0606 ADDITIONAL MATHEMATICS

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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS Intenational Geneal Cetificate of Seconday Education MARK SCHEME fo the Octobe/Novembe 011 question pape fo the guidance of teaches 0606 ADDITIONAL MATHEMATICS 0606/11 Pape 1, 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. Cambidge will not ente into discussions o coespondence in connection with these mak schemes. Cambidge is publishing the mak schemes fo the Octobe/Novembe 011 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, 1, 0 means that the candidate can ean anything fom 0 to. Univesity of Cambidge Intenational Examinations 011

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 1 A penalty of MR 1 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 1, This is deducted fom A o B maks when essential woking is omitted. PA 1 S 1 EX 1 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. Univesity of Cambidge Intenational Examinations 011

4 Page Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (a) (i) 7 and 0 B fo each. (ii) and 15 B fo each. (b) 3 sets enclosed in a ectangle [] fo set P and set Q sepaate fo set R contained within set P ( ) f : a+ b= 8 fo substitution of a coect value of x f : a+ b= fo each coect equation (allow unsimplified) a= 33, b= 18, fo solution to obtain a and b f( 1) = 19 [6] on thei a and b 3 (i) Gadient m = lg c = 0.6 fo a valid attempt to obtain lg c fo attempt to deal with lg c c = 0.51 (ii) N = 0.51t on thei m and c (i) 6! = 70 (ii) 5! = 0 (iii) 5! = 80 (iv) Even fist and last:! () Odd fist and even last: x! (1) Total: 7! = 168 [3] Univesity of Cambidge Intenational Examinations 011

5 Page 5 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) v= cos t when t = 0, v = (ii) cos t = 0, t = t = ( 0.785) [] [] fo attempt to diffeentiate fo attempt to solve and deal with t (iii) when t =, x = when t = 0, x = 3 distance moved = 1 [] fo thei 3 (iv) a= sint when 3 t =, a= [] 6 (a) 5 = p + 3 tan fo use of, 5 1 p = 1 = ' p' + 3tan 3q tan 3q = 1 q = 1 fo use of thei p and (q, 1) (b) amplitude a = b = 5 When f = 11, x = 0, so c = 7 O when f = 3, x =, so c = 7 3 fo use of eithe max and x = 0, o min and x = 3 Univesity of Cambidge Intenational Examinations 011

6 Page 6 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) ( ) n n 1 3 = 5 5 fo coect tem n n n 30 = 0 o = 15 equating 3 d tem to 3 5 attempt to solve quadatic o ealising n that = 15 when n = 6 n = 6 (ii) nx + x + 5 x x tem : fo 1n (1.) fo nd tem 5 ( ) 0.18 n n (5.) fo 3 d tem = 5 e x x 8 (a) + e + 1dx fo expansion 0 e x x + e + x 0 fo each coect tem = 1.6, [6] fo coect use of limits y = 1 x+ 1 ( + c ) fo attempt to integate (b) ( ) 1 fo ( x + 1) 1 1 x + 1 fo ( ) 1 when y =.5, x =, c = 3 fo attempt to find c, must be fom integation y = 1 ( ) 1 x [5] fo c = 3 Univesity of Cambidge Intenational Examinations 011

7 Page 7 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe (i) cosec x = 8sin x fo use of coect identity o equivalent sin x = fo dealing with cosec o equivalent sin x = fo attempt to solve x = 30, 150, [5] Withhold last if exta solutions (ii) tan ( 0.3) 5 y =, fo attempt to get in tems of tan y 0.3 =.55, fo dealing with ode coectly y = 1.7,.8 (allow 1.8 and.85), [5] Univesity of Cambidge Intenational Examinations 011

8 Page 8 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe EITHER 1 1 (i) ( ) ( 3θ) θ = 5 fo use of secto aea fo attempt to equate aea to 5 1 θ = ( θ) P = θ fo use of ac length leading to 8 P = + fo attempt to get P in tems of and θ (answe given) [6] dp 8 (ii) = + d fo attempt to diffeentiate and equate to zeo. when d P d = 0, = P = 8 fo attempt to obtain P (iii) d P 16 d =, + ve minimum fo coect method and conclusion 3 when =, 1 θ = [] Univesity of Cambidge Intenational Examinations 011

9 Page 9 Mak Scheme: Teaches vesion Syllabus Pape IGCSE Octobe/Novembe OR (i) OC = 10 (ii) sinθ =, sinθ = OC 10 fo attempt to use sinθ leading to 10sinθ = 1 + sinθ [] fo coect attempt to simplify to given answe (iii) d 10cosθ = dθ 1 sin ( + θ ) fo coect attempt to diffeentiate a quotient A, 1, 0 1 each eo when =, sin θ =, cosθ = 3 fo attempt to find sin o cos fo substitution d 0 3 = (3.85) dθ 9 [6] (iv) d dt =, when θ =, d θ 3 3 = 6 d 0 dθ d dθ = fo coect use of ates of change dt dt d leading to d θ 3 3 = dt 10 (0.50) [3] Univesity of Cambidge Intenational Examinations 011

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

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