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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.