# Mathematics Today - August 2018

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内容提示： 3 AUGUST ’18 MATHEMATICS TODAY 4AUGUST ’18 MATHEMATICS TODAY 5 AUGUST ’18 MATHEMATICS TODAY 6AUGUST ’18 MATHEMATICS TODAY MATHEMATICS TODAY | AUGUST‘187CONTENTSSubscribe online at www.mtg.inVol. XXXVI No. 8 August 2018Corporate Of f i ce:Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200e-mail : info@mtg.in website : www.mtg.inRegd. Of f i ce:406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor : Mahabir SinghEditor : Ani...

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MATHEMATICS TODAY | AUGUST‘187CONTENTSSubscribe online at www.mtg.inVol. XXXVI No. 8 August 2018Corporate Of f i ce:Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200e-mail : info@mtg.in website : www.mtg.inRegd. Of f i ce:406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor : Mahabir SinghEditor : Anil Ahlawat8 Maths Musing Problem Set - 18810 Concept Boosters23 CBSE Drill (Series 4)32 MPP-446 Concept Map34 Concept Boosters47 Concept Map48 CBSE Drill (Series 4)58 MPP-417 Olympiad Corner60 Challenging Problems63 Math Archives65 Target JEE71 Mock Test Paper JEE Main 2019 (Series 3)77 Brain @ Work84 Maths Musing SolutionsClass XI Class XII Competition Edge10347763716523488601746 47Printed and Published by Mahabir Singh on behalf of MTG Learning Media Pvt. Ltd. Printed at HT Media Ltd., B-2, Sector-63, Noida, UP-201307 and published at 406, Taj Apartment, Ring Road, Near Safdarjung Hospital, New Delhi - 110029.Editor : Anil AhlawatReaders are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine. Focus/Infocus features are marketing incentives. MTG does not vouch or subscribe to the claims and representations made by advertisers. All disputes are subject to Delhi jurisdiction only.Copyright© MTG Learning Media (P) Ltd.All rights reserved. Reproduction in any form is prohibited.Send D.D/M.O in favour of MTG Learning Media (P) Ltd.Payments should be made directly to : MTG Learning Media (P) Ltd,Plot 99, Sector 44 Institutional Area, Gurgaon - 122 003, Haryana.We have not appointed any subscription agent.Individual Subscription RatesRepeaters9 monthsClass XI27 monthsClass XII15 monthsMathematics Today 300 500 850Chemistry Today 300 500 850Physics For You 300 500 850Biology Today 300 500 850Combined Subscription RatesRepeaters9 monthsClass XI27 monthsClass XII15 monthsPCM 900 1400 2500PCB 900 1400 2500PCMB 1200 1900 3400
MATHEMATICS TODAY | AUGUST‘18 8JEE MAIN1. Five digit numbers are formed using the digits 0, 1, 2, 3, 4, 6, 7 without repetition. Find the probability that a number is divisible by 3.(a) 600 (b) 700 (c) 800 (d) 9002. If p = +⋅⋅+⋅ ⋅⋅ ⋅+ ∞131 33 61 3 53 6 9... , then(a) p 2 – 2p + 2 = 0 (b) p 2 + 2p – 2 = 0(c) p 2 – 2p – 2 = 0 (d) none of these3. Te integral dxx x x ( ) 12+ −∫is equal to(where C is a constant of integration.)(a) −+−+ 211xxC (b) −−++11xxC(c) −−++ 211xxC (d) 211+−+xxC4. If f(x) =−+−−2 22 2x xx x, then f–1 (x) is (a) 12 12logxx − (b) 12112log+−xx(c) 1212log+ xx (d) 12222log+−xx5. Equation of a line which is tangent to both the curves y = x 2 + 1 and y = –x 2 is (a) yx = − 212 (b) yx = + 212(c) y x = − + 212 (d) yx = − − 212JEE ADVANCED6. Let ? ? ?x y z , and be three vectors each of magnitude 2 and the angle between each pair of them is p/3. If ?a is a non-zero vector perpendicular to Set 188? ? ?x y z and × and ?b is non-zero vector perpendicular to ? ? ?y z x and × , then (a) ? ?? ? ?b b z z x = ⋅ − ( )( ) (b) ? ? ? ? ?a a y y z = ⋅ − ( )( )(c) ??? ???a b a y b z ⋅ = − ⋅ ⋅ ( )( ) (d) ? ? ? ? ?a a y z y = ⋅ − ( )( )COMPREHENSIONA variable straight line is drawn through the point A(–1, 1) to intersect the parabola y 2 = 4x at the points B and C. Let P be a point on the chord BC.7. If AB, AP, AC are in H.P., then locus of P is(a) a straight line (b) a pair of lines (c) a circle (d) a parabola8. If AB, AP, AC are in A.P., then the locus of P is(a) a straight line (b) a pair of lines(c) a circle (d) a parabolaINTEGER TYPE9. If I(x, y) is the incentre of the triangle formed by the points (3, 4), (4, 3), (1, 2), then x isMATRIX MATCH10. If abc = 1 and Aa b cc a bb c a= is an orthogonal matrix. Ten, match the following columns.Column-I Column-IIP. Te least value of a + b + c is 1. –1Q. Te value of ab + bc + ca is 2. 