{"id":170439,"date":"2024-05-23T16:52:45","date_gmt":"2024-05-23T11:22:45","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170439"},"modified":"2024-05-23T16:54:51","modified_gmt":"2024-05-23T11:24:51","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-1-ex-1-3\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.3"},"content":{"rendered":"

Students often turn to ML Aggarwal Class 11 ISC Solutions<\/a> Chapter 1 Sets Ex 1.3 to clarify doubts and improve problem-solving skills.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.3<\/h2>\n

Question 1.
\nIf A = {1, 2, 3, 4, 5, 6) and B = {2, 4, 6, 8}, find
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A
\nSolution:
\nGiven A = {1, 2, 3, 4, 5, 6}
\nand B = {2, 4, 6, 8}
\n(i) A \u222a B = {1, 2, 3, 4, 5, 6, 8}
\n(ii) A \u2229 B = {2, 4, 6}
\n(iii) A – B = {1, 3, 5}
\n(iv) B – A = {8}<\/p>\n

Question 2.
\nIf A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 5, 6, 8, 9, 10), then find A \u0394 B.
\nSolution:
\nGiven A = {1, 2, 3, 4, 5, 6}
\nand B = {2, 4, 5, 6, 8, 9, 10}
\n\u2234 A – B = {1, 3)
\nand B – A = {8, 9, 10}
\nThus A \u0394 B = (A – B) \u222a (B – A)
\n= {1, 3} \u222a {8, 9, 10}
\n= {1, 3, 8, 9, 10}<\/p>\n

Question 3.
\nIf A = {x : x is a natural number and 1 < x \u2264 6} and B = {x : x is a natural number and 6 < x < 10}, find
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A.
\nSolution:
\nGiven
\nA= {x : x is a natural number and 1 < x \u2264 6}
\n\u2234 A = {2, 3, 4, 5, 6}
\nand B = {x : x is a natural number and 6 < x < 10}
\n= {7, 8, 9)<\/p>\n

(i) A \u222a B = {2, 3, 4, 5, 6, 7, 8, 9}
\n(ii) A \u2229 B = { } or \u03a6
\n(iii) A – B = {2, 3, 4, 5, 6} = A
\n(iv) B – A = {7, 8, 9} = B<\/p>\n

\"ML<\/p>\n

Question 4.
\nWhich of the following pairs of sets are disjoint?
\n(i) {a, e, i, o, u) and {c, d, e, f}
\n(ii) {2, 6, 10, 14, 18} and {3, 7, 11, 15}
\n(iii) {1, 2, 3, 4} and {x : x is a natural number and 4 \u2264 x \u2264 6}
\n(iv) {x : x is an even integer} and {x : x is an odd integer}.
\nSolution:
\n(i) Let A = (a, e, i, o, u}
\nand B = {c, d, e, f}
\n\u2234 A \u2229 B = {e} \u2260 \u03a6
\nThus A and B are not disjoint sets.<\/p>\n

(ii) Let A = {2,6, 10, 14, 18}
\nand B = {3, 7, 11, 15}
\n\u2234 A \u2229 B = \u03a6
\nThus, A and B are disjoint sets.<\/p>\n

(iii) Let A = {1, 2, 3, 4)
\nand B = {4, 5, 6)
\n\u2234 A \u2229 B = {4} \u2260 \u03a6
\nThus sets A and B are disjoint sets.<\/p>\n

(iv) Let A = {x : x is an even integer}
\n= {……………….., – 4, – 2, 0, 2, 4, 6, …………….}
\nand B = {x : x is an odd integer
\n= {………………., – 3, – 1, 3, 5, ……………………}
\nClearly A \u2229 B = \u03a6
\nThus A and B are disjoint sets.<\/p>\n

