A-level Mathematics. Paper 3 Mark scheme. Practice paper Set 1. Version 1.0

Transcription

1 A-level Mthemtics Pper 3 Mrk scheme Prctice pper Set 1 Version 1.0

2 Mrk schemes re prepred by the Led Assessment Writer nd considered, together with the relevnt questions, by pnel of subject techers. This mrk scheme hs been prepred for prctice ppers nd hs not, therefore, been through the process of stndrdising tht would tke plce for live ppers. Further copies of this mrk scheme re vilble from llboutmths.q.org.uk Copyright 017 AQA nd its licensors. All rights reserved. AQA retins the copyright on ll its publictions. However, registered schools/colleges for AQA re permitted to copy mteril from this booklet for their own internl use, with the following importnt exception: AQA cnnot give permission to schools/colleges to photocopy ny mteril tht is cknowledged to third prty even for internl use within the centre.

3 Mrk scheme instructions to exminers Generl The mrk scheme for ech question shows: the mrks vilble for ech prt of the question the totl mrks vilble for the question mrking instructions tht indicte when mrks should be wrded or withheld including the principle on which ech mrk is wrded. Informtion is included to help the exminer mke his or her judgement nd to delinete wht is creditworthy from tht not worthy of credit typicl solution. This response is one we expect to see frequently. However credit must be given on the bsis of the mrking instructions. If student uses method which is not explicitly covered by the mrking instructions the sme principles of mrking should be pplied. Credit should be given to ny vlid methods. Exminers should seek dvice from their senior exminer if in ny doubt. Key to mrk types M R A B E F mrk is for method mrk is for resoning mrk is dependent on M or m mrks nd is for ccurcy mrk is independent of M or m mrks nd is for method nd ccurcy mrk is for explntion follow through from previous incorrect result Key to mrk scheme bbrevitions CAO CSO ft their AWFW AWRT ACF AG SC OE NMS PI SCA sf dp correct nswer only correct solution only follow through from previous incorrect result Indictes tht credit cn be given from previous incorrect result nything which flls within nything which rounds to ny correct form nswer given specil cse or equivlent no method shown possibly implied substntilly correct pproch significnt figure(s) deciml plce(s) Ve rsion of 1

4 Exminers should consistently pply the following generl mrking principles No Method Shown Where the question specificlly requires prticulr method to be used, we must usully see evidence of use of this method for ny mrks to be wrded. Where the nswer cn be resonbly obtined without showing working nd it is very unlikely tht the correct nswer cn be obtined by using n incorrect method, we must wrd full mrks. However, the obvious penlty to cndidtes showing no working is tht incorrect nswers, however close, ern no mrks. Where question sks the cndidte to stte or write down result, no method need be shown for full mrks. Where the permitted clcultor hs functions which resonbly llow the solution of the question directly, the correct nswer without working erns full mrks, unless it is given to less thn the degree of ccurcy ccepted in the mrk scheme, when it gins no mrks. Otherwise we require evidence of correct method for ny mrks to be wrded. Digrms Digrms tht hve working on them should be treted like norml responses. If digrm hs been written on but the correct response is within the nswer spce, the work within the nswer spce should be mrked. Working on digrms tht contrdicts work within the nswer spce is not to be considered s choice but s working, nd is not, therefore, penlised. Work ersed or crossed out Ersed or crossed out work tht is still legible nd hs not been replced should be mrked. Ersed or crossed out work tht hs been replced cn be ignored. Choice When choice of nswers nd/or methods is given nd the student hs not clerly indicted which nswer they wnt to be mrked, only the lst complete ttempt should be wrded mrks. Ve rsion of 1

5 1 Circles correct nswer B1 y = ln( 5x+ ) Circles correct nswer AO. R () Writes the expression in form tht my be expnded Correctly expnds t lest three terms Obtins correct expnsion with ll terms simplified x 16 + x = x x 3 1 x ! 16 3! 16 3 x x x = x x x = Totl 3 3 (b) Sttes correct rnge of vlues AO.3 B1 x < 16 Ve rsion of 1

6 4 () Tkes logs of both sides Uses n pproprite log rule y = b log x y = log x ( b ) x log y = log + log b log y = log + xlog b Uses second pproprite log rule to complete demonstrtion of liner form. Only wrd if cler with no slips in lgebr AO.1 R1 If we let y = log y nd rewrite 10 y = ( log b) x+ log This hs the form Y = mx + c, which is liner reltionship Explins clerly how the derived expression gives liner reltionship. This could include comprison with the structure of y = mx + c or using substitution Y = log y 10 AO.4 E1 Totl 4 Ve rsion of 1

7 4 (b)(i) Clcultes correct vlues for log y 10 At lest 5 correct PI by correct points x y Plots points nd drws pproprite stright-line grph Totl 4 (b)(ii) Obtins verticl intercept c nd ttempts 10 c Condone AWFW 0.4 to 0.45 for intercept Or Grdient = 0.49 Intercept = 0.4 b 0.4 = 10 = = 10 = 1.8 Obtins grdient m nd ttempts 10 m Condone AWRT 0.5 for grdient Obtins correct vlue for Obtins correct vlue for b Totl 3 Ve rsion of 1

