0606 ADDITIONAL MATHEMATICS 0606/01 Paper 1, maximum raw mark 80

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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Intenational Geneal Cetificate of Seconday Education MARK SCHEME fo the Octobe/Novembe 009 question pape fo the guidance of teaches 0606 ADDITIONAL MATHEMATICS 0606/0 Pape, maximum aw mak 80 This mak scheme is published as an aid to teaches and candidates, to indicate the equiements of the examination. It shows the basis on which Examines wee instucted to awad maks. It does not indicate the details of the discussions that took place at an Examines meeting befoe making began, which would have consideed the acceptability of altenative answes. Mak schemes must be ead in conjunction with the question papes and the epot on the examination. CIE will not ente into discussions o coespondence in connection with these mak schemes. CIE is publishing the mak schemes fo the Octobe/Novembe 009 question papes fo most IGCSE, GCE Advanced Level and Advanced Subsidiay Level syllabuses and some Odinay Level syllabuses.

2 Page Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe Mak Scheme Notes Maks ae of the following thee types: M A B Method mak, awaded fo a valid method applied to the poblem. Method maks ae not lost fo numeical eos, algebaic slips o eos in units. Howeve, it is not usually sufficient fo a candidate just to indicate an intention of using some method o just to quote a fomula; the fomula o idea must be applied to the specific poblem in hand, e.g. by substituting the elevant quantities into the fomula. Coect application of a fomula without the fomula being quoted obviously eans the M mak and in some cases an M mak can be implied fom a coect answe. Accuacy mak, awaded fo a coect answe o intemediate step coectly obtained. Accuacy maks cannot be given unless the associated method mak is eaned (o implied). Accuacy mak fo a coect esult o statement independent of method maks. When a pat of a question has two o moe method steps, the M maks ae geneally independent unless the scheme specifically says othewise; and similaly when thee ae seveal B maks allocated. The notation DM o DB (o dep*) is used to indicate that a paticula M o B mak is dependent on an ealie M o B (asteisked) mak in the scheme. When two o moe steps ae un togethe by the candidate, the ealie maks ae implied and full cedit is given. The symbol implies that the A o B mak indicated is allowed fo wok coectly following on fom peviously incoect esults. Othewise, A o B maks ae given fo coect wok only. A and B maks ae not given fo fotuitously coect answes o esults obtained fom incoect woking. Note: B o A means that the candidate can ean o 0. B,, 0 means that the candidate can ean anything fom 0 to. UCLES 009

3 Page 3 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe The following abbeviations may be used in a mak scheme o used on the scipts: AG BOD CAO ISW MR PA SOS Answe Given on the question pape (so exta checking is needed to ensue that the detailed woking leading to the esult is valid) Benefit of Doubt (allowed when the validity of a solution may not be absolutely clea) Coect Answe Only (emphasising that no follow though fom a pevious eo is allowed) Ignoe Subsequent Woking Misead Pematue Appoximation (esulting in basically coect wok that is insufficiently accuate) See Othe Solution (the candidate makes a bette attempt at the same question) Penalties MR A penalty of MR is deducted fom A o B maks when the data of a question o pat question ae genuinely misead and the object and difficulty of the question emain unalteed. In this case all A and B maks then become follow though maks. MR is not applied when the candidate miseads his own figues this is egaded as an eo in accuacy. OW, This is deducted fom A o B maks when essential woking is omitted. PA S EX This is deducted fom A o B maks in the case of pematue appoximation. Occasionally used fo pesistent slackness usually discussed at a meeting. Applied to A o B maks when exta solutions ae offeed to a paticula equation. Again, this is usually discussed at the meeting. UCLES 009

4 Page 4 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) a 3 7a + 7a + 6 = 0 leading to a 3 = 8, a = 3 (ii) = [] [] fo use of x = a and equated to zeo, maybe implied fo substitution of x = into thei expession o f(x) (i) (ii) = , [] B,, 0 [] fo each matix, must be in coect ode fo each eo 3 4(k + ) = 4(k + ) 4k + 3k = 0 fo use of b 4ac Coect quadatic expession leading to k = 4, fo coect attempt at solution fo both values 4 (3 3y) + 3y = 43 (o x (3 x) + = 43) 3 6(y 3y + ) = 0 (o (x 3x + 0) = 0) (y 7)(y 3) = 0 (o (x 5)(x 4) = y = 3 o x = o (o x = 4 o y = o 3 ) D, fo eliminating one vaiable fo coect quadatic D fo coect attempt at solving quadatic fo each coect pai 5 (i) ( 3 + ) + ( 3 ) = fo use of Pythagoas Use of decimals, A0 AC = [] (ii) tan A = fo coect atio ( 3 )( 3 ) ( 3 + )( 3 ) = 6 7, fo ationalising tem denominato UCLES 009

