Date Morning/Afternoon Time allowed: 2 hours 40 minutes

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1 A Level Further Mathematics B (MEI) Y40 Core Pure Sample Question Paper Date Morning/Afternoon Time allowed: hours 40 minutes You must have: Printed Answer Booklet Formulae Further Mathematics B (MEI) Scientific or graphical calculator en OCR supplied materials: Printed Answer Booklet Formulae Further Mathematics B (MEI) * Sp ec im * INSTRUCTIONS Use black ink. HB pencil may be used for graphs and diagrams only. Complete the boxes provided on the Printed Answer Booklet with your name, centre number and candidate number. Answer all the questions. Write your answer to each question in the space provided in the Printed Answer Booklet. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s). 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 to a degree of accuracy appropriate to the context. INFORMATION The total number of marks for this paper is 44. The marks for each question are shown in brackets [ ]. 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 used. You should communicate your method with correct reasoning. The Printed Answer Booklet consists of 4 pages. The Question Paper consists of 8 pages. OCR /364/X Y40 B004/4. Turn over

2 Section A (33 marks) Answer all the questions. Find the acute angle between the lines with vector equations 3 r 0 and 3 r 5. [3] 3 (i) On an Argand diagram draw the locus of points which satisfy z arg 4i. [] 4 (ii) Give, in complex form, the equation of the circle which has centre at 6 4i and touches the locus in part (i). [4] 3 3 Transformation M is represented by matrix M. 4 (i) On the diagram in the Printed Answer Booklet draw the image of the unit square under M. [] x x (ii) (A) Show that there is a constant k such that M 5 for all x. [] kx kx (B) Hence find the equation of an invariant line under M. [] (C) Draw the invariant line from part (ii) (B) on your diagram for part (i). [] 4 You are given that z i is a root of the equation Find the other roots, the value of q. 3 z z qz 5 5 0, where q. 5 (i) Express in partial fractions. [] ( r)( r3) (ii) Hence find n, expressing your answer as a single fraction. [5] r ( r)( r3) [5] OCR 07 Y40

3 3 6 (i) A curve is in the first quadrant. It has parametric equations x cosh t sinh t, y cosh t sinh t where t. Show that the cartesian equation of the curve is xy. [] Fig. 6 shows the curve from part (i). P is a point on the curve. O is the origin. Point A lies on the x-axis, point B lies on the y-axis and OAPB is a rectangle. y B O A P Fig. 6 (ii) Find the smallest possible value of the perimeter of rectangle OAPB. Justify your answer. [4] x OCR 07 Y40 Turn over

4 4 Section B ( marks) Answer all the questions 7 (i) Use the Maclaurin series for ln( x) up to the term in 3 x to obtain an approximation to ln.5. [] (ii) (A) Find the error in the approximation in part (i). [] (B) Explain why the Maclaurin series in part (i), with x, should not be used to find an approximation to ln3. [] (iii) Find a cubic approximation to x ln x. [] (iv) (A) Use the approximation in part (iii) to find approximations to ln.5 and ln3. (B) Comment on your answers to part (iv) (A). [] 8 Find the cartesian equation of the plane which contains the three points (, 0, ), (,,) and (,, ). 9 A curve has polar equation r asin3 for, where a is a positive constant. (i) Sketch the curve. 3 3 (ii) In this question you must show detailed reasoning. Find, in terms of a and, the area enclosed by one of the loops of the curve. [5] 0 (i) Obtain the solution to the differential equation dy x 3y, where x 0, dx x [3] [5] [] given that y when x. [7] (ii) Deduce that y decreases as x increases. [] OCR 07 Y40

5 5 (i) It is conjectured that 3 n b... a,! 3! 4! n! n! where a and b are constants, and n is an integer such that n. By considering particular cases, show that if the conjecture is correct then ab. [] (ii) Use induction to prove that 3 n... for n. [7]! 3! 4! n! n! In this question you must show detailed reasoning. dy (i) Given that y arctan x, show that. [3] dx x Fig. shows the curve y x. y Fig. (ii) Find, in exact form, the mean value of the function f( x) x for x. [3] (iii) The region bounded by the curve, the x-axis, and the lines x and x is rotated through radians about the x-axis. Find, in exact form, the volume of the solid of revolution generated. [7] x OCR 07 Y40 Turn over

