CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME f the October/November 013 series 0580 MATHEMATICS 0580/ Paper (Extended), maximum raw mark 70 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe 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 Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper IGCSE October/November 013 0580 Abbreviations cao crect answer only cso crect solution only dep dependent ft follow through after err isw igne subsequent wking equivalent SC Special Case www without wrong wking soi seen implied Qu. Answers Mark Part Marks 1 19% 0.719 5 0. 038 sin 11.4 1/5 B1 f decimals [0.19], [0.], 0.194, 0.197, 0.19 seen (a) 447 1 (b) 1 Or f four in crect der 3 15.7 15.70 to 15.71 M1 f π.5 8 4 160 M1 f 360 18 5 (a) 1 (b) Some possible answers: 1 6 [±] y 4 final answer M1 f first move completed crectly M1 f second move completed crectly on answer line 1 7 170 M1 f (1 + ) 10 8 3619 to 360 M1 f 1 4 3 π 1 better 3 9 decagon 3 M1 f 360 36 A1 f 10 10 10.1[0] 3 M1 f 1.3199 and 1.3401 seen and M1 f 500 1.3199 500 1.3401 f 500 (their highest their lowest) k 11 10 3 M1 f v = d A1 f k = 600 Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 1 p = 71.405 cao q = 73.105 cao 3 B1 f 8.45 and 8.55 seen M1 f their LB [π] their UB [π] If 0 sced, SC1 f one crect. 13 10[.00] 3 M f 1.90 and.90 and 5.0 only M1 f two of 1.90,.90, 5.0 in a list of three two values from the table 3.40 + 5.0 SC1 FOR 1.90,.90, 4.30 from 14 5 3 B f AOB = 104 B1 f OAB OBA = 38 15 (8, ) 3 M1 f crectly eliminating one variable A1 f x = 8 A1 f y = If 0 sced, SC f crect substitution and crect evaluation to find the other value. 16 x <6.8 4 B3 f 6.8 with wrong inequality equal as answer. Or M1 f first move completed crectly and M1 f second move completed crectly and M1 f third move completed crectly 11 5 17 (a) 6 30 SC1 f one crect row column (b) 1 6 8 4 1 6 1 B1 f k 4 1 a b B1 f 8 c d 18 (a) (1.5, 1.5) B1 f either codinate (b) y = 3x + 8 3 B f y = mx + 8 y = 3x + c 3x + 8 B1 f gradient ( m) = 3 and B1 f c = 8 If 0 sced, SC1 f 3 = their m 5 + c f = their m + c f 1.5 = their m 1.5 + c (c) Most common methods: Crectly substituting P (3, 17) into y = 3x + 8 Showing the gradient of AP BP = 3 Other methods possible. 1 Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 19 (a) a c M1 f BO = a c f any crect route crect unsimplified expression (b) a + c M1 f any crect route crect unsimplified expression (c) a c FT FT their (a) crect answer Or M1 f a crect non direct route from O to E f crect unsimplified expression f crect FT unsimplified 0 (a) 4.05 to 4. 1 (b).6 to.75 B1 f 9.6 seen (c).05 to.5 B1 f [UQ] 5.0 to 5.1 and [LQ].85 to.95 seen (d) 5 48 M1 f 5 1 (a) 37. 37.17 to 37.19 3 4 sin 65 M f sin[ ] = 6 4 6 M1 f = sin[] sin 65 (b) 11.7 11.7 to 11.74 3 M1 f [B =] 160 65 their (a) M1 f 1 4 6 sin their 77.8 Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME f the October/November 013 series 0580 MATHEMATICS 0580/3 Paper (Extended), maximum raw mark 70 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe 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 Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper IGCSE October/November 013 0580 3 Abbreviations cao crect answer only cso crect solution only dep dependent ft follow through after err isw igne subsequent wking equivalent SC Special Case www without wrong wking Qu. Answers Mark Part Marks 1 39 M1 f 5 45 60 Any two of (0, 8) ( 4, 0) (1, 4) B1 f one crect 1 3 8 M1 f x = 16 + x = 7. 