GCE Mathematics A H230/02: Pure Mathematics and Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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GCE Mathematics A H230/02: Pure Mathematics and Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021 Oxford Cambridge and RSA Examinations GCE Mathematics A H230/02: Pure Mathematic ... s and Mechanics Advanced Subsidiary GCE Mark Scheme for Autumn 2021Oxford 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 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 2021H230/02 Mark Scheme October 2021 2 Text Instructions 1. Annotations and abbreviations Annotation in RM assessor Meaning 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 BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark, indicated by *. The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which rounds to BC By Calculator DR This question included the instruction: In this question you must show detailed reasoning.H230/02 Mark Scheme October 2021 3 2. Subject-specific Marking Instructions for A Level Mathematics A a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. 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 always 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.H230/02 Mark Scheme October 2021 4 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, 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 method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. 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. 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. 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, 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.H230/02 Mark Scheme October 2021 5 f 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. • When a value is not given in the paper accept any answer that agrees with the correct value to 3 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 s.f”. Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts: • If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others. • If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete. • if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. 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 or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers, provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” or “Determine”. 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. j If in any case the scheme operates with considerable unfairness consult your Team Leader.H230/02 Mark Scheme October 2021 6 Question Answer Marks AO Guidance 1 2 2 2 4 0 ( )3 + − = k M1 1.1 Sets f 2 ( ) equal to 0 Or any other complete method k = −6 A1 1.1 [2] 2 y x ≤ − + 2 4 B1 1.1 SC All 3 but with (correct) strict inequalities B2 2 y x ≥ − 4 B1 1.1 x ≥ 0 B1 1.1 Only 2 but with (correct) strict inequalities B1 [3] 3 (a) V at b = + 2375 6 , 2825 12 = + = + a b a b M1* 3.3 Using a linear model to set up two equations with the values given Or: correct attempt to find gradient M1dep* 3.1a Solving their two equations to find a linear model, possibly BC Using their gradient to find a linear model V t = + 75 1925 A1 1.1 cao [3] 3 (b) When t = 0, V = 1925 B1ft 3.4 Uses model and states initial investment is their ‘b’ Only ft a linear model from part (a) Compare their “1925” with 1900 and make a sensible comment about whether the straight-line model in part (a) is supported or not B1ft 3.5a [2]H230/02 