A-Level History Exam Questions

Y533/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 Y533/01 Mark Scheme June 2023 4 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. 5. 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). Y533/01 Mark Scheme June 2023 5 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 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 Y533/01 Mark Scheme June 2023 6 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. 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. Y533/01 Mark Scheme June 2023 7 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. Y533/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 (a) 2.4m ( + 0(3m)) = (m + 3m)v M1 1.1 Conservation of momentum with A and B having some post-‘collision’ velocity and initial momentum > 0 Treat consistent calculation using mB = m (possibly combined with mA = 3m) as MR. Can be awarded if seen in (b) 4v = 2.4 => v = 0.6 so speed of B is 0.6 (m s–1) A1 1.1 mB = mA = m => v = 1.2 mB = m, mA = 3m => v = 1.8 [2] (b) Impulse on B = change in B’s momentum = 3m0.6 ( – 0(3m)) M1 1.1 Using I = mv so magnitude of impulse is 1.8m (Ns) A1 1.1 Do not allow -1.8m mB = mA = m => I = 1.2m mB = m, mA = 3m => I = 1.8m [2] Y533/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 2 (a) Initial KE = ½  3u2 B1 3.4 Used in solution 1.5u2 PE when at rest = 3g(3.2 – 3.2cos60) M1 3.1b 24g/5 or 47.04 or 4.8g. Attempt to use mgh to find the PE at the instant that P comes to rest Use of suvat is M0 Assuming zero PE level at lowest point. Otherwise this mark is for attempting to find the difference between PE when at rest and PE at the bottom 3 2 𝑢2 = 24𝑔 5 M1 3.4 Using energy conservation and their expressions to set up an equation in u2 u = 5.6 A1 1.1 No need to eliminate negative value explicitly here [4] (b) Assume that air resistance is negligible B1 3.3 Any sensible specific assumption E.g. no air resistance P is a particle/has no dimensions “No resistance to motion” “No energy is lost to the surroundings” “No other forces acting” are insufficient Ignore references to assumptions stated in the question, e.g. no friction between P and the surface, “other forces”, “does not come off the surface” [1] Y533/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 3 (a) F = 80cos40 = 45a => a = 80cos40 / 45 M1 1.1 Using F = ma with a resolved component of the pulling force and m (not mg) to find a 1.36185… Or 𝑚𝑣= 𝐹𝑡 (= 306.4) s = 0.85 + ½ “1.36185...”52 M1 1.1 Using suvat equation(s) with their value of a to find the distance travelled (21.0232... m)... ...or final velocity (v = 0.8 + ½1.36185...5 (=7.609283...)) Or 𝑣= 𝐹𝑡 𝑚+ 0.8 (= 306.4 45 + 0.8) WD by pull = 80cos4021.0232... M1 1.1 Their component of force  their distance If a = 0 used (e.g. 245Nm) then only this mark can be awarded. ...or change in KE = ½15(7.609283...2 – 0.82) = awrt 1290 (J) A1 1.1 1288.377… [4] (b) (i) v = 0.8 + 1.36185...5 = 7.609283... and use in KE = ½ mv2 M1 1.1 Attempt to use suvat equation with their a to find the velocity after 5 seconds (with no vertical component) and using this to attempt to find the KE (may be seen in (a)) Or WD + initial KE (= 14.4) Must be initial and not final KE, and combined with WD from part (a) so KE = ½457.609283...2 = awrt 1300 (J) A1 1.1 𝑂𝑟 14.4 + 1288 1302.777… [2] (b) (ii) Work done should equal the increase in (kinetic) energy so no account has been made of the fact that the crate has some initial energy. B1 2.4 Ignore reference to any kind of resistive force [1] (c) Av Power = 80cos4021.0232... / 5 = awrt 258 (W) B1FT 1.1 Their (1290)/5 Or average velocity × resolved force (= 0.5(0.8 + 7.609283. . . ) × 80 cos 40) 257.675… B0 if 𝑎= 0 used [1] (d) Impulse = Their (horizontal) Force  time [= 80cos405] M1 1.1 or = change in crate’s momentum = 45(“7.609283”... – 0.8) Must lead to an answer > 0 = awrt 306 (N s) A1 1.1 [2] Y533/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 4 (a) 450/0.5 – 150 = 240a M1 3.3 Using F = ma with m substituted in and a force derived from P = Fv and the resistance force as negative a = 3.125 so the maximum acceleration is awrt 3.13 (m s –2) A1 1.1 [2] (b) 210 = Dv M1 3.4 Use of “P = Fv” or 𝑝= 𝐹𝑑 𝑡 with 210 substituted in Constant speed => a = 0 => 210/v = 150 M1 2.2a Using F = ma with a = 0 to deduce the required force (soi) 210 = 150𝑑 𝑡 v = 1.4 A1 1.1 Or 𝑡= 150×350 210 oe t = 350 / 1.4 = 250 so 250 seconds A1 1.1 4 minutes 10 seconds [4] (c) The model assumes that the power and hence driving force is constant but in practice this will not be the case (since the oars go in and out of the water periodically) Or: Rower may get tired (& reduce power output). Or: Speed may vary, hence power will vary (if the force/resistance is constant). B1 3.5b Detailed knowledge of the mode of propulsion of a rowing boat is not required. If mentioning change of resistance, force or speed, this must be linked to power output. Allow any response along the lines that any human way of providing power will not in practice be constant. [1] Y533/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 5 (a) (i) 5m + (–3)m = (–2)m + mvB M1 3.4 Conservation of momentum 𝑢𝐴 must be > 𝑢𝐵 oe vB = 4 A1 1.1 in the direction of motion of A before the collision oe A1 1.1 Must be clearly stated or shown, e.g. consistent with arrow on diagram Direction of B is reversed [3] (ii) e = (4 – (–2)) / (5 – (–3)) M1 3.4 Attempt at restitution - condone sign error as long as consistent 0 ≤(±)𝑒≤1 Must have sufficient detail, e.g. 6/8 on its own is M0. = 6/8 = ¾ A1 1.1 AG [2] (iii) Initial KE = ½452 + ½4(–3)2 = 68 (J) Final KE = ½4(–2)2 + ½442 = 40 (J) M1 3.4 Attempt to calculate total initial or final KE Both values must be positive Or KE loss for A = ½452 – ½4(– 2)2 = 42 J or KE gain for B = ½442 – ½432 = 14 J so loss is 68 – 40 = 28 (J) A1 1.1 42 J – 14 J = 28 J [2] (b) e = 4 / 4 = 1 B1FT 1.1 FT their vB provided that 0 < e  1 (using their values in (a)) Allow e = 1 without working, provided (a)(i) is correct The collision is perfectly elastic. B1FT 1.2 FT their e provided that 0 < e  1 (using their values in (a)) Do not accept phases such as “completely elastic” [2] (c) (–2)m + (–4)m = mVA + mVB M1* 3.1b Conservation of momentum with consistent signs. 2VA + 2VB = –12 (if “positive” direction reversed: 2m + 4m = mVA + mVB) M0 if approach speed < 0 ¾ = (VB – VA) / ((–2) – (–4)) M1* 3.1b Restitution with consistent signs 2VB – 2VA = 3 Allow use of e (e.g. 𝑉𝐵 – 𝑉𝐴= ±2𝑒) 2VA + 2VB = –12 2VB – 2VA = 3 VB = … or VA = … M1de p 1.1 Attempt to solve both their equations simultaneously and reach a solution for VA or VB Allow use of e (e.g. 𝑉𝐴= ±(3 + 𝑒) or 𝑉𝐵= ±(3 −𝑒)) VB = –2.25 or VA = –3.75 A1 1.1 A numerical value is required here (may be implied by a correct final answer). Impulse on A = change in A’s momentum = 4(–3.75 – (–2)) = –7 ⇒7(Ns) A1 1.1 Ignore wrong units ISW e.g. any statements regarding direction of travel [5] Y533/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 6 (a) [P] = MLT–2 / L2 = ML–1T–2 B1 1.1 Penalise wrong dimensional symbols only once as accuracy mark. Penalise + instead of × when combining dimensions only once as accuracy mark [1] (b) [½mu2] or [½mv2] or [W] = ML2T–2 M1 1.1 ½ not necessary [mP] = M2L–1T–2 so the equation is dimensionally inconsistent A1 2.1 Correct dimensions for mP and conclusion [2] (c) [RHS] = [M0]L +  T–( +  - ) oe M1 3.4 eg RHS has no M while LHS has M1 so the equation must be dimensionally inconsistent A1ft 2.4 or “some M” oe Or “no M” compared to LHS Ignore one minor slip in L or T and [W] = ML2T–2 so comparing indices leads to a contradiction Allow incorrect expression for [W] from part (b), provided it includes an element of M [2] (d) (i) Because there are 4 unknowns and DA can only give us a maximum of 3 equations B1 2.4 Must be specific comparison. Or 3 equations in 4 unknowns seen (condone one slip), with appropriate comment about the equations in L and T at least Must show all 3 equations, not just the two that involve 𝛼, 𝛽 and 𝛿 [1] (d) (ii) ML2T–2 = L3T–3LT–2M T B1ft 3.4 Correct dimensional expansion of both sides with  = 3 substituted at some point (using their [W]) M L3 +  T – 3 – 2 Award if seen in part (i) (with  = 3) M: 1 = , L: 2 = 3 + , T: –2 =  – 3 – 2 M1 1.1 Correctly comparing indices for all three dimensions. Allow 1 slip M0 if M does not appear on both sides of the dimensional equation  = –1,  = 1,  = –1 A1 1.1 [3] Y533/01 Mark Scheme June 2023 14 (d) (iii) The resultant formula is 𝑊= 𝑘𝑢3𝑚 𝑎𝑡. This is unlikely to be correct since it suggests, for example, that the total work done becomes (very) small or negligible B1 3.5a Correct conclusion from their 𝛿≤0 Condone “not correct/incorrect” Answers referring to t = 0 are not valid for this question Condone statements such as “… because work done decreases as time increases” oe, (which contradicts the expectation of a positive relationship between W and t) [1] Y533/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) For P: ↔ 𝑇𝑆1 = 1.5 × 5𝜔2 (= 7.52) M1 3.1b NII for particle P using a = r2 𝑇𝑆1and 𝑇𝑆2must be in terms of 𝜔 (may be seen later e.g. by using 𝜔= 𝑣𝑃 5 or 𝜔= 𝑣𝑄 5 sin 𝜃 oe) Condone use of m and r SC1 only for omitting the element of 𝑚, or for using a specific value of 𝜔 to get the result that 𝑇𝑆1 = 𝑇𝑆1. For Q: ↔ 𝑇𝑆2 𝑠𝑖𝑛𝜃= 1.5𝑟𝜔2 M1 1.1 Resolving tension for Q and using NII in the horizontal with a = r2 Allow sin/cos confusion Allow use of specific value of 𝜃 Allow use of 𝑟sin 𝜃 or 𝑟cos 𝜃 𝑟= 5 𝑠𝑖𝑛𝜃⇒𝑇𝑆2 = 1.5 × 5 𝑠𝑖𝑛𝜃× 𝜔2 𝑠𝑖𝑛𝜃 = 7.5𝜔2 ∴𝑇𝑆`1 = 𝑇𝑆2 A1 1.1 AG Two identical expressions clearly seen. A0 if a specific value of 𝜃 has been used [3] (b) P: 1 2 × 1.5𝑣2 = 39.2, so 𝑣2 = 784 15 M1 1.1 Using kinetic energy is 39.2 to find 𝑣 or 𝑣2 𝑣2 = 52.666 … , 𝑣= 28√15 15 = 7.229 … (allow 7.22) 𝑡𝑃= 2𝜋 𝜔𝑃 = 2𝜋× 5 𝑣 = 5√15𝜋 14 A1 1.1 Use of "𝜔= 2𝜋 𝑡" and finding the time using “v = r” for P. NB 𝑡𝑃= 4.3454 … , 𝜔𝑃≈1.45 Allow 4.33-4.35 Allow unsimplified. Penalise inexact value only at the end Q: 𝜔𝑄= 𝑣 5 sin 𝜃 (= 28√15 75 ) B1 1.1 Use of “v = r” for Q with correct radius Or 𝑡𝑄= 2𝜋×5 sin 𝜃 𝑣 (may be seen later) Or 𝜔𝑄= 𝜔𝑃sin 𝜃 May be seen later as 𝜔𝑄= 𝑣 4 = 7 √15 or 𝑡𝑄= 2𝜋 𝜔𝑄= 4√15𝜋 14 B0 for assuming specific value of 𝜃 (and for subsequent marks) 𝑄: ↕𝑇𝑆2 cos 𝜃= 1.5𝑔 M1 1.1 Resolving the tension vertically and balancing with weight (condone missing g here) 4g = 39.2 Condone use of 𝑚 Y533/01 Mark Scheme June 2023 16 𝑄: ↔ 𝑇𝑆2 sin 𝜃= 1.5𝑎= 1.5𝑣2 5 sin 𝜃 M1 1.1 Resolving tension horizontally and using NII and 𝑎= 𝑣2 𝑟 with correct radius Or: 𝑇𝑆2 = 7.5 ( 𝑣 5 sin 𝜃) 2 Could see use of 𝑣2 = 784 15 here so 𝑇𝑆2 𝑠𝑖𝑛𝜃= 1.5𝑎= 1176 75 𝑠𝑖𝑛𝜃 Condone use of 𝑚 and 𝑟 𝑠𝑖𝑛2 𝜃 𝑐𝑜𝑠𝜃= (1.5𝑣2 5 ) 1.5𝑔= 𝑣2 5𝑔= 784 15×5𝑔= 16 15 15(1 −𝑐𝑜𝑠2 𝜃) = 16 𝑐𝑜𝑠𝜃 (⇒15𝑐2 + 16𝑐−15 = 0) M1 2.3 Finding an equation in , using 𝑣2 = 784 15 and substituting sin2 𝜃= 1 −cos2 𝜃 to get an equation in cos 𝜃 only ((5c – 3)(3c + 5) = 0 =>) cos = 3/5 since cos cannot be –5/3) 𝛥𝑡= 5√15𝜋 14 − 4√15𝜋 14 = √15 14 𝜋, so difference in time periods is √15 14 𝜋 (s) oe A1 1.1 Allow cos = 3/5 or sin = 4/5 to appear without working. Use of "𝜔= 2𝜋 𝑡" and finding the time difference. (For reference: 0.86909…) or 𝑡𝑄= 2𝜋𝑟 𝑣= 2𝜋×5 sin 𝜃 𝑣 = 2𝜋×5×4 5 √784 15 = 4√15𝜋 14 [7] 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. 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H074/01 Mark Scheme June 2023 5 11. Annotations Annotation Meaning Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response. Positive Recognition Assessment Objective 1 Assessment Objective 2 Assessment Objective 3 Assessment Objective 4 Assessment Objective 5 Attempted or insecure Expression Answering the question Relevant but broad, general or implicit

