Friday 6 June 2014 – Afternoon - MEImei.org.uk/files/papers/Add-maths-2014-June.pdf · Friday 6 June 2014 – Afternoon FSMQ ADVANCED LEVEL 6993/01 Additional Mathematics QUESTION

Friday 6 June 2014 – AfternoonFSMQ ADVANCED LEVEL

6993/01 Additional Mathematics

QUESTION PAPER

*1264621551*

INSTRUCTIONS TO CANDIDATES

These instructions are the same on the Printed Answer Book and the Question Paper.

• The Question Paper will be found inside the Printed Answer Book.• Write your name, centre number and candidate number in the spaces provided on the

Printed Answer Book. Please write clearly and in capital letters.• Write your answer to each question in the space provided in the Printed Answer

Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).

• Use black ink. HB pencil may be used for graphs and diagrams only.• Answer all the questions.• Read each question carefully. Make sure you know what you have to do before starting

your answer.• Do not write in the bar codes.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given correct to three significant figures where appropriate.

INFORMATION FOR CANDIDATES

This information is the same on the Printed Answer Book and the Question Paper.

• The number of marks is given in brackets [ ] at the end of each question or part question on the Question Paper.

• You are advised that an answer may receive no marks unless you show sufficient detail of the working to indicate that a correct method is being used.

• The total number of marks for this paper is 100.• The Printed Answer Book consists of 20 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

INSTRUCTION TO EXAMS OFFICER / INVIGILATOR

• Do not send this Question Paper for marking; it should be retained in the centre or recycled. Please contact OCR Copyright should you wish to re-use this document.

OCR is an exempt CharityTurn over

© OCR 2014 [100/2548/0]DC (LK/SW) 73742/4

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Printed Answer Book 6993/01

Other materials required:• Scientific or graphical calculator

Duration: 2 hours

2

6993/01 Jun14© OCR 2014

Formulae Sheet: 6993 Additional Mathematics

In any triangle ABC

Cosine rule a2=b2+c2–2bccosA

C

b a

c BA

Binomial expansion

Whennisapositiveinteger

(a+b)n=an+(n1)an–1b+(n

2)an–2b2+…+(nr )an–rbr+…+bn

where

(nr )=nCr=

n!––––––––r!(n–r)!

3

6993/01 Jun14 Turn over© OCR 2014

Section A

1 Solvethefollowing.

x6 2 1 7– –1 1 [3]

2 Thegradientfunctionofacurvethatpassesthroughthepoint(1,2)isgivenby

.xy

x x3 4 7d

d2= - +

Findtheequationofthecurve. [4]

3 (i) Findtheareaenclosedbetweenthecurve y x8 3= ,thex-axisandthelinex=2. [3]

(ii) Hence, or otherwise, deduce the area between the x-axis, the y-axis, the line x 2= and the curvey x 58

3= + . [1]

4 A train travels from station A to station B. It starts from rest at A and comes to rest again at B.The displacement of the train from A at time t seconds after starting from A is s metres where

. . .s t t0 09 0 0001–

2 3=

(i) FindthevelocityattimetsecondsafterleavingAandhencefindthetimetakentoreachB.Givetheunitsofyouranswer. [4]

(ii) FindthedistancebetweenAandB.Givetheunitsofyouranswer. [2]

5 Ashipismovingonabearingof025ºat14knots(1knot=1nauticalmileperhour).AsitpassespointA,alighthouseLisseenonabearingof340º.After30minutes,theshippassespointBfromwherethelighthouseisseenonabearingof320º.

N

N

Not to scale

L

A

B

25°

(i) FindtheangleBALandtheangleALB. [3]

(ii) Hence,orotherwise,calculatethedistanceBLinnauticalmiles. [3]

4

6993/01 Jun14© OCR 2014

6 Thefunction x x x ax b4f3 2= - + +^ h issuchthat

• x=3isarootoftheequationf(x)=0, • whenf(x)isdividedby(x–1)thereisaremainderof4.

