H630/01 Mark Scheme June 2023 7 12. Subject Specific Marking Instructions 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. 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 H630/01 Mark Scheme June 2023 8 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. 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. 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. H630/01 Mark Scheme June 2023 9 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. 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. • When a value is not given in the paper accept any answer that agrees with the correct value to 2 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 A the rubric specifies 3 s.f. as standard, so this statement reads “3 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. E marks are lost unless, by chance, the given results are established by equivalent working. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. H630/01 Mark Scheme June 2023 10 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. H630/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 1 d 2 1.2 d v a t t = = + m s–2 M1 A1 1.1a 1.1b Attempt to differentiate. Note 2 + 0.6𝑡 is obtained by division so award M0A0 if seen. cao [2] Question Answer Marks AO Guidance 2 (a) Substitute x = 40, 17 15cos240 9.5 h = + = m B1 1.1b cao [1] 2 (b) Maximum when 6 0 x = or 360 so 0, 60 x = M1 2.1 Attempt to use the period of the cos 6𝑥 function. Must include a reference to either 60 or 240 or a sketch illustrating the x-direction stretch. Allow if wrong conclusion reached. Do not allow for an argument based on mechanics principles alone but the model is only valid for 0 40 x   So Tom’s argument is invalid. E1 2.3 clear argument [2] H630/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 3 (a) position vector of C is 2 2 0 1 2 1 −      + =      −      B1 2.5 Correct column vector notation. ISW if the modulus of the vector is given as well [1] 3 (b) AB ⃗⃗⃗⃗⃗ = ( 5 −2 4 −(−1)) , BC ⃗⃗⃗⃗⃗ = (0 −5 1 −4), M1 2.1 attempt to calculate one of vectors AB ⃗⃗⃗⃗⃗ , BA ⃗⃗⃗⃗⃗ , CB ⃗⃗⃗⃗⃗ or BC ⃗⃗⃗⃗⃗ soi AB = √32 + 52 [= √34] BC =√52 + 32 [= √34] M1 2.1 Attempts to find both lengths. Also allow for argument without distances based on matching components distances equal, so B is equidistant from A and C E1 2.2a Complete argument www [3] H630/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 DR ( ) 2 6 1 sin sin 5 x x − + = M1 3.1a Uses the identity 2 2 sin cos 1 x x + = 2 6sin sin 1 0 x x − −= M1 1.1a Collects terms and attempts to solve their quadratic 1 1 sin , 2 3 x = − A1 1.1b soi. May be BC. FT their quadratic When 1 sin 2 x = , 30 ,150 x =   A1 1.1b At least one correct root in the interval for either value of sin x FT their valid value for sin 𝑥 When 1 sin 3 x = − , 19.5 , 160.5 x = − −  A1 1.1b All roots seen from complete working – no extras in the range FT their other valid value for sin 𝑥 Ignore additional answers outside the range. [5] H630/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 5 (a) gradient from (0, 4) to (2, 7) is 1.5 So 1.5 4 s t = + M1 A1 3.3 3.3 attempt to find the gradient – award if 3 2 seen must be s in terms of t [2] 5 (b) 1.5 -3.5 B1* B1 (dep) 3.4 3.4 Three horizontal lines, above, on and below the t-axis. Also allow if the third line segment is above the first line segment because speed is given instead of velocity. their 1.5 and –3.5 seen and 2 t = and 5 t = clear [2] 5 (c) The changes in velocity are instantaneous. In reality, the velocity changes over a period of time E1 3.5b Accept “suddenly stops” or similar Also accept that the velocity at 𝑡= 2 or 𝑡= 5 is ambiguous [1] H630/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 6 B1 1.1b Attempt to divide the cubic by M1 1.1a Allow grid method or long division as far as the linear term of the quotient A1 1.1b May be embedded in final expression A1 1.1b May be embedded in final expression FT their 𝑏 B1 1.1b May also be found using the remainder theorem. May be embedded in final expression [So ] Need not be given explicitly if all coefficients seen By inspection B1 B1 for 𝑎 embedded in their expression even if incomplete M1 Method may be implied by correct 𝑏, or correct FT for 𝑐 A1 𝑏= −5 A1 𝑐= 4 FT their 𝑏 with 𝑐= −6 −2𝑏 B1 May also be found using the remainder theorem. May be embedded in their expression [5] H630/01 Mark Scheme June 2023 16 Alternative method B1 expanding and equating coefficients quadratic term linear term 2 6 b c + = − constant term 2 5 c d + = − M1 expanding and equating coefficients for at least the quadratic or linear term A1 May be embedded in final expression A1 May be embedded in final expression FT their 𝑏 B1 May also be found using the remainder theorem. May be embedded in final expression So Need not be given explicitly if all coefficients seen [5] H630/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 7 DR Points of intersection with x-axis when −3𝑥2 + 7𝑥−2 = 0 M1 3.1a Attempt to find intersection with x-axis A1 1.1b Both exact roots seen Area = M1* A1 1.1a 1.1b Allow for indefinite integral also Correct indefinite integral (−8 + 7×4 2 −2 × 2) −(−( 1 3) 3 + 7 2×9 − 2 3) M1 (dep) 1.1b Substitution of their limits into their cubic expression must be seen A1 