H074/01 Mark Scheme June 2023 6 12. Subject Specific Marking Instructions Candidates answer one question from Section A and one question from Section B. Assessment objectives AO1, AO2, AO3 and AO4 are assessed in Section A. Assessment objectives AO3 and AO5 are assessed in Section B. For each section the level descriptors are organised with the dominant assessment objective first. The question-specific guidance on the tasks provide an indication of what candidates are likely to cover in terms of AOs 1, 2, 3, 4 and 5. The guidance and indicative content are neither prescriptive nor exhaustive: candidates should be rewarded for any relevant response which appropriately addresses the Assessment Objectives. H074/01 Mark Scheme June 2023 7 Awarding Marks (i) Section A has one question worth 30 marks. In Section B candidates choose one question worth 20 marks. (ii) For each answer, award a single overall mark out of 30 (Section A) and 20 (Section B), following this procedure: • refer to the question-specific Guidance for descriptions of Higher and Lower response and indicative content • using ‘best fit’, make a holistic judgment to locate the answer in the appropriate level descriptor • place the answer precisely within the level and determine the appropriate mark out of 30 (Section A) and 20 (Section B) considering the relevant AOs • bear in mind the weighting of the AOs, and place the answer within the level and award the appropriate mark out of 30 (Section A) and 20 (Section B) • if a candidate does not address one of the assessment objectives tested in the question, they cannot achieve all of the marks in the given level. Mark positively. Use the lowest mark in the level only if the answer is borderline / doubtful. Use the full range of marks, particularly at the top and bottom ends of the mark range. (iii) When the complete script has been marked: • if necessary, follow the instructions concerning rubric infringements; • add together the marks for the two answers, to arrive at the total mark for the script. Rubric Infringement Candidates may infringe the rubric in one of the following ways: • only answering one question; • answering two or three questions from Section B; If a candidate has written three or more answers, mark all answers and award the highest mark achieved in each Section of the paper. H074/01 Mark Scheme June 2023 8 USING THE MARK SCHEME Study this Mark Scheme carefully. The Mark Scheme is an integral part of the process that begins with the setting of the question paper and ends with the awarding of grades. Question Papers and Mark Schemes are developed in association with each other so that issues of differentiation and positive achievement can be addressed from the very start. This Mark Scheme is a working document; it is not exhaustive; it does not provide ‘correct’ answers. The Mark Scheme can only provide ‘best guesses’ about how the question will work out, and it is subject to revision after we have looked at a wide range of scripts. The Examiners’ Standardisation Meeting will ensure that the Mark Scheme covers the range of candidates’ responses to the questions, and that all Examiners understand and apply the Mark Scheme in the same way. The Mark Scheme will be discussed and amended at the meeting, and administrative procedures will be confirmed. Co-ordination scripts will be issued at the meeting to exemplify aspects of candidates’ responses and achievements; the co- ordination scripts then become part of this Mark Scheme. Before the Standardisation Meeting, you should read and mark in pencil a number of scripts, in order to gain an impression of the range of responses and achievement that may be expected. In your marking, you will encounter valid responses which are not covered by the Mark Scheme: these responses must be credited. You will encounter answers which fall outside the ‘target range’ of levels for the paper which you are marking. Please mark these answers according to the marking criteria. Please read carefully all the scripts in your allocation and make every effort to look positively for achievement throughout the ability range. Always be prepared to use the full range of marks. H074/01 Mark Scheme June 2023 9 These are the Assessment Objectives for the English Language and Literature specification as a whole. AO1 Apply concepts and methods from integrated linguistic and literary study as appropriate, using associated terminology and coherent written expression. AO2 Analyse ways in which meanings are shaped in texts. AO3 Demonstrate understanding of the significance and influence of contexts in which texts are produced and received. AO4 Explore connections across texts informed by linguistic and literary concepts and methods. AO5 Demonstrate expertise and creativity in the use of English to communicate in different ways. WEIGHTING OF ASSESSMENT OBJECTIVES The relationship between the components and the assessment objectives of the scheme of assessment is shown in the following table: % of AS level Component AO1 AO2 AO3 AO4 AO5 Total Non-fiction written and spoken texts (01) 8% 7% 13% 7% 15% 50% The language of literary texts (02) 14% 20% 8% 8% 0% 50% Total 22% 27% 21% 15% 15% 100% H074/01 Mark Scheme June 2023 10 Component 1 Section A (Non-fiction anthology texts) 30 marks The weightings for the assessment objectives are: AO1 8.0% AO4 8.0% AO2 7.0% AO3 7.0% Total 30% In Section A the dominant assessment objectives are AO1 Apply concepts and methods from integrated linguistic and literary study as appropriate, using associated terminology and coherent written expression and AO4 Explore connections across texts, informed by linguistic and literary concepts and methods. Answers will also be assessed for AO2 and AO3. Candidates should apply concepts and methods as appropriate, using relevant linguistic terminology and fluent expression (AO1). They should explore connections across the two anthology texts, comparing and contrasting details, informed by linguistic and literary concepts and methods (AO4). They should analyse ways in which meanings are shaped in the texts (AO2) and develop their answer with reference to the significance and influence of the contexts in which texts are produced and received (AO3). The criteria below are organised to reflect the order of the dominant assessment objectives. A response that does not address any one of the four assessment objectives targeted cannot achieve all of the marks in the given level. Level 6: 26–30 marks AO1 Excellent application of relevant concepts and methods from integrated linguistic and literary study as appropriate. Consistently coherent and fluent written expression and apt and consistent use of terminology relevant to the task and texts. AO4 Excellent and detailed exploration of connections across texts informed by linguistic and literary concepts and methods. AO2 Excellent, fully developed and detailed critical analysis of ways in which meanings are shaped in texts. AO3 Perceptive understanding of the significance and influence of the contexts in which texts are produced and received H074/01 Mark Scheme June 2023 11 Level 5: 21–25 marks AO1 Secure application of relevant concepts and methods from integrated linguistic and literary study as appropriate. Consistently clear written expression and appropriate use of terminology relevant to the task and texts. AO4 Clearly developed exploration of connections across texts informed by linguistic and literary concepts and methods. AO2 Clear and well developed critical analysis of ways in which meanings are shaped in texts. AO3 Clear and relevant understanding of the significance and influence of the contexts in which texts are produced and received. Level 4: 16–20 marks AO1 Competent application of relevant concepts and methods from integrated linguistic and literary study as appropriate. Generally clear written expression and mainly appropriate use of terminology relevant to the task and texts. AO4 Competent exploration of connections across texts informed by linguistic and literary concepts and methods. AO2 Competent analysis of ways in which meanings are shaped in texts. AO3 Some understanding of the significance and influence of the contexts in which texts are produced and received. Level 3: 11–15 marks AO1 Some application of relevant concepts and methods selected appropriately from integrated linguistic and literary study. Generally clear written expression with occasional inconsistencies and some appropriate use of terminology relevant to the task and texts. AO4 Some attempt to explore connections across texts informed by linguistic and literary concepts and methods. AO2 Some analysis of ways in which meanings are shaped in texts. AO3 Some awareness of the significance and influence of the contexts in which texts are produced and received. H074/01 Mark Scheme June 2023 12 Level 2: 6–10 marks AO1 Limited attempt to apply relevant concepts and methods from integrated linguistic and literary study appropriately. Some inconsistent written expression and limited use of terminology relevant to the task and texts. AO4 Limited attempt to make connections across texts informed by linguistic and literary concepts and methods. AO2 Limited analysis of ways in which meanings are shaped in texts. AO3 Limited awareness of the significance and influence of the context in which texts are produced and received. Level 1: 1-5 marks AO1 Very little attempt to apply relevant concepts and methods from integrated linguistic and literary study appropriately. Inconsistent written expression and little use of terminology relevant to the task and texts. AO4 Very little attempt to make connections across texts informed by linguistic and literary concepts and methods. AO2 Very little analysis of ways in which meanings are shaped in texts. AO3 Very little awareness of the significance and influence of the contexts in which texts are produced and received. 0 marks: no response or response not worthy of credit. H074/01 Mark Scheme June 2023 13 Question Response Mark Guidance Text A is Text A is an extract from the graphic memoir I Was Their American Dream. Text B is an extract from the podcast This American Life. 1 Compare the ways in which the writers or speakers use language to respond to the events they are describing. In your answer you should consider: • context • mode and genre • purpose and audience. A higher level response (levels 4 –6) will: AO1 Use vocabulary and terminology appropriately, referring to a range of language levels, including grammar and discourse, e.g. conjunctions, examples of elision and contractions. Express ideas fluently and coherently, with a wide vocabulary. AO4 Make comparisons between texts, e.g. analysing similarities (both present a range of formality in lexis) and differences (the interplay between visuals and linguistics in the graphic memoir versus the more spontaneous nature of conversation in the podcast.) AO2 Explore the ways the speakers use language to achieve their purposes for their respective audiences, e.g. the use of captions in the graphic memoir to aid the reader’s understanding of the visual images that follow; or Joffe-Walt’s more formal constructions when providing narration directly to the listener in the podcast. 30 The indicative content shows an integrated approach to the four assessment objectives. AO1, AO4, AO2 and AO3. Context/audience/ purpose e.g. • The audience for Text A’s graphic memoir is Americans and Gharib wrote this after having been disturbed by anti-immigrant rhetoric in America in 2016. She makes many references to her American readership (American’s ate Hamburger Helper), and she uses American idioms and cultural references (Mom, $5, trash) alongside cultural references to her mixed heritage (monggo). • Text B has also been created for an American audience as acknowledged by the title of the podcast series (This American Life) and is broadcast on public radio across the US. This is also evidenced by American contextual references (Wall Street), it is an American transcript using American English spellings (organized, color), and also contains American slang (ratchet-ass, hood), and American educational terms (ninth grade). • Text A’s purpose for the graphic memoir is to present a picture of immigrants and their experiences in America, and to show how children struggle to feel American even though they are born in the States. Gharib wants to show her cultural isolation and the difficulties she had integrating because of her mixed heritage. However, she also wants to present elements of her culture that Americans can enjoy or learn about such as the recipe for monggo. • For Text B, the purpose is to inform and entertain. Each episode for the podcast is built around a certain theme and this episode is about the divisions between private and the public-school children in America, looking at 2 schools only 3 miles apart from each other in New York. The episode aims to show the gap between rich and poor and the racial divides, and how these children struggled to integrate with each other when they were brought together (I felt like I didn’t belong there). Mode e.g. • Text A adheres to the conventions of a graphic novel. The writer has employed the use of both linguistic and visual elements. Created in panels, the writer adopts the use of first-person captions usually in the top left corner (on TV, kids got allowances for doing H074/01 Mark Scheme June 2023 14 AO3 Understand the significance of a range of contextual factors, e.g. Gharib’s mixed heritage versus the podcast’s messages about racial divides in America; shared purposes; intended/implied audiences. A lower level response (levels 1 –5) will: AO1 Use some appropriate terminology, mainly at level of word choice, e.g. adjective use, or syntax, such as complex sentences. Expression is clear but may lack precision. AO4 Make general comparisons between language use, e.g. formal versus informal. AO2 Examine some ways the writers use language in each text, e.g. use of swear words to show anger versus the use of interjections in dialogue bubbles. AO3 Recognise and show an understanding of the differences between contexts. extra chores) to provide narration. The positioning of the caption at the top left encourages the reader to look at this first before looking at the accompanying image. • In Text B it is the presenter of the podcast who takes on the role of the interviewer, they control the conversation and try to prompt extra information to aid listener understanding and to control the order in which they hear information. The presenter, Joffe-Walt, has the listener in mind and controls the conversation and she usually interjects and ask