Question 5.
\nIf \u03be = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 5, 6} and B = {2, 3, 4, 7, 8}, find the following:
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A
\n(v) A \u2229 B
\n(vi) (A \u222a B)’
\n(vii) A’ \u2229 B’
\n(viii) A’ \u222a B
\n(ix) (A – B)’
\nSolution:
\nGiven \u03be = {1, 2, 3, 4, 5, 6, 7, 8}
\nA = {1, 2, 3, 5, 6}
\nand B = {2, 3, 4, 7, 8}<\/p>\n

(i) A \u222a B = {1, 2, 3, 4, 5, 6, 7, 8} = \u03be<\/p>\n

(ii) A \u2229 B = {2, 3}<\/p>\n

(iii) A – B = {1, 5, 6}<\/p>\n

(iv) B – A = {4, 7, 8}<\/p>\n

(v) Now B’ = \u03be – B = {1, 5, 6}
\n\u2234 A – B\u2019 = {1, 2, 3, 5, 6) \u222a {1, 5, 6}
\n= {1, 5, 6}<\/p>\n

(vi) A \u222a B = {1, 2, 3, 4, 5, 6, 7, 8}
\n(A \u222a B)’ = \u03be – (A \u222a B) = \u03a6<\/p>\n

(vii) Now, A’ = \u03be – A
\n= {4, 7, 8}
\nand B’ – B = {1, 5, 6}
\n\u2234 A’ \u2229 B’ = {4, 7, 8} \u2229{1, 5, 6} = \u03a6<\/p>\n

(viii) A’ \u222a B = {4, 7, 8} \u222a {2, 3, 4, 7, 8}
\n= {2, 3, 4, 7, 8}<\/p>\n

(ix) A – B = {1, 5, 6}
\n\u2234 (A – B)\u2019 = \u03be – (A – B) = {2, 3, 4, 7, 8}<\/p>\n

\"ML<\/p>\n

Question 6.
\nIf A and B are two sets such that Ac B, then what is
\n(i) A \u222a B?
\n(ii) A \u2229 B?
\nSolution:
\nGiven A \u2282 B
\n(i) A \u222a B = B
\n[\u2235 B be the larger set and all members of A are members of B]<\/p>\n

(ii) A \u2229 B = A
\n[\u2235 A be the smaller set]<\/p>\n

Question 7.
\nTaking the set of natural numbers as the universal set., write the complements of the following sets :
\n(i) A = {x : x is an odd natural number}
\n(ii) B = {x : x is an even natural number}
\n(iii) C = {x : x is a multiple of 5}
\n(iv) D = {x : x is divisible by 2 and 3}
\n(v) E = {x : x is a perfect cube}
\n(vi) F = {x : x + 5 = 8}
\n(vii) G = {x : x \u2265 7}
\n(viii) H = {x : 3x – 1 < 14}.
\nSolution:
\nHere \u03be = N
\n= {1, 2, 3, 4, ………………….}<\/p>\n

(i) A = {x : x is an odd natural number}
\n= {1, 3, 5, ……………..}
\n\u2234 A’ = \u03be – A
\n= {2, 4, 6, ………………….}
\n= {x : x be an even natural number)<\/p>\n

(ii) Given B = {x : x is an even number}
\n= {2, 4, 6……………..}
\n\u2234 B’ = \u03be – B
\n= {1, 3, 5, …………………}
\n= {x : x is an odd natural number}<\/p>\n

(iii) Given C = {x : x is a multiple of 5}
\n\u2234 C\u2019 = \u03be – C
\n= {x : x \u2208 N, x is not a multiple of 5}<\/p>\n

(iv) Given D = {x : x is divisible by 2 and 3}
\n= {x : x is divisible by 2 \u00d7 3 = 6}
\n\u2234 D’ = \u03be – D
\n= {x : x \u2208 N, x is not divisible by 6}<\/p>\n

(v) Given E = {x : x is a perfect cube}
\n\u2234 E’ = – E
\n= {x : x \u2208 N, x is not a perfect cube}<\/p>\n

