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PROBLEMS AND SOLUTIONS IN ENGINEERING …

PROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSINENGINEERINENGINEERINENGINEERI NENGINEERINENGINEERINGGGGG MATHEMATICMATHEMATICMATHEMATICMATHEMATIC MATHEMATICSSSSSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSINENGINEERINENGINEERINENGINEERI NENGINEERINENGINEERINGGGGG / 1st Year (I & II Semesters)(Volume-I)ByDr. , Ph. DDeputy Director-General (Retired),Indian Council of Medical ResearchAnsari Nagar, New DelhiFormerly Scientist,Defence Research and Development Organisation,New DelhiUNIVERSITY SCIENCE PRESS(An Imprint of Laxmi Publications Pvt. Ltd.)BANGALOREl CHENNAIl COCHINl GUWAHATIl HYDERABADJALANDHARl KOLKATAl LUCKNOWl MUMBAIl PATNARANCHI l NEW DELHIC opyright 2012 by Laxmi Publications Pvt. Ltd. All rights reserved. Nopart of this publication may be reproduced, stored in a retrieval system, ortransmitted in any form or by any means, electronic, mechanical, photocopying,recording or otherwise without the prior written permission of the by :UNIVERSITY SCIENCE PRESS(An Imprint of Laxmi Publications Pvt.)

ENGINEERING MATHEMATICS For B.E./B.Tech. 1st Year (I & II Semesters) (Volume-I) By Dr. T.C. GUPTA M.A., Ph. D Deputy Director-General (Retired), ... A α Alpha I ι Iota P ρ Rho B β Beta K κ Kappa ΣσSigma ΓγGamma ΛλLambda T τ Tau D δ Delta M µ Mu Y υ Upsilon E ε Epsilon N ν Nu ΦϕPhi

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Transcription of PROBLEMS AND SOLUTIONS IN ENGINEERING …

1 PROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSINENGINEERINENGINEERINENGINEERI NENGINEERINENGINEERINGGGGG MATHEMATICMATHEMATICMATHEMATICMATHEMATIC MATHEMATICSSSSSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSPROBLEMS AND SOLUTIONSINENGINEERINENGINEERINENGINEERI NENGINEERINENGINEERINGGGGG / 1st Year (I & II Semesters)(Volume-I)ByDr. , Ph. DDeputy Director-General (Retired),Indian Council of Medical ResearchAnsari Nagar, New DelhiFormerly Scientist,Defence Research and Development Organisation,New DelhiUNIVERSITY SCIENCE PRESS(An Imprint of Laxmi Publications Pvt. Ltd.)BANGALOREl CHENNAIl COCHINl GUWAHATIl HYDERABADJALANDHARl KOLKATAl LUCKNOWl MUMBAIl PATNARANCHI l NEW DELHIC opyright 2012 by Laxmi Publications Pvt. Ltd. All rights reserved. Nopart of this publication may be reproduced, stored in a retrieval system, ortransmitted in any form or by any means, electronic, mechanical, photocopying,recording or otherwise without the prior written permission of the by :UNIVERSITY SCIENCE PRESS(An Imprint of Laxmi Publications Pvt.)

2 Ltd.)113, Golden House, Daryaganj,New Delhi-110002 Phone : 011-43 53 25 00 Fax : 011-43 53 25 : ` Edition : 2007 ; Second Edition : 2012 OFFICES&Bangalore080-26 75 69 30&Kolkata033-22 27 43 84&Chennai044-24 34 47 26&Lucknow0522-220 99 16&Cochin0484-237 70 04, 405 13 03 &Mumbai022-24 91 54 15, 24 92 78 69&Guwahati0361-251 36 69, 251 38 81&Patna0612-230 00 97&Hyderabad040-24 65 23 33&Ranchi0651-221 47 64&Jalandhar0181-222 12 72 UPS-9659-550-PROB SOL ENGG MATH I&II-GUPT ypeset at : Goswami Associates, at : CONTENTSC haptersPages1. Infinite Matrices and Its Applications of Partial Applications of Single Multiple Vector Ordinary Differential Equations and Its Laplace Transforms and Its Partial Differential Applications of Partial Differential (v)PREFACEI have no words to express my gratitude towards my worthy students on account ofwhose keen interest and continuous suggestions this book is appearing in its present form asthe new Revised Second Edition (Part-I) Keeping in view the changes done by Some Universitiesin the Syllabus of First and Second Semesters, I have revised it thoroughly to make aComprehensive Book by rearranging some topics/chapters, adding new and important problemsand all the questions asked/set in the previous university response to the First Edition of this book (All the three Volumes for respectiveSemesters)

