ∈ B and y ∈ A
⇒ x ∈ A and y ∈ B….. [A = B]
= (x, y) ∈ (A × B)
⇒ (B × A) ⊆ (A × B) —- (ii)
From (i) and (ii), we can conclude that, A × B = B × A (Proved)

Q. 4. (i) If A ⊆ B, prove that A × C ⊆ B × C for any set C.

Answer: Given: A ⊆ B
Let us consider,
(x, y) ∈ (A × C)
⇒ x ∈ A and y ∈ C
= x ∈ B and y ∈ C ….. [A ⊆ B]
⇒ (x, y) (B × C)
∴ A × C ⊆ B × C (Proved)

Q. 4. (ii) If A ⊆ B and C ⊆ D then prove that A × C ⊆ B × D.

Answer: Given: A ⊆ B and C ⊆ D
Let us consider,
(x, y) ∈ (A × C)
⇒ x ∈ A and y ∈ C
= x ∈ B and y ∈ D ….. [A ⊆ B and C ⊆ D]
⇒ (x, y) ∈ (B × D)
∴ A × C ⊆ B × D (Proved)

Q. 5. If A × B ⊆ C × D and A × B ≠ φ, prove that A ⊆ C and B ⊆ D.

Answer: Given: A × B ⊆ C × D and A × B ≠ φ
Let us consider,
(x, y) ∈ (A × B)
⇒ x ∈ A and y ∈ B —- (i)
Again
 (x, y) ∈ (A × B)
⇒ (x, y) ∈ (C × D) …….. [ A × B ⊆ C × D]
⇒x ∈ C and y ∈ D —- (ii)
Comparing (i) and (ii) we can say that,
⇒ x ∈ A and x ∈ C
⇒ A ⊆ C
Again,
⇒ y ∈ B and y ∈ D
⇒ B ⊆ D (Proved)

RS AGGARWAL CLASS 11 MATHS SOLUTION RELATIONS

Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(A × B)

Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
∵  n(A × B) = n(A) × n(B)
⇒ n(A × B) = 3 × 4
⇒ n(A × B) = 12

Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(B × A)

Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
∵ n(B × A) = n(B) × n(A)
⇒ n(B × A) = 4 × 3
⇒ n(B × A) = 12

Q. 6. (i) If A and B be two sets such that n(A) = 3, n(B) = 4 and n(A ∩ B) = 2 then find. n(A × B) ∩ (B × A)

Answer: Given: n(A) = 3, n(B) = 4 and n(A ∩ B) = 2
   n((A × B) ∩ (B × A))
= n[(A ∩ B) × (B ∩ A)]
= n(A ∩ B) × n(B ∩ A)
⇒ 2×2 = 4

Q. 7. For any two sets A and B, show that A × B and B × A have an element in common if and only if A and B have an element in common.

Answer: Let us consider A ∩ B ≠ φ
Then  (A ∩ B) ∩ (B ∩ A)  ≠ φ
Let x be an element of A ∩ B
⇒ x ∈ A and x ∈ B
⇒ (x, x) ∈ A × B
Similarly
    x ∈ B and x ∈ A
⇒ (x, x) ∈ B × A
Therefore
  (x, x) ∈ A × B ∩ B ∩ A
⇒ (A × B) ∩ (B × A)  ≠ φ
That means A × B and B × A have an element in common if and only if A and B have an element in common. (Proved)

Q. 8. Let A = {1, 2} and B = {2, 3}. Then, write down all possible subsets of A × B.

Answer: Given: A = {1, 2} and B = {2, 3}
(A × B) = {(x, y): x ∈ A and y ∈ B}
∴   (A × B)
= {1, 2} × {2, 3}
= {(1, 2), (1, 3), (2,2), (2,3)}
So, all the possible subsets of A × B are:
φ, {(1, 2)}, {(1, 3)}, {(2,2)}, {(2,3)}, {(1, 2), (1, 3)}, {(1, 2), (2,2)}, {(1, 2), (2,3)}, {(1, 3), (2,2)}, {(1, 3), (2,3)}, {(2,2), (2,3)}, {(1, 2), (1, 3), (2,2)}, {(1, 2), (1, 3), (2,3)}, {(1, 3), (2,2), (2,3)}, {(2,2), (2,3), (1, 2)}, {(1, 2), (1, 3), (2,2), (2,3)}