5 better 4 tan 100, cos 100, 1/100, 100 0.1 B1 f decimals 0.1[[7..], 5.[67..], [0.01], 0.6[3..] f three in the crect der 5 (a) 600 000 1 (b) 79. M1 f 60 60 1000 6 5[.00] 3 100 M f 30 10 M1 f 30 associated with 10% e.g. 1.x = 30 7 5 3 M f (x 5)(x 1) M1 f evidence of a factisation which gives the crect cfficient of x positive prime constant term e.g. (x 7)(x + 1), (x 4)(x ), (x 3)(x 1) 8 1.6 3 M1 f m = kx 3 A1 f k = 5 9 (a) a + ab + b B1 f a [+] ab [+] ab [+] b better seen (b) 1 [] 10 160 3 M1 f sin 15 = 68 better Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 3 3 1 11 (a) 4 1 (b) 1 1 1 a b B1 f 10 4 B1 f 10 c d 1 k 4 3 1 (a) 7.5 10 3 6 M1 f 0.075 0.75 10 1 40 80 (b) 9.3 10 7 M1 f 93 000 000 93 10 6 0.93 10 8 13 (a) 4 M1 f MOC = 48 (b) 4 M1 f ACM = 66 B1 f 48 their (a) 14 (a) 8q 1 q 8 B1 f 8q k kq 1 (b) 1/5 0. M1 f 5 1, [0].04 seen 5 15 (a) Circle, radius 3 cm, centre A, not inside the rectangle (b) One line of symmetry with crect arcs. E.g.: M1 f arc full circle centre A radius 3 cm f an increct size circle at A outside rectangle B1 f crect ruled line (must reach cross two sides) B1 f pairs of intersecting arcs 1 16 (a) 8.61 8.609 to 8.610 4 M1 f 3 π sin10 M1 f 30 π 3 [ ] 360 M1 f area of triangle + sects (b) 430 431 430.4 to 430.41 1FT FT their (a) 50 Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 3 17 (a) triangle at (0, 3) (, 3) and (, 4) 3 B1 f each crect vertex If 0 sced then M1 f crect reflection in the y axis crect translation of their first stage 3 right up (b) reflection in y axis B1 f reflection B1 f y axis x = 0 18 (a) 19 19.1 1 (b) 3 M1 f 47 seen (c) 4.9 to 5.7 B1 f [UQ] 1.7 to. and [LQ] 16.5 to 16.8 (d) 45 B1 f 45 seen 50 5 SC1 f isw 50 19 (a) 75 B1 f [g(6) =] 36 (b) 3.5 6.5 3 M1 f (x + 3) = 100 M1 f x + 3 = [±]10 If 0 sced, SC1 f one crect value as answer (c) x 3 final answer M1 f x = y + 3 y 3 = x better y = x + 3 (d) 5 1 Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME f the October/November 013 series 0580 MATHEMATICS 0580/41 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 which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe 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 Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Abbreviations cao crect answer only cso crect solution only dep dependent ft follow through after err isw igne subsequent wking equivalent SC Special Case www without wrong wking art anything rounding to soi seen implied Qu Answers Mark Part Marks 1 (a) (i) cao 1 5 (ii) 3 : cao 1 (b) (i) 1. M1 f 86.38 8 1.56 (ii) 1.3 [0] nfww 3 M f 1.56 1. M1 f 1.56 = 10% soi (c) 33.6[0] M1 f (667 314.) 