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### Relations and Functions, Exercise No. 1.1, 10th New Syllabus, Mathematics, Tamil Nadu syllabus.

Relations and Functions,

Exercise No. 1.1

(1)  Find A x B , A x A and B x A
(i) A = {2, −2, 3} and B = {1, −4}
(ii) A = B = {p, q}
(iii) A = {m, n} ; B = ∅
Solution :
(i) A = {2, −2, 3} and B = {1, −4}
A x B  = {(2, 1)(2, -4)(-2, 1)(-2, -4)(3, 1)(3, -4)}
A = {2, −2, 3}, A = {2, −2, 3}
A x A
=  {(2, 2)(2, -2)(2, 3)(-2, 2)(-2, -2)(-2, 3)(3, 2)(3, -2)(3, 3)}
To find B x A from A x B, we have to interchange the first and second elements.
B x A  = {(1, 2)(-4, 2)(1, -2)(-4, -2)(1, 3)(-4, 3)}
(ii) A = B = {p, q}
A = {p, q} and B = {p, q}
A x B  = {(p, p) (p, q) (q, p) (q, q)}
A = {p, q}, A = {p, q}
A x A
=  {(p, p) (p, q) (q, p) (q, q)}
To find B x A from A x B, we have to interchange the first and second elements. Since the elem
B x A  = {(p, p) (q, p) (p, q) (q, q)}
(iii) A = {m, n} ; B = ∅
Since B = ∅, the value of A x B and B x A = ∅.
A = {m, n} and A = {m, n}
A x A  = {(m, m) (m, n) (n, m) (n, n)}

(2)  Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A x B and B x A.
Solution :
A = {1, 2, 3}
B = {x | x is a prime number less than 10}.
B  = {2, 3, 5, 7}
A x 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 x 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) }

(3)  If B × A = {(−2, 3),(−2, 4),(0, 3),(0, 4),(3, 3),(3, 4)} find A and B.
Solution :
In the Cartesian product B x A, first terms are set of elements of B and the second terms are set of elements of A.
B = {-2, 0, 3} and A = {3, 4}

(4)  If A = {5, 6} , B = {4, 5, 6} , C = {5, 6, 7} , Show that A × A = (B × B) n (C × C).
Solution :
A = {5, 6} , B = {4, 5, 6} , C = {5, 6, 7}
L.H.S
A = {5, 6} and A = {5, 6}
A x A   = {(5, 5) (5, 6) (6, 5) (6, 6)  ----(1)
R.H.S
B = {4, 5, 6} and B = {4, 5, 6}
B × B
= {(4, 4) (4, 5) (4, 6)(5, 4) (5, 5) (5, 6)(6, 4) (6, 5) (6, 6)}
C = {5, 6, 7} and C = {5, 6, 7}
C × C
= {(5, 5)(5, 6)(5, 7)(6, 5)(6, 6)(6, 7)(7, 5)(7, 6)(7, 7)}
(B × B) n (C × C)  = {(5, 5)(5, 6)(6, 5)(6, 6)}  -----(2)
(1)  = (2)
L.H.S  = R.H.S

(5)  Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A n C) × (B n D) = (A × B)n(C × D) is true?
Solution :
In order to check if the given statement is true, let us find values of L.H.S and R.H.S
A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}
A n C  means common elements of sets A and C.
A n C  = {3}
B n D means common elements of sets B and D.
B n D  = {3, 5}
L. H.S
(A n C) × (B n D)  = { (3, 3)(3, 5) }  -----(1)
(A × B)
A = {1, 2, 3}, B = {2, 3, 5}
A x B= {(1, 2)(1, 3)(1, 5)(2, 2)(2, 3)(2, 5) (3, 2)(3, 3)(3, 5)}
(C × D)
C = {3, 4} and D = {1, 3, 5}
C x D= {(3, 1)(3, 3)(3, 5)(4, 1)(4, 3)(4, 5)}
R.H.S
(A × B)n(C ×D)  = {(3, 3)(3,5)}  ------(2)
(1)  = (2)
Hence the given statement is true.

(6)  Let A = {x ∈ W | x < 2} , B = {x ∈|1 < x ≤ 4} and C = {3, 5} . Verify that
(i) A × (B U C) = (A × B) U (A × C)
(ii) A × (B n C) = (A × B) n (A × C)
(iii) (A U B) × C = (A × C) U (B × C)

Solution :
A = {x W | x < 2} , B = {x |1 < x 4} and C = {3, 5} .
A = {0, 1} , B = {2, 3, 4} and C = {3, 5} .
(i) A × (B U C) = (A × B) U (A × C)
L.H.S
(B U C)  = {2, 3, 4, 5}
A × (B U C)
= {(0, 2) (0, 3) (0, 4) (0, 5)(1, 2) (1, 3) (1, 4) (1, 5)}  --(1)
R.H.S
(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 x B) U (A x C)  = {(0, 2) (0, 3) (0, 4) (0, 5)(1, 2) (1, 3) (1, 4) (1, 5)}  --(2)
(1)  = (2)
Hence proved
(ii) A × (B n C) = (A × B) n (A × C)
A = {0, 1} , B = {2, 3, 4} and C = {3, 5} .
L.H.S
(B n C)  = {3}
A × (B n C)  = {(0, 3) (1, 3)}  ---(1)
R.H.S
(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) n (A × C)  = {(0, 3) (1, 3)} ---(2)
(1)  = (2)
Hence proved.
(iii) (A U B) × C = (A × C) U (B × C)
A = {0, 1} , B = {2, 3, 4} and C = {3, 5} .
A U B  = {0, 1, 2, 3, 4}
(A U B) × C  = { (0, 3)(0, 5)(1, 3)(1, 5)(2, 3)(2, 5)(3, 3)(3, 5)(4, 3)(4, 5) }  -----(1)
(A × B) = {(0, 2) (0, 3) (0, 4)(1, 2) (1, 3) (1, 4)}
(B × C) = {(2, 3) (2, 5) (3, 3)(3, 5) (4, 3) (4, 5)}
(A × C) U (B × C) =  { (0, 3)(0, 5)(1, 3)(1, 5)(2, 3)(2, 5)(3, 3)(3, 5)(4, 3)(4, 5) }  -----(2)
(1)  = (2)
Hence proved,

(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 n B) × C = (A × C) n (B × C)
(ii) A × (B −C) = (A × B) − (A × C)
Solution :
A = The set of all natural numbers less than 8
Natural number starts with 1.
A  = {1, 2, 3, 4, 5, 6, 7}
B = The set of all prime numbers less than 8
A number which is divisible by 1 and itself are known as prime numbers.
B = {2, 3, 5, 7}
C = The set of even prime number
C = {2}   2 is the one and only even prime number.
(i) (A n B) × C = (A × C) n (B × C)
L.H.S
(A n B)  = {2, 3, 5, 7}
(A n B) × C  = {(2, 2) (3, 2) (5, 2) (7, 2)}  ----(1)
R.H.S
(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) n (B × C)  = {(2, 2) (3, 2) (5, 2) (7, 2)}  ----(2)
(1)  = (2)
Hence proved.
(ii) A × (B − C) = (A × B) − (A × C)
A  = {1, 2, 3, 4, 5, 6, 7}, B = {2, 3, 5, 7} and C = {2}
Solution :
(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)
(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)
Hence proved.

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