{"id":162729,"date":"2023-10-31T17:54:33","date_gmt":"2023-10-31T12:24:33","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=162729"},"modified":"2023-10-31T17:54:33","modified_gmt":"2023-10-31T12:24:33","slug":"op-malhotra-class-12-maths-solutions-chapter-3-ex-3a","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/op-malhotra-class-12-maths-solutions-chapter-3-ex-3a\/","title":{"rendered":"OP Malhotra Class 12 Maths Solutions Chapter 3 Binary Operations Ex 3(a)"},"content":{"rendered":"

Utilizing OP Malhotra Class 12 Solutions<\/a> Chapter 3 Binary Operations Ex 3(a) as a study aid can enhance exam preparation.<\/p>\n

S Chand Class 12 ICSE Maths Solutions Chapter 3 Binary Operations Ex 3(a)<\/h2>\n

Question 1.
\nLet * be a binary operation on N given by a * b = HCF (a, b), a, b \u2208N. Write the value of 22 * 4.
\nSolution:
\n(i) Given * be a binary operation on N
\ngiven by a * b = HCF(a, b) \u2200a,b \u2208 N
\n\u2234 22 * 4 = HCF (22, 4) = 2<\/p>\n

Question 2.
\nLet * be a binary operation defined by a * b = 2a + b – 3, find 3 * 4.
\nSolution:
\nGiven * be a binary operation defined by
\na * b = 2a + b – 3
\n\u2234 3 * 4 = 2 x 3 + 4 – 3
\n= 6 + 4 – 3 = 7.<\/p>\n

Question 3.
\nIf a * b = 3a + 4b – 2, find the value of 8 * 2.
\nSolution:
\nGiven a * b = 3a + 4b – 2
\n\u2234 8 * 2 = 3 * 8 + 4 * 2 – 2
\n= 24 + 8 – 2 = 30.<\/p>\n

\"OP<\/p>\n

Question 4.
\nIf the binary operation * on the set of integers Z, is defined by a * b = a + 3b\u00b2, then find the value 2 * 4.
\nSolution:
\nGiven binary operation * on the set of integers Z is defined by
\na * b = a + 3b\u00b2
\n\u2234 2 * 4 = 2 + 3 * 4\u00b2 = 2 + 48 = 50<\/p>\n

Question 5.
\nLet * be a binary operation on N given by a* b = LCM of (a, b) for a, b \u2208 N. Find 5 * 7.
\nSolution:
\nLet * be a binary operation on N given by
\na* b = LCM of (a, b)
\nThus, 5 * 7 = LCM of (5, 7) = 35<\/p>\n

Question 6.
\nThe binary operation * : R * R \u2192 R is defined as a * b = 2a + b. Find (2 * 3) * 4.
\nSolution:
\nThe binary operation * : R * R \u2192 R is defined as
\na * b = 2a + b
\n\u2234 (2 * 3) * 4 = (2 * 2 + 3) * 4
\n= 7 * 4 = 2 * 7 + 4 = 18<\/p>\n

Question 7.
\nLet * be a binary operation, on the set of all non-zero numbers, given by a * b = \\(\\frac { ab }{ 5 }\\) for all a, b \u2208 R – {0}. Find the value of *, given that 2* (** 5) 10.
\nSolution:
\nLet * be a binary operation on set of all non-zero numbers given by
\na * b = \\(\\frac { ab }{ 5 }\\) \u2200a, b \u2208 R – {0}
\nand 2 * (x * 5) = 10
\n\u21d2 2 * \\(\\frac { 5x }{ 5 }\\) = 10 \u21d2 2 * x = 10
\n\u21d2 \\(\\frac { 2x }{ 5 }\\) = 10 \u21d2 x = 25<\/p>\n

\"OP<\/p>\n

Question 8.
\nCheck whether the following are binary operation or not.
\n(i) S = {0, 1} under
\n(a) multiplication, (b) addition
\n(ii) S = {1, 3, 5, 7,….} undergo
\n(a) multiplication (b) addition
\n(iii) S = {2, 4, 6, 8,…} under
\n(a) multiplication (b) addition
\nSolution:
\n(i) Given S = {0, 1}
\nSince 0 x 0 = 0; 0 x 1 = 1 x o = 0; 1 x 1 = 1
\nThus the operation x is defined by
\na x b = ab \u2208 S \u2200a, b \u2208 S
\n\u2234 operation * is binary under multiplication.
\nSince 1 * 1 = 1 + 1 = 2 \u2209 S
\n\u2234 operation * is defined by
\na * b = a + b \u2209 S \u2200 a b \u2208 S
\nThus * is not a binary operation under addition.<\/p>\n

(ii) Given S = {1, 3, 5, 7,….}
\nwe know that product of two odd natural numbers is an odd natural number
\n\u2234 a * b = a * b \u2208S
\nThus * is a binary operation under multiplication.
\nWe know that sum of two odd numbers is even.
\nNow 1 * 3 = 1 + 3 = 4 \u2209 S,
\nwhere 1, 3 \u2208 S
\nThus * is not a binary operation under addition.<\/p>\n

(iii) Given S = {2, 4, 6, 8,…}
\nclearly product of two even numbers be an even number
\n\u2234 a * b = a * b \u2208 S
\nThus * is a binary operation under multiplication in given set S.
\nclearly sum of two even nuumbers be an even number
\n\u2234 a * b = a + b \u2208 S
\nThus * is a binary operation under addition in S.<\/p>\n

Question 9.
\nExamine whether the following are binary operations.
\n(i) Subtraction on N.
\n(ii) Addition on irrationals.
\n(iii) * on R defined by a * b = max {a, b)
\n(iv) * on R defined by a* b = min {a, b)
\nSolution:
\n(i) Clearly operation * on N defined by
\na * b = a – b \u2200a b \u2208N
\nNow 1, 2 \u2208 N but 1 * 2 = 1 – 2 = – 1 \u2209 N
\nThus * is not a binary operation under subtraction in N.<\/p>\n