6 6 3 Matrix M is given by k 5 M 3 3, where k is a constant. (i) Show that det M ( k 3). [] (ii) Find a solution of the following simultaneous equations for which x z. 4x y 5z 6 x 3y 3z 6 x y z 6 (iii) (A) Verify that the point (, 0, ) lies on each of the following three planes. 3x y 5z x 3y 3z x y z 0 (B) Describe how the three planes in part (iii) (A) are arranged in 3-D space. Give reasons for your answer. (iv) Find the values of k for which the transformation represented by M has a volume scale factor of 6. [3] [] [4] [3] OCR 07 Y40

7 7 4 (i) Starting with the result i e cos isin, show that n (A) cos isin cos n isin n []. i i (B) cos e e [] (ii) Using the result in part (i) (A), obtain the values of the constants a, b, c and d in the identity cos6 cos cos cos 6 4 a b c d. [6] (iii) Using the result in part (i) (B), obtain the values of the constants P, Q, R and S in the identity (iv) Show that cos 64 P Q R S. [5] 6 cos cos6 cos 4 cos 5 In this question you must show detailed reasoning. Show that [3] arsinh xdx ln 3. [8] 3 3 OCR 07 Y40 Turn over

8 8 6 A small object is attached to a spring and performs oscillations in a vertical line. The displacement of the object at time t seconds is denoted by x cm. Preliminary observations suggest that the object performs simple harmonic motion (SHM) with a period of seconds about the point at which x 0. (i) (A) Write down a differential equation to model this motion. (B) Give the general solution of the differential equation in part (i) (A). [3] [] Subsequent observations indicate that the object s motion would be better modelled by the differential equation d x dx (*) k k 9 x 0 dt dt (ii) (A) Obtain the general solution of (*). en where k is a positive constant. Sp ec im (B) State two ways in which the motion given by this model differs from that in part (i). [3] [] The amplitude of the object s motion is observed to reduce with a scale factor of 0.98 from one oscillation to the next. (iii) Find the value of k. [3] At the start of the object s motion, x 0 and the velocity is cm s in the positive x direction. (iv) Find an equation for x as a function of t. [4] (v) Without doing any further calculations, explain why, according to this model, the greatest distance of the object from its starting point in the subsequent motion will be slightly less than 4 cm. [] END OF QUESTION PAPER Copyright Information: OCR is committed to seeking permission to reproduce all third-party content that it uses in the 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 ( after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB GE. 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. OCR 07 Y40

9 day June 0XX Morning/Afternoon A Level Further Mathematics B (MEI) en Y40 Core Pure ec 44 Sp MAXIMUM MARK im SAMPLE MARK SCHEME This document consists of 4 pages B004/4. Duration: hours 40 minutes

10 Y40 Mark Scheme Text Instructions Annotations and abbreviations Annotation in scoris and BOD FT ISW M0, M A0, A B0, B SC ^ MR Highlighting Meaning Other abbreviations in mark scheme E dep* cao oe rot soi www AG awrt BC DR Meaning im en Benefit of doubt Follow through Ignore subsequent working Method mark awarded 0, Accuracy mark awarded 0, Independent mark awarded 0, Special case Omission sign Misread ec Mark for explaining a result or establishing a given result Mark dependent on a previous mark, indicated by * Correct answer only Or equivalent Rounded or truncated Seen or implied Without wrong working Answer given Anything which rounds to By calculator This indicates that the instruction In this question you must show detailed reasoning appears in the question. Sp. June 0XX

11 Y40 Mark Scheme June 0XX. Subject-specific Marking Instructions for A Level Further Mathematics B (MEI) a b c d 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. 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, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader. 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, e.g. 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 A cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. 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. 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. 3

12 Y40 Mark Scheme June 0XX e f g h i j k 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, what is acceptable will be detailed in the mark scheme. If this is not the case please, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. 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. Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so. When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question. 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. 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; mark is generally appropriate, though this may differ for some papers. This is achieved by withholding one A mark in the question. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. Fresh starts will not affect an earlier decision about a misread. Note that a miscopy of the candidate s own working is not a misread but an accuracy error. If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. If in any case the scheme operates with considerable unfairness consult your Team Leader. Anything in the mark scheme which is in square brackets [ ] is not required for the mark to be earned on this occasion, but shows what a complete solution might look like 4

