{"id":170428,"date":"2024-05-23T16:52:42","date_gmt":"2024-05-23T11:22:42","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170428"},"modified":"2024-05-23T16:55:36","modified_gmt":"2024-05-23T11:25:36","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-2\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.2"},"content":{"rendered":"

Accessing ML Aggarwal Class 11 Solutions ISC<\/a> Chapter 1 Sets Ex 1.2 can be a valuable tool for students seeking extra practice.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.2<\/h2>\n

Question 1.
\nIn the following, state whether A = B or not :
\n(\u00ee) A = {x : x + 2 = 3), B = {x : x \u2208 N and is less than 2}
\n(ii) A = {x : x \u2208 N and 3x – 1 < 2}, B = {x : x \u2208 W and 3x – 1 < 2}
\n(iii) A = (x : x \u2208 N and a prime factor of 36), B = {1, 2, 3, 4, 6, 9, 12}
\n(iv) A = {x : x \u2208 I and x2<\/sup> \u2264 4), B = {x : x \u2208 R and x2<\/sup> – 3x + 20}
\n(v) A = {2, 3} and B = {x : x is a solution of x2<\/sup> + 5x + 6 = 0}
\n(vi) A = {x : x is a letter in the word LOYAL} and B = {y : y is a letter in the world ALLOY}
\n(vii) A = {x : x is a letter in the word WOLF} and B = {y : y is a letter in the world FOLLOW)
\n(viii) A = {10, 20, 30, 40, 50, ………………} and B = {x : x is a multiple of 10}.
\nSolution:
\n(i) Given A = {x : x + 2 = 3}
\nSince x + 2 = 3
\n\u21d2 x = 3 – 2 = 1
\n\u2234 A{1}
\nand B = {x : x \u2208 N and is less than 2} = {1}
\nSince sets A and B have same element
\n\u2234 A and B are equal sets i.e. A = B.<\/p>\n

(ii) Given A = {x : x \u2208 N and 3x – 1 < 2}
\nSince 3x – 1 < 2
\n\u21d2 3x < 3
\n\u21d2 x < 1, x \u2208 N
\n\u2234 A = \u03a6 i.e. A be an empty set.
\nand B = {x : x \u2208 W and 3x – 1 < 2}
\nSince 3x – 1 < 2
\n\u21d2 3x < 3
\n\u21d2 x < 1, x \u2208 W
\n\u2234 x = 0
\n\u2234 B = {0} and B be a singleton set.
\nHence A and B do not have same elements.
\n\u2234 A \u2260 B.<\/p>\n

\"ML<\/p>\n

(iii) Given A = {x : x \u2208 N and is a prime factor of 36)
\nSince prime factors of 36 are 2, 3
\n\u2234 x = 2, 3
\n\u2234 A = {2, 3}
\nand B = {1, 2, 3, 4, 6, 9, 12}
\nClearly n(A) = 2
\nand n(B) = 7
\n\u2234 n(A) \u2260 n(B)
\nMoreover A and B does not have same elements
\n\u2234 A and B are not equal sets.
\nThus A \u2260 B.<\/p>\n

(iv) Given A = {x : x \u2208 I and x2<\/sup> \u2264 4}
\nSince x2<\/sup> \u2264 4
\n\u21d2 |x| \u2264 2
\n\u21d2 – 2 \u2264 x \u2264 2, x \u2208 I
\n\u2234 x = – 2, – 1, 0, 1, 2
\nThus, A = {- 2, – 1, 0, 1, 2}
\nand B = {x : x \u2208 R and x2<\/sup> – 3x + 20}
\nSince x2<\/sup> – 3x + 2 = 0
\n\u21d2 (x – 1) (x – 2) = 0
\n\u21d2 x = 1, 2.
\nThus. B = {1, 2)
\nClearly the elements o\u00edA and B are not same.
\n\u2234 A and B are not equal sets
\n\u2234 A \u2260 B.<\/p>\n

(v) Given A = {2, 3)
\nand B = {x : x is a solution of x2<\/sup> + 5x + 6 = 0)
\nSince x2<\/sup> + 5x + 6 = 0
\n\u21d2 (x + 2) (x + 3) = 0
\n\u21d2 x = – 2, – 3
\nThus B = {- 2, – 3}
\nClearly the elements of set A and B are not same
\n\u2234 A \u2260 B.<\/p>\n