1.1b must be exact. Allow mixed number or recurring decimal 2.3148 www [6] H630/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 8 (a) DR M1 A1 3.1a 1.1b Attempt to complete the square for at least one variable Fully correct. Need not be simplified Centre (1, –2) A1 1.1b FT their completed square form Radius 5 A1 1.1b cao [4] 8 (b) DR Rewrite equation of the line B1 3.1a soi Substitute M1 1.1a Attempt to form quadratic in 𝑦 only Allow either form of the equation used M1 1.1a Attempt to simplify the quadratic to 3 terms A1 1.1b Both roots seen So points of intersection at (4, 2) and (1,3) A1 1.1b FT their y-values. No extra points ISW (2, 4) and (3,1) if 𝑥= 4 and 𝑥= 1 seen matched to their 𝑦 Alternative method Rewrite equation of the line B1 soi Substitute into equation of the circle M1 Attempt to form quadratic in 𝑥 only Allow either form of the equation used M1 Attempt to simplify the quadratic to 3 terms A1 Both roots seen So points of intersection at (1,3) and (4, 2) A1 FT their x-values No extra points Do not allow for (2, 4) and (3,1) [5] H630/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 9 (a) Stretch in the x-direction B1 1.2 Stretch scale factor ½ B1 1.1b [2] 9 (b) B1 B1 1.1b 1.1b General shape of exponential graph less steep than the given graph for positive x (note red graph is printed) Horizontal asymptote above the x-axis and intersection with y-axis must be above that for the given graph [2] 9 (c) The graphs intersect when 2e e x x k = + So when 2e e 0 x x k − − = M1 2.1 Attempts to solve simultaneously. Allow 𝑘= − 1 4 substituted discriminant ( ) ( ) 2 1 4 k − − − M1 2.1 Uses discriminant of the equation is negative for 1 4 k − so no real roots and no points of intersection E1 2.1 must state no real roots or no points of intersection [3] 9 (d) When 2 k = , 2e e 2 0 x x − − = gives e 1, 2 x = − M1 2.1 Evaluates ex from their quadratic and attempts to use natural logs So ln2 x = as e 1 x = − is not possible A1 2.1 must state that ln2 is a root and that there are no others. Allow SC1 for substituting 𝑥= ln 2 in both equations and concluding it must be a root H630/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance [2] Question Answer Marks AO Guidance 10 (a) (i) When 0, 2800 t L = = She invests £2800 B1 3.1b cao [1] 10 (a) (ii) Each year the amount is multiplied by 1.023 which is 2.3% annual interest B1 3.1b cao [1] 10 (b) 3000 1.02t A=  So 𝑎= 3000 B1 1.1b Allow for a and b given explicitly or embedded in an exponential expression And 𝑏= 1.02 B1 1.1b [2] H630/01 Mark Scheme June 2023 21 10 (c) Equal amounts if 3000 1.02 2800 1.023 t t  =  ln 3000 + 𝑡log 1.02 = ln 2800 + 𝑡ln 1.023 𝑡= ln 3000 −ln 2800 ln 1.023 −ln 1.02 = 23.5 M1 M1 3.1b 1.1a Use of laws of logarithms leading to a linear equation in 𝑡 using their values of 𝑎 and 𝑏 Collecting terms So they have equal amounts after 23.5 years A1 1.1b Cao must be 1 d.p. Alternative method Equal amounts if 3000 1.02 2800 1.023 t t  =  3 1.023 2.8 1.02 t   =     so 3 log2.8 23.5 1.023 log 1.02 t = = M1 M1 Equating and attempt to collect terms using their values of 𝑎 and 𝑏 leading to an equation in which 𝑡 appears only once Uses logarithms leading to a value for 𝑡 allow log1.0031.07 or log 1.07 log 1.003 or better for the method mark So they have equal amounts after 23.5 years A1 Cao must be 1 d.p. Note this is obtained from exact values or using 1.00294 and 1.0714 or better Allow full credit for trial and improvement that gives 23.5 and £4778 to the nearest pound If M0M0 given, allow SC2 for 23.5 seen, without £4778 If M0M0 given, allow SC1 for at least 2 trials clearly seen even if a root not found [3] H630/01 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 11 (a) B1 B1 3.3 3.3 Both forces on the sphere correct and labelled All forces correct on the block. Tension must be marked the same or if T1 and T2 used, equated to each other elsewhere in the question. The horizontal force F must be in the correct direction [2] 11 (b) For the system to be in equilibrium sphere gives 1.2 T g = N M1 1.1b soi for the block 1.2 11.76 F T g = = = A1 1.1b any form [2] 11 (c) B1 3.4 cao [1] 11 (d) For the block 𝑇= 3𝑎 Add the equations M1 1.1a Attempt to solve simultaneous equations leading to a value for a. Do not award if their equation 11(c) does not have T. ms-2 Substitute A1 1.1b Cao. Allow 6 7 𝑔 [2] T T 1.2g N 3g N R F H630/01 Mark Scheme June 2023 23 Question Answer Marks AO Guidance 12 For AB, Newton’s second law acceleration is 1.6 ms-2 B1 3.1b Uses Newton’s second law to calculate acceleration for AB m M1 1.1a Uses suvat equation(s) and their 𝑎 leading to a value for s velocity at B ms-1 M1 3.1b Uses suvat equation(s) and their 𝑎 leading to a value for velocity at B for BC Newton’s second law M1 3.1b Uses Newton’s second law to calculate acceleration. Condone missing 8N force. Allow sign errors. acceleration is ms-2 A1 1.1b soi for BC M1 1.1a Uses suvat equation(s) and their 𝑎 leading to a value for s. m A1 1.1b FT their negative a and their positive velocity at B distance AC is m A1 1.1b Allow 10 m [8] Need to get in touch? If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in touch with our customer support centre. 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