0580 W12 Ms 42 - CIE Notes

CAMBRIDGE INTERNATIONAL EXAMINATIONSInternational General Certificate of Secondary EducationMARK SCHEME for the October/November 2012 series0580 MATHEMATICS0580/42Paper 4 (Extended), maximum raw mark 130This mark scheme is published as an aid to teachers and candidates, to indicate the requirements ofthe examination. It shows the basis on which Examiners were instructed to award marks. It does notindicate the details of the discussions that took place at an Examiners’ meeting before marking began,which would have considered the acceptability of alternative answers.Mark schemes should be read in conjunction with the question paper and the Principal ExaminerReport for Teachers.Cambridge will not enter into discussions about these mark schemes.Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE,GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Levelcomponents.

Page 2Mark SchemeIGCSE – October/November 2012Syllabus0580Paper42Abbreviationscaocorrect answer onlycsocorrect solution onlydepdependentftfollow through after erroriswignore subsequent workingoeor equivalentSCSpecial Casewww without wrong workingartanything rounding tosoiseen or impliedQu.1AnswersMark(a) (i) 5(ii) 108M1 for2M1 for 60 M1Correct equalising of weightse.g.M12[0]3[0]or A 3[0]2[0]or J 3 and A 2 or J 30 andA 2097 to 98 or 201[.39 ] and Ann48.9[4. ] and 48.2[0] and Annor 68[.16] to 68.[2] and 67[.13] andAnn4.88 to 4.9 and 4.82 and Annor 6.8[1.] to 6.82 and 6.7[1 ] andAnnwww3 15(5 3 1)2(b) Correct conversion of moneyJ 0.718 or A 0.718J Part Marks9oe5Correct conversion of moneysoi by 146.83[1] rounded or truncated to 3sf or134.26[1 ] rounded or truncated to 3 sf ifdone 1stCorrect equalising of weights or moneyAccept other methods that give a pair ofcomparable values for method and accuracymarksThis mark can be implied by values seencorrect to 3 sf or betterThe underlined values imply M1 for themoney conversionA2Or A1 for 97 to 98 or 201[.39 ]or a correct pair of values with wrong/noconclusion(c) 302 Final answer3M1 for 60 60 4 soi by 14400 or figs 6048or figs 3024and M1 for (1000 20) soiAnswer 302.4 implies M2(d) 13.6[0]3M2 for(e) 12115.3[0]oe1.125or M1 for 15.3[0] associated with 112.5% Cambridge International Examinations 2012

Page 32Mark SchemeIGCSE – October/November 2012(a) (i) [cosA ]322 642 4322 32 64M2A2(ii) 616 or 616.2 to 616.4 264 sin 55soi by7048.49 rounded or truncatedor x² –(73.41 to 73.42) x – 804 [ 0](b) [Sin ADC ]70 sin(125 their 48.5)sin 55or 642 702 – 2 64 70cos(125 –their 48.5)or solving their 3 term quadraticequation3www(a) (i) 2(2x 1)(x – 5) final answer(ii) –1/2oe , 5(b)Paper42M1 for correct implicit version432 32 2 64 2 2 32 64cos A37.00[.]228 or 228.0 to 228.1Syllabus0580[ ]7 ([ ]7) 2 4(2)( 10)2( 2)A1 for3271or 0.798 to 0.7994096M1 for ½ 32 64 sin 37 oeM2M1 for correct implicit version of sine rule orcosine rule with xM2M1 for implicit sine rule or cosine ruleor for one error in quadratic solutionIgnore negative solutionsA23A1 for 83.0 to 83.1B1 for 2(2x² – 9x – 5)and B1 for (2x 1) (x – 5)or SC2 for expansion of brackets gives 3correct terms e.g. (2x 1) (2x – 10)or (4x 2)(x – 5)or SC1 for expansion of brackets gives 2correct terms e.g. (2x – 1)(2x 10)or (4x – 2)(x – 4)1ftCorrect or ft their 2 bracketsB2B1 for([ ]7) 2 4(2)( 10) [ 129 ]If in formp qp qor,rrB1 for – – 7 and 2(2) or better–1.09 , 4.59final answersB1B1If B0, SC1 for –1.1 and 4.6 as final answers or–1.089. and 4.589. as final answersor – 1.09 and 4.59 seen Cambridge International Examinations 2012

Page 4(c)Mark SchemeIGCSE – October/November 2012 10 10or2(3 x 1)( x 2)3x 7 x 23Paper42M1 for 6(x – 2) – 2(3x – 1) or better.Allow recovery after missing bracket[s]as final answer4Syllabus0580and B1 for (3x – 1)(x – 2) as commondenominator seen (may be as two fractions)(a) (i) 1482B1 for tangent/radius 90 seen.May be on diagram(ii) 741ft(iii) 212ft their (a)(i) 2 dep on (a)(i) 180M1 for 360 – 90 – 143 – 32 – their (ii) oee.g. using quadrilateral AOCD(iv) 20.9 or 20.92 3M2 for 6 tan 74 oe or explicit sine ruleOr M1 for implicit version(b) (i) 512M1 for ABC 90 . May be on diagram.(ii) 562M1 for 39 17 or 180 – (73 their 51)or [AXB ] 180 – (39 17)(iii) Angle at centre twice oe angle atcircumference1(iv) 221(v) 68.3 or 68.27 to 68.293Allow326π as final answer15M2 for360 34 2π 12360or 2π 12 –or π 12 34 2π 12360180 34 2π 12360or M1 for use ofθ 2π 12360for θ multiples of 90 o Cambridge International Examinations 2012