(vi) Given A = {x : x is a letter in the word LOYAL)
\n= {L, O, Y, A}
\nand B = {y : y is a letter in the word ALLOY)
\n= {A, L, O, Y}
\nClearly sets A and B have same elements and hence A and B are equal sets
\n\u2234 A = B<\/p>\n

(vii) Given A = {x : x is a letter in the word WOLF)
\n= {W, O, L, F)
\nand B = {y : y is a letter in the word FOLLOW)
\n= {W, O, L, F)
\nClearly sets A and B have same elements and hence both sets are equal
\n\u2234 A = B<\/p>\n

(viii) Given A = {10, 20, 30, 40, 50, …………..)
\nB = {x : x is a multiple of 10}
\nClearly multiple of 10 are 0, 10, 20, 30, 40, 50.
\n\u2234 B = {0, 10, 20, 30, 40, 50}
\nThus set A and set B does not have same elements and hence Aand B are not equal sets
\n\u2234 A \u2260 B.<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf A = {1, 2, 3, 4, 5), B = {2, 3, 4) and C = {2, 4, 5); state whether the following statements are true or false.
\n(i) A \u2282 B
\n(ii) A \u2282 C
\n(iii) B \u2282 A
\n(iv) B \u2282 C
\n(y) C \u2282 A
\n(vi) C \u2282 B
\n(vii) B = C
\n(viii) \u03a6 \u2282 B
\n(ix) A \u2194 B
\n(x) B \u2194 C
\n(xi) A \u2194 C
\nSolution:
\nGiven, A = {1, 2, 3, 4, 5} ;
\nB = {2, 3, 4}
\nand C = {2, 4, 5}<\/p>\n

(i) since 1 \u2208 A but 1 \u2209 B
\n\u2234 A \u2284 B
\n\u2234 given statement is false<\/p>\n

(ii) False, since 1 \u2208 A but 1 \u2209 C
\n\u2234 A \u2284 C<\/p>\n

(iii) True, since every member of B is a member of A.<\/p>\n

(iv) False. since 3 \u2208 B and 3 \u2209 C
\n\u2234 B \u2284 C<\/p>\n

(v) True, since every member of C is a member of A
\n\u2234 C \u2282 A<\/p>\n

\"ML<\/p>\n

(vi) False, As 5 \u2208 C and 5 \u2209 B
\n\u2234 C \u2284 B<\/p>\n

(vii) False, since 3 \u2208 B but 3 \u2209 C
\nThus sets B and C does not have same elements
\n\u2234 B and C are not equal seis.
\n\u2234 B \u2260 C<\/p>\n

(viii) True, since empty set is a subset of every set.
\n\u2234 C \u2282 A<\/p>\n

(ix) False, since cardinal number of A = no. of distinct elements in A = n(A) = 5
\nwhile n(B) = cardinal number of set B = 3
\n\u2234 n(A) \u2260 n(B)
\n\u2234 both sets A and B are not equivalent sets.
\n\u2234 A \u21ae B.<\/p>\n

(x) True; since cardinal number of B = n(B) = 3
\nand cardinal number of C = n(C) = 3
\n\u2234 n(A) = n(B)
\n\u21d2 B \u2194 C<\/p>\n

(xi) False, since cardinal no. of A = n(A) = 5
\nand cardinal no. of C = n(C) = 3
\n\u2234 n(A) \u2260 n(C)
\nThus, A and C are not equivalent sets.
\n\u2234 A \u21ae\u00a0 C.<\/p>\n

\"ML<\/p>\n

Question 3.
\nUse appropriate symbol \u2018\u2208, \u2209, \u2282, \u2283, =\u2018 to fill in the blanks:<\/p>\n

(i) 4 …………… {1, 2, 3, 4}
\n(ii) – 5 ……………. (2, 3, 4, 5, 6)
\n(iii) (2) ……………. (2, 3, 4)
\n(iv) [a, b, c] …………. {a, b, b, a, c}
\n(vi) {a, i, u} ……………. {a, a, i}
\n(vii) MUMBAI …………………. {x : x is a capital city of countries in Asia}.
\nSolution:
\n(i) Since 4 be a member of {1, 2, 3, 4}
\n\u2234 4 \u2208 {1, 2, 3, 4}<\/p>\n

(ii) Since – 5 is not a member of {2, 3, 4, 5, 6)
\n\u2234 – 5 \u2208 {2, 3, 4, 5, 6}<\/p>\n

(iii) Since {2} is not a member of {2, 3, 4}
\n\u2234 {2} \u2209 {2, 3, 4}
\nalso {2} be a subset of {2, 3, 4}
\n\u21d2 {2} \u2282 {2, 3, 4}<\/p>\n