0R. Te value of a 2 + b 2 + c 2 is 3. 1S. Te value of a 3 + b 3 + c 3 can be 4. 2 P Q R S(a) 3 2 1 4(b) 4 3 2 1(c) 1 2 3 4(d) 1 3 2 4Maths Musing was started in January 2003 issue of Mathematics Today. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benef i tting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.See Solution Set of Maths Musing 187 on page no. 84
MATHEMATICS TODAY | AUGUST‘1810SETSA set is well defned class or collection of objects. z Roster method or Listing method : In this method, a set is described by listing elements, separated by commas, within braces {}. Te set of vowels of English alphabet may be described as {a, e, i, o, u}. z Set-builder method or Rule method : In this method, a set is described by a characterizing property P(x) of its elements x. In such a case, the set is described by {x : P(x) holds} or {x | P(x) holds}, which is read as ‘the set of all x such that P(x) holds’. Te set A = {0, 1, 4, 9, 16, ...} can be written as A = {x 2 : x ? Z}.TYPES OF SETS z Singleton set : A set consisting of a single element is called a singleton set. Te set {5} is a singleton set. z Null set or Empty set : Te set which contains no element at all is called the null set. Tis set is also called the ‘empty set’ or the ‘void set’. It is denoted by the symbol ? or {}. z Finite set : A set is called a fnite set if it is either void set or its elements can be listed (counted) by a certain natural number. Cardinal number of a fnite set : Te number n in the above defnition is called the cardinal number or order of a fnite set A and is denoted by n(A) or O(A). z Infnite set : A set whose elements cannot be listed (counted) by a certain natural number (n) is called an infnite set. z Equivalent sets : Two finite sets A and B are equivalent, if their cardinal numbers are same i.e. n(A) = n(B). If A = {1, 3, 5, 7}; B = {10, 12, 14, 16} then A and B are equivalent sets, as O(A) = O(B) = 4. z Equal sets : Two sets A and B are said to be equal if every element of A is an element of B and also every element of B is an element of A. Symbolically, A = B if x ? A ? x ? B. If A = {2, 3, 5, 6} and B = {6, 5, 3, 2}, then A = B because each element of A is an element of B and vice-versa. Note : Equal sets are always equivalent but equivalent sets need not to be equal sets.SUBSETS (SET INCLUSION) z Let A and B be two sets. If every element of A is an element of B, then A is called a subset of B. If A is subset of B, we write A ? B, which is read as “A is a subset of B” or “A is contained in B”. Tus, A ? B i.e., a ? A ? a ? B. Te total number of subsets of a fnite set containing n elements is 2 n . z Proper and improper subsets : If A is a subset of B and A ? B, then A is a proper subset of B. We write this as A ? B.Te null set ? is subset of every set and every set is subset of itself, i.e., ? ? A and A ? A for every set A. Tey are called improper subsets of A. It should be noted that ? has only one subset ? which is improper. * Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.* ALOK KUMAR, B.Tech, IIT KanpurThis column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc. and be also in command of what is being covered in their school as part of NCERT syllabus. The problems here are a happy blend of the straight and the twisted, the simple and the diff i cult and the easy and the challenging.Sets and Relations
