10th Maths Book Back Relations and Functions Ex 1.1

Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 1 – Relations and Functions Ex 1.1 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 1 Relation and Functions consist of 6 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 1.1 Answers in English.




10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Chapter 1 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

Chapter 1

Exercise 1.1 Relations and Functions

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)
Solu.:
(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 = { }

2.Let A= {1,2,3} and B = {× | x is a prime number less than 10}. Find A × B and B × A.
Solu.:
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)}

3.If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4),(3, 3), (3, 4)} find A and B.
Solu.:
B × A ={(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}
A = {3, 4), B = { -2, 0, 3}

4.If A= {5, 6}, B = {4, 5 ,6}, C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C)
Solu.:

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)




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?
Solu.:
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.

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)
Solu.:

(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)

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 numbers. 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}
Solu.:
(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.

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