(iv) Since {b, a, e} and {a, b, c} have same elements namely a, b and c respectively
\n\u2234 {a, b, c} = {b, a, c}<\/p>\n

(v) Since the sets {a, b, c} and {a, b, b, a, c} have same elements namely a, b and e respectively and
\nhence both sets are equal.
\n\u2234 {a, b, c} = {a, b, b, a, c}<\/p>\n

(vi) Since, all members of {a, a, i} are the members of {a, j, u}
\n\u2234 {a, a, i} be a subset of {a, i, u} or {a, i, u) is a super set or contains (a, a, iI).
\nThus, {a, i, u) \u2283 (a, a, i).<\/p>\n

(vii) Since MUMBAI is not a separate capital of any country in Asia.
\n\u2234 MUMBAI \u2209 {x : x is a capital city of countries in Asia}.<\/p>\n

Question 4.
\nExamine whether the following statements are true or false:
\n(i) {a, b} \u2284 {b, c, a)
\n(ii) {1, 2, 3) \u2282 {1, 3, 5}
\n(iii) {a, e} \u2282 {a, b, c}
\n(iv) {a} \u2208 (a, b, c}
\n(v) {a, e) \u2282 {x : x is a vowel in the English alphabet)
\n(vi) (x : x is an even natural less than 8) \u2282 (x : x is a natural number which divides 36}.
\nSolution:
\n(i) Clearly (a, b) be a subset of (b, c, a)
\n\u2234 {a, b} \u2282 {b, c, a}
\nThus, given statement is false.<\/p>\n

(ii) Clearly (1, 2, 3) be not a subset of (1, 3, 5}
\n\u2234 {1, 2, 3} \u2282 {1, 3, 5}
\nThus, given statement is false.<\/p>\n

(iii) True, since {a} be a subset of {a, b, c}.<\/p>\n

(iv) False, clearly a, b, c \u2208 {a, b, c} but {a} \u2208 {a, b, c}<\/p>\n

(v) True, clearly {a, e} be a subset of {a, e, i, o, u}
\n\u2234 {a, e} \u2282 {a, e, i, o, u}<\/p>\n

(vi) Since even natural numbers less than 8 are 2, 4, 6.
\n\u2234 A = {x : x is a even natural number less than 8} – {2, 4, 6}
\nand B = {x \ud83d\ude21 isanatural numberwhichdivides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}
\nClearly {2, 4, 6)} \u2282 {1, 2, 3, 4, 69, 12, 18, 36}
\nHence the given statement is true.<\/p>\n

\"ML<\/p>\n

Question 5.
\nState whether each of the following statements is true or false:
\n(i) 2 \u2282 {1, 2, 3}
\n(ii) {2} \u2208 {1, 2, 3}
\n(iii) {\u03a6} \u2208 (1, 2, 3}
\n(iv) 0 \u2208 \u03a6
\n(v) {1, 2} \u2282 {1, {2, 3}}
\n(vi) \u03a6 \u2282 {\u03a6}.
\nSolution:
\n(i) False, since {2} \u2282 {1, 2, 3} while 2 \u2284 {1, 2, 3}<\/p>\n

(ii) False, since subset of {1, 2, 3} are \u03a6, {1}, (1), {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
\n\u2234 {2} \u2282 {1, 2, 3}
\nbut (2) \u2209 {1, 2, 3}<\/p>\n

(iii) False, since empty set is a subset of every set
\n\u2234 \u03a6 \u2282 {1, 2, 3}.
\nThus \u03a6 \u2209 (1, 2, 3) since \u03a6 is not a member of {1, 2, 3}.<\/p>\n

(iv) False, 0 \u2209 \u03a6 as \u03a6 = { } i.e. an empty set contains no clement.<\/p>\n

(v) False, since the subsets of {1, {2, 3}} are \u03a6, {1}, {{2, 3}}, {1, (2, 3}}
\nThus {1, 2} \u2284 {1, {2, 3}}
\nTrue, since empty set \u03a6 be a subset of every set.
\n\u2234 \u03a6 \u2282 {\u03a6}<\/p>\n