5 Page 5 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) 3x 0x 8 = 0 (3x + )(x 4) = 0 fo attempt to solve quadatic citical values 3, 4 fo citical values A = {x : 3 Y x Y 4} Follow though on thei citical values. (ii) B = {x : x [ 3} A B = {x : 3 Y x Y 4} [] fo values of x that define B. (bewae of fotuitous answes) 7 (i) 3 C 8 = 87, [] fo coect C notation (ii) 7 teaches, student : 6 6 teaches, students 7 C 6 6 C : 05 5 teaches, 3 students 7 C 5 6 C 3 : (i) When t = 0, N = 000 (ii) dn dt = 000ke kt when t = 0, k = 50 dn dt = 0 leading to [] D fo diffeentiation dn D fo use of = ±0 dt (iii) 500 = 000e kt t = 50ln leading to 34.7 mins fo attempt to fomulate equation using half life fo a coect attempt at solution (bewae of fotuitous answes) 9 (i) 0 ( x) 9, [] fo 0 and ( x) 9 fo povided ( x) 9 is pesent (ii) x + x ln x ISW (iii) x (sec (x + )) tan(x + ) x ISW fo attempt to diffeentiate a poduct. fo x all othe tems coect fo attempt to diffeentiate a quotient. fo diffeentiation of tan (x + ) all othe tems coect UCLES 009

6 Page 6 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) d y = 9x 4x + dx at P gad = 7 tangent y 3 = 7(x ) D fo diffeentiation fo gadient = 7 and y = 3 D fo attempt to find tangent equation. (ii) at Q, 7x 4 = 3x 3 x + x leading to 3x 3 x 5x + 4 = 0 (x )(3x + x 4) 0 (x )(3x + 4)(x ) = 0 leading to x = 3 4, y = 3 40 D D fo equating tangent and cuve equations fo ealising (x ) is a facto D attempt to factoise cubic D fo attempt to solve quadatic fo both sinθ (a) tan θ + cot θ = cosθ sin θ + cos θ = cosθ sinθ = cosθ sinθ = cos ec θ sec θ cosθ + sinθ fo attempt to obtain one faction fo use of an appopiate identity fo simplification Scheme follows fo altenative poofs (b) (i) tan x = 3sin x sin x = 3sin x cos x sin x 3sin x cos x = 0 leading to cos x =, sin x = 0 3 fo use of tan x = attempt to solve sin x cos x and coect x = 70.5, 89.5 and x = 80 on thei x = 70.5 fo x = 80 (ii) cot y + 3 cosecy = 0 (cosec y ) + 3 cosecy = 0 cosec y + 3 cosecy = 0 ( cosecy )(cosecy + ) = 0 7π π leading to sin y =, y =, 6 6 allow y = 3.67, 5.76, fo use of coect identity fo attempt to solve quadatic fo dealing with cosec/cot Scheme follows fo altenative solutions UCLES 009

7 Page 7 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe EITHER (i) π h = 000, leading to 000 h = π [] fo attempt to use volume (ii) A = πh + π leading to given answe A = π [] fo A = πh + π GIVEN ANSWER (iii) (iv) da 000 = 4π d da 000 when = 0, 4π = d leading to = 5.4 d A 4000 = 4π + 3 d + ve when = 5.4 so min value A min = 554 D fo attempt to diffeentiate given A all coect D fo solution = 0 fo second deivative method o gadient method fo minimum, must be fom coect wok OR (i) y = x + cos x d y = sin x dx d y when = 0, sin x = dx π 5π leading to x =, 5π (ii) Aea = x + cos x.dx π x = + sin x π = 5π π D D, [6], D fo attempt to diffeentiate D fo setting to 0 and attempt to solve D fo coect ode of opeations fo attempt to integate fo each tem coect D fo coect use of limits must be in adians (Tig tems cancel out) UCLES 009

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