(i) Findthevalueofaandthevalueofb. [4]

(ii) Solvetheequationf(x)=0. [3]

7 ThepointsAandBhavecoordinates(3,7)and(5,11)respectively.

(i) FindtheexactlengthofAB. [2]

(ii) FindtheequationofthecirclewithdiameterAB. [3]

8 FourpointshavecoordinatesA(−5,−1),B(0,4),C(7,3)andD(2,−2).

(i) Usinggradientsoflines,provethatABCDisaparallelogram. [2]

(ii) Usinglengthsoflines,provefurtherthatABCDisarhombus. [2]

(iii) ProvethatABCDisnotasquare. [2]

9 (i) Showthat .sin

costan

xx x

1

1

–

–2

22= [1]

(ii) Hencesolvetheequationsin

costan

xx x

1

13 2

–

–

2

2

= - forvaluesofxintherange x0 180G Gc c. [4]

10 (i) FindthecoordinatesofthepointPonthecurve y x x2 52= + - wherethegradientofthecurveis5.

[3]

(ii) FindtheequationofthenormaltothecurveatthepointP. [3]

5

6993/01 Jun14 Turn over© OCR 2014

Section B

11 Kalaismakinganopenboxoutofarectangularpieceofcardmeasuring30cmby14cm.Shecutssquaresofsidex cmoutofeachcornerandturnsupthesidestoformthebox.

30 cm

x

x

x

x

x

x

x

x

14 cm

(i) Find an expression in terms of x for the volume,V cm3, of the box and show that this reduces to

V x x x4 88 4203 2= - + . [4]

(ii) Findthetwovaluesofxthatgive xV

0d

d= . [5]

(iii) Explainwhyoneofthesevaluesshouldberejectedandfindthemaximumvolumeoftheboxusingtheothervalue. [3]

12 Paulwalked fromAnytown toNexttown, adistanceof15km.Whenhegot therehe thenwalkedback.Hisaveragespeedonthereturnjourneywas2kmperhourlessthanontheoutwardjourney.

LetPaul’saveragespeedontheoutwardjourneybexkmhr–1.

(i) Writedownanexpressionforthetime,inhours,takenforthewholejourney. [2]

ThetimetakenbyPaulforthewholejourneywas6hours.

(ii) Useyourexpressionin(i)toformanequationinxandshowthatitsimplifiesto

x x7 5 02 - + = . [4]

(iii) SolvethisequationtofindPaul’saveragespeedontheoutwardjourney. [3]

(iv) Findthedifferenceintimebetweentheoutwardandreturnjourneys.Giveyouranswertothenearestminute. [3]

6

6993/01 Jun14© OCR 2014

13 Acompanyneeds tobuysomestorageunits.Thereare two typesofunitavailable, typeXand typeY.Thecostofeachtypeofunit,thefloorspacerequiredandthevolumeforstoragearegiveninthefollowingtable.

Cost per unit (£)

Floor space required (m2)

Volume for storage (m3)

X 100 2 3.5Y 120 1.5 3

Themaximumcostallowedforthepurchaseoftheunitsis£1200andthemaximumfloorspaceavailableis18m2.

Thecompanywantstomaximisethevolumeforstorage.

Letxandybethenumberofeachtypeofunit,XandY,respectively.

(i) Writedownaninequalityforthetotalcostandaninequalityforthetotalfloorspacerequired. [3]

(ii) Drawtheinequalitiesyougavein(i)onthegridprovidedintheanswerbook.Giventhatx 0H andy 0H ,indicatetheregionforwhichtheinequalitiesholdbyshadingtheareathatisnotrequired. [4]

(iii) Writedowntheobjectivefunctionforthevolumeforstorageandfindthecombinationofunitsthatshouldbeboughttomaximisethevolumeforstorage.

Writedownthismaximumvolume. [5]

14 Mugsarepackedinboxesof10.Onaverage,5%ofthemugsareimperfect.Aboxofmugsisclassifiedas“unsatisfactory”ifitcontainstwoormoreimperfectmugs.

(i) Statetwoconditionsthatmustbesatisfiedforthenumberofimperfectmugsinaboxtohaveabinomialdistribution. [2]

(ii) Assumingthatthesetwoconditionsaresatisfied,calculatetheprobabilitythataboxchosenatrandomis“unsatisfactory”. [6]

Ashopreceivesadeliveryofalargenumberofboxesofmugs.Thedeliveryischeckedasfollows.

Aboxischosenatrandom. • Iftherearenoimperfectmugsintheboxthenthewholedeliveryisaccepted. • Iftheboxis“unsatisfactory”thenthewholedeliveryisrejected. • Ifthereisexactlyoneimperfectmugintheboxthenasecondboxischosenatrandom.Thedeliveryis

acceptedonlyifthisboxcontainsnoimperfectmugs.