(vi) Given F = {x : x + 5 = 8}
\nSince x + 5 = 8
\n\u21d2 x = 8 – 5 = 3
\n\u2234 F = {3) and {x : x \u2208 N and x \u2260 3} = F’<\/p>\n

(vii) Given G = {x : x \u2265 7}
\n\u2234 G’ = \u03be – G
\n= {x : x \u2208 N, x < 7}
\n= {1, 2, 3, 4, 5, 6}<\/p>\n

(viii) Given H’ = {x : 3x – 1 < 14}
\nsince 3x – 1 < 14
\n\u21d2 3x < 15
\n\u21d2 x < 5
\nand x \u2208 N
\n\u2234 H = {1, 2, 3, 4)
\n\u2234 H’ = {5, 6, 7, 8, 9}
\nThus, H’ = \u03be – H = {x : x \u2208 N ; x \u2265 5}<\/p>\n

\"ML<\/p>\n

Question 8.
\nFrom the adjoining Venn diagram, determine the following sets :<\/p>\n

\"ML<\/p>\n

(i) A \u222a B
\n(it) A \u2229 B
\n(iii) A – B
\n(iv) A – B
\n(v) (A \u2229 B)\u2019.
\nSolution:
\nHere \u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
\nHere A = {1, 2, 4, 7}
\nand B = {0, 1, 3, 7, 8}
\n(i) A \u222a B = {0, 1, 2, 3, 4, 7, 8}<\/p>\n

(ii) A \u2229 B = {1, 7}<\/p>\n

(iii) A – B = {2 ,4}<\/p>\n

(iv) (A \u2229 B)’ = – (A \u2229 B)
\n= {0, 2, 3, 4, 5, 6, 8, 9}<\/p>\n

Question 9.
\nFrom the adjoining Venn diagram, write the following:
\n(i) A’
\n(ii) B’
\n(iii) (A \u2229 B)’<\/p>\n

\"ML<\/p>\n

Solution:
\nHere \u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
\nA = {0, 2, 3, 4, 8)
\nand B = {0, 3, 5, 6, 8}<\/p>\n

(i) A\u2019 = {1, 5, 6, 7, 19}<\/p>\n

(ii) B\u2019 = {1, 2, 4, 7, 9}<\/p>\n

(iii) A \u2229 B = {3, 0, 8}
\n\u2234 (A \u2229 B)\u2019 = {1, 2, 4, 5, 6, 7, 9}<\/p>\n

Question 10.
\nLet \u03be = {x : x \u2208 N and x < 10}, A = {x : x = 2y + 1 and y \u2208 N} and B = {x : x = 3y- 1 and y \u2208 N), list the elements of A\u2019 \u222a B.
\nSolution:
\nGiven \u03be = {x : x \u2208 N, x < 10}
\n= {1, 2, 3, 4, 5, 6, 7, 8, 9}
\nand A = {x : x = 2y + 1 and y \u2208 N}
\nwhen y = 1
\n\u21d2 x = 2 + 1 = 3
\nwhen y = 2
\n\u21d2 x = 4 + 1 = 5
\nwhen y = 3
\n\u21d2 x = 6 + 1 = 7
\nwhen y = 4
\n\u21d2 x = 8 + 1 = 9
\nwhen y = 5
\n\u21d2 x = 10 + 1 = 11 \u2209 \u03be
\n\u2234 A = {3, 5, 7, 9}
\nand B = {x : x = 3y – 1 and y \u2208 N}
\nwhen y = 1
\n\u21d2 x = 3 – 1 = 2
\nwhen y = 2
\n\u21d2 x = 6 – 1 = 5
\nwhen y = 4
\n\u21d2 x = 12 – 1 = 11
\nwhen y = 3
\n\u21d2 x = 9 – 1 = 8
\n\u2234 B = {2, 5, 8}
\n\u2234 A’ = \u03be – A = {1, 2, 4, 6, 8}
\nThus, A’ \u222a B = {1, 2, 4, 6, 8} \u222a {2, 5, 8}
\n= {1, 2, 4, 5, 6, 8}<\/p>\n

