A-LEVEL Mathematics. Paper 3 Mark scheme. Specimen. Version 1.2

Transcription

1 A-LEVEL Mathematics Paper 3 Mark scheme Specimen Version.

2 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.org.uk Copyright 07 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

3 Mark scheme instructions to examiners Mark scheme instructions to examiners General The mark scheme for each question shows: the marks available for each part of the question the total marks available for the question marking instructions that indicate when marks should be awarded or withheld including the principle on which each mark is awarded. Information is included to help the examiner make his or her judgement and to delineate what is creditworthy from that not worthy of credit a typical solution. This response is one we expect to see frequently. However credit must be given on the basis of the marking instructions. If a student uses a method which is not explicitly covered by the marking instructions the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their senior examiner if in any doubt. Key to mark types M dm R A B E F mark is for method mark is dependent on one or more M marks and is for method mark is for reasoning mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation follow through from previous incorrect result Key to mark scheme abbreviations CAO CSO ft their AWFW AWRT ACF AG SC OE NMS PI SCA sf dp correct answer only correct solution only follow through from previous incorrect result Indicates that credit can be given from previous incorrect result anything which falls within anything which rounds to any correct form answer given special case or equivalent no method shown possibly implied substantially correct approach significant figure(s) decimal place(s)

4 Examiners should consistently apply the following general marking principles No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to students showing no working is that incorrect answers, however close, earn no marks. Where a question asks the student to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. Diagrams Diagrams that have working on them should be treated like normal responses. If a diagram has been written on but the correct response is within the answer space, the work within the answer space should be marked. Working on diagrams that contradicts work within the answer space is not to be considered as choice but as working, and is not, therefore, penalised. Work erased or crossed out Erased or crossed out work that is still legible and has not been replaced should be marked. Erased or crossed out work that has been replaced can be ignored. Choice When a choice of answers and/or methods is given and the student has not clearly indicated which answer they want to be marked, only the last complete attempt should be awarded marks.

5 Circles correct answer AO.b B 8 (a) Makes clear attempt to use the cosine rule Total Uses trig identity with their cos AO.a M AO3.a M cos cos 30 5 Constructs rigorous argument leading to correct result AG AO. R sin 5 Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips sin (AG) (b) Writes down correct angle AO.a B.64 (c) Uses their angle in r AO.a M A Correct area AO.b AF = 0.5 m FT use of incorrect obtuse angle provided both M marks awarded in part (a) and M awarded in (c) Total 6 5 of 8

6 3(a) Translates proportionality into a AO3.3 M da differential equation involving A dt da, A and a constant of da dt ka dt proportionality. da k dt A Separates variables AO.a M Integrates both of their sides AO.b AF ln A kt c correctly Constructs a rigorous AO. R mathematical argument that A e kt c A Be kt AG supports use of the given model. AG Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips. (b)(i) States correct value of B AO.b B B = 0.5 or B = 4 (b)(ii) Uses t = 0 and A = 0.5 to find k Finds correct value of k AO3.b AO.b M A When t = 0, A = e k Substitutes their k to get A in 0k ln AO.a M terms of t k Constructs rigorous and AO. R ln 0 convincing argument to show t t ln 0 A (e ) 0 A 4 Using correct notation throughout. AG t 0 A t 0 A AG (b)(iii) Uses the model to set up correct AO3.4 M equation and attempt to find t Finds correct value of t AO.b A t 0 π t = days (c) States any sensible and relevant limitation of the model that is specified in terms of the pond, area, weed, rate of change or time. (d) Any sensible and relevant refinement to the model that is specified in terms of the pond, area, weed, rate of change or time AO3.5b E Model predicts that the area of weed will increase without limit and this is not possible since the area of the pond is 4 AO3.5c E Introduce a limiting factor such as fish eating weed or rate of growth decreases as surface area covered Total 3 6 of 8

7 4 Selects a method of integration, AO3.a M dv 3 which could lead to a correct u ln x; x dx solution. Evidence of integration by 4 parts du x ; v dx x 4 OR an attempt at integration by 4 3 x x inspection. ln ( x) dx 4 4 Applies integration by parts formula correctly OR correctly differentiates an expression of the form Ax 4 ln x Obtains correct integral, condone missing limits. AO.b AO.b A A 4 4 x x ln ( x) = ln( 4) ln( ) ln so p q 4 6 Substitutes correct limits into their integral Obtains correct p and q FT use of incorrect integral provided both M marks have been awarded AO.a AO.b M AF ALT d ( x 4 lnx ) 4x 3 ln xx 4. dx x x x lnxd x ( x lnx ) = ln( 4) ln( ) ln p q 4 6 Total 5 7 of 8

