{"id":170468,"date":"2024-05-23T17:05:08","date_gmt":"2024-05-23T11:35:08","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170468"},"modified":"2024-05-23T17:06:44","modified_gmt":"2024-05-23T11:36:44","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-5","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-5\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.5"},"content":{"rendered":"

Utilizing Understanding ISC Mathematics Class 11 Solutions<\/a> Chapter 1 Sets Ex 1.5 as a study aid can enhance exam preparation.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.5<\/h2>\n

Question 1.
\nState whether the following statemcnts are true or false :
\n(i) A \u222a A = A
\n(ii) A \u2229 A = \u03a6
\n(iii) A \u222a A’ = \u03be
\n(iv) A \u2229 A’ = \u03a6
\n(v) A \u2282\u03a6 = A \u21d2 A \u222a B = B
\n(vi) A \u2282 B \u21d2 A \u2229 B = A
\n(vii) A \u2229 \u03a6 = A
\n(viii) A \u222a \u03a6 = A
\nSolution:
\n(i) A \u222a A = {x : x \u2208 A or x \u2208 A}
\n= {x : x \u2208 A} = A
\n\u2234 Given given statement is true.<\/p>\n

(ii) A \u2229 A = {x : x \u2208 A and x \u2208 A}
\n= {x : x \u2208 A) = A
\n\u2234 given statement is false.<\/p>\n

(iii) A \u222a A’ = {x : x \u2208 A or x \u2208 A’}
\n= {x : x \u2208 A or x \u2209 A}
\n= {x : x \u2208 \u03be} = \u03be<\/p>\n

(iv) A \u2229 A’ = {x : x \u2208 A and x \u2208 A’}
\n= {x : x \u2208 A and x \u2209 A} = \u03a6
\n\u2234 given statement is true.<\/p>\n

(v) Given A \u2282 B …………………(1)
\nT.P. A \u222a B = B
\n\u2200 x \u2208 A \u2229 B
\n\u21d2 x \u2208 A or x \u2208 B
\n\u21d2 x \u2208 B or x \u2208 B [\u2235 by (1) ; A \u2282 B]
\n\u21d2 x \u2208 B
\n\u21d2 A \u222a B \u2282 B ……………..(2)
\n\u2200 x \u2208 B and A \u2282 B
\nx \u2208 B or x \u2208 A
\nx \u2208 A \u222a B
\nB \u2282 A \u222a B
\nFrom (2) and (3) ; we have
\nA \u222a B = B.
\nThus given statement is true.<\/p>\n

(vi) Given A \u2282 B …………………(1)
\nT.P. A \u2229 B = A
\n\u2200 x \u2208 A \u2229 B
\n\u21d2 x \u2208 A and x \u2208 B (\u2235 A \u2282 B)
\n\u21d2 x \u2208 A [\u2235 A \u2282 B]
\n\u2234 A \u2229 B \u2282 A …………………(2)
\n\u2200 x \u2208 A \u21d2 x \u2208 A and x \u2208 B (\u2235 A \u2282 B)
\n\u21d2 x \u2208 A \u2229 B
\n\u21d2 A \u2282 A \u2229 B ………………..(3)
\n\u2234 from (2) and (3) ; we have
\nA \u2229 B = A
\n\u2234 given statement is true.<\/p>\n

(vii) A \u2229 \u03a6 = {x : x \u2208 A \u2229 \u03a6}
\n= {x : x \u2208 A and x \u2208 \u03a6}
\n= {x : x \u2208 A} = A
\n\u2234 given statement is false.<\/p>\n

\"ML<\/p>\n

Question 2.
\nFill in the blanks to make each of the following a true statement :
\n(i) \u03a6’ \u2229 A = …………..
\n(ii) \u03be \u2229 A = ……………..
\nSolution:
\n(i) Now \u03a6’ = {x : x \u2208 \u03a6’}
\n= {x : x \u2209 \u03a6}
\n= {x : x \u2208 \u03be} = \u03be
\nAlso,
\n\u03a6’ \u2229 A = \u03be \u2229A
\n= {x : x \u2208 \u03be, and x \u2208 A}
\n= {x : x \u2208 A) = A<\/p>\n