questions that prompt further information (Wait. What did you look like?). • In Text B, it is not a sequential conversation between two people. Snippets of the pre-recorded conversation between Melanie and Joffe-Walt have been provided but the presenter interjects with summary and expositions (For so many University Heights kids I talked to, seeing Fieldston was shocking) before playing more of the conversation with Melanie. There are also moments when the presenter introduces Melanie’s feelings to the listener before Melanie has mentioned them as if preparing the listener or wanting to draw their attention to significant feelings before they hear the testimony (For Melanie it wasn’t that. Melanie had imagined it.) • In Text A rectangular panels are used to present the images. Some panels are open (without any borders), whereas some are closed with borders and these borders can help to symbolise feelings of cultural isolation. An open panel is used for the image of her entire family perhaps to emphasise the size of the family or how different they are to the families presented on American television-the image of the family on TV does have a border around it and is much smaller. Images sometimes stray outside the panel border such as the image with the caption ‘My family didn’t look like the ones on TV’. This could be to emphasise the fact that she feels her family do not fit in. • In Text B, it is the interviewee’s snippets of testimony that provides the listener with information about her feelings. Melanie’s first- person account conveys her strong feelings of social isolation (I felt like a ratchet-ass girl from the hood), and this is an adjacency pair. By asking a question (what did you feel like that day?), the presenter provokes a responding utterance to establish how the person felt. Whilst it may seem very conversational, it is not completely spontaneous. Joffe-Walt has planned some of these questions in advance, and this is an edited recording. H074/01 Mark Scheme June 2023 15 • At points in Text A, the graphic memoir strays into other genres such as instructional writing with the insertion of the recipe for monggo. This could help to show that Gharib wants to share her heritage and show that Americans can learn and enjoy something from her culture. • In Text A, captions and speech bubbles are all written using the same uppercase childish handwriting to emphasise the idea of it being a child’s perspective of events. However, a different text type is used to signify sound coming from the television in the first panel-this uses lower case handwriting. Again, this could emphasise how different Gharib felt from the culture around her. • In Text B, we hear perspectives from both Melanie about her own personal experience of the event, but also from Joffe-Walt about her other research into the event (for so many University Heights kids I’ve talked to) which reminds us of Joffe-Walt’s role as a researcher or journalist and the fact that this extract is part of wider episode that involves other people’s experiences. • In Text A motion lines add to the visual effect of a static image such as the lines coming out of the overflowing rubbish bin to emphasise the idea of smells emanating from the rubbish or bubbles rising from the washing bowl. • Text B contains moments that are typical of an aural text type. There are moments when the speech is inaudible. Lexis/ Pragmatics e.g. • In Text B Melanie uses cultural references from 80s films such as the Goonies to imply how socially different the public-school children were from the private school children and the divide between them (I would say we looked like the goonies walking in like a Wall Street building). • In Text A the writer uses colours within the visual images to try to imply the cultural differences and cultural isolation. The style of art is child-like, more cartoonish than realistic and has a use of symbolic colours – mainly oranges and blues. The colours of characters’ hair in particular helps to symbolise how different Gharib’s family are to those families presented on American television. On TV, the family of 4 all have the same colour hair: orange. However, the picture of Gharib’s family shows a large family containing a mixture of hair colours and head dresses with the use of blue, orange, grey and white. This mixture of colours not only emphasises how different Gharib feels, but also reflects her H074/01 Mark Scheme June 2023 16 mixed heritage because of her Egyptian father and Filipino mother. The visual depiction of Gharib as a child has her consistently wearing the same orange and blue t-shirt throughout all panels perhaps to reflect the dual heritage she has and to incorporate the mother’s orange hair, and the father’s blue hair that we see in the family picture. This mixture of two colours on her t-shirt sharply contrasts with the American family on TV-each member of this family wears clothes that contain one colour only, they do not mix colours. • The speaker in Text B recognises that there is a social divide because of race and whilst she once dreamed of having the same opportunities as the private school children (This is what I wanted to see myself going to as a high school experience) she implies that she realised, even as a child, this equality is not something she will ever have. The past tense emphasises the fact that, for her, this dream is over, and she moves to present tense, to her reality, where equality does not exist (OK, this is not free. This is not available for kids of color). It is this realisation that angers her and her anger can be seen with the use of swearing (everything kind of like is a fucking lie that you see your whole life growing up on TV shows or movies). • The writer of Text A also recognises, through TV, that her family are different to other American families, but she thinks that she can try to be more like the others (If I mow the lawn can you give me 5$). However, there is comedy to this line and it shows a childish or naive belief that she will be able to share the same experiences when her mother states ‘we don’t have a lawn.’ • In Text A, the writer implies that, as an adult, she has become much more accepting of her family’s cultural differences and this is reflected in the choice to include the recipe. This is now something she wants to share, and she admits, ‘I actually like it now!’ This shows the more mature, adult perspective and presents a positive message to the reader that, as children get older, they feel more accepting and even proud of their cultural differences. • In Text B, the interviewer seems surprised by Melanie’s recount and the fact that, as a child, Melanie was already predicting how her adult life would be filled with racial inequalities (you thought that when you were at Fieldston?) and Joffe-Walt again emphasises her shock at this with the colloquialism, ‘Whoa. That’s wild.’ This implies the presenter has not experienced this especially not during her own childhood. H074/01 Mark Scheme June 2023 17 Grammar/syntax e.g. • Text B contains several fronted conjunctions and colloquialisms to reflect the fact that many moments between Melanie and Joffe- Walt are spontaneous speech and therefore informal (So it was like when the shit hits the fan.) There are moments of overlapping speech which also mirrors natural conversation, and they even complete each other’s sentences (So you like imagined this is – MELANIE The future, yeah.) • The writer of Text A also tries to capture the informality of speech within the dialogue bubbles. Several of the dialogue starts with interjections (Ah no reason, Hey Ma! Oh man). • The mode of interviewer and interviewee in Text B is reflected in some of the syntax choices. When Joffe-Walt asks a question (so what happened when you went there?) Melanie sometimes starts her answers using the same words from the question she has just heard (When we went there). • In Text A, the captions often start with the prepositional phrase ‘On TV’. This signifies how much Gharib drew her knowledge of what it was to be American from the television. (On TV, Moms basked cookies.). However, the end of the caption always reflects her reality, which was far from the picture presented on television (I never did that with my parents). • In Text B, there are slight grammatical changes in the part when Joffe-Walt provides narrative or is the presenter in contrast to when she is taking on the role of interviewer asking Melanie questions. Her constructions in the narrative part are more formal with complex constructions. Sentences start with subordinate clauses (For so many University Heights kids I talked to). In this part, Joffe- Walt also switches to perfect tense such as past perfect (Melanie had imagined it) or present perfect (They could not have imagined a place like Fieldston.) These more complex and formal constructions reflect the nature of the podcast-the fact that some parts are scripted and rehearsed, others are more informal, spontaneous speech. • In Text A there are also grammatical changes when the genre changes and the writer follows the conventions of instructional writing with the recipe for monggo. For this, sentences start with imperatives (Serve immediately), and the writer also uses brackets with information directed specifically at the reader (see recipe below). H074/01 Mark Scheme June 2023 18 Component 1 Section B (Non-fiction writing) 20 marks The weightings for the assessment objectives are: AO5 15.0% AO3 5.0% Total 20% In Section B the dominant assessment objective is AO5 Demonstrate expertise and creativity in the use of English to communicate in different ways. Answers will also be assessed for AO3. Candidates should demonstrate expertise and creativity in their own original non-fiction writing (AO5) showing understanding of the significance and influence of the contexts in which texts are produced and received (AO3). The criteria below are organised to reflect the order of the dominant assessment objectives. A response that does not address any one of the two assessment objectives targeted cannot achieve all of the marks in the given level. Level 6: 17–20 marks AO5 • Flair, originality and a high degree of control demonstrated in the use of English to communicate in different ways. AO3 • Perceptive understanding of the significance and influence of the contexts in which texts are produced and received. Level 5: 14–16 marks AO5 • Control and creativity demonstrated in the use of English to communicate in different ways. AO3 • Clear and relevant understanding of the significance and influence of the contexts in which texts are produced and received. H074/01 Mark Scheme June 2023 19 Level 4: 11–13 marks AO5 • Competence and engaging effects demonstrated in the use of English to communicate in different ways. AO3 • Some understanding of the significance and influence of the contexts in which texts are produced and received. Level 3: 8–10 marks AO5 • Some accuracy and an attempt to create effects demonstrated in the use of English to communicate in different ways. AO3 • Some awareness of the significance and influence of the contexts in which texts are produced and received. Level 2: 5–7 marks AO5 • Limited accuracy and some attempt to create effects demonstrated in the use of English to communicate in different ways. AO3 • Limited awareness of the significance and influence of the context in which texts are produced and received. Level 1: 1-4 marks AO5 • Little accuracy and little attempt to create effects demonstrated in the use of English to communicate in different ways. AO3 • Very little awareness of the significance and influence of the contexts in which texts are produced and received. 0 marks: no response or response not worthy of credit. H074/01 Mark Scheme June 2023 20 Question Response Mark Guidance 2 Write a speech about a time when you felt that you didn’t fit in, either real or imagined. Your audience is students about to start a new school or college. 20 Candidates will show awareness of the ways language varies according to contextual factors by demonstrating understanding of generic conventions in their own text. For example: • Use techniques effective for providing information and commenting in an engaging way; • show awareness of the style and approach of a speech/an article/a script • adapt language as appropriate for a speech, formal or informal piece of writing. Candidates will establish some interaction with their audience as appropriate, e.g. through personal anecdote or use of humour. 3 Write an article for a magazine entitled ‘Food That Has Made Me.’ The article should be about a dish or food that you associate with your upbringing and the memories you have of this. 4 Write an introductory script to be read by the presenter of a podcast. It will be for an episode entitled, ‘Are school days really the best days of our lives?’ A higher level response (levels 4–6) will: AO5 Demonstrate expertise in the use of English to create an effective speech/article/script, with a high degree of control over the techniques that have been chosen. AO3 Demonstrate understanding of the influence of context on how texts are produced and received. A lower level response (levels 1–3) will: AO5 Show some ability to shape an effective speech/article/script, drawing on a range of different techniques. AO3 Show some awareness of the influence of context on how texts are produced and received. H074/01 Mark Scheme June 2023 21 Appendix 1 Assessment Objective weightings are given as percentages. Assessment Objectives Grid Anthology Question AO1% AO2% AO3% AO4% AO5% Total% 1 8 7 8 7 0 30% Totals 8% 7% 8% 7% 0% 30% H074/01 Mark Scheme June 2023 22 Original non-fiction writing Question AO1% AO2% AO3% AO4% AO5% Total% 2 0 0 5 0 15 20% 3 0 0 5 0 15 20% 4 0 0 5 0 15 20% Totals 0% 0% 5% 0% 15% 20% 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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H230/02 Mark Scheme June 2023 3 4. Annotations Annotation 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 H230/02 Mark Scheme June 2023 4 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 June 2023 5 5. 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. H230/02 Mark Scheme June 2023 6 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. H230/02 Mark Scheme June 2023 7 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. 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 June 2023 8 Question Answer Marks AO Guidance 1 ( )( ) 2 3 4 k k = − B1 1.1 Correct discriminant of given quadratic equation (may be implied) Not given if just seen in quadratic formula 2 9 4 ( 0) k −  leading to two c.v. of k M1 1.1 Finds two critical values of k from their discriminant Or for one correct inequality (may come from only one c.v.) 3 2 k  [or] 3 2 k − A1 1.1 oe e.g. set notation     3 3 2 2 : : k k k k −   [3] H230/02 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 2 DR 5 32 45 2 x x x + = + ( ) 45 2 5 32 x + − = M1* 1.1 Re-arranging and factorising out x 45 3 5 = B1 1.1 Replacing 45 3 5 = (Or 45 5 15 if multiplying through by 5)  = Could appear at any point 32 2 5 2 2 5 2 2 5 2 x   −   =      + −    M1dep* 1.1a Correct method for rationalising the surd of the denominator with x taking the form 1 2 3 o.e. 5 k k k + ( ) 32 2 5 2 4 5 4 20 4 x − = = − − A1 1.1 cao where 4, 4 a b = = − Need to see some correct working [4] ALTERNATIVE SCHEME 45 3 5 = 2 2 2 (32 2 ) (2 5 ) 16 128 1024( 0) x x x x − =  + − = 4 5 4 only x = − B1 M1* M1dep* A1 [4] Replacing 45 3 5 = (Or 45 5 15 if multiplying through by 5)  = Rearranging and squaring leading to a 3TQ Solve by completing the square or using quadratic formula Could appear at any point H230/02 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 3 (a) (i) 29 (m) B1 1.1 [1] 3 (a) (ii) 1 (m) B1 1.1 [1] 3 (b) ( ) ( ) 15 14cos 150 1 cos 150 1 k k − =  = M1 3.1b Setting given expression equal to (a)(ii) and re- arranging to get ( ) cos 150 ... k = or for stating 150 360 k = Must substitute/use t = 150 for this method mark. May use t = 75 with h = 29 150k = 2π, (i.e. using radians) M1 150 360 2.4 k k =  = A1 1.1 The correct answer only implies the M mark so “k = 2.4” without working is M1A1. [2] 3 (c) ( ) ( ) 5 14 15 14cos 20 cos kt kt − =  = − M1* 3.1b Setting given expression equal to 20 to obtain an equation of the form ( ) 1 cos kt k = Could use inequalities Need −1  ⩽ 𝑘1  ⩽ 1 46.2186....,103.7813 t = M1dep* 1.1 Obtaining at least one value of t correctly from their equation above 2.4 110.9248...,249.0751... t = 103.7813... 46.2186... or 150 (2 46.2186) o.e. − −  M1 1.1 Subtracting their two positive values of t (both obtained correctly from their equation above) or 150 (2 their )t −  Dependent on previous two M marks Therefore above 20 m for 57.6 (s) A1 3.2a Must be to 1 decimal place 57.562639… A0 if using radians. [4] H230/02 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 4 (a) ( ) ( ) ( ) 6 2 6 6 5 6 4 1 2 1 1 1 3 3 3 2 2 2 2 ... kx C kx C kx + = + + + M1 1.1a Attempt at least 2 of these terms – products of binomial coefficients and correct powers of 2 and 1 3 kx Using kx rather than ⅓kx mark as MR -2 64 64kx + A1 1.1 + 2 2 80 3 k x A1 1.1 [3] 4 (b) ( )( ) 2 2 80 3 3 4 64 64 x kx k x − + + + ( ) 2 2 192 ... 80 256 k k x = + + − M1* 3.1a Using two terms from the expansion in (a) to find the coefficient of 2 x 2 5 16 12 0 ... k k k − − =  = M1dep* 2.1 Forming a 3TQ in k Using 3 their constant term from (a)  8 2 31 5 k + = A1 2.2a BC must be positive root only [3] H230/02 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 5 (a) scale factor 4 B1 1.1 Allow s.f. or ‘factor’ but not just 4 or 4 units or ‘scale factor 4 units’ Comments MUST refer to a stretch otherwise B0B0 parallel to the y-axis B1 2.5 oe e.g. allow in the y-direction, vertically, vertically upwards, vertical stretch, in the vertical direction, positive y-direction, parallel to the positive y-axis but not just upwards or in/on/about/across/through/along/towards the y-axis If B0B0 then SCB1 for re-writing as ( ) 3 ( )4 2 x y = [2] 5 (b) 3 2 3 2 2 36 x x + − = B1 1.1 For the correct equation in x Or 4 36 12 A A A y y y − =  = ( ) 3 3 2 12 log 2 log12 x x =  = M1 1.1 Simplify to 3 2 x k = where k> 0 and take logs of both sides (any base) 1 2 3 log 12 x = A1 1.1 oe, e.g.x = log2(121/3) [3] H230/02 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 6 (a) ( ) ( ) ( ) 2 2 3 5 1 0 AB = −− + − or ( ) ( ) ( ) 2 2 5 9 0 7 BC = − + − M1 1.1 Correct formula for the distance between two points for either AB or BC (or these distances squared) 3 out of 4 values correct for either distance 65 AB BC = = A1 1.1 Correctly showing that AB = BC, exact values needed [2] 6 (b) ( ) ( ) ( ) 2 2 3 9 1 7 AC = −− + − M1 2.1 Attempt to find AC (or its square) – 3 out of 4 values correct Or find gradients of both line segments 1 0 3 5 AB m − −− = and 7 0 9 5 BC m − − = ( ) ( ) ( ) ( ) 2 2 2 65 65 130 180 AC + =  = so angle ABC is not a right angle Or 5 cos , which is not =0 therefore angle ABC is not a right angle 13 ABC = − Or o Angle 112.62... which is not a right angle ABC = A1 2.4 Show correctly that Pythagoras does not hold in triangle ABC Using cosine rule Or ( ) 7 7 1 8 4 32 1 − = − − So angle ABC is not a right angle o.e. [2] 6 (c) (3, 4) B1 [1] 1.1 H230/02 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (d) 0 5 4 0 3 5 y x − − − − = M1 1.1 Correct formula for the equation of the line between B and their midpoint of AC from (c) Or using ( ) 1 1 BM y y m x x − = − Or using BM y m x c = + 2 10 x y + = A1 1.1 o.e. required form. [2] 6 (e) ( ) ( ) 2 2 3 1 65 x y + + − = B1 B1FT 1.1 1.1 B1 for correct LHS B1FT for their 2 AB on RHS Must be an equation to gain marks [2] 6 (f) (1, 8) B1 2.2a [1] H230/02 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) ( ) d 8 10 d y x x = − B1 1.1 Correct derivative At ( ) 1 2 61 ,3 : 6 T N m m = − = M1* 1.2 Substitutes x = 0.5 into their two-term derivative and using product of gradients is 1 − ( ) 1 2 1 6 3 y x − = − M1dep* 1.1 Using ( ) 1 2 3 y m x −= − with 6 m −or their tangent gradient (so must have attempted normal gradient) Or using y mx c = + 2 12 35 0 x y − + = A1 1.1 Must = 0 and integer coefficients All terms on one side [4] (b) 𝑥  ≥ 1.25 B1 1.1 ( 0 1.25) y x =  = 𝑦  ≥ 4𝑥2 −10𝑥+ 7 B1 2.2a 2𝑥−12𝑦+ 35  ⩾ 0 B1FT 2.2a o.e. Follow through their (a) SCB2 if all “correct” (including FT from (a)), but either all strict or a mix of strict and non-strict inequalities used [3] H230/02 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 8 DR 1 2 8 3 x x + M1* A1 2.1 1.1 M1 for either term integrated correctly ( ) ( ) 1 2 8 3 16 12 7 a a + − + = M1dep* 1.1 Correct use of correct limits and equating to 7 – allow one substitution error 1 2 3 8 35 0 a a + − = M1 1.1 Forming a 3TQ in 1 2 a Any three-term form (so terms do not need to be on the same side) ( )( ) 1 1 2 2 3 7 5 0 a a − + = M1 3.1a Dependent on all previous M marks – correct method for solving for 1 2 a Or 1 2 8 35 3 a a = − ( )( ) 2 9 274 1225 0 9 49 25 0 a a a a − + = − − = 1 2 5 a −as 1 2 a can’t be negative A1 2.3 Explicit rejection of 5 − No specific justification required Explicit rejection of a = 25 No specific justification required 1 2 7 49 3 9 a a  = = A1 2.2a Correct value only [7] H230/02 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 9 (a) 50 (s) B1 3.4 [1] 9 (b) B1 1.1 Correct (t, v) graph – no values on axes required [1] 9 (c) ( )( ) ( )( ) 1 2 1 2 15 20 15 15 '50' 1950 or T + + = ( ) 1 2 20 '50' ) 15 1950 o.e. T T + + +  = M1 3.4 “Correct” equation for finding required time T using their (a) May not be earned until e.g. "1425" seen 15 95 (s) A1 1.1 [2] H230/02 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 (a) ( ) ( ) 2 4 9 d 2 9 v t t t t c = − = − +  M1* 1.1 Integrate given expression for a with at least one term (unsimplified) correct +c not required for this first M mark ( ) 1,2 2 2 9 ... c c  = − + = M1dep* 3.4 Using given conditions to find +c 2 2 9 9 v t t = − + A1 1.1 Condone ‘v =’ missing. [3] 10 (b) 2 2 9 9 0 ... t t t − + = = M1 3.4 Setting their 3-term quadratic for v, from (a), equal to zero and solving for t 1 2 ( ) 1.5, ( ) 3 t t = = A1 1.1 BC [2] 10 (c) ( ) 1.5 2 0 45 8 2 9 9 d t t t − + =  B1FT 3.1b BC – correct value for their 1 0 d t v t  Only FT when their 3-term quadratic in (b) leads to positive values for 1 2 and t t ( ) 3 2 1.5 9 8 2 9 9 d t t t − + = −  B1FT 1.1 BC – correct value for their 2 1 d t t v t  Total distance travelled is 6.75 (m) B1 3.2a cao [3] H230/02 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 11 (a) ( ) 1 4.5 0.6 3.5 T − = M1 3.3 N2L for P – correct number of terms and dimensionally consistent – allow sign confusion 1 6.6 T = (N) A1 1.1 [2] 11 (b) For Q: ( ) 2 1 0.4 0.4 3.5 T g T + − = For R: ( ) 2 3.5 mg T m − = M1 M1 A1 3.3 3.3 1.1 M1 for N2L for Q and M1 for N2L for R – correct number of terms and dimensionally consistent – allow sign confusion A1 for both correct (allow with their tension from (a)) Must be using a = 3.5 Or (by considering Q and R together) 1 (0.4 ) (0.4 ) 3.5 m g T m + − = +  scores M2 A1 m = 0.648 A1 1.1 3sf required (0.6476190…) (m = 68 105) [4] H230/02 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 11 (c) Before string breaks P moves ( )( ) ( ) 2 0.5 3.5 0.4 0.28 = B1 3.4 Correct (unsimplified) expression When string breaks the speed of P is ( ) ( ) 3.5 0.4 1.4 = B1 3.4 Correct (unsimplified) expression For P: 4.5 0.6 T a − = For Q: 0.4 0.4 g T a − = M1* A1 3.3 1.1 M1 for attempt at N2L for both P and Q after string breaks – correct number of terms and dimensionally consistent- but allow sign confusion For reference if solved correctly then a is 0.58 − When string breaks P travels a distance s where ( ) 2 0 1.4 2 0.58 s = + − M1dep* 3.1b Use of 2 2 2 v u as = + with v = 0 and their values for u and a M0 if a = 3.5 used For reference 49 29 1.689655... s = = Total distance is 0.28 + 1.689… = 1.9696… < 2 or Total distance is 0.28 + 1.689… = 1.9696… so P does not reach the pulley A1 2.2a AG [6] 11 (d) Both P and the pulley are modelled as having negligible size rather than as objects with dimensions and therefore this could account for why P does reach the pulley See Appendix for some allowable responses B1 3.5a An answer that refers to the dimensions of P and/or the pulley If more than one factor given then B1 if all are acceptable, B0 if not. Identifying a relevant factor is sufficient [1] H230/02 Mark Scheme June 2023 21 APPENDIX Exemplar responses for Q11(d) Response Mark Frictional force may not be constant B1 String not light B1 String not inextensible B1 String may be elastic B1 Elasticity B1 String may be extensible B1 Friction is constant B1 Friction (as a one word answer) B1 Surface being smooth B0 Air resistance B0 Particle may be smooth B0 Frictional force on pulley B0 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. 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H230/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 H230/01 Mark Scheme June 2023 4 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. H230/01 Mark Scheme June 2023 5 5 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 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 H230/01 Mark Scheme June 2023 6 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. 