FSMQ. Additional FSMQ. Mark Scheme for June Free Standing Mathematics Qualification. 6993: Additional Mathematics

Transcription

1 FSMQ Additional FSMQ Free Standing Mathematics Qualification 699: Additional Mathematics Mark Scheme for June 01 Oxford Cambridge and RSA Examinations

2 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, OCR Nationals, 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 01 Any enquiries about publications should be addressed to: OCR Publications PO Box 5050 Annesley NOTTINGHAM NG15 0DL Telephone: Facsimile: publications@ocr.org.uk

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

4 Subject-specific Marking Instructions a b Annotations should be used whenever appropriate during your marking. 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 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. 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.

5 d e 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. 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 g 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 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. or 4 significant figures where 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. 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. 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.

6 Viewing tips for this paper In general, set your screen to fit width. You may find it helpful to set to fit height for the some questions: [if you set a view, it stays for subsequent scripts]. If the writing is too small, you may wish to zoom in. 4

7 Section A 1 (i) (x ± 1)(x ± ) ( 0) Attempt to factorise oe SC Test integers and select 1 Correct (or sight of 1 & ) and (ii) 1 x www Alternative: Draw curve for parabola the right way up Correct points on x-axis answer 1 [] [1] Answer Correct answer Or: Follow through their ans to (i). Accept x and 1 x Or : from 1 to inclusive (must imply inclusion of end points). Filled in circles must be evident. SC if correct but M0 in (i). Accept alternative conventions. Answer must be a range (ie just a set of points is 0). 5

8 (i) p 5 p does not have to be 0.8 for 1 this mark but the power must 5 be 5. (ie p could be 0.) () 15 Alternative: P(1) P(5) 5 terms added, each term with powers correct Answer 10 pq (ii) www 65 [] Condone missing coeffs for Must include powers of p and q and 5 5 or C (which need not be evaluated Powers Coefficient soi Accept but not 0.05 Terms are: , 0.048, 0.051, , Can be obtained by listing. [4] 6

9 (i) f() = 1 7+a + 6 = 1 a = 1 a = 7 [] Alternative: Substitute a = 7 and show that R = 1 (ii) f(1) = 0 or (x 1) seen f(x) = (x 1)(x )(x + ) [] Divide, try factor theorem for at least one value, or obtain a -term quadratic factor by inspection. Using or getting a correct factor or root Answer If this method is used then if long division is used then x x must be seen. NB Answer given so long division must be totally correct for Divide means you need to see the x in the quotient and x and x terms correct in the initial dividing line. 7

10 4 u v In any order using any valid formulae s t s www Ignore units eg s vuat a www [4] eg a Alternative order: a s MR u = 0 and v = 10 gives s = 50, a = 1 Or u= 0 and v= 16 gives s = 80 and a = 1.6 A0 A0 8

11 5 (i) sin = sin + 1 sin + sin = 0 www Sight of and use of cos = 1 sin Must see = 0 [] NB answer given (ii) (sin )( sin + 1) = 0 Solve to obtain sin 1 or sin SC sin θ 1, sin θ 1 or sin θ Sight of both values θ 90,18., 1.8 (only) (Allow 18 and ) θ 70, 41.8, 18. A [4] All with no extras in range Ignore 90 for one or two values Or: all values correct but extra values in range. Anything that rounds to 41.8 and 18 Allow 18 but not 4 9

12 6 (i) dy Differentiation 6x 18x1 All three terms dx dy When x, dep Sub x = into or factorise their dx derived function. Get 0 or set = 0 and get. (ii) d y 1x 18 dx d y When x, 0 dx giving a minimum [4] [] Diffn their derived function correctly. BOD no arithmetic computations seen. At least terms with powers reduced by 1 (NB: beware division by x). Do not condone division by 6 before substituting for x. NB answer given. Numerical values must be seen Second dep on first Using the function x can earn A0. Alternative: Sign of gradient either side of x = Or: Values of y either side of x = and the value of y at stationary point. Correct answer (provided l.h. x > 1) BOD no arithmetic computations seen. For LH x greater than 1 Allow sketch of function indicating left stationary value is maximum and right one is minimum. 10

13 Question Answer Mark Guidance 7 (i) (CB ) cos0 8, 9 must be used, any angle Ignore units CB.11 (ii) sin ABC sin their 0 8 their.11 sin ABC ABC Bearing = 15 Alternative methods: 9 their CB 8 Cosine Rule: cosabc 9 their CB ft Then angle and bearing ft OR: Perpendicular from C and use of sin twice h 8sin their0.76 ft.76 SinABC theircb Then angle and bearing ft Or: Find other angle by sine rule Angle ACB = giving ABC = Bearing = 180 ( ) = 15 ft ft [] ft [4] soi Anything that rounds to.11 Correct application of sine rule Must be same angle as used in (i) and their CB Anything that rounds to 6 www Anything that rounds to 15 Correct application of cos rule Must be same angle as used in (i) and their CB 90 + their ABC NB Question asks for ABC so if not found /4 Angle = can earn A0 A0 (for ABC) ft only 11