Page 55Mark SchemeIGCSE – October/November 2012(a) 20, 60, 100, 140, 180, 220(6 20 10 60 28 100 76 140 22 180 16 220)Syllabus0580Paper42M1At least 5 correct mid - values soiM1 fm where m is in the correct interval, alloweither end of interval as m( 21640)allow one further slip 158or f137 or 136.9 to 137.0(b) (i) 16, 126M1Depend on second methodA1SC2 for 137 or better ww1, 1(ii) rectangular bar of height 0.2rectangular bar of height 1.05correct widths of 80 and 120with no gaps(c) 1351ft1ftStrict ft from their 16Strict ft from their 12613M2 for15 136 3 13015 3or M1 for 15 136 and 3 130[2040] and [390]6(a) 5.83 or 5.830 to 5.8312Allow 34 as final answerM1 for (3² ([–]5)²)(b) (i) Vector drawn from P to Q at(14, 3)1Must have arrow in correct direction(ii) Points at (8, 11) and (13, 14)1, 1(c) 3a – 2b2 7 (d) 6 1(e) (i) b – c oe1SC1 for points at (8, 5) and (3, 2)M1 for a – 3b 2a b or CD DE oeAllow mixtures of vector notation.1Allow unsimplified Cambridge International Examinations 2012

Page 6Mark SchemeIGCSE – October/November 2012(ii) MX MB BXM1 ¼ or ¾ used¾ c – ¼b or ¼ (3c – b) or7Syllabus0580Paper42Any order for the M marksFor a correct routeM13c b 4 4A2(a) (i) x 5A1 for ½ b ¾ (c – b) oeAny correct unsimplifiedAfter 0 scored SC2 for 2/3c –1/6bB1 for each correct inequalityPenalise the first occurrence only when strictinequalities usedy 8x y 148y ½ x oe4(ii) x 5 ruledy 8 ruledx y 14 ruledy ½ x ruledregion indicated11111depEach line long enough to be boundary ofregionCheck at interceptsCheck at (10, 5)Dependent on 4 lines correct(b) (i) 4802M1 for 20 x 45 y where x and y areintegers and (x, y) is in their quadrilateral(ii) 6, 81In correct order1reasonable tangent at correct point, nodaylight, or chord, crossing x-axis between 1.7,2.0 when extended if necessary2depDependent on correct tangent or close attemptat tangent at x 2.5(a) (i) Tangent drawn at x 2.5(ii) 1.55 to 2.2M1dep attempts y step / x stepwith correct scales(b) 1.42 to 1.45 and 2.8 to 2.82(c) (i) 4.4, 2.5, 1.51, 12B1 for 2 correct values Cambridge International Examinations 2012

Page 7Mark SchemeIGCSE – October/November 2012(ii) 6 correct points plottedcurve through all 6 points andcorrect shape(iii) 0.75 to 0.99P1ft for 4 or 5 correct plotsC1Smooth curve but last 3 points may be ruled.In absence of plot[s], allow curve to implyplot[s]1.6 to 1.712.6 to 2.712FS511Paper42P2ft1(a) (i)Syllabus0580Solutions may be in any orderB1 for 2 outside of circles in diagramor all three of 5, 11, 7 correctly placed72ft their 2 their 7(ii) 91ft(iii) 14111(iv) 251ftft their 11 from diagram / 252ftisw incorrect cancelling(v)427oe 100600ft their 7 from diagram for numeratorM1 fortheir 7 their (7 1) 2524After 0 scored, SC1 for Cambridge International Examinations 2012their 7 their (7) 2525

Page 8Mark SchemeIGCSE – October/November 2012Syllabus0580Paper42(b) (i)FS57G124FG544712B1 for any correct diagram with blanks orzeroswhereneededandlabelledunambiguouslyB1 for 4 in correct placeB1 for 12 in correct placeB1 for 5 and 7 in correct placeSS12F7G54(ii) 28101ft(a) (i) 201(ii) n – 4 oen 4 oen 6 oeAccept unsimplified2(iii) (n – 4)(n 4) – (n – 6)(n 6)M1n2 – 4n 4n – 16 – (n2 – 6n 6n– 36) or better20(b) (i) 24Correct or ft from their diagramB1 for two correctft from their algebraic expressions can beimplied by n2 – 4n 4n – 16 – (n2 – 6n 6n –36) or n2 – 16 – (n2 – 36)Must have a line of algebraE1With no errors or omission of brackets1 Cambridge International Examinations 2012

Page 9Mark SchemeIGCSE – October/November 2012(ii) (n – 5)(n 5) – (n – 7)(n 7)isw2Syllabus0580M1 for n – 5, n 5, n – 7, n 7 seenor n2 – 25 – (n2 – 49) iswor n2 – 25 – n2 49 isw(c) (11 23) – (9 25)253 – 225[ 28](d) 4t oeAllow algebraic solution fromE11(n – 6)(n 6) – (n – 8)(n 8)Accept unsimplifiede.g. n2 – (t – 1)2 – [n2 – (t 1)2](e) c 28 and d 3052Paper4211 Cambridge International Examinations 2012

0580 MATHEMATICS 0580/42 Paper 4 (Extended), maximum raw mark 130 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of . 21 2 M1 for 360 – 90 – 143