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. H230/01 Mark Scheme June 2023 7 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. H230/01 Mark Scheme June 2023 8 Question Answer Mark AO Guidance 1 (a) LHS = sin cos cos sin x x x x +  M1* 1.1a Use of tan = sin/cos = 2 2 cos sin cos x x x + dM1 1.1 Attempt common denominator (by multiplying numerator) OR apply sin2 𝑥+ cos2 𝑥≡1 in correct working to reach cos 𝑥+ 1−cos2 𝑥 cos 𝑥 oe = 1 cos x AG A1 2.2a www, Must see previous line and answer [3] 1 (b) tan2 x = 1 2 M1 1.1a or 3sin2 x = 1 or 2 = 3cos2 x and attempt to solve tan x = ± 1 2 sin x = ± 1 3 cos x = ± 2 3 x = 35.3° or x = 145° (Both) A1 1.1 (35.26…°, 144.7…°) A0 if any additional solutions within range (isw any outside) [2] Question Answer Mark AO Guidance 2 (a) (i) AB = 3 6 −       − 4 3 −       or 1 3    M1 1.1 One of these. Attempt b − a or c − b or similar BC = 1 12 −       − 3 6 −       or 2 6    BC = 2 AB or BC is a multiple of AB A1 2.1 Dep correct AB and BC Multiple (2) not required, but if given must be correct. Hence B lies on AC Alternative method 1: AB = 3 6 −       − 4 3 −       or 1 3    M1 One of these. Attempt b − a and c − a or similar H230/01 Mark Scheme June 2023 9 AC = 1 12 −       − 4 3 −       or 3 9    AC = 3 AB or AC is a multiple of AB A1 Dep correct AB and AC Multiple (3) not required, but if given must be correct. Hence B lies on AC Alternative method 2: Gradient of line AC is m = 3 Equation of line AC is 𝑦−(3) = 3(𝑥−(−4)) (𝑦= 3𝑥+ 15) M1 Find (gradient and) equation of line AC – need not be simplified At 𝑥= −3, 𝑦= 3(−3) + 15 = 6 (i.e. B) Hence B lies on AC A1 Substituting in x-coordinate of B (or both x,y) to show consistent Dep on correct equation Alternative method 3: Gradient of line AB is 3 AND Gradient of line BC is 3 M1 Must both be explicitly stated for this method. As B lies on both AB and BC, and AB and BC have the same gradient, B lies on AC. (OR therefore A,B,C are colinear) A1 Must make a convincing argument (not just conclude directly from two gradients) www. [2] 2 (a) (ii) AB : BC = 1 : 2 B1 1.1 Must be a ratio (but may be equivalent e.g. 2 : 4) [1] 2 (b) Q marked at (4, 2) or (4, 2) stated B1 3.1a May be implied by correct magnitude or direction Magnitude = 2 2 or 8 or 2.83 (3 sf) B1 1.1 Direction = −45o or 315o B1 1.1 Accept any unambiguous indication of the direction of 𝑃𝑄 ⃗⃗⃗⃗⃗ e.g. “towards the x-axis along 𝑥+ 𝑦= 6” OR an arrow on diagram OR stating direction together with the column vector 𝑃𝑄 ⃗⃗⃗⃗⃗ = ( 2 −2) Condone 135° as a bearing (but must state “bearing”) [3] H230/01 Mark Scheme June 2023 10 2 (c) 4 3    and 0 5    B1 B1 3.1a 1.1 SC. Either or both “correct” but coordinates: max SCB1 [2] Question Answer Mark AO Guidance 3 (a) Straight line starting at (0, k) where k < 0 B1 1.1a Line must intersect y-axis for this mark with positive gradient throughout B1 1.2 cutting or touching x-axis vertically below minimum B1 2.2a Mark the intention [3] All three marks are independent 3 (b) 0 B1 1.1 Condone “there would be no gradient” [1] Question Answer Mark AO Guidance 4 (a) 10 = e3x 3x = ln 10 M1 1.1a Attempt to take logs of 10 = 𝑒3𝑥 x = 1 3 ln 10 or 0.768 A1 1.1 0.767528… Allow answer in range [0.767, 0.768] Answer only (without working) SCB1 [2] 4 (b) Gradient = 3e3x M1 1.1a soi. Allow ke3x for this mark (𝑘≠1) or sight of 3𝑒3(2) Gradient of tangent at x = 2 is 3e6 or 1210 A1 1.1 1210.286… isw if numerical form inaccurate (but do not accept -1/m if the perpendicular gradient is given as the final answer) Answer only (without working) SCB1 [2] H230/01 Mark Scheme June 2023 11 Question Answer Mark AO Guidance 5 DR x(x2 − 4) = 0 B1 3.1a Evidence of factorising or otherwise attempting to solve (=0 not required for this mark) x = 0,−2, 2 B1 1.1 This mark may be implied by correct limits ( ) 2 3 1 0 4 d A x x x = −  M1* 1.1 Ignore limits for this mark 2 4 2 0 2 4 x x   = −     = −4 A1 1.1 Must be seen for this mark (or clear indication of taking modulus) Condone area from [0,-2] as -4 or from [2,0] as +4 but must be consistent with their limits A2 = |A1| = 4 or A2 = −A1 = 4 dM1 2.1 By symmetry: Total area = 2 × (their A1) or adding together two areas of the same sign from their two integrals (or just 4 + 4) Total area = 8 A1 1.1 www, Area must be positive Alternative method for final M1*A1dM1A1 ( ) ( ) 0 2 3 3 2 0 4 d 4 d x x x x x x − − − −   M1* Ignore limits for this mark = 4 − (−4) A1 dM1 Correct area of -4 seen Attempt combine the two areas, with correct signs = 8 A1 www, Area must be positive NB ( ) 2 3 2 4 d x x x − −  = 0 scores B1B1M1A0M0A0 if working seen SC, no working or inadequate working: One area = 4: SCB3 or Total area = 8: SCB4 [6] H230/01 Mark Scheme June 2023 12 Question Answer Mark AO Guidance 6 (a) x3 = –1 or 1 8 M1 1.1 For attempting to solve as a quadratic in x3 (May be implied by these correct answers for x3) x = –1 A1 1.1 (Must be explicitly, separately, stated following x3 = –1) or x = 1 2 A1 1.1 Some working must be seen Answers only (no working or inadequate working) SCB1 for each (max 2/3) [3] 6 (b) d d y x = 56x6 + 49x3 – 7 M1 A1 3.1a 1.1 Attempt differentiate, at least two terms correct All correct 56x6 + 49x3 – 7 = 0 M1 1.1 Setting their d d y x = 0 and attempting to solve (must show attempt at solving the quadratic in x3 or reach x=… from part (a)) Must see =0 here (may be implied by fully correct solutions only) (8x6 +7x3 – 1 = 0, x = –1 or x = 1 2 ) SPs are (–1, 11.25) and (0.5, 171 64 − ) or (0.5, –2.67) (3sf) A1 1.1 (Accept 11.3 or 45 4 ) Coordinates must be correctly paired [4] 6 (c) d2𝑦 d𝑥2 = 336x5 + 147x2 M1* 1.1 Attempt differentiate their d𝑦 d𝑥 , allow one error only x = 1 2  d2𝑦 d𝑥2 > 0 d2𝑦 d𝑥2 = 189 4 = 47.25 (not required but if given then must be correct for their value of x) Hence x = 1 2 gives a minimum A1ft 1.1 Must see both statements; no need to see calculation ft their 2nd derivative and their positive x x = –1  d2𝑦 d𝑥2 < 0 dM1 1.1 Substitute their −ve x into their d2𝑦 d𝑥2; may be implied by –189 Hence x = –1 gives a maximum A1ft 2.2a www in this part, ft their -ve x value (-189 not required but if given then must be correct for their value of x) [4] NB differentiating 8x6 +7x3 – 1 can achieve M1A0dM1A0 (max 2/4) H230/01 Mark Scheme June 2023 13 Question Answer Mark AO Guidance 7 (a) (i) ( ) 2 2 2 2 2 , a b a b + + B1 1.1 Both oe (NOT (𝑎+𝑏)2 2 ) [1] isw any multiplying out in this part 7 (a) (ii) ( ) 2 2 2 2 2 a b a b + + − M1* 3.1a Difference between two expressions from (a)(i) = 2 2 2 2 2 2 ( 2 ) 4 a b a ab b + − + + dM1 1.1 oe, attempt collect over denominator of 4 = 2 ( ) 4 a b − > 0 A1 2.2a Must see 2 ( ) 4 a b − and "> 0" Alternative method: (𝑎+ 𝑏 2 ) 2 = 𝑎2 + 2𝑎𝑏+ 𝑏2 4 For the statement to be true: 𝑎2 + 𝑏2 2 ≥𝑎2 + 2𝑎𝑏+ 𝑏2 4 2𝑎2 + 2𝑏2 ≥𝑎2 + 2𝑎𝑏+ 𝑏2 𝑎2 + 𝑏2 ≥2𝑎𝑏 𝑎2 + 𝑏2 −2𝑎𝑏≥0 Which is true because (𝑎−𝑏)2 ≥0 M1* dM1 A1 Multiplying out their expression for the square of the mean of the form (𝑐+ 𝑑)2 to reach 𝑐2 + 2𝑐𝑑+ 𝑑2 (need not include the denominator). May be seen in part (a)(i). If two such expressions are present they must both be correctly multiplied out. Comparing their two expressions (condone <,>,= but expressions must be of the correct form i.e. not Σ) and attempting to manipulate (must include denominators) (for any real values of a and b) Must see this or an equivalent statement [3] 7 (b) Variance is > 0 B1 3.2a Condone Variance is positive or ‘never negative’ (because Variance is the difference between the mean of the squares and the square of the mean) [1] Question Answer Mark AO Guidance H230/01 Mark Scheme June 2023 14 8 (x – 3)2 + (y – 2)2 = (1) M1 3.1a DR Attempt rearrange to this form Centre (3, 2), A1 1.1 soi but nfww radius 1 A1 1.1 soi but nfww M1 2.1 Attempt at a correct diagram seen (allow e.g. a slip in labelling) or geometrical understanding clearly implied (e.g. by giving gradient of AC or the gradient of the radius along AC as 1 3 oe) Diagram must show the circle, tangents and either axes or labelled coordinates to be adequate on its own (but may be implied by later correct working). (AC = 10 ) sin α = 1 10 B1 1.1 or tan α = 1 3 2 × sin−1 1 10 M1 2.1 or tan 2α = ( ) 1 3 2 1 3 2 1  − (= 3 4 ) Angle between tangents = 36.9o (3 sf) A1 3.2a Angle between tangents = tan−1 3 4 or 36.9o (3 sf) Alternative method y = mx + 1 B1 soi x2 + (mx + 1)2 – 6x – 4(mx + 1) + 12 = 0. M1* Substitute their line equation into circle equation (circle equation may be rearranged first) (1 + m2)x2 – (6 + 2m)x + 9 = 0 A1 (6 + 2m)2 – 36(1 + m2) = 0 dM1 Attempt b2 – 4ac = 0 (must be =0 or ≥0) 24m – 32m2 = 0 dM1 Rearrange and attempt to solve quadratic equation in m (must reach a value for m) 1 2 3 1 2 3 4 0 x y A C H230/01 Mark Scheme June 2023 15 m = 0 or 3 4 A1 Both needed Angle between tangents = tan−1 3 4 or 36.9o A1 or 0.644 (radians) [7] (If two partial solutions given, apply whichever scheme gains the most marks) Question Answer Mark AO Guidance 9 (a) 20 50 or 2 5 or 0.4 B1 1.1 isw [1] 9 (b) 20 45 oe B1 1.1a May be implied e.g. by 5 45 + 15 45 20 19 45 44  M1 2.1 M1 for subtracting 1 from numerator and denominator and multiplying = 19 99 or 0.192 (3 sf) A1 1.1 oe (e.g. 380 1980) [3] Question Answer Mark AO Guidance 10 (a) eg 120 × 6 24 or 120 × 150 600 M1 3.1a Attempt 120 × Area of 60-65 block Total area , using any units (must see 120 used for this mark, but may be implied by ‘30’) = 30 A1 1.1 cao [2] 10 (b) (i) Attempt areas of other blocks M1 1.1 May be implied by 𝑥̅ ∈[55,60] OR 𝜎∈[5,7] OR 𝜎2 ∈[30,40] 57.7 (3 sf) A1 2.1 BC 6.20 (3 sf) Allow 6.2 A1 1.1 BC [3] 10 (b) (ii) Distribution of masses in classes unknown B1 3.2b Not just “Because midpoints used.” Or “actual masses not known” or similar. [1] H230/01 Mark Scheme June 2023 16 10 (c) 57.7 – 2 × 6.20 M1 2.1 Attempting 𝜇± 2𝜎 = 45.3 or 45, hence 4 outliers A1ft 2.2a Allow 4 outliers ft their values from (b)(i) but must be consistent Allow ‘whole classes’ or appropriate interpolation here (NB frequencies by class are: 4,10,20,22,20,30,14) Allow 14 from the sum of the first two classes (10+4) [2] 10 (d) Can obtain actual frequencies from histogram B1 1.2 or similar; eg pie chart only shows relative frequencies or “histogram has figures while pie chart shows a percentage” Acceptable answers must be specific to the question (finding actual frequencies to estimate mean/standard deviation) Not just “show spread of data” or “easier to read” or “easier to see distribution” etc. [1] Question Answer Mark AO Guidance 11 (a) Need to find P(X > 50) B1 1.2 OR 1 −P(𝑋≤49) Condone just 𝑋≥50 Ignore all else (but do not accept just ‘cumulative’) [1] 11 (b) X~B(60, 0.75) and X > 50 M1 3.3 Allow X > 50, X < 50, X < 50, X = 50 May be implied by P(X > 50) = 0.0859 (or P(X > 50) = 0.0452) P(X > 50) = 0.0859 A1 3.4 BC (accept awrt 0.086 (2sf)) 0.0859 > 0.05 A1ft 1.1 ft correct comparison of their value with 0.05 (OR 0.95 as appropriate – but must see the value being compared) Do not reject H0 M1 1.1 Must be correct based on their value comparison Condone ‘Accept H0’ or ‘Reject H1’ Insufficient evidence (at 5% level) that more than 75% have BR as favourite band A1 2.2b In context, not definite Condone ‘no evidence’, ‘not likely that’ etc. From correct working only [5] H230/01 Mark Scheme June 2023 17 11 (c) (i) B(60, 0.75) B1 3.3 OR Binomial, n=60, p=0.75 (must have all three) [1] 11 (c) (ii) Yes. Whether a student’s favourite is BR is not independent of other students. B1 3.5b Must be in context or As students are chosen in turn, the probability changes Allow any of: • P(a chosen student’s favourite is BR) is affected by previously chosen students • P(a student’s favourite is BR) changes Do not accept: • “Risk of picking the same student again” [1] H230/01 Mark Scheme June 2023 18 Question Answer Mark AO Guidance 12 (a) Fig 1: Positive. Fig 2: Negative B1 1.2 Ignore all else [1] 12 (b) Fig. 1: 25-44s have children so proportion of 0-4 is directly related to proportion of 25- 44. B1 2.4 oe e.g. “25-44s more likely to be parents” Must be in context so do not accept generic descriptions of the correlation. Fig. 2: High proportion of older people means lower proportion of families with young children B1 2.4 Oe e.g. “>60s less likely to live with young children” Must be in context so do not accept generic descriptions of the correlation. [2] In both parts: - Need not refer to proportion but may not refer to amount (i.e. do not accept answers including references to ‘amount’ or ‘number’) - SC Max [1/2] for correct description in context but referencing ‘amount’ or ‘number’ or ‘population’ - Ignore irrelevant statements Condone ‘children’ as shorthand for the 0-4 age group in this question (and ‘elderly’ as shorthand for the >60 age group) 12 (c) This LA has a similar number of >60s to many other LAs, but these represent by far the lowest proportion of its population. OR This LA has a large number of 0-4s, but its proportion of 0-4s is similar to several other LAs. (Hence must have large number in total.) B1 2.4 oe. Must make a correct connection between number and proportion and compare to other LAs to show that this is one of the largest Do not accept statements referring to 25-44 (from Fig. 1) [1] Need to get in touch? 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H630/02 Mark Scheme June 2023 5 11. Annotations Annotation 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting H630/02 Mark Scheme June 2023 6 Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result 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. BP Blank Page Seen Highlighting