(ii) Let IR<\/sub> be the set of all irrational numbers
\nAn operation * on IR<\/sub> defined by
\na * b = a + b \u2208 a b \u2208 IR<\/sub>
\nHere \\(\\sqrt{2}\\), – \\(\\sqrt{2}\\) \u2208 IR<\/sub>.
\nBut \\(\\sqrt{2}\\) + (-\\(\\sqrt{2}\\)) = 0 \u2209 IR<\/sub>
\nThus, * is not a binary operation under addition an irrational numbers.<\/p>\n

(iii) Given * be an operation on R defined by
\na* b = max {a, b}
\nSince a* b = max {a, b} be either a or b \u2208 R \u2200 a, b \u2208 R
\nThus * be an binary operation.<\/p>\n

(iv) \u2200 a, b \u2208 R s.t a * b = min {a, b} be
\neither a or b \u2200 a, b \u2208 R
\nThus * is binary operation on R.<\/p>\n

\"OP<\/p>\n

Question 10.
\nShow that:
\n(i) Multiplication is a binary operation on S = {1, -1} but not on 7= {-1, 2};
\n(ii) Addition is a binary relation on S = {x : x \u2208 Z, x < 0} but multiplication is not.
\nSolution:
\n(i) Given S = {1, -1}
\nSince 1 x (- 1) = – 1 ; – 1 x 1 = – 1 ;
\n1 x 1 = 1 & (-1) x (- 1) = 1
\nThus the operation * on S defined by
\na * b = a * b \u2200 a b\u2208S
\n\u2234 given operation is binary under multiplication on S.
\n– 1 x 2 = – 2 \u2209 T
\na * b = a * b \u2209 T.
\nThus the operation * on T is not a binary operation under multiplication.<\/p>\n

(ii) Since sum of two negative integers is a negative integer
\n\u2234 a * b = a + b \u2200 a b \u2208 S
\n\u2200a b\u2208S then a + b\u2208S \u21d2 a * b\u2208S
\n\u2234 addition is a binary operation S
\n\u2200 a b\u2208S, a * b = ab\u2209S
\nSince product of two negative integers is a positive integer, – 1, – 2\u2208S
\nbut (- 1) * (- 2) = (- 1)(- 2) = 2\u2209S
\nThus multiplication is not a binary operation on S.<\/p>\n

Question 11.
\nDetermine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this
\n(i) On Z+<\/sup>, defined * by a * b = a – b
\n(ii) On Z+<\/sup>, defined * by a * b = ab
\n(iii) On R, defined by a * b = ab\u00b2
\n(iv) On Z+<\/sup>, defined * by a * b = |a – b|
\n(v) On R, defined * by a * b = |a – b|.
\n(vi) On Z+<\/sup>, defined * by a * b = a
\n(vii) On Z+<\/sup>, defined * by a* b = 2ab<\/sup>.
\nSolution:
\n(i) Given * on Z+<\/sup>, defined by
\na * b = a – b \u2209 Z+<\/sup>.
\nAs, 1, Z\u2208Z+2<\/sup> but 1 * 2 = 1 – 2 = – 1 \u2209 Z+<\/sup>
\n\u2234 * is not a binary operation on Z+<\/sup><\/p>\n

(ii) Given operation * on Z+<\/sup> defined by a * b = ab \u2209 Z+<\/sup>
\nSince product of two positive integers is a positive integer.
\nThus * be a binary operation on Z+<\/sup><\/p>\n

(iii) Given operation * on R defined by a * b = ab\u00b2
\n\u2200a b\u2208R \u21d2 ab\u00b2\u2208R \u21d2 a * b \u2208R
\nThus * is a binary operation on R.<\/p>\n

(iv) Given operation * on Z+<\/sup> defined by a * b = | a – b |
\nNow a, a\u2208Z+<\/sup>
\na * a = | a – a | = 0 \u2209 Z+<\/sup>
\nThus a * b = | a – b | \u2209 Z+<\/sup>
\n\u2234 * is not a binary operation on Z+<\/sup>.<\/p>\n

(v) Given operation * on R defined by
\na * b = | a – b |
\n\u2200a b \u2208 R, a * b = |(a – b)|\u2208R
\n\u2234 * be a binary operation on R.<\/p>\n

(vi) Operation * on Z+<\/sup> defined by
\na * b = a.
\n\u2200 a b\u2208Z+<\/sup> s.t a * b = a\u2208Z+<\/sup>
\nThus * be a binary operation on Z+<\/sup><\/p>\n

(vii) \u2200 a, b \u2208 Z+<\/sup>, a * b = 2ab \u2209 Z+<\/sup>
\n[\u2235 a, b \u2208 Z+<\/sup> \u21d2 ab\u2208 Z+<\/sup> \u2234 2ab \u2208 Z+<\/sup>]
\nThus * be a binary operation on Z+<\/sup>.<\/p>\n","protected":false},"excerpt":{"rendered":"

Utilizing OP Malhotra Class 12 Solutions Chapter 3 Binary Operations Ex 3(a) as a study aid can enhance exam preparation. S Chand Class 12 ICSE Maths Solutions Chapter 3 Binary Operations Ex 3(a) Question 1. Let * be a binary operation on N given by a * b = HCF (a, b), a, b \u2208N. …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[115871],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162729"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=162729"}],"version-history":[{"count":2,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162729\/revisions"}],"predecessor-version":[{"id":162732,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/162729\/revisions\/162732"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=162729"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=162729"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=162729"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}