13 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 3 M. 7 cos 7 M or 0.70 radians A [3] (i) B a half line from 4i Allow 4i included or Im excluded B direction above the 4 4 [] positive x axis (ii) z (6 4i) 3 M correct form Re A centre correct M 3.a Attempt to find distance from centre to line A radius correct [4] 5

14 Mark Scheme Question 3 (i) 3 (ii) (A) Answer parallelogram vertices (0, 0), (, ), (5, 5), (3, 4) 3 x x 5 4 kx kx Solve Obtain k twice (C) Hence y x is an invariant line Draw y x. [] B Conjugate root i [because q ] i i correct coordinates calculated or drawn all correct.a [] B.a M a A either Guidance a im 4 (ii) (B) A [] M ec 3 (ii) AOs A [] E Sp 3 Marks M June 0XX en Y40 Alternative solution or i i 5 Conjugate root i [because q ] B ( z i)( z i) z z 5 z 3 5 z qz 5 ( z z 5)( z 3) ( i)( i) 3( i) 3( i) q M 3.a q A or q 5 0 M q A [5] 6 FT MA q 6 5 MA The other roots are 3 and ( i)

15 Y40 Marks AOs A(r 3) B(r ) M A =, B = r r 3 r r 3 A ec im [] June 0XX Guidance Correct form plus some progress en Answer A B r r 3 r r 3 Sp Question 5 (i) Mark Scheme 7

16 Y40 Answer r ( r )( r 3) n n n n 3 Marks M AOs 3.a A. n 3 n n 3 A 5n 5n 30 n 30 (n )(n 3) 5n 3n n n 3 ec M Guidance Write out series to identify: Cancelling terms, denominators 4, 5, n+ Ignore absence of initial. Non-cancelling terms at the beginning. Ignore absence of initial.. Non-cancelling terms at the end: Ignore absence of initial.. Attempt at writing as single fraction, with product of their linear terms. Completely correct argument; each of numerator and denominator can be either factorised or multiplied out. im 3(n )(n 3) (n )(n 3) 6(n 3) 6(n ) (n )(n 3) June 0XX en A Sp Question 5 (ii) Mark Scheme [5] 8

17 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 6 (i) xy cosh t sinh t M Substituting cosh tsinh t So xy 6 (ii) Perimeter xy (cosh t sinh t) (cosh t sinh t) M 4cosht A The minimum value of cosht is B 3.a So the minimum value of the perimeter is 4 A 3.a [4] 7 (i) ln A []. M Substitute x 0.5 into series for ln x up to x A Obtain [] 7 (ii) (A) error 0.0 A Compare ln.5 = to 4 d.p. [] 7 (ii) (B) ln 3 would require x =, beyond the range of convergence of the series E [].3 9

18 Y40 Mark Scheme Answer 7 (iv) (A) (B) 3.a Attempt at series for ln x x x3 x x 3 A [] M x.5 so x 0.. x ln Using x 0.5, ln (Much) better approximation to ln.5 ln3: Inside range of convergence A B [3] E E [] en (iv) x x3 3 ec 7 M Guidance x ln( x) ln( x) x ln( x) x ln AOs.a im ln Marks Sp Question 7 (iii) June 0XX 0.3.3

19 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 8 Need a vector perpendicular to the plane M.4 Or similar language e.g. need vector perpendicular to [ calculated vectors] or Vector product gives vector perpendicular to plane or Find normal vector using vector product 0 4 B 3.a Use vector product with any 3 two vectors in the plane. 3 Plane has equation 4x 3y z d M Passes through (, 0, ) so 4 d A Equation of plane is 4x 3y z 3. A 3.a [5]

20 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 9 (i) B.5 B for loop in first quadrant B for loop in second quadrant with indication (e.g. dotted line) of r negative 9 (ii) DR Area is 3 0 a sin 3d [] M*. Must be seen Use cos6 sin 3 to obtain M* 3.a A Must be seen 3 a cos 6 d 4 0 A FT Must be seen a 3 sin a 0 A dep* [5]

21 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 0 (i) dy 3 M Write equation in correct y dx x x form (may be implied) and integrating factor is exp(3 ln x) attempt to find integrating factor integrating factor is exp(ln(x 3 )) = x 3 A 3 dy x 3x y x dx M. Multiply by integrating factor 0 (ii) 3 d xy x dx 3 x y x c x, y c x y 3 x y = 3 x x Explain that y is the sum of two decreasing functions and hence decreasing AA A each side of equation B Use condition A.5 Expressing y as a function of x (can be in part (ii)) [7] B.a Write y as sum of two fractions OR differentiate E.4 OR show that the derivative is negative hence decreasing [] 3

