Exaguru

TERM - 1 10 CBSE 2021-22 MATHEMATICS (BASIC) Visit https://telegram.me/booksforcbse for more books. Full Marks Pvt Ltd (Progressive Educational Publishers) An ISO : 9001-2015 Company New Delhi-110002

Published by: An ISO : 9001-2015 Company 9, Daryaganj, New Delhi-110002 Phone: 011- 40556600 (100 Lines) Website: www.fullmarks.org E-mail: [email protected] © Publishers All rights reserved. No part of this publication may be reproduced or transmitted, in any form or by any means, without permission. Any person who does any unauthorised act in relation to this publication may be liable to criminal prosecution and civil claims for damages. Branches: • Chennai • Guwahati Marketing Offices: • Ahmedabad • Bengaluru • Bhopal • Dehradun • Hyderabad • Jaipur • Jalandhar • Kochi • Kolkata • Lucknow • Mumbai • Patna • Ranchi NEW EDITION “This book is meant for educational and learning purposes. The author of the book has taken all reasonable care to ensure that the contents of the book do not violate any existing copyright or other intellectual property rights of any person in any manner whatsoever. In the event the author has been unable to track any source and if any copyright has been inadvertently infringed, please notify the publisher in writing for corrective action.” Printed at:

Note from the Publishers Mathematics Basic-10 (Term-1) is based on the latest curriculum released by CBSE in July 2021. It will certainly prove to be a torch-bearer for those who toil hard to achieve their goal. This All-in-one Question Bank has been developed keeping in mind all the requirement of the students for Board Examinations preparations like learning, practicing, revising and assessing. Salient Features of the Book: • Each chapter is designed in ‘Topic wise’ manner where each topic is briefly explained with sufficient Examples and Exercise. • In all chapters, Topic wise exercises cover MCQs, Assertion-Reason and Case Study Questions as per the Special Scheme of Assessment suggested by CBSE vide Circular No. Acad-51/2021. Answers and sufficient hints are also provided separately at the end of exercise. • Common Errors by the students are provided to make students aware what errors are usually committed by them unknowingly. • Chapterwise Important Formulae and Quick Revision Notes have been prepared for Quick Revision. • Experts’ Opinion has been provided to suggest the students which type of questions are important for examination point of view. • 3 Sample Papers (including Latest CBSE Sample Question Paper with Marking Scheme) for mock test are given with answers and OMR Sheets for self assessment. • The book has been well prepared to build confidence among students. Suggestions for further improvement of the book, pointing out printing errors/mistakes which might have crept in spite of all efforts, will be thankfully received and incorporated in the next edition. –Publishers (iii)

E:\\Working\\EG_Mathematics-10_(Term-1)\\Open_Files\\Ch-1\\Ch-1 Mathematics 10 \\ 30-Jul-2021   Praveen Kumar  Proof-3 GLIMPSE OF A CHAPTER Reader’s Sign _______________________ Date __________ . Exercise related to each topic dealt separately and 1 Real Numbers Questions included segregated into Multiple Choice Topics covered Questions, Assertion-Reason Type Questions andE:\\Working\\EG_Mathematics-10_(Term-1)\\Open_Files\\Ch-1\\Ch-1 1. The Fundamental Theorem of Arithmetic \\ 30-Jul-2021   Praveen Kumar  Proof-3 2. Decimal Representation of Rational Numbers Reader’s Sign _______________________ Date __________ Case Study Based Questions. IntroductIon Exercise 1.1 We know that all the rational and irrational numbers together make up the collection of real numbers. A. Multiple Choice Questions (MCQs) In this chapter, we shall study divisibility of integers and expressing positive integers as a product of prime integers. Let us recall divisibility of integers and product of prime integers. Choose the correct answer from the given options: Divisibility of integers: Any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b. 1. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers then HCF Product of prime numbers: Every composite number can be expressed as a product of primes. (a, b) is: We shall also study some more topics in this chapter such as Fundamental Theorem of Arithmetic, rational and irrational numbers, etc. (a) xy (b) xy2 (c) x3y3 (d) x2y2 1. the Fundamental theorem oF arIthmetIc 2. The LCM of smallest two-digit composite number and smallest composite number is: Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a (a) 12 (b) 4 (c) 20 (d) 44 product of primes and this expression (factorisation) is unique, apart from the order in which the prime factors occur. 3. 325 can be expressed as a product of its primes as: For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. Example 1. The values of A and B in the following factor tree, respectively are (a) 52 × 7 (b) 52 × 13 (c) 5 × 132 (d) 2 × 32 × 52 A 4. The total number of factors of a prime number is [CBSE Standard 2020] B (a) 1 (b) 0 (c) 2 (d) 3 (a) 42, 21 (b) 21, 42 (c) 12, 10 (d) 10, 12 5. The HCF and the LCM of 12, 21, 15 respectively are: [CBSE Standard 2020] Solution. The factor tree with missing entries filled is given below. (a) 3, 140 (b) 12, 420 (c) 3, 420 (d) 420, 3 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: Hence, option (a) is the correct answer. (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). Example 2. When 2120 is expressed as the product of its prime factors we get (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (a) 2 × 53 × 53 (b) 23 × 5 × 53 (c) 5 × 72 × 31 (d) 52 × 7 × 33 9 (A).E:\\Working\\EG_Mathematics_Basic-10_(Term-1)\\Open_Files\\Ch-1\\Ch-1 Reader’s Sign _______________________ Date __________ \\ 01-(Acug)-20A21s s e rtPiroavneen(AKum)air s  trPuroeof-b2 ut reason (R) is false. .Each chapter is divided into topics and (d) Assertion (A) is false but reason (R) is true. explained separately. Reader’s Sign _______________________ Date __________ 1. Assertion (A): The number 6n, n being a natural number, ends with the digit 5. E:\\Working\\EG_Mathematics-10_(Term-1)\\Open_Files\\Ch-1\\Ch-1 \\ 30-Jul-2021   Praveen Kumar  Proof-3 Reason (R): The number 9n cannot end with digit 0 for any natural number n. 2. Assertion (A): If m and n are odd positive integers, then m2 + n2 is even but not divisible by 4. B. AssReertaisoonn-R(Rea):s3on× T5y×p7e +Qu7eisstaiocnosmposite numb2e.r.(a) Both assertion (A) and reason (R) are true 1. (b) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of but reason (R) is not the cCorarescteexSpltaunadtiyonBased Qausseerstitoino(nAs) I. Toofeanshsaenrtcioenth(Ae )r.eading skills of grade X students, the school nominates you and two of your friends to set up iancsleacstsiolinbrAarayn.dCT3ha6esrseetuaSdreetntuwtsdoiynsesBcetcaitosinoesnd—BQ. suecetisotnioAnasnd section B of grade X. There are 32 students 21. (d) Planes take off from five different runways B. Assertion-Reason Type Questions 1. (d) Assertion (A) is false but reason (R) is true. at 3, 4, 8, 12, 15 minutes intervals. To find 2. (b) Both assertion (A) and reason (R) are true I. The Army Day is celebrated on 15th January every year in India. The day is celebrated in the form of but reason (R) is not the correct explanation parades and other military shows in the national capital New Delhi as well as in all headquarters of when all planes will take off together again of assertion (A). army. if they take off at 7 : 30 a.m. once, we must find the LCM of 3, 4, 8, 12 and 15. Parade I: An Army contingent of 616 members is to march behind an army band of 32 members Now 3 = 3 × 1 in parade. The two groups are to march in the same number of columns. 4 = 22 8 = 23 Case Study Based Questions 12 = 22 × 3 I. 1. (c) 288 2. (b) 4 3. (a) 22 × 32 15 = 3 × 5 Parade II: An Army contingent of 1000 members is to march behind an army band of 56 members \\ LCM = 23 × 3 × 5 4. (b) composite number in parade. The two groups are to march in the same number of columns. = 40 × 3 = 120 minutes = 2 hours 5. (b) a2b2 fi Plane will took off together again after 2 II. 1. (d) 22 × 33 2. (c) 12 112.R..dW(eNT(aifas)hheu)terarm1i2tHtb4b1oiu4eCs×PtrFet16adh4o1re1aef6qd×m3cue2a2iannI2alinl1bmydeua6emm1x6pornnie(sugs(bmsb)se)btd2ue12adr2×se8ona1ft1psb1rooo×fdo1usk4ecs1tctyoifoouintsAwpi(rolcilmr)(acse2ce)3fqca×2utci8io7tro81ner×sBfoa1?sr11the hours 3. (b) 432 4. (a) 8:27:12 hrs class library, so that they can be (d) 2(d4 )× 2772 2× 111 i.e. 9:30 a.m. 22. (c) Now, number of rooms required for 5. (d) 5184 participants in Hindi = 60 =5 III. 1. (b) 23 × 71 × 111 2. (c) 2464 (a) 8 (b) 16 (c) 18 (d) 12 12 Number of rooms required for participants 3. (b) 8 4. (d) 8 Refer to Parade II Real NumbeRs 11 5. (a) 23 × 53 in English = 84 = 7 3. The LCM of 56 and 1000 is 12 IV. 1. (b) 13915 2. (c) 11 (a) 6000 (b) 7000 (c) 8000 (d) 9000 Number of rooms required for participants 3. (b) 23 4. Number 1000 can be expressed as a product of its prime factors as in Mathematics = 108 = 9 4. (a) Composite number (a) 23 × 53 (b) 22 × 54 (c) 24 × 52 (d) 23 × 54 12 5. (c) 5 × 112 × 23 5. The maximum number of columns in which army can march is fi Number of rooms needed = 21. (a) 6 (b) 10 (c) 12 (d) 8 2. Decimal RepResentation of Rational numbeRs Ans. 1. (c) 23 × 71 × 111 2. (a) 8 3. (b) 7000 4. (a) 23 × 53 5. (d) 8 We know that rational numbers and irrational numbers together are known as real numbers. II. Traffic Lights are used to control movement of traffics. They are installed at crossings and intersections of roads. Usually three different colours of lights (Red, Irrational Number Yellow and Green) are used to tell commuters what to do. A number is called irrational if it cannot be written in the form p , where p and q are integers and q ≠ 0. Examples of irrational numbers are 2 , 3 , 15 , p, etc. q The traffic lights at different road crossings change after every 48 seconds, 72 Theorem 1. Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. This theorem is based on Fundamental Theorem of Arithmetic. . Topic wise concepts areseconds and 120 seconds respectively. Rational Numbers and their Decimal Expansions 1. 120 can be expressed as a product of its prime factors as p (b) 22 × 32 × 51 (d) 23 × 32 × 51 A number is called rational if it can be expressed in the form q , where p and q are both integers and q π 0. presented to remember them(a) 23 × 31 × 51 Examples of rational numbers are 2 , 10 , 9, 3.5, 2.045, etc. (c) 22 × 31 × 52 5 15 2. The HCF of 48, 72 and 120 is Theorem 2. Let x = p , where p and q are co-primes, be a rational number whose decimal expansion terminates. (b) 16 easily. (c) 18 q (a) 12 (d) 24 Then, the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. 3. The LCM of 48, 72 and 120 is Theorem 3. Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, where (a) 432 (b) 420 (c) 720 (d) 840 q 4. If all the traffic lights change simultaneously at 7 : 30 : 00 hours, they will change again simultaneously n, m are non-negative integers. Then x has a decimal expansion which terminates. at (a) 7 : 42 : 00 hrs (b) 7 : 52 : 00 hrs (c) 7 : 36 : 00 hrs (d) 7 : 46 : 00 hrs 16 mathematics–10 16 MatheMatics–10 . Each topic is well explained with relevant example for better understanding. (iv)