3 , has been overwhelming and very encouraging which amply indicates that thisbook has proved extremely useful and helpful to all the students of Engineeringcolleges and Institutes throughout the country. Obviously it has helped them to be betterequipped and more confident in solving the PROBLEMS asked in several university the PROBLEMS have been solved systematically and logically so that even an averagestudent can become familiar with the techniques to solve the mathematical problemsindependently. mathematics has always been a problematic subject for the students, hencethey have been depending and relying upon private tuitions and coaching academies. It ishoped that this book in its new form will provide utmost utility to its readers. Author(vii)SYMBOLSG reek AlphabetsA AlphaI IotaP RhoB BetaK Kappa Sigma Gamma LambdaT TauD DeltaM MuY UpsilonE EpsilonN Nu PhiZ Zeta XiX ChiH EtaO Omicron Psi Theta Pi Omega There existsVFor allMetric Weights and MeasuresLENGTHCAPACITY10 millimetres= 1 centimetre10 millilitres= 1 centilitre10 centimetres= 1 decimetre10 centilitres= 1 decilitre10 decimetres= 1 metre10 decilitres= 1 litre10 metres= 1 decametre10 litres= 1 dekalitre10 decametres= 1 hectometre10 decalitres= 1 hectolitre10 hectometres= 1 kilometre10 hectolitres= 1 kilolitreVOLUMEAREA1000 cubic centimetres = 1 centigram100 square metres= 1 are1000 cubic decimetres = 1 cubic metre100 ares= 1 hectare100 hectares= 1 square KilometreWEIGHTABBREVIATIONS10 milligrams= 1 centigramkilometrekmtonnet10 centigrams= 1 decigrammetremquintalq10 decigrams= 1 gramcentimetrecmkilogram kg10 grams= 1 decagrammillimetremmgramg10 dekagrams= 1 hectogramkilolitreklarea10

4 Hectograms= 1 kilogramlitrelhectare ha100 kilograms= 1 quintalmillilitremlcentiareca10 quintals= 1 metric ton (tonne)IMPORTANT DEFINITIONS AND FORMULAE1. Convergent, Divergent and Oscillating Sequences:A Sequence {an} is said to be convergent or divergent if Ltnna is finite or not example, consider the sequence 2311 1,,, ..222 Herean = 11,LtLt22nnnnna = = 0 which is finite. The sequence {an} is the sequences {n2} or { 2n}.Herean =n2or 2nLtnna = or . Both these sequences are a sequence {an} neither converges to a finite number nor diverges to + or , it is called anOscillatory sequences are of 2 types:(i) A bounded sequence which does not converge, is said to oscillate example, consider the sequence {( 1)n}.Here an = ( 1)n. It is a bounded sequence because there exist two real numbers k and K (k K)such that k an K n N.{an} = { 1, 1, 1, 1, 1, ..} 1 an 1 Now2 Ltnna =2Lt ( 1)nn = 121 Ltnna+ =21Lt ( 1)nn+ = 1 Thus Ltnna does not exist The sequence does not converge.

5 Hence this sequence oscil-lates finitely.(ii) An unbounded sequence which does not diverge, is said to oscillate : When we say Ltnna = l, it means2 Ltnna =21 Ltnna+ = Series2 PROBLEMS AND SOLUTIONS IN ENGINEERING MATHEMATICS2. Infinite Series: If {un} is a sequence of real numbers, then the expression u1 + u2 + .. + un + ..[ , the sum of the terms of the sequence, which are infinite in number] is called an infiniteseries, usually denoted by 1nnu = or more briefly, by Partial Sums: If un is an infinite series where the terms may be +ve or ve, then Sn = u1 + u2+ .. + un is called the nth partial sum of un. Thus, the nth partial sum of an infinite series is thesum of its first n terms. S1, S2, S3, .. are the first, second, third, .. partial sums of the series. Sincen N (set of natural numbers), {Sn} is a sequence called the sequence of partial sums of theinfinite series un. Therefore, to every infinite series un, there corresponds a sequence {Sn} of itspartial Behaviour of an Infinite Series: An infinite series un converges, diverges or oscillates (finitelyor infinitely) according as the sequence {Sn} of its partial sums converges, diverges or oscillates(finitely or infinitely).

6 5. Geometric Series: The geometric series 1 + x + x2 + x3 + .. to (i) converges if 1 < x < ,|x| < 1(ii) diverges if x 1(iii) oscillates finitely if x = 1(iv) oscillates infinitely if x < 16. Theorem: If a series un is convergent, thenLtnnu = , converse of the above theorem is not always true , the nth term may tend to zero asn even if the series is not Ltnnu = 0 un may or may not be Ltnnu 0 un is not positive term series either converges or diverges to + . are six different comparison tests which can be used to examine the nature of infiniteseries. These are described in detail in question number 18 of this General procedure for testing a series for convergence is given under question 127, dependingupon the type of series whether it is alternating, positive term series or a power series. an example of a monotonic increasing sequence which is (i) convergent (ii) PROBLEMSSol.(i) Consider the sequence 123.

