GCE Mathematics. Mark Scheme for June Unit 4721: Core Mathematics 1. Advanced Subsidiary GCE. Oxford Cambridge and RSA Examinations

Transcription

1 GCE Mathematics Unit 7: Core Mathematics Advanced Subsidiary GCE Mark Scheme for June 05 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, 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 05

3 7 Mark Scheme June 05 Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M Method mark awarded 0, A0, A Accuracy mark awarded 0, B0, B Independent mark awarded 0, SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in Meaning mark scheme E Mark for explaining U Mark for correct units G Mark for a correct feature on a graph M 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

4 7 Mark Scheme June 05 Subject-specific Marking Instructions for GCE Mathematics Pure strand 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 A cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.

5 7 Mark Scheme June 05 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, 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. 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 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. 5

6 7 Mark Scheme June 05 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; 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

7 7 Mark Scheme June 05 Question Answer Marks Guidance 8 M Multiply top and bottom by or evidence of multiplying out needed Alternative: M Correct method to solve simultaneous equations formed from 8 8 A Either numerator or denominator correct equating expression to a b A Either a or b correct A Final answer cao A Both correct [] (i) B Excellent curve in both quadrants: correct shape, symmetrical, not touching axes asymptotes clearly the axes not finite allow slight movement away from asymptote at one end but not more. N.B. Ignore feathering now that answers are scanned. B only correct shape in nd and th quadrants only. Graph must not touch axes more than once. Finite plotting condoned. [] (ii) M ( y ) or ( y ) x x y oe x A Fully correct, must include y [] (iii) Stretch B Stretch or stretched etc.; do not accept squashed, compressed, enlarged etc. Scale factor parallel to the x-axis (or y-axis) B [] Correct description Condone just factor but no reference to units. Must not follow e.g. reflection ( y ) or( y ) is M0 x x 0/ if more than one type of transformation mentioned ISW non-contradictory statements For parallel to the x/y axis allow vertically, in the x/y direction. Do not accept in/on/ across/up/along/to/towards the x/ y axis 7

8 7 Mark Scheme June 05 Question Answer Marks Guidance (i) 5 8 B cao [] (ii) 5 Fourth root soi (iii) 9 5 k x k k 6 0 ( k )( k ) 0 A [] M A [] M* cao www ( 5 ) or 5 5 or other correct product of two simplified powers of 5 oe cao www Use a substitution to obtain a quadratic, or factorise into brackets each containing Mdep Attempt to solve resulting three-term quadratic see guidance in appendix x No marks if whole equation cubed/ rooted etc. No marks if straight to quadratic formula with no evidence of substitution at start and no cube rooting/cubing at end. k, k A Correct values of k Spotted solutions: If M0 DMO or M DM0 x, x M Attempt to cube at least one value SC B x = 7 www x 7, x 8 A Final answers correct ISW SC B x = 8 www [5] (Can then get 5/5 if both found www and exactly two solutions justified) 5 (i) AB (5 ) ( ) M Attempt to use Pythagoras theorem / numbers substituted correctly and attempt to square root AB 5 A Final answer correct, must be fully processed. ±5 is A0. [] 8

9 7 Mark Scheme June 05 Question Answer Marks Guidance (ii) 5 M Correct method to find mid-point of line Alternative using general point on the, perpendicular (.5, ) A M States P (x, y) a point on the Gradient of AB = B Processed perpendicular and attempts PA = PB or PA = PB Perpendicular gradient = Bft processed A At least one of PA, PB correct their gradient A Both correct M Expands and simplifies A Correct equation found A Correct equation in required form 7 M Equation of straight line through their midpoint, any non-zero gradient in any form y ( x ) A 6x8y 9 0 A cao Must be correct equation in required form i.e. k( 6x 8y 9) 0 for integer k. [7] Must have =0 6 x (5 x) = M* Substitute for x/y or valid attempt to If y eliminated: eliminate one of the variables y + 0y = 0 x 0x + 8 = 0 A Three term quadratic in solvable form (y + )(x ) = 0 (x )(x ) = 0 Mdep Correct method to solve three term quadratic see appendix x, x y, y Spotted solutions: If M*0 SC B x =, y = www A Both x values correct SC B www A [5] 7 (a) (x + )(5 x) = 5x x + 5 x M dy dx = 0x x A M A [] Both y values correct. Allow A mark for one correct pair of x and y from correct factorisation. Attempt to multiply out brackets, Must have four terms, at least three correct Fully correct expression. Do not ISW if signs then changed. Max /. Attempt to differentiate their expression, (power of at least one term involving x reduced by one) Must show on both line and curve (Can then get 5/5 if both found www and exactly two solutions justified) Alternative using product rule: Clear attempt at correct rule M* Both expressions fully correct A Expand brackets of both parts M*dep Fully correct expression A 9