\"ML<\/p>\n

Question 11.
\nIf \u03be = {1, 2, 3, ……………., 10}, A = {x : x is prime} and B = {x : x is even integer}, then write the following :
\n(i) A – B
\n(ii) A \u2229 B\u2019.
\nSolution:
\nGiven \u03be = {1, 2, 3, ……………….., 10}
\nA = {x : x is prime} = {2, 3, 5, 7}
\nB = {x : x is even integer) = {2, 4,6, 8, 10}
\n(i) A – B = {3, 5, 7}
\nHere B’ – B = {1, 3, 5, 7, 9}<\/p>\n

(ii) A \u2229 B\u2019 = {2, 3, 5, 7} \u2229 {1, 3, 5, 7, 9}
\n= {3, 5, 7}<\/p>\n

Question 12.
\nIf \u03be = {all digits in our decimal system}, A = {x : x is prime}, and B = {x : x2<\/sup> < 25}, then write the following:
\n(i) B – A
\n(ii) A \u222a B
\n(iii) (A \u222a B)\u2019.
\nSolution:
\nGiven \u03be = {0, 1, 2, 3 ,9}
\nA = {x : x is prime} = {2, 3, 5, 7}
\nand B = {x : x2<\/sup> < 25} = {1, 2, 3, 4}
\n[Since 12<\/sup> = 1 < 25;
\n22<\/sup> = 4 < 25 ;
\n32<\/sup> = 9 < 25
\nand 42<\/sup> = 16 < 25]<\/p>\n

(i) B – A = {1, 4}<\/p>\n

(ii) A \u222a B = {1, 2, 3, 4, 5, 7}<\/p>\n

(iii) (A \u222a B)’ = \u03be – (A \u222a B) = {0, 6, 8, 9}<\/p>\n

Question 13.
\nOn the real axis, if A = (0, 3] and B = [2, 6], then write the following:
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(Iv) B – A
\nSolution:
\nGiven A = (0, 3]
\nand B = [2, 6]
\n(i) \u2234 \u03bb \u22a5 \u21d2 \u03c3
\nA \u222a B = (0, 3] \u222a [2, 6] = (0, 6]<\/p>\n

(ii) A \u2229 B = [2, 3]<\/p>\n

(iii) A – B = (0, 2)<\/p>\n

(iv) B – A = (3, 6]<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 14.
\nOn the real axis, if A = – [ 1, 1) and B = [0, 4], then find the following:
\n(i) A’
\n(ii) B’
\n(iii) A \u222a B
\n(iv) A \u2229 B
\n(v) A – B
\n(vi) B – A
\nSolution:<\/p>\n

Given A = [- 1, 1)
\nand B = [0, 4]<\/p>\n

(i) A’ = R – A
\n= R – [- 1, 1]
\n= (- \u221e, – 1) \u222a [1, \u221e)<\/p>\n

(ii) B’ = R – B
\n= (- \u221e, \u221e) – [0, 4]
\n= (- \u221e, 0) \u222a (4, \u221e)<\/p>\n

\"ML<\/p>\n

(iii) A \u222a B = [- 1, 1) \u222a [0, 4]
\n= [- 1, 4]<\/p>\n

(iv) A \u2229 B = [- 1, 1) \u2229 [0, 4]
\n= [0, 1)<\/p>\n

(v) A – B = [- 1, 1) – [0, 4]
\n= [- 1, 0)<\/p>\n

(vi) B – A = [0, 4] – [- 1, 1)
\n= [1, 4]<\/p>\n

Question 15.
\nIf A = the set of letters in the word \u2018JAIPUR\u2019 and B = the set of letters in the word \u2018JODHPUR\u2019, find the following.
\n(i) A \u222a B
\n(ii) A \u2229 B
\n(iii) A – B
\n(iv) B – A.
\nAlso verify the following results :
\n(a) n (A \u222a B) = n (A) + n (B) – n (A \u2229 B)
\n(b) n(A – B) = n (A) – n (A \u2229 B)
\n= n (A \u222a B) – n (B)
\nSolution:
\nA = {J, A, I, P, U, R}
\nB = {J, O, D. H, P, U, R}<\/p>\n