Case Study Based Questions I. 1. (b) Irrational 2. (d) Non-terminating repeating 3. (d) 2 and 5 5. (d) four decimal placesReader’s Sign _______________________ Date __________ E:\\Working\\EG_Mathematics_Basic-10_(Term-1)\\Open_Files\\Ch-2\\Ch-2 4. (c) (3 + 7 )(3 − 7 )\\ 01-Aug-2021   Amit  Proof-1 II. 1. (d) All of these 2. (c) 15/4 3. (c) 11.10203040... 4. (a) One decimal place 5. (b) 21 × 51 ExpErts’ OpiniOn 2. The nFuomllboewr oinf gzertoyepseosf tohfe qpouleynsotimoinasl ya=rep(vx)eirsy important for examination point of view. So, students are (a) 1advised to revise(bt)he2m thoroughly. (c) 3 (d) 4 3. T(ah)e–z4e1,r.o–e2As,p2op,fl4tichaetpioonlysnoomf(bFi)aul–na4dr,ea–m1,e2n, t4al Theore(mc)o–f4A, r–i2th, m0,e4tic. (d) –2, 0, 2, 4 4. The ex2p.reEsxsipolnoroinf gtheraptoiolynnaol mnuiaml ibsers and their decimal expansions. (a) x4 + 20x2 + 64 (b) x4 – 20x2 + 64 QU(cI)CxK4 –R2E0xV2I–S6I4ON NO(dT)ExS4 + 20x2 – 64 5. The value of the polynomial when x = 2 is (a) 1•44The Fundamen(bt)al–1T2h8eorem of Arit(hc)m0etic: Every compo(dsi)te32number can be expressed (factorised) Ans. 1.. (b)Apaasrsabatophlareodsucet otfoppriimcess, fainnddt2h.ips(del)ax4pcreessiionn (tfhacetoreisxataiomn) ismunoi3qs.u(tea,)oa–pf4a,tr–ht2f,er2o,m4 the other in which the 4. (b)•txi4pFm–roirm2e0aexn2ifyta+tcw6itso4orspsooucsictguivrg.eeinstetgeerds5.at(ach)na0d tb,sHtCuFd(ae,nb)t×sLhCoMu(lad, b)p=uat×mb.ore ExpErts•’eOLmpeitnppiOhnbeaaspirsimoe nnumthbeer.sIef ptdoivpidiecssa.2, then p divides a, where a is a positive integer. tQhueemsttihoonrsob•uagsLheleydt.oxn=folqlpow, iwnghetyrpeeps aarnedveqryarime pcoor-tparnimt foers,exbaemas.raStoio, sntauldneuntms baerer awdhviosseeddtoecreimvisael expansion terminates. 1. Finding zerToheesno,f tthhee qpuraimdreatficacptoolyrnisoamtiiaolnanodf qtheisreolfattihone bfoetrwmee2nn5thme, wzehroeeres ann,dmthaereconeoffnic-iennetgsa. tive integers. 2. Forming a Lqthueeat dzxera=roticesqppoolfbyaneopamolriyaanltoiwomitnhiaaglliifnveiutsnmssboumemre,oszf uezrceohreosethsaaraetntgdhiveperpondr. iumcteoff azcetrEo\\o:\\3reW0is-osJur.akl-i2nt0gi\\2oE1Gn _M oa fthPeqrmavaietsiecnso-K10uf_m(tTaherr em -1f)Po\\rOorpomef-n3_F2ilens5\\Cmh,-1\\wChh-1ere 3. n, m Finding •all Reader’s Sign _______________________ Date __________ • If a, b•arLaerezeetrnxooe=ns -onqpfetghbeaetqiavuaerdairtnaittoieIcngMapeloPrlnsOy.unRTmoThmbeAienaNrl,TsaxuxFh2cOha+sRtbhaMxad+tUetcLch,AietmhEpeanrlimexepfaancstioorniswathioicnhotfeqrmisinnaotet so.f the form 2n5m, where n, .beQ seusfioeterneceeeffxeoacfmttihvsee. cfHoharavpaeteqtruh. ieckcoremvpisleiotenCOMMON ERRORSE\\:\\3W0-oJurkl-i(2ni0g)\\2E1G a_M +a thbPemrma=vaeatiecrns−e-aK1b0un_m(Toaerr nm- -1n)P\\erOogpoefa-n3_tFiivlees\\Cihn-1t\\Cehg-1ers. Then, x has aanbon=-tcerminating repeating (recurring) decimal expansion. (ii) a ReaderC’s Saigsn e___S__t__u__d__y__B___a__s__e_d__ QDautee__s_t__i_o__n__s • If a, b are zeroes of the quadratic polynomial p (x), then p (x) = x2 – (a + b)Ix. +1.ab(b) Irrational 2. (d) Non-terminating repeating 3. (d) 2 and 5 Errors ( )( )4. (c) C3 +orr7ect3io−ns 7 5. (d) four decimal places (i) Incorrectly using theQreUsIuCltKfoRrEthVrIeSeInOuNmNbeOrTsES (i) The corIrIe.c1t.re(sdu)lAtslal roef:these 2. (c) 15/4 3. (c) 11.10203040... ( )( )C••• IaAAT.sh41pqee..ouhSla((yibdcgtnr))uhoaedmtIsipr3yctri(aaxp+pBlHt)ooipoaw=lC(yn7sxenFaa)reonliodx(m3npfn−iQo,+xanqliuaen,in7env–rps1a)(xxrtxn×iiw)–ao1ibLinst+lheCsc.a.52xrM.le..l+iaes((lddd(aapc))2ntox,hfN2eaeqofl+o,gfduinerearcb)g-1idtexrr=eeane+rcietmcpsiaomeii0×fsnxawtaplohqthrfepieen×ltsprahgseocreiorlaeyefnnsonp,iroaenmman–xt1iiaan,oxl.gf.LH2p.t,+h(CCaxe0M)Fb.fa3xo(r.pr(e+mp,r(caF,edq,adqo),lw4v,lrnl2E.ihor)su)xe=aw(merapn=edibE)dnLreatgCOH5ort,ssMnbtCrwy’,eeFpOi(cvdt(ehpipeassp,ircaene,qinpomiqtOp)≠rqfhe)naq⋅r0aeql⋅Lrl.muLHpCHelCCatsMChcMtFeoiF(o((rq(qnop,p,su,r,rgq)aq)5h,r⋅,⋅.elLryHr()bC.)vC)eMFr2y(1( × 51 for examination point of view. So, students are p, r) important p, r) IIn. u1m. b(edr)s(wAiiil)tlhMoafiπtshu0en.sdeerstanding t2h.a(tct)h1e5p/4rime factors of (ii) If an3y.on(ce1)o. 1fA12p.p1ol0ric25a0to3ioc0nc4su0ro..sf.,Fthunendaamlseonittahl TasheteorrmeminoaftiAnrgithmetic. • Le4t.p((xa)) bOe naedpdeoenlyconimmomianliaapltloianrcexshaondulk5db.he(baa)vney21bre×oatl5hn1 u2mabnedr,5thfeonrthe valudeeocbimtaianleed2xb.pyEarxnepsplioloarcnini.nggraxtbioynkalinnumbers and their decimal expansions. • p(x)EixspcEarltlesd’ttOhepivdnaeilOcuniemoaf lpe(xx)paatnxs=ioknatnodbisedteenrmoteindabtyinpg(.k). QUICK REVISION NOTES can be expressed (factorised) • FaAAdovlprleioosalwleyndninu(otigmmoiibi)trayeelDprvmekeissaciesyiodshtfiahanieqvdgemutoenintosbhctezoiooearrrnorozesue,crgoaotnhlryeoelyfoa.varbemporoyoulrytienmaothpmanoniuratmolanpnbe(txez)rfeoirforqppee(skx.)a(=mii0ii)n. aFtiirosnt mpoaik•netToqhpfevcFioeu-wnp.driSammoe,esantntuaddlteThnehtnseodarerecemideofwAhreitthhemr eittiics: Every composite number • A1.pAolpypnloicmatiiaoltnops(xho)afovFfeudnetdgearremmeeinnhtaaatlsiTnahgtmeoorrsetmnnozonefr-Aoteerisrt.hmminetaicti.ng terminatingasoar pnroond-utectrmofinparitminegs,eaxnpdanthsiisonex. pression (factorisation) is unique, apart from the other in which the • T2h.eEzxeprlooerisngoefrxaaptiaopnosalyilonnnoumjmuisbatlebrpys(xlao)noadkretihnpegrireactditeshceelimydeathnl eeoxmxp-icanonaostroiodrninsa.tes of the pointsprwimheerefatchteorgsraopchcuorf. • •Iyf=Tapha(enxd2)Fib2untnaerdreswaetmchittehseznothetureatolxem-TsaahoxkfiesMiot.nhrgeeamqptuhaaodenfrMdaAtaqiQrcticUtpichooIm-slCpy–eKnr1itomimRc0:eEia.EVlvaIeSxr2Iy+OcbNoxmN+pOco,TstihEteeSnnaum+bber=•••c−aLFLanboeettr,bapexanbbe=yxe=ptaqpwracepo,sr.wsipmeohdseei(rtnfeiauvpcmetoaibnnreitdrse.egqIdef)rapsreadicavoni-ddperbsi,maH2e,Cst,hFbe(enaa,pbrda)itv×ioidLneaCslMan,u(wma,hbbeer)re=waahios×saebp.doesciitmivaelienxtepgaenrs.ion terminates. as a product of primes, and this expression (factorisation) is unique, apartTfrhoemn, tthhee optrhimereinfawcthoircihsathioen of q is of the form 2n5m, where n, m are non-negative integers. prime factors occur. LePtoxl=ynop mbiaelas 29 • For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.• q rational number, such that the prime factorisation of q is of the form 2n5m, where n, m • Let p be a prime number. If p divides a2, then p divides a, where a is a posaitrievneoinnt-engeegra.tive integers. Then, x has a decimal expansion which terminates. p • Let x = q , where p and q are co-primes, be a rational number whose d•ecLiemtaxl =expqanbseioanratetiromnianlanteusm. ber such that the prime factorisation of q is not of the form 2n5m, where n, Then, the prime factorisation of q is of the form 2n5m, where n, m are non-nmegaaretivneonin-nteeggeartsiv. e integers. Then, x has a non-terminating repeating (recurring) decimal expansion. • Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, where n, m q COMMON ERRORS are non-negative integers. Then, x has a decimal expansion which terminates. • Let x = p Errors Corrections q be a rational number such that the prime factorisation of q is n(io)tIonfcothrreecfotlrymus2inn5gmt,hwe rheesrueltnf,or three numbers (i) The correct results are: m are non-negative integers. Then, x has a non-terminating repeating (recurHrinCgF) (dpe,cqim, ra)l ×exLpCanMsio(pn,. q, r) = p × q × r LCM (p, q, r) = pqr HCF( p, q, r) HCF( p, q) ⋅ HCF(q, r) ⋅ HCF( p, r) COMMON ERRORS HCF (p, q, r) = pqr LCM( p, q, r) Errors Corrections LCM( p, q) ⋅ LCM(q, r) ⋅ LCM( p, r) (i) Incorrectly using the result for three numbers (i) The correct results are(i:i) Misunderstanding that the prime factors of .(ii)C Ifoamny omneoofn2 eorr5roocrcsursh, tahevnealsbo eit heans tetramgingatiengd HCF (p, q, r) × LCM (p, q, r) = p × q × r dpeqnroHmCinFa(topr, sqh,oru)ld have both 2 and 5 for todeccimlealaexrpcanosionnf.usions with cautions LCM (p, q, r) = HCF( p, tqh)e⋅dHeCciFm(aql, erx) p⋅ aHnCsiFo(npt,orb) e terminating. HCF (p, q, r) = (iii) pDqercLidCinMg(ipn,cqo,rrre)ctly about a number p (iii)laeFniarsdrt nmiaankegnqp.s wers for productive LCM( p, q) ⋅ LCM(q, r) ⋅ LCM( p, r) q co-prime and then decide whether it is (ii) Misunderstanding that the prime factors of (ii) dIfeacnimy aolneexopfa2nsoiro5n.occurswte,oxtihpthehanaonvusaietlosmnotejaiurtkmshitnaibgnsyatpeltoiranmongkidinnoqagrticannotgot-hpner-idtmeernem.ominiantaitnogr terminating or non-terminating expansion. denominator should have both 2 and 5 for the decimal expansion to be terminating. (iii) Deciding incorrectly about a number p (iii) First make p co-pr2im2e and then decidMe awthheethMear titiciss–10 q q to have terminating or non-terminating terminating or non-terminating expansion. expansion just by looking at the denominator without making p and q co-prime. 22 MatheMatics–10 (v)

SYLLABUS Units FIRST TERM Marks I. Number Systems 06 II. Algebra Unit Name 10 III. Coordinate Geometry 06 IV. Geometry Total 06 V. Trigonometry 05 VI. Mensuration 04 03 VII. Statistics & Probability 40 UNIT I: NUMBER SYSTEMS 1. Real Number Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals. UNIT II: ALGEBRA 2. Polynomials Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials only. 3. Pair of Linear Equations in Two Variables Pair of linear equations in two variables and graphical method of their solution, consistency/ inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution and by elimination. Simple situational problems. Simple problems on equations reducible to linear equations. UNIT III: COORDINATE GEOMETRY 4. Lines (In Two-dimensions) Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division). (vi)

UNIT IV: GEOMETRY 5. Triangles Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. 4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar. 6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other. 7. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. 8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. 9. (Prove) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to the first side is a right angle. UNIT V: TRIGONOMETRY 6. Introduction to Trigonometry Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined). Values of the trigonometric ratios of 30°, 45° and 60°.. Relationships between the ratios. Trigonometric Identities Proof and applications of the identity sin2 A + cos2 A = 1. Only simple identities to be given. UNIT VI: MENSURATION 7. Areas Related to Circles Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60° and 90° only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.) UNIT VII: STATISTICS AND PROBABILITY 8. Probability Classical definition of probability. Simple problems on finding the probability of an event. (vii)

CONTENTS 1. Real Numbers ........................................................................................................................ 9 2. Polynomials ........................................................................................................................... 18 3. Pair of Linear Equations in Two Variables ........................................................................... 31 4. Coordinate Geometry ............................................................................................................ 62 5. Triangles ................................................................................................................................ 76 6. Introduction to Trigonometry ................................................................................................ 91 7. Areas Related to Circles ........................................................................................................ 102 8. Probability ............................................................................................................................. 117 • Sample Paper 1 [Issued by CBSE on 2nd September, 2021] ................................................ 125 Marking Scheme ................................................................................................................... 132 • Sample Paper 2 ..................................................................................................................... 136 • Sample Paper 3 ..................................................................................................................... 142 • OMR Sheets (viii)

1 Real Numbers Topics Covered 1. The Fundamental Theorem of Arithmetic 2. Decimal Representation of Rational Numbers Introduction We know that all the rational and irrational numbers together make up the collection of real numbers. In this chapter, we shall study divisibility of integers and expressing positive integers as a product of prime integers. Let us recall divisibility of integers and product of prime integers. Divisibility of integers: Any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b. Product of prime numbers: Every composite number can be expressed as a product of primes. We shall also study some more topics in this chapter such as Fundamental Theorem of Arithmetic, rational and irrational numbers, etc. 1. The Fundamental Theorem of Arithmetic Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes and this expression (factorisation) is unique, apart from the order in which the prime factors occur. For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. Example 1. The values of A and B in the following factor tree, respectively are A B (a) 42, 21 (b) 21, 42 (c) 12, 10 (d) 10, 12 Solution. The factor tree with missing entries filled is given below. Hence, option (a) is the correct answer. Example 2. When 2120 is expressed as the product of its prime factors we get (a) 2 × 53 × 53 (b) 23 × 5 × 53 (c) 5 × 72 × 31 (d) 52 × 7 × 33 9