8 Marking Instructions AO Marks Typical Solution 5(a) Uses binomial expansion, with at least two terms correct, may be un-simplified AO.a M ( 6x) 3 ( 6x) 6x Obtains correct simplified answer AO.b A 3 ( 6x) x4 x (b) Determines the correct value for x and substitutes this into their answer to part (a) AO3.a M x 0.03 Obtains correct approximation for their answer to part (a) AO.b AF (0.03) 4(0.03).0564 FT allowed only if M from part (a) and M from part (b) have been awarded (c) Explains the limitation of the expansion found in part (a) AO.4 E Although with reference to x x cannot be used since the expansion is only valid for x 6 Total 5 8 of 8

9 6 Uses partial fractions with linear denominators 6x A B 6x 7x axb cxd AO3.a M 6x A B 6x 7x 3x x A( x) B( 3x) 6x Obtains correct linear denominators AO.b B Obtains at least one numerator correct (using any valid method, eg equating coefficients or substitution of values) AO.b A x, A 5 so A5 3 3, B x 4 so B dx 3x x 5ln ( 3x ) 4ln ( x ) Obtains partial fractions completely correct AO.b A 5ln( 4 ) 4ln( 3) ( 5ln() 4ln()) 0ln( ) 4ln( 3 ) Integrates their partial fractions, must include logs pln( axb) qln ( cxd) AO.a M Their integral correct (ignore limits) AO.b AF Substitutes limits into their integral AO.a M Correct final answer in correct form CAO AO.b A Total 8 9 of 8

10 7(a) Finds nd derivative and sets up an AO3.a M dy x inequality xe dx Obtains correct first derivative AO.b A d y x Obtains second derivative correct AO.b AF e 4 x e from their first derivative dx x x e Deduces correct final inequality AO.a A 4x e 0 (could use set notation) 4x 0 (b) Uses trapezium rule AO.a M Trapezium rule entries all correct AO.b A Finds correct value AO.b A (c) References area being completely within concave section So AO.4 E x x 0.5 e x 0. d x (e 0.0 e (e e e )) , 0.5, area is completely within concave section Trapezia all fall completely underneath the curve therefore underestimate (only award this mark if previous E has been awarded) (d) Uses suitable rectangle to obtain over-estimate Explains that this rectangle lies above the curve Constructs rigorous mathematical argument about accuracy, which leads to required result Only award if they have a completely correct solution, which is clear, easy to follow and contains no slips. AO.4 E Hence trapezia lie below curve and give an under-estimate for the area AO3.a AO.4 B E Using a rectangle with the left hand edge the same height as the curve will produce an overestimate Area of rectangle = e AO. R A 0.40 So A = 0.4 to dp Total 0 of 8

11 8(a) Circles correct answer AO. B The 065 households in the village (b) Circles correct answer AO. B Simple random Total 9(a) Finds value for p AO.a M 680 p Finds correct probability from calculator AO.b A Using X ~ B(5,0.7), P(X = ) = (b) Explains the reason why the model may no longer apply in context AO3.5b E It is likely that all the houses (and gardens) will be of similar types, and hence similar owners, so not likely to be independent as binomial model requires. Total 3 of 8

12 0(a) Indicates that Semiskimmed milk is part of the whole Skimmed milks category. AO.4 E In Figure, the vertical axis represents Skimmed milks whereas in Figure the vertical axis represents Semiskimmed milk that is just one part of the Skimmed milks category. Hence, for each region, the recorded values for Semi-skimmed milk will always be lower that the ones for Skimmed milks. (b)(i) States: strong positive correlation between purchases of Semiskimmed milk and Liquid wholemilk, full price for most regions. AO.5 B Regions A, B, D, F, G, H and J indicate strong positive correlation between purchases of Semi-skimmed milk and Liquid wholemilk, full price. Identifies regions C and E as outliers AO3.b E Regions C and E do not follow the same pattern as the other regions in terms of purchases of skimmed milk and wholemilk or regions C and E can be excluded as they appear to be outliers. ALT Allow mark if candidate states: weak negative correlation between consumption of Semiskimmed milk and Liquid wholemilk, full price. AO.5 AO3.b B E0 of 8

13 0(b)(ii) States that there is evidence of negative correlation between purchases of Mineral or spring water and Skimmed milk. AO3.b B Scatter graph indicates evidence of negative correlation between consumption of Mineral or spring water and Skimmed milks. The correlation is between water and milk purchases within the regions and not between purchases made by individual people so Bilal s claim is not proven AO.3 E Bilal s claim is not supported because the correlation in figure is between water and milk purchases within the regions and he is assuming that individuals will follow a similar pattern but the data does not tell us anything about individuals, it refers to average purchases made within the nine regions. (c) Identifies London AO.b B London Gives valid reason based on knowledge of the LDS AO.4 E Having studied the LDS I have identified a clear trend: London is a frequent outlier in most categories Total 7 3 of 8