(ii) Now \u03be’ = {x : x \u2208 \u03be’}
\n= {x : x \u2209 \u03be}
\n= {x : x \u2208 \u03a6} = A
\n\u2234 \u03be’ \u2229 A = \u03a6 \u2229 A
\n= {x : x \u2208\u03a6 and x \u2208 A}
\n= {x : x \u2208 \u03a6} = \u03a6<\/p>\n

Question 3.
\nFor any two sets A and B, is it truc that P (A) \u222a P (B) = P (A \u222a B) ? Justfy your answer.
\nSolution:
\nLet us take A = {1, 2}
\nand B = {2, 3}
\n\u2234 P(A) = {\u03a6, (1), {2}, {1, 2})
\nP (B) = {\u03a6, {2}, {3}, (2, 3))
\nThus P (A) \u222a P (B)
\n= {\u03a6, {1), {2}, (3), (1,2). {2,3}}
\nA \u222a B = {1, 2, 3}
\nP (A \u222a B) = {\u03a6 {1}, {2}, {3}, {1, 2} {1, 3}, {2, 3}, {1, 2, 3}}
\nClearly P (A) \u222a P (B) \u2260 P (A \u222a B).<\/p>\n

\"ML<\/p>\n

Question 4.
\nLet A, B be two sets. Prove that (A – B) \u222a B = A if and only if B \u2282 A.
\nSolution:
\nLet (A – B) \u222a B = A
\nT.P B \u2282 A
\n\u21d2 (A \u2229 B’) \u222a B = A
\n\u21d2 (A \u222a B) \u2229 (B’ \u222a B) = A
\n\u21d2 (A \u222a B) \u2229 U = A
\n\u21d2 A \u222a B = A
\n\u21d2 B \u2282 A
\nConverse :
\nB \u2282 A
\nT.P (A – B) \u222a B = A
\n(A – B) \u222aB = (A \u2229 B\u2019) \u222a B
\n= (A \u222a B)\u2229 (B’ \u222a B)
\n= (A \u222a B) \u2229 U
\n= A \u222a B = A [\u2235 B \u2282 A]<\/p>\n

Question 5.
\nFor any sets A and B, prove that P(A) \u222a P(B) \u2282 P(A \u222a B).
\nSolution:
\nLet X \u2208 P(A) \u222a P(B)
\n\u21d2 X \u2208 P(A) or X \u2208 P(B)
\n\u21d2 X \u2282 A or X \u2282 B
\n\u21d2 X \u2282 A \u222aB
\n\u21d2 X \u2208 P(A \u222a B)
\nThus P (A) \u222a P (B) \u2282 P (A \u222a B)<\/p>\n

\"ML<\/p>\n

Question 6.
\nShow that the following four conditions are equivalent:
\n(i) A \u2282 B
\n(ii) A – B =
\n(iii) A \u2229 B = A
\n(iv) A \u222a B = B.
\nSolution:
\nGiven A \u2282 B
\n\u2234 A \u2229 B = A
\n\u2200 x \u2208 A – B
\n\u21d4 x \u2208 A and x \u2209 B
\n\u21d4 x \u2208 B and x \u2209 B
\n[\u2235 A \u2282 B]
\n\u21d4 x \u2208 B and x \u2208 Bc<\/sup>
\n\u21d4 x \u2208 B \u2229 Bc<\/sup> = \u03a6
\n\u2234 A – B = \u03a6<\/p>\n

Aliter:
\nLet A \u2282 B T.P. A – B = \u03a6
\n\u21d2 A \u2229 B’ \u2282B \u2229 B’ = \u03a6
\n\u21d2 A – B \u2282 \u03a6
\nalso \u03a6 be a subset of every set.
\n\u03a6 \u2282 A – B
\nThus, A – B = \u03a6<\/p>\n