(iii) Calculatetheprobabilitythatthedeliveryisaccepted. [4]

7

6993/01 Jun14© OCR 2014

BlAnk pAge

Friday 6 June 2014 – AfternoonFSMQ ADVANCED LEVEL

6993/01 Additional Mathematics

PRINTED ANSWER BOOK

INSTRUCTIONS TO CANDIDATESThese instructions are the same on the Printed Answer Book and the Question Paper.• The Question Paper will be found inside the Printed Answer Book.• Write your name, centre number and candidate number in the spaces provided on the

Printed Answer Book. Please write clearly and in capital letters.• Write your answer to each question in the space provided in the Printed Answer

Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).

• Use black ink. HB pencil may be used for graphs and diagrams only.• Answer all the questions.• Read each question carefully. Make sure you know what you have to do before starting

your answer.• Do not write in the bar codes.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given correct to three significant figures where appropriate.

INFORMATION FOR CANDIDATESThis information is the same on the Printed Answer Book and the Question Paper.• The number of marks is given in brackets [ ] at the end of each question or part question

on the Question Paper.• You are advised that an answer may receive no marks unless you show sufficient detail

of the working to indicate that a correct method is being used.• The total number of marks for this paper is 100.• The Printed Answer Book consists of 20 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

* 6 9 9 3 0 1 *

OCR is an exempt CharityTurn over

© OCR 2014 [100/2548/0]DC (LK/SW) 73743/4

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Question Paper 6993/01 (inserted)

Other materials required:• Scientific or graphical calculator

Duration: 2 hours

*1264661704*

2

© OCR 2014

Section A

1

3

Turn over© OCR 2014

2

4

© OCR 2014

3 (i)

3 (ii)

5


4 (i)

4 (ii)

6

© OCR 2014

5 (i)

5 (ii)

7


6 (i)

6 (ii)

8

© OCR 2014

7 (i)

7 (ii)

9


8 (i)

8 (ii)

8 (iii)

10

© OCR 2014

9 (i)

9 (ii)

11


10 (i)

10 (ii)

12

© OCR 2014

Section B

11 (i)

13


11 (ii)

11 (iii)

14

© OCR 2014

12 (i)

12 (ii)

15


12 (iii)

12 (iv)

16

© OCR 2014

13 (i)

13 (ii)

2 4 6 x

y

8 10 12

12

10

8

6

4

2

0

There is a spare copy of this grid on page 20. If you wish to offer a second attempt, then you must cross through the attempt on this page.

17


13 (iii)

18

© OCR 2014

14 (i)

14 (ii)

19


14 (iii)

20

© OCR 2014

Copyright Information

OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.

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OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

This is a spare copy of the grid for Question 13(ii).Only write on this page if you want to offer a second attempt at the graph.If you do so, then you must cross through the first attempt on page 16.

2 4 6 x

y

8 10 12

12

10

8

6

4

2

0

Oxford Cambridge and RSA

Oxford Cambridge and RSA Examinations

FSMQ

Additional Mathematics

Unit 6993: Additional Mathematics Free Standing Mathematics Qualification

Mark Scheme for June 2014

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2014

6993 Mark Scheme June 2014

1

Annotations and abbreviations

Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning

E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working


2

Subject-specific Marking Instructions for 6993 a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


3

Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument. Unless otherwise stated (by for instance, cao usually apply isw

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.


4

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


5

Section A Question Answer Marks Guidance

1 6 2 1 7x 55 2 8 4

2x x isw B1

B1 One end point The other

Condone incorrect signs (including equals) for first 2 marks. Ignore listings.