10.5 (a) 3 crect lines on grid (0, 0) to (40, 5) (40, 5) to (100, 5) (100, 5) to (10, 0) Allow good freehand SC1FT f lines crect, FT from an increct line (b) 5 1 40 (c) 3.75 4 M f 0.5 40 5 + 60 5 + 0.5 0 5 [450] M1 f evidence of a relevant area = distance and M1dep their area ( distance) 10 Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Qu Answers Mark Part Marks 3 (a) (i) 04 04. to 04.3 M1 f π 5 13 implied by answer in range 04.1 to 04.3 (ii) 1 cao 3 M f 13 5 states 5, 1, 13 triangle M1 f 13 = 5 + h better (iii) 314 314.1 to 314. M1 f 5 3 answer in range 314 to 314.3 FT FT their (a) (iii) 100 3 crectly evaluated and (iv) 3.14 10 4 3.141 to 3.14 10 4 given in standard fm to 3 sig figs better M1 FT f their (a) (iii) 100 3 SC1 f conversion of their m 3 into standard fm only if negative power (b) 138 138.3 to 138.5 4 10 π M3 f 360 6π π 5 13their (a)(i) 360 π 13 M f a crect fraction without 360 M1 f π 13 [81.6 to 81.8] seen π 13 [530.6 to 531.] seen 4 (a) 45.[0] 45.01 to 45.0 nfww 4 M f 55 + 70.55.70 cos 40 M1 f crect implicit equation A1 f 06.. (b) 84.9 84.90 to 84.9 4 B1 f angle BDC = 40 soi 70sin ( their 40) M f sin 3 M1 f crect implicit equation (c) (i) 4060 4063 to 4064 nfww 3 M f 1 1 ( 55 70 sin 40) + ( 70 ( b)sin (180 their 40 3) ) their M1 f crect method f one of the triangle areas (ii) 100 1015 to 1016 FT FT their (c) (i) 4 crectly evaluated M1 their (c) (i) figs 4 (d) 35.4 35.35 nfww distance M1 f sin 40 = better 55 1 f (55 70 sin 40) = (70 distance) better Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Qu Answers Mark Part Marks 5 (a) (i) Crect reflection to (4, 8) (, 9) (4, 9) SC1 f reflection in line x = 5 reflection in y = k Igne additional triangles (ii) Crect rotation to (4, ), (4, 3) (6, 3) (iii) Shear, x-axis invariant, [fact] SC1 f rotation 180 with increct centre Igne additional triangles 3 B1 each (independent) (iv) 1 0 1 FT FT their shear fact B1FT f one crect column row in by matrix but not identity matrix 1 0 SC1FT f 1 (b) (i) p + s final answer M1 f recognising OQ as position vect soi 1 (ii) s + p final answer 1 B1 f s + kp ks + p crect route (k 0) (c) parallel and OQ = SR 1 6 (a) (i) 1.4 to 1.6 1 (ii) 1.15 to 1.5 1 (iii) 1 1 (iv).5 to.1 0.9 to 0.75. to.35 3 B f crect B1 f one crect B1 f y = x drawn ruled to cut curve 3 times (b) (i) 15 B1 f [h(3) =] 8 seen M1 f 1 (x 1) better (ii) 1 x 1 x final answer M1 f x = 1 y x = 1 y better (iii), 3 M1 f x 1 = 3 better B1 f one answer (iv) 1 3 M f 8x = 1 8x 1 = 0 nfww 8 M1 f 1 (3x) [= x] Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Qu Answers Mark Part Marks 7 (a) 4.7 4.66 to 4.67 4 M1 f midpoints soi (condone 1 err omission) (5, 15, 5, 35, 45, 55) and M1 f use of fx with x in crect interval including both boundaries (condone 1 further err omission) and M1 (dependent on second M) f fx 10 (b) (i) 50, 90, 114 B1 f crect (ii) Crect curve ruled polygon 3 Igne section to left of t = 10 B1 f 6 crect hizontal plots and B1FT f 6 crect vertical plots If 0 sced SC1 f 5 out of 6 crect plots and B1FT f curve polygon through at least 5 of their points dep on an increasing curve/polygon that reaches 10 vertically (iii) 1.5 to 3 15 to 16.5 4 to 6 4 B1 B1 B B1 f 7 7.6 seen (c) (i) 50, 30 B1 each (ii) Crect histogram 3FT B1 f blocks of widths 0 0, 30 60 (no gaps) B1FT f block of height.5 their 50 0 and B1FT f