Solution. Using the factor tree for prime factorisation, we get: \\ 2120 = 2 × 2 × 2 × 5 × 53 = 23 × 5 × 53 Hence, option (b) is the correct answer. Example 3. [HCF × LCM] for the numbers 100 and 190 is (a) 190 (b) 1900 (c) 19000 (d) None of these Solution. HCF × LCM = Product of two numbers = 100 × 190 = 19,000 Hence, option (c) is the correct answer. Example 4. The HCF of the smallest composite number and the smallest prime number is (a) 1 (b) 2 (c) 3 (d) 5 Solution. The smallest composite number is 4 and the smallest prime number is 2. The prime factorisation of 4 = 2 × 2 = 22 and the prime factorisation of 2 = 21 Now, the HCF of 2 and 4 is the product of smallest power of each common prime factor in the numbers. HCF (2, 4) = 21 = 2 Hence, option (b) is the correct answer. Example 5. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. The minimum distance each should walk so that each can cover the same distance in complete steps is (a) 1260 cm (b) 1920 cm (c) 2242 cm (d) 2520 cm Solution. Step measures of three persons are 40 cm, 42 cm and 45 cm. The minimum distance each should walk so that each can cover the same distance in complete steps is the LCM of 40 cm, 42 cm and 45 cm. Prime factorisation of 40, 42 and 45 gives 40 = 23 × 5, 42 = 2 × 3 × 7, 45 = 32 × 5 LCM (40, 42, 45) = Product of the greatest power of each prime factor involved in the numbers = 23 × 32 × 5 × 7 = 8 × 9 × 35 = 72 × 35 = 2520 cm. Hence, option (d) is the correct answer. Example 6. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then the other number is (a) 20 (b) 28 (c) 60 (d) 80 Solution. According to the question, LCM + HCF = 600 Since LCM = 14 × HCF \\ 14 × HCF + HCF = 600  ⇒  15 × HCF = 600  ⇒  HCF = 600 ÷ 15 = 40 \\ LCM = 600 – HCF = 600 – 40 = 560 10  Mathematics–10

We know that HCF (a, b) × LCM (a, b) = a × b ⇒ Other number = HCF × LCM = 40 × 560 = 80 Given number 280 Hence, option (d) is the correct answer. Example 7. Four bells toll at an interval of 8, 12, 15 and 18 seconds respectively. All the four begin to toll together. The number of times they toll together in one hour excluding the one at the start will be (a) 5 (b) 8 (c) 10 (d) 12 Solution. Four bells toll at an interval of 8, 12, 15 and 18 seconds. All the four bells begin to toll together. To find how many times will they toll together in one hour excluding the one at the start, we first find the LCM of the numbers 8, 12, 15 and 18. Now, prime factorisation of the given numbers are: 8 = 2 × 2 × 2 = 23 12 = 2 × 2 × 3 = 22 × 31 15 = 3 × 5 = 31 × 51 18 = 2 × 3 × 3 = 21 × 32 LCM (8, 12, 15 and 18) = 23 × 32 × 51 = 8 × 9 × 5 = 360 sec = 6 min \\  Four bells toll together in one hour = 60 ÷ 6 = 10 times. Hence, option (c) is the correct answer. Exercise 1.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. The LCM of smallest two-digit composite number and smallest composite number is: (a) 12 (b) 4 (c) 20 (d) 44 2. 325 can be expressed as a product of its primes as: (a) 52 × 7 (b) 52 × 13 (c) 5 × 132 (d) 2 × 32 × 52 3. HCF (a, b) × LCM (a, b) is equal to (a) a + b (b) a – b (c) a × b (d) a ÷ b 4. The values of x and y in the given figure respectively are (a) x = 84, y = 21 (b) x = 21, y = 84 (c) x = 42, y = 24 (d) x = 24, y = 42 5. If a and b are co-prime, then a2 and b2 are (a) primes (b) composites (c) co-primes (d) None of these 6. When 429 is expressed as a product of its prime factors, we get (a) 2 × 5 × 29 (b) 33 × 13 × 1 (c) 3 × 11 × 9 (d) 3 × 11 × 13 7. The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, the other number is (a) 31 (b) 71 (c) 61 (d) 91 8. When 156 is expressed as the product of primes, we get (a) 22 × 3 × 13 (b) 22 × 3 × 11 (c) 2 × 32 × 13 (d) 2 × 32 × 11 Real Numbers  11

9. The HCF and LCM of 404 and 96 respectively are (a) 2, 9696 (b) 4, 9696 (c) 8, 3636 (d) 10, 2020 10. The LCM of 150 and 200 is (a) 320 (b) 400 (c) 550 (d) 600 11. 3 bells ring at an interval of 4, 7 and 14 minutes. All three bells rang at 6 am. When the three bells will ring together next? (a) 6:20 am (b) 6:24 am (c) 6:28 am (d) 6:30 am 12. The LCM and the HCF of 15, 18, 45 respectively are (a) 3, 30 (b) 4, 40 (c) 5, 50 (d) 3, 90 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): The number 6n, n being a natural number, ends with the digit 5. Reason (R): The number 9n cannot end with digit 0 for any natural number n. 2. Assertion (A): If m and n are odd positive integers, then m2 + n2 is even but not divisible by 4. Reason (R): 3 × 5 × 7 + 7 is a composite number. Answers and Hints A. Multiple Choice Questions (MCQs) 10. (d) Given numbers are 150 and 200. 1. (c) 20 2. (b) 52 × 13 \\ 150 = 2 × 3 × 5 × 5 2 150 = 2 × 3 × 52 3 75 3. (c) a × b 4. (b) x = 21, y = 84 5. (c) co-primes 5 25 6. (d) 429 = 3 × 11 × 13  3 429 5 11 143 \\ 200 = 2 × 2 × 2 × 5 × 5 2 200 13 13 = 23 × 52 2 100 1 2 50 7. (d) F or any two positive integers a and b, 5 25 a × b = HCF (a, b) × LCM (a, b) 5 fi 26 × b = 13 × 182 LCM of 150 and 200 = 23 × 3 × 52 = 8 × 3 × 25 = 600 fi b = 13 × 182 = 91 11. (c) 4 = 2 × 2 26 7 = 7 × 1 14 = 2 × 7 Hence, the other number is 91. LCM = 2 × 2 × 7 = 28 8. (c) 156 = 22 × 3 × 13 The three bells will ring together again at 6 : 28 am 9. (b) 404 = 2 × 2 × 101 = 22 × 101 12. (d) 3, 90 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3 B. Assertion-Reason Type Questions 1. (d) A ssertion (A) is false but reason (R) is true. \\ HCF of 404 and 96 = 22 2. (b) Both assertion (A) and reason (R) are true = 4 but reason (R) is not the correct explanation LCM of 404 and 96 = 101 × 25 × 3 of assertion (A). = 9696 12  Mathematics–10

2. Decimal Representation of Rational Numbers We know that rational numbers and irrational numbers together are known as real numbers. Irrational Number A number is called irrational if it cannot be written in the form p , where p and q are integers and q ≠ 0. Examples of irrational numbers are 2 , 3 , 15 , p, etc. q Theorem 1. Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. This theorem is based on Fundamental Theorem of Arithmetic. Rational Numbers and their Decimal Expansions p A number is called rational if it can be expressed in the form q , where p and q are both integers and q π 0. 2 10 Examples of rational numbers are , , 9, 3.5, 2.045, etc. 5 15 Theorem 2. Let x = p , where p and q are co-primes, be a rational number whose decimal expansion terminates. q Then, the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. Theorem 3. Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, where q n, m are non-negative integers. Then x has a decimal expansion which terminates. Theorem 4. Let x = p be a rational number such that the prime factorisation of q is not of the form 2n5m, q where n, m are non-negative integers. Then, x has a non-terminating repeating decimal expansion. Example 1. The decimal expansions (without actual division) and its nature (terminating or non- terminating) of 987 will be 10500 __ (b) 0.094, terminating (a) 0.094, non-terminating (c) 0.094, non-terminating (d) 0.049, terminating Solution. We have 987 47 × 3 × 7 47 10500 = 500 × 3 × 7 = 500 Here, the prime factorisation of q = 500 is 2 × 2 × 5 × 5 × 5 = 22 × 53 which is of the form 2n5m, where n, m are non-negative integers. Hence, 987 will have a terminating decimal expansion. 10500 Now, 987 47 = 47 = 47 × 2 = 94 = 0.094 10500 = 500 22 × 53 (2 × 5)3 1000 987 Hence, decimal expansion of 10500 terminates after 3 places. Hence, option (b) is the correct answer. Real Numbers  13

Example 2. The decimal expansion of the rational number 43 will terminate after how many places of decimals? 2453 (a) 2 places (b) 3 places (c) 4 places (d) 5 places fSoorlmuti2onn5.m,Hwerheerweenh, amveaarerantoionn-anlegnautmivbeerinotfegtheers,fosromdeqpcimanadl the prime factorisation of q is of the expansion of given rational number terminates. Now, 43 = 43 × 5 = 215 = 0.0215 2453 2454 104 So, given rational number will terminate after 4 decimal places. Hence, option (c) is the correct answer. Example 3. Rational number between 2 and 3 is [Delhi 2019] (d) All of these (a) 1.5 (b) 1.6 (c) 1.7 Solution. We have 2 = 1.4142135.... and 3 = 1.7320508... Since every terminating decimal or repeating decimal represents a rational number. So, 1.666666... is a rational number between 2 and 3 . Also 1.5, 1.6, 1.7 are rational numbers between 2 and 3 . Hence, option (d) is the correct answer. Exercise 1.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. A rational number can be expressed as a terminating decimal if the denominator has factors (a) 2, 3 or 5 p (b) 2 or 3 (c) 3 or 5 (d) 2 or 5 q 2. Rational number , q ≠ 0, will be terminating decimal if the prime factorisation of q is of the form (m and n are non-negative integers) (a) 2m × 3n (b) 2m × 5n (c) 3m × 5n (d) 3m × 7n 3. Which of the following is the decimal expansion of an irrational number? (a) 4.561 (b) 0.12 (c) 5.010010001… (d) 6.03 4. 2.35 is (a) an integer (b) a rational number (c) an irrational number (d) a natural number 5. A rational and an irrational number lying between 0.25 and 0.32 are respectively. (a) 0.30, 0.3010203040... (b) 0.20, 0.2010203040... (c) 0.33, 0.3510203040... (d) None of these 6. The decimal expansion (without actual division) and its nature (terminating or non-terminating) of 17 will be 8 (a) Terminating after 2 places (b) Non-terminating but repeating (c) Non-terminating but non-repeating (d) Terminating after 3 places 7. The decimal expansion (without actual division) and its nature (terminating or non-terminating) of 15 will be 1600 (a) Terminating after 6 places (b) Non-terminating but repeating (c) Non-terminating and non-repeating (d) Terminating after 2 places 14  Mathematics–10

8. The decimal expansion (without actual division) and its nature (terminating or non-terminating) of 64 will be 455 (a) Terminating after 2 places (b) Non-terminating but repeating (c) Non-terminating but non-repeating (d) Terminating after 3 places 3 9. The 2n5m (where n and m are non-negative integers) from of denominator of and its decimal 5 expansion respectively are (a) 20 × 51, 0.6 (b) 21 × 50, 0.5 (c) 21 × 51, 0.6 (d) 22 × 50, 0.8 13 10. The 2n5m (where n and m are non-negative integers) form of denominator of and its decimal 80 expansion respectively are (a) 23 × 52, 0.1248 (b) 22 × 53, 0.1698 (c) 24 × 51, 0.1625 (d) None of these B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): The sum or difference of a rational number and an irrational number is irrational. Reason (R): Negative of an irrational number is rational. 2. Assertion (A): 5 + 3 is an irrational number. Reason (R): The sum or difference of a rational and an irrational number is always irrational. Answers and Hints A. Multiple Choice Questions (MCQs) 15 = 3×5 = 3 7. (a) 26 × 52 26 × 51 1. (d) 2 or 5 2. (b) 2m × 5n 1600 3. (c) 5.010010001… 4. (b) arationalnumber Denominator 26 × 51 is of the form 2n5m. 5. (a) Rational number = 0.30 Irrational number = 0.3010203040… Now, 3 55 = 3 × 55 = 3 × 55 26 × 51 × 55 26 × 56 106 Or any other correct rational and irrational number Hence 15 has a terminating decimal 17 = 17 17 1600 6. (d) 23 = 23 × 50 expansion with 6 decimal places. 8 Clearly, 17 and 8 are co-prime. 64 26 8. (b) 455 = 5 × 7 × 13 17 Clearly, 64 and 455 are co-prime. \\  is in the simplest form. Denominator 8 = 23 × 50 is of the form 2n 5m. 64 \\  455 is in its simplest form. 17 = 17 ×125 Also, Denominator 455 = 5 × 7 × 13 is not of the 8 8 ×125 form 2n5m. 2125 Hence 64 has a non-terminating repeating = = 2.125 455 1000 decimal expansion. 17 9. (a) 20 × 51, 0.6 10. (c) 24 × 51, 0.1625 Hence, has a terminating decimal 8 expansion with 3 decimal places. Real Numbers  15

B. Assertion-Reason Type Questions 2. (a) Both assertion (A) and reason (R) are true 1. (b) B oth assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A) but reason (R) is not the correct explanation of assertion (A). Case Study Based Questions I. The Army Day is celebrated on 15th January every year in India. The day is celebrated in the form of parades and other military shows in the national capital New Delhi as well as in all headquarters of army. Parade I: An Army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns. Parade II: An Army contingent of 1000 members is to march behind an army band of 56 members in parade. The two groups are to march in the same number of columns. Refer to Parade I 1. Number 616 can be expressed as a product of its prime factors as (a) 21 × 141 × 221 (b) 22 × 111 × 141 (c) 23 × 71 × 111 (d) 24 × 72 × 111 2. The HCF of 32 and 616 is (a) 8 (b) 16 (c) 18 (d) 12 Refer to Parade II 3. The LCM of 56 and 1000 is (a) 6000 (b) 7000 (c) 8000 (d) 9000 4. Number 1000 can be expressed as a product of its prime factors as (a) 23 × 53 (b) 22 × 54 (c) 24 × 52 (d) 23 × 54 5. The maximum number of columns in which army can march is (a) 6 (b) 10 (c) 12 (d) 8 Ans. 1. (c) 23 × 71 × 111 2. (a) 8 3. (b) 7000 4. (a) 23 × 53 5. (d) 8 II. Traffic Lights are used to control movement of traffics. They are installed at crossings and intersections of roads. Usually three different colours of lights (Red, Yellow and Green) are used to tell commuters what to do. The traffic lights at different road crossings change after every 48 seconds, 72 seconds and 120 seconds respectively. 1. 120 can be expressed as a product of its prime factors as (a) 23 × 31 × 51 (b) 22 × 32 × 51 (c) 22 × 31 × 52 (d) 23 × 32 × 51 2. The HCF of 48, 72 and 120 is (a) 12 (b) 16 (c) 18 (d) 24 3. The LCM of 48, 72 and 120 is (a) 432 (b) 420 (c) 720 (d) 840 4. If all the traffic lights change simultaneously at 7 : 30 : 00 hours, they will change again simultaneously at (a) 7 : 42 : 00 hrs (b) 7 : 52 : 00 hrs (c) 7 : 36 : 00 hrs (d) 7 : 46 : 00 hrs 16  Mathematics–10