14 (a) Finds correct value of k AO.b B k 6 (b) Selects relevant probability AO.a M P( checkouts staffed) 3 k 6 = = 4 Finds correct probability AO.b AF ALT FT their value of k found in part (a) P( checkouts staffed) = 3 4 = 4 Total 3 4 of 8

15 (a) States both hypotheses correctly for one-tailed test AO.5 B X = number of Christmas holidays without illness since January 007 X ~ B 7, p H 0 p 0.65 H p 0.65 States model used (condone rather than 0.056) PI AO.b M Under null hypothesis, X B( 7, 0.65 ) Using calculator, or better AO.b A P( X ) Evaluates binomial model by comparing P( X ) with 0.05 PI AO3.5a M >0.05 Infers H 0 accepted PI AO.b A Accept H 0 Concludes correctly in context. not sufficient evidence or equivalent required (b) States one correct assumption(s) regarding validity of model States corresponding correct description(s) of likelihood of validity in context States second correct assumption(s) regarding validity of model States corresponding correct description(s) of likelihood of validity in context Max two assumptions with description of validity AO3.a AO3.5b AO.4 AO3.5b E E E E There is not sufficient evidence that the John s rate of illness has decreased Assumption The probability of illness remains constant throughout one s life Validity Not fully valid, as age has an impact on the immune system OR AO.4 E Assumption Annual results (of illness) are independent of one another Validity (Largely) valid. Trials are sufficiently far apart that an illness spanning two Christmases is unlikely. OR Total 0 Assumption 3 There are only two states, well and ill Validity Unclear. Grey area exists. eg does a mild sore throat count as ill? 5 of 8

16 3(a)(i) Finds mean AO.b B Mean X Finds standard deviation awfw ( ) Allow s = = 4.79 AO.b B Standard deviation s = = 4.0 (ii) States max range for a normal distribution Explains why normal model acceptable AO. M X 3 s 43 X 3 s 8 AO.4 E Normal dist model acceptable because data can be regarded as continuous and the range of amounts is within X 3 s (iii) Finds probability awfw ( ) AO.b B P( X < 5) = (b) Finds z from calculator AO.b B Z Uses Normal model to form equation AO3.4 M P = = = Finds correct sd Correct accuracy required for this mark Accept 3.5 or 3.7 AO.b A Total 8 6 of 8

17 4(a)(i) States both hypotheses using AO.5 B H0 : 3 correct language H : 3 Finds test statistic AO.a M 7 3 Test statistic Obtains correct test statistic AO.b A = 6.30 Infers H 0 rejected by comparison of ts with cv AO.b A Critical z values >.96 Concludes correctly in context, including some evidence ALT States both hypotheses using correct language AO3.a E Reject H 0 there is evidence (at the 5% level) to suggest the mean expenditure on bread had changed from 0 to 03 AO.5 B H 0 : 3 H : 3 Attempts to find p value for z-test AO.a M From calculator, P(X < 7 ) Finds correct p value AO.b A < 0.05 Infers H 0 rejected by comparison AO.b A of p with 0.05 Concludes correctly in context, including some evidence (a)(ii) Uses Normal model to find critical values PI Obtains correct critical values Correct accuracy required for this mark Disallow integer answers (b)(i) States valid reason for statement not supported Infers that model/test used would not imply the statement (b)(ii) States valid reason for statement not supported Infers that model/test used would not imply the statement Total 0 AO3.a E Reject H 0 - there is evidence (at the 5% level) to suggest the mean expenditure on bread had changed from 0 to 03 AO3.4 M AO.b A min=.75 and max=4.5 AO.4 R The conclusion implies that the mean changed, not that it increased by a specific amount, so the statement is not supported AO.b R AO.4 AO.b R R The conclusion implies that there is evidence that the mean has changed, but expenditure increase may be due to price changes, so statement is not supported 0 7 of 8

18 5 Uses conditional probability, either or Obtains both equations and correctly Evaluates P ( A B) correctly PI AO3.b M P( A B) P( A) 4 P( A) 4 P( AB) AO.b A P( A B) P( B) 0 P( B) 0P ( AB) AO.b B 39 P( AB) Uses addition law AO.a M Combines the three equations Obtains correct probability, as a fraction or decimal ALT Produces a relevant Venn diagram AO.a AO.b AO3.b M A M P( A) P( B) P( AB) P( AB) 0P( AB) P( AB) P( AB) 3 00 OR 00 B A 9x x 3x Labels Venn diagram correctly Forms correct equation to find x PI AO.b A AO.b B 9x + x + 3x = 00 Combines terms AO.a M 3x = 78 Solves equation AO.a M x = 6 Obtains correct AO.b A probability P( AB) 6 or Total 6 TOTAL 00 8 of 8