Conversely:
\nLet A – B = \u03a6
\nT.P. A \u2282 B
\nNow,
\nA = A \u2229 \u03be
\n= A \u2229 (B \u222a B’)
\n= (A \u2229 B) \u222a (A \u2229 B’)
\n[\u2235 A – B = \u03a6
\n\u21d2 A \u2229 B\u2019 = \u03a6]
\nA = A \u2229 B …………………….(1)
\nSince A \u2229 B \u2282 B = A \u2282 B [using(1)]
\n\u2234 A – B = \u03a6 iff A \u2282 B
\n\u2234 (i) \u2194 (ii)
\nFrom (1) ;
\nIf A – B = \u03a6 then A \u2229 B = A<\/p>\n

Conversely:
\nWhen A \u2229 B = A T.P. A – B = \u03a6
\nNow A – B = A \u2229 B’
\n= A \u2229 B \u2229 B’
\n= A \u2229 \u03a6
\n= \u03a6 [using given]
\nThus A – B = \u03a6 iff A \u2229 B = A
\n\u2234 (ii) \u2194 (iii)
\nGiven A \u2229 B = A T.P. A \u222a B = B
\nA \u222a B = (A \u2229 B) \u222a B
\n[\u2235 A \u2229 B = A]
\n= (A \u222a B) \u2229 B = B
\n[\u2235 A \u222a B \u2283 B]
\nClearly B \u2282 A \u222a B<\/p>\n

Conversely :
\nA \u222a B = B T.P. A \u2229 B = A
\nClearly A \u2229 B \u2282 A ……………….(2)
\nNow, A \u2229 B = A \u2229 (A \u222a B)
\n= (A \u2229 A) \u222a (A \u2229 B)
\n= A \u222a (A \u2229 B)
\n[\u2235 A \u2229 B \u2282 A]
\nThus, (iii) \u2194 (iv)
\nGiven, A \u222a B = B T.P. A \u2282 B
\n\u2200 x \u2208 A
\n\u21d2 x \u2208 A or x \u2208 B
\n\u21d2 x \u2208 A \u222aB
\n\u21d2 x \u2208 B
\n\u2234 A \u2282 B.<\/p>\n

Conversely:
\nA \u2282 B T.P. A \u222a B = B
\n\u21d2 A \u222a B \u2282 B \u222a B
\n\u21d2 A \u222a B \u2282 B ………………(3)
\nand B be a subset of A \u222a B always
\n\u2234 B \u2282 A \u222a B ……………………..(4)
\nFrom (3) and (4) ; we have
\nA \u222a B = B
\nThus (iv) \u2194 (i)
\nHence all the four conditions are equivalent.<\/p>\n

\"ML<\/p>\n

Long Answer questions (6 to 10) :<\/p>\n

Question 7.
\nFor any two sets A and B, prove that
\n(i) (A – B) \u222a B = A \u222a B
\n(ii) (A – B) \u2229 B = \u03a6
\nSolution:
\n(i) (A – B) \u222a B = (A \u2229 B’) \u222a B
\n= (A \u222a B) \u2229 (B’ \u222a B)
\n= (A \u222a B) \u2229 U
\n[distributive law]
\n[where U = universal set]
\n= A \u222a B<\/p>\n

(ii) (A – B) \u2229 B = (A \u2229 B’) \u2229 B
\n= A \u2229 (B’ \u2229 B) [Associative law]
\n= A \u2229 \u03a6 = \u03a6<\/p>\n

Question 8.
\nFor any two sets A and B, show that
\n(i) (A \u222a B) – B = A – B
\n(ii) A – (A \u2229 B) = A – B.
\nSolution:
\n(i) (A \u222a B) – B = (A \u222a B) \u2229 B\u2019
\n= (A \u2229 B’) \u222a (B \u2229 B\u2019) [distributive law]
\n= (A \u2229 B\u2019) \u222a \u03a6
\n= A \u2229 B\u2019
\n= A – B<\/p>\n

