பத்தாம் வகுப்பிற்கு கணித பாடத்தில் கற்பிக்கும் ஆசிரியர்களும் மற்றும் மாணவ மாணவியர்களுக்கும் அன்பான வணக்கம்...,
நமது
www.kanimaths.com website ஆசிரியர்களுக்கும் மாணவ மாணவியர்களுக்கும் கணித பாடத்தில்
பயன்படும் வகையில்
Chapter - 1 : Relations & Functions EXERCISE : 1.1 அனைத்து கணக்குகளும் தனித்தனியாக தொகுக்கப்பட்டு
விடைகளுடன் solved செய்து கீழே கொடுக்கப்பட்டுள்ளது.
ஆசிரியர்களுக்கு பாடம் நடத்தும் வேலையில் ஏதேனும்
சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை refer செய்து கொள்ளலாம்.Chapter
-1 பயிற்சி கணக்குகள் அனைத்தும் தனித்தனியாக தொகுக்கப்பட்டு கீழே கொடுக்கப்பட்டுள்ளது.
மேலும் மாணவர்களுக்கும் சந்தேகங்கள் இருந்தால்
கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை பார்த்து சந்தேகங்களை சரி பார்த்துக் கொள்ளலாம்.
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) A = {2,-2,3}, B = {1,-4}
A × B = {(2, 1), (2, -4), (-2, 1), (-2, -4), (3,1) , (3,-4)}
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) 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} × Φ
A × B = { }
A × A = {(m, m), (m, n), (n, m), (n, n)}
B × A = { }
Question 2.
Let A= {1,2,3} and B = {× | x is a prime number less than 10}. Find A × B and B × A
Answer:
A = {1,2,3}, B = {2, 3, 5, 7}
A × B = {1,2,3} × {2, 3, 5, 7}
= {(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, 3, 5, 7} × {1,2,3}
= {(2, 1)(2, 2)(2, 3)(3, 1)(3, 2)(3, 3) (5, 1)(5, 2)(5, 3) (7, 1) (7,2)(7, 3)}
Question 3.
If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.
Solution:
B × A ={(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}
A = {3, 4), B = { -2, 0, 3}
Question 4.
If A= {5, 6}, B = {4, 5 ,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)
Answer:
A ={5,6}, B = {4,5,6}, C = {5, 6,7}
A × A = {5, 6} × {5,6}
= {(5, 5) (5, 6) (6, 5) (6, 6)} ….(1)
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) we get
A × A = (B × B) ∩ (C × C)
Question 5.
Given A ={1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A ∩ C) x (B ∩ D) = (A × B) ∩ (C × D) is true?
Solution:
LHS = {(A∩C) × (B∩D)
A ∩C = {3}
B ∩D = {3, 5}
(A ∩ C) × (B ∩ D) = {(3, 3) (3, 5)} ………….. (1)
RHS = (A × B) ∩ (C × D)
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)
∴ (1) = (2) ∴ It is true.
Question 6.
Let A = {x ∈ W | x < 2},
B = {x ∈ N | 1 < 1 < × < 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)
Answer:
(i) A = {0, 1}
B = {2,3,4}
C = {3,5}
(i) A × (B ∪ C) = (A × B) ∪ (A × c)
B ∪ C = {2, 3,4} ∪ {3,5}
= {2, 3, 4, 5}
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)} ….(1)
A × B = {0, 1} × {2,3,4}
= {(0,2) (0,3) (0,4) (1,2) (1,3) (1,4) }
A × C = {0, 1} × {3, 5}
{(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) we get
A × (B ∪ C) = (A × B) ∪ (A × C)
(ii) A × (B n C) = (A × B) n (A × C)
B ∩ C = {2,3,4} ∩ {3,5}
= {3}
A × (B ∩ C) = {0, 1} × {3}
= {(0,3) (1,3)} ….(1)
A × B = {0,1} × {2,3,4}
= {(0, 2) (0, 3) (0, 4) (1,2) (1,3) (1,4)}
A × C = {0,1} × {3,5}
{(0, 3) (0, 5) (1,3) (1,5)}
(A × B) n (A × C) = {(0, 3) (1, 3)} ….(2)
From (1) and (2) we get
A × ( B n C) = (A × B) n (A × C)
(iii) (A ∪ B) × C = (A × C) ∪ (B × C)
A ∪ B = {0, 1} ∪ {2,3,4}
= {0,1, 2, 3, 4}
(A ∪ B) × C = {0, 1,2, 3,4} × {3,5}
= {(0, 3) (0, 5) (1, 3) (1, 5)(2, 3) (2, 5) (3, 3)(3, 5) (4, 3)(4, 5)} ….(1)
A × C = {0, 1} × {3,5}
= {(0,3) (0,5) (1,3) (1,5)}
B × C = {2,3,4} × {3,5}
= {(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) we get
(A ∪ B) × C = (A × C) ∪ (B × C)
Question 7.
Let A = The set of all natural numbers less than 8, B = The set of all prime numbers 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)
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7}
C = {2}
Solution:
(i)(A ∩ B) × C = (A × c) ∩ (B × C)
LHS = (A ∩ B) × C
A ∩ B = {2, 3, 5, 7}
(A ∩ B) × C = {(2, 2), (3, 2), (5, 2), (7, 2)} ………… (1)
RHS = (A × C) ∩ (B × C)
(A × C) = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (7, 2)}
(B × C) = {2, 2), (3, 2), (5, 2), (7, 2)}
(A × C) ∩ (B × C) = {(2, 2), (3, 2), (5, 2), (7, 2)} ……….. (2)
(1) = (2)
∴ LHS = RHS. Hence it is verified.
(ii) A × (B – C) = (A × B) – (A × C)
LHS = A × (B – C)
(B – C) = {3,5,7}
A × (B – C) = {(1, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7) , (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3) , (6, 5), (6, 7), (7, 3), (7, 5), (7, 7)} …………. (1)
RHS = (A × B) – (A × C)
(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),
(4, 2), (4, 3), (4, 5), (4, 7),
(5, 2), (5, 3), (5, 5), (5, 7),
(6, 2), (6, 3), (6, 5), (6, 7),
(7, 2), (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, 3), (1, 5), (1, 7), (2, 3), (2, 5), (2, 7), (3, 3), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 3), (5, 5), (5, 7), (6, 3), (6, 5), (6, 7), (7, 3), (7, 5), (7,7) } ………….. (2)
(1) = (2) ⇒ LHS = RHS.
Hence it is verified.
பத்தாம்
வகுப்பு கணித பாடத்திற்கு அனைத்து பயிற்சி கணக்குகளுக்கும் அனிமேஷன் மூலம் எடுக்கப்பட்டுள்ளது.மேலும்
மாணவர்களுக்கு சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள CLICK HERE என்ற ஆங்கில வார்த்தையை
பயன்படுத்தி சந்தேகங்களை சரிபார்த்துக் கொள்ளலாம்.
.jpg)
.jpg)
.jpg)
0 Comments