8 4 (c) Substitutes x = 6.5 into their model AO3.4 y = = Follow through their nd b Explins tht the model predicts lrger vlue, suggesting tht it would need to be modified AO3.b F Totl Ve rsion of 1

9 5 Trnsltes finding the re into definite integrl Solves y = 0 to find limits of integrtion AO3.1 AO. R1 y = x sin x x sin xx d u = x u' = x v' = sin x v= cos x Uses integrtion by prts correctly Obtins correct integrl fter first ppliction ( ) x sin xx d = x cos x x cos xdx = x cos x+ xcos xdx u = x u' = 1 v' = cos x v= sin x Uses integrtion by prts for second time correctly ( d ) I = x cos x+ xsin x sin x x = x cos x+ x sin x+ cos x All nottion including dx correct Obtins fully correct integrl must hve correct limits AO.5 B1 π re = x sin xdx 0 = x cos x+ x sin x+ cos x ( π π+ π π+ π) ( ) cos sin cos 0 cos0 0 sin0 cos0 = + + π 0 Substitutes their limits into their integrl ( π ) ( ) = = π 4 Completes rgument to show the required result AO.1 R1 Totl 9 Ve rsion of 1

10 6 Identifies nd clerly defines vribles AO3.1b B1 Let t = time in yers M = mss Sets up n exponentil model, including the initil mss nd constnt Uses given condition to determine the constnt Obtins correct constnt Uses their model to find t AO3.3 AO3.1b AO3.4 M = M e kt 0 When t = 10, M = 0.8M0 10k 0.8 = e 10k = ln0.8 1 k = ln k = when M = 0.1M t 0.1= e 1 t = ln0.1= Obtins correct number of yers ft their model AO3. F It will tke nother 93. yers Totl 6 Ve rsion of 1

11 7 () Sketches nd lbels C 1 correctly AO1. B1 Sketches nd lbels C correctly Lbels correct y-intercept Totl 3 Ve rsion of 1

12 7 (b) Uses correct method to find the correct re This could be integrting nd subtrcting Or Subtrcting then integrting Or Finding ¼ of the required re nd multiplying by 4 Integrtes one term correctly Obtins correct expression AO3.1 Obtins correct limits Substitutes their limits Obtins correct re through completely correct rgument AG AO.1 R1 A = x dx x = 4 3 = 4 3 = = 3 ALT 3 A= x x dx = x dx 3 = x x 3 0 ( ) ( ) 3 3 = = = = Totl 6 7 (c) Forms nd solves correct eqution = 7 Obtins correct vlue for CAO 3 = 7 = 9 Totl Ve rsion of 1

13 8 () Squres the terms of the sequence PI by stting correct first term nd rtio AO3.1, r, r,... initil sequence,( r),( r ),... terms squred, r, r 1st term, rtio r 4 Obtins/sttes correct first term nd rtio Forms n eqution using both sums to infinity AO3.1 = 1 r 1 r ( 1 r) = ( 1 r ) ( 1 r) = 1 ( r)( 1+ r) Assuming 0 = ( 1+ r) AG Since r < 1 1 r 0 Shows cler nd correct lgebric resoning to rech the required result. Fctorising must be seen AG AO.1 R1 Explins why cncelling 1 r nd is llowed AO.4 E1 Totl 5 8 (b) Uses 1< r < 1to strt to find n inequlity for Obtins correct set for AO.1 AO. 1< r < 1 0< r + 1< 0< ( r + 1) < 4 ( 04, ) Totl Ve rsion of 1

14 9 Ticks the correct box AO1. B1 Approximtely 95% of the vlues re within three stndrd devitions of the men 10 Ticks the correct box AO1. B1 The dt for pper is negtively skewed 11 () Gives two criticisms in context. Allow exmples tht illustrte the issues Any one of A, B or C AO.3 E1 A: Not every member will be llocted 3 digits becuse > 1000 B: Only rndom numbers with 3 different digits will be chosen so not every member cn be chosen. Any two of A, B or C AO.3 E1 C: Repeted numbers re possible so sme member could be selected twice. Condone incorrect extrs Totl 11 (b) Gives cler method to chieve simple rndom smple. The method suggested here uses the Rnint function on clcultor Any one of A, B or C oe AO3.5c E1 A: Allocte every member number from 1 to 350 B: Use rndom number genertor to choose rndom integer from 1 to 350 C: Continue until 0 different numbers hve been identified nd select the corresponding members Any two of A, B or C AO3.5c E1 Any three of A, B or C Condone incorrect extrs AO3.5c E1 Totl 3 Ve rsion of 1

15 1 Uses re of rectngles to find either totl number of crs or number of crs 30 mph correctly Number of crs = (10 5) + (5 15) + (5 0) + (5 10) + (5 5) = 400 Number crs 30mph = (5 10) + (5 5) = 175 Obtins correct probbility ccept ACF P(speed 30) 175 = 400 Totl 13 () Finds the probbilities for x = 1 to 3 using given f(x) t lest 1 correct Uses Σp i = 1 (t lest correct p i s) Obtins correct exct vlue for k CAO P( X = 1) = + = P( X = ) = P( X = 3) = k = = 79 Totl 3 13 (b) Correctly finds P(3 X 4) ACF B Ve rsion of 1

