MARK SCHEME For The October/November 2009 Question

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONSInternational General Certificate of Secondary EducationMARK SCHEME for the October/November 2009 question paperfor the guidance of teachers0580 MATHEMATICS0580/04Paper 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 must be read in conjunction with the question papers and the report on theexamination. CIE will not enter into discussions or correspondence in connection with these mark schemes.CIE is publishing the mark schemes for the October/November 2009 question papers for most IGCSE,GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Levelsyllabuses.

Page 2Mark Scheme: Teachers’ versionIGCSE – October/November oeSCsoiwww1correct answer onlycorrect solution onlydependentfollow throughignore subsequent workingor equivalentSpecial Caseseen or impliedwithout wrong working(a) (i) 8.4(0)(ii)their (i) 100 oe2042 ft www2A1ft6B2(c)2.4 3 oe ( 3.6 seen)2or their (a) (i) 7 3M1ft their 8.4 5After 0 scored SC1 ft for20 their (i) 100 correctly given2058%orM1 for 9 or 8 (1 8 3) soiM1A12.40oe1.251.92 www2M1(a) (i) Reflection (M), x 1B1,B1(d)B1 for 1.2 or 3.6 seenor SC1 for figs 84 in answerM1(b)3 9 oe ( 2.25 seen)121.6(0) cao www32B2Implied by figs 192A1[11]If extra transformations given in part (a) thenzero scored(ii) Rotation (R)180(centre) (1, 0)B1B1B1Must be “rotation”.Allow half turn for 180.Allow other clear forms of (1, 0)(iii) Enlargement (E)(centre) (6, 4)(scale factor) 3B1B1B1Must be “enlargement”Allow other clear forms of (6, 4) e.g. vectorAccept 3 : 1 or 1 : 3(iv) Shear (H)y-axis invariant oeB1B1Must be “shear”Allow other explanation for invariant but not“parallel to”isw after y-axis invariant seen(factor) –1B1 UCLES 2009

Page 3(b) (i)(ii)3Mark Scheme: Teachers’ versionIGCSE – October/November 2009 1 0 0 1 0 1 1 ft 1 (a) (i) 1(ii)3(b) (i)230oe www2(ii) 6–12 and 12–6 and 7–11 and 11–7 soik 430(iii)(c)(d)16 15for k integerB1 for correct right–hand column in 2 by 2matrixB2ftFt only their factor in (a) (iv) provided not zeroB1ft for left-hand column in 2 by 2 matrixprovided shear factor is not zero 1 0 if not ftor SC1 for 1 1 [15]B1Penalty of –1 in question if any answers givenas decimals or percentages (to 3sf) alone, butisw cancelling/conversion after correct answer261B2M1 forM1Evidence of all pairs adding up to 18 but noextras e.g. 4/6 1/6Without seeing the first M, 4 1 oe scoresM1 56A126 155oe scores M1M1www2242 6 5 oe618oe cao304B2M2,oe www342 568oe30Paper04B1oe6Syllabus0580A1M1www226 their (b) (iii)A1B2 UCLES 2009M1 for (1 1 6 7 11 12 x) 7 6 orbetter[13]

Page 44Mark Scheme: Teachers’ versionIGCSE – October/November 2009(a) (i) Accurate triangle with 2 arcs seen,2 mm accuracy for lines AC and BCSyllabus0580Paper04B2SC1 if accurate but no arcs or one arc or if ACand BC are wrong way round with arcs(ii) Accurate bisector of angle ACB, 2 accuracy and both pairs of arcs shown(accept equidistant marks on edges for 1stset of arcs) must meet ABB2ftFt their triangleSC1ft if accurate but no/one pair of arcs orshort with arcsIn both (ii) and (iii) isw(iii) Accurate perpendicular bisector of AD2 mm accuracy at mid-point and 2 forright angle and shows both sets of arcs must meet ACB2ftft their D, which must be on ABSC1ft if accurate but no/one pair of arcs orshort with arcsB1Dependent on correct triangle, accuratebisectors of angle ACB and side AD with correctDM2(–5600/50400 or –14/126)Allow use of 7, 9 and 12M1 for correct implicit statementVerification using 96.4 scores M2 maxE1Accept M1A1(s 280), allow use of 7, 9 (31.3 )M2Allow use of 7, 12M1 for correct implicit statementSC2 for correct answer by other method(iv) Correct region shaded cao(b) (i)(cos C) 140 2 180 2 240 22 140 180oe– 0.111(1) or better or 96.37 to 96.38(ii) 0.5 140 180 sin (their 96.4) oe12521 to 12523 or 12 500 or 12520 caowww2(iii)(Sin B )140 sin(their 96.4)240oe35.4 or 35.42 to 35.44 cao www3A119but not a non-reduced fraction[15]5(a) (i) (x 3)(2x 5) – x(x 4) 59 oe2x2 6x 5x 15 –x2 – 4x 59 oex2 7x – 44 0(ii) (x 11)(x – 4)(iii) –11, 4 www ft(iv)(their ve root ) 32(their ve root ) 528.3 (00 ) ft www2tan oeM1A1E1Implies M1 (allow 11x for 6x 5x)Correct conclusion – no errors or omissionsB2SC1 any other (x a)(x b) wherea b – 44 or a b 7B1ftStrict ft dep on at least SC1 in (ii)allow recovery if new working seenM1A1ft UCLES 2009Could be alt trig methodoe M1 where trig function is explicitft one of their positive roots(27.4 (27.40 – 27.41) from x 11)

