{"id":2329,"date":"2021-04-03T09:48:37","date_gmt":"2021-04-03T04:18:37","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=2329"},"modified":"2021-04-03T10:21:06","modified_gmt":"2021-04-03T04:51:06","slug":"samacheer-kalvi-11th-maths-solutions-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-1-ex-1-1\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1<\/h2>\n

11th Maths Exercise 1.1 Answers Question 1.<\/strong>
\nWrite the following in roster form.
\n(i) {x \u2208 N : x2<\/sup> < 121 and x is a prime}.
\n(ii) the set of all positive roots of the equation (x – 1)(x + 1)(x2<\/sup> – 1) = 0.
\n(iii) {x \u2208 N : 4x + 9 < 52}.
\n(iv) {x : \\(\\frac{x-4}{x+2}\\) = 3, x \u2208 R – {-2}}
\nSolution:
\n(i) A = {2, 3, 5, 7}
\n(ii) B = {1}
\n(iii) 4x + 9 < 52
\n4x + 9 – 9 < 52 – 9
\n4x < 43
\nx < \\(\\frac{43}{4}\\) (i.e.) x < 10.75 4
\nBut x \u2208 N
\n\u2234 A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
\n\"11th
\n(i.e.) x – 4 = 3(x + 2)
\nx – 4 = 3x + 6
\n– 4 – 6 = 3x – x
\n2x = -10 \u21d2 x = -5
\nA = {-5}<\/p>\n

11th Std Maths Exercise 1.1 Answers Question 2.<\/strong>
\nWrite the set {-1,1} in set builder form.
\nSolution:
\nA = {x: x2<\/sup> = 1}<\/p>\n

11th Maths Book Exercise 1.1 Answers Question 3.<\/strong>
\nState whether the following sets are finite or infinite.
\n(i) {x \u2208 N : x is an even prime number}
\n(ii) {x \u2208 N : x is an odd prime number}
\n(iii) {x \u2208 Z : x is even and less than 10}
\n(iv) {x \u2208 R : x is a rational number}
\n(v) {x \u2208 N : x is a rational number}
\nSolution:
\n(i) Finite set
\n(ii) Infinite set
\n(iii) Infinite
\n(iv) and
\n(v) infinite<\/p>\n

Exercise 1.1 Class 11 Maths State Board Question 4.<\/strong>
\nBy taking suitable sets A, B, C, verify the following results:
\n(i) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C).
\n(if) A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C).
\n(iii) (A \u00d7 B) \u2229 (B \u00d7 A) = (A \u2229 B) \u00d7 (B \u2229 A).
\n(iv) C – (B – A) = (C \u2229 A) \u222a (C \u2229 B).
\n(v) (B – A) \u2229 C = (B \u2229 C) – A = B \u2229 (C – A).
\nSolution:
\nTo prove the following results let us take U = {1, 2, 5, 7, 8, 9, 10}
\nA = {1, 2, 5, 7}
\nB = {2, 7, 8, 9}
\nC = {1, 5, 8, 7}
\n(i) To prove: A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\nB \u2229 C = {8}; A = {1, 2, 5, 7}
\nSo A \u00d7 (B \u2229 C) = {1, 2, 5, 7} \u00d7 {8}
\n= {(1, 8), (2. 8), (5, 8), (7, 8)}
\nNow A x B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9), (7, 2), (7, 7), (7, 8), (7, 9)} …. ( 1)
\nA \u00d7 C = {(1, 1), (1, 5),(1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
\n(A \u00d7 B) \u2229 (A \u00d7 C) = {(1, 8), (2, 8), (5, 8), (7, 8)} ……… (2)
\n(1) = (2)
\n\u21d2 A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n