14 8 (i) x x x dx Integrate Test for integration is are there at x x All three terms least two terms with the power 0 0 increased by 1? 8 Care that the process is not just oe multiplying each term by x. Completion to Working must be seen as the www []. answer is given. Ignore absence of 0. (ii) Because the curve crosses the x-axis in the range Because one bit is +ve and the other is ve. (iii) 1 x x x x or x x or 1 Total area = 1 4 [1] [] Calculation of their integral between 0 & 1 or 1& One of the areas Any reference to x= - will be 0. If there is an additional statement give 0. 1

15 9 (i) h = 7 5 cos0 = [1] (ii) Set cos = 1 h =7 ( 5) = 1 (iii) 9 7 5cos(480 t) cos(480 t) 0.4 oe 480t t 0.66 time = 0.66mins 14 sec [] [4] Substitute h = 9 soi Allow 114 leading to t =

16 Section B 4,6 [1] (ii) Attempt to find radius or diameter by Answer given with no working 47 6 Distance MC: pythagoras. then bod B 5 soi (ie r = 5, r = 5, d = 10, d = 100) 10 (i) Equation of circle: x4 y6 5 5 [4] Must include their M and their r Can be expanded form. Alternative: Equation of circle on AC as diameter: x1 x7 y10 y 0 x x y y x y isw (iii) B lies on circle as Working must be convincing (iv) gradient of AM = gradient of BM = [1] One gradient (need not be simplified) Second gradient (need not be simplified) Labelling does not need to be specific. SC Both gradients upside down or signs the wrong way round B0 B0 4 Since 1 the lines are 4 perpendicular Demonstration that m 1 m = 1 is satisfied and all working to derive gradients shown. [] 14

17 Alternative: Use of Pythagoras Attempt to find all three lengths 5, 5, 50 seen and used Arithmetic correct and final statement (v) 4 4 Idea of BM = MD soi B to M = M to D = D is 0,, Each value [] Alternative: Centre as midpoint: Idea 8 x 4 x 0 9 y Each value 6 y Alternative: Equation BM is y x 4 Sub in eqn for circle x 8x0 x 0 Sub to give y Idea Each value 15

18 11 (i) dy Differentiation If no differentiation then 0/5 x dx At A gradient of tangent = 1 ft Follow through their gradient of so gradient of normal =. tangent. dep Using (, ) and their normal 1 Eqn of AB is y x gradient terms only y x6 oe [5] (ii) line meets curve when x x 6 Equate their straight line to given curve. x x60 Quadratic xx0 9 At B x, y [] (iii) Area between = Area under line area under curve dep [4] Attempt to evaluate area under curve by integration soi Attempt to evaluate area under their straight line by trapezium or integration soi Subtracting areas, dep on both M marks Answer Seen by power increased by 1. Care not to multiply by x Ignore absence of limits for first marks 16

19 1 (i) Substitute: 75 = 900a + 0b 40 = 600a + 60b Allow unsimplified coefficients 1 1 Solve: a,b1d v v 0 0 Solve a b (ii) D (iii) 1 Substitute: 50 v v or v 0v [5] [] Calculation at each value and subtraction attempted For either 15 or 76.5 soi Allow 8.8 Substitute Quadratic (in any form) isw ie equal coefficients and subtract or correct substitution. NB Answers given so algebra for first value found must be convincing. Or.75 or 5 Correct application of completion of square is (v + 10) = k seen v mph [4] Solving their quadratic using correct formula or completion of square or B answer with no working SC for trial and improvement with values between 0 and 5. ans correct to sf SCB If answer given with no quadratic. Final answer is anything that rounds to. Ignore negative values 17

20 1 (i) (ii) ( + h) ( + h) h 8.4h.h h 8 1h6h h [] For each coefficient or term that is correct Ignore incorrect identification of coefficients after expansion Mark final line Change in y Change in x ie allow answer left in simplified expansion form. Accept description in words Gradient = h h 8 8 h h [] Only award if you are satisfied that the algebra is correct (iii) h 8 81h6h h 8 h h 1h6h h 1 6hh h (iv) Their 1 in (iii) (v) 4 4 h 16h4h 8h h [] [1] Or using their part (i) Dependent on (iii) being a polynomial. Allow 16 + h + (higher orders of h) This answer must be consistent with (iii) Gradient of chord = + 4h + 8h + h Giving www [] Allow + (higher orders of h) Dependent on previous work 18

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