MATHEMATICS TODAY | AUGUST‘1811
MATHEMATICS TODAY | AUGUST‘1812All other subsets of A are called its proper subsets. Let A = {1, 2}. Ten A has ?, {1}, {2}, {1, 2} as its subsets out of which ? and {1, 2} are improper and {1} and {2} are proper subsets.UNIVERSAL SETA set that contains all sets in a given context is called the universal set.It should be noted that universal set is not unique.POWER SETIf S is any set, then the set of all the subsets of S is called the power set of S.Te power set of S is denoted by P(S). Symbolically, P(S) = {T : T ? S}. Obviously ? and S are both elements of P(S). Let S = {a, b, c}, then P(S) = {? , {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. Note: Power set of a given set is always non-empty.DISJOINT SETSTwo sets A and B are said to be disjoint, if A ? B = ?. If A ? B ? ?, then A and B are said to be non-intersecting or non-overlapping sets.VENN DIAGRAMSTe combination of rectangles and circles are called Venn diagrams. The universal set is usually represented by a rectangle and its subsets by circles.Note : If A and B are not equal but they have some common elements, then to represent A and B we draw two intersecting circles. Two disjoints sets are represented by two non-intersecting circles.OPERATIONS ON SETS z Union of sets : Let A and B be two sets. Te union of A and B is the set of all elements which are either in set A or in B. We denote the union of A and B by A ? B (read as “A union B”). Symbolically, A ? B = {x : x ? A or x ? B}. Shaded portion in the given fgure represents A ??B. z Intersection of sets : Let A and B be two sets. Te intersection of A and B is the set of all those elements that belong to both A and B. AUBTe intersection of A and B is denoted by A ? B (read as “A intersection B”). Tus, A ? B = {x : x ? A and x ? B}. z Diference of sets : Let A and B be two sets. Te diference of A and B written as A – B, is the set of all those elements of A which do not belong to B. Tus, A – B = {x : x ? A and x ? B} Similarly, the diference B – A is the set of all those elements of B that do not belong to A ? B – A = {x : x ? B and x ? A}.Note : For three sets A, B and C,(A – B) – C ? A – (B – C). z Symmetric diference of two sets : Let A and B be two sets. Te symmetric diference of sets A and B is the set (A – B) ? (B – A) and is denoted by A ? B. Tus, A?B = (A – B) ? (B – A) = {x : x ? A ? B}.COMPLEMENT OF A SETLet U be the universal set and let A be a set such that A ? U. Then, the complement of A with respect to U is denoted by A? or A c or U – A and isdefned the set of all those elements of U which are not in A. Tus, A? = {x ? U : x ? A}. LAWS OF ALGEBRA OF SETS z Idempotent laws : For any set A, we have (i) A ? A = A (ii) A ? A = A z Identity laws : For any set A, we have (i) A ? ? = A(ii) A ? U = A i.e., ? and U are identity elements for union and intersection respectively. z Commutative laws : For any two sets A and B, we have (i) A ? B = B ? A (ii) A ? B = B ? A (iii)A ? B = B ? A z Associative laws : If A, B and C are any three sets, then (i) (A ? B) ? C = A ? (B ? C) AU A?
MATHEMATICS TODAY | AUGUST‘1813 (ii)A ? (B ? C) = (A ? B) ? C (iii)(A ? B) ? C = A ? (B ? C)i.e., union, intersection and symmetric diference of two sets are associative. z Distributive laws : If A, B and C are any three sets, then (i) A ? (B ??C) = (A ? B) ? (A ? C) (ii) A ? (B ? C) = (A ? B) ? (A ? C)i.e., union and intersection are distributive over intersection and union respectively. z De-Morgan’s law : If A, B and C are any three sets, then (i) (A ? B)? = A? ? B? (ii) (A ? B)? = A? ? B? (iii) A – (B ? C) = (A – B) ? (A – C) (iv) A – (B ? C) = (A – B) ? (A – C) z If A and B are any two sets, then (i) A – B = A ? B?; (ii) B – A = B ? A? (iii) A – B = A ? A ? B = ? (iv) (A – B) ? B = A ? B (v) (A – B) ? B = ? (vi) A ? B ? B??? A? (vii) (A – B) ? (B – A) = (A ? B) – (A ? B) z If A, B and C are any three sets, then (i) A ? (B – C) = (A ? B) – (A ? C) (ii) A ? (B ? C) = (A ? B) ? (A ? C)SOME IMPORTANT RESULTSIf A, B and C are fnite sets and U be the fnite universal set, then z n(A ? B) = n(A) + n(B) – n(A ? B) z n(A ? B) = n(A) + n(B) ? A, B are disjoint non-void sets. z n(A – B) = n(A) – n(A ? B) i.e., n(A – B) + n(A ? B) = n(A) z n(A ? B) = Number of elements which belong to exactly one of A or B = n((A – B) ? (B – A)) = n(A – B) + n(B – A) [ ?(A – B) and (B – A) are disjoint sets] = n(A) – n(A ? B) + n(B) – n(A ? B) = n(A) + n(B) – 2n(A?? B) z n(A ? B ? C) = n(A) + n(B) + n(C) – n(A ? B) – n(B ? C) – n(A ? C) + n(A ? B ??C) z Number of elements in exactly two of the sets A, B, C = n(A ? B) + n(B ? C) + n(C ? A) – 3n(A ??B ??C) z Number of elements in exactly one of the sets A, B, C = n(A) + n(B) + n(C) – 2n(A ? B) – 2n(B ? C)– 2n(A ? C) + 3n(A ? B ? C) z n(A? ? B?) = n(A ? B)? = n(U) – n(A ? B) z n(A? ? B?) = n(A ? B)? = n(U) – n(A ? B) CARTESIAN PRODUCT OF SETSLet A and B be any two non-empty sets. Te set of all ordered pairs (a, b) such that a ? A and b ? B is called the cartesian product of the sets A and B and is denoted by A × B. Tus, A × B = [(a, b) : a ? A and b ? B] If A = ? or B = ?, then we defne A × B = ?.Let A = {a, b, c} and B = {p, q}.Ten, A × B = {(a, p), (a, q), (b, p), (b, q), (c, p), (c, q)} Also, B × A = {(p, a), (p, b), (p, c), (q, a), (q, b), (q, c)}Note : z Cartesian product of two sets is not commutative (in general). z Let A and B two non-empty sets having n elements in common, then A × B and B × A have n 2 elements in common.IMPORTANT THEOREMS ON CARTESIAN PRODUCT OF SETSFor any three sets A, B and C z A × (B ? C) = (A × B) ? (A × C) z A × (B ? C) = (A × B) ? (A × C) z A × (B – C) = (A × B) – (A × C) z (A × B) × C ? A × (B × C) z A × (B? ? C?)? = (A × B) ? (A × C) z A × (B? ? C?)? = (A × B) ? (A × C)If A and B are any two non-empty sets, thenA × B = B × A ? A = B z If A ??B, then A × A ? (A × B) ? (B × A) z If A ? B, then A × C ? B × C, for any set C. z If A ? B and C ? D, then A × C ? B × D z For any sets A, B, C and D (A × B) ? (C × D) = (A ? C) × (B ? D)RELATIONSLet A and B be two non-empty sets, then every subset of A × B defnes a relation from A to B i.e., every relation from A to B is a subset of A × B.Let R ? A × B and (a, b) ? R. Ten we say that a is related to b by the relation R and write it as a R b.Total number of relations : Let A and B be two non-empty fnite sets consisting of m and n elements respectively. Ten A × B consists of mn ordered pairs. So, total number of subsets of A × B is 2 mn . Since each subset of A × B defnes a relation from A to B, so total number of relations from A to B is 2 mn . Among these 2 mn relations, the void relation ??and the universal relation A × B are trivial relations from A to B.
MATHEMATICS TODAY | AUGUST‘1814DOMAIN AND RANGE OF A RELATIONLet R be a relation from a set A to a set B. Ten the set of all frst components or coordinates of the ordered pairs belonging to R is called the domain of R, while the set of all second components or coordinates of the ordered pairs in R is called the range of R. Tus, Dom (R) = {a : (a, b) ? R} Range (R) = {b : (a, b) ? R}.Note : z Te whole set B is called co-domain of R. z Range ? Co-domainPROBLEMSSingle Correct Answer Type1. Which of the following is the empty set ?