FSMQ Advanced FSMQ Additional Mathematics 699 Mark Scheme for June Oxford Cambridge and RSA Examinations
OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of pupils 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 Any enquiries about publications should be addressed to: OCR Publications PO Box Annesley NOTTINGHAM NG DL Telephone: 87 77 66 Facsimile: 6 E-mail: publications@ocr.org.uk
699 Mark Scheme June OCR - ADDITIONAL MATHEMATICS 699 Marking instructions. The total mark for the paper is. Marks for method are indicated by an M. A method that is dependent on previous work is DM. Marks for accuracy are of two kinds: (i) A mark indicates correct work only and (ii) F mark indicates that a "follow through" is allowed. If an M mark is not gained then nor do any of the accuracy marks associated with it. Marks not associated with a method are denoted B, which should be treated as "correct only", and E which may be wrong because of a previous error. Marks are not divisible except as indicated. e.g. A, means that are awarded for a correct answer and for an answer that is only partially correct, as agreed at the meeting of Examiners. When the method of solution is not one that has been discussed and does not fit the existing scheme then an alternative scheme should be devised which maintains the same number of M, A, F, B and E marks. You should also bring this to the attention of the Principal Examiner. The rubric says that the norm is for answers to be given to s.f. except where indicated. Where this rubric is ignored then mark should be deducted once in the paper, at the point where it is first met. This should be indicated -, TMSF or -TFSF. Details will be discussed at the meeting of examiners. Misreading of a question (including the candidate's own working) should normally be penalised by the loss of the relevant accuracy mark or two marks (whichever is less); but if the question is made substantially easier then further penalties may be imposed. Sub-marks should be shown near to the relevant work. If these are individual marks then the appropriate letter should be given. Sub-marks are given in the question paper and the mark scheme. For substantially correct solutions a number of sub-marks may be combined, even up to the total mark for the question for a totally correct question. The sum of the sub-marks are then added and ringed at the end of the question. (This means that a totally correct question has the total mark written twice - once as a "sum of sub-marks" and unringed and once ringed as the total for the question.) The total mark for the paper should be given on the front page, top right and ringed. Work that is crossed out and not replaced should be marked. If work has been crossed out and replaced then the replacement work should be marked even if it is incorrect and the crossed out work correct. Any notation that is understandable may be used to support your marking. In particular: isw ignore subsequent working www without wrong working soi seen or implied
699 Mark Scheme June An independent person should be used to check the summation of marks. You should add the ringed marks on the paper to check the addition and the independent checker should add the unringed marks. There is a fee paid for this checking - if you are unable to find anyone to do this work the Board and the Principal Examiner must be informed. Please mark in red. If examiners have any doubt about the interpretation of any instructions or if any point of difficulty arises during the marking of scripts, they should communicate with the Principal Examiner.
699 Mark Scheme June Section A x < ( x ) x< x 7< x 7 x > Sight of x Sight of ax and b where either a = or b= 7 oe Final answer WWW = x + x x = x+ 66x x Signs and powers out of coefficients worked out All coefficients and Ignore terms of higher power (i) Remainder is f( ) = + 8 = For long division x + x in working and x in quotient must be seen for Or by inspection (x + )(x + ) for Or long division must be seen or implied (ii) + + 8= x x x ( )( ) x+ x 6x+ 8 = ( )( )( ) x+ x x = x =,, D Factorise cubic to give (x + )(ax + bx + c) Solve their quadratic Allow ans with no working Alt: Trial to find one root: x =,, x =,,
