MARK SCHEME for the October/November 2014 series 0580 MATHEMATICS

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME f the October/November 04 series 050 MATHEMATICS 050/43 Paper 4 (Extended), maximum raw mark 30 This mark scheme is published as an aid to teachers cidates, 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 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 04 series f most Cambridge IGCSE, Cambridge International A AS Level components some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Examinations.

2 Page Mark Scheme Syllabus Paper Cambridge IGCSE October/November Abbreviations cao crect answer only dep dependent FT follow through after err isw igne subsequent wking equivalent SC Special Case nfww not from wrong wking soi seen implied Qu. Answers Mark Part Marks (a) (i) 5.37[...] M f [AD = ] better (ii) to M f tan [BCD =] (.6) B f 3.4 seen + 7 (iii) 65. M f 4. 7 (b) FT FT their (a)(iii) 4 crectly evaluated M f their (a)(iii) M f [scale fact =] soi (a) (i) 7 [=05] 7 [= 05] 6 (ii) to M f ( + + 7) [ 00] (b) final answer 5 B4 f [0] [area A total sales] B3 f 040 [area B] [area C] [area B + area C] B f 040 [area B] [area C] M f 736 [B tickets] M f 43 [C tickets] After 0 sced SC f answer of 96 SC f 3754 (A total sales) Cambridge International Examinations 04

3 Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (c) M f 0.95 M f 3534 associated with 95[%] 3 (a) (i) 5 Angles in same segment dep Accept same arc, same side of same chd (ii) 04 Angle at centre is twice angle at circumference (iii) 34 Angle between tangent radius = 90 Accept double, but not middle, edge Accept right angle, perpendicular (b) (i) 7.65 to M f cos56 M f crect implicit fmula A f 5.5 to 5.6 (ii) to M f [sinbec =] M f sin56 7 sin56 sin BEC 7 4 (a) (i) Ariven with comparable fm f both shown difference between the two fractions shown Accept probabilities changed to decimals percentages (to sf better) (ii) (iii) 6 3 M f M f + their (a)(ii) M f seen (b) (i) Completes tree diagram crectly 3 B f 5 values crect B f value crect (ii) M f Cambridge International Examinations 04

4 Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (iii) 5 (a) (i) (ii) (iii) M f their M f their identifies the 7 routes attempt to add 7 probabilities with at least 5 crect B f three crect elements 0 (iv) 3 B f either crect in this fm 5 (b) 3 M f understing to find the inverse of Q 0 M f det = f k k 0 0 Alternative a b 0 0 c d 0 Leading to a c = c = 0 then a = b d = d = then b = M all four equations, M f a pair of crect equations 6 (a) (i) x 3 final answer (ii) 5x 7 y 3 final answer M f elements crect (iii) 6x final answer M f 6x k kx Cambridge International Examinations 04

5 Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (b) [ ] better p = [ ]7 r = (3) B B f x Must see f 7 6 ± 7 6 p + r q p r q both 3.4,.5 cao BB After B0, SC f answer to.4 seen f 3.4,.5 seen f answer x + 5 (c) x nfww 5 + final answer x x 3 B f (x + 5)(x 5) B f x (x 5) 7 (a) sin to 6.53 M A [½ ] sin cos [½ ] (b) (i) 7.[0] 7.7 to M f 6.5/( π 6.5 ) x M f π better (ii)...4 to.7 3 M f π their(a) M f π (c) (i) π r r sin 30 π r r 4 r π 4 3 M A A M f 30 π r r sin 30 Dep on M A no errs seen (ii) to M f [r 5 =] ( π ) 4 3 M f one crect rearrangement step to r from r π = Cambridge International Examinations 04

6 Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (a) (i) (, ) + (ii) y = 3x cao final answer 3 4 M f gradient = 3 M f substituting (3, ) (, 4) into their y = 3x + c f finding y-intercept is (b) (i) (x + 5)(x ) isw solutions SC f (x + a)(x + b) where ab = 0 a + b = 3 (ii) [a =] 5 [b =] [c =] 0 3FT BFT f each of their 5 their from (b)(i) B f c = 0 (iii) x =.5 FT FT x = (their (a + b))/ (c) Inverted parabola B x-axis intercepts at 9 y-axis intercept at B B B f each After B0 allow SC f (9 x)( + x) (d) (i) p = 6 q = 43 3 B f (x + 6) 43 p = 6 q = 43 M f (x + 6) x + px + px + p M f 7 (their 6) p q = 7 p = (ii) 43 FT FT their q 9 (a) (i) p M f p 7. 7 better M f sum of two crect products better f [total =] p B f 5 + p = 7.7 (33 + p) (ii) 7 FT STRICT FT median f their p if integer (b) (i) 64 M f (ii) 40 M f (iii).6[0] FT FT / their (b)(ii) evaluated crectly to dp M f / their (b)(ii) Cambridge International Examinations 04

7 Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE October/November (c) 9.95 cao 5 B4 f answer (a) (i) 5x + 4 final answer M f 5x + k kx + 4 M f 5 to to 30 B f 3.5 used B f mins 5 secs used M indep f any crect conversion seen m to km (ii) 4. 3 M f 5x = 3 4 FT their expression in (a)(i) AFT f x = 3.6 (b) a 3b + 4 = 3.5 better 5a + 4b = better B B a 3b =.5 5a + 4b = 6.5 Equates cfficients of either a b 40a 5b = a + 3b = 0 3a b = 74 5a + b = 7.75 M rearranges one of their equations to make a b the subject 3 b +.5 e.g. a Adds subtracts to eliminate 47b = a = 5.75 [a =] 3.5 [b =].5 M A A Dep on previous method crectly substitutes into the second equation 5( 3b +.5) e.g. + 4b 6.5 After M0 sced SC f crect values with no wking f two values that satisfy one of their iginal equations Cambridge International Examinations 04

0580 MATHEMATICS. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers.

0580 MATHEMATICS. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME f the May/June 205 series 0580 MATHEMATICS 0580/2 Paper 2 (Extended), maximum raw mark