Q. 9. (i) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
A × (B ∩ C) = (A × B) ∩ (A × C)

Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
   (B ∩ C) = {d, e}
L.H.S.
= A × (B ∩ C)
⇒ {a, b, c, d} × {d, e}
= {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)}

Now
   (A × B)
= {a, b, c, d} × {c, d, e}
= {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
  (A × C)
= {a, b, c, d} × {d, e, f, g}
= {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)}
R.H.S.
= (A × B) ∩ (A × C)
= {(a, d), (a, e), (b, d), (b, e), (c, d), (c, e), (d, d), (d, e)} = L.H.S.
∴ A × (B ∩ C) = (A × B) ∩ (A × C) (Proved)

Q. 9. (ii) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
A × (B – C) = (A × B) – (A × C)

Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
   (B – C) = {c}
L.H.S.
= A × (B – C)
⇒ {a, b, c, d} × {c}
= {(a, c), (b, c), (c, c), (d, c)}

Now
   (A × B)
= {a, b, c, d} × {c, d, e}
= {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
  (A × C)
= {a, b, c, d} × {d, e, f, g}
= {(a, d), (a, e), (a, f), (a, g), (b, d), (b, e), (b, f), (b, g), (c, d), (c, e), (c, f), (c, g), (d, d), (d, e), (d, f), (d, g)}
R.H.S.
    (A × B) – (A × C)
= {(a, c), (b, c), (c, c), (d, c)} = L.H.S.
∴ A × (B – C) = (A × B) – (A × C) (Proved)

Q. 9. (iii) Let A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}. Then verify each of the following identities:
(A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

Answer: Given: A = {a, b, c, d}, B = {c, d, e} and C = {d, e, f, g}
   (A × B)
= {a, b, c, d} × {c, d, e}
= {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, c), (c, d), (c, e), (d, c), (d, d), (d, e)}
   (B × A)
= {c, d, e} × {a, b, c, d}
= {(c, a), (c, b), (c, c), (c, d), (d, a), (d, b), (d, c), (d, d), (e, a), (e, b), (e, c), (e, d)}

L.H.S.
= (A × B) ∩ (B × A)
= {(c, c), (c, d), (d, c), (d, d)}
Now
(A ∩ B) = {c, d}
R.H.S.
= (A ∩ B) × (A ∩ B)
= {(c, c), (c, d), (d, c), (d, d)} = L.H.S.
∴ (A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B) (Proved)

Exercise 2C

Q. 1. Let A and B be two nonempty sets.
(i) What do you mean by a relation from A to B?
(ii) What do you mean by the domain and range of a relation?

Answer: (i) If A and B are two nonempty sets, then any subset of the set (A × B) is said to a relation R from set A to set B. 
Therefore R ⊆ (A × B).

(ii) Let R be a relation from A to B. Then, the set containing all the first elements of the ordered pairs belonging to R is called Domain. 
For the relation R,
Dom(R) = {x: (x, y) ∈R}
The set containing all the second elements of the ordered pair belonging to R is called Range. 
For the relation R,
Range(R) = {y: (x, y) ∈ R}

Q. 2. (i) Find the domain and range of each of the relations given below:
R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)}

Solution: Given: R = {(–1, 1), (1, 1), (–2, 4), (2, 4), (2, 4), (3, 9)}
∴ Dom(R)
= {x: (x, y) ∈ R}
= {-2, -1, 1, 2, 3}
∴ Range(R)
= {y: (x, y) ∈ R}
= {1, 4, 9}

Q. 2. (ii) Find the domain and range of each of the relations given below:
R = {(x, ¹/_x): x is an integer, 0 < x < 5} 

Solution: Given:
R = {(x, ¹/_x): x is an integer, 0 < x < 5}
⇒ R = {(1, 1), (2, ¹/₂), (3, ¹/₃), (4, ¹/₄)
∴ Dom(R)
= {x: (x, y) ∈ R}
= {1, 2, 3, 4}