(a) {x : x is a real number and x 2 – 1 = 0}(b) {x : x is a real number and x 2 + 1 = 0}(c) {x : x is a real number and x 2 – 9 = 0}(d) {x : x is a real number and x 2 = x + 2}.2. If a set A has n elements, then the total number of subsets of A is(a) n (b) n 2 (c) 2 n (d) 2n3. In a town of 10,000 families, it was found that 40% families buy newspaper X, 20% buy newspaper Y and 10% families buy newspaper Z, 5% families buy both X and Y, 3% buy both Y and Z and 4% buy both X and Z. If 2% families buy all the three newspapers, then number of families which buy newspaper X only is(a) 3100 (b) 3300 (c) 2900 (d) 14004. In a city, 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Ten persons travelling by car or bus is (a) 80 percent (b) 40 percent(c) 60 percent (d) 70 percent5. In a class of 55 students, the number of students studying diferent subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. Te number of students who have taken exactly one subject is(a) 18 (b) 20(c) 12 (d) none of these6. If A, B and C are any three sets, then A × (B ? C) is equal to (a) (A × B) ? (A × C) (b) (A ? B) × (A ? C)(c) (A × B) ? (A × C) (d) none of these7. If A = {2, 4, 5}, B = {7, 8, 9} then n(A × B) is equal to(a) 6 (b) 9 (c) 3 (d) 08. Te smallest set A such that A ? {1, 2} = {1, 2, 3, 5, 9} is(a) {2, 3, 5} (b) {3, 5, 9}(c) {1, 2, 5, 9} (d) none of these9. If A and B are two sets, then A ? B = A ? B if(a) A ? B (b) B ? A(c) A = B (d) none of these 10. If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12}, C = {4, 5, 6, 12, 14}, then (A ? B) ? (A ? C) is equal to (a) {3, 4, 10} (b) {2, 8, 10}(c) {4, 5, 6} (d) {3, 5, 14}11. If A and B are two sets, then A A B ∩ ∪ ′ ( ) is equal to(a) A (b) B(c) ? (d) none of these12. If N an n Na= ∈ [ : }, then N 5 ? N 7 =(a) N 7 (b) N 10 (c) N 35 (d) N 513. Te shaded region in the given fgure is (a) A ? (B ? C)(b) A ? (B ? C) (c) A ? (B – C) (d) A – (B ? C) 14. Let U be the universal set and A ??B ??C = U. Ten {( ) ( ) ( )} A B B C C A − ∪ − ∪ − ′ is equal to(a) A ? B ? C (b) A ??(B ? C)(c) A ? B ? C (d) A ? (B ??C)15. In a battle, 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x% lost all the four limbs. Te minimum value of x is(a) 10 (b) 12(c) 15 (d) none of these16. Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. Te number of boys who did not play any game is (a) 128 (b) 216 (c) 240 (d) 16017. In a class of 100 students, 55 students have passed in Mathematics and 67 students have passed in Physics. Ten the number of students who have passed in Physics only is(a) 22 (b) 33 (c) 10 (d) 4518. If A, B and C are any three sets, then A – (B ??C) is equal to(a) (A – B) ? (A – C) (b) (A – B) ? (A – C)(c) (A – B) ? C (d) (A – B) ? C
MATHEMATICS TODAY | AUGUST‘181519. If A, B, C are three sets, then A ? (B ? C) is equal to(a) (A ? B) ? (A ? C) (b) (A ? B) ? (A ? C) (c) (A ? B) ? (A ? C) (d) none of these20. In a class of 30 pupils, 12 take English, 16 take physics and 18 take history. If all the 30 pupils take at least one subject and no one takes all three then the number of pupils taking exactly 2 subjects is(a) 16 (b) 6 (c) 8 (d) 2021. A class has 175 students. Te following data shows the number of students obtaining one or more subjects. Mathematics–100, Physics–70, Chemistry–40; Mathematics and Physics–30, Mathematics and Chemistry–28, Physics and Chemistry –23, Mathematics, Physics and Chemistry–18. How many students have ofered Mathematics alone?(a) 35 (b) 48 (c) 60 (d) 2222. Given n(U) = 20, n(A) = 12, n(B) = 9, n(A ??B) = 4, where U is the universal set, A and B are subsets of U, then n((A ??B) c ) = (a) 17 (b) 9 (c) 11 (d) 323. Te relation R defned on the set of natural numbers as {(a, b) : a difers b by 3}, is given by (a) {(1, 4), (2, 5), (3, 6),.....}(b) {(4, 1), (5, 2), (6, 3),.....