(ii) To prove A \u00d7 (B \u222a C) = (A \u00d7 B) (A \u00d7 C)
\nB = {2, 7, 8, 9}, C = {1, 5, 8, 10)
\nB \u222a C = {1, 2, 5, 7, 8, 9, 10}
\nA = {1, 2, 5, 7}
\nA \u00d7 (B \u222a C) = {(1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)) …. (1)
\nA \u00d7 B = {(1, 2), (1, 7), (1, 8), (1, 9), (2, 2), (2, 7), (2, 8), (2, 9), (5, 2), (5, 7), (5, 8), (5, 9),
\n(7, 2), (7, 7), (7, 8), (7, 9)}
\nA \u00d7 C = {(1, 1), (1, 5), (1, 8), (1, 10), (2, 1), (2, 5), (2, 8), (2, 10), (5, 1), (5, 5), (5, 8), (5, 10), (7, 1), (7, 5), (7, 8), (7, 10)}
\n(A \u00d7 B) \u222a (A \u00d7 C) = (1, 1), (1, 2), (1, 5), (1, 7), (1, 8), (1,9), (1, 10), (2, 1), (2, 2), (2, 5), (2, 7), (2, 8), (2, 9), (2, 10), (5, 1), (5, 2), (5, 5), (5, 7), (5, 8), (5, 9), (5, 10), (7, 1), (7, 2), (7, 5), (7, 7), (7, 8), (7, 9), (7, 10)} …… (2)
\n(1) = (2) \u21d2 A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)<\/p>\n

(iii) A \u00d7 B = {(1, 2), (1, 7), (1, 8), (1, 9) (2, 2), (2, 7), (2, 8), (2, 9) (5, 2), (5, 7), (5, 8), (5, 9) (7, 2), (7, 7), (7, 8), (7, 9)}
\nB \u00d7 A = {(2, 1), (2, 2), (2, 5), (2, 7) (7, 1), (7, 2), (7, 5), (7, 7) (8, 1), (8, 2), (8, 5), (8, 7) (9,1), (9, 2), (9, 5), (9, 7)}
\nL.H.S. (A \u00d7 B) \u2229 (B \u00d7 A) = {(2, 2), (2, 7), (7, 2), (7, 7)} …. (1)
\nR.H.S. A \u2229 B = {2, 7}
\nB \u2229 A = {2, 7}
\n(A \u2229 B) \u00d7 (B \u2229 A) = {2, 7} \u00d7 {2, 7}
\n= {(2, 2), (2, 7), (7, 2), (7, 7)} ……… (2)
\n(1) = (2) \u21d2 LHS = RHS<\/p>\n

(iv) To prove C – (B – A) = (C \u2229 A) \u222a (C \u2229 B)
\nB – A = {8, 9}
\nC = {1, 5, 8, 10}
\n\u2234 LHS = C – (B – A) = {1, 5, 10} …… (1)
\nC \u2229 A = {1}
\nU = {1, 2, 5, 7, 8, 9, 10}
\nB = {2, 7, 8, 9} \u2234 B’ = {1, 5, 10}
\nC \u2229 B = {1, 5, 10}
\nR.H.S. (C \u2229 A) \u222a (C \u2229 B) = {1} \u222a {1, 5, 10}
\n= {1, 5, 10} ……. (2)
\n(1) = (2) \u21d2 LHS = RHS<\/p>\n

(v) To prove (B – A) \u2229 C = (B \u2229 C) – A = B \u2229 (C – A)
\nA= {1, 2, 5, 7}, B = {2, 7, 8, 9}, C = {1, 5, 8, 10}
\nNow B – A = {8, 9}
\n(B – A) \u2229 C = {8} ……. (1)
\nB \u2229 C = {8}
\nA = {1, 2, 5, 7}
\nSo (B \u2229 C) – A = {8} …… (2)
\nC – A = {8, 10}
\nB = {2, 7, 8, 9}
\nB \u2229 (C – A) = {8} …. (3)
\n(1) = (2) = (3)<\/p>\n

(vi) To prove (B – A) \u222a C ={1, 5, 8, 9, 10}
\nB – A = {8, 9},
\nC = {1, 5, 8, 10}
\n(B – A) \u222a C = {1, 5, 8, 9, 10} ……. (1)
\nB \u222a C = {1, 2, 5, 7, 8, 9, 10}
\nA – C = {2, 7}
\n(B \u222a C) – (A – C) = {1, 5, 8, 9, 10} ……… (2)
\n(1) = (2)
\n\u21d2 (B – A) \u222a C = (B \u222a C) – (A – C)<\/p>\n

