{"id":25,"date":"2020-01-30T10:44:30","date_gmt":"2020-01-30T10:44:30","guid":{"rendered":"https:\/\/tnboardsolutions.com\/?p=25"},"modified":"2021-07-08T03:51:49","modified_gmt":"2021-07-08T09:21:49","slug":"samacheer-kalvi-10th-maths-guide-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/tnboardsolutions.com\/samacheer-kalvi-10th-maths-guide-chapter-1-ex-1-1\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 1 Relations and Functions Ex 1.1"},"content":{"rendered":"

Students can download Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.1<\/h2>\n

1. Find A \u00d7 B, A \u00d7 A and B \u00d7 A
\n(i) A = {2, -2, 3} and B = {1, -4}
\n(ii) A = B = {p, q}
\n(iii) A – {m, n} ; B = \u03a6
\nAnswer:
\n(i) A = {2, -2, 3} and B = {1, -4}
\nA \u00d7 B = {2,-2, 3} \u00d7 {1,-4}
\n= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}
\nA \u00d7 A = {2,-2, 3} \u00d7 {2,-2, 3}
\n= {(2, 2)(2, -2)(2, 3)(-2, 2)
\n(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}
\nB \u00d7 A = {1,-4} \u00d7 {2,-2, 3}
\n= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3)}<\/p>\n

\"Samacheer<\/p>\n

(ii) A = B = {p, q}
\nA \u00d7 B = {p, q) \u00d7 {p, q}
\n= {(p,p),(p,q)(q,p)(q,q)}
\nA \u00d7 A = {p,q) \u00d7 (p,q)
\n= {(p,p)(p,q)(q,p)(q,q)
\nB \u00d7 A = {p,q} \u00d7 {p,q}
\n= {(p,p)(p,q)(q,p)(q,q)<\/p>\n

(iii) A = {m, n} \u00d7 B = \u03a6
\nNote: B = \u03a6 or {}
\nA \u00d7 B = {m, n) \u00d7 { }
\n= { )
\nA \u00d7 A = {m, n) \u00d7 (m, n)}
\n= {(m, m)(w, w)(n, m)(n, n)}
\nB \u00d7 A = { } \u00d7 {w, n}
\n= { }<\/p>\n

Question 2.
\nLet A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A \u00d7 B and B \u00d7 A.
\nSolution:
\nA = {1, 2, 3}, B = {2, 3, 5, 7}
\nA \u00d7 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)}
\nB \u00d7 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)}<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nIf B \u00d7 A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.
\nAnswer:
\nB \u00d7 A = {(-2, 3)(-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}
\nA = {3,4}
\nB = {-2,0,3}<\/p>\n

Question 4.
\nIf A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A \u00d7 A = (B \u00d7 B) \u2229 (C \u00d7 C).
\nSolution:
\nA = {5,6}, B = {4, 5, 6},C = {5, 6, 7}
\nA \u00d7 A = {(5, 5), (5, 6), (6, 5), (6, 6)} \u2026…….. (1)
\nB \u00d7 B = {(4, 4), (4, 5), (4, 6), (5, 4),
\n(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} \u2026(2)
\nC \u00d7 C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),
\n(6, 7), (7, 5), (7, 6), (7, 7)} \u2026(3)
\n(B \u00d7 B) \u2229 (C \u00d7 C) = {(5, 5), (5,6), (6, 5), (6,6)} \u2026(4)
\n(1) = (4)
\nA \u00d7 A = (B \u00d7 B) \u2229 (C \u00d7 C)
\nIt is proved.<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nGiven A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if
\n(A \u2229 C) \u00d7 (B \u2229 D) = (A \u00d7 B) \u2229 (C \u00d7 D) is true?
\nAnswer:
\nA = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}
\nA \u2229 c = {1,2,3} \u2229 {3,4}
\n= (3}
\nB \u2229 D = {2,3, 5} \u2229 {1,3,5}
\n= {3,5}
\n(A \u2229 C) \u00d7 (B \u2229 D) = {3} \u00d7 {3,5}
\n= {(3, 3)(3, 5)} ….(1)
\nA \u00d7 B = {1,2,3} \u00d7 {2,3,5}
\n= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}
\nC \u00d7 D = {3,4} \u00d7 {1,3,5}
\n= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}
\n(A \u00d7 B) \u2229 (C \u00d7 D) = {(3, 3) (3, 5)} ….(2)
\nFrom (1) and (2) we get
\n(A \u2229 C) \u00d7 (B \u2229 D) = (A \u00d7 B) \u2229 (C \u00d7 D)
\nThis is true.<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nLet A = {x \u2208 W | x < 2}, B = {x \u2208 N |1 < x\u00a0<\u00a04} and C = {3, 5}. Verify that
\n(i) A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)
\n(ii) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\n(iii) (A \u222a B) \u00d7 C = (A \u00d7 C) \u222a (B \u00d7 C)
\n(iv) A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)
\nSolution:
\nA = {x \u2208 W|x < 2} = {0,1}
\nB = {x \u2208 N |1 < x\u00a0<\u00a04} = {2,3,4}
\nC = {3,5}
\nLHS =A \u00d7 (B \u222a C)
\nB \u222a C = {2, 3, 4} \u222a {3, 5}
\n= {2, 3, 4, 5}
\nA \u00d7 (B \u222a C) = {(0, 2), (0, 3), (0,4), (0, 5), (1, 2) , (1, 3), (1, 4),(1, 5)} \u2026……. (1)
\nRHS = (A \u00d7 B) \u222a (A \u00d7 C)
\n(A \u00d7 B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
\n(A \u00d7 C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
\n(A \u00d7 B) \u222a (A \u00d7 C)= {(0, 2), (0, 3), (0,4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5)} \u2026.(2)
\n(1) = (2), LHS = RHS
\nHence it is proved.<\/p>\n