\"ML<\/p>\n

Question 6.
\nLet A = {1, 2, {3, 4}, 5}. Which of the following statements are not true and why?
\n(i) {3, 4} \u2282 A
\n(ii){3, 4} \u2208 A
\n(iii) {{3, 4}} \u2282 A
\n(iv) {1, 2, 5} \u2282 A
\n(v) {1, 2, 5} \u2208 A
\n(vi) 1 \u2208 A
\n(vii) 1 \u2282 A
\n(viii) \u03a6 \u2208 A
\n(ix) \u03a6 \u2208 A
\n(x) \u03a6 \u2208 A
\n(xi) {\u03a6} \u2282 A.
\nSolution:
\nGiven A = {1, 2, {3, 4}, 5}
\n(i) Clearly {3, 4} be not a subset of A
\n\u2234 {3, 4} \u2284 A
\n\u2234 given statement is not true.<\/p>\n

(ii) Clearly {3, 4} be a member of A
\nThus {3, 4} \u2208 A is true.<\/p>\n

(iii) Clearly {{3, 4}} be a subset of A
\nThus given statement is true.<\/p>\n

(iv) Clearly {1, 2, 5} be a subset of A
\n\u2234 Given statement is true.<\/p>\n

(v) Since 1, 2, 5 \u2208 A but {1, 2, 5} A
\nAs { 1, 2, 5) be a subset of A.
\nThus, given statement is not true.<\/p>\n

(vi) Clearly {1, 2, 3} be not a subset of A
\nThus, given statement is not true.<\/p>\n

(vii) Clearly 1 be a member of A
\n\u2234 1 \u2208 A
\nThus given statement is true.<\/p>\n

(viii) Clearly 1 \u2208 A but {1} \u2209 A as {1} \u2282 A.
\n\u2234 given statement is not true.<\/p>\n

(ix) Clearly \u03a6 \u2282 A but is not a member of set A.
\nThus given statement is false.<\/p>\n

(x) Since empty set be a subset of every set
\n\u2234 \u03a6 \u2282 A is a true statement<\/p>\n

(xi) Clearly \u03a6 \u2282 A but {\u03a6} be a set containing element \u03a6
\n\u2234 {\u03a6} \u2284 A
\nThus given statement is not true.<\/p>\n

\"ML<\/p>\n

Question 7.
\nIf \u03be = {1, 2, 3, ……………., 10} and A = {|x| x is a prime factor of 66}. List the set A.
\nSolution:
\nGiven = {1, 2, 3, …………………., 10}
\nSince prime factors of 66 are 2, 3, 11.
\nbut A \u2282 \u03be
\n\u2234 A = {2, 3}
\nSince 11 \u2209 \u03be .<\/p>\n

Question 8.
\nWrite down ati the subsets of the following sets:
\n(i) {a}
\n(ii) {a, b}
\n(iii) \u03a6
\n(iv) {\u03a6, 1).
\nSolution:
\n(i) \u03a6, {a}<\/p>\n

(ii) \u03a6, {a}, {b}, {a, b}<\/p>\n

(iii) \u03a6<\/p>\n

(iv) \u03a6, \u03a6 {1}, {\u03a6, 1}<\/p>\n

Question 9.
\nWrite the power set of A, where A = {- 1, 0, 2}.
\nSolution:
\nGiven A = {- 1, 0, 2}
\nSo power set of A = P(A) = set of all subsets of A
\n= {\u03a6, {- 1}, {0}, {2}, {- 1, 0}, {0, 2}, {- 1, 2}, {- 1, 0, 2}}<\/p>\n

Question 10.
\nWrite the number of proper subsets of A, where A = {- 3, – 1, 0, 1, 4}.
\nSolution:
\nGiven A = {- 3, – 1, 0, 1, 4}
\nSo cardinal number of set A = n(A)
\n= no. of distinct elements of A = 5
\nThus the no. of subsets of A = 2n(A)<\/sup>
\n= 25<\/sup> = 32
\nHence the total no. of proper subsets of A = 32 – 1 = 31
\n[Since every set is a subset of itself so subset
\nA is not included in proper subset]<\/p>\n

\"ML<\/p>\n

Question 11.
\nIf \u03be = {1, 2, 3, ……………., 40} and A = {x : x is a factor of 42}, then write n(A).
\nSolution:
\nGiven \u03be = {1, 2, 3, …………………., 40}
\nand A = {x : x is a factor of 42}
\n= {1, 2, 3, 6, 7, 14, 21, 42}
\nSince A \u2282\u03be
\n\u2234 A = {1, 2, 3, 6, 7, 14, 21}
\nThus n(A) = Cardinal number of A = 7<\/p>\n