B1 Both together - accept 2 separate inequalities linked by "and" cao but not "or" or comma or nothing

B3 Final answer seen www [3] S.C. If solution only given as number line

with two clear open circles at ends then B2

Question Answer Marks Guidance

2 2d 3 4 7dy x xx

M1 Integrate (at least 2 powers increased by 1)

Beware just multiplying by x

3 22 7y x x x c oe A1 Ignore lack of y = and c Substitute (1, 2) 2 1 2 7 c M1 Dependent on 1st M 4c 3 22 7 4y x x x A1 Final equation must be seen Must include y = and 4

Condone f(x) = [4]


6

Question Answer Marks Guidance 3 (i) 2

23 4

00

Area = 8 d 2x x x M1

M1

Integrate correct function (power of 4 soi). Ignore wrong or no limits Substitute x = 2. Dependent on first M

Beware just multiplying by x i.e. 8x4

32 ( 0) 32 A1 [3] (ii) Add a rectangle of area 10 Total area = 42 B1 ft Addition of 10 must be correct! Can be by integration [1]

Question Answer Marks Guidance 4 (i) 2 30.09 0.0001 s t t 20.18 0.0003v t t isw M1

A1 Diffn (both powers reduced by 1).

Beware division by t.

0.18When 0, 0 or 6000.0003

v t t M1 Set v = 0 and any attempt to solve. Dependent on first M.

Condone division by t or a constant when expression not set to 0

Time = 600 seconds or 10 mins isw A1 Units required. [4] (ii) Substitute their t into s M1 s = 10800 m or 10.8 km isw A1 Units required, but only withhold this

mark for units wrong or missing if not already withheld in part (i).

[2]


7

Question Answer Marks Guidance 5 (i) Angle BAL = 20 + 25 = 45 B1 BAL and ALB must be correctly

identified (not from use in (ii)) Angle ABL = 180 65 = 115 soi

OR exterior angle = 65 soi OR angle LBN = 40 soi

B1

Angle ALB = 20 B1 Or B2 for ALB www [3] (ii) AB = 7 soi

LB AB 7sin 45 sin 20 sin 20

B1 M1

For their AB and their angles.

7sin 45LB 14.5sin 20

A1 Anything that rounds to 14.5

[3]

Question Answer Marks Guidance 6 (i) f(3) = 0 27 36 + 3a + b = 0 or better B1 e.g. 3a + b = 9.

Powers need to be evaluated

f(1) = 4 1 4 + a + b = 4 or better

B1 e.g. a+ b = 7

Powers need to be evaluated

Solve their simultaneous eqns from above M1 Attempt to find a and b from their eqns Their working need not be correct a = 1, b = 6 A1 [4] (ii) Sight of (x – 3)(x2 + px + q) for any p, q

Or algebraic division M1 Or attempt to find another root of their

cubic by remainder theorem soi Algebraic division seen by x2 + ... in quotient and x3 3x2 in division

(x – 3) (x + 1)(x – 2) = 0 A1 For correct complete factorisation soi by final answer

(x =) 3, 2, 1 A1 Correct solution [3]


8

Question Answer Marks Guidance 7 (i) 2 22Distance = 5 3 11 7 ( 20) M1 Soi e.g. by 4.47….. d = 20 is M0 distance 20 2 5 A1 Must be exact isw

[2] (ii) Centre = midpoint = (4, 9) B1 Centre Radius = 5 or decimal equivalent B1 Radius soi by for e.g. r2 = 5 2 24 9 5x y B1 ft their identified centre (but don't

accept A or B) isw Rhs must be 5, not 25

2 2OR 8 18 92 0x y x y Alternative: 11 7. 1

5 3y yx x

B1 Use of m1m2 = −1

11 7 5 3 0y y x x B1 2 2 8 18 92 0x y x y B1 [3]


9

Question Answer Marks Guidance 8 (i) 4 1 2 3Grad AB = Grad CD 1 and

0 5 2 7

oe B1 For showing one pair of

gradients equal and correct www

bod no working but care about

seeing xy

1 3 4 2 1Grad BC = Grad AD = and 7 7 0 2 5

Two pairs of parallel sides ( means ABCD parallelogram )

B1 For showing other pair of gradients equal and correct plus completion

Final statement is necessary. Condone 2 pairs of equal gradients (providing they are correct)

[2] (ii) 2 2 2AB 5 5 (=50) oe for any side B1 One length (or squared length) 2 2 2BC 1 7 (=50) 2 2AB = BC ( 50)

Equal sides ( means rhombus) B1 For adjacent length plus

completion www Final statement is necessary

[2] (iii) Gradients do not fulfil m1. m2 = 1 oe M1 For use of m1. m2 = 1 i.e. 2nd gradient not the negative

reciprocal of the other 1ie 1 1

7

Therefore lines not perpendicular

A1

Gradients must be correct.