10 7 Mark Scheme June 05 Question Answer Marks Guidance (b) dy M x Attempt to differentiate i.e. x k soi for dx positive integer k A Fully correct When x = 8 d y dx ( 8) B ( 8) www Must use 8 6 dy A Final answer dx 6 8 [] 8 (i) (x )(x + ) = 0 M Correct method to find roots see appendix A Correct roots x, x Aft Good curve: Correct shape, symmetrical positive quadratic Minimum point in the correct quadrant for their roots (ft) their x intercepts correctly labelled (ft) x misread as x earns max /: dy x M A0 MR dx ( 8) B Final answer A0 MR B y intercept at (0, ). Must have a graph. 8 (ii) x, x [] M Aft [] Chooses the outside region Follow through x-values in (i). Allow x, x, x or x but do not allow x and x If restarted, fully correct method for solving a quadratic inequality including choosing outside region needed for M NB e.g. x scores MA0 Must be strict inequalities for A mark 0

11 7 Mark Scheme June 05 Question Answer Marks Guidance 8 (iii) b ac = ( + k) M Rearrangement and use of b ac < 0, must involve and k in constant term (not k) 5 + 8k < 0 A p + 8k < 0 oe found, any constant p. p need not be simplified 5 k 8 A [] Correct final answer Alt for first two marks: M Attempt to find turning point and form inequality k < y min 5 A turning point correct (, ) 8 If M0 (either scheme) SC B or seen 9 (i) dy dx = M Attempt to differentiate, at least two nonzero terms correct 6x ax + 8 A Fully correct When x =, d y dx = 0 8a M Substitutes x = into their d y dx These Ms may be awarded in either order dy = 0 gives a = M Sets their d y to 0. Must be seen dx A dx [5] (ii) d y dx =x 6 M Correct method to find nature of stationary Alternate valid methods include: point e.g. substituting x = into second ) Evaluating gradient at either side derivative (at least one term correct from their first derivative in (i) ) and consider the of ( x ) e.g. at, 6 at 5, 8 sign ) Evaluating y = 6 at and either side of ( x ) e.g. (, 7), (5, ) When x =, d y dx > 0 so minimum A www If using alternatives, working must be [] fully correct to obtain the A mark (iii) 6x 6x + 8 = 0 M Sets their derivative to zero (x )(x ) = 0 M Correct method to solve quadratic (appx ) Could be (6x )(x ) = 0 A oe or (x )(x 8) = 0 x []

12 7 Mark Scheme June 05 Question Answer Marks Guidance 0 (i) C = (5, ) B Correct centre (x 5) + (y + ) 5 = 0 M (x ± 5) 5 and (y ± ) seen (or Or attempt at r = f + g c implied by correct answer) Radius = 5 A Correct radius do not allow A mark from ±5 or A0. [] (x + 5) and/or (y ) 0 (ii) Gradient PC = M 8 5 A Gradient of tangent = Bft y ( x 8) M Attempt to find gradient of radius (/ correct) their gradient processed Equation of straight line through P, using their perpendicular gradient (not from rearrangement) See also alternative methods on next page Do not allow use of gradient of radius instead of tangent yx A Rearrange to required form www AG Ignore order of terms [5] PLEASE SEE NEXT PAGE FOR 0ii ALTERNATIVE METHODS (iii) Q = (0, ) B Q found correctly For the M mark, allow splitting into R = (0, 8) B R found correctly two triangles 68 8 Area = (8 ) 8 M Attempt to find area of triangle with their If using PQ as base then expect to see Q, R and height 8 i.e. ( y R yq ) www 0 A []

13 7 Mark Scheme June 05 Alternative methods for 0(ii) Alternative by rearrangement Alternative for equating given line to circle Alternative for implicit differentiation: Gradient of radius = Substitute for x/y or attempt to get an equation in M* Attempt at implicit differentiation as MA 8 5 variable only M evidenced by dy y term Attempts to rearrange equation of line to find gradient of line = and compares with gradient of radius M Multiply gradients to get B Check (8, ) lies on line B k(x 6x + 6) = 0 or k(y y + ) = 0 A Correct method to solve quadratic see appendix M x = 8, y = found A States one root implies tangent B dx dy dy A x y 0 0 dx dx A Substitution of (8, ) to obtain Then as main scheme OR Attempts to rearrange equation of line to find gradient of line = Mdep Check (8, ) lies on line B

14 7 Mark Scheme June 05 APPENDIX Solving a quadratic This is particularly important to mark correctly as it features several times on the paper. Consider the equation: x x 0 = 0 ) If the candidate attempts to solve by factorisation, their attempt when expanded must produce the correct quadratic term and one other correct term (with correct sign): (x + 5)(x ) M x and 0 obtained from expansion (x )(x ) M x and x obtained from expansion (x + 5)(x + ) M0 only x term correct ) If the candidate attempts to solve by using the formula a) If the formula is quoted incorrectly then M0. b) If the formula is quoted correctly then one sign slip is permitted. Substituting the wrong numerical value for a or b or c scores M0 ( ) 0 ( ) 0 ( ) ( ) earns M (minus sign incorrect at start of formula) earns M (0 for c instead of 0 is the only sign slip) M0 ( sign errors: initial sign and c incorrect) M0 (c on the denominator instead of a) Notes for equations such as x x 0 = 0, then b = would be condoned in the discriminant and would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0. c) If the formula is not quoted at all, substitution must be completely correct to earn the M ) If the candidate attempts to complete the square, they must get to the square root stage involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions! x x 0 0 x x 0 0 x 6 x 6 x This is where the M is awarded arithmetical errors may be condoned provided x seen or implied 6 If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last full attempt.

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