Standardising Meeting Version MARK SCHEME

ADVANCEDGeneral Certificate of Education2019MathematicsAssessment Unit A2 1assessingPure Mathematics[AMT11]TUESDAY 28 MAY, MORNINGStandardising Meeting VersionMARKSCHEME11864.01 F

GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICSIntroductionThe mark scheme normally provides the most popular solution to each question. Other solutions givenby candidates are evaluated and credit given as appropriate; these alternative methods are not usuallyillustrated in the published mark scheme.The marks awarded for each question are shown in the right-hand column and they are prefixed by theletters M, W and MW as appropriate. The key to the mark scheme is given below:Mindicates marks for correct method.Windicates marks for working.MW indicates marks for combined method and working.The solution to a question gains marks for correct method and marks for an accurate working based onthis method. Where the method is not correct no marks can be given.A later part of a question may require a candidate to use an answer obtained from an earlier part of thesame question. A candidate who gets the wrong answer to the earlier part and goes on to the later part isnaturally unaware that the wrong data is being used and is actually undertaking the solution of a parallelproblem from the point at which the error occurred. If such a candidate continues to apply correct method,then the candidate’s individual working must be followed through from the error. If no further errors aremade, then the candidate is penalised only for the initial error. Solutions containing two or more workingor transcription errors are treated in the same way. This process is usually referred to as “follow-throughmarking” and allows a candidate to gain credit for that part of a solution which follows a working ortranscription error.Positive marking:It is our intention to reward candidates for any demonstration of relevant knowledge, skills orunderstanding. For this reason we adopt a policy of following through their answers, that is, havingpenalised a candidate for an error, we mark the succeeding parts of the question using the candidate’svalue or answers and award marks accordingly.Some common examples of this occur in the following cases:(a) a numerical error in one entry in a table of values might lead to several answers being incorrect, butthese might not be essentially separate errors;(b) readings taken from candidates’ inaccurate graphs may not agree with the answers expected butmight be consistent with the graphs drawn.When the candidate misreads a question in such a way as to make the question easier only a proportion ofthe marks will be available (based on the professional judgement of the examining team).11864.01 F2[Turn over

1x3 3y2 11Differentiate to give 3x2 6ydyx2 dx – 2y2t AVAILABLEMARKSdy 0dxMW3MW1y3a4M1 W1y2 x a b 9a2 lM1 y2 9axW1Alternative Solutiony2 9a2t23M1 W1 y2 9a(at2)M1 y2 9axW1D60(i) tan θ 40 θ 0.98279 60Bθ404M1 W1W1C EBD π – 2 0.98279M1 1.17600 1.18 radiansW1M1(ii) BD 602 402 20 13W1Area of sector 1 20 13h2 1.17600.2M1 3057.6 Area of 2 triangles 2 W11 60 402M1 2400W1Total area 5457.6 cm2MW1 5460 cm2 (3sf)11864.01 F123

4(i) cosec 2θ – cot 2θ tan θAVAILABLEMARKS1 – cos 2θsin 2θ sin 2θ 1 – cos 2θsin 2θ MW2MW12 1 – (1–2 sin θ)sin 2θ 2 sin2 θ2 sin θ cos θ tan θ (ii) tanMW1 W1 MW1 W1πππ cosec – cot844M1 2 – 111864.01 FW149[Turn over

5(a) (i)f(x) – 8MW1(ii) y x2 – 8M1 x2 y 8MW1 x y 8MW1 f –1 : x x 8, x [ R, x – 8MW1AVAILABLEMARKSy(iii)33OxMW1 W1(iv) gf: x (x2 – 8) –3 M1 gf: x x2 – 11 , x [ R, x 0(b) (i)y2MW1W1Q'P'1O(ii)x1y3Q′′11864.01 FW1MW1W1P′′x3513

6(i) 8 sin x 15 cos x R(sin x cos α cos x sin α) R cos α 8 and R sin α 15 tan α MW1AVAILABLEMARKSM115 8 α W1 Also R 82 152 M1 R 17W1(ii) Function can be re-written as1817 sin (x α) 23M1Maximum occurs when denominator is minimum.This occurs when sin (x α) – 1Maximum value M118–17 23 3W1 x α 270 x 208 MW1(Alternative answer x – 152 )11864.01 F106[Turn over

AVAILABLEMARKS7(i)xx2(x 3)(x – 1)MW124 0.85MW12.525 0.7575 .3333 0.754W1 1 2M1 0.766W11x2x2 2x – 3–2x 3(ii)x2 2x – 3 M1W1x2–2x 3 1 (x 3)(x – 1)(x 3)(x – 1)W1–2x 3AB (x 3)(x – 1) x 3 x – 1M1 –2x 3 A(x –1) B(x 3)MW1Let x 1 B 1 B 4 W1Let x –3 9 – 4A A –943 b1–2 94x 3MW1 14x–1l dx [x – 94 ln x 3 14 ln x – 1 ] 32MW3 [3 – 94 ln 6 14 ln 2] – [2 – 94 ln 5 14 ln 1] 1 – 94 ln 6 94 ln 5 14 ln 2W1 0.763 (3 sf)(iii) Use more strips (or smaller intervals) to improve the approximation11864.01 F7MW118