22 Mark Scheme Result holds for n 3 k Assume! 3! 4! k! k! 3 k k! 3! 4! k! k! k k k k! k! k! which is k! A [] B.a. Assume true for n = k. Add correct next term to both sides M minus fraction with correct denominator on RHS A.a Showing that this is the given result with k + replacing k E.5 Complete argument E The result is true for n =. If true for n = k it is also true for k + hence true for n =, 3, 4, Guidance Two values of n to give two simultaneous equations MA k k! k! RHS is AOs 3.a im (ii) Marks M ec Answer b E.g. n = gives a b 5 and n = 3 gives a 6 6 Verify (or solve) a b Sp Question (i) June 0XX en Y40 [7] 4

23 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance (i) DR dy M Must be seen tany x so sec y d x dy A. Evidence of tan y dx sec y tan y must be dy x dx dy dx tan y AG seen A. Use x tan y to obtain given answer [3] (ii) DR M M for arctan must be seen A. arctanx A A0 for a decimal answer 4 [3] 5

24 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance (iii) DR x dx ( tan u) cos udu sec udu cos u du M. Must be seen M 3.a Substitute x tanu, A A. d x sec u du One intermediate step required Integrand New limits M 3.a Must be seen A. Integrate u sin u A oe, but A0 for a decimal 4 answer [7] 6

25 Y40 Mark Scheme Answer detm k Simplify to k 3 AG x 6 y z One of A [] M A or x 6i, y 0, z 6i Guidance Answer given so method must be clear Use of inverse matrix with k = 4. BC 3.a A0 if any solution given as final answer with x z. B Convincing substitution of the point into all three equations [] B B E E [4].4.4..a This may be implied A x 6i, y 0, z 6i 3.a im (ii) AOs ec 3 Marks M en Question 3 (i) June 0XX 3 3 (iii) (iii) (A) (B) Sp [3] The coefficients are the matrix M with k 3 and determinant of M is zero when k 3 Hence no unique solution The planes are distinct, so the planes form a sheaf 7

26 Mark Scheme Question 3 (iv) Answer k 3 6 k= 4 (i) (A) cos or 6 or k 3 i sin n e i n ein cos n isin n AG (i) e i cos isin AG c is 6 c6 5 c4 s 5 c s 4 s6 c6 5 c 4 c 5 c c 3c6 48c4 8c [3] M. c 3 Guidance or statement relating determinant to volume scale factor A. [] M 3.a A. MA. AA.a [6] 8 Answer given so working must be convincing ec cos6 Re 3.a [] M Sp (ii) M A A Hence obtain given result 4 AOs im 4 (B) Marks M June 0XX en Y40 Answer given so working must be convincing A any two terms correct, A all four a, b, c, d implicit or explicit

27 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 4 (iii) M 3.a set up expansion i cos z where z e z z z z z z z cos6 cos 4 cos (iv) Use result in part (iii): cos cos cos cos cos 64 6 A M. AG A. result of expansion collecting terms A.a A for any two coefficients correct, A all four P, Q, R, S implicit or explicit [5] M 3.a A Putting correct trig values into their (iii) [3] 9

28 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 5 DR dv M 3.a Use parts and arsinh x Let uarsinh x, dx v x M derivative of arsinh x is 4x 3 x AA A for each term 3 xarsinh x 0 dx 0 must be seen 4x A. 3 xarsinh x 4x 4 5 arsinh ln = ln3 AG A M A [8]. converting to logarithms AG so must be convincing 0

29 Mark Scheme Question 6 (i) (A) Answer T with T d x x dt 6 (i) (B) x a cos t b sin t 6 (ii) (A) Auxiliary equation k (k 9) 0 6 (iii) e 3 k 0.98 k [3] E Sp and period is a bit more than s [3] B Guidance or e.g. x Asin( t ) a.a 3.5b E [] M 3.5b B A [3] 3.3 ec Motion is damped [SHM] 3.3 im Hence x e ( A cos3t B sin3t ) (B) A B A A kt (ii) AOs 3.b [] M Roots k 3i 6 Marks M June 0XX en Y40 3.b must be comments about motion Attempt may use incorrect period for correct period