7 ,, ..234+1nnSince <<<, the sequence is monotonic = 1, limlimlim1111nnnnnnannn ==+++ = 1, which is finite. The sequence is convergent.(ii) Consider the sequence 1, 2, 3, .., n, ..Since 1 < 2 < 3 < .. < n < .., the sequence is monotonic increasing,INFINITE SERIES3an = n, limlimnnnan = = The sequence diverges to + . an example of a monotonic decreasing sequence which is (i) convergent (ii) (i) Consider the sequence 1, 11 1,, .., ..23nSince ,23>>> the sequence is monotonic = 11, limlim0nnnann == The sequence converges to 0.(ii) Consider the sequence 1, 2, 3, .., n, ..Since 1 > 2 > 3 > .., the sequence is monotonic = n, limlim ()nnnan = = The sequence diverges to . the convergence of the sequence {an}, where(i)an = 2nn1+(ii)an = ++ ++(iii)an = 1nn+.Sol.(i) Here,an=21nn+ an + 1 an=221(1)11nnnn+ ++ + = 2222(1)( 1)( 22)(22)(1)nnnnnnn n++ ++++ +=22210(22)(1)nnnnn n +< ++ + an + 1 < an {an} is a decreasing sequenceAlso,an=201nnn> + {an} is bounded below by 0.

8 {an} is decreasing and bounded below, it is =221limlim0111nnnnnn ==++ The sequence {an} converges to zero.(ii) Here,an= ++ ++= sum of (n + 1) terms of a whose first term is 1 andcommon ratio is 13= 11113113n+ (1)1nnarSr = 4 PROBLEMS AND SOLUTIONS IN ENGINEERING mathematics =131123n+ Now,an + 1=21111 +++ ++ + an + 1 an=1103nn+> an + 1 > an n {an} is an increasing ,an=13131223nn+ < {an} is bounded above by 32. {an} is increasing and bounded above, it is =1313 3lim1(1 0)2223nn+ = = The sequence {an} converges to 32.(iii)an=1nn+ an + 1 an=21 101(1)nnnnnnn++ = < ++ an + 1<an n {an} is a decreasing ,an=1111nnnn+=+ > {an} is bounded below by 1. {an} is decreasing and bounded below, it is =1lim 11nn += The sequence {an} converges to is an infinite series ? When does it converge, diverge or oscillates (finitely or infinitely) ? {un} is a sequence of real numbers, then the expression u1 + u2 + u3 +.

9 + un + .. ( , thesum of the terms of the sequence, which are infinite in number) is called an infinite series. Theinfinite series u1 + u2 + .. + un + .. is denoted by 1nnu = or more briefly, by every infinite series un, there corresponds a sequence {Sn}, where Sn = u1 + u2 + u3 + .. + un iscalled the partial sum of its first n infinite series un converges, diverges or oscillates (finitely or infinitely) according as thesequence {Sn} of its partial sums converges, diverges or oscillates (finitely or infinitely)(i) Series nu is convergent if limnnS = finite.(ii) Series nu is divergent if limnnS =+ or (iii) Series nu oscillates finitely if {Sn} is bounded and neither converges nor diverges.(iv) Series nu oscillates infinitely if {Sn} is unbounded and neither converges nor whether the following series converges or otherwise, ..() n 1++++ + + ,un=111(1)1nnn n= ++Putting n = 1, 2, 3, ..n, we haveu1=112 , u2 = 1123 , u3 = 1134 u4=1145.

10 Un = 111nn +Adding,Sn=111n +limnnS = 1 0 = 1 {Sn} converges to 1 un converges to the convergence of the series 11(2)nnn =+ . ,un = 111111(2)2 2(2)22nnnnn n = = +++ = 11 1112112nnnn + +++ Putting n = 1, 2, 3, .. n, we obtainu1=111112223 + ,u2 = 111 11223 34 + u3=111 11234 45 + ,..un=11 1112112nnnn + +++ Adding,Sn=111112122nn + ++ limnnS =1110022 + =34, a finite quantity. the given sequence <Sn> converges to 34. Hence the given infinite series 1nnu = convergesto AND SOLUTIONS IN ENGINEERING that the series 1134nn = converges to thenSn=u1 + u2 + u3 + .. + un= ++ ++ = 3114314n = 3414n limnnS = 4[1 0][ if |x| < 1, then xn 0 as n ]= 4, a finite quantity. the sequence <Sn> converges to 4. Hence the given series 1nnu = converges to convergence or otherwise of the series, 12 + 22 + 32 + .. + n2 + .. + 22 + 32 + .. + n2 = (1)(21)6nnn++limnnS =+ <Sn> diverges to + the given series diverges to +.


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