Final statement is necessary.

[2] Alternatives: A: Use of cosine rule

Does not give 900 M1 A1

www

B: Use of Pythagoras Not satisfied therefore not 900

M1 A1

www

C: Use of pythagoras to find length of diagonals ( i.e. 160 and 40 ) Diagonals not equal

M1

A1

www


10

Question Answer Marks Guidance 9 (i) 2 2

22 2

1 cos sin tan1 sin cos

x x xx x

B1 Use of sin tan

cosx xx

and use of 2 2sin cos 1x x

[1] (ii) 2

2

1 cos 3 2 tan1 sin

x xx

2tan 3 2 tanx x M1 Correct use of (i) 2tan 2 tan 3 0x x tan 3 tan 1 0x x M1 Factorise their three term quadratic

or insertion of their values into correct formula Dep on 1st M

(Check the two linear factors by whether they multiply out to give the first and last terms of their quadratic.)

tan 3 or tan 1x x 108(.4...) or 45x

A1A1 1 for any other values inside range, ignore extra values outside range

[4]

Question Answer Marks Guidance 10 (i) 2 d2 5 4 1

dyy x x xx

B1

5 1, 2x y

M1 A1

Equating to 5 and solving For both

[3] (ii) 1gradient normal =

5 B1 Gradient of normal soi

1( 2) 15

y x

M1

Equation using their (1, 2) and their normal gradient (which may

only be 1 or 55

)

5 9 0y x A1 oe, but only 3 terms isw [3]


11

Section B

Question Answer Marks Guidance 11 (i) Length = 30 – 2x B1 Soi Breadth = 14 – 2x B1 Soi (Height = x) (30 2 )(14 2 )V x x x M1 Product of their length, breadth and x N.B. dimensionally correct 2

3 2

4 88 420

4 88 420

x x x

x x x

oe

A1

www; must show at least one product of any two lengths step (N.B. Answer given)

2e.g (30 2 )(14 2 )x x x Length and breadth must be functions of x

[4] (ii) 3 24 88 420V x x x M1 Diff (at least two powers reduced by 1) Beware division by x.

Condone prem div by a constant. 2d 12 176 420

dV x xx isw A1

20 when 12 176 420 0x x M1 Set their function = 0 Dep on 1st M

Soi by solution

23 44 105 0x x 3 35 3 0x x M1 Factorise three term quadratic or

insertion of their values into correct formula

(Check the two linear factors by whether they multiply out to give the first and last terms of their quadratic.)

35 or anything that rounds to 11.7, 33

x x A1 Both

S.C. Answers only B1, B1 [5] (iii) 35

3x should be rejected as it is over half

of 14

B1 ft from their incorrect x Explanation necessary (e.g. one length is -ve) Alt: use of second derivative acceptable.

"V is -ve" as the only explanation not accepted. Nor is "x is too big".

Substitute an acceptable x into V ( 0 < x < 7) M1 Volume = 576 A1 Ignore units [3]


12

Question Answer Marks Guidance 12 (i)

Time out = 15x

, Time back = 152x

B1 For one

Total time = 15 15

2x x

B1 Addition of two correct terms isw

[2] (ii) 15 15 6

2x x

B1 Equate their time to 6 3 might be divided throughout

here 15 2 15 6 ( 2)x x x x M1

Multiply throughout by LCM

LCM implies 2 different algebraic denominators

215 30 15 6 12x x x x oe A1 Brackets cleared N.B. Algebra might have been done in (i)

26 42 30 0x x 2 7 5 0x x A1 www At least one interim step must be

seen. N.B. Answer given

[4] (iii) 2 7 5 0x x 7 49 20 7 29

2 2x

or 6.19 and 0.807

M1 A1

Use of correct formula with given equation

Condone 0.81 or 0.8 but not 6.2

(Paul’s speed) = 6.19 km hr –1 A1 + units but only if their 0.807 is discarded

N.B. If 3 marks not awarded then look for the S.C.