block of height 1 their 30 30 Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Qu Answers Mark Part Marks 8 (a) ( 11) 4( 8)( 11) better p = ( 11), r = (8) better B1 B1 11 Seen anywhere f x 16 Must be in the fm p + r q p r q B1 f 11 11 + 8 16 + 11 16 0.67,.05 final answers B1B1 SC1 f 0.7 0.67 to 0.671 and.0.046 to.047 answers 0.67 and.05 (b) 13 3 M1 f y = k x x = ky A1 f k = 6 better f k = 0.1666 to 0.167 [k = 6 implies M1A1] (c) 0 with suppting algebraic wking 6 B f 14.5 + x = 19.5 0.5 x x 14.5 B1 f.5. 5 M1dep on B f first completed crect move to clear both fractions M1 f second completed crect move to collect terms in x to a single term M1 f third completed crect move to collect numeric term[s] leading to ax = b SC1 f 0 with no algebraic wking 9 (a) y = y = x y = 1 x + 5 (b) y y x 1 M1 f y = kx, k 0 gradient soi M1 f gradient ½ soi y = kx + 5 x + y = k k 0 If L and L 3 both crect but interchanged then SC3 y 1 x + 5 3 B1 f each crect inequality, allow in any der After 0 sced, SC1 f all inequalities reversed (c) (i) 4 [bushes], 3 [trees] M1 f any crect trial using integer codinates in region 30x + 00y = 70 seen (ii) [bushes], 4 [trees] 860 1 M1 f any crect trial using integer codinates in region Cambridge International Examinations 013
Page 7 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 41 Qu Answers Mark Part Marks 10 (a) (i) 1 + + 3 + 4 + 5 = 15 1 (ii) Crect substitution equating to sum ( + 1) e.g. = 3 and k = stated k with no errs seen M1 f using a value of n in ( + 1) e.g. = 3 k f a verification using k = ( + 1) e.g. = 3 ( n + ) n 1 k (iii) 1830 1 (iv) 30 M1 f n( n + 1) = 465 better (v) n 8 1 (b) (i) 5, 15 B1 either (ii) n ( n + ) 4 1 1 (iii) 36100 M1 f 19 ( 19 + 1) 4 190 Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME f the October/November 013 series 0580 MATHEMATICS 0580/4 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 which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe 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 Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 Abbreviations cao crect answer only cso crect solution only dep dependent ft follow through after err isw igne subsequent wking equivalent SC Special Case www without wrong wking art anything rounding to soi seen implied Crect answer Mark Part marks 1 (a) (i) 316 Final answer M1 f (18900 5500) 0.4 (ii) 1307 Final answer FT FT (18900 their (a)(i)) 1 crectly evaluated M1 f (18900 their (a)(i)) 1 (b) 4.5[%] nfww 19750.50 [ 18900] M1 f 100 18900 19750.50 18900 18900 (c) A by 31.05 31.04 to 31.05 31.[0] 31.1[0] 5 M1 f 1500 4.1/100 3 [+ 1500] M1 f 1500 1.033 3 [ 1500] A1 f 1684.5 184.5 1653[.45..] 153[.45..] and M1dep f subtraction of their amounts their interests (a) 36.9 36.86 to 36.87 M1 f tan[dbc] = 1.8/.4 (b) (i) 1.8² +.4² leading to 9 M1 f 1.8² +.4² better 6.46 + 3 8.6 (ii) [cosabd) =] 6.46 3 17 16.8 M A M1 f crect cos rule but implicit version A1 f 0.599 After 0 sced, SC nfww f answer 17 16.8 to 16.96 from other methods no wking shown (c) 39.6 39.7 39.59 to 39.68 3 M f ½ (.4 + 8.6) 1.8 4 1.8 Or M1 f (.4 + 8.6) soi by 9.9 to 9.9 Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 