}(c) Both (a) and (b) (d) none of these24. If R is a relation from a fnite set A having m elements to a fnite set B having n elements, then the number of relations from A to B is(a) 2 mn (b) 2 mn – 1 (c) 2mn (d) m n25. Te relation R defned on the set A = {1, 2, 3, 4, 5} by R = {(x, y) : |x 2 – y 2 | < 16} is given by (a) {(1, 1), (2, 1), (3, 1), (4, 1), (2, 3)}(b) {(2, 2), (3, 2), (4, 2), (2, 4)}(c) {(3, 3), (3, 4), (5, 4), (4, 3), (3, 1)}(d) none of these26. If A is the set of even natural numbers less than 8 and B is the set of prime numbers less than 7, then the number of relations from A to B is (a) 2 9 (b) 9 2 (c) 3 2 (d) 2 9 – 127. Te power set of A = {?, {?}}(a) {?, {?}, {{?}}} (b) {?, {?}, {{?}}, {?, {?}}}(c) {{?}, {{?}}, {?, {?}}}(d) None of these28. (A ??B) ? (A ??B C ) equals(a) A (b) B (c) A ??B? (d) A ??B?29. In a survey of 100 persons, it was found that 28 read magazine A, 30 read magazine B, 42 read magazine C, 8 read magazines A and B, 10 read magazines A and C, 5 read magazines B and C and 3 read all the three magazines. Ten, number of persons who read none of the three magazines and magazine C only respectively are(a) 20, 30 (b) 30, 20 (c) 25, 35 (d) 25, 4030. Which of the following statements is true?(a) A = B ? A ? B and B ? A (b) A ? B ? C ?? A ? C (c) A ? ?? ? A = ? (d) all of theseSOLUTIONS1. (b) : Since, x 2 + 1 = 0, gives x 2 = –1 ? x = ±i ? x is not real but x is real (given) ? No value of x is possible.2. (c) : Number of subsets of A having n elements = 2 n 3. (b) : Let sets A, B and C represents families who buy newspaper X, Y and Z respectively.n(A) = 40% of 10000 = 4000n(B) = 20% of 10000 = 2000n(C) = 10% of 10000 = 1000n(A ? B) = 5% of 10000 = 500 n(B ? C) = 3% of 10000 = 300n(C ? A) = 4% of 10000 = 400 n(A ? B ? C) = 2% of 10000 = 200We want to fnd n(A ? B c ? C c ) = n[A ? (B ? C) c ]= n(A) – n[A ? (B ? C)]= n(A) – n[(A ? B) ? (A ? C)]= n(A) – [n(A ? B) + n(A ? C) – n(A ? B ? C)] = 4000 – [500 + 400 – 200] = 4000 – 700 = 3300.4. (c) : Let B and C represents the set of population who travel by bus and car respectively.Given, n(C) = 20, n(B) = 50, n(C ? B) = 10Now, n(C ? B) = n(C) + n(B) – n(C ? B) = 20 + 50 – 10 = 60.Hence, required number of persons = 60%.5. (d) : n(M) = 23, n(P) = 24, n(C)= 19n(M ? P) = 12, n(M ? C)= 9, n(P ? C)= 7n(M ? P ? C) = 4We have to fnd n(M ? P? ? C?), n(P ? M? ? C? ), n (C ? M? ? P?) Now, n(M ? P? ? C?) = n[M ? (P ? C)?]= n(M) – n(M ? (P ? C)) = n(M) – n[(M ??P) ? (M ??C)]= n(M) – n(M ? P) – n(M ? C) + n(M ? P ? C) = 23 –12 – 9 + 4 = 27 –21 = 6n(P ? M? ? C?) = n[P ? (M ? C)?] = n(P) – n[P ? (M ? C)] = n(P) – n[(P ? M) ? (P ? C)]
MATHEMATICS TODAY | AUGUST‘1816= n(P) – n(P ? M) – n(P ? C) + n(P ? M ? C)= 24 – 12 – 7 + 4 = 9Similarly, n(C ? M????P?) = n(C) – n(C ??P) – n(C ??M) + n(C ??P ??M) = 19 – 7 – 9 + 4 = 23 – 16 = 7.Tus, the number of students who have taken exactly one subject = 6 + 9 + 7 = 226. (a)7. (b) : A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9)}? n(A × B) = n(A) · n(B) = 3 × 3 = 9.8. (b) : Given A ? {1, 2} = {1, 2, 3, 5, 9}. Hence, A = {3, 5, 9}.9. (c) : Let x ? A ? x ? A ? B [ A ? A ? B] ? x ? A ? B [ ?A ? B = A ??B] ? x ? A and x ? B ? x ? B ? A ? B Similarly, x ? B ? x ? A ? B ? A Now, A ? B, B ? A ? A = B 10. (a) : A ? B = {2, 3, 4, 8, 10} ? {3, 4, 5, 10, 12}= {3, 4, 10}Also, A ? C = {4}? (A ? B) ? (A ? C) = {3, 4, 10}.11. (c) : A ? (A ? B)? = A ? (A? ? B?) (De-Morgan's law)= (A ? A?) ? B? (By associative law)= ? ? B??= ?12. (c) : N 5 ? N 7 = N 35 , [ 5 and 7 are relatively prime numbers].13. (d)14. (c) : From Venn diagram, it is clear that,{(A – B) ? (B – C)? ? ? ? ? ? ? (C – A)}?