[3] S.C (Paul's speed) is 6.19 km hr –1 B2

or x = 6.19 B1

(iv) 15 15 15 152 4.19 6.19x x

M1 Sub their x into correct expression Or sub their x into the two separate correct expressions for time and then subtract

Or the simplified expression Accept other way round giving a negative value

Sight of 3.58 or 2.42 or 1.16 B1 69 or 70 mins or 1hr 9 mins or 1 hr 10 mins A1 Answer correct in minutes [3]


13

Question Answer Marks Guidance 13 (i)

100x + 120y ≤ 1200 oe M1 A1

Attempting to use information to create an inequality

Seen by one LH side Condone use of <

2x + 1.5y ≤ 18 oe A1 Ignore extra inequalities [3] (ii)

B1

B1

B1

B1

One line (allow intercepts ±0.1) Other line (allow intercepts ±0.1) Shading one line – 1st quad only Shading other line – 1st quad only N.B. Shading below a line gets B0

N.B. Candidates may get inequalities wrong in (i) but get the shading correct in (ii) - this should be allowed. Allow ft for shading of wrong line but only if the gradient is -ve Allow ft for shading of wrong line but only if gradient is -ve and the two lines intersect in the 1st quadrant (not on axes)

[4] (iii) (P = ) 3.5x + 3y B1 Ignore any equating to a number M1 test at least two integer points in correct

feasible region in correct OF

e.g. (9, 0) gives 31.5, (0, 10) gives 30 (4, 6) gives 32, (5, 5) gives 32.5

A1 Both points correct. Ignore any others

(3, 7) gives 31.5 (6, 4) gives 33 (6, 4) A1 For (6, 4) chosen i.e. the point must be identified as

the maximum. gives 33 A1 For 33 [5] S.C. for last 4 marks. (6,4) gives 33 B2

Either 33 or (6,4) B1


14

Question Answer Marks Guidance 14 (i) Probability remains constant

Imperfection of mugs independent B1 B1

Not "Random"

[2] (ii) 1 19,

20 20p q B1 Soi

10 919 19 1P(0 or 1) = 1020 20 20

M1

A1

q10 + kpq9 attempted, k an integer > 0 p + q = 1 k = 10 soi

For k Not 101

10 or C

1

0.5987 0.3151 = 0.9139 A1 soi accept rounding to 3dp P( 2) 1 0.9139

0.086

M1

A1 Dependent on previous M (Anything that rounds to 0.086)

[6] (iii) P(accepted) = P(0 imperfect) M1 Correct plan soi by both correct terms Words only sufficient + P(1 imperfect) P(0 imperfect)

10 9 1019 19 1 191020 20 20 20

A1 A1

Correct 1st term (as an expression) soi Correct 2nd term (as an expression including the 10) soi

soi by final answer

0.5987 0.1887 0.787(4...) A1 Accept 0.788 Anything that rounds to 0.787 or

0.788 [4] Alternative:

P(accept) = 1 (Ans to (ii) + P(1)×P(at least 1)) = 1 (ans to (ii) + P(1)×(1P(0)) = 1 - (0.0861+ 0.3151 × 0.4013) = 1 - (0.0861+ 0.1264) =1 - 0.213 = 0.787(4...)

M1

A1 A1

A1

Correct plan soi by both correct terms taken from 1 Ft Using their ans to (ii) 2nd term as an expression soi by final answer

Words only sufficient

Oxford Cambridge and RSA

Oxford Cambridge and RSA Examinations

FSMQ

Additional Mathematics

Unit 6993: Paper 1 Free Standing Mathematics Qualification

OCR Report to Centres June 2014

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS / A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching / training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This report on the examination provides information on the performance of candidates which it is hoped will be useful to teachers in their preparation of candidates for future examinations. It is intended to be constructive and informative and to promote better understanding of the specification content, of the operation of the scheme of assessment and of the application of assessment criteria. Reports should be read in conjunction with the published question papers and mark schemes for the examination. OCR will not enter into any discussion or correspondence in connection with this report. © OCR 2014

CONTENTS

Additional Mathematics FSMQ (6993)

