10th Maths Guide Exercise 1.1 English Medium
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10th maths Guide English Medium
10th maths Exercise 1.1 Solution (Answers)
10th maths Example 1.1 Solutions
Question:
If A = {1, 3, 5} and B = {2, 3} then
(i) find A × B and B × A.
(ii) Is A × B = B × A? If not why ?
(iii) Show that n(A × B) = n(B × A) = n(A) × n(B).
Solution :
Given that A = {1, 3, 5} and B = {2, 3}
(i) A × B = {1, 3, 5} × {2, 3}
= {(1, 2), (1, 3), (3, 2), (3, 3), (5, 2), (5, 3)}
B × A = {2, 3} × {1, 3, 5}
= {(2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5)} .
(ii) From (1) and (2) we conclude that
A × B ≠ B × A as (1, 2) ≠ (2, 1)
and (1, 3) ≠) (3, 1), etc.
(iii) n(A) = 3; n(B) = 2.
From (1) and (2) we observe that,
n(A × B) = n(B × A) = 6;
we see that, n(A) × n(B) = 3 × 2 = 6 and
n(B) × n(A) = 2 × 3 = 6Hence,
n(A × B) = n(B × A) = n(A) × n(B) = 6.
Thus, n(A × B) = n(B × A) = n(A) × n(B).
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10th Maths Example 1.2 solutions
If A × B = {(3, 2), (3, 4), (5, 2), (5, 4)} then find A and B.
Solution :
A × B = {(3, 2), (3, 4), (5, 2), (5, 4)}
We have
A = {set of all first coordinates of elements of A × B}. Therefore A = {3, 5}
B = {set of all second coordinates of ele-ments of A × B}. ThereforeB = {2, 4}
Thus A = {3, 5} and B = {2, 4}.
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Samacheer kalvi 10th maths Example 1.3 solution
Let A = {x ∈ N | 1 < x < 4}, B = {x ∈ W | 0 ≤ x < 2} and C = {x ∈ N | x < 3}.Then verify that (i) A×(B∪C)= (A×B) ∪ (A×C) (ii) A × (B ∩ C) = (A × B) ∩ (A × C)
Solution :
A = {x ∈ N | 1 < x < 4} = {2, 3},
B = {x ∈ W | 0 ≤ x < 2} = {0, 1}.
C = {x ∈ N | x < 3} = {1, 2}
(i) A × (B ∪ C) = (A × B) ∪ (A × C) B ∪ C = {0, 1} ∪ {1, 2} = {0, 1, 2} A × (B ∪ C)
= {2, 3} × {0, 1, 2}
= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)}
A × B = {2, 3} × {0, 1}
= {(2, 0), (2, 1), (3, 0), (3, 1)}
A × C = {2, 3} × {1, 2}
= {(2, 1), (2, 2), (3, 1), (3, 2)} (A × B) ∪ (A × C)
= {(2, 0), (2, 1), (3, 0), (3, 1)} ∪ {(2, 1), (2, 2), (3, 1), (3, 2)}
= {(2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2)} ... (2)
From (1) and (2), A × (B ∪ C)
= (A × B) ∪ (A × C) is verified.
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(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
B ∩ C = {0, 1} ∩ {1, 2}= {1}
A × (B ∩ C) = {2, 3} × {1} = {(2, 1), (3, 1)} ... (3)
A × B = {2, 3} × {0, 1} = {(2, 0), (2, 1), (3, 0), (3, 1)}
A × C = {2, 3} ×{1, 2} = {(2, 1), (2, 2), (3, 1), (3, 2)}
(A × B) ∩ (A × C)
= {(2, 0), (2, 1), (3, 0), (3, 1)} ∩ {(2, 1), (2, 2), (3, 1), (3, 2)}
= {(2, 1), (3, 1)} ... (4)
From (3) and (4),
A×(B ∩ C) = (A ×B) ∩ (A × C) is verified.
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10th maths EXERCIsE 1.1 solution
10th maths Exercise 1.1 Question 1.
Find A × B, A × A and B × A (i) A = {2, – 2, 3} and B = {1, – 4} (ii) A = B = {p, q} (iii) A = {m, n} ; B = φ
Solution:
(i) Given A = {2, – 2, 3}, B = {1, – 4}.
A × B = {(2, 1), (2, – 4), (– 2, 1), (–2, – 4), (3, 1), (3, – 4)}i
A × A = {(2, 2), (2, – 2), (2, 3), (– 2, 2), (– 2, – 2), (– 2, 3), (3, 2), (3, – 2), (3, 3)}
B × A = {(1, 2), (1, – 2), (1, 3), (– 4, 2), (– 4, – 2), (– 4, 3)}
(ii) Given
A = B = {p, q} A × B = {(p, p), (p, q), (q, p), (q, q)}
A × A = {(p, p), (p, q), (q, p), (q, q)}
B × A = {(p, p), (p, q), (q, p), (q, q)}
(iii) A = {m, n}, B = φ
If A = φ (or) B = φ, then A × B = φ.
and B × A = φ
A × B = φ and B × A = φ
A × A = {(m, m), (m, n), (n, m), (n, n)}
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10th maths Exercise 1.1 Question 2.
Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.
Solution :
Given A = {1, 2, 3}, B = {x |
x is a prime number less than 10}.