(i) A \u222a B = (J, A, J. P, U, R, O, D, H)<\/p>\n

(ii) A \u2229 B = {J, P, U, R}<\/p>\n

(iii) A – B = {A, 1}<\/p>\n

(iv) B – A = {O, D, H}
\nHere n(A) = no. of distinct elements in A = 6
\nn (B) = 7 ;
\nn (A \u222a B) = 9 ;
\nn (A \u2229 B) = 4<\/p>\n

(a) n (A \u222a B) = 9 = L.H.S.
\nand R.H.S. = n (A) + n (B) – n (A \u2229 B)
\n= 6 + 7 – 4 = 9
\n\u2234 n (A \u222a B) = n (A) + n (B) – n(A \u2229 B)<\/p>\n

(b) Here n (A – B) = 2 ;
\nn (B – A) = 3
\nNow n (A) – n (A \u2229 B) = 6 – 4
\n= 2 = n (A – B)
\nand n (A \u222a B) – n (B) = 9 – 7
\n= 2 = n (A – B)<\/p>\n

\"ML<\/p>\n

Question 16.
\nIf \u03be = {0, 1, 2, 3, 4, 5, 6, 7), A = {2, 5, 7}, B = {0, 2, 3, 7} and C = {0, 3, 4, 6}, form the following sets :
\n(i) (A \u222a B)’
\n(ii) A – C
\n(iii) A \u2229 (B \u222a C).
\nSolution:
\nGiven \u03be = {0, 1, 2, 3, 4, 5, 6, 7} ;
\nA = {2, 5, 7} ;
\nB = {0, 2, 3, 7}
\nand C = {0, 3, 4, 6}<\/p>\n

(i) A \u222a B = {2, 5, 7, 0, 3}
\n\u2234 (A \u222a B)’ = \u03be – AuB
\n= {1, 4, 6}<\/p>\n

(ii) A – C = {2, 5, 7)<\/p>\n

(iii) B \u222a C = {0, 2, 3, 4, 6, 7}
\n\u2234 A \u2229 (B \u222a C) = {2, 7}<\/p>\n

Question 17.
\nLet A = {2x : x \u2208 N and 1 \u2264 x < 4}, B = {x + 2 ; x \u2208 N and 2 \u2264 x < 5) and C = {x : x \u2208 N and 3 < x < 6}, determine the following :
\n(i) A \u2229 B
\n(ii) A \u222a B
\n(iii) (A \u222a B) \u2229 C
\nAlso verify that n (A \u222a B) = n (A) – n (A \u2229 B)
\nSolution:
\nGiven A = {2x : x \u2208 N and 1 \u2264 x < 4}
\nsince 1 \u2264 x < 4
\n\u2234 x = 1, 2, 3 as x \u2208 N
\n\u2234 A = {2, 4, 6}
\nB = {x + 2 ; x \u2208 N and 2 \u2264 x < 5}
\nsince x \u2208 N and 2 \u2264 x < 5
\n\u2234 x = 2, 3, 4
\n\u2234 B = {2 + 2, 3 + 2, 4 + 2} = {4, 5, 6}
\nand C = {x : x \u2208 N and 3 < x < 6} = {4, 5}<\/p>\n