OCR REPORT TO CENTRES

Content Page Additional Mathematics – 6993 1

OCR Report to Centres – June 2014

1

Additional Mathematics – 6993

Once again, the overall statistics of achievement by candidates is marred by a small proportion who gains very few marks. It is particularly distressing to see candidates scoring only in single figures. The specification is intended as an enrichment specification for bright students and we feel that scoring very few marks is not a good experience for these candidates. The mean mark for this paper was 51%, a reduction of 3 marks on the entry for last year. This does not mean that the content of the paper was harder, but it may be that there were some tricky applications required of some basic topics. Comments on individual questions Section A Q1 (Inequality) Most candidates found this a good start to the paper. A few were unable to cope with the double inequality (for instance, adding 1 to the right hand side but not the left) and a few were clearly working on the ideas of similar questions in previous years by writing down a list of integers that satisfied the inequality. Q2 (Calculus – finding the equation of a curve) A surprising number of candidates were unable to understand this question. They substituted x = 1 into the derived function and then found the equation of the line with this gradient. This of course is the equation of the tangent at the given point which was not the question. A few failed to finish the question by not finding the constant of integration and a few failed to get their simple arithmetic right. The question asked for the equation of the curve. We therefore expected to see “y = .....”. Q3 (Calculus – area under a curve) There were, as usual, a proportion of candidates who differentiated instead of integrated to find the area in part (i); again there were some simple arithmetic errors. In part (ii) it was intended that candidates realised that moving the curve up the y-axis meant that the area was increased by the area of a rectangle. A number simply added 10 to their answer to part (i), but the majority started again with the new curve, not seeing any connection. It was disappointing to see some able students failing to understand this, particularly in light of the fact that by this means part (ii) was longer than part (i) and yet only carried one mark. It is worth noting at this point that here, and elsewhere in the paper, the presentation of an integral was very poor. Often limits were omitted from the first integral statement, though used correctly later, and the lack of “dx” was worryingly prevalent. Q4 (Rate of change) Many candidates, when faced with such a question, are unable to discern whether they are being tested on constant acceleration (requiring the constant acceleration formulae) or rate of change requiring calculus. Consequently, many candidates spent a lot of time getting nowhere. Others were unable to interpret the question in terms of the need to set the velocity equal to zero to find the time taken in part (i) and then with this time to find the distance between the stations. Some candidates took the rather simple quadratic in t (there is no constant term) and attempted to solve using the formula. Without the constant term candidates often became confused and decided on incorrect and often highly unrealistic answers. Use of the formula is, of course, a correct method but it was surprising how many candidates were unable to write down the correct formula.


2

Those without a value of t from the first part were unable to continue with the question. Q5 (Trigonometry and the Sine Rule) The ability to sort out the angles in the triangle was distressingly weak in part (i). Bearings were not properly understood and some unrealistic answers were seen. The lack of the correct angles had a knock-on effect for part (ii), though marks were available for using the sine rule correctly and for calculating the distance travelled in 30 minutes. Q6 (Cubic functions and equations) This was perhaps the least well answered question in Section A. Candidates were expected to develop two simultaneous equations in a and b and solve simultaneously. Doing an algebraic long division in this situation caused complications over how to determine the remainder. Those who did it this way usually found one equation and then made a wild guess. Some who appeared to know what to do took f(3) = 0 rather than f(3) = 0. In part (ii) many ignored the information of part (i) and started again with arbitrary factors; these sometime worked and sometimes they did not. The idea that the product of roots was –b and so the only possibilities to try were ±1, ±2, ±3 was hardly ever seen. Those that used the fact that (x – 3) was a factor of the cubic function did not need this of course as their problem was to factorise a simple quadratic. Q7 (Coordinate geometry and the circle) While most candidates were able to work out the distance between two points, not all understood the meaning of the word “exact”. It was pleasing to see so many candidates write the

answer as 2 5 , which is a better answer than 20 , but disappointing that they then went on to write 4.472.... In part (ii) most were able to recover by using exact values to get the correct equation for the circle, but a few approximated their length in part (i), divided by 2 to get the radius, then squared to get a number that was different to 5. Q8 (Coordinate geometry of a quadrilateral) There was some GCSE knowledge required here and a significant number of candidates did not seem to have that knowledge. Those that did understand what was required often fell down in their explanations which were usually quite poor. In part (i), quite apart from those that could not find gradients, many thought that the definition of a parallelogram required one pair of sides to be parallel, not two. In part (ii) likewise, many thought that all that was required was to see that one pair of sides were equal in length. The fact that the quadrilateral had already been shown to be a parallelogram meant of course that it was sufficient to show that one pair of adjacent sides were equal, but where this was seen it was not explained carefully enough that as the quadrilateral had been shown to be a parallelogram then it was a property that opposite sides were also equal. In part (iii) many lost a mark by not explaining how they knew the sides were not perpendicular. It was necessary to state the property of perpendicular lines ( 1 2 1m m ) to earn full marks.