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 4 Considers f (2 ) f (2) h h + − B1 2.1 Or considers f ( ) f ( ) x h x h + − with x = 2 substituted at some point ( ) ( )2 2 f 2 2 2 3 2 8 8 3 h h h h + = + − = + + − M1 1.1 Considers f 2 ( + h) and attempts to expand Or considers f (x h + ) and attempts to expand f 2 f (2) 2 8 5 5 2 8 ( + − = + + − = + h h h h h ) ( 2 2 ) A1 1.1 Correct simplified expression for f 2 f (2) ( + − h) Correct simplified expression for f f ( ) ( x h x + − ) f (2 ) f (2) h 2 8 h h + − = + A1 1.1 Correct simplified expression for f (2 ) f (2) h h + − Correct simplified expression for f ( ) f ( ) x h x h + − ( ) ( ) 0 f 2 f (2) f 2 lim 8 h h h → + − ′ = = A1 2.2a cao – must be explicit that the limit (and not simply h = 0) is considered [5]H230/02 Mark Scheme October 2021 8 Question Answer Marks AO Guidance 5 (a) 2cos tan 3 1 cos x x x 2 = + ( ) ( ) 2 2 sin 2cos 3 1 cos cos x x x x = + M1 3.1a Replaces tan2 x with 2 2 sin cos x x ( ) 2 2 1 cos 2cos 3 1 cos cos x x x x   −     = +   M1 3.1a Replaces sin2 x with 1 cos − 2 x 2 1 cos 3cos 1 cos ( − = + 2 x x x ) ( ) 2 2 2 2cos 3cos 3cos − = + x x x 2 5cos 3cos 2 0 x x + − = A1 2.1 AG – correct working throughout Must show enough working to justify the given answer [3] 5 (b) DR (5cos3 2 cos3 1 0 θ θ − + = )( ) M1 1.1a Attempt to solve 3-term quadratic cos3 1 θ = − and 2 cos3θ = 5 A1 2.1 May be implied θ = − 13 arccos 1 ( ) ,θ = 1 2 3 5 arccos( ) , 60 M1 A1 1.1 1.1 Correct order of operation to find one value of θ (or all values of 3θ correct) (3 66.42..., ) 180, 293.57... θ = 22.1 A1 1.1 Correct value (to at least 1 dp) 97.9 A1 1.1 Correct value (to at least 1 dp) Any additional values in the range loses final A mark if earned [6]H230/02 Mark Scheme October 2021 9 Question Answer Marks AO Guidance 6 (a) 3 2 0 x2 + > for all values of x therefore stationary point is a minimum B1 2.4 [1] 6 (b) y x x x x k ′ = + = + + ∫(3 2 d 2 2 3 ) M1* 2.1 Attempt to integrate (at least one of the terms in x correct) Condone with no +k y′ = 0 at x = −1 ⇒ − + − + = ( 1 2 1 0 )3 ( ) k leading to k =… M1dep* 1.1 Uses correct conditions to find the value of k (candidates may use the fact that when x = 0, y′ = 3) If correct k = 3 y x x x x x x c = + + = + + + ∫( 3 2 '3' d ) 14 4 2 '3' M1 1.1 Integrates their y′ correctly (allow with k = 0) Condone with no +c (allow use of same letter for second constant) (− ⇒ − + − + − + = 1, 1 1 '3' 1 1 1 4 4 ) ( )4 2 ( ) ( ) c 14 leading to c = … M1 1.1 Uses correct conditions to find the value of c 1 4 2 y x x x = + + + 4 3 2 A1 2.5 cao (must include y =) [5]H230/02 Mark Scheme October 2021 10 Question Answer Marks AO Guidance 7 (a) x y x y 2 2 + − + + = 6 9 19 0 ( x y − − + + − + = 3 9 )2 ( 9 81 2 4 )2 19 0 M1 1.1 ( x y ± ± + ± ± + = 3 ... )2 ( 92)2 ... 19 0 C (3,− 92 ) A1 1.1 cao Radius is 41 2 A1 1.1 cao (oe) [3] 7 (b) 55 4 11 y x = −11 4 11 5 y x     ⇒ = −   B1 2.1 Equation of line AB (any equivalent form) – allow unsimplified 2 2 4 4 11 6 9 11 19 0 5 5 x x x x     + − − + − + =         M1* 3.1a Substitute equation of line into equation of circle Or 5 mCD = − 4 ( Use of m m 1 2 = −1 with their gradient of AB) 41 82 2 41 0 25 5 x x − + = M1dep* 1.1 Simplify to three-term quadratic in x (or y) Equation of CD is y x + = − − 9 5 2 4 ( 3) (using their C from (a)) x-coordinate of D is 5 (or y-coordinate of D is -7) A1 1.1 BC Area of OBD = 1 (11 '5' )( ) 2 M1 3.2a 12(11 -coordinate of )(x D) or other complete method Dependent on both previous M marks = 27.5 A1 1.1 [6]H230/02 Mark Scheme October 2021 11 Question Answer Marks AO Guidance 8 (a) 6 y x 1 x = − + leading to y′ = ... M1 2.1 Derivative of the form 32 1 kx − − + 32 y x 1 3 − ′ = − − A1 1.1 At 1 x m m = = − ⇒ = 1, 4 T N 4 M1* 1.2 Substitutes x = 1 into their derivative and correct use of mm′ = −1 y x − = − 6 1 14 ( ) M1dep* 1.1 Use of y m x − = − 6 1 N ( ) − + = x y 4 23 A1 1.1 oe [5] 8 (b) 6 4, 1 4 4 x y = = − + = 0 B1 1.1 AG – must show sufficient working and must see = 0 [1] 8 (c) DR ⌠  ⌡ (1 d − + = x x 6x ) M1* 2.1 Attempt to integrate with at least two terms correct 1 2 12 2 = − + x x x A1 1.1    4 4 12 4 1 12 − + − − + 1 1 2 2 ( 2 )          = … M1dep* 1.1 Use of correct limits (1 and 4) If correct, then expect to see 7.5 1 23 6 (1) 2 4     +   B1ft 3.1a Any correct numerical expression for the area of the trapezium between x = 0 and x = 1 using their result from (a) 107 8 A1 2.2a Or exact equivalent (e.g. 13.375) [5]H230/02 Mark Scheme October 2021 12 Question Answer Marks AO Guidance 9 R F F (↑ + = ):3 30 2 1 M1 3.4 Attempt to resolve vertically and horizontally with at least one correct equation seen R F F (→ + = + ) : 2 60 28 1 2 A1 1.1 Both equations correct F F 1 2 = = 42, 4 A1 1.1 BC [3] 10 (a) B1 B1 1.1 1.1 Correct shape Correct labels on axes v,t (or velocity, time), 6, 10, 25 and 150 need to be seen Must be correct shape [2] 10 (b) 0.24 (m s−2) B1 1.1 oe [1] 10 (c) T T = − − 150 25 1 where 1 10 6 0.05 T = − M1 3.1b Complete method to find T T = 45 A1 1.1 [2] 10 (d) 1 1 1 (6 25 10 6 ( ) 10 )( ) ( ) 1 ( ) 2 2 2 s = + + + T T M1* 3.3 Complete method to find total distance travelled Av. Speed = '940' 150 M1dep* 3.4 Divides their distance travelled by 150 = 6.27 (m s−1) A1 1.1 cao (oe e.g. exact answer is 15 94 ) [3]H230/02 Mark Scheme October 2021 13 Question Answer Marks AO Guidance 11 (a) M1 3.4 Use of 1 2 s ut at = + 2 12950 15 0.5 0.01 = + t t ( ) 2 leading to t = … A1 1.1 Correct equation and attempt to solve Consistent values used 11.7 (minutes) A1 1.1 BC cao (exact value is 11 2 3 ) [3] 11 (b) D − − = 1250 9000 70000(0.01) M1 3.3 N2L applied to the whole system, correct number of terms (allow sign errors) 10 950 (N) A1 1.1 cao [2] 11 (c) '10950 ' 9000 50000(0.01) − − = TC or TC − = 1250 20000 0.01 ( ) M1 3.3 N2L applied correctly to either A or B, correct number of terms (allow sign errors) 1450 (N) A1 1.1 cao [2]H230/02 Mark Scheme October 2021 14 Question Answer Marks AO Guidance 12 − + = 6 24 0 t M1 3.1b Either attempt to differentiate and set a = 0 or attempt to complete the square − − + 3 4 ... (t )2 to find t Need bt + = 24 0 T = 4 A1 1.1 ∫(− + + 3 24 d t t k t 2 ) M1* 1.1 Attempt to integrate v (at least two terms correct) = − + + t t kt 3 2 12 (+c) A1 1.1 (− + + − − + + = 4 12 4 4 1 12 1 297 3 3 ( )2 2 k k ) ( ( ) ) leading to k =… M1dep* 2.1 Setting up an equation in k using 297, and t = 1 and their t = 4 and attempt to solve for k If correct k = 60 2 t t − − = 8 20 0 (t t − + = 10 2 0 )( ) M1 1.1 Setting v = 0 and attempt to solve three-term quadratic in t (perhaps BC) Dependent on both previous M marks As t ≥ 0, t ≠ − ∴ 2 t = 10 A1 2.3 Must see explicit rejection of negative value of t [7]OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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