8(i)dP kPdt1 dP PAVAILABLEMARKSkdtM1 W1 ln P kt cMW1When t 0, P P0M1 ln P0 cW1 ln P kt ln P0P ln ktP0 P P0 ektMW1(ii) P P0 ekt P0 P0 e5kM1 e5k 2W11ln 25W1Alternative solutionWhen t 5, P 2P0M1 k ln (2P0) 5k ln P0 ln 2 5k1 k ln 25W1W1(iii) When P 3P0M11MW1 3P0 P0 e(5ln 2)t1 e(5ln 2)t 3ln3 t 15ln 2 t 7.92Time is 8 yearsW1Alternative solutionWhen P 3P01 ln(3P0) a 5 ln 2kt ln P0ln3 t 1ln25M1MW1 t 7.92Time is 8 yearsW1(iv) Population cannot grow indefinitely since the number of people wholive in each house (and the number of houses) is finite.11864.01 F8MW113[Turn over

9(i)y (x – 5) ln x dy (x – 5) 1 ln xdxx dy 1 – 5 ln xdxx(ii) f (x) 1 –AVAILABLEMARKSM1 W2MW15 ln xxf (2) – 0.80685 M1f (3) 0.43194 W1Since the gradient of y (x – 5) ln x has a change of sign betweenx 2 and x 3, and is continuous in this region, then the curve hasa turning point between x 2 and x 35 ln xx5 1 f ʹ(x) 2 xxMW1(iii) f (x) 1 –M1 W15 x1 2.4 – 1 – 2.4 ln 2.4 j51 2.42 2.4M1 W1x1 2.56 (3 sf)11864.01 FMW1912

–1x 2 ln x dx10 (a)1u ln x–1dv x 2dxdu 1 dx xv 2 x211 2 x 2 ln x – 1 2 x 2 dxx1–12 2 x 2 ln x – 2 x1 5 1MW1u2 x2 4 2udu 2xdxM1u dx duxW1 x 5 u 3x 032MW1 u 23(u2 – 4)2u 1 duu2(u – 4)2 (u23MW2dxx3dx x2 4 AVAILABLEMARKSMW2 2 x 2 ln x – 4 x 2 c(b)M1 W12M1 W1– 4)duMW13 9 1 u3 – 4uC32MW1 99 – 12C – 9 8 – 8C3 211864.01 F13MW11015[Turn over

11 (i) sin 2x cos 2xM1 tan 2x 1AVAILABLEMARKSMW1π 5π x 4 4π 5π x 8 8W1W1(ii) Hence area is given by5π8π(sin 2x – cos 2x) dxM1 W1 MW185π8 :– 1 cos 2x – 1 sin 2xCπ228 ;MW21111–D D – ;– 2 2 2 2 2 MW1y(iii)2O12 xy2 dxM1 (4 – x ) dxW2 π :4x – 1 x3C3 01 π :4 – C311π 3MW2V π π120111864.01 FMW11116

12 (a) (i)Sn a (a d ) (l – d ) lM1AVAILABLEMARKSAlso Sn l (l – d ) (a d ) a Sn (a l ) (a l ) (a l ) (a l ) M1W1 Sn n(a l )1 Sn 2 n(a l )MW11(ii) 2 n(7 79) 1075 M1 W186n 2150 n 25MW1(iii) 7 24d 79M1 W1 24d 72 d 3MW1(b) (i) Year 1: 400Year 2: 400 1.02 400M1 W1Year 3: 400 1.022 400 1.02 400Hence he has 1,224.16W1(ii) Year n: 400 1.02n–1 400 1.02n–2 400 1.02 400This is a GP with a 400, r 1.02 Sn MW1W1 W1400(1.02n–1)1.02 – 1M1 W1 Sn 20 000 (1.02n–1)MW1(iii) 20 000 (1.02n–1) 7 000 1.02n – 1 0.35 1.02n 1.35M1W1 n ln 1.02 ln 1.35 n 15.2Hence it will take 16 years to exceed 7,00011864.01 FMW112M1W124Total150[Turn over

Pure Mathematics [AMT11] TUESDAY 28 MAY, MORNING MARK SCHEME Standardising Meeting Version 11864.01 F. 2 11864.01 F [Turn over GCE ADVANCED/ADVANCED SUBSIDIARY (AS) MATHEMATICS Introduction The mark scheme normally provides the most popular solution to each question. Other solutions given by candidates are evaluated and credit given as appropriate; these alternative methods are not usually .