Q9 (Trigonometry) A number of candidates lost a lot of time on part (i) and might have realised that the result was very short, due to there being only one mark allocated. Examiners needed evidence of


3

Pythagoras being used. Some poor algebra was evident here, which combined with some

incorrect trigonometry, resulted in answers such as 2 2 2

22 2 2

1 cos cos costan

1 sin sin sin

x x x xx x x

.

In part (ii) a number did not use the identity of part (i) and usually got lost. Some who obtained values for tanx thought that tanx = 3 was not a valid result and so rejected it, missing a part of the solution or were unable to find the appropriate angle from their calculator display. Q10 (The normal to a curve) Part (i) was usually done well. In part (ii), however, many candidates saw the derived function of

4x + 1 and decided that the normal gradient was therefore 1

4 .

Section B Q11 (Maximum and minimum) This was a standard development of the volume of the box and most candidates achieved the result in only a few steps. It was necessary only to have sight of the lengths of the sides and to see the product being found. It was pleasing to see that a number of candidates who failed to obtain the result in part (i) were able to use the result given to earn marks in part (ii). This part was usually done well. Even those who did not complete part (i) were able to start with the differentiation to find the two stationary values. Many candidates failed to simplify their quadratic equation by dividing throughout by common factors leading (for those using the formula) to some pretty hefty (and unnecessary) arithmetic. Since arithmetic is not always a strong point for candidates, many errors were seen in what should have been a straightforward process. In part (iii) it was not satisfactory to assert that the rejected value gave a negative volume as the reason for the rejection. The reason required was to be able to say why one of the values should not be used to find V so it was the fact that one side became negative with the larger value. It was acceptable to find the second derivative and to substitute and use the fact that a negative value indicated a maximum value even though this method of determining the nature of stationary values is not in the specification. It was, however, quite a lot of work for only one mark. There were many spurious and wild suggestions as to why one value should be rejected, such as “you cannot use a recurring decimal to calculate the volume as it is too hard”. Q12 (Constant acceleration) There were some very odd expressions given for the time and in particular many candidates were confused over units, not aware that x has no units. So we saw expressions such as

15 15

km/hr 2km/hrx x

resulting in a muddle over the next part. Many simplified the addition of

fractions in part (i) which was expected in part (ii) (and marks assigned here) but full marks were given for correct simplification wherever we saw it. A common misunderstanding of the question was to find an average time for the whole journey,

giving the answer as 15 15 30

2 2 2x x x

.

In part (iii) one of the roots of the quadratic equation has to be rejected, though in this question we did not demand a reason for it, and then the answer had to be given in correct units – in other words we needed to know Paul’s average speed and not just the value of x. Q13 (Linear programming) The majority of candidates achieved the correct inequalities in part (i) and were able then to draw the correct lines and shade appropriately to show the feasible region in part (ii).


4

Part (iii) caused more of a problem. Some candidates did not actually state an objective function, though subsequent working indicated that they knew what it was. Some candidates wasted time finding the point of intersection of two lines; because this was not an integer point it was not required. What was required was evidence that some of the possible points were tested to find a maximum value. Candidates failed here in many ways including not testing at least two points, not including the correct point in the list of points being tested or performing some incorrect arithmetic. Usually it is the point of intersection but when there is a practical context and the point of intersection is not an integer point then candidates need to be aware that some different work needs to be done. Nonetheless, this question was a good source of marks for most candidates. Q14 (Probability) The two conditions required in part (i) were the constant value of p and the independence of each event from all others. Very few candidates were able to give these conditions. Parts (ii) and (iii) were often very muddled and it was difficult to discern what candidates were trying to find out. A simple statement such as “ P(unsatisfactory) = P(two or more imperfect mugs) = 1 – P(0 or 1 mugs imperfect)” would have clarified for the assessor what was being found and must surely have clarified the plan for the candidate. A significant number in both parts seem to have set out in the right way, but got lost along the way.

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Friday 6 June 2014 – Afternoon - MEImei.org.uk/files/papers/Add-maths-2014-June.pdf · Friday 6 June 2014 – Afternoon FSMQ ADVANCED LEVEL 6993/01 Additional Mathematics QUESTION