5. The [HCF × LCM] for the numbers 48, 72 and 120 is (a) 17480 (b) 17280 (c) 12280 (d) 18280 3. (c) 720 Ans. 1. (a) 23 × 31 × 51 2. (d) 24 4. (a) 7 : 42 : 00 hrs 5. (b) 17280 Experts’ Opinion Following types of questions are very important for examination point of view. So, students are advised to revise them thoroughly. 1. Applications of Fundamental Theorem of Arithmetic. 2. Exploring rational numbers and their decimal expansions. QUICK REVISION NOTES •• The Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this expression (factorisation) is unique, apart from the other in which the prime factors occur. •• For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. •• Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. •• Let x = p , where p and q are co-primes, be a rational number whose decimal expansion terminates. q Then, the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. •• Let x = p be a rational number, such that the prime factorisation of q is of the form 2n5m, where n, m q are non-negative integers. Then, x has a decimal expansion which terminates. •• Let x = p be a rational number such that the prime factorisation of q is not of the form 2n5m, where n, q m are non-negative integers. Then, x has a non-terminating repeating (recurring) decimal expansion. COMMON ERRORS Errors Corrections (i) Incorrectly using the result for three numbers (i) The correct results are: HCF (p, q, r) × LCM (p, q, r) = p × q × r LCM (p, q, r) = pqr HCF( p, q, r) HCF( p, q) ⋅ HCF(q, r) ⋅ HCF( p, r) HCF (p, q, r) = pqr LCM( p, q, r) LCM( p, q) ⋅ LCM(q, r) ⋅ LCM( p, r) (ii) Misunderstanding that the prime factors of (ii) If any one of 2 or 5 occurs, then also it has terminating denominator should have both 2 and 5 for decimal expansion. the decimal expansion to be terminating. (iii) Deciding incorrectly about a number p (iii) First make p co-prime and then decide whether it is qq to have terminating or non-terminating terminating or non-terminating expansion. expansion just by looking at the denominator without making p and q co-prime. Real Numbers  17

2 Polynomials Topics Covered 1. Geometrical Meaning of the Zeroes of a Polynomial 2. Relationship between Zeroes and Coefficients of a Polynomial Introduction Polynomial: An expression of the form p(x) = anxn + an – 1xn – 1 + ... + a1x + a0, where n is a non-negative integer, a1, a2, ..., an are constants (real numbers) and an π 0, is called a polynomial in x of degree n. Degree: If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x). Type of Polynomials: Polynomials of degree 1, 2 and 3 are called linear, quadratic and cubic polynomials respectively. 1. Geometrical Meaning of the Zeroes of a Polynomial Zeroes of a Polynomial: A real number ‘a’ is said to be a zeroes of a polynomial p(x) if p(a) = 0. Note: A polynomial may have no zero, one or more than one zeroes. If p(a) ≠ 0, then ‘a’ is not a zero of the polynomial p(x). Example 1. The value of k for which (–4) is a zero of the polynomial x2 – x – (2k + 2) is (a) 2 (b) –6 (c) 9 (d) 8 Solution. Given that –4 is a zero of the polynomial p(x) = x2 – x – (2k + 2). \\ p(–4) = 0  ⇒ (–4)2 + 4 – 2k – 2 = 0 ⇒ 16 + 4 – 2k – 2 = 0  ⇒  18 = 2k  ⇒  k = 18 ÷ 2 = 9 Thus, for k = 9, –4 is a zero of the given polynomial. Hence, option (c) is the correct answer. Example 2. The solution of x2 + 6x + 9 = 0 is (a) –1 (b) 3 (c) –3 (d) 1 Solution. Let p(x) = x2 + 6x + 9 \\ p(– 3) = (– 3)2 + 6 × (– 3) + 9 = 9 + (– 18) + 9 = 18 – 18 = 0 \\ x = –3 is a solution of x2 + 6x + 9 = 0. Hence, option (c) is the correct answer. Example 3. The number of polynomials having zeroes as –2 and 5 is (a) 1 (b) 2 (c) 10 (d) infinite Solution. As –2 and 5 are zeroes of the polynomial p(x), so (x + 2) and (x – 5) are the factors of p(x). \\ p(x) = k(x + 2) (x – 5)  ⇒  p(x) = k(x2 – 3x – 10), where k is a real number. Now for different values of k, we get different polynomials. T hus, the number of polynomials having zeroes as –2 and 5 is infinite. Hence, option (d) is the correct answer. 18

Example 4. The graph of y = f (x) is given in figure (i), for some polynomial Y f (x). The number of zeroes of f (x) is (a) 1 (b) 2 (c) 3 (d) many X′ O X Solution. Polynomial f (x) has 3 zeroes as graph intersects the x-axis at three distinct points. Y′ Hence, option (c) is the correct answer. Figure (i) Example 5. The graph of y = f (x) is given in figure (ii). How many zeroes Y are there of f (x)? (a) 0 (b) 1 (c) 2 (d) many X′ X Solution. We observe that the graph of y = f (x) intersects x-axis in one point only. Therefore, the number of zeroes of f (x) is one. Hence, option (b) is the correct answer. Y′ Example 6. The graph of y = f (x) is given in figure (iii). How many zeroes Figure (ii) Y are there of f (x)? (a) 1 (b) 2 (c) 3 (d) None X′ X Solution. The given graph y = f (x) does not intersect x-axis. So, it has no zeroes. Hence, option (d) is the correct answer. Y′ Figure (iii) Exercise 2.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. Which of the following is not the graph of a quadratic polynomial? (a) (b) (c) (d) 2. The value of p, for which (–4) is a zero of the polynomial x2 – 2x – (7p + 3) is (a) 0 (b) 2 (c) 3 (d) None of these 3. If 2 is a zero of the polynomial ax2 – 2x, then the value of ‘a’ is (a) 3 (b) 1 (c) 2 (d) 5 4. What number should be added to the polynomial x2 – 5x + 4, so that 3 is the zero of the polynomial? (a) 0 (b) 1 (c) 2 (d) 3 5. For what value of k, (– 4) is a zero of p(x) = x2 – x – (2k – 2)? (a) 8 (b) 11 (c) 13 (d) 15 6. A teacher asked 10 of his students to write a polynomial in one variable on a paper and then to handover the paper. The following were the answers given by the students: 2x + 3, 3x2 + 7x + 2, 4x3 + 3x2 + 2, x3 + 3x + 7, 7x + 7 , 5x3 – 7x + 2, 2x2 + 3 – 5 , 5x – 1 ax3 + bx2 + cx + d, x + 1 . , x x2 Polynomials  19

How many of the above ten, are not polynomials? (a) 1 (b) 2 (c) 3 (d) 4 (d) 3 7. How many of the above ten (in question 12), are quadratic polynomials? (a) 0 (b) 1 (c) 2 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): x2 + 4x + 5 has two zeroes. Reason (R): A quadratic polynomial can have at the most two zeroes. 2. Assertion (A): A quadratic polynomial whose zeroes are 5 + 2 and 5 – 2 is x2 – 10 x + 23. Reason (R): If a and b are zeroes of the quadratic polynomial p(x), then p(x) = x2 – (a + b) x + ab. Answers and Hints A. Multiple Choice Questions (MCQs) fi 20 – 2k + 2 0 fi –2k = –22  fi  k = 11 1. (d) 2. (c) 3 6. (c) 3 x3 + 3x + 7, 2x2 + 3 – 5 and x + 1 3. (b) 1 f(x) = x2 – 5x + 4 f(x) = 32 – 5 × 3 + 4 = –2 xx 4. (c) Let 7. (b) 1 Then 3x2 + 7x + 2 For f(x) = 0, 2 must be added to B. Assertion-Reason Type Questions f(x) 1. (d) Assertion (A) is false but reason (R) is true. 5. (b) If – 4 is a zero of p(x) = x2 – x – (2k – 2) 2. (a) Both assertion (A) and reason (R) are true then p(–4) = 0 and reason (R) is the correct explanation of assertion (A). fi 16 + 4 – (2k – 2) = 0 2. Relationship between Zeroes and Coefficients of a Polynomial • If a, b are zeroes of a quadratic polynomial p(x) = ax2 + bx + c, where a ≠ 0, then (i) Sum of zeroes = a + b = −b   ⇒  a+b= − (Coefficient of x) a Coefficient of x2 (ii) Product of zeroes = ab = c   ⇒  ab = Constant term a Coefficient of x2 • If a, b are zeroes (or roots) of a quadratic polynomial p(x), then p(x) = x2 – (a + b) x + ab ⇒  p(x) = x2 – (Sum of zeroes) x + (Product of zeroes) Example 1. A quadratic equation x2 – 2x – 8 is given. The zeroes of it are (a) –2 and 4 (b) 3 and 5 (c) 1 and 6 (d) None of these Solution. We have p(x) = x2 – 2x – 8 x2 – 2x – 8 = x2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4) (x + 2) Now, So, p(x) = 0 fi (x – 4) (x + 2) = 0 fi x = 4, x = –2 are the zeroes of p(x) Hence, option (a) is the correct answer. 20  Mathematics–10

Example 2. The sum and product of the zeroes of the quadratic equation given in example 1 are respectively (a) 2, 4 (b) 5, –8 (c) 6, 8 (d) 2, –8 Solution. sum of its zeroes = 4 + (–2) = 4 – 2 = 2 = −(−2) = −(Coefficient of x) 1 Coefficient of x2 and product of its zeroes = 4 × (–2) = –8 = −8 = Constant term 1 Coefficient of x2 Hence, option (d) is the correct answer. Example 3. The zeroes of the quadratic equation 4s2 – 4s + 1 are 11 11 11 11 (a) 2 , 4 (b) 2 , 2 (c) 4 , 14 (d) 3 , 4 Solution. We have p(s) = 4s2 – 4s + 1 Since 4s2 – 4s + 1 = 4s2 – 2s – 2s + 1 = 2s(2s – 1) – 1(2s – 1) = (2s – 1) (2s – 1) So, p(s) = 0 11 fi (2s – 1) (2s – 1) = 0 fi s = 2 , 2 are the zeroes of p(s) Hence, option (b) is the correct answer. Example 4. The sum and product of zeroes of the quadratic equation given in example 3 are respectively 31 1 (a) 2, 4 (b) 0, 8 (c) 1, 4 (d) None of these Solution. Sum of its zeroes = 1 + 1 =1= −(−4) = −(Coefficient of s) 2 2 4 Coefficient of s2 1 1 1 Constant term ×= 4 = Coefficient of s2 and product of its zeroes = 22 Hence, option (c) is the correct answer. Example 5. The set of the zeroes of the polynomial x2 – 25, their sum and product is (a) 4, 3; 7; 12 (b) –3, 3; 0; –9 (c) 5, –5; 0; –25 (d) None of these Solution. We know that a2 – b2 = (a – b) (a + b), so we can write x2 – 25 = (x – 5) (x + 5) Thus value of p(x) = x2 – 25 is zero, when either x – 5 = 0 or x + 5 = 0 fi x = 5 or x = –5 – (Coefficient of x) So, sum of the zeroes = 5 + (–5) = 0 = Coefficient of x2 Constant term and product of the zeroes = (5) (–5) = –25 = Coefficient of x2 Hence, option (c) is the correct answer. Example 6. If the product of the zeroes of the polynomial ax2 – 6x – 6 is 4, then value of a is 1 1 –5 –3 (a) 8 (b) – 4 (c) 3 (d) 2 Solution. Given polynomial is ax2 – 6x – 6 Comparing it with ax2 + bx + c, we get product of the zeroes = c = −6 = 4 (given) a a Polynomials  21

\\   −6 = 4 ⇒ a = −6 = −3 a 42 Hence, option (d) is the correct answer. Example 7. The value of k, if the sum of the zeroes of the polynomial x2 – (k + 6) x + 2 (2k – 1) is half of their product is (a) 7 (b) 11 (c) 12 (d) None of these Solution. Comparing x2 – (k + 6)x + 2(2k – 1) with ax2 + bx + c, we get −{−(k + 6)} = k + 6 1 Sum of the zeroes of the quadratic polynomial = Product of the zeroes of the quadratic polynomial = c = 2 (2k – 1) = 2 (2k – 1) a 1 According to the question, we get 1 k + 6 = × 2(2k − 1) ⇒ k + 6 = 2k – 1 fi k=7 2 Hence, option (a) is the correct answer. Example 8. If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, the values of p and q are respectively (a) 5, 6 (b) 4, 7 (c) –5, –6 (d) –4, –7 Solution. Let a and b be the zeroes of 2x2 – 5x – 3. \\  5 ab = −3 a + b = ,  2 2 As per the question, It is given that 2a, 2b are the zeroes of x2 + px + q \\ 2a + 2b = –p 5 fi 2(a + b) = –p  fi  2× = –p  fi  p = –5 2 Also, (2a) (2b) = q fi  4ab = q fi ab = q ⇒ −3 = q   fi  q = – 6 4 2 4 Hence, option (c) is the correct answer. Example 9. A quadratic polynomial whose zeroes are 1 and –3 is (a) x2 + 3x – 2 (b) x2 + 5x – 5 (c) x2 + 2x – 3 (d) None of these Solution. Given that zeroes of the polynomial are 1 and –3. Thus sum of the zeroes = 1 + (–3) = –2 and product of the zeroes = 1 × (–3) = –3 Required quadratic polynomial is p(x) = x2 – (Sum of the zeroes) x + Product of the zeroes = x2 + 2x – 3 Hence, option (c) is the correct answer. Example 10. The quadratic polynomial, sum of whose zeroes is 8 and their product is 12, is given by (a) x2 – 8x + 12 (b) x2 + 8x – 12 (c) x2 – 5x + 7 (d) x2 + 5x – 7 22  Mathematics–10