3 (a) 4x 7 10 final answer nfww 3 5(x 1) (3x + 1) M f 5 5(x 1) (3x + 1) 5 5 M1 f attempt to convert to common denominat of 10 multiple of 10 with one err in numerat (b) x² + 9 final answer nfww 4 B3 f 4x² 6x 6x + 9 3x² +1x crect answer given and then spoilt B1 f 4x² 6x 6x + 9 seen and B1 f 3x² +1x (3x² 1x) seen (c) (i) (x 1)(x + 3) isw solving M1 f (x + a)(x + b) where ab = 3 b + a = 5 with integers a and b (ii) x 1 x 1 ( x 3) x 6 final answer nfww 3 M f (x + 3)(x 3) (x 6)(x + 3) (x + 6)(x 3) seen M1 f (x² 9) seen 4 (a) (i) 90 (4/360 π 8 ) o.e. 3.836 to 3.837 M3 A1 M f 4/360 π 8 h = 90 M1 f 4/360 π 8 (ii) 131 130.75 to 130.9 nfww 5 M f 4/360 π 8 3.84 [.48 to.53] M1 f 4/360 π 8 soi [5.86 to 5.87] and M1 f (8 3.84) [61.37 to 61.44] and M1 f (4/360 π 8 ) [46.88 to 47] (b).4.416 to.419 3 M f 3.84 3 M1 f 3 3.84 3 h 3 =.5 90.5 3 90 90.5 3 3.84.5 h = 3 90 90 seen.5 Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 5 (a) 7, 11.5, 4.5 1,1,1 (b) Crect curve cao 5 B3FT f 10 crect plots, on crect vertical grid line and within crect mm square vertically Or BFT f 8 9 crect plots Or B1FT f 6 7 crect plots and B1 indep f two separate branches on either side of y-axis (c) (i) 0.69 < x < 0.81 1 (ii).3 < x <. 0.8 < x < 0.6 0.35 < x < 0.5 3 B1 f each crect After 0 sced, allow SC1 f drawing line y = 7.5 long enough to cross curve at least once (d) (i) y = 10 3x ruled crectly B long enough to cross curve twice. B1 f ruled line gradient 3 y intercept at 10 but not y = 10 Or B1 f crect but freehand 0.55 < x < 0.45 0.35 < x < 0.45 B1dep B1dep Dependent on at least B1 sced f line After 0 sced, SC f 0.5 and 0.4 [from solving equation] (ii) 10 1 10 1 3 B f x 10x [= 0] 1 Or B1 f 10 = 0 Crectly x x eliminating 3x Or B1 f x 3x 3 = 10x 3x 3 Crectly clearing fractions Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 6 (a) (i) 1 110 M1 f 1 1 11 10 (ii) 6 110 3 55 M1 f 3 11 10 (iii) 8 110 4 55 1 FT FT their (a)(ii) + crectly evaluated 11 10 1 M1 their (a)(ii) + 11 10 (b) (i) 6 990 1 165 M1 f 3 11 10 1 9 (ii) 336 56 990 165 M1 f 8 11 7 10 6 9 (iii) 198 990 1 5 5 3 8 1 9 M4 f 3 + 3 11 10 9 11 10 9 3 8 1 9 M3 f 3 3 11 10 9 11 10 9 Or 3 8 M1 f seen and M1 f 11 10 9 1 9 11 10 9 seen Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 7 (a) 14 10 10 pm final answer M1 f (0)8 10 answer 14 hours and 10 minutes answer 10 [am] (b) 5 hours 45 minutes cao M1 f 345 [mins] seen f 805 /7 3 5.75 seen (c) (i) 798 798. to 798.4. M1 f 1071 / 5 13 1071 13.4 60 (ii) 1.8 10 5 1.815 10 5 to 1.816 10 5 4 B3 f 18000 181500 to 181600 seen M f 1071000/59 M1 f figs 1071/figs 59 soi by figs 18 figs 1815 to 1816 and B1 FT f their number of litres crectly converted to standard fm rounded to 3sf better (d) 8600 3 M f 10148 1.18 M1 f 10148 associated with 118[%] 8 (a) (i) 6 1 (ii).75 M1 f [g(x) =] 0.5 7/14 7 7 Or + 5 x + 1 x + 1 (b) x 3 4 4 x 4 3 Final answer M1 f y 3 = 4x better x = 4y + 3 better 4 y = 4 3 + x flowchart with 3 then 4 (c) (i) 5 M1 f 4x = 3 3 x + 4 3 = 4 3 better (ii) x² + 5x 7 = 0 B1 May be implied by crect values in fmula 5 ± 5 4(1)( 7) (1) B1 B1 B1 f 5 4(1)( 7) better [53] p + q p q If in fm, B1 f 5 and r r (1) better No recovery of full line unless seen 1.14 and 6.14 final answers B1 B1 Or SC1 f 1.1 1.140. and 6.1 6.140 Or answers 1.14 and 6.14 Cambridge International Examinations 013