Question 12.
\nIf \u03be = {all digits in our number system} and A = {x : x is a multiple of 3}, then write n(A).
\nSolution:
\nGiven \u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} as A \u2282 \u03be
\nand A = {0, 3, 6, 9)
\nThen n(A) = Cardinal number of A = 4.<\/p>\n

\"ML<\/p>\n

Question 13.
\nWrite the following sets as intervals:
\n(i){x : x \u2208 R, – 3 < x \u2264 5}
\n(ii) {x : x \u2208 R, – 5 < x< – 1}
\n(iii) {x : x \u2208 R, 2 \u2264 x \u2264 7}
\n(iv) {x : x \u2208 R, 0 \u2264 x < 3}
\n(v) {x : x \u2208 R, x \u2264 5}
\n(vi) (x : x \u2208 R, x < – 3}
\n(vii) {x : x \u2208R x – 2}
\nSolution:
\n(i){x : x \u2208 R, – 3 < x \u2264 5} = {- 3, 5]<\/p>\n

(ii) {x : x \u2208 R, – 5 < x < – 1} = {- 5, – 1}
\nsince – 5 and – 1 both are excluded<\/p>\n

(iii) {x : x \u2208 R, + 2 \u2264 x \u2264 7} = [2, 7]<\/p>\n

(iv) {x : x \u2208 R, 0 \u2264 x < 3} = [0, 3)<\/p>\n

(v) {x : x \u2208 R, x \u2264 5} = {- \u221e, 5)<\/p>\n

(vi) {x : x \u2208 R, x < – 3} = (- \u221e, – 3)<\/p>\n

(vii) {x : x \u2208 R, x \u2265 – 2} = [- 2, \u221e)<\/p>\n

Question 14.
\nWrite the following intervals in set builder form:
\n(i) (- 2, 0]
\n(ii) (2, 7)
\n(iii) [- 5, – 2]
\n(iv) [- 9, 4)
\n(v) (3, \u221e)
\n(vi) (- \u221e, – 1]
\n(vii) (- \u221e, 4).
\nSolution:
\n(i) (- 2, 0] = {x : x \u2208 R, – 2 < x \u2264 0}<\/p>\n

(ii) (2, 7) = {x : x \u2208 R, 2 < x < 7}<\/p>\n

(iii) [- 5, – 2] = [x : x \u2208 R, – 5 \u2264 x < – 2}<\/p>\n

(iv) [- 9, 4) = {x : x \u2208 R, – 9 \u2264 x < 4}<\/p>\n

(v) (3, – \u221e) = {x : x \u2208 R, x > 3}<\/p>\n

(vi) (- \u221e, – 1] = {x : x \u2208 R, x \u2264 – 1}<\/p>\n

(vii) (- \u221e, 4) = {x : x \u2208 R, x < 4}<\/p>\n

\"ML<\/p>\n

Question 15.
\nIf \u03be = {1, 2, 3, …………………., 12), A = {x |2x + 3 \u2264 18} and B = {x |x2<\/sup> \u2264 40) ; write A and B in the roster form.
\nSolution:
\nGiven \u03be = {1, 2, 3, …………………. 12}
\nand A = {x |2x + 3 \u2264 18}
\nSince 2x + 3 \u2264 18
\n\u21d2 2x \u2264 15,
\n\u21d2 x \u2264 \\(\\frac{15}{2}\\) = 7.5
\n\u2234 A = {1, 2, 3, 4, 5, 6, 7}
\nand B = {x |x2<\/sup> \u2264 40} since x2<\/sup> \u2264 40
\nNow 12<\/sup> \u2264 40 ;
\n22<\/sup> = 4 \u2264 40 ;
\n32<\/sup> = 9 \u2264 40 ;
\n42<\/sup> = 16 \u2264 40 ;
\n52<\/sup> \u2264 40 ;
\n62<\/sup> = 36 \u2264 40
\nThus, B = {1, 2, 3, 4, 5, 6}<\/p>\n

