MARK SCHEME for the October/November 2015 series 0580 MATHEMATICS

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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME f the October/November 05 series 0580 MATHEMATICS 0580/4 Paper 4 (Extended), maximum raw mark 0 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 does 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 05 series f most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Examinations.

2 Page Mark Scheme Syllabus Paper Abbreviations cao crect answer only dep dependent FT follow through after err isw igne subsequent wking oe equivalent SC Special Case nfww not from wrong wking soi seen implied Question Answer Mark Part marks (a) (i).9[0] M f.6 (ii) cao B f any crect unsimplified fraction 8 (iii) 4 M f oe M f associating 9 with (00 6.5)% (b) 09 cao B f 08.5 to 08.6 M f oe 00 (a) (i) Image at (, 5), (, 5), (, 7) SC f translation k k4 4 crect vertices plotted but not joined (ii) Image at (, ), (5, ), (5, 5) SC f a reflection in a hizontal line in the line x = crect vertices plotted but not joined (b) Rotation 80 oe Alt Enlargement SF (, 0) (, 0) Not as column vect (c) (i) Reflection y = x oe (ii) 0 SC f a crect row column 0 Cambridge International Examinations 05

3 Page Mark Scheme Syllabus Paper (a) 4 00 M f 0.5 (5 + 5) 0 oe M f 0.5 (5 + 5) oe (b) (i) 0.5 (5 + 0) 6 0 [= 9 800] M Dep on a valid method f obtaining the width of 0 cm B f 0.5 (5 + 5) oe (ii) FT FT f their (a) (c) hr 9 min 4 B f.65 [h] 99 mins 0 M f M f oe If zero sced then SC f figs 65 and B f converting their time (in hours) into hours and minutes (d).8.80 to M f π M f πr r = (e) [.0] M f Cambridge International Examinations 05

4 Page 4 Mark Scheme Syllabus Paper 4 (a).5, 0.5 B, B (b) Crect curve 5 B FT f 0 points BFT f 8 9 points BFT f 6 7 points and B independent f two branches (c).5 to.5 (d) (e) (i) x SC4 f crect curve but branches joined (ii) Ruled line with gradient through (0, ) and fit f purpose.5 to.5 cao FT SC f ruled line, with gradient through (0, ), but not y = FT their y = mx + c from (e)(i), if m 0 SCFT f ruled line either with crect gradient through (0, c), but not y = c 5 (a) nfww 4 M f cos 65 oe M f crect implicit cosine fmula A f to (b) M f M f 80sin = oe sin 40 sin M (c) FT FT 0 + their (b) (d) (i) 9 oe BFT 50 + their (b) f [i.e. f C from M] (ii) 650 M f 560 journey time Cambridge International Examinations 05

5 Page 5 Mark Scheme Syllabus Paper 6 (a) to 0.6 nfww 4 M f 55, 90, 0, 60 soi M f Σ fm with frequencies and each m in on a boundary of a crect interval 750, 700, 4400, 6400 (b) Crect histogram drawn with crect widths and heights,.5 and (no gaps) M dep on nd M f 60 B f each crect block If zero sced, SC f crect heights frequency densities (c) (d) (i) (ii) 40 oe oe M f oe M f + oe M f one of these products soi 7 (a) 8 nfww 4 B f 7x = 4 7x = 4. oe in fm ax = b final answer of 0.8 B f 6x + x 55 = 56 oe 6x + x [0.] 55 = [.]56 M f 6x + (x [0.0]5) = [.]56 (b) oe nfww 4 M f y ( y + ) oe ( y + )( y + ) oe and B f y + 6y = y + y + y + oe better B f (y + )( y + ) = y + y + y + soi Cambridge International Examinations 05

6 Page 6 Mark Scheme Syllabus Paper (c) 5 nfww 4 4[.]80 7[.]80 M f w w 4[.]80 7[.]80 M f = oe w w M f 480(w ) = 780( w ) oe ALT M f n ( w ) = 4[.]80 n(w ) = 7[.] 80 M f wn n = 7[.] 80 wn n = 9[.]60 oe M f 9 n = 80 oe better ALT M f n ( w ) = 4[.]80 n(w ) = 7[.] 80 4 [.]80 + n 7[.]80 + n M f = n n M f 9 n = 80 oe better (d) (i) u (u ) =.5 One further crect step leading to u u 5 = 0 with no errs M A First step must involve u (u ) (ii) ( u 5)( u + ) SC f ( u + a)( u + b) where ab = 5 a + b = [a, b integers] (iii) M f tan = their 5 5 their M f substituting their positive value of u into [u and] u 8 (a) (i) Angle A is common to both triangles oe ADB = ABC Third angle of triangles equal oe dep Accept DAB = CAB oe Dep on previous mark (ii) Similar (iii) 8.5 M f 6 = oe better BD (b) (i) 8 (ii) 8 (iii) 78 (iv) 6 Cambridge International Examinations 05

7 Page 7 Mark Scheme Syllabus Paper (c) 6 nfww 5 B4 f an equation in m that simplifies to 5 m = 80 B f each of of the listed angles expressed in terms of m, in it s simplest fm, stated labelled on diagram Angle PQO = m Angle QOR = m Angle OQR = m m Angle PQR = m 80 m (a) 8 Angle POR = 80 m 4m 60 6m Reflex angle POR = 60 4m 6m 80 + m (b) B f [g(0.5) =] soi M f better x (c) x + final answer M f x = y y + = x better y = x (d) 4x M f (x ) x = 4x x x+ (e) 4x 4x + 7 B f ( ) (f) x (g) g ( x ) = g( x) (h) fh(x) Cambridge International Examinations 05

8 Page 8 Mark Scheme Syllabus Paper 0 A, 0 B 7n + oe 9 0, n + 4 oe n + 7 C 6, 7 n + oe D 6, 486 SC f 7n + k kn + oe B f n + 4 oe n + 7 oe seen, but not in wrong position n oe n+ p SC f k [k, p integers] Accept n Cambridge International Examinations 05

Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education. Published

Cambridge Assessment International Education Cambridge International General Certificate of Secondary Education MATHEMATICS 0580/43 Paper 4 (Extended) 017 MARK SCHEME Maximum Mark: 130 Published This mark