Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/ Paper May/June 06 MARK SCHEME Maximum Mark: 80 Published 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 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 Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 06 series for 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. This document consists of 7 printed pages. UCLES 06 [Turn over
PagP ge MarM k Schemee CamC mbridgee IGCSEE May/June 06 Syllabus 0606 Paper Abbreviations awrt caoo depp FTT isww oe rott SCC soii ww www answers whichh round too correctt answerr only dependent followw throughh afterr errorr ignore subsequent workw king or equivalent rounded or truncatedd Special Casee seenn orr impliedd without wrongg workingg QueQ estionn Answer Markss Guidancee x x M expandss and rearranges to formm a term quadratic criticall values and A not fromm wronw ng workingg x < x > A markk final inequality; ; A00 iff spurious attempt too combine e.g. < x < (a) B A C It must be clear howw the sets are nested (b) (i) h P Alloww { m, a, t, h, s } for P n(p n Q) = cao { t, h, s} (i) n ) lg logl 00 = =[(lg y) y ] or l lg 0 lgg 4 = [(lg y) lg g ] M One logg laww used correctlyy ) correctt completion too (lg isw A answer only does not scoree (iv)) [ logg ]6x r = [log ]600 r Condone baseb e missingg x = 0 only Cambridgee International Examinationss 066
Page Mark Scheme Syllabus Paper Cambridge IGCSE May/June 06 0606 4 (i) π isw [Area triangle ABC =] 0 sin their π oe M seen or implied by or 4.(0...) [Area sector = ] their π oe their60 or π 60 M seen or implied by 6π or.0(8...) or.09 Complete correct plan M e.g. their triangle (their sector) (a) π 4.0(...) or isw A Units not required ( 7 + ) 8 7 7 + and attempt to M multiply 6 + 40 oe A not from wrong working 4 + 0 A (b) q + 4q + soi 8 = q + oe M can be implied by 4 and 6 or 4 and 6 q = 4, 4 p = 6, 6 A all values 6 (i) 4( x + ) 9 B,,,0 one mark for each of p, q, r correct in a correctly formatted expression; allow correct equivalent values; If B0 then SC for ( x + ) 4 9 or SC for correct values seen in 4 x+ x 9 or incorrect format e.g. 4 x + 9 or for a correct completed square form of the original expression in a different but correct format. e.g. ( x + ) 9 Cambridge International Examinations 06
Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE May/June 06 0606 (, 9) BFT FT ( q, r) r < 0 for each correct coordinate 0 9 8 7 6 4 Correct symmetric W shape with cusps on x-axis y-intercept marked at only or coords indicated on graph x-intercepts marked at. and 0. only x-axis or coords indicated on graph or close by -. 0. 7 (i) (a) q p (b) q p or ( q p ) The points are collinear oe PQ is a (scalar) multiple of QR and they have a point in common. oe Condone PQ is parallel to QR and [ OR =] 4i j oe soi 4 + ( ) (=) M condone 4 + ; may be implied by correct answer or correct FT answer 4 i j oe A 8 (a) (i) 4 4 a + 4a b+ 6a b + 4ab + b final answer B,,0 each error/omission 6 ( x) x soi M Could be in full expansion 4 or 0.96 isw A Must be explicitly identified (b) nn ( n) n = soi leading to a 8 6 cubic or quadratic ( n n 8= 0) M Must attempt to expand and remove fractions Solves their quadratic [(n 6)(n + )] M must have come from a valid attempt [n =] 6 only, not from wrong working A Must be n if labelled Cambridge International Examinations 06
Page Mark Scheme Syllabus Paper Cambridge IGCSE May/June 06 0606 9 (a) a = b = 4 c = B for each correct value (b) (i) B,,, 0 - sinusoidal curve symmetrical about y-axis clear intent to have amplitude of cycles If not fully correct max B π, π, π 6 6, π, π, π cao B for any 4 correct 0 (a) (i) 4! or! oe M 48 A P or!! or 4 oe M 60 A (b) (i) 4 [!] oe M Correct first step implied by a correct product of two elements 4 A! or seen M 8 A (i) ( c) x x + isw + (9, 0) oe Not just x = 9 Substitute (, 9) into both lines Or solves simultaneously (6x = 7 x oe) to get x =, y = 9 = 9 and 7 = 9 Cambridge International Examinations 06
Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE May/June 06 0606 Area AOB = 9 9 oe ( 8 or 40.) M Uses their. May split into triangles (. and 7). May integrate. Must be a complete method. (iv) [ ] (9) (9) their [ 0] (= 4.) M lower limit may be omitted but must be correct if seen 8 4 their their M must be from genuine attempts at area 0 of triangle and area under curve 6. A (i) ( x ) ( x ) ( x ) dy = dx MA Allow slips in d u dx and d v but must dx be explicit. Allow ( x ) = x x+ isw ALT using 4x + 4 x + 4x y = x d y ( x)( 4x+ 4) ( x + 4x) = d x ( x ) d y dx = k( x) k = 6 isw M AFT M A FT on their derivative of term quadratic No additional terms Cambridge International Examinations 06
Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE May/June 06 0606 their = 0 and find a value for x M x 4x + 9 = 0 oe ( x ) (x )(x ) = 0 oe x = 0. and x =. A y = and y = A if A0 A0 then A for a correct (x, y) pair 6 ( 0.) > 0 therefore min when x = 0. oe 6 or ( 0.) x = 0. oe = 48 therefore min when 6 (0.) < 0 therefore max when x =. oe 6 or = 48 therefore max (0.) when x =. oe MA is possible from other methods Cambridge International Examinations 06