= A ? B ? C .15. (a) : Minimum value of n = 100 – (30 + 20 + 25 + 15) = 100 – 90 = 1016. (d) : Given, n(C) = 224, n(H) = 240, n(B) = 336n(H ? B) = 64, n(B ? C) = 80n(H ? C) = 40, n(C ? H ? B) = 24Number of boys who did not play any game= n(C c ? H c ? B c ) = n[(C ? H ? B) c ] = n(U) – n(C ? H ??B)= 800 – [n(C) + n(H) + n(B) – n(H ? C)–n(H ? B) – n(C ? B) + n(C ? H ? B)]= 800 – [224 + 240 + 336 – 64 – 80 – 40 + 24] = 160.17. (d) : n(M) = 55, n(P) = 67, n(M ? P) = 100 Now, n(M ? P) = n(M) + n(P) – n(M ? P) 100 = 55 + 67 – n(M ? P)? n(M ? P) = 122 – 100 = 22 Now, n(P only) = n(P) – n(M ? P) = 67 – 22 = 45.18. (a) : From De-morgan’s law, we haveA – (B ? C) = (A – B) ? (A – C).19. (b) : From distributive law, we haveA ? (B ? C) = (A ? B) ? (A ? C)20. (a) : Given, n(E) = 12, n(P) = 16, n(H) = 18and n(E ? P ? H) = 30 n(E ? P ? H) = n(E) + n(P) + n(H) – n(E ? P) – n(P ? H) – n(E ? H) + n(E ? P ? H)? n(E ??P) + n(P ??H) + n(E ??H) = 16Now, number of pupils taking exactly two subjects= n(E ??P) + n(P ??H) + n(E ??H)– 3n(E ??P ??H) = 16 – 0 = 1621. (c) : n(M alone)= n(M) – n(M ??C) – n(M ??P) + n(M ??P ??C) = 100 – 28 – 30 + 18= 60. 22. (d) : n(A ? B) = n(A) + n(B) – n(A ? B) = 12 + 9 – 4 = 17Now, n((A ? B) c ) = n(U) – n(A ? B) = 20 – 17 = 3.23. (c) : R = {(a, b) : a, b ? N, | a – b| = 3} = {(4, 1), (5, 2), (6, 3), ....} or {(1, 4), (2, 5), (3, 6) ...}.24. (a) : Number of relations from set A to set B having m and n elements respectively is 2 mn .25. (d) : Here R = {(x, y) : |x 2 – y 2 | < 16}and given A = {1, 2, 3, 4, 5}? R = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (5, 4), (5, 5)}26. (a) : Here, A = {2, 4, 6}; B = {2, 3, 5}?? A × B contains 3 × 3 = 9 elements.Hence, number of relations from A to B = 2 9 .27. (b) : P(A) = {?, {?}, {{?}}, {?, {?}}}28. (a)29. (a) : n(A) = 28, n(B) = 30, n(C) = 42, n(A ? B) = 8, n(A ? C) = 10, n(B ? C) = 5, n(A ? B ? C ) = 3 Since, n(A ? B ? C) = n(A) + n(B) + n(C) – (n(A ? B) + n(A ? C) + n(B ??C)) + n(A ? B ??C)= 100 – 23 + 3 = 80 ? Number of persons who read none of the three magazines = 100 – 80 = 20Also, n(read magazine C only)= n(C) – (n(A ? C) + n(B ? C)) + n(A ? B ? C) = 42 – 10 – 5 + 3 = 3030. (d)
MATHEMATICS TODAY | AUGUST‘18178. If for every positive integer n, f(n) is defined asf n nf nnn( )( )=−=≥1112forforthen prove that : 1992 1992431992 < < f( ) .9. We consider regular n-gons with a fxed circumference 4. We call the distance from the centre of such a n-gon to a vertex r n and the distance from the centre to an edge a n . (a) Determine a 4 , r 4 , a 8 , r 8(b) Give an appropriate interpretation for a 2 and r 2 .(c) Prove: aa r r a rn n n n n n 2 2 212= + = ( ) . and (d) Let u 0 , u 1 , u 2 , u 3 , … be defned as follows : u 0 = 0, u 1 = 1, u n = 122 1u un n − −+ ( ) for even n and u n = u un n − − 2 1 for odd n. Determine: limn→∞u n10. Tere are real numbers a, b, c such that a ? b ??c > 0. Prove that abcc baa cb2 2 2 2 2 2−+−+− ??3a – 4b + c.SOLUTIONS1. PTR – r???Op Oq P' Q'QS?First, we note that triangle ?TOpOq has a right angle a nr n1. Two unequal circles of radii R and r touch externally, and P and Q are the points of contact of a common tangent to the circles, respectively. Find the volume of the frustum of a cone generated by rotating PQ about the line joining the centres of the circles.2. Let n ≥ 2 be a natural number. Show that there exists a constant C = C(n) such that for all real x 1 ,..., x n ≥ 0 we have x x Ckknkkn≤ += =∑ ∏1 1( ). Determine the minimum C(n) for some values of n.3. Find all real coefcients polynomials p(x) satisfying (x –1) 2 p (x) = (x–3) 2 p(x+2) for all x.4. Prove that : 0 ? yz + zx + xy – 2xyz ? 727, where x, y, z are non-negative real numbers for which x + y + z = 1.5. Find the value of the continued root: 4 27 4 29 4 31 4 33 + + + + …6. A hexagon is inscribed in a circle with radius r. Two of its sides have length 1, two have length 2 and the last two have length 3. Prove that r is a root of the equation 2r 3 – 7r – 3 = 0.7. Let k ≥ 2 be an integer. Te sequence (x n ) is defned by x 0 = x 1 = 1 and xxxnnnkn+−=+≥1111 for .(a) Prove that for each positive integer k ≥ 2 the sequence (x n ) is a sequence of integers.(b) If k = 2, show that x n+1 = 3x n – x n–1 for n ≥ 1.