699 Mark Scheme June (i) 6 = =.8 6 96 (ii) 6 6 6 6 = =.8.88 96 96 7 9 = = =.9 96 Either form or.8 isw terms soi Powers Ans in either form or. Alt: Add three terms 6 + + 6 6 6 6 6 =.7 +. +.77 =.9 both coeffs powers ans
699 Mark Scheme June (i) dy = x 6x 9 dx = x x = x x = when 6 9 ( )( ) x x+ = x=, When x=, y = Diffn and set = Derived fn Stationary point d y dx = 6x 6 < when x= so maximum To find nature of turning points Allow SC for (, ) with no working Alternative ways to demonstrate maximum at x = Value of y + y < y = y < Gradient of tangent + dy dx > dy dx = dy dx < Allow at most one integer either side (typically, x =, if turning point is correct) (ii) General shape: turning points in correct quadrants Intercept on y axis in [,] Does not turn back on itself. 6 (i) u = 9, v= 6, s = 6 = + a 6 9 6 9 6 86 a = = = m s (ii) u = 9, v= 6, a = 6= 9 t 8 t = = secs The two parts can be the other way round Using correct formula Correct substitution Using correct formula
699 Mark Scheme June 7 (i) sinθ cosθ sin θ + cos θ + = cosθ sinθ sinθcosθ = sinθcosθ Alt: sin θ+ cos θ = cos θ sin θ + = sin θ sin θ sin θ cosθ + = cosθ sin θ sin θcosθ (ii) sinθ cosθ sinθcosθ = + = cosθ sinθ tanθ + = tanθ (iii) tanθ + = tan θ + = tanθ tanθ t t+ = Using (i) and tan Clear fractions to give term quadratic ± 6 t = = ± =.7 and.68 θ = and 7 ( ) Sub numbers into correct quadratic sf or more Rounds to these 8 Sp Case for and for 7 with no supporting working v= 6 t t + t ( ) ( ) s = 6t 6t + t dt = t t + t A, Integrate Terms each error = 6 m If 6 is left out then / only. D Sub t = Cao 6
699 Mark Scheme June 9 (i) + + Centre is, = (8, ) For 8 WWW For WWW Nb Working with vectors to give diameter = [,] and so radius = [7,] giving centre ( 7, ) is correct. (ii) PC = (8 ) + ( ) = = Alt: Length of diameter = + = + ( ) ( ) For = = (iii) (i) Sub (,) Radius = ( x 8) + ( y ) = x + y 6x y+ 6 + = x + y 6x y+ 8 = Gives k = (ii) Sub (, ) (iii) Gives c = ( ) ( ). When x = y = = for cubic Or when x=, y > for cubic Correct use of formula including and using their midpoint. John s model is better D 7
699 Mark Scheme June Section B Allow sf in this question (i) AF BF = = sin7 sin6 sin AF = sin 7 ( =.7 m) sin BF = sin 6 =. m oe sin Sin rule applied Sight of and 7 Correct sine rule to find BF (ii) Alt: Cosine rule for BF: BF = +.7.7 cos6 = 78 BF =. FT = AF tan =.7 tan =.6 m Anything that rounds to.6 (iii) CF =.7sin6 =6. m Or: = their BF sin7 Accept 6. or 6 TheirFT tanθ = TheirCF θ =. Alt: to find CF. Area of triangle = AF. ABsin 6 = CF = CF = 6. F Using tan correctly Substituting correctly Accept or 8
699 Mark Scheme June (i) (ii) (iii) = y.x.x dy =.6x. dx When x= g =. g = n AB: y = x+ y = t ( x ) Solve simultaneously: y+ x= y+ x= 6y+ x= 6y+ 9x= x = x=, y = 6 SC: answer with no working Area of triangle = their y = ( ) Area under curve =.x.x dx =.x.7x = 6. Area of card = + 6. =. Other methods, follow scheme iee Area of triangle area as integral Integrand value for area Final answer F E Derivative Find g t and use of m m = For g n Line in any simplified form Method to eliminate one variable x and y. Might appear anywhere in this part Ignore limits here Condone lack of ve sign 9
699 Mark Scheme June (i) 7 7 Ali: Beth: t t+ (ii) 7 7 = t t+ 7( t+ ) 7t = t( t+ ) 7t+ 7t = t( t+ ) tt ( + ) = Accept 7 Beth: t Subtraction of their terms = Multiply out and simplify Alternative (based on alternative answer to (i) (iii) 7 = t + 7 t 7t = 7 t ( t+ ) ( ) 7 7 6 t = t t + t 6 ( ) t + t = t t+ = tt ( + ) = 6 t + t = t t 8 + = ( t+ 8)( t 6) = t = 6 For quadratic in simplified form. (See below) www Ali takes 6 hours and Beth takes 8 hours. SC for answer with no working What is simplified form? Either a quadratic with all three terms on left = ready for the use of the formula OR: Divide through by giving t + t = 8 ready for solving by the completion of the square.
699 Mark Scheme June (i) x + y oe (ii) y x (iii) Deriving a linear inequality E One line Other line Shading for both, ft their inequalities No Scales: B, B, E Condone scales not as instructed. (iv) C = 8x + 6y Correct point is (, ) Cost = 7 In absence of OF, 8 + 6 must be seen (v) Now minimum cost is at (7, ) Giving 6 Nb (8, ) gives 6 Sub in OF
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