30 Y40 Mark Scheme June 0XX Question Answer Marks AOs Guidance 6 (iv) Starting with e kt x ( A cos3 t B sin3 t ), with k as in (iii). t 0, x 0 gives A 0 B a dx kt e (3B cos3t kb sin 3 t) dt M dx A.4 Must see reasoning for where t 0, gives B 4 dt B comes from. kt x 4e sin 3twith k as in (iii) B 3.3 [4] 6 (v) Without damping ie k 0 the amplitude is 4 E 3.4 In fact there is slight damping, so amplitude slightly E 3.5a less than 4. []

31 Y40 Mark Scheme June 0XX Question AO AO AO3(PS) AO3(M) Totals i ii i iiA iiB iiC i ii i 0 0 6ii i iiA iiB iii 0 0 7ivA ivB i 0 0 9ii i ii i 0 0 ii i ii iii i ii iiiA iiiB iv iA 0 0 4iB 0 0 4ii iii 0 5 4iv

32 Y40 Mark Scheme June 0XX Question AO AO AO3(PS) AO3(M) Totals 6iA 0 3 6iB iiA iiB iii 0 3 6iv 0 4 6v Totals

33 A Level Further Mathematics B (MEI) Y40 Core Pure Printed Answer Booklet Date Morning/Afternoon Time allowed: hours 40 minutes OCR supplied materials: Printed Answer Booklet Formulae Further Mathematics B (MEI) Centre number ec Last name * * im First name en You must have: Printed Answer Booklet Formulae Further Mathematics B (MEI) Scientific or graphical calculator Candidate number Sp INSTRUCTIONS Use black ink. HB pencil may be used for graphs and diagrams only. Complete the boxes provided on the Printed Answer Booklet with your name, centre number and candidate number. Answer all the questions. Write your answer to each question in the space provided in the Printed Answer Booklet. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s). 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 to a degree of accuracy appropriate to the context. INFORMATION 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 used. You should communicate your method with correct reasoning. The Printed Answer Booklet consists of 4 pages. The Question Paper consists of 8 pages. OCR 07 QN: 603/364/X Y40 B004/4.0 Turn over

34 Section A (33 marks) (i) OCR 07 Y40

35 3 (ii) 3 (i) (ii) (C) A spare copy of this grid for question 3 (i) and (ii) can be found on page OCR 07 Y40 Turn over

36 4 3 (ii) (A) 3 (ii) (B) 4 OCR 07 Y40

37 5 5(i) 5(ii) (answer space continues on next page) OCR 07 Y40 Turn over

38 6 5(ii) (continued) 6 (i) 6 (ii) OCR 07 Y40

39 7 Section B ( marks) 7 (i) 7 (ii) (A) 7 (ii) (B) 7 (iii) 7 (iv) (A) OCR 07 Y40 Turn over

40 8 7 (iv) (B) 8 OCR 07 Y40

41 9 9 (i) 9 (ii) OCR 07 Y40 Turn over

42 0 0 (i) OCR 07 Y40

43 0 (ii) (i) DO NOT WRITE IN THIS SPACE OCR 07 Y40 Turn over

44 (ii) OCR 07 Y40

45 3 (i) (ii) OCR 07 Y40 Turn over

46 4 (iii) OCR 07 Y40

47 5 3 (i) 3 (ii) OCR 07 Y40 Turn over

48 6 3 (iii) (A) 3 (iii) (B) 3 (iv) 4 (i) (A) OCR 07 Y40

49 7 4 (i) (B) 4 (ii) OCR 07 Y40 Turn over

50 8 4 (iii) 4 (iv) OCR 07 Y40

51 9 5 OCR 07 Y40 Turn over

52 0 6 (i) (A) 6 (i) (B) 6 (ii) (A) 6 (ii) (B) OCR 07 Y40

53 6 (iii) 6 (iv) OCR 07 Y40 Turn over

54 6 (v) Spare copy of grid for question 3 (i) and (ii) OCR 07 Y40

55 3 DO NOT WRITE ON THIS PAGE OCR 07 Y40 Turn over

56 4 DO NOT WRITE ON THIS PAGE Copyright Information: OCR is committed to seeking permission to reproduce all third-party content that it uses in the 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 ( after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB GE. OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations OCR 07 Y40