Page 7 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 4 9 (a) (i) Reflection x = B1 f either (ii) Translation 7 B1 f either (iii) Stretch x-axis invariant [fact] 3 3 B1 f each (b) (i) Triangle with cods at (8, ) (7, 3) and (7, 5) (ii) Triangle with cods at (, 5) ( 6, 5) and ( 8, 7) (iii) Triangle with cods at (1, 1) (4, 6) and (3, 5) B1 f rotation about (6, 0) but 90 anticlockwise Or f rotation 90 clockwise around any point B1 f crect points f enlargement of SF any centre B1 f crect points codinates of points shown (c) 1 0 1 B1 f one row one column crect but not identity matrix. 1 Or SC1 f 0 1 10 (a) 48 and 57, 9n + 3 1 B1 f 9n + k (b) 56 and 50, 86 6n 1 B1 f k 6n (c) 15 and 16, n 3 1 1 (d) 130 and n 3 + n 1 1FT FT their (c) + n dep on expression in n in (c) Cambridge International Examinations 013
CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME f the October/November 013 series 0580 MATHEMATICS 0580/43 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 which Examiners were instructed to award marks. It ds not indicate the details of the discussions that took place at an Examiners meeting befe 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 Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the October/November 013 series f most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 Abbreviations cao crect answer only cso crect solution only dep dependent ft follow through after err isw igne subsequent wking equivalent SC Special Case www without wrong wking art anything rounding to soi seen implied Qu. Answers Mark Part Marks 1 (a) (i) 45 M1 f 5 63 7 (ii) 0 M1 f 5 56 14 (iii) 3.4 3.38 to 3.41 3 13 4.9 48.8 M f 13 4.9 100 4.9 48.8 13 4.9 100 Or 13 4.9 48.8 48.8 M1 f 13 4.9 13 4.9 100 76.6[ ] (b) 18 4 Using fractions (percentages / decimals): 3 3 9 M1 f 4 8 = 37. 5 3 100 9 A1 f 3 8.15[%] M1 f 36 3 9 36 100 8.15 Partial percentages M1 f (Remaining) A1 f 96 75 M1 f 96 100 SC1 f 88 100 36 37.5 [= 96] Cambridge International Examinations 013
Page 3 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 6 sin1 (a) 119.94[ ] nfww 3 M f sin 6 AC 6 M1 f = sin1 sin 6 SC f crect answer from alternative methods (b) 109 108.7 to 108.8 nfww 4 M f 119.9.. + 55 119.9.. 55cos65 A1 f 1187[ ] 11834 to 11835[ ] M1 f implicit version (c) 1970 1969 to 1970.4 M1 f ½ 119.9.. 6 sin 3 (d) 300 310 to 30 3 M f (their (c) + 0.5 55 119.9.. sin65) 4.5 M1 f their (c) + 0.5 55 119.9.. sin65 3 (a) 9 x, 7 x B1 f each, accept in any der (b) x(9 x)(7 x) 4x 3 3x + 63x M1FT A1 Crect expansion and simplification with no errs (c) 4 0 B1 f each crect value (d) Crect curve 3 BFT f 5 crect plots B1FT f 3 4 crect plots (e) 0.65 to 0.75 x B1 f 0.65 to 0.75 seen (f) (i) 36 to 37 1 (ii) 1. to 1.4 1 4 (a) 48 and 84 66 and 66 B1 f each pair (b) 540 M1 f 3 180 ( 5 4) 90 5 (180 360 5) (c) 160 M1 f 7 360 their 540 360 (d) (i) x + 5 + 3y 0 + 4x 5 + x + y 10 = 360 1 Allow partial simplification but not 7x + 4y 30 = 360 (ii) x + 5 + 3y 0 = 180 1 (iii) [x =] 30, [y =] 45 nfww 4 M1 f crect multiplication M1 f crect elimination A1 x = 30 y = 45 (iv) 65, 115, 115, 65 1 Accept in any der If 0 sced SC1 f crect substitution to find the other variable Cambridge International Examinations 013