(ii) A – (A \u2229 B) = A \u2229 (A \u2229 B)\u2019
\n= A \u2229 (A\u2019 \u222a B\u2019) [using Demorgan\u2019s law]
\n= (A \u2229 A\u2019) \u222a (A \u2229 B\u2019)
\n= \u03a6 \u222a (A \u2229 B\u2019)
\n= A \u2229 B’
\n= A – B<\/p>\n

\"ML<\/p>\n

Question 9.
\nFor any sets A, B and C, using properi les of sets, prove that
\n(i) A – (A – B) = A \u2229 B
\n(ii)(A – B) \u222a (A – C) = A – (B \u2229 C)
\n(iii) (A – B) \u2229 (C – B) = (A \u2229 C) – B
\n(iv) A – (B – C) = (A – B) \u222a (A \u2229 C).
\nSolution:
\n(i) A – (A – B) = A \u2229 (A – B)\u2019
\n= A \u2229 (A \u2229 B\u2019)\u2019
\n= A \u2229 (A\u2019 \u222a (B\u2019)\u2019) [using demorgan\u2019s law]
\n= A \u2229 (A\u2019 \u222a B)
\n= (A \u2229 A\u2019) \u222a (A \u2229 B)
\n= \u03a6 \u2229 (A \u2229 B)
\n= A \u2229 B<\/p>\n

(ii) (A – B) \u222a (A – C) = (A \u2229 B\u2019) \u222a (A \u2229 C\u2019)
\n= A \u2229 (B\u2019 \u222a C\u2019) [distributive law]
\n= A \u2229 (B \u2229 C)\u2019
\n= A – (B \u2229 C) [using Demorgan\u2019s law]
\n(A – B) \u2229 (C – B) = (A \u2229 B\u2019) \u2229 (C \u2229 B\u2019)
\n= (B\u2019 \u2229 A) \u2229 (B\u2019 \u2229 C)<\/p>\n

(iv) A – (B – C) = A \u2229 (B – C)\u2019 = A \u2229 (B \u2229 C\u2019)\u2019
\n= A \u2229 (B\u2019 \u222a (C\u2019)\u2019) [Demorgan law]
\n= A \u2229 (B\u2019 \u222a C)
\n= (A \u2229 B\u2019) \u222a (A \u2229 C)
\n= (A – B) \u222a (A \u2229 C)<\/p>\n

\"ML<\/p>\n

Question 10.
\nFor any two sets A and B, prove that (A – B) \u222a (B – A) = (A \u222a B) – (A \u2229 B).
\nSolution:
\n(A – B)u(B – A) = (A \u2229 B\u2019) \u222a (B \u2229 A\u2019)
\n= {(A \u2229 B\u2019) \u222a B} \u2229 {(A n B\u2019) \u222a A\u2019
\n= {(A \u222a B) \u2229 (B\u2019 \u222a B)} \u2229 {(A \u222a A\u2019) \u2229 (B\u2019 \u222a A\u2019)}
\n= {(A \u222a B) \u2229 U) \u2229 {U \u2229(B\u2019 \u222a A\u2019)}
\n= (A \u222a B) \u2229 (B\u2019 \u222a A\u2019)
\n= (A \u222a B) \u2229 (B \u2229 A)\u2019 [Dernorgan\u2019s law]
\n= (A \u222a B) – (B \u2229 A)<\/p>\n","protected":false},"excerpt":{"rendered":"

Utilizing Understanding ISC Mathematics Class 11 Solutions Chapter 1 Sets Ex 1.5 as a study aid can enhance exam preparation. ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.5 Question 1. State whether the following statemcnts are true or false : (i) A \u222a A = A (ii) A \u2229 A …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170468"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170468"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170468\/revisions"}],"predecessor-version":[{"id":170471,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170468\/revisions\/170471"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170468"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170468"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170468"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}