H630/02 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 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. H630/02 Mark Scheme June 2023 8 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. 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. H630/02 Mark Scheme June 2023 9 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. 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 H630/02 Mark Scheme June 2023 10 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. H630/02 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 1 (a) The given data (on the graph) is continuous, not discrete oe E1 2.4 Need a comment on the suitability of the histogram for the given data – we need a comment on the type(s) of data. The following score -The given data was discrete, but this diagram represents continuous data -The diagram is for continuous data, but the original data was discrete -‘It is continuous data’ as the question asked about the histogram - The data is discrete so there should be gaps between the bars -‘The histogram uses ranges e.g. 6- 7 but the data is discrete and cannot have values between 6-7’ etc Comments such as ‘there are no gaps between the bars’ or ‘it should be a bar chart’ or FD should be used or ‘it is discrete data’ (ambiguous) score 0 [1] 1 (b) (i) 3.183… to 3.184 or 3.18 BC Mark at most accurate B1 1.1 Exact value is 3.183673469 or 3 9 49 or 156 49 o.e. Allow if seen in (b)(ii) if not seen in (b)(i) SC: 3.2 allow B1 [1] 1 (b) (ii) 1.61… to 1.62 BC OR 1.59… to 1.60 B1 1.1 Exact value is 1.615980989 (sample) Exact value is 1.599406387 (population) [1] H630/02 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 2 (a) (𝑥−3)2 seen M1 1.1 eg in (𝑥−3)2 −9 + 1 (𝑥−3)2 −8 A1 1.1 [2] 2 (b) (3, ‒8) B1FT 1.1 FT their completed square [1] 3 6n ‒ 1 evaluated for any positive integer M1 1.1 eg 6×1 ‒ 1 = 5 n 1 2 3 4 5 6 7 8 9 6n - 1 5 11 17 23 29 35 41 47 53 n 10 11 12 13 14 15 16 17 18 6n – 1 59 65 71 77 83 89 95 101 107 n 19 20 21 22 23 24 25 26 27 6n – 1 113 119 125 131 137 143 149 155 161 eg 6×6 ‒ 1 = 35 = 5×7 which is not prime A1 1.1 may see eg n = 11, 17, 23, 29- sight of any value in the table would imply the M1 Must show it’s factorisation so show it’s not prime and give a concluding comment e.g ‘not prime’ If they say ’35 is divisible by 5’ so not prime etc then A1 BUT ’35 isn’t prime’ is A0 [2] H630/02 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 (a) 4 B1 1.1 [1] 4 (b) B1 Dep B1 1.1 1.1 Correct shape in both quadrants; Must not cut either axis or bend away excessively from either axis Reasonably Symmetrical about y-axis Condone slight feathering, and slight asymmetry along the y - axis [2] 5 √(11 −5)2 + (−1 −2)2 o.e. OR √(5 −11)2 + (2 −(−1))2 M1 2.1 allow one sign error √45 A1 1.1 3√5 A1 2.2a Condone √45 for A1A1 and √45 or 3√5 only implies full marks [3] 6 (a) Population because all the available data are used E1 2.4 This is an ‘explain’ question, so we do need ‘population’ and a correct justification. Accept ’population as it is data from every single day the phone was used’ scores ‘Population as it is every day the phone is used’ scores [1] 6 (b) Negative skew B1 1.2 ‘Negative’ is B0 [1] H630/02 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (c) Q3 = 58 or Q1 = 42 identified B1 1.1 IQR = 16 B1 1.1 ‘IQR = 16’ implies B1B1 [2] 6 (d) 42 −1.5 × 16 = 18 M1 FT (c) 1.1a Ignore checking of upper tail For calculating 𝑄1 −1.5 × 𝐼𝑄𝑅 for their values Smallest value is 19 which is not an outlier, so no outliers in lower tail A1 1.1 Comparison of lower bound with 19 and conclusion e.g. ’18 < 19 or 19 > 18 (or equivalent in words) so no’ etc If they calculate 18 then and mention 19 and ‘no’ then A1 ‘..18 so no as all values > 18’ is A1 ’18 so no as all values more than 18’ is A1 ’18 so no etc’ is A0 as they need to compare to smallest value (directly or indirectly) and explain why. Command word ‘determine’- justification needed [2] H630/02 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) 23 + 6 × 22 −2 −30 =0 so (𝒙−𝟐) is a factor B1 1.1 Factor theorem must be used, and a concluding statement needed Statement might be at the start e.g 𝑓(2) = 0 ⇒(𝑥−2) a factor e.g. Synthetic Division or Long Division is B0 Must see evidence of the substitution- ‘show that’ so e.g simply 𝑓(2) = 0 is B0 [1] 7 (b) (𝑥−2)(𝑥2 + 8𝑥+ 15) M1 A1 By inspection or from long division, allow sign errors only Fully correct linear × quadratic (𝑥−2)(𝑥+ 5)(𝑥+ 3) A1 1.1 Fully correct and fully factorised [3] Alternatively f(k) evaluated, where k is -3 and -5 M1 1.1a Allow a slip with either but not both (𝑥+ 3) 𝑜𝑟 (𝑥+ 5) identified as a factor A1 1.1 (𝑥−2)(𝑥+ 5)(𝑥+ 3) A1 8 (a) Remove any data where #N/A is in the column, as there is no data available B1 2.4 LDS advantage Comments such as exclude 20/81 year old female are B0 as this is only an extract of the full set of data Must refer to N/A/missing data but accept ‘not available/non- existent’ Remove data without a pulse reading is B0 as we could have missing BMI data too H630/02 Mark Scheme June 2023 16 Question Answer Marks AO Guidance [1] 8 (b) (62.77, 84) ringed and no others B1 1.1 [1] 8 (c) No evidence of a linear relationship oe so unlikely to be reliable B1 2.2b Any comment relating to interpolation or extrapolation is B0 as we want a comment on the appropriateness of the model considering the scatter diagram Condone ‘there appears to be little correlation/weak positive correlation’ etc ‘No/Zero correlation’ is B0 e.g the PMCC could be 0.1 etc [1] H630/02 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 8 (d) None of the pulse rates are that unusual so should not be removed B1 2.2b LDS advantage Need ‘no/keep’ and reason Condone ‘No as they are not outliers’ Accept ‘Higher pulse rates are not uncommon’ [1] 9 (a) 0.1 + 0.3 + 𝑞+ 2𝑞+ 3𝑞= 1 M1 1.2 Setting sum of values equal to 1 𝑞= 0.1 A1 1.1 [2] 9 (b) (0.1 + 0.3 + 0.1 + 0.2 =) 0.7 B1 1.1 Or 1 −𝑝(𝑋= 5) [1] 9 (c) 0.1 × 0.3 seen or 0.03 seen M1 1.1 0.1 × 0.3 + 0.1 × 0.3 = 0.06 A1 1.1 0.06 o.e 6 100 or 3 50 etc [2] 9 (d) 0.098314…correct to 2 or more sf BC B1 1.1 By using e.g 𝑋~𝐵(50,0.3) and finding 𝑃(𝑋= 17) [1] H630/02 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 Cos A = 3.52+3.92−4.52 2×3.9×3.5 oe M1 2.1 Or Cos B = 4.52+3.92−3.52 2×3.9×4.5 or Cos C = 3.52+4.52−3.92 2×4.5×3.5 Correct use of cosine rule – might not see the ‘Cos A’ etc till next line, but must be a correct use so no ‘sin’ etc Can be in form 𝑎2 = 𝑏2 + 𝑐2 −2𝑏𝑐Cos𝐴 e.g. 4.52 = 3.52 + 3.92 −2 × 3.5 × 3.9 × Cos 𝐴 OR 3.52 = 4.52 + 3.92 −2 × 4.5 × 3.9 × Cos 𝐵 OR 3.92 = 3.52 + 4.52 −2 × 3.5 × 4.5 × Cos 𝐶 cos A = 0.2641… correct to 2 or more sf soi A1 1.1 cos B = 0.66125… or cos C = 0.5488889 A = 74.686° correct to 2 or more sf soi A1 1.1 B = 48.604° or C = 56.709° 1 2 × 3.5 × 3.9 × sin74.686 M1FT 3.1a or 1 2 × 4.5 × 3.9 × sin′48.604′ or 1 2 × 3.5 × 4.5 × sin′56.709′ Must be using their included angle for their two adjacent sides For the Final two marks: They could also find an altitude, h, using one of the angles e.g using angle at C ‘ℎ= 3.5 their sin 𝐶= 2.9256 … ‘ then 1 2 𝑏ℎ= 1 2 (4.5)(2.9256. . ) = 6.58267.. awrt 6.58 or 6.6 A1 3.2a H630/02 Mark Scheme June 2023 19 Question Answer Marks AO Guidance [5] 11 𝑑𝑦 𝑑𝑥= 6𝑥2 + 18𝑥+ 24 B1 2.1 their derivative 𝑑𝑦 𝑑𝑥= 0 M1 1.1 The 𝑑𝑦 𝑑𝑥= 0 may be implied by their concluding statement e.g if they say ‘no real roots’ Use of discriminant implies this mark their 182 −4 × 6 × 24 calculated (may be seen embedded in an attempt to solve their quadratic with QF etc. If no formula quoted then the the solutions must be correct for the method mark. If the formula is quoted, we can allow one error with the substitution of values) M1 3.1a Most common quadratics seen are: 6𝑥2 + 18𝑥+ 24 = 0 or 3𝑥2 + 9𝑥+ 12 = 0 or 2𝑥2 + 6𝑥+ 8 = 0 or 𝑥2 + 3𝑥+ 4 = 0 May have to check for their quadratic NOTE: They could also use a sketch method here: sketch their quadratic and then complete the square to show that the TP is above the x – axis- will need to check their work carefully ‘−252 < 0’ o.e. for their quadratic A1 1.1 -252 < 0 or -63 < 0 or -28 < 0 or -7 < 0 etc May be implied by correct solutions to their quadratic e.g. −3±√7 𝑖 2 Hence 𝑑𝑦 𝑑𝑥= 0 has no solutions and therefore there are no stationary points on the curve A1* 3.2a Must give a concluding statement e.g. ‘therefore no stationary points’. Depends on all previous marks. Condone SPs or ‘turning points’ or TPs for stationary points H630/02 Mark Scheme June 2023 20 Question Answer Marks AO Guidance [5] 12 (a) Opportunity/Convenience sampling B1 1.2 Condone ‘Opportunistic Sampling’ [1] 12 (b) Because every sample (of size n) does not have the same probability of being selected B1 2.4 Accept ‘all adult males registered at the surgery do not have an equal chance of being selected’ Accept ‘everyone registered at the surgery does not have an equal chance of being selected OR ‘For a SRS each element from the SF must have an equal chance of selection’ OR ‘A subset of the population cannot form a complete sampling frame’ OR ‘The sampling frame would be incomplete’ ‘No random method employed in the process’ B0 (need to have the idea that the SF is incomplete) ‘Only collected data from one week’ is B0 [1] H630/02 Mark Scheme June 2023 21 Question Answer Marks AO Guidance 12 (c) w 50- 65- 70- 80- 90- 100- 120 f 6 8 8 11 6 6 B1 1.1 [1] 12 (d) 2 3 × 𝑡ℎ𝑒𝑖𝑟 6 + 1 2 × 𝑡ℎ𝑒𝑖𝑟 6 Or (10 x 0.4) +(10 x 0.3) M1 1.1 IF part (c) is correct then this could be implied by ‘4 +3’ 7 45 or 0.15̇ or 0.15555 to 0.156 Mark at most accurate A1FT 1.1 FT their 6, 6 and 45. May need to check their calculation. May see interpolation methods, which lead to the same calc. [2] 12 (e) The distribution of the weights within each class is unknown E1 2.4 Accept ‘we assume that the values (individual weights) are equally distributed in each class interval’ Accept ‘the individual values (weights) are not known’ Accept ‘the number of people (frequency) in each category of the histogram may not be spread out equally across the category’ Accept ‘we don’t know exactly how many were less than 60kg and how many were more than 110kg’ (idea of correct frequency at both ends for correct probability calculation) [1] H630/02 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 13 (a) H0 : p = 0.37 and H1 : p < 0.37 B1 1.1 Allow equivalent in words Do not allow percentages [1] 13 (b) p is the probability that an adult selected at random in the United Kingdom never exercises (or plays sport) B1 2.5 Accept ‘proportion’ but not number/amount etc B1B1 in (a)(b) if another symbol instead of p used if correctly defined Underlined words needed [1] 13 (c) P(X) ≤ 35 = 0.058… B1 1.1 their 0.058.. compared with 0.05 M1 1.1 may see 𝑃(𝑋< 35) =0.0386 or 𝑃(𝑋≤36) = 0.0847 using 𝑋~𝐵(118, 0.37) – SC 1 mark Use of 𝑷(𝑿= 𝟑𝟓) = 𝟎. 𝟎𝟏𝟗𝟔𝟏..is M0 There is another approach using Normal approx to Binomial to find 5% CR 𝑌~𝑁(43.66, 27.5058) which gives CV 35.033… Or finding value from ND 𝑃(𝑌≤35.5) = 0.059867.. (must be using a continuity correction so 𝑃(𝑌≤35) = 0.0493465.. is M0) 0.058 > 0.05 or 35.033… > 35 or 0.059867 > 0.05 and ‘so do not reject H0’ A1 2.2b ‘accept H0’ is ok H630/02 Mark Scheme June 2023 23 Question Answer Marks AO Guidance There is no evidence (or insufficient evidence) to suggest/support at the 5% level that the percentage of adults (selected at random in the United Kingdom) who never exercises (or plays sport) is less than 37% A1* 2.4 Fully correct contextual conclusion No assertive statements such as ‘proves that’ or ‘shows that’ ‘concludes that’ etc Accept percentage/proportion or probability with 0.37 Dependent on award of all other marks in (c) [4] 14 𝑦= 16𝑥 1 2 + 8𝑥−1 B1 3.1a May be implied by correct derivative 𝑑𝑦 𝑑𝑥= 8𝑥−1 2 −8𝑥−2 M1 A1 1.1 1.1 At least one term of the form 𝛼𝑥−1 2 or 𝛽𝑥−2 obtained All correct 𝑥= 4, 𝑑𝑦 𝑑𝑥= 7 2 B1FT 1.1 FT their 𝑑𝑦 𝑑𝑥, dep on award of M1 x = 4, y = 34 B1 1.1 y ‒ their 34 = (their 7 2)(x ‒4) oe e.g. sub (4, ‘34’) into their 𝑦= 𝑚𝑥+ 𝑐 to find their ‘c’ M1 FT 1.1 Their 7/2 must come from substituting x = 4 into their derivative H630/02 Mark Scheme June 2023 24 Question Answer Marks AO Guidance 𝑦= 7 2 𝑥+ 20 o.e. A1 3.2a All correct. Depends on all previous marks. We can accept any form of the equation of the line: 7𝑥−2𝑦+ 40 = 0 or 𝑦−34 = 7 2 (𝑥−4) o.e. Once the correct equation is seen in any form we can ISW if they simplify incorrectly etc NOTE: Final answer can be obtained from incorrect working- check their derivative [7] 15 (a) c = 1.14 B1 3.3 [1] 15 (b) 1.20 = 4𝑎+ 2𝑏+ 1.14 oe 1.25 = 16𝑎+ 4𝑏+ 1.14 oe M1 3.3 both equations. FT their c 𝑎= −0.00125, 𝑏= 0.0325 A1 1.1 Fractional equivalents are 𝑎= − 1 800 and 𝑏= 13 400 Equivalents in standard form is acceptable [2] 15 (c) 1.29 = 1.14 + 0.0325𝑡−0.00125𝑡2 M1 3.1b FT their a,b,c (Can be > etc) t = 6 and 20 A1 3.4 6 ≤𝑡≤20 A1 3.5a Set notation such as 𝑡∈[6, 20] is fine but must not be soft brackets 𝑡≥6 and 𝑡≤20 or 𝑡≥6 ⋂ 𝑡≤20 but NOT 𝑡≥6 , 𝑡≤20 [3] H630/02 Mark Scheme June 2023 25 Question Answer Marks AO Guidance 15 (d) It will eventually predict a negative exchange rate oe (will fall below zero etc) B1 3.5a ‘Exchange rate tends to zero’ is B0 Must mention the variable ‘exchange rate’ Underlined words needed [1] 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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H630/01 Mark Scheme June 2023 5 11. Annotations Annotation 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting H630/01 Mark Scheme June 2023 6 Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result 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. BP Blank Page Seen Highlighting

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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Y531/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 Y531/01 Mark Scheme June 2023 4 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.