∴ Range(R)
= {y: (x, y) ∈ R}
= {1, ¹/₂, ¹/₃, ¹/₄}

Q. 2. (iii) Find the domain and range of each of the relations given below:
R = {(x, y) : x + 2y = 8 and x, y ∈ N}

Solution: Given: R = {(x, y): x + 2y = 8 and x, y ∈ N}
⇒ R = {(2, 3), (4, 2), (6, 1)}
∴ Dom(R)
= {x: (x, y) ∈ R}
= {2, 4, 6}
∴ Range(R)
= {y: (x, y) ∈ R}
= {1, 2, 3}

Q. 2. (iv) Find the domain and range of each of the relations given below:
 R = {(x, y), : y = |x – 1|, x ∈ Z and |x| ≤ 3}

Solution: Given: R = {(x, y): y = |x – 1|, x ∈ Z and |x| ≤ 3}
⇒ R =  {(-3, 4), (-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1),  (3, 2)}
∴ Dom(R)
= {x: (x, y) ∈ R}
= {-3, -2, -1, 0, 1, 2, 3}
∴ Range(R)
= {y: (x, y) ∈ R}
= {0, 1, 2, 3, 4}

Q. 3. Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8}.
Let R = {(x, y), : x ∈ A, y ∈ B and x > y}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
(iii) Depict R by an arrow diagram.

Solution: Given: A = {1, 3, 5, 7} and B = {2, 4, 6, 8}
(i) R = {(x, y) : x ∈ A, y ∈ B and x > y}
∴ R in Roster Form:
R = {(3, 2), (5, 2), (5, 4), (7, 2), (7, 4), (7, 6)}
(ii) Dom(R) = {x: (x, y) ∈ R} = {3, 5, 7}
Range(R) = {y: (x, y) ∈ R} = {2, 4, 6}
(iii) Aarrow diagram:

ARROW DIAGRAM:

Q. 4. Let A = {2, 4, 5} and b = {1, 2, 3, 4, 6, 8}.
Let R = {(x, y) x ∈ A, y ∈ B and x divides y}.
(i)Write R in roster form.
(ii) Find dom (R) and range (R).??????

Solution: Given: A = {2, 4, 5} and b = {1, 2, 3, 4, 6, 8}
(i) R = {(x, y): x ∈ A, y ∈ B and x divides y}
∴ R in Roster Form:
R = {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8)}

(ii) Dom(R)
= {x: (x, y) ∈ R}
= {2, 4,}
Range(R)
=  {2, 4, 6, 8}

Q. 5. Let A = {2, 3, 4, 5} and B = {3, 6, 7, 10}.
Let R = {(x, y): x ∈ A, y ∈ B and x is relatively prime to y}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: Given: A = {2, 3, 4, 5} and B = {3, 6, 7, 10}
(i) R = {(x, y), : x ∈ A, y ∈ B and x is relatively prime to y}
∴ R in Roster Form:
R = {(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}

(ii) Dom(R)
= {x: (x, y) ∈ R}
= {2, 3, 4, 5}
Range(R)
= {y: (x, y) ∈ R}
= {3, 6, 7, 10}

Q. 6. Let A = {1, 2, 3, 5} And B = {4, 6, 9}.
Let R = {(x, y): x ∈ A, y ∈ B and (x – y) is odd}.
Write R in roster form.

Solution: Given: A = {1, 2, 3, 5} AND B = {4, 6, 9}
R = {(x, y): x ∈ A, y ∈ B and (x – y) is odd}
∴ R in Roster Form:
R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}

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Q. 7. Let A = {(x, y): x + 3y = 12, x ∈ N and y ∈ N}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: Given: A = {(x, y): x + 3y = 12, x ∈ N and y ∈ N}
∵ x + 3y = 12
⇒ 3y = 12 – x
⇒ y = ^{12 – x}/₃
If x= 1, y = ^{12 – 1}/₃ = ¹¹/₃ ∉ N
If x= 2, y = ^{12 – 2}/₃ = ¹⁰/₃ ∉ N
x= 3, y = ^{12 – 3}/₃ = ⁹/₃ = 3 ∈ N
If x= 6, y = ^{12 – 6}/₃ = ⁶/₃ = 2 ∈ N
If x= 9, y = ^{12 – 9}/₃ = ³/₃ = 1 ∈ N
x= 12, y = ^{12 – 12}/₃ = ⁰/₃ = 0 ∉ N
(i) ∴ R in Roster Form: R = {(3, 3), (6, 2), (9, 1)}
(ii) Dom(R) = {x: (x, y) ∈ R} = {3, 6, 9} Range(R) = {y: (x, y) ∈ R} = {1, 2, 3}