16 14 ()(i) Sttes wek/some/moderte (nothing stronger thn moderte) negtive correltion, condone no context. Allow no correltion AO.5 B1 Wek negtive correltion between purchsed quntities of brown bred nd purchsed quntities of white bred 14 ()(ii) Infers Jmes is incorrect AO.b E1 Jmes is incorrect becuse the sctter grph does not show trend over the time period of the LDS Uses their knowledge of the LDS to explin tht trend vries over time AO3.b de1 Trends vry over time Totl 14 (b) Clcultes relevnt rtio to give comprison of 014 nd 011 B1 Chnge = = Chnge rtio ( 13 11) = Clcultes rtio with rounding considered in both numertor nd denomintor AO.3 Using rounding ( ) ( ) ( ) Produces convincing rgument to show tht 11% is justified or sttes clcultion tht gives 11% AO3.1b 1 Min % = 100 = 8.70% Mx % 100 = 8.6% % is within this rnge so possible vlue nd hence justified. Totl 3 Ve rsion of 1

17 15 ()(i) Uses clcultor to find correct probbility AWRT B ()(ii) Uses clcultor to find correct probbility AWRT B (b) Identifies tht 30 is discrete vlue or T is continuous/norml distribution AO.4 E1 30 is discrete vlue but T is continuous distribution So P(T = 30) = 0 s T tkes n infinite number of possible vlues Expresses clerly the ide tht P(exct vlue) = 0 s there re n infinite number of possible vlues or re bove tht point would be zero (OE) AO.4 E1 Totl 15 (c) Sttes correct nswer (using inverse norml function on clcultor) 41. to 41.3 AWFW B1 y = 41. Ve rsion of 1

18 15 (d) Finds P(T < 0) from clcultor B1 P(T < 0) = As P(T < 0) is so smll the model could be vlid, lthough in relity it should be 0 Mkes pproprite comment supporting the model, compring the probbility found to zero AO3.5 E1 Totl 15 (e) Recognises the need to find P(30 8 T ) AO3.1b P( T 38) = Obtins correct probbility from clcultor (AWRT 0.683) X ~ B(9, 0.683) P(X 5)= Selects nd uses binomil model use of B(9, p) for their p seen AO3.3 Obtins correct vlue from clcultor (AWRT 0.879) Totl 4 Ve rsion of 1

19 16 () Sttes both hypotheses using correct lnguge AO.5 B1 H0: p = 0.7 H1: p 0.7 ( tiled test, 5% level) Sttes model used PI by or seen X ~ B(8, 0.7) (X = no. who rte pp s excellent) P(X 15) = Evlutes probbility by clcultor, AWRT Evlutes Binomil model by compring P(X 15) with 0.05 nd infers H 0 is ccepted/not rejected AO.b B1 As > 0.05 we do not reject H 0 There is insufficient evidence to suggest tht the proportion of customers who rte the pp s excellent hs chnged. Concludes correctly in context requires not sufficient or no oe AO3. E1 Totl 5 16 (b) Sttes why Derek thinks tht the vlue of p hs fllen to below 55% Sttes vlid reson why Derek s conclusion is incorrect AO.4 E % = 53.8% < 55% 8 AO.4 E1 Test only confirms tht the vlue of p is unlikely to hve chnged, not tht it hs fllen or fllen below specific vlue So Derek is incorrect Totl Ve rsion of 1

20 17 ()(i) Sttes clerly the mening of A Bin context oe AO3. B1 Ellie does not order nchos s strter nd does order chicken burrito s min course. 17 ()(ii) Sttes the event in terms of A nd B oe AO3.1b B1 A B 17 (b)(i) Uses P(B A) formul nd solves for P(A) ACF B1 P(B A) ( B A) P( A) P 9 = = 13 9 P(A) = 0 13 = = (b)(ii) Uses expression to find P(B) or shows probbilities on Venn digrm or Obtins correct vlue of P(B) Uses P(A B) formul nd chieves correct result ACF P( B) = P( A B) + P( A B) = + = P(A B) = 0 9 = = Totl 3 17 (c) Sttes no with vlid reson, some comprison of numericl vlues required AO.4 E1 P P 9 = = = ( A B) ( A) P( B) P(A B) P(A)P(B) so A nd B re not independent Ve rsion of 1

21 18 Sttes both hypotheses using correct terminology AO.5 B1 H 0 : µ = 8 H 1 : µ > 8 (1 tiled test, 1% level) Finds smple men B x = = Considers X N 8, P(X > 8.05) = < 0.01 we reject H 0 The evidence suggests tht the men sugr content of the br hs incresed Obtins correct probbility P(X > 8.05) Infers H 0 is rejected by comprison of probbility nd significnce level AO.b Concludes correctly in context, not too definitive AO3. E1 Totl 6 TOTAL 100 Ve rsion of 1