Y531/01 Mark Scheme June 2023 5 5. 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. Y531/01 Mark Scheme June 2023 6 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. Y531/01 Mark Scheme June 2023 7 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. Y531/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 u = x + 2 B1 1.1 Correct substitution stated or used (u – 2)3 = u3 – 3×2u2 + 3×22u – 23 M1 1.1 Attempt to expand their (u – 2)3 following a linear substitution. 4 terms using ቀ𝑛𝑛 𝑟𝑟ቁ2𝑟𝑟𝑢𝑢𝑛𝑛−𝑟𝑟 u3 – 6u2 + 12u – 8 Binomial coefficients might not be correct or evaluated. Might be by expanding brackets. (u – 2)4 = u4 – 4×2u3 + 6×22u2 – 4×23u + 24 M1 1.1 Attempt to expand their (u – 2)4 following a linear substitution. 5 terms using ቀ𝑛𝑛 𝑟𝑟ቁ2𝑟𝑟𝑢𝑢𝑛𝑛−𝑟𝑟 u4 – 8u3 + 24u2 – 32u + 16 Binomial coefficients might not be correct or evaluated. Might be by expanding brackets. 4(u – 2)4 – 2(u – 2)3 – 3(u – 2) + 2 (= 0) M1 1.1 Forming (LHS of) equation in u (could be using their expansions) ∴4(u4 – 8u3 + 24u2 – 32u + 16) – 2(u3 – 6u2 + 12u – 8) – 3(u – 2) + 2 = 0 ∴ 4u4 – 34u3 +108u2 – 155u + 88 = 0 A1 1.1 Final answer can be in terms of x Final answer needs to be an equation with coefficients simplified. ISW attempts to cancel down following correct equation [5] Y531/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 2 (a) –5 + 5λ = 24 + 3µ & 6 – 2λ = 1 + µ B1 1.1 Forming 2 correct equations in λ and µ. Third equation is 15 – 2λ = –5 – 4µ 5λ – 3µ = 29 & 6λ + 3µ = 15 M1 1.1 Attempt to solve (eg scaling one equation and adding; or rewriting to a standard form for solution BC) If scaling or substituting method must result in correctly eliminating one variable (Other coefficients may be incorrect). λ = 4 & µ = –3 A1 1.1 Both λ = 4 & µ = –3 => LHS = 15 – 2×4 = 7 and RHS = –5 – 4×–3 = 7 = LHS so all 3 equations are satisfied so L1 and L2 do intersect A1 1.1 Convincing justification but could be by finding the same r from both equations 15 2 7     −       A1 1.1 Condone coordinates. Can be awarded even if previous A mark not awarded (i.e. if not checked third equation) [5] (b) 5 3 10 2 1 14 2 4 11           − × =           − −      B1 3.1a Could be BC. SOI 15 10 2 14 7 11 ν         = − +             r B1FT 1.1 FT their point of intersection from (a) and their attempt at direction vector. SOI Condone use of λ or µ. No need to see “ r= ” Must be a recognisable attempt at a vector perpendicular to both L1 and L2 15 2 7 10 14 11 x y z − + − = = B1FT 1.1 FT their vector equation. This is for correctly turning a vector equation into a cartesian one. Correct equation here implies the other two marks. [3] Y531/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 3 (a) DR z1z2 = (3 + 4i)(–5 + 12i) = –15 + 36i – 20i – 48 M1 1.1 Attempt at expansion (4 terms soi) using i2 = –1 DR – Need to see at least one line of expanded terms before answer = –63 + 16i A1 1.1 [2] (b) DR |𝑧𝑧2| = (ඥ(−5)2 + 122 = √169) = 13 B1 1.1 Not ± unless later corrected. Allow modulus of 13 for the B1 as long as no incorrect working Treat attempt to write z1 or z1z2 in mod/arg form as MR so B0M1A1 available 1 12 tan 5 −    −   M1 1.1 Evidence of using trigonometry towards finding the correct angle, perhaps by finding a related angle. Treat 1 12 tan 5 −      as such evidence for M1 but not 1 5 tan 12 −  −     or tan−1 ቀ 5 12ቁ unless supported eg by a diagram or by working leading to correct answer. 2 13(cos1.97 isin1.97) z ∴ = + (3 sf) A1 2.5 For argument accept awrt 1.97 only. Do not accept answers not written correctly in mod-arg form. Do not accept –1.18 or –4.32 as argument. Answer must be in radians for A1. Accept [r, θ] or r cis θ Is 1.965587446… eg do not accept the following 13cos1.97 13isin1.97 13(cos4.32 isin 4.32) + − NB 𝑧𝑧1 = 5(cos0.927 + isin0.927) 𝑧𝑧1𝑧𝑧2 = 65(cos2.89 + isin2.89) [3] Y531/01 Mark Scheme June 2023 11 (c) DR (arg(𝑧𝑧1𝑧𝑧2) =)tan−1 ቀ 16 −63ቁ M1 1.1 Using trigonometry to find the argument. Do not accept any other form unless supported by clear evidence (eg diagram) This mark may be awarded if z1z2 was incorrect from (a). 1 1 2 4 arg( ) arg( ) tan 1.965587... 3 0.927295... 1.965587... 2.89288... z z −  + = +     = + = M1 2.1 Attempt to calculate RHS using their values (either value could have been found earlier but both must be in [0, 2π)). Could accept 0.927… as evidence of arctan(4/3) arg(𝑧𝑧1𝑧𝑧2) 0.2487099... 2.892882... π = − + = so they are equal A1 2.2a Accept rounding to 3 sf or better but rounding must be correct (e.g. 0.927 + 1.96 = 2.89 would score A0). Answer must be in radians for A1. If MR z1 or z1z2 in part (b) then full credit available for a correct solution here. [3] Question Answer Marks AO Guidance 4 2 6 0 3    =     −   p. B1 2.2a Knowledge that perpendicularity implies that scalar product = 0, used anywhere in solution 2a2 + 6(a – 5) – 3×26 = 0 M1 3.1a Using the scalar product to set up a quadratic equation in a => 2a2 + 6a – 108 = 0 => a = 6 or a = –9 A1 1.1 Correctly solving the equation (could be BC). Might only see a = 6 here. a2 + 3a – 54 = (a – 6)(a + 9) = 0 2 2 3 225 3 54 2 4 a a a   + − = + −     a = –9 leads to negative y component (–14) so a = 6 is the only solution A1 2.3 a = –9 or brackets (a-6)(a+9) must be seen in solution and then a = –9 explicitly rejected with rationale Accept “all components must be positive” for rationale. Do NOT accept “a must be positive” as sole rationale [4] Y531/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 5 DR 𝛼𝛼+ 𝛽𝛽= 3 5 B1 1.1 12 5 αβ = B1 1.1 2 2 2 2 2 2 2 2 1 1 ( ) 2 ( ) β α α β αβ α β α β αβ + + − + = = M1 3.1a Rewrite the expression in terms of the standard symmetric functions Need to see 𝛼𝛼2 + 𝛽𝛽2 = (𝛼𝛼+ 𝛽𝛽)2 −2𝛼𝛼𝛼𝛼oe No need to see 𝛼𝛼2𝛽𝛽2 = (𝛼𝛼𝛼𝛼)2 = ቀ3 5ቁ 2 −2×12 5 ቀ12 5 ቁ 2 = − 37 48 A1 1.1 cao Accept eg 0.77083 −  but not rounded, incorrect or incomplete decimal form. SC B1 for correct answer if B0B0M0 Accept any equivalent fraction [4] Y531/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 6 n = 0, 4×1 + 66 = 70 = 5×14 is divisible by 14 B1 2.1 Basis case. Must explicitly see 70 and state, or show, divisibility. If n = 1 (leading to 98 = 7×14) used and not corrected then allow this mark but withhold final mark. Assume true for n = k ie that 4×8k + 66 is divisible by 14 oe M1 2.1 Statement of inductive hypothesis. Allow “= 14p” without further qualification Considering 4×8k +1 + 66 and rewriting it as 4×8×8k + 66 or 32×8k + 66 oe M1 1.1 Uses law of indices correctly to obtain expression in terms of 8k and no other exponential term Could consider f(k+1)—f(k) or similar = 8(14p – 66) + 66 from inductive hypothesis M1 1.1 Uses inductive hypothesis properly. Do not allow if eg 32k (ie law of indices must have been correctly used) Allow for substitution of their n=k case into their n=k+1 case = 112p – 462 or 14×8p – 14×33 oe =14(8p – 33) which is divisible by 14 A1 2.2a Simplification with sufficient working to establish and state divisibility for k + 1. Must either show 14× explicitly in each term or 14 is a factor So true for n = k ⇒ true for n = k + 1. But true for n = 0. So true for all integers n ≥ 0 A1 2.4 Clear conclusion for induction process. See note for basis case above. Do not allow “true for all positive integers”. Need to see if true for n=k, then true for n=k+1 i.e. k and k+1th case linked to n. Could see Proposition notation (𝑃𝑃𝑘𝑘⟹𝑃𝑃𝑘𝑘+1) A formal proof by induction, with no gaps in logic, is required for full marks. Full marks can be gained by using induction to prove that 2×8n + 33 is divisible by 7 for all n ≥ 0 and observing that 4×8n + 66 = 2(2×8n + 33) is divisible by 2 and 2×8n + 33 and hence by both 2 and 7 and hence by 14 [6] Y531/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 7 DR detA = a(a(a – 1) – 4×(–13)) – (–6)((a + 9)(a – 1) – 0) + (a – 3)((a + 9)(–13) – 0) M1 3.1a Attempt to expand the determinant. If using standard method must see at least two terms, at least one of which comprises ± a number multiplied by the residual determinant. eg a(a(a – 1) – 4×(–13)) – (a + 9) ((–6)(a – 1) – (a – 3)(–13)) This mark can be awarded with a = 2 substituted (ie attempt to find 2 6 1 11 2 4 0 13 1 − − − ) but working must be shown. eg 2(2 + 52) – 11(–6 – 13) (= 317) Allow other correct methods. = a3 – 8a2 + 22a +297 A1 1.1 a3 – 8a2 + 22a +297 = 23 – 8×22 + 22×2 +297 (or a3 – 8a2 + 22a +297 = 317) M1 2.2a Setting up equation equating their determinant to their specific determinant value for a = 2 a3 – 8a2 + 22a – 20 = 0 a2(a – 2) – 6a(a – 2) +10(a – 2) = 0 a2 – 6a +10 = 0 M1 1.1 Rearranging to = 0 and use of factor theorem to derive a quadratic equation in a by dividing by (a – 2) Need some evidence of how to solve cubic (Not just cubic written and then roots BC). Need to see =0 on one side of cubic (but it might disappear after this, e.g. when dividing by (a-2)) (a – 3)2 – 9 + 10 = 0 (a – 3)2 = –1 a – 3 = ±i M1 1.1 Attempt to solve quadratic involving √–1 = i oe 2 6 6 4 1 10 6 4 6 2i 2 1 2 2 ± − × × ± − ± = = × (a =) 3 ± i (and a = 2) A1 1.1 Both. No need to mention a = 2 Don’t need to see “a=” explicitly SC – if third and/or fourth method mark not awarded then allow SC B1 for sight of 3 ± i. [6] Y531/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 8 (a) ω = a + ib oe M1 3.1a Writing ω in a form which allows 2 equations to be found oe (eg taking Re and Im of both sides) a + 2 = 3a b + 7 = –3b – 1 M1 1.1 Equating real and imaginary parts. Aef This might happen after some simplification. a = 1 or b = –2 A1 1.1 (ω =) 1 – 2i A1 1.1 Need to see answer as a complex number. Ignore presence of ω*=1+2i as long as 1-2i identified as ω. [4] (b) If z is purely imaginary, so z = ki for some real k, then z* = –ki = –z as required B1 2.1 <=. This could, with care, be included in the => proof. If z = r + si (r, s real) then z* = r – si so z = – z* => r + si = –(r – si) = –r + si => r = 0 (so s ≠ 0 since z is non-zero) so z is purely imaginary B1 2.1 =>. z being non-zero does not have to be rigorously dealt with. Could instead consider if z is not purely imaginary and show this means 𝑧𝑧∗≠−𝑧𝑧 [2] (c) (i) Reflection in the real axis B1 1.2 Must be real, rather than x, axis. If mention of real axis, can ignore x [1] (ii) z = z* means that A and B are coincident so A is an invariant point so A must lie on the mirror line, which is the real axis, so A must represent a purely real number so z is purely real. B1 2.4 =>. Could, with care, be included in the <= proof. Needs to be a geometric explanation. If z is purely real then A lies on the real axis so it is invariant under a reflection in the real axis so the conjugate z* is also represented by the same point so z = z* B1 2.4 <=. [2] Y531/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 9 (a) 0 0 0 1 0 0 1 0 0 0 0 0 0 1 1 0 0 0 a b a b b a b a a a b b ab ab ab ab b b a a − −                × + −× − −     = + × +     + −× + ×   M1 1.1 Some indication of knowledge of how to multiply 3 by 3 matrices Can be implied by 3 non-zero correct terms 2 2 2 2 0 2 0 1 0 2 0 a b ab ab a b   − −   =     −   A1 1.1 Final solutions must have all terms simplified. [2] (b) ( ) ( ) ( ) 2 2 2 1 3 1 3 1 3 1 4 4 16 4 ab = + − = − = or 2 2 2 2 2 2 3 ( )( ) 4 4 1 3 2 a b a b a b − = − + = × = B1 3.1a Explicitly finding an expression for either ab (or 2ab) or a2 – b2 𝑅𝑅2 = ൮ 1 2 √3 0 −2 × 1 4 0 1 0 2 × 1 4 0 1 2 √3 ൲ = ⎝ ⎜ ⎛ 1 2 √3 0 − 1 2 0 1 2 × 2 0 1 2 0 1 2 √3⎠ ⎟ ⎞= 1 2 ቌ √3 0 −1 0 2 0 1 0 √3 ቍ so k = ½ B1 2.2a AG. Explicitly finding the expression for the other substituting into expression for R2 (or carrying out the matrix multiplication again). k can be embedded. If k embedded need to see 1 2 × 2 or 2 2. [2] Y531/01 Mark Scheme June 2023 17 (c) 4 1 1 1 1 3 0 3 0 2 2 2 2 0 1 0 0 1 0 1 1 1 1 0 3 0 3 2 2 2 2 1 1 0 3 2 2 0 1 0 1 1 3 0 2 2    − −       =               −     =         R For reference – does not need to be seen in working. SC – if answers left in terms of k then award B1 if one or two correct and B2 if all three correct 6 1 1 1 1 3 0 0 3 2 2 2 2 0 1 0 0 1 0 1 1 1 1 0 3 3 0 2 2 2 2 0 0 1 0 1 0 1 0 0    − −       =             −     =       R B1 1.1 For correct R6 By calculator expected, so intermediate steps might not be shown. 12 0 0 1 0 0 1 0 1 0 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 1 − −       =          −     =     −   R B1 1.1 For correct R12 Y531/01 Mark Scheme June 2023 18 24 1 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 0 1 − −       =       − −        =       R B1 1.1 For correct R24 [3] (d) Rotation B1 3.1a (360°/24 =) 15° B1 2.2a Also allow 345° (Rotation in opposite sense) Also allow in radians 𝜋𝜋 12 Clockwise about the y-axis. B1 3.2a Both sense and axis must be correct. If correct transformation is combined with an incorrect one (such as correct rotation combined with a reflection) then maximum mark is B2. Could be a rotation of 345° anticlockwise about y-axis [3] 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. 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Y532/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 Y532/01 Mark Scheme June 2023 4 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.