\"Samacheer<\/p>\n

(ii) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\nLHS = A \u00d7 (B \u2229 C)
\n(B \u2229 C) = {3}
\nA \u00d7 (B \u2229 C) = {(0, 3), (1, 3)} \u2026(1)
\nRHS = (A \u00d7 B) \u2229 (A \u00d7 C)
\n(A \u00d7 B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}
\n(A \u00d7 C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
\n(A \u00d7 B) \u2229 (A \u00d7 C) = {(0, 3), (1, 3)} \u2026…….. (2)
\n(1) = (2) \u21d2 LHS = RHS.
\nHence it is verified.<\/p>\n

(iii) (A \u222a B) \u00d7 C = (A \u00d7 C) \u222a (B \u00d7 C)
\nLHS = (A \u222a B) \u00d7 C
\nA \u222a B = {0, 1, 2, 3, 4}
\n(A \u222a B) \u00d7 C = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} …………. (1)
\nRHS = (A \u00d7 C) \u222a (B \u00d7 C)
\n(A \u00d7 C) = {(0, 3), (0, 5), (1, 3), (1, 5)}
\n(B \u00d7 C) = {(2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)}
\n(A \u00d7 C) \u222a (B \u00d7 C) = {(0, 3), (0, 5), (1, 3), (1, 5), (2, 3), (2, 5), (3, 3), (3, 5), (4, 3), (4, 5)} \u2026……… (2)
\n(1) = (2)
\n\u2234 LHS = RHS. Hence it is verified.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nLet 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
\n(i) (A \u2229 B) \u00d7 C = (A \u00d7 c) \u2229 (B \u00d7 C)
\n(ii) A \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C)
\nAnswer:
\nA = {1,2, 3, 4, 5,6, 7}
\nB = {2, 3, 5,7}
\nC = {2}<\/p>\n

(i) (A \u2229 B) \u00d7 C = (A \u00d7 C) \u2229 (B \u00d7 C)
\nA \u2229 B = {1, 2, 3, 4, 5, 6, 7} \u2229 {2, 3, 5, 7}
\n= {2, 3, 5, 7}
\n(A \u2229 B) \u00d7 C = {2, 3, 5, 7} \u00d7 {2}
\n= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)
\nA \u00d7 C = {1,2, 3, 4, 5, 6, 7} \u00d7 {2}
\n= {(1,2) (2, 2) (3, 2) (4, 2)
\n(5.2) (6, 2) (7, 2)}
\nB \u00d7 C = {2, 3, 5, 7} \u00d7 {2}
\n= {(2, 2) (3, 2) (5, 2) (7, 2)}
\n(A \u00d7 C) \u2229 (B \u00d7 C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)
\nFrom (1) and (2) we get
\n(A \u2229 B) \u00d7 C = (A \u00d7 C) \u2229 (B \u00d7 C)<\/p>\n

\"Samacheer<\/p>\n

(ii) A \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C)
\nB – C = {2, 3, 5, 7} – {2}
\n= {3,5,7}
\nA \u00d7 (B – C) = {1,2, 3, 4, 5, 6,7} \u00d7 {3,5,7}
\n= {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5)
\n(2, 7) (3, 3) (3, 5) (3, 7) (4, 3)
\n(4, 5) (4, 7) (5, 3) (5, 5) (5, 7)
\n(6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ………….(1)
\nA \u00d7 B = {1,2, 3, 4, 5, 6, 7} \u00d7 {2, 3, 5,7}
\n= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3)
\n(2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7)
\n(4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5)
\n(5, 7) (6, 2) (6, 3) (6, 5) (6, 7)
\n(7, 2) (7, 3) (7, 5) (7, 7)}
\nA \u00d7 C = {1,2, 3,4, 5, 6, 7} \u00d7 {2}
\n= {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6.2) (7,2)}
\n(A \u00d7 B) – (A \u00d7 C) = {(1, 3) (1, 5) (1, 7)
\n(2, 3) (2, 5) (2, 7) (3, 3) (3, 5)
\n(3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5)
\n(5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ….(2)
\nFrom (1) and (2) we get
\nA \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C)<\/p>\n

\"Samacheer<\/p>\n

Relations
\nLet A and B be any two non-empty sets. A “relation” R from A to B is a subset of A \u00d7 B satisfying some specified conditions.<\/p>\n

Note:<\/p>\n

    \n
  1. The domain of the relations R = {x \u2208 A\/xRy, for some y \u2208 B}<\/li>\n
  2. The co-domain of the relation R is B<\/li>\n
  3. The range of the ralation<\/li>\n<\/ol>\n

    R = (y \u2208 B\/xRy for some x \u2208 A}<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can download Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams. Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/25"}],"collection":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/comments?post=25"}],"version-history":[{"count":0,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/25\/revisions"}],"wp:attachment":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/media?parent=25"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/categories?post=25"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/tags?post=25"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}