Solution. Sum of the zeroes is 8 and product of zeroes is 12. So, the required polynomial p(x) = x2 – (Sum of zeroes) x + Product of zeroes = x2 – 8x + 12 Hence, option (a) is the correct answer. Example 11. If a, b are the zeroes of the polynomial 2x2 – 5x + 7, then a polynomial whose zeroes are 2a + 3b, 3a + 2b is (a) k  x2 − 3 x + 21 (b) k  x2 − 25 x + 41 (c) k  x2 + 9 x − 45 (d) None of these  5  2  2 Solution. Since a, b are the zeroes of 2x2 – 5x + 7 SSRTSSSSSSSprodsuumct WWWXWWVWWWW \\ a + b = −(−5) = 5 and αβ = 7  of the zeroes = –b 2 2 2 of the zeroes = a c a The given zeroes of required polynomial are 2a + 3b and 3a + 2b Sum of the zeroes = 2a + 3b + 3a + 2b = 5a + 5b = 5(a + b) = 5 × 5 25 = 2 2 Again, product of the zeroes= (2a + 3b) (3a + 2b) = 6 (a2 + b2) + 13ab = 6 [(a + b)2 – 2ab)] + 13 ab = 6(a + b)2 + ab = 6  5 2 + 7 = 75 + 7 = 82 = 41  2 2 2 2 2 Now, required polynomial is given by 25 2 k [x2 – (Sum of the zeroes) x + Product of the zeroes] = k [ x2 − x + 41 ], where k is any non-zero real number. Hence, option (b) is the correct answer. Example 12. A quadratic polynomial whose product and sum of zeroes are 13 and 3 respectively. − , 55 (a) k(x2 + 12x + 5) (b) k[x2 – (8x) + (–9)) (c) k  x2 − 1 x +  − 7 (d) k  x2 − 3 x +  − 13    2  5    5  5   Solution. Product of zeroes of a quadratic polynomial = – 13 5 3 and sum of zeroes of quadratic polynomial = . 5 Thus, required quadratic polynomial = k  x2 −  3 x +  −513  , where k is a non-zero real number.  5  Hence, option (d) is the correct answer. Exercise 2.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. Sum of the zeroes of the polynomial x2 + 7x + 10 are (a) 7 (b) – 7 (c) 10 (d) – 10 Polynomials  23

2. The quadratic polynomial, the sum of whose zeroes is –5 and their product is 6, is (a) x2 + 5x + 6 (b) x2 – 5x + 6 (c) x2 – 5x – 6 (d) –x2 + 5x + 6 3. Quadratic polynomial having zeroes a and b is (a) x2 – (ab)x + (a + b) (b) x2 – (a + b)x + ab (c) x2 − α x + αβ (d) None of these β 4. If one zero of the quadratic polynomial x2 – 5x – 6 is 6, then other zero is (a) 0 (b) 1 (c) –1 (d) 2 5. A quadratic polynomial, the sum and product of whose zeroes and (–3) and 2 respectively is (a) x2 + 3x + 2 (b) x2 – 3x + 2 (c) x2 + 3x – 2 (d) None of these 6. If the sum of the zeroes of the quadratic polynomial 3x2 – kx + 6 is 3, then the value of k is (a) 3 (b) 6 (c) 9 (d) 0 7. A quadratic polynomial, whose sum of zeroes is 2 and product is –8 is (a) x2 – 3x – 3 (b) x2 + 2x + 8 (c) x2 + 3x + 3 (d) x2 – 2x – 8 8. A quadratic polynomial whose zeroes are 5 − 3 2 and 5 + 3 2 is (a) x2 – 10x + 7 (b) x2 + 10x + 7 (c) x2 – 5x + 9 (d) x2 + 5x – 9 9. The zeores of the quadratic polynomial 6x2 – 3 – 7x are (a) 3 , −1 (b) 2 , –3 (c) 3 −3 (d) None of these 23 3 , 57 10. The zeroes of the quadratic polynomial 4x2 – 4x – 3 are (a) 3 , −1 (b) 3 , −1 (c) 2 , −2 (d) None of these 23 22 55 11. The zeroes of the quadratic polynomial x2 + 7x + 10 are (a) –2, –5 (b) 2, 5 (c) –3, –8 (d) 3, 8 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): If one zero of polynomial p(x) = (k2 + 4)x2 + 13x + 4k is reciprocal of each other, then k = 2. Reason (R): If (x – a) is a factor of p(x), then p(a) = 0, i.e. a is a zero of p(x). 2. Assertion (A): If both zeroes of the quadratic polynomial x2 –2kx + 2 are equal in magnitude but 1 opposite in sign, then value of k is . 2 Reason (R): Sum of zeroes of a quadratic polynomial ax2 + bx + c is −b . a Answers and Hints A. Multiple Choice Questions (MCQs) 5. (a) Let a and b be the zeroes of the required 1. (b) – 7 2. (a) x2 + 5x + 6 polynomial f(x). 3. (b) x2 – (a + b)x + ab 4. (c) –1 Here, a + b = –3 and ab = 2 24  Mathematics–10

\\ f(x) = x2 – (a + b)x + ab So the two zeroes are x = 3 −1 = x2 – (–3)x + 2 , 3 = x2 + 3x + 2 2 10. (b) 3 −1 , k 22 3 6. (c) a + b = 11. (a) We have p(x) = x2 + 7x + 10 p(x) = x2 + 2x + 5x + 10 k 3 = 3 = x(x + 2) + 5(x + 2) \\ k = 9 = (x + 2) (x + 5) 7. (d) Quadratic polynomial is given by The zeroes of polynomial p(x) is given by x2 – (a + b)x + ab =x 2 – 2x – 8 p(x) = 0 ⇒ (x + 2) (x + 5) = 0 8. (a) Sum of zeroes = 5 − 3 2 + 5 + 3 2 = 10  Either x + 2 = 0  or  x + 5 = 0 ( )( ) Product of zeroes = 5 − 3 2 5 + 3 2 = 7 ⇒ x = –2, x = –5 Therefore   P(x) = x2 – 10x + 7 ⇒ x = –2, –5 9. (a) p(x) = 6x2 – 7x – 3 ...(i) Thus, the zeroes of x2 + 7x + 10 are a = –2 and b = –5 = 6x2 – 9x + 2x – 3 B. Assertion-Reason Type Questions = 3x(2x – 3) + 1(2x – 3) 1. (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of = (2x – 3) (3x + 1) assertion (A). The zeroes of p(x) is given by p(x) = 0. 2. (d) Assertion (A) is false but reason (R) is true. Case Study Based Questions I. Due to heavy storm an electric wire got Y bent as shown in the figure. It followed a mathematical shape. Answer the 6 following questions below: 5 1. Name the shape in which the wire is bent. 4 (a) spiral (b) ellipse 3 (c) linear (d) parabola 2 2. How many zeroes are there for the 1 polynomial (shape of the wire)? X' X –6 –5 –4 –3 –2 –1 (a) 2 (b) 3 –1 1 2 34 5 6 7 (c) 1 (d) 0 3. The zeroes of the polynomial are –2 (a) –1, 5 (b) –1, 3 –3 (c) 3, 5 (d) –4, 2 –4 4. What will be the expression of the –5 Y' polynomial? (b) x2 – 2x + 3 (a) x2 + 2x – 3 (d) x2 + 2x + 3 (d) 0 (c) x2 – 2x – 3 5. What is the value of the polynomial if x = –1? (a) 6 (b) –18 (c) 18 Ans. 1. (d) parabola 2. (a) 2 3. (b) –1, 3 4. (c) x2 – 2x – 3 5. (d) 0 Polynomials  25

II. A child was playing with a rope. He laid the rope on two straight lines which were perpendicular to each other. He marked the horizontal line from –3 to 3 and vertical line from –2 to 2. The rope forms some mathematical shape which can be answered below: 1. What is the name of the shape from A, B to C? (a) parabola (b) linear (c) elliptical (d) spiral 2. How many zeroes are there for the polynomical (shape of the rope)? (a) 3 (b) 1 (c) 2 (d) 4 3. The zeroes of the polynomial are (a) 3, –1, 2 (b) –3, –1, 1 (c) 3, 1, –2 (d) –3, 2, –2 (d) x3 + 3x2 + x – 3 4. What will be the expression of the polynomial? (d) 0 (a) x3 + 3x2 – x – 3 (b) x3 – 3x2 + x + 3 (c) x3 + 3x2 + x + 3 5. What is the degree of the polynomial so formed? (a) 1 (b) 2 (c) 3 Ans. 1. (a) parabola 2. (a) 3 3. (b) –3, –1, 1 4. (a) x3 + 3x2 – x – 3 5. (c) 3 III. Due to heavy storm and rainfall, electric wire broke down on a chauraha and it disturbed the traffic. One piece of the wire is shown in the figure which followed some mathematical shape. Answer the following questions below: 1. Find the coordinates where does the graph cut the x-axis. (a) (–5, 0), (–2, 0), (2, 0) (b) (–5, 0), (–2, 0), (0, 2) (c) (–2, 0), (0, 2), (2, 0) (d) (–5, 0), (0, 2), (2, 0) 2. Find number of zeroes of the graph y = f(x) (a) 0 (b) 2 (c) 1 (d) 3 3. Find the zeroes of the graph y = f(x) (a) –5, 0, 2 (b) –5, –2, 2 (c) –2, 0, 2 (d) –5, –2, 0 26  Mathematics–10

4. What will be the expression of the polynomial? (b) x3 + 5x2 + 4x – 20 (a) x3 + 5x2 – 4x – 20 (d) x3 – 5x2 + 4x + 20 (c) x3 + 5x2 + 4x + 20 5. What is the value of the polynomial if x = –5? (a) –125 (b) 125 (c) 0 (d) 1 Ans. 1. (a) (–5, 0), (–2, 0), (2, 0) 2. (d) 3 3. (b) –5, –2, 2 4. (a) x3 + 5x2 – 4x – 20 5. (c) 0 IV. A student of class Xth are interested in drawing so he draw some mathematical shape on the ground with chalk. He has some questions in his mind which he wants to know. Answer the following questions: Y B (0, 2) A (-2, 0) C (2, 0) X′ -4 -3 -2 -1 O 1 2 3 4 X Y′ 1. What is the name of the shape of the figure? (a) Ellipse (b) parabola (c) linear (d) spiral 2. How many zeroes are there for the polynomial? (a) 1 (b) 2 (c) 3 (d) 4 3. The zeroes of the polynomial are (a) –2, 0 (b) 0, 2 (c) –1, 2 (d) –2, 2 4. The expression of the polynomial is (b) x2 – 4x + 4 (a) x2 + 4x + 4 (d) x2 – 4x – 4 (c) x2 – 4 5. The value of the polynomial if x = –2 is (a) 16 (b) 12 (c) 8 (d) 0 Ans. 1. (b) parabola 2. (b) 2 3. (d) –2, 2 4. (c) x2 – 4 5. (d) 0 Polynomials  27

V. In a game to entertain themselves, students of class-10th have drawn following figure with chalk on the ground. They have some questions in their mind which they want to solve. Please answer and solve their question. 1. The number of zeroes of the polynomial P(x) is (a) 1 (b) 2 (c) 3 (d) 4 2. The zeroes of the polynomial are (a) 1, 3 (b) –1, 3 (c) 1, –3 (d) –1, –3 3. The expression of the polynomial is (c) x2 + 4x – 3 (d) x2 – 4x – 3 (a) x2 + 4x + 3 (b) x2 – 4x + 3 4. The value of the polynomial if x = 3 i (a) 24 (b) 0 (c) 18 (d) –6 5. What is the degree of the polynomial? (a) 0 (b) 1 (c) 2 (d) 3 Ans. 1. (b) 2 2. (a) 1, 3 3. (b) x2 – 4x + 3 4. (b) 0 5. (c) 2 VI. To scare his friends in a street, a naughty boy spread a rope which looks in some mathematical shape. Now few questions arises in his mind which he wants you to answer. Y 4 3 2 y = p(x) 1 (–4, 0) (–2, 0) (2, 0) (4, 0) X′ B O A X Y′ 1. Name the shape between points A and B. (c) ellipse (a) linear (b) parabola (d) spiral 28  Mathematics–10