B = {2, 3, 5, 7}
A × B = {(1, 2), (1, 3), (1, 5), (1, 7), (2, 2), (2, 3), (2, 5), (2, 7), (3, 2), (3, 3), (3, 5), (3, 7)}
B × A = {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (5, 1), (5, 2), (5, 3), (7, 1), (7, 2), (7, 3)}
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Exercise 1.1 Question 3.
If B × A = {(– 2, 3), (– 2, 4), (0, 3), (0, 4), (3, 3), (3, 4)} find A and B.
Solution :
Given B × A = {(–2, 3), (– 2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}
∴ B = {– 2, 0, 3}, A = {3, 4}
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Exercise 1.1 Question 4. .
If A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}. Show that A × A = (B × B) ∩ (C × C).
Solution :
Given A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}
LHS : A × A = {5, 6} × {5, 6}
= {(5, 5), (5, 6), (6, 5), (6, 6)} ...(1)
RHS : B ×B = {4, 5, 6} ×{4, 5, 6}
= {(4, 4), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
C × C = {5, 6, 7} × {5, 6, 7}
= {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6), (6, 7), (7, 5), (7, 6), (7, 7)}
∴ (B × B) ∩ (C × C) = {(5, 5), (5, 6), (6, 5), (6, 6)} ...(2)
∴ From (1) and (2). LHS = RHS
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10th maths Exercise 1.1 Question 5
Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true ?
Solution : Given A = {1, 2, 3}, B = {2, 3, 5},
C = {3, 4}, D = {1, 3, 5}
A ∩ C = {3}, B ∩ D = {3, 5}
∴ (A ∩ C) × (B ∩ D) = {(3, 3), (3, 5)} ... (1)
A × B = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5)}
C × D = {(3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5)}
∴ (A × B) ∩ (C × D) = {(3, 3), (3, 5)} ...(2)
∴ From (1) and (2) LHS = RHS.
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10th maths Exercise 1.1 Question 6.
Let A = {x ∈ W | x < 2}, B = {x ∈ N | 1 < x ≤ 4} and C = {3, 5}.
Verify that
(i) A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B ∩ C) = (A × B) ∩ (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
Solution :
Given A = {x ∈ W | x < 2} ⇒ A = {0, 1}
B = {x ∈ N | 1 < x ≤ 4} ⇒ B = {2, 3, 4}
C = {3, 5}
(i) To verify :
A × (B ∪ C) = (A × B) ∪ (A × C)
B ∪ C = {2, 3, 4, 5}
∴ A × (B ∪ C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} ...(1)
A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
∴ (A × B) ∪ (A × C) = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)} ...(2)
∴ From (1) and (2) LHS = RHS.
(ii) To verify :
A × (B ∩ C) = (A × B) ∩ (A × C) B ∩ C = {3}
∴ A × (B ∩ C) = {(0, 3), (1, 3)} ...(1)
A × B = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
∴ (A × B) ∩ (A × C) = {(0, 3), (1, 3)} ...(2)
∴ From (1) and (2), LHS = RHS.
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A ∪ B = {0, 1, 2, 3, 4}
∴ (A ∪ B) × C = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} ...(1)
A × C = {(0, 3), (0, 5), (1, 3), (1, 5)}
B × C = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}
∴ (A × C) ∪ (B × C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5) (3, 3), (3, 5), (4, 3), (4, 5)} ...(2)
∴ From (1) and (2) LHS = RHS.
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10th maths Exercise 1.1 Question 7. Let A = The set of all natural numbers less than 8,
B = The set of all prime num-bers less than 8,
C = The set of even prime number. Verify that
(i) (A ∩ B) × C = (A × C) ∩ (B × C)
(ii) A × (B – C) = (A × B) – (A × C)
Solution :
Given A = {1, 2, 3, 4, 5, 6, 7}
B = {1, 3, 5, 7}
C = {2}
(i) To verify :
(A ∩ B) × C = (A × C) ∩ (B × C)
A ∩ B = {1, 3, 5, 7}
∴ (A ∩ B) × C = {(1, 2), (3, 2), (5, 2), (7, 2)} ... (1)
A × C = {(1, 2), (2, 2),(3, 2),(4, 2),(5, 2),(6, 2), (7, 2)}
B × C = {(1, 2), (3, 2), (5, 2), (7, 2)} ... (2)
∴ From (1) and (2), LHS = RHS.
(ii) To verify :
A × (B – C) = (A × B) – (A × C)
B – C = {1, 3, 5, 7}
∴ A × (B – C) = {(1, 1), (1, 3), (1, 5), (1, 7),
(2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),
(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),}
A × B = {(1, 1), (1, 3), (1, 5), (1, 7),
(2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),
(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),}
A × C = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
∴ (A × B) – (A × C) = {(1, 1), (1, 3), (1, 5), (1, 7), (2, 1), (2, 3), (2, 5), (2, 7),
(3, 1), (3, 3), (3, 5), (3, 7),
(4, 1), (4, 3), (4, 5), (4, 7),
(5, 1), (5, 3), (5, 5), (5, 7),
(6, 1), (6, 3), (6, 5), (6, 7),
(7, 1), (7, 3), (7, 5), (7, 7),} ...(2)
∴ From (1) and (2), LHS = RHS.
10th maths Exercise 1.2 solutions Click here
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