11th Maths Exercise 1.1 Question 5.<\/strong>
\nJustify the trueness of the statement.
\n\u201cAn element of a set can never be a subset of itself.\u201d
\nSolution:
\nA set itself can be a subset of itself (i.e.) A \u2286 A. But it cannot be a proper subset.<\/p>\n

11th Maths Chapter 1 Exercise 1.1 Question 6.<\/strong>
\nIf n(P(A)) = 1024, n(A \u222a B) = 15 and n(P(B)) = 32, then find n(A \u2229 B).
\nSolution:
\nn(P( A)) = 1024 = 210<\/sup> \u21d2 n( A) = 10
\nn(A \u222a B) = 15
\nn(P(B)) = 32 = 25<\/sup> \u21d2 n(B) = 5
\nWe know n(A \u222a B) = n{A) + n(B) – n(A \u2229 B)
\n(i.e.) 15 = 10 + 5 – n(A \u2229 B)
\n\u21d2 n(A \u2229 B) = 15 – 15 = 0<\/p>\n

11 Maths Exercise 1.1 Question 7.<\/strong>
\nIf n(A \u2229 B) = 3 and n(A \u222a B) = 10, then find n(P(A(A \u2206 B)).
\nSolution:
\nn(A \u222a B) = 10; n(A \u2229 B) = 3
\nn(A \u2206 B) = 10 – 3 = 7
\nand n(P(A \u2206 B)) = 27 = 128<\/p>\n

11th Maths 1.1 Exercise Question 8.<\/strong>
\nFor a set A, A \u00d7 A contains 16 elements and two of its elements are (1, 3) and (0, 2). Find the elements of A.
\nSolution:
\nA \u00d7 A = 16 elements = 4 \u00d7 4
\n\u21d2 A has 4 elements
\n\u2234 A = {0, 1, 2, 3}<\/p>\n

11th Maths Exercise 1.1 Answers In Tamil Question 9.<\/strong>
\nLet A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A \u00d7 B, find A and B, where x, y, z are distinct elements.
\nSolution:
\nn(A) = 3 \u21d2 set A contains 3 elements
\nn(B) = 2 \u21d2 set B contains 2 elements –
\nwe are given (x, 1), (y, 2), (z, 1) are elements in A \u00d7 B \u21d2 A = {x, y, z} and B = {1, 2}<\/p>\n

11th Maths Exercise 1.1 Answers State Board Question 10.<\/strong>
\nIf A \u00d7 A has 16 elements, S = {(a, b) \u2208 A \u00d7 A : a < b} ; (-1, 2) and (0, 1) are two elements of S, then find the remaining elements of S.
\nSolution:
\nn(A \u00d7 A) = 16 \u21d2 n( A) = 4
\nS ={(-1, 0), (-1, 1), (0, 2), (1, 2)}<\/p>\n

Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 Additional Questions<\/h3>\n

Samacheer Kalvi 11th Maths Example Sums Question 1.<\/strong>
\nWrite the following sets in roster form
\n(a) {x \u2208 N; x3<\/sup> < 1000}
\n(b) {The set of positive roots of the equation (x2<\/sup> – 4) (x3<\/sup> – 27) = 0}
\nSolution:
\n(a) A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
\n(b) B = {2, 3}<\/p>\n

11th Maths Exercise 1.1 4th Sum Question 2.<\/strong>
\nBy taking suitable sets A, B, C verify the following results
\n(i) A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)
\n(ii) (B – A) \u222a C = (B \u222a C) – (A – C)
\nSolution:
\nProve by yourself<\/p>\n

11th Maths 1st Chapter Exercise 1.1 Question 3.<\/strong>
\nGiven n(A) = 7; n(B) = 8 and n(A \u222a B) = 10 find n[P(A \u2229 B)].
\nSolution:
\nn(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n(i.e.,) 10 = 7 + 8 – n(A \u2229 B)
\n\u21d2 n(A \u2229 B) = 7 + 8 – 10 = 5
\nSo n[P(A \u2229 B)] = 25 = 32<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 1 Sets Ex 1.1 11th Maths Exercise 1.1 Answers Question 1. Write the following in roster form. (i) {x \u2208 …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/2329"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=2329"}],"version-history":[{"count":0,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/2329\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=2329"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=2329"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=2329"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}