Page 4 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 5 (a) (i) 3.81 3.81 to 3.813 3h 49min nfww 4 M1 f midpoints soi (condone 1 err omission and M1 f use of fx with x in crect interval including both boundaries (condone 1 further err omission) and M1 (dep on nd M1) f fx 80 (305 80) (ii) Crect histogram 4 B1 f each crect block and B1 f crect widths (b) (i) (ii) 1 3 1,,, 5 4 4 4 18 9 nfww 0 10 1 B1 f both s in crect place 5 4 3 1 M FT f 1 their their 5 4 3 3 3 1 + their 5 4 5 4 3 + their 5 4 1 M1 FT f their their 5 4 3 1 their 5 4 3 + their 5 4 (iii) 7 3 3 3 [0.16] M1 f 15 5 5 5 6 (a) 39.7 to 330 3 M f ½π(1 + 8.75 3.5 ) M1 f ½π1 ½π8.75 ½π3.5 SC f answer 1318 to 130 (b) 970 967 to 969.[ ] 4 M3 f ½π(4 + 17.5 + 6.5) 35 + their (a) M f ½π(4 + 17.5 + 6.5) 35 M1 f ½π 4 ½π 17.5 ½π 6.5 SC3 f 3955 to 3960 dep on SC in (a) (c) 11.5 11.6 11.53 to 11.55 3FT M1 f their (a) 35 A1 f 11500 11530 to 11550 Cambridge International Examinations 013
Page 5 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 (d) (i) r h = 0 40 r 0 h = 1 40 Accept 0 : 40 = r : h leading to 40r = 0h [r = h/] 0 1 r 1 = and = 40 h (ii) 35.3 35.31 to 35.34 3 M f 3 their 11545 1 π M1 f their 11545 = their 11545 = their r 1 h π 3 h 1 π r r 3 7 (a) (i) 3 1.5 M1 f 14 ( 4) 8 ( 4) (ii) y = 3 3 x + B1 f y = their x + c o.e. y = mx +, m 0 3 SC1 f x + (iii) 1 1 18 (iv) 1.6 1.63[ ] M1 FT f their 1 + their 18 (b) (i) (a) 3b 4a 1 (b) 1 1 (6b 8a) simplified M1 f (1a + 6b) 4a AR = AO + OR 5 5 (c) 6a + 3b simplified 1 (ii) OR is parallel to OT 1 Dep on OT crect (iii) 9.5 M1 f 4 3 Cambridge International Examinations 013
Page 6 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 8 (a) ( s ut) t nfww 3 M1 f a crect rearrangement to isolate the a term and M1 f a crect multiplication by and M1 f a crect division by t (b) 36.75 cao 3 M f 15.5 +.5 8.5 B1 f two of 15.5,.5, 8.5 seen (c) (i) 16 5 better [3.] 1 (ii) 11. 4 M f ½(5 + 10)16 (= 80) M1 f appreciation of distance from area and M1 f their 80 5 (dep on M1) 9 (a) 15 18 3n + 3 3(n + 1) 6 10 5 36 (n + 1) 9 B f 15, 6, 5 B1 f two crect values B3 f 18, 10, 36 B1 f each crect value B f 3n + 3 M1 f 3n + k, f any k B f (n + 1) M1 f a quadratic expression (b) 14 M1 f (n + 1)(n + ) = 40 better 15 16 = 40 (c) (i) ½ + p + q = 9 1 (ii) [p = ] 3 11 [q = ] 5 B f 4p + q = 3 B1 f ½ 3 + p + q M1 f crect multiplication and subtraction of their equations 11 A1 f [p = ] 3 [q = ] If 0 sced then SC1 f either crect Cambridge International Examinations 013
Page 7 Mark Scheme Syllabus Paper IGCSE October/November 013 0580 43 10 (a) x x + 3 cao 3 B1 f (x + 3)(x 3) B1 f x(x 3) (b) 3 and 5 7 M f 15(x + 1) 0x = x(x + 1) M1 f multiplication by one denominat only 15( x + 1) 0x x( x + 1) and B f x + 7x 15 [= 0] B1 f 15x + 15 0x x + x and M f (x 3)(x + 5) their crect facts fmula M1 f (x + a)(x + b) where ab = 15 a + b = 7 A1 f x = 3 and 5 Cambridge International Examinations 013