Page 5(b) (i)Mark Scheme: Teachers’ versionIGCSE – October/November 20092x 5x 3 oex 4xx2 4x 3x 12 2x2 5xx2 – 2x – 12 0(ii)Must be seen. Allow ratio or correct productsA1Correct expansion of brackets seen (allow 7x for4x 3x)Correct conclusion – no errors or omissions M1must be seenB1,B12(1)or (x – 1)² –12 – 1 (B1)and x – 1 13 (B1)– 2.61, 4.61 final answers www4B1,B1(iii) 26.4 (26.42 . to 26.44 .) ftPaper04M1E1 ( 2) ( 2) 2 4(1)( 12)Syllabus0580B1ftIn square root B1 for (–2)2 – 4(1)(–12) or betterp qp qIf in formor,rrB1 for – (–2) and 2(1) or betterIf B0, SC1 for –2.6 and 4.6 or both answerscorrect to 2 or more dps rot– 2.6055 , 4.6055 .ft 4 a positive root 8[16]6(a) (i) –16B1(ii) 18 to 19B1(b) (i) –4.3 to –4.2, 1.5 to 1.6B1,B1(ii) –4.5 to –4.4 , 1.3 to 1.4B1,B1(iii) –4.5 to –4.4 x 1.3 to 1.4 ft(c) 307B1ftB2oe isw conversionFt their (ii). Allow clear worded explanationsand condone Y signs2Accept 4 7 , 30/–7M1 for 30/7 oe fracts, isw conversion or for–30/7 oe soi(d)Ruled line passing within 2 mm of(–5, 30) and (2, 0)(e) (i) Ruled horizontal line through (–3, –27)(ii) y –27(f)B2B1 for ruled line parallel to g(x). By eye (21 to 25 to horizontal if in doubt) allow brokenlineB1No daylight, not chord (allow broken)B1Ruled lines x –3, x –2, y 40B1Long enough to be boundary of region – allowbroken or solid ruled linesRegion enclosed by lines x –3,x –2, y 40 and y g(x)B1Allow any clear indication[15] UCLES 2009

Page 67(a) (i)(ii)Mark Scheme: Teachers’ versionIGCSE – October/November 200960 π 2 24 oe36025.1 (25.12 to 25.14) www2M1A1Accept 8 π60 π 242 oe360301 or 302 or 301.4 to 301.7 www2M1A1Accept 96 π(b) (i) πd their (a) (i) oe4 (3.99 – 4.01) cao www213Paper04M1A1(ii) 242 – (their radius)223.7 (23.66 to 23.67) cao www2(iii)Syllabus0580 π (their r)2 (their h)M1A1Alt trig method for h explicitAccept 560 ,2 140 ,4 35M1Not for h 24A1394 – 398 cao www2(c) (i) 27WB1(ii) 4WB1If B0, B0 in (c), SC1 for 27 and 4 alone[12]8(a)5.5 t Y 6B1Condone poor notation(b)4.25, 4.75, 5.25, 5.75, 6.25, 6.75(2 4.25 7 4.75 8 5.25 18 5.75 10 6.25 5 6.75) ( 283.5)M1M1At least 5 correct mid-values seenfx where x is in the correct interval allow 50 or their5.67 www4 f(c) (i) 17, 15(ii) Rectangular bars of heights 11.3and 15Correct widths of 1.5 and 1 – no gaps(iii) 2.5 caoM1A1 one further slipDepend on second methodAfter M3 allow 5.7isw conversion to mins/secs and reference toclassesB1B1ftB1ftft their 17 divided by 1.5ft their 1511.3 plot between 11 and 12 include lines and15 to be touching the 15 lineB1B1 UCLES 2009[10]

Page 79(a)Mark Scheme: Teachers’ versionIGCSE – October/November 20093(m – 3) 4(m 4) –7 12M23m – 9 4m 16 –84A1–13 www4A1(b) (i) 0.5 oe(ii) 3( x 3) 2( x 1)( x 1)( x 3)x 11final answer( x 1)( x 3)(iii)(c)10 (a)(b)x( x 11) 1 ft or( x 1)( x 3)1(x – 1)(x 3) or better ftx 11 xSyllabus0580Paper04Allow all over 12 at this stageM1 for 3(m – 3) 4(m 4) seenAllow all over 12 at this stageMay be seen in stagesB1M1A1If brackets not seen allow3x 9 – 2x 2 as numerator with a correctdenominatorisw incorrect expansion of denominator ifcorrect brackets seenx2 11x x2 3x – x – 31 3 oe cso www3M1Must clear one denominator correctlyFt their (b)(ii) dep on fraction in (ii) with(x –1)(x 3) oe as denominatorDepend on previous M1A1– 0.33(33 )p(q – 1) t oepq t pt poe final answer www3pM1M1Multiplying by (q – 1)Ft their first stepe.g. pq only term on one sideFt their 2nd stepe.g. dividing by pttNote: q – 1 is M2 and then q 1 isppM1[13]21 23 25 27 29 12531 33 35 37 39 41 216B1B1CubesB1(c) (i) n oe(ii) n3 oeM1M1B1B1(d)42 – 4 1 13 wwwE1Allow 16 for 42, otherwise all must be seen(e)7 43 2 4 6 8 10 12B1All must be seen(f)n(n – 1) final answer oeB1(g)n(n2 – n 1) their (f)n3 – n2 n n2 – n n3M1E1All must be seen, no errors or omissions[10] UCLES 2009

0580 MATHEMATICS 0580/04 Paper 4 (Extended), maximum raw mark 130 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on wh