2. The number of zeroes of the polynomial y = p(x) is (a) 1 (b) 2 (c) 3 (d) 4 3. The zeroes of the polynomial are (d) –2, 0, 2, 4 (d) x4 + 20x2 – 64 (a) –4, –2, 2, 4 (b) –4, –1, 2, 4 (c) –4, –2, 0, 4 (d) 32 4. The expression of the polynomial is 3. (a) –4, –2, 2, 4 (a) x4 + 20x2 + 64 (b) x4 – 20x2 + 64 (c) x4 – 20x2 – 64 5. The value of the polynomial when x = 2 is (a) 144 (b) –128 (c) 0 Ans. 1. (b) parabola 2. (d) 4 4. (b) x4 – 20x2 + 64 5. (c) 0 Experts’ Opinion Questions based on following types are very important for exams. So, students are advised to revise them thoroughly. 1. Finding zeroes of the quadratic polynomial and the relation between the zeroes and the coefficients. 2. Forming a quadratic polynomial with given sum of zeroes and product of zeroes. 3. Finding all the zeroes of a polynomial if its some zeroes are given. IMPORTANT FORMULAE • If a, b are zeroes of the quadratic polynomial ax2 + bx + c, then (i) a + b = −b (ii) ab = c aa • If a, b are zeroes of the quadratic polynomial p (x), then p (x) = x2 – (a + b) x + ab QUICK REVISION NOTES •• A polynomial p(x) in one variable x is an algebraic expression in x of the form p(x) = anxn + an–1xn–1 + ... + a2x2 + a1x + a0 where an, an–1, ..., a0 are real numbers with an ≠ 0. •• The highest power of x in p(x) is called the degree of the polynomial p(x). •• A quadratic polynomial in x with real coefficients is of the form ax2 + bx + c, where a, b, c are real numbers with a π 0. •• Let p(x) be a polynomial in x and k be any real number, then the value obtained by replacing x by k in p(x) is called the value of p(x) at x = k and is denoted by p(k). •• A real number k is said to be a zero of a polynomial p(x) if p(k) = 0. •• A polynomial may have no zero, one or more than one zeroes. •• A polynomial p(x) of degree n has atmost n zeroes. •• The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. •• If a and b are the zeroes of the quadratic polynomial ax2 + bx + c, then a + b = −b , ab = c . aa Polynomials  29

COMMON ERRORS Errors Corrections (i) Finding incorrectly p(x) at x = k. (i) To find p(x) at x = k, replace x by k in p(x) carefully and get the answer. (ii) Doing mistakes in finding zeroes of a polynomial (ii) Remember that x-coordinate(s) of point(s), where from graph. the graph of given polynomial intersects the x-axis are zeroes of the polynomial. (iii) Splitting the middle term of a polynomial (iii) To split b of the polynomial ax2 + bx + c, write b incorrectly to find the zeroes. as the sum of two numbers whose product is ac. (iv) Formating incorrectly a quadratic polynomial (iv) Form a quadratic polynomial with zeroes a, b as when zeroes are a, b as (x – a) (x – b). k (x – a) (x – b), where k is real. (v) Finding the incorrect value of y for various values (v) Transpose y term to either side so as to get of x. positive coefficient. (vi) Finding the incorrect value of y for various values (vi) Transpose y term to either side so as to get of x. positive coefficient. 30  Mathematics–10

3 Pair of Linear Equations in Two Variables Topics Covered 1. Pair of Linear Equations in Two Variables and Finding its Solution by Graphical Method 2. Solving a Pair of Linear Equations by Algebraic Methods 3. Solving a System of Equations Reducible to Simultaneous Linear Equations 4. Solving Word Problems on Pair of Linear Equations in Two Variables Introduction An equation which can be expressed in the form ax + by + c = 0, where a, b and c are real numbers, and a π 0, b π 0 is called a linear equation in two variables x and y. A solution of a linear equation in two variables is a pair of values of x and y, which satisfy the equation. The graph of a linear equation in two variables is a straight line. Every linear equation in two variables has infinitely many solutions and each one of them is represented by a point on the graph of the equation. 1. Pair of Linear Equations in Two Variables and Finding its Solution by Graphical Method Pair of Linear Equations in Two Variables Two linear equations in the same two variables are called a pair of linear equations in two variables. General form of a pair of linear equations is a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 where a1, a2, b1, b2, c1, c2 are real numbers such that a12 + b12 ≠ 0, a22 + b22 ≠ 0. Some examples of a pair of linear equations in two variables are: (i) x + 2y = 17, 3x – 5y = 0 (ii) 2a + 3b – 2 = 0, a + b – 2 = 0 The solution of a given pair of linear equations in x and y is a pair of values of x and y, which satisfy each of the equations. A pair of linear equations in two variables is said to be consistent, if it has at least one solution. A pair of linear equations in two variables is said to be inconsistent, if it has no solution. Graphical Method of Solution of a Pair of Linear Equations In order to find a solution of a pair of linear equations in two variables by graphical method, we draw the graph of each of the given linear equations. When we represent a pair of linear equations graphically as two lines, we can observe that the lines (i) may intersect. (ii) may be parallel. (iii) may coincide. The number of solutions for a system of two linear equations in two variables is given by one of the following observations. What this Means Graphically Number of Solutions Type of Equations The two lines intersect at a single point. Unique common solution. Consistent The two lines are parallel. No common solution. Inconsistent The two lines are coincident. Infinitely many solutions. Dependent Consistent 31

Conditions for Consistency of a Pair of Linear Equations Again, consider a pair of two linear equations a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 where a1, b1, c1 and a2, b2, c2 denote the coefficients of the equations given above in the general form. By comparing the values of a1 , b1 and c1 , we can observe that if the lines represented by the above a2 b2 c2 stated equations are intersecting or parallel or coincident. For the system of two linear equations a1x + b1y + c1 = 0; a2x + b2y + c2 = 0 S.No. Compare the ratios Graphical representations Algebraic interpretation 1. a1 ≠ b1 Intersecting lines • Exactly one solution (Unique) 2. a2 b2 • Consistent system of equations • Equations are independent a1 = b1 = c1 Coincident lines • Infinitely many solutions a2 b2 c2 • System is consistent • Equations are dependent 3. a1 = b1 ≠ c1 Parallel lines • No solution a2 b2 c2 • System is inconsistent Example 1. Given: 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. The pair of linear equations representing the above situation and their solution, respectively are (a) x + y = 10, x – y = –4; (3, 7) (b) x – y = 10, x + y = 4; (3, 5) (c) 2x + y = 10, 2x – y = 4; (2, 5) (d) 2x – y = 10, 2x + y = 4 Solution. Let the number of boys and girls who took part in Mathematics quiz be x and y respectively. lY 10 (0, 10) Then, x + y = 10 ...(i) m and y = x + 4 ...(ii) y=x+4 Now, we draw the graph for equations (i) 8 (3, 7) and (ii). 10 6 x+y= We find two solutions of each of the equations which are given in table: 3 4 (0, 4) Table for x + y = 10 x0 2 y = 10 – x 10 7 X' O X 2 4 6 8 10 Table for y = x + 4 x 0 3 Y' y=x+4 4 7 32  Mathematics–10

From graph, we observe that the two lines representing two equations intersect each other at point (3, 7). So, the solution of the given pair of linear equations is x = 3 and y = 7. Thus, 3 boys and 7 girls took part in Mathematics quiz. Hence, option (a) is the correct ansswer. Example 2. Given: 5 pencils and 7 pens together cost ` 50, whereas 7 pencils and 5 pens together cost ` 46. The pair of linear equations representing the above situation and the cost of one pencil and that of one pen, respectively are (a) 5x – 7y = 50, 7x – 5y = 46; (3, 5) (b) 5x + 7y = 50, 7x + 5y = 46; (3, 5) (c) 3x – 8y = 48, 3x + 8y = 48; (5, 3) (d) None of these Solution. Let cost of one pencil be ` x and cost of one pen be ` y. Then, 5x + 7y = 50 ...(i) and 7x + 5y = 46 ...(ii) Now, we draw the graph of equations (i) and (ii). Y We will now complete the tables as follows: Table for 5x + 7y = 50 3 10 x 10 5 8 50 − 5x 0 y= 7 6 (3, 5) Table for 7x + 5y = 46 45 = 46 x+7+ 5y x8 3 7x 5 2 y = 50 y= 46 – 7x –2 5 (10, 0) X' X O 2 4 6 8 10 Plotting the values of x and y from these –2 (8, –2) tables on the graph, we observe that two lines Y' intersect each other at point (3, 5). So, the solution of the pair of linear equations is x = 3 and y = 5. Thus, the cost of 1 pencil is ` 3 and that of 1 pen is ` 5. Hence, option (b) is the correct answer. Example 3. The area of the quadrilateral formed by the lines x = 3, x = 6, 2x – y – 4 = 0 and x-axis is (a) 8 sq. units (b) 12 sq. units (c) 15 sq. units (d) None of these Solution. First we find three solutions of the equation 2x – y – 4 = 0 (see table) and then plot these solutions on the graph. Pair of Linear Equations in Two Variables  33

Table for 2x – y – 4 = 0 ⇒ y = 2x – 4 Y x 02 6 8 D (6, 8) y –4 0 8 Drawing the graph of x = 3, x = 6, 7 x=3 we get the lines parallel to y-axis and 6 2x – passing through the points (3, 0) and 5 y – 4 =0 (6, 0) respectively. x=6 Area of the trapezium ABCD = 1 (sum of parallel sides) × height 4 2 = 1 (2 + 8) × 3 3 2 = 5 × 3 = 15 sq. units 2 A Hence, option (c) is the correct B (3, 0) answer. 1 34 Example 4. If the pair of equations (6, 0) x + y = 5 and 2x + 2y = 10 is consistent, the two solutions obtained graphically X' (2, 0) C O 12 X 56 7 are –1 (a) (0, 4), (4, 0) (b) (7, –2), (2, 7) –2 (c) (0, 5), (5, 0) (d) None of these [Imp.] –3 Solution. The given equations can be written –4 (0, –4) x + y – 5 = 0 ...(i) Y' and 2x + 2y – 10 = 0 ...(ii) Here, a1 = 1, b1 = 1, c1 = –5 and a2 = 2, b2 = 2, c2 = –10 Now, a1 1 b1 = 1 c1 = −5 1 =, , = a2 2 b2 2 c2 −10 2 Clearly, a1 = b1 = c1 a2 b2 c2 So, the linear equations represented by (i) and (ii) are coincident. The system of equations has infinitely many solutions. The system is consistent and the equations are dependent. Now, we draw the graph of equations. First we find two solutions of each equation (see tables). Table for x + y – 5 = 0 Table for 2x + 2y – 10 = 0 fi x + y – 5 = 0 x0 5 x05 0 y=5–x 5 y=5–x 5 0 34  Mathematics–10

The graph of the given system of equations is shown in the figure. There are infinitely many solutions. Hence, option (c) is the correct answer. Example 5. The nature of the system of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0 is (a) Unique (b) Consistent (c) Inconsistent (d) None of these Solution. We have 2x – 2y – 2 = 0 ...(i) and 4x – 4y – 5 = 0 ...(ii) Here, a1 = 2, b1 = –2, c1 = –2 and a2 = 4, b2 = –4, c2 = –5 Now, a1 2 1 , b1 = −2 = 1 , c1 = −2 = 2 == a2 4 2 b2 −4 2 c2 −5 5 Clearly, a1 = b1 ≠ c1 a2 b2 c2 So, the lines are parallel, i.e., they have no point in common. So, the given system of equations has no solution, i.e., it is inconsistent. Hence, option (c) is the correct answer. Example 6. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. The dimensions of the garden are (use graphical method) (a) Length = 20 m, width = 16 m (b) Length = 16 m, width = 24 m (c) Length = 18 m, width = 12 m (d) Length = 24 m, width = 15 m Solution. Let the length and width of the garden be x m and y m respectively. Then, x = y + 4 ...(i) Also x + y = 36 ...(ii) Now, we find two solutions of each equation and then draw their graphs. Table for x = y + 4 x04 y=x–4 –4 0 Table for x + y = 36 x 16 20 y = 36 – x 20 16 From the graph, we observe that the lines intersect each other at the point (20, 16). Therefore, x = 20 and y = 16 are the required solutions. Thus, the length of the garden is 20 m and the width of the garden is 16 m. Hence, option (a) is the correct answer. Pair of Linear Equations in Two Variables  35

Example 7. The solution of the following system of equations when solved graphically is: 2x – 3y – 6 = 0 2x + y + 10 = 0 (a) (3, 4) (b) (–3, –4) (c) (4, 5) (d) (–4, –5) Solution. We have, 2x – 3y – 6 = 0 ...(i) ...(ii) 2x + y + 10 = 0 Let us find the table of values for each equation. Table for 2x – 3y – 6 = 0 ⇒ y = 2x − 6 Table for 2x + y + 10 = 0 ⇒ y = –10 – 2x 3 x360 x –5 –6 –4 y 0 2 –2 y 0 2 –2 We plot these values to get the following graph: Y m 3 l (–6, 2) 2 2x + y + 10 = 0 (6, 2) 1 2x – 3y – 6 = 0 –2 –1 0 X' (–5, 0) –3 (3, 0) –6 –5 –4 X 123456 –1 (–4, –2) –2 (0, –2) –3 (–3, –4) –4 Y' We observe that the point (–3, –4) is an intersecting point of graphs of the equations (i) and (ii). Thus, the required solution is x = –3, y = –4. Hence, option (b) is the correct answer. Example 8. The coordinates of vertices of a triangle formed by the equations x – y + 1 = 0, 3x + 2y – 12 = 0 and x-axis are (a) (1, 3), (–2, 3), (3, 0) (b) (3, 4), (–3, 5), (2, 3) (c) (2, 5), (–3, 0), (5, 0) (d) (2, 3), (–1, 0), (4,0) 36  Mathematics–10

Solution. We find the tables of values to draw the graph of the equations, Table for x – y + 1 = 0 mY l x 0 –1 (0, 6) x– 6 y+ 1 y=x+1 1 0 = 0 4 P (2, 3) Table for 3x + 2y – 12 = 0 2 x 20 (0, 1) A (–1, 0) B (4, 0) 246 X' 0 3x + 2y – 12 = 0 X –2 8 y = 12 − 3x 3 6 –2 2 Y' From the graph we observe that the two lines intersect each other at the point P(2, 3). So, x = 2 and y = 3 is the solution of the given system of equations. The triangle formed by these lines and the x-axis is DPAB whose vertices are P(2, 3), A(–1, 0) and B(4, 0). Hence, option (d) is the correct answer. Example 9. The coordinates of the vertices of a triangle formed by the equations of sides are: y = x; y = 2x; x + y = 6 are (a) (0, 0), (3, 3), (2, 4) (b) (0, 1), (5, 5), (2, 5) (c) (4, 4), (3, 0), (1, 6) (d) None of these Solution. We find the table of values for drawing the graph of equations. Table of values for y = x Table of values for y = 2x nY m x02 x01 y02 y02 (0, 6) 6 Now, we find the table of values for (1, 5) l x + y = 6 5 fi y = 6 – x B 4 (2, 4) y = 2x Table for x + y = 6 3 y= x01 (1, 2) x y65 (3, 3) 2 A 6 x+y= (2, 2) From the graph, we observe that a triangle is formed, X' 1 namely OAB and the vertices of triangle OAB are O(0, 0), X A(3, 3) and B (2, 4). (0, 0) 1 23 4 0 Hence, option (a) is the correct answer. Example 10. The vertices of the triangle formed by the –1 graphs of the equations 4x – 3y – 6 = 0, x + 3y – 9 and Y' y-axis are (a) (1, 3), (2, 4), (–3, 5) (b) (0, 3), (3, 2), (0, –2) (c) (0, 4), (3, 3), (2, 6) (d) None of these Pair of Linear Equations in Two Variables  37