(i) A \u2229 B = {2, 4, 6) \u2229 (4, 5, 6) = {4, 6}<\/p>\n

(ii) A \u222a B = (2, 4, 6) \u222a {4, 5, 6) = {2, 4, 5, 6}<\/p>\n

(iii) (A \u222a B) \u2229 C = {2, 4, 5, 6} \u2229 {4, 5} = {4, 5}
\n\u2234 n(A \u222a B) = 4 ;
\nn (A) = 3 ;
\nn (B) = 3 ;
\nn (A \u2229 B) = 2
\n\u2234 R.H.S. = n (A) + n (B) – n (A \u2229 B)
\n= 3 + 3 – 2
\n= 4 = n (A \u222a B)<\/p>\n

\"ML<\/p>\n

Question 18.
\nIf \u03be = the set of all digits in our decimal system, A = {x : x is prime} and B = {x : x is a perfect square}, then verify the following results :
\n(i) A \u222a A’ = \u03be
\n(ii) A \u2229 A\u2019= \u03a6
\n(iii) A-B = A \u2229 B’
\n(iv) (A \u2229 B)’ = A’ \u222a B’
\n(v) (A \u222a B)’ = A’ \u2229 B’
\n(vi) (A’)’ = A
\nSolution:
\nGiven \u03be = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
\nA = {x : x is prime}
\n= {2, 3, 5, 7}
\nand B = {x : x is perfect square}
\n= {1, 4, 9}<\/p>\n

(i) A’ = \u03be – A
\n= {0, 1, 4, 6, 8, 9}
\n\u2234 A \u222a A’ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = \u03be<\/p>\n

(ii) A \u2229 A’ = {2, 3, 5, 7} \u2229 {0, 1, 4, 6, 8, 9} = \u03a6<\/p>\n

(iii) A – B = {2, 3, 5, 7}
\nAlso B’ = \u03be – B\u2019
\n= {0, 2, 3, 5, 6, 7, 8}
\n\u2234 A \u2229 B’ = {2, 3, 5, 7} ………………(2)
\nFrom (1) and (2) ; we have
\nA – B = A \u2229 B’<\/p>\n

(iv) A \u2229 B = \u03a6
\n\u2234 (A \u2229 B)’ = \u03a6’ = \u03be ………………..(1)
\n\u2234 A’ \u222a B’ = {0, 1, 4, 6, 8, 9} \u222a {0, 2, 3, 5, 6, 7, 8}
\n= {0, 1, 2, 3, 4. 5, 6, 7, 8, 9}
\n= \u03be ………………….(2)
\nFrom (1) and (2) ; we have
\n(A \u2229 B)’ = A’ \u222a B’<\/p>\n

(v) L.H.S. = (A \u222a B)’ – (A \u222a B)
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} – {1, 2, 3, 4, 5, 7, 9}
\n= {0, 6, 8} ……………………..(1)
\nR.H.S. = A’ \u2229 B’
\n= {0, 1, 4, 6, 8, 9} \u2229 {0, 2, 3, 5, 6, 7, 8}
\n= {0, 6, 8}
\nFrom (1) and (2) ; we have
\n(A \u222a B)’ = A’ \u2229 B’<\/p>\n

(vi) A’ = \u03be – A
\n= {0, 1, 4, 6, 8, 9}
\n\u2234 (A’)’ = \u03be A’
\n= {0, 1, 2, 3, 4, 5, 6, 7, 8, 0} – {0, 1, 4, 6, 8, 9}
\n= {2, 3, 5, 7}
\n= A<\/p>\n","protected":false},"excerpt":{"rendered":"

Students often turn to ML Aggarwal Class 11 ISC Solutions Chapter 1 Sets Ex 1.3 to clarify doubts and improve problem-solving skills. ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.3 Question 1. If A = {1, 2, 3, 4, 5, 6) and B = {2, 4, 6, 8}, find (i) …<\/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\/170439"}],"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=170439"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170439\/revisions"}],"predecessor-version":[{"id":170463,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170439\/revisions\/170463"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170439"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170439"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170439"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}