MATHEMATICS TODAY | AUGUST‘1818angle at T, with OpOq = R + r and OpT = R – r. Hence OqT = 2 Rr .Because of parallel and perpendicular lines, all of angles ∠PSOp, ∠TOqOp, ∠OpPP' and ∠OqQQ' are equal. We denote the common value by ?? From triangle ?OpOqT, we note that sin ??= R rR r−+and cos ????2 RrR r + Using various right triangles, we obtain:OpP? = R sin ? = R R rR r( ) −+, PP? = R cos ? = 2R RrR r +P?S = PP' cot ? =2 2 R RrR rRrR r + −= 422 2rRR r −,QqQ? = r sin ? = r R rR r( ) −+QQ? = r cos ? = 2r RrR r +Q?S = QQ? cot ? = 2 2 r RrR rRrR r + −= 422 2r RR r −Hence, V = π32 2( ) ( ) PP P S QQ Q S ′ ′ − ′ ′=+ −−+ −π34 4 4 43222 23222 2R rR rrRR rRrR rr RR r ( ) ( )=−+ −1632 2 3 32 2 2πR r R rR r R r( )( ) ( ) =+ ++1632 2 2 23πR r R Rr rR r( )( )Which is the required volume.2. We show that the inequality is valid for an aggregate of values of C of which the least isC nnnnn n( ) , . =−≥− −121 2Let us frst do the easier task of proving the existence of C's which make the inequality valid. Of course this part will be redundant as soon as we improve the technique to fnd the least C.Setting x i = y i 2 where y i ≥ 0 (i = 1,…,n), we are to show, equivalently, that for some C we havey y Ciiniin= =∑ ∏≤ +1221 ...(i)Treating the right hand side of (1) as a polynomial in C, we observe that all coefcients are non-negative and that the coefcient of C n–1 is ?y 2i .Tus, y C y Ci iinnin2 2111+ ≥=−=∑ ∏But by the Cauchy–Schwarz inequality, we havey n yiiniin= =∑ ∑≤1221,So inequality (i) will be valid if we choose C = n 1 /(n–1) or larger. Tis completes the easier task.It turns out that n 1/(n-1) is only a slight overestimate of the minimum C, which we now seek. for any C for which (i) is valid, set w i = yi nC −1/, so that (i) becomeswCnw niin nniin=−−=∑ ∏≤−+ −12112111( )( )or equivalentlywC nnwniin n nniin=−−=∑ ∏≤−−+121121111( ) ...(ii)To fnd the minimum C we shall frst show that the following inequality is valid:w nwniiniin= =∑ ∏≤−+1222111 ...(iii)we shall use the weierstrass inequality( ) 1 11 1+ ≥ += =∑ ∏a ai iimimWhich holds if all a i ? 0 or if –1 < a i < 0 for all i. Without loss of generality let w 1 ,…w t ? 1 and 0 ? w t+1 ,…, w n < 1, where t ?{0, 1,…,n}. Tenwnwnwniiniinii t21212111111−+ =−+−+= = = +∏ ∏.1 1n∏ ? 11112121+−+−= = +∑ ∑wnwniitii tn= 122121nn t w t wiitii tn− ++= = +∑ ∑= 11 1221212121nw wiiti tnitii tn= = + = = +∑ ∑ ∑ ∑++ ? 121nw iin=∑(the last inequality by the Cauchy-Schwarz inequality), which proves (iii). Note that equality occurs for w 1 = … w n = 1. We conclude that (ii) is valid for any C with C n–1 n n /(n–1) n–1 ? n 2 , i.e., withCnnnn n≥−≥− −11 2,2.
MATHEMATICS TODAY | AUGUST‘1819Te minimum value C(n) We seek is then as stated at the beginning, since forx y cwnCni ii= =−=−22111.the original inequality reduces to equality.Remark : Te above shows C(2) = 1,C C ( ) . , ( ) . , 3231 1547 4341 190523= ≈ = ≈C( ) . , 5451 196334= ≈ and generally C (n) ≈ n 1 / (n–1) which approaches 1 in the limit.3. We consider polynomials p(x) with coefcients in a feld F of arbitrary characteristic and fnd as follows:(i) If char (F) = 0, (in particular, if F = R), then p(x) = a(x–3) 2 , where a is any scalar (possibly 0) in F;(ii) If char (F) = 2, then every p(x) satisfes the equation (clear);(iii) If char (F) = 2 an odd prime, l, then there are infnitely many solutions, including all p(x) = a(x – 3) 2 (x l v – x + c) with a, c ? F, and v = 0, 1, 2, … (Note that p(x) has the form a(x – 3) 2 if v = 0.To prove this, observe that if char (F) ? 2, then x – 1 and x – 3 are coprime, whence p(x) = (x – 3) 2 q(x) in F[x].Tus our equation becomes(x l v – 1) 2 (x – 3) 2 q(x) = (x – 3) 2 (x – 1) 2 q(x + 2) ( * )whence q(x) = q(x+2), as polynomials; that is, elements of F[x].Now if char (F) = 0, then ( * ) has only constant solution.(Te most elementary proof of this: without loss of generally, q(x) = x n + ax n–1 + …. Ten q(x + 2) – q(x) = 2nx n–1 + …, and this is non-zero if n ????? Another proof: ( * ) implies that q(x) is periodic, which forces equations q (x) = c to have infnitely many roots x, a contradiction).Tis establishes the assertion (i).Re: assertion (iii). Let char (F) = l andq(x) = x l v – x + c.Ten for x = 0, 1,…, l – 1, (that is for each element of the prime feld), we have q(x) = c and so q(x) = q(x+1) = q(x + 2) = …, yielding polynomials of degree greater than or equal to l which satisfy. ( * ). Tis establishes the assertion (iii).4. In this problem, we will prove that for x, y, z ? 0, 0 ? (yz + zx + xy) (x + y + z) – 2xyz ? 727(x + y + z) 3 .Tis is the homogeneous version of the o...