Solution. We first find the table of solutions of each of the equations, (see table). Table for 4x – 3y – 6 = 0 Y m 3 A (0, 3) x + 3y – 9 = 0 P (3, 2) x 30 2 y = 4x − 6 2 –2 3 4x – 3y – 6 = 0 l Table for x + 3y – 9 = 0 1 x 0 3 X' 0 X y= 9−x 3 2 –1 123 4 3 –1 Now, we plot the points of both tables in the graph which B (0, –2) is shown alongside. –2 We observe that there is a point (3, 2) common to both Y' the lines l and m, so the solution of linear equations is x = 3 and y = 2, i.e., the given system of equations is consistent. The vertices of the triangle formed by lines l, m and the y-axis are A(0, 3), P(3, 2) and B(0, –2). Hence, option (b) is the correct answer. Example 11. The nature of graphs of equations x + 4y = 3, 2x + 8y = 6 and the number of their solutions are (a) Consistent, one (b) Consistent, two (d) Inconsistent, no solution (c) Dependent, many Solution. We have x + 4y = 3 ...(i) and 2x + 8y = 6 ...(ii) To draw the graph for given equations, we find the tables of values which are shown below: Table for x + 4y = 3 Table for 2x + 8y = 6 fi x + 4y = 3 x 7 –1 –5 x –1 –5 –3 2 y= 3– x –1 1 y= 3– x 1 2 1.5 4 4 We plot the points of both the tables to get the following graph. Y (–5, 2) 2 (–3, 1.5) (–1, 1) 1 x + 4y = 3 X' X –5 –4 –3 –2 –1 0 123 45 67 –1 2x + 8y = 6 (7, –1) Y' We observe that graphs of given two equations coincide. So, the given equations are pair of dependent equations and they have infinitely many solutions. Hence, option (c) is the correct answer. 38  Mathematics–10

Example 12. The value of k so that the following system of equations has no solution is 3x – y – 5 = 0, 6x – 2y + k = 0 [Imp.] (a) 10 (b) –10 ...(i) ...(ii) (c) Both 10 and –10 (d) All real values of k except –10 Solution. We have 3x – y – 5 = 0 and 6x – 2y + k = 0 Here, a1 = 3, a2 = 6, b1 = – 1, b2 = – 2, c1 = – 5, c2 = k For no solution, a1 = b1 ≠ c1 a2 b2 c2 \\ 3 = −1 ≠ −5   ⇒  1 ≠ − 5   ⇒  k ≠ – 10 6 −2 k 2k \\  For all real values of k except –10, the system of equations has no solution. Hence, option (d) is the correct answer. Example 13. If the pair of equations x sin q + y cos q = 1 and x + y = 2 has infinitely many solutions, then the value of q is (a) 30° (b) 45° (c) 60° (d) 90° Solution. We have x sin q + y cos q = 1 ...(i) and x + y = 2 ...(ii) Here, a1 = sin q, a2 = 1, b1 = cos q, b2 =1, c1 = – 1, c2 = – 2 For infinitely many solutions, a1 = b1 = c1   ⇒  −1 ⇒  q = 45° a2 b2 c2 sin q = cos q = − 2   Hence, option (b) is the correct answer. Example 14. The number of solutions of the following pair of linear equations is x + 2y – 8 = 0 2x + 4y = 16 (a) No solutions (b) One solution (c) Two solutions (d) Infinitely many solutions Solution. We have x + 2y – 8 = 0 and 2x + 4y = 16 Here, a1 = 1, a2 = 2, b1 = 2, b2 = 4, c1 = –8, c2 = –16 a1 Now, a2 1 b1 = 2 = 1 , c1 = −8 1 = 2 , b2 4 2 c2 = −16 2 Since a1 = b1 = c1 = 1 a2 b2 c2 2 \\ The pair of linear equations is consistent and has infinitely many solutions. Hence, option (d) is the correct answer. Example 15. The lines represented by the equations 5x – 4y + 8 = 0, 7x + 6y – 9 will (a) intersect at a point (b) be parallel (c) be coincident (d) None of these Pair of Linear Equations in Two Variables  39

Solution. We have 5x – 4y + 8 = 0 ...(i) ...(ii) and 7x + 6y – 9 = 0 Here, a1 = 5, b1 = –4, c1 = 8 and a2 = 7, b2 = 6, c2 = –9 Since, a1 = 5 and b1 = −4 = 2 . So, a1 ≠ b1 a2 7 b2 − a2 b2 63 Thus, the lines representing the given equations are intersecting. The system of equations is consistent and has a unique solution. Equations are independent. Hence, option (a) is the correct answer. Example 16. The lines represented by the equations 9x + 3y + 12 = 0 and 18x + 6y + 24 will (a) intersect at a point (b) be parallel (c) be coincident (d) None of these Solution. We have 9x + 3y + 12 = 0 ...(i) and 18x + 6y + 24 = 0 ...(ii) Here, a1 = 9, b1 = 3, c1 = 12 and a2 = 18, b2 = 6, c2 = 24 Since, a1 91 b1 31 and c1 12 1 So, a1 = b1 = c1 a2 = =, b2 == c2 == a2 b2 c2 18 2 62 24 2 Thus, the lines representing the given equations are coincident. The system of equations is consistent and the equations are dependent. There are infinitely many solutions. Hence, option (c) is the correct answer. Example 17. The lines represented by the equations 6x – 3y + 10 = 0 and 2x – y + 9 = 0 will (a) intersect at a point (b) be parallel (c) be coincident (d) None of these Solution. We have 6x – 3y + 10 = 0 ...(i) and 2x – y + 9 = 0 ...(ii) Here, a1 = 6, b1 = –3, c1 = 10 and a2 = 2, b2 = –1, c2 = 9 Since, a1 = 6 = 3, b1 = −3 = 3 and c1 10 a1 = b1 ≠ c1 a2 2 b2 −1 c2 = . So, a2 b2 c2 9 Thus, the lines representing the given equations are parallel. The system of equations is inconsistent and has no solution because the lines never intersect. Hence, option (b) is the correct answer. Example 18. On comparing a1 , b1 , c1 , the graphical representation of equations 3x + 2y = 5, 2x – 3y = 7 will be a2 b2 c2 (a) Intersecting (b) Coincident (c) Parallel (d) None of these Solution. We have ...(i) ...(ii) 3x + 2y – 5 = 0 and 2x – 3y – 7 = 0 Here, a1 = 3, b1 = 2, c1 = –5 and a2 = 2, b2 = –3, c2 = –7 Now, a1 = 3 b1 = 2 = −2  fi  a1 ≠ b1 , a2 b2 a2 2 b2 −3 3 40  Mathematics–10

Thus, the lines representing the given equations are intersecting. The system of equations is consistent and has a unique solution. Equations are independent. Hence, option (a) is the correct answer. Example 19. On comparing a1 , b1 , c1 , the graphical representation of equations 2x – 3y = 8 and a2 b2 c2 4x – 6y – 9 = 0 will (a) Intersecting lines (b) Coincident lines (c) Parallel lines (d) None of these Solution. We have 2x – 3y – 8 = 0 ...(i) and 4x – 6y – 9 = 0 ...(ii) Here, a1 = 2, b1 = –3, c1 = –8 and a2 = 4, b2 = –6, c2 = –9 Now, a1 = 2 = 1 b1 = −3 = 1 and c1 = −8 = 8 fi  a1 = b1 ≠ c1 a2 4 , b2 −6 2 c2 −9   a2 b2 c2 2 9 Thus, the lines representing the given equations are parallel. The system of equations is inconsistent and has no solution. Hence, option (c) is the correct answer. Exercise 3.1 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. The pair of linear equations 3x + 5y = 7 and 9x + 10y = 14 is 23 (a) consistent (b) inconsistent (c) consistent with one solution (d) consistent with many solutions 2. The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is 14 2 (d) 10 (a) − (b) (c) 5 35 3. If a pair of linear equations is consistent, then the line represented by them are (a) parallel (b) intersecting or coincident (c) always coincident (d) always intersecting 4. The value of k for which the system of equations kx + 4y = k – 4, 16x + ky = k have infinite number of solutions is (a) k = 2 (b) k = 4 (c) k = 6 (d) k = 8 5. The value of k for which the system of linear equations 3x + y = 1, (2k –1)x + (k –1)y = 2k + 1 have no solution is (a) k = 2 (b) k = 4 (c) k = 6 (d) k = 8 6. If the equations kx – 2y = 3 and 3x + y = 5 represent two intersecting lines at unique point, then the value of k is (a) Only 4 (b) Only 5 (c) Only 6 (d) Any number other than –6 7. The value of k for which the given system has unique solution 2x + 3y – 5 = 0, kx – 6y – 8 = 0 is (a) k = 2 (b) k ≠ 4 (c) k = 4 (d) k ≠ 4 Pair of Linear Equations in Two Variables  41

8. For what value of k, the following system of equations have infinite solutions: [Imp.] 2x – 3y = 7, (k + 2)x – (2k + 1)y = 3 (2k – 1)? (a) k = 2 (b) k = 3 (c) k = 4 (d) k = 8 9. The value of m for which the pair of linear equations 2x + 3y – 7 = 0 and (m – 1) x + (m + 1) y = (3m – 1) has infinitely many solutions is (a) 5 (b) 8 (c) – 5 (d) 8 10. For what values of p, the pair of equations 4x + py +8 =0 and 2x +2y +2 = 0 have unique solution? (a) p = 4 (b) p ≠ 4 (c) p = 7 (d) p ≠ 7 11. What type of straight lines will be represented by the system of equations 2x + 3y = 5 and 4x + 6y = 7?  (a) Intersecting (b) Parallel (c) Conincident (d) None of these 12. For what value of p, the following pair of linear equations have infinitely many solutions? (p – 3)x + 3y = p, px + py = 12 (a) 4 (b) 6 (c) 9 (d) 11 13. The value of k for which the following pair of linear equations have infinitely many solutions: 2x + 3y = 7, (k – 1)x + (k + 2)y = 3k is (a) 2 (b) 4 (c) 7 (d) 9 14. The value(s) of k for which the pair of linear equations kx + y = k2 and x + ky = 1 have infinitely many solutions is (a) 1 (b) 2 (c) 3 (d) 4 B. Assertion-Reason Type Questions In the following questions, a statement of assertion (A) is followed by a statement reason (R). Choose the correct choice as: (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). (b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A). (c) Assertion (A) is true but reason (R) is false. (d) Assertion (A) is false but reason (R) is true. 1. Assertion (A): Pair of linear equations x + y = 14, x – y = 4 is consistent. Reason (R): By comparing a1 and b1 if we get a1 ≠ b1 , then given system of equations is consistent. a2 b2 a2 b2 2. Assertion (A): For k = 6, the system of linear equations x + 2y + 3 = 0 and 3x + ky + 6 = 0 is inconsistent. Reason (R): The system of linear equations a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 is inconsistent if a1 = b1 = c1 . a2 b2 c2 Answers and Hints A. Multiple Choice Questions (MCQs) 8. (c) k = 4 9. (a) 5 1. (b) inconsistent 2. (d) 10 For infinitely many solutions the condition is 3. (b) intersecting or coincident 4. (d) k = 8 5. (a) k = 2 a1 = b1 = c1 ⇒ 23 7 == 6. (d) any number other than –6 a2 b2 c2 m − 1 m + 1 3m − 1 7. (d) k ≠ –4 Now, 2(m + 1) = 3(m – 1)  ⇒  m = 5 42  Mathematics–10

and 3(3m – 1) = 7(m + 1)  ⇒  m = 5 ⇒ p – 3 = 3  ⇒  p = 6 and 3 = p Hence, for m = 5, the system has infinitely p 12 many solutions. 10. (b) p ≠ 4 ⇒ p2 = 36  ⇒  p = ± 6 a1 ≠ b1 is the condition for the given pair Q For p = 6, (i) is true. So, for p = 6, system a2 b2 of equations to have a unique solution. has infinitely many solutions. 4≠ p 13. (c) 7 22 ⇒     For infinitely many solutions the condition is ⇒     p ≠ 4 a1 = b1 = c1 a2 b2 c2 Therefore, for all real values of p except 4, the given pair of equations will have a unique ⇒ 2 37 solution. k −1 = k + 2 = 3k 11. (b) Parallel Now, 2k + 4 = 3k – 3  ⇒  k = 7 Here, a1 = 2 = 1 and 9k = 7k + 14  ⇒  k = 7 a2 42 Hence, the value of k is 7. b1 3 1 and c1 = 5 14. (a) 1 b2 = 6 = 2 c2 7 For pair of equations kx + y = k2 and x + ky = 1 1 15 a1 = k , b1 = 1 c1 = k 2 2 = 2 ≠ 7  , a2 1 b2 k c2 1 a1 b1 c1 For infinitely many solutions, = = a1 = b1 ≠ c1 a2 b2 c2 a2 b2 c2 is the condition for which \\ k = 1 1k the given system of equations will represent ⇒ k2 = 1  ⇒  k = 1, –1 ...(i) parallel lines. So, the given system of linear equations will and k = k 2 ...(ii) represent a pair of parallel lines. ⇒ 11 12. (b) 6 k = 1 For infinitely may solutions, From (i) and (ii), k = 1 p − 3 = 3 = p ...(i) B. Assertion-Reason Type Questions p p 12 1. (a) B oth assertion (A) and reason (R) are true Now p − 3 = 3 and reason (R) is the correct explanation of p p assertion (A). 2. (c) Assertion (A) is true but reason (R) is false. 2. Solving a Pair of Linear Equations by Algebraic Methods The solution of a pair of linear equations can be obtained by using any one of the following algebraic methods: (i) Substitution Method  (ii) Elimination Method   A. To find the solution of a pair of linear equations using Substitution Method, we follow these steps: Step 1: From one equation, find the value of one variable (say y) in terms of other variable, i.e. x. Step 2: Substitute the value of variable obtained in step 1 in the other equation to get an equation in one variable. Step 3: Solve the equation obtained in step 2 to get the value of one variable. Step 4: Substitute the value of variable so obtained in any given equation to find the value of other  variable. Pair of Linear Equations in Two Variables  43