Question 16.
\nLet \u03be = {0, 1, 2, 3 , 50}, A = {x : x = 6n, n \u2208 W}, B = {x : x = 7n, n \u2208 W} and C = {x : x is a factor of 72}, then
\n(i) write the sets A, B and C in roster form
\n(ii) state n(A), n(B) and n(C).
\nSolution:
\nGiven \u03be = {0, 1, 2, 3, ……………….. 50}
\nA = {x : x = 6n, n \u2208 W}
\nsince n \u2208 W
\n\u2234 n = 0, 1, 2, 3, ………………………
\nwhen n = 0 \u21d2 x = 0
\nwhen n = 1 \u21d2 x = 6
\nwhen n = 2 \u21d2 x = 12
\nwhen n = 3 \u21d2 x = 18
\nwhen n = 4 \u21d2 x =24
\nwhen n = 5 \u21d2 x = 30
\nwhen n = 6 \u21d2 x = 36
\nwhen n = 7 \u21d2 x = 42
\nwhen n = 8 \u21d2 x = 48
\nwhen n = 9 \u21d2 x = 54
\n\u2234 A = {0, 6, 12, 18, 24, 30, 36, 42, 48}
\nB = {x : x = 7n, n \u2208 W) = {0, 7, 14, 21, 28, 35, 42, 49}
\nand C = {x : x is a factor of 72) = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36)
\nas 72 \u2209 \u03be<\/p>\n

(ii) Thus cardinal number of A = no. of distinct elements in A
\n= n(A) = 9
\nCardinal no. of B = no. of distinct elements in B
\nn(B) = 8
\nand No. of distinct elements in C = n(C) = 11.<\/p>\n

\"ML<\/p>\n

Question 17.
\nGiven A = {x : x is a letter in the word ACCUMULATOR}
\n(i) Express A in roster form.
\n(ii) Find the cardinal number of the set of vowels in A.
\n(iii) Write down the power set of the set of vowels in A.
\nSolution:
\n(i) In roster form, A = {A, C, U, M, L,T, O, R}<\/p>\n

(ii) Let B = {all vowels in A) = {A, U, O}
\n\u2234 n(B) = Cardinal no. of B = 3<\/p>\n

(iii) Power set of set of vowels in A = {\u03a6, {A}, {U), {O}, {A, U), {A. O), {U, O), {A, U, O}}<\/p>\n

Question 18.
\nList all the proper subsets of {0, 1, 2, 3}.
\nSolution:
\nLet A = {0, 1, 2, 3}
\nThus, the required proper subsets of A is given by
\n\u03a6, (0), (1), {2}, {3), {O, I), {0, 2), {0, 3}, {1, 2}, {1, 3), {2, 3}, {0, 1, 2), {0, 1, 3), {0, 2, 3}, {1, 2, 3}<\/p>\n

\"ML<\/p>\n

Question 19.
\nIf the set A has five elements, how many subsets will A have ? If A has six elements, how many proper subsets will A have?
\nSolution:
\nGiven set A has five elements
\n\u2234 Cardinal no. of A = n(A) = 5
\nThus, No. of subsets of A = 2n(A)<\/sup>
\n= 25<\/sup> = 32
\nGiven A has six elements
\n\u2234 n(A) = 6
\nthus, no. of proper subsets of A = 2n(A)<\/sup> – 1
\n= 26<\/sup> – 1 = 63
\n[Since the subset A of set A is excluded]<\/p>\n

Question 20.
\nTwo finite sets have m and k elements. If the number of subsets of the first set is 112 more than the number of subsets of the second set, then find the values of m and k.
\nSolution:
\nLet the two given sets be A and B
\nThen n(A) = m
\nand n(B) = k, n > k
\nNow, n(P(A)) = 2m<\/sup> ;
\nn(P(B)) = 2k<\/sup>
\naccording to given condition ; we have
\n2m<\/sup> = 12 + 2k<\/sup>
\n\u21d2 2m<\/sup> – 2k<\/sup> = 112
\n\u21d22k<\/sup> (2m – k<\/sup> – 1) = 112 = 24<\/sup> \u00d7 7
\n\u21d2 2k<\/sup> = 24<\/sup> and 2m – k<\/sup> – 1 = 7
\n\u21d2 k = 4 and 2m – k<\/sup> = 8 = 23<\/sup>
\n\u21d2 k = 4 and m – k = 3
\n\u21d2 k = 4 and m = 7
\nHence, m = 7 and k = 4.<\/p>\n","protected":false},"excerpt":{"rendered":"

Accessing ML Aggarwal Class 11 Solutions ISC Chapter 1 Sets Ex 1.2 can be a valuable tool for students seeking extra practice. ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.2 Question 1. In the following, state whether A = B or not : (\u00ee) A = {x : x + …<\/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\/170428"}],"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=170428"}],"version-history":[{"count":5,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170428\/revisions"}],"predecessor-version":[{"id":170462,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170428\/revisions\/170462"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170428"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170428"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170428"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}