Q. 8. Let A = {1, 2, 3, 4, 5, 6}.
Define a relation R from A to A by R = {(x, y): y = x + 1}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).
(iii) What is its co-domain?
(iv) Depict R by using arrow diagram.

Solution: Given: A = {1, 2, 3, 4, 5, 6}
(i) R = {(x, y): y = x + 1}
∴ R in Roster Form:
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

(ii) Dom(R)
= {x: (x, y) ∈ R}
= {1, 2, 3, 4, 5}
Range(R)
= {y: (x, y) ∈ R}
= {2, 3, 4, 5, 6}

(iii) Here, y = x + 1
∴ Co-domain(R)
= {1, 2, 3, 4, 5, 6}

(iv) Aarrow diagram:

ARROW DIAGRAM:

Q. 9. Let R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: Given: R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
(i) R in Roster Form:
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

(ii) Dom(R) = {0, 1, 2, 3, 4, 5}
Range(R) = {5, 6, 7, 8, 9, 10}

Q. 10. Let A = {1, 2, 3, 4, 6} and R = {(a, b) : a, b ∈ A, and a divides b}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: Given: A = {1, 2, 3, 4, 6}
(i) R = {(a, b) : a, b ∈ A, and a divides b}
R in Roster Form:
R = {(1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)

(ii) Dom(R)
= {1, 2, 3, 4, 6}
Range(R)
= {2, 3, 4, 6}

Q. 11. Define a relation R from Z to Z, given by R = {(a, b): a, b ∈ Z and (a – b) is an integer}.
Find dom (R) and range (R).

Solution: Given: R = {(a, b): a, b ∈ Z and (a – b) is an integer}
The difference between any two integers is always an integer.
The domain of a relation is the set of all first elements of the ordered pairs.
Here first element is a and a is an integer.
∴ Dom (R) = Z
The range of a relation is the set of all second elements of the ordered pairs.
Here second element is b and b is an integer.
∴ Range (R) = Z

12. Let R = {(x, y): x, y ∈ Z and x² + y² ≤ 4}.
(i) Write R in roster form.
(ii) Find dom (R) and range (R).

Solution: Given: R = {(x, y): x, y ∈ Z and x² + y² ≤ 4}
(i) R in Roster Form:
R = {(-2, 0), (-1, -1), (-1, 0), (-1, 1), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (1, -1), (1, 0), (1, 1), (2, 0)}

(ii) Dom(R)
= {-2, -1, 0, 1, 2}
Range(R)
= {-2, -1, 0, 1, 2}

Q. 13. Let A = {2, 3} and B= {3, 5}
(i) Find (A × B) and n(A × B).
(ii) How many relations can be defined from A to B?

Solution: Given: A = {2, 3} and B= {3, 5}
(i) (A × B) = {(2, 3), (2, 5), (3, 3), (3, 5)}
Therefore, n(A × B) = 4

(ii) No. of relation from A to B is a subset of Cartesian product of (A × B).
Here no. of elements in A = 2 and
no. of elements in B = 2.
So, (A × B) = 2 × 2 = 4
So, the total number of relations can be defined from A to B is
= 2⁴ = 16

Q. 14. Let A = {3, 5} and B = {7, 9}. Let R = {(a, b): a ∈ A, b ∈ B and (a – b) is odd}.
Show that R is an empty relation from A to B.

Solution: Given: A = {3, 5} and B = {7, 9}
R = {(a, b): a ∈ A, b ∈ B and (a – b) is odd}
So, R = {(4, 7), (4, 9)}
The elements of set A and set B are odd.
The difference between two odd numbers is always even.
So (a – b) is always even.
There is not any elements in Relation R from A to B.
An empty relation means there is no elements in the relation set.
Therefore R is an empty relation from A to B.