B. To find the solution of a pair of linear equations using Elimination Method, we follow these steps: Step 1: Multiply the given equations to make the coefficient of one of the unknown variables (either x or y), numerically equal. Step 2: Add the equations obtained after multiplication, if the numerically equal coefficients are opposite in signs or else subtract them. Step 3: Solve the resulting linear equation in one unknown variable to find its value. Step 4: Substitute this value in any one of the equations and find the value of the other variable. The values of the two variables constitute the solution of the given pair of linear equations. Example 1. In figure, ABCD is a rectangle. The values of x and y, respectively are (a) x = 12, y = 16 (b) x = 16, y = 10 (c) x = 22, y = 8 (d) x = 15, y = 18 Solution. AB = DC and BC = AD ⇒ x + y = 30 ...(i) ...(ii) and x – y = 14 We have x + y = 30  ⇒  x = 30 – y. Substituting x = 30 – y in (ii), we get x – y = 14 ⇒ 30 – y – y = 14 ⇒ 30 – 14 = 2y 16 ⇒ y = = 8 2 \\ x = 14 + 8 = 22 Thus, x = 22 and y = 8 Hence, option (c) is correct answer. Example 2. The solution of the system of equations x + y = 5, x – y = 2 using substitution method is: 73 31 31 25 (a) x = 2 , y = 2 (b) x = 5 , y = 2 (c) x = 5 , y = 4 (d) x = 5 , y = 2 Solution. We have x + y = 5  ⇒  y = 5 – x Substituting y = 5 – x in x – y = 2, we get x – (5 – x) = 2 7 ⇒ 2x – 5 = 2  ⇒ 2x = 7  ⇒  x = 2 \\ y = 5 – 7 3 = 22 Thus, 73 x = 2  and y = 2 Hence, option (a) is correct answer. 44  Mathematics–10

Example 3. The solution of given system of equations: x + y = a + b, ax – by = a2 – b2 is (a) x = 2a, y = b (b) x = a, y = 2b (c) x = a, y = b (d) x = 1 , y= 1 Solution. We have a b x + y = a + b ...(i) and ax – by = a2 – b2...(ii) From (i), we get x = a + b – y...(iii) Substituting this value of x in (ii), we get a(a + b – y) – by = a2 – b2 a2 + ab – ay – by = a2 – b2  ⇒  – (a + b)y = a2 – b2 – a2 – ab ⇒ – (a + b) y = – b2 – ab  ⇒  – (a + b)y = – b(b + a) ⇒ y = −b(a + b) = b − (a + b) Substituting this value of y in (iii), we get x = a + b – b = a Thus, the solution of the given system is x = a, y = b. Hence, option (c) is correct answer. 11 31 Example 4. When 3x + 2y = 3 and –7x + 5y = 3 are solved by elimination method, we get (a) x = 5 , y= 111 9 , y= 160 −4 , y= 5 x = −7 , y = 170 19 37 (b) x = 85 (c) x = 71 28 (d) 87 87 27 Solution. We have 11 ...(i) 3x + 2y =   ⇒ 9x + 6y = 11 3 and 31 ...(ii) –7x + 5y =   ⇒ –21x + 15y = 31 ...(iii) 3 ...(iv) Multiplying equation (i) by 7 and equation (ii) by 3, we get 63x + 42y = 77 – 63x + 45y = 93 Adding (iii) and (iv), we get 87y = 170  ⇒  y = 170 87 Substituting this value of y in equation (iii), we get 170 2380 63x + 42 × 87 = 77  ⇒  63x + 29 = 77 63x = 77 − 2380 2233 − 2380 = 147 147 7 ⇒ 29 = 29 −   ⇒  x= − = − 29 × 63 87 29 Thus, the required solution is x = 7 and y = 170 − . 87 87 Hence, option (d) is correct answer. Example 5. Solving 3x – 5y – 4 = 0 and 9x = 2y + 7 by the elimination method, we get the values of x and y as 9 −5 11 15 17 16 (a) x = 13 , y = 13 (b) x = 24 , y = 23 (c) x = 25 , y = 9 (d) None of these Pair of Linear Equations in Two Variables  45

Solution. We have 3x – 5y – 4 = 0 ...(i) ...(ii) and 9x – 2y – 7 = 0 ...(iii) Multiplying (i) by 3, we get 9x – 15y – 12 = 0 Subtracting (iii) from (ii), we get 13y + 5 = 0  ⇒  y= − 5 13 Substituting this value of y in (i), we get 3x − 5  − 153 − 4 = 0  ⇒  3x + 25 − 4 = 0 13 ⇒ 3x − 27 = 0  ⇒  x = 9 13 13 Thus, the solution is x = 9, y = − 5. 13 13 Hence, option (a) is correct answer. Exercise 3.2 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 1. Solution of the simultaneous linear equations: 2x – y = – 1 and x + 2y = 3is y 2 6 2 3 (a) x = 2, y = – 3 (b) x = – 2, y = 3 (c) x = 2, y = 3 (d) x = – 2, y = – 3 2. The value of x satisfying both the equations 4x – 5 = y and 2x – y = 3, when y = –1 is (a) 1 (b) –1 (c) 2 (d) –2 3. Which of the following is not a solution of the pair of equations 3x – 2y = 4 and 6x – 4y = 8? (a) x = 2, y = 1 (b) x = 4, y = 4 (c) x = 6, y = 7 (d) x = 5, y = 3 Answer the questions (Q 4 to Q 9) using best suitable algebric method. 4. If 2x + 5y – 1 = 0, 2x + 3y – 3 = 0, then (a) x = 1, y = – 3 (b) x = 3, y = –1 (c) x = 2, y = 5 (d) x = 5, y = – 3 5. If x + 2y – 3 = 0, 3x – 2y + 7 = 0, then (a) x = –1, y = 2 (b) x = 1, y = 2 (c) x = 2, y = 3 (d) x = – 2, y = – 3 6. If 2x = 5y + 4, 3x – 2y + 16 = 0, then (a) x = 2, y = –2 (b) x = 3, y = –3 (c) x = 4, y = 5 (d) x = – 8, y = – 4 7. If 4 + 3y = 8; 6 − 4 y = − 5, then xx (a) x = 2, y = 2 (b) x = 1, y = –1 (c) x = 2, y = –2 (d) x = 3, y = – 3 8. If 2x + 3y = 11 and 2x – 4y = –24, then the value of ‘m’ for which y = mx + 3 is (a) 0 (b) 1 (c) –1 (d) – 2 9. If 2x + 3y = 11 and x – 2y = –12, then the value of ‘m’ for which y = mx + 3 is (a) 1 (b) –1 (c) 2 (d) – 2 46  Mathematics–10

Answer the questions (Q 10 & Q 11) by elimination method. 10. If 7(y + 3) – 2(x + 2) = 14, 4(y – 2) + 3(x – 3) = 2, then (a) x = 1, y = 4 (b) x = 3, y = 5 (c) x = 5, y = 1 (d) None of these 4 + 5y = 7; 3 + 4y = 5, then 11. If x x (a) x = 1 , y = –1 (b) x = 8, y = 3 (c) x = 4, y = 7 (d) x = 5, y = 9 3 Answer the following by the method of substitution. 12. If 3 – (x – 5) = y + 2, 2(x + y) = 4 – 3y, then (a) x = 13 , y = 9 (b) x = 7 , y = 5 (c) x = 4 , y = 9 (d) x = 26 , y = −8 4 10 16 8 9 12 3 3 Answers and Hints A. Multiple Choice Questions (MCQs) Substituting y = 5 in (iii), we get 1. (c) x = 2, y = 3 x = 11 − 3(5) 2 2. (a) 1 3. (d) x = 5, y = 3 fi x = –2 4. (b) x = 3, y = – 1 \\ x = –2  and  y = 5 5. (a) x = –1 and y = 2. Now, y = mx + 3 x + 2y = 3  ...(i) fi 5 = m(–2) + 3 3x – 2y = –7  ...(ii) Thus, m = –1 On adding the equations (i) and (ii), we get 9. (b) –1 4x = –4 10. (c) x = 5, y = 1 fi x = –1 7(y + 3) – 2(x + 2) = 14 Putting x = –1 in eq. (i), we get fi –2x + 7y + 3 = 0 ...(i) –1 + 2y = 3 and4 (y – 2) + 3(x – 3) = 2 fi 2y = 4 fi 3x + 4y – 19 = 0 ...(ii) fi       y = 2 Solving (i) and (ii) by elemination method, Hence, solution of the system is x = –1 and y = 2. x = 5, y = 1 6. (d) x = –8, y = –4 11. (a) x = 1 ,  y = –1 3 7. (a) x = 2,  y = 2 26 ,  y = −8 8. (c) –1 12. (d) x = 3 3 2x + 3y = 11  ...(i) ...(ii) 2x – 4y = –24 3 – (x – 5) = y + 2 11 − 3y  fi y + 2 – 8 + x = 0  2  From (i), x = ...(iii) fi x + y – 6 = 0 ...(i) 2(x + y) = 4 – 3y Substituting this value of x in (ii), we get fi 2x + 2y – 4 + 3y = 0 11 − 3 y  fi 2x + 5y – 4 = 0 ...(ii)  2  2 − 4 y = –24 Solving by elimination method, fi 11 – 7y = –24 fi  y = 5 x = 26  and y = − 8 33 Pair of Linear Equations in Two Variables  47

3. Solving a System of Equations Reducible to Simultaneous Linear Equations Now, we shall learn the method to find the solutions of such pairs of equations which are not linear but can be reduced to linear form by making some suitable substitutions. Consider following examples. Example 1. When 1 + 1 =2 and 1 + 1 = 13 are solved by reducing them to a pair of linear equations, we get 2x 3y 3x 2y 6 (a) x = 1 , y = 1 (b) x = 1 , y = 1 (c) x = 1 , y = 1 (d) x = 1 , y = 1 23 45 54 57 Solution. We have 1+1 =2 ...(i) and 1 + 1 = 13 ...(ii) 2x 3y 3x 2y 6 Let 1 =u and 1 = v , then the given equations (i) and (ii) become x y u + v = 2  ⇒ 3u + 2v = 12 ...(iii) 2 3 ...(iv) and u + v = 13   ⇒ 2u + 3v = 13 3 2 6 Solving (iii) and (iv), we get x = 1 ,y= 1 2 3 Hence, option (a) is the correct answer. Example 2. The value of x and y for the following system of equations: 2+ 3 =2 and 4 − 9 = − 1, respectively are x y xy (a) x = 4 , y = 9 (b) x = 5 , y = 10 (c) x = 8 , y = –9 (d) None of these Solution. We have 2+ 3 =2 ...(i) and 4 − 9 = − 1 ...(ii) x y xy Let 1 = u and 1 = v, then equations (i) and (ii) become ...(iii) xy ...(iv) 2u + 3v = 2 ...(v) and 4u – 9v = –1 Multiplying (iii) by 2, we get 4u + 6v = 4 Subtracting (iv) from (v), we get 5 1 15 3 15 v = 5  ⇒  v = = From (iii), we get 2u + 3  13 = 2  1 ⇒ 2u = 1 ⇒ u = 2 Now, 1 1 = 1   ⇒  x=2 ⇒x=4 u = 2   ⇒  x 2 48  Mathematics–10

and 1 1 1 ⇒  y =3⇒y=9 v =   ⇒  =  3 y3 Thus, the solution is x = 4, y = 9. Hence, option (a) is the correct answer Example 3. Given the following system of equations: 10 2 and 15 − 5 = − 2  + =4 x+y x−y x+y x−y Then which of the following is true? (a) x = 2 , y = 5 (b) x = 1 , y = 2 (c) x = 3 , y = 2 (d) None of these 10 2 =4 15 − 5 = − 2 ...(ii) Solution. We have + ...(i) and x+y x−y x+ y x− y Let x 1 y = u and 1 = v , then equations (i) and (ii) become + x− y 10u + 2v = 4 ...(iii) and 15u – 5v = –2 ...(iv) Solving (iii) and (iv), we get x = 3, y = 2. Hence, option (c) is the correct option 51 63 Example 4. If + = 2 and − =1 x −1 y−2 x −1 y−2 (a) x = 1 , y = 2 (b) x = 2 , y = 3 (c) x = 4 , y = 5 (d) x = 5 , y = 7 51 Solution. We have + y−2=2 ...(i) x −1 63 and − = 1 ...(ii) x −1 y−2 Let 1 =u and 1 x −1 y − 2 = v, the equations (i) and (ii) become 5u + v = 2 ...(iii) ...(iv) and 6u – 3v = 1 Solving (iii) and (iv), we get x = 4, y = 5 Hence, option (c) is the correct answer. Exercise 3.3 A. Multiple Choice Questions (MCQs) Choose the correct answer from the given options: 23 54 1. If x + y = 13 and x − y = –2, then x + y equals 1 (b) −1 (c) 5 (d) −5 (a) 66 6 6 Pair of Linear Equations in Two Variables  49