Y532/01 Mark Scheme June 2023 5 5. 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 Y532/01 Mark Scheme June 2023 6 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. 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. Y532/01 Mark Scheme June 2023 7 • 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. Y532/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 (a) Flaws occur independently of one another B1 3.3 At least one contextualised reason, ignore “singly”. and at constant average rate (or “uniform rate” which is equivalent) B1 [2] 3.3 A second assumption, not “singly”. Allow “fixed average rate”. Not: “constant rate” or “average constant rate” or “constant average number” or “constant probability” or “probability same for each 1 km”. More than 2 different assumptions (ignore “singly”): max B1. (b) P( 11) – P( 7) M1 1.1 Allow M1 for 0.109(24)... Not, eg, P( 11) – (1 – P( 7)) = 0.77 = 0.202 (0.20171) A1 [2] 1.1 Awrt 0.202 (c) Po(5.7  5) (= Po(28.5)) M1 3.3 Stated or implied their P( 29)(1 – their P( 29))2 (= 0.586  0.414  2) M1 1.1 Or 1 – [their P( 29)]2 – [1 – their P( 29)]2 (= 1 – 0.5862 – 0.4142) Allow M1 for 2 omitted or e.g. P( 29)(1 – P( 30))2 (both: M0) = 0.485(19) A1 [3] 1.1 Awrt 0.485 Question Answer Marks AO Guidance 2 (a) Class Baroque CDs as single unit M1 3.1b e.g. 23!7! seen, with or without other terms, or 24! (with 7! omitted) 24!  7! (= 6.21023  5040) A1 1.1 These, and no other terms, in numerator (allow even if no denominator)  30! = 1.1810–5 = 0.000 011 8 A1 [3] 1.1 Awrt 1.210–5, or 1 84825 (b) 6: 7C6  23C4 (= 78855 or 61 985) M1* 2.1 Clear attempt at one (allow for 7C6  23C4  other things), allow 10C6  … 7: 1  23C3 (= 1771) A1 1.1 Both expressions correct Add, and divide by 30C10 (= 30 045 015) depM1 3.1b [ 7 1 3393 16 965 = + ] Needs two terms, allow dividing by 30P10 if consistent = 4 1885 or 0.002 12 (0.002 122 ...) A1 [4] 1.1 Any equivalent exact fraction, or 0.00212 or better SC: 7 30 B(10, ) , 0.014(0): B1 max SC: (7P6  23P4) + (7P7  23P3), M1;  20!/30!, M1 (same as  30P10) OR 7C6 [10C4 + 10C313 + 10C213C2 + 1013C3 + 13C4] + [10C3 + 10C213 + 1013C2 + 13C3] = 7(210+1560+3510+2860+715) + (120+585+780+286) Y532/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 3 (a) 64282/32 – (1340/32)2 = 255(.297) B1 [1] 1.1 Awrt 255. Allow 263.52 from n/(n – 1). Don’t give ISW for 255 (b) y = 8.02 + 0.265(2)x [ 131039 4333 16339 16339 x + ] B2 [2] 1.1 1.1 Coefficients exact or correct to 3 sf, allow 8.03, letters correct. One error: B1 (c) 8.02 + 0.265248 = £20 700 (3 sf) (20749) B1 [1] 1.1 Awrt 20700 (not 20.7) or in range [20740, 20 750]. Ignore absence of £ NB: can be obtained from calculator even if (b) is wrong; B1 for this (d) SD is 255  16 and 48 is less than 6 away from x so extremely likely that range includes 48 B1 B1 [2] 1.1 2.3 Relevant calculation, e.g. 1340/32  2255, or difference is 0.383 SD or variance mentioned and nuanced conclusion e.g. “very likely that Tom is wrong” or more extreme, but not “Tom is wrong” SC: Only variance mentioned: max (B0)B1 (e) (48 almost certainly within range but) correlation only moderate so not very reliable. M1 A1 [2] 2.4 2.4 Comment on size of PMCC, allow comparison with CV Nuanced conclusion, but not from “significant evidence of correlation” OE (a significance test asks “is there evidence that  > 0?”, but here the issue is “how close is  to 1?”, so a significance test is irrelevant) (d) (A) The standard deviation is  16, so Tom is likely to be right B0 (B) Variance is large so very likely that Tom is wrong (SC – but not “variance is very large so results inaccurate”) B0B1 (C) Less than 2 SD above mean, so Tom is incorrect (B1, but not nuanced so B0) B1B0 (D) Variance is large so results vary a lot, so likely to be data above 48, so unlikely that Tom’s claim is correct B0B1 (E) Less than one standard deviation away from mean [consistent with (a)], so Tom is very unlikely to be right (minimum for B1B1) B1B1 (e) (F) PMCC shows quite strong correlation and probably within range, so reliable M1A0 (G) PMCC shows quite strong correlation so fairly reliable M1A1 (H) Not very reliable as PMCC is low and might be extrapolating M1A1 (I) Not very reliable as PMCC is low M1A1 Y532/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 4 a + b = 0.4 B1 1.1 Allow a + b + 0.6 = 1, or use both 2a + b= 0.55 and 58a + 59b = 23.45 –2a – b + 0.2 + 0.2 + 0.3 = 0.15 M1 1.1 Use (W – 60)P(w) = 0.15 2a + b = 0.55 A1 1.1 Correct simplified equation a = 0.15, b = 0.25 A1 1.1 4a + b + 10.2 + 40.1 + 90.1 (= 2.35) M1 2.1 Find (W – 60)2P(w) – 0.152 and 16 M1 1.1 (independent of previous M1) = 37.24 A1 [7] 1.1 Or 931 25 Or a + b = 0.4 B1 Allow a + b + 0.6 = 1 58a + 59b + 600.2 +…+ 630.1 = 60.15 M1 Use WP(w) = 0.15 + 60 58a + 59b = 23.45 A1 Correct simplified equation a = 0.15, b = 0.25 A1 16(582a + 592b + … + 6320.1) (= 57925.6) M1 16  w2P(w) – (460.15)2 M1 – (4  their w)2 (independent of previous M1) = 37.24 A1 [7] Or 931 25 Or (last 3 marks) 582a + 592b … + 6320.1 – 60.152 M1 = 3620.35 – 3618.0225 = 2.3275 16 M1 Allow for 16 w2P(w) without having subtracted 60.152 = 37.24 A1 SC If B0M0, give SC B1 for Var(4W – 60) = 16Var(W) used B1 Y532/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 5 (a) H0:  = 0, H1:  > 0 where  is the population product-moment correlation coefficient between the test scores. B2 1.1 2.5 One error, e.g. two-tailed, or  not defined: B1. Allow H0:   0 Symbols: Definition of  needs “population” or context or both, and “correlation”. Allow r in place of . Do not allow “association” here Verbal: H0: no correlation between openness and creativity, H1: positive correlation: max B1 unless “population” explicit. Needs “positive”. Allow “association” here.  = 717  (1075.6  2016) = 0.487 (0.48691) M1 A1 1.1 1.1 Art 0.487 seen gets M1A1. Else allow M1 for correct subs into formula, or any two of 1075.6, 2016 and 717, or any two of 71.7, 134.3 and 47.8 0.487 > 0.4409 so reject H0. M1ft 1.1 Compare their r with 0.4409 or 0.441 and reject (ft on TS) There is significant evidence of positive association between openness and creativity. A1ft [6] 2.2b Correct contextualised conclusion, not too assertive, allow omission of “positive”, FT on their r, no FT for hypotheses wrong way round (b) Points lie fairly (but not very) close to straight line B1 2.4 Must refer to diagram of points, not just to correlation. Not “points lie close to a line” – some level of nuance needed. Allow general statement, e.g. “it shows how close to a straight line the points are” … with positive gradient B1 [2] 2.4 Any wrong statement: max B1B0 (c) Disagree as  is unchanged by linear scaling B1 [1] 1.2 “Disagree” oe and correct reason, allow omission of “linear”. Allow “It wouldn’t affect the value” (b) (A) Points will lie roughly in an ellipse (that is a necessary condition for validity of a test, not a consequence of the value of ) B0 (B) Points are vaguely scattered with weak indication of positive correlation (no mention of line) B0 (C) Positively correlated but points are not very close to a straight line B1B0 (D) It shows its gradient and how close to a straight line the points are B1B0 (E) The closer  is to +1/–1, the closer the points are to a straight line, with positive/negative gradient B2 Y532/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 6 (a) Either 6p = 3.35 ⇒ p = 0.558(33) ⇒ variance should be 1.48 (1.47958) M1 A1 3.4 1.1 Use np and npq Attempt to use Poisson: M0 Correct relevant calculation, e.g. q = 1.025..., or p = –0.0125 or solve 6p2 – 6p + 3.392 to get both p  1.4 or –0.4, but not from p = 0.5 Not close to 3.392 so B(6, p) not a good model A1 [3] 3.2b Validly deduce that B(6, p) not valid, e.g. 0 < p < 1, and state conclusion SC: 0.5 used: M1A0A1 Or npq > np; so q > 1 which is impossible Hence B(6, p) not a good model M1A1 A1 (qualitative argument) (b) Expected frequencies 10, 12, 14, 16, 14, 12, 10 B1 3.3 Use (O – E)2/E M1 1.1 Allow from at least one of 0.083(…) and 1.6 correct 0.083(3…), 1.6 and total 3.3362 or 3.3363 A1 [3] 1.1 Allow 3.34, 3.336 or better. If total omitted, or “0”, in (b), can be recovered from (c) (“0” probably comes from misunderstanding “Total”) (c) H0: data consistent with proposed model, H1: not so B1 1.1 Allow “data follows …” but not “data is in ratio ...” nor “evidence that ...” 3.336(2) < 10.64 B1ft 1.1 Compare their 3.336 with correct CV (3.336 may be from calculator) Do not reject H0 M1ft 1.1 Correct first conclusion, FT on their TS and on CV 9.236 or 12.59 Insufficient evidence that proposed model does not fit data A1ft [4] 2.2b Contextualised, not over-assertive. Needs ‘double negative’, not “significant evidence that data is consistent”, etc. A0 if hypotheses wrong way round (d) Inferences from a hypothesis test are not “definite” B1 2.2b “Definite” stated to be too strong, oe (not just “Rosa is wrong”) All we have is evidence / Sample size is small / other experiments might produce different results B1 [2] 3.5a Relevant valid comment, e.g. “data might be misleading”, “second model likely to be correct”, “either could be correct”, and no wrong extras “Neither/both good” etc, from wrong conclusion to (a) or (c): max B1B0 Hypotheses (A) H0: data (results) are in ratio 5:6:7:8:7:6:5, H1: they are not (the data are definitely not in that ratio!) B0 (B) H0: model follows ratio, H1: it doesn’t (the model is known; hypotheses concern the population) B0 (C) H0: follows given ratio, H1: doesn’t follow given ratio (BOD); also (D) H0: data fits ratio, H1: it doesn’t B1 (d) (E) Wrong: first model could be either and second model shows definitely independent B0 (F) Cannot be definite as first model has mean  variance so binomial is a good model B1B0 (G) Not enough to be certain as it could change with a different significance level B1B0 (H) Not definite as the second shows only that the proposed model is not a good fit for the data (from wrong conclusion in (c)) B1B0 (I) Not definite as correlation does not imply causation (the second clause is a ‘wrong extra’ and so B0) B1B0 (J) I agree with Rosa as it is likely that the second model is correct. B0B1 (K) Cannot say “definitely” as there is a 10% chance that the second test is wrong (condone this inaccurate second clause) B2 (L) Neither is definitely correct but the second model is quite likely to be correct B2 Y532/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 7 (a) Geo(0.1) M1 3.3 Geo(0.1) stated or implied, e.g. by 0.10.913 P(R > 13) = (1 – p)13 M1 1.1 Or sum. Allow 0.914 (= 0.229) or 0.912 (= 0.282) or 1 term omitted or extra in sum, but not 1 – (0.9)anything = 0.254 A1 [3] 1.1 Awrt 0.254, needs one of first two lines (“Determine”). Assume that, e.g. 1 – 0.746 is from calculator. SC no working: 0.254 B2 (b) pq2 – 0.4pq – ap = 0 M1 3.1b Use formula for Geo(p) to get equation involving 3 terms p2 – 1.6p + 0.6 – a = 0 or q2 – 0.4q – a = 0 A1 1.1 Correct quadratic stated or implied, e.g. by correct use of ‘b2 – 4ac’ 1.6 0.16 4 2 a p  + = or 2 0.4 0.4 4 2 a q  + = M1 1.1 Obtain explicit formula for p or q, e.g. from (q – 0.2)2 = a + 0.04, needn’t be simplified p > 0 ⇒ 0.8 0.04 0 a − +  or q < 1 ⇒ 0.2 0.04 1 a + +  M1 2.2a Use p > 0 from negative sign or q < 1 from positive sign (ignore other combinations of inequality and sign). Allow p  0 or q  1. 0.04 0.8 a +   a < 0.6 A1 1.1 Obtain a < 0.6, allow a  0.6 here p =0.8 0.04 a − + decreases with a (so any small positive a gives a valid value of p) (or similar for q) B1 2.2a Reason why the lower limit is (not greater than) 0, e.g. sketch of p against a, or other valid justification other than just “a > 0 is given” 0 < a < 0.6 (strict inequalities only) A1 [7] 3.2a Fully correct, allow just a < 0.6, needs all previous marks apart from B1 OR a = p2 – 1.6p + 0.6 (or a = q2 – 0.4q) M1 A1 M1 A2 Write a in terms of p or q and draw graph Correct parabolic shape, and intersections at (0.6, 0) and (0, 0.6) clear Identify range of a for which 0 < p < 1 0 < a < 0.6 ( used, or –0.04 < a < 0.6: A1) Max 7/7 Intersections (0, 0.6), (0.6, 0) (and (1, 0)) 0 < p < 1  –0.04 < a < 0.6 But a > 0 so a satisfies 0 < a < 0.6 Y532/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance OR T&I: Test values of p or q, or a, from 0 to 1 by increments of at most 0.25; at least 3 correct values Reject p > 0.6 or q < 0.4, as giving a < 0 p decreases as a increases (hence lower limit is 0) Conclude 0 < a < 0.6 M1 A1 A1 B1 A1 (p, a) = (0, 0.6), (0.1, 0.45), (0.2, 0.32), (0.25, 0.2625), (0.3, 0.21), (0.4, 0.12), (0.5, 0.05), (0.6, 0) Clear rejection of wrong values OE, as above Fully correct, allow just a < 0.6, needs all previous marks apart from B1 OR Let f(a, p) = p2 – 1.6p + 0.6 – a f(a, 0)  f(a, 1) < 0  f(a, p) = 0 for some 0 < p < 1  (0.6 – a)(–a) < 0  a2 – 0.6a < 0  0 < a < 0.6 M1 A1 M1 A1 This method is not fully valid as it does not consider the possibility of there being a solution to f(a, p) = 0 even when f(a, 0)  f(a, 1) > 0 (which occurs for –0.04 < a < 0 and it needs to be established that there are no positive values of a for which this happens) Max 6/7 SC p = 0 (or q = 1)  a = 0.6; p = 1 (or q = 0)  a = 0 so 0 < a < 0.6 M1 A1 Both 0 and 1 substituted and a found Same marks if insufficient working. Max 4/7 SC Use ‘b2 – 4ac’ only: no more marks Max 2/7 (the issue is not “are there any solutions for p?” but “are there any solutions in the range 0 < p < 1?”) 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. 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Y534/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 Y534/01 Mark Scheme June 2023 4 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.

Y534/01 Mark Scheme June 2023 5 5. 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. Y534/01 Mark Scheme June 2023 6 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. Y534/01 Mark Scheme June 2023 7 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. 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. Y534/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 1 (a) Find a bus or train (or both) that can be used to travel from home to town B1 1.1 Any appropriate problem, request or question, in context, that would require a construction for its solution [1] 1 (b) 7 = 6 + 2 – n(BT) 8 – 7 M1 2.5 Use inclusion-exclusion (soi) Maybe shown as a Venn diagram o.e. 5 1 1 Bus Train 1 journey by both bus and train A1 2.2a 1 seen as answer (not picked out from Venn diagram, unless clearly indicated) [2] Y534/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 2 (a) A 1 0 B 3 4 E 2 3 4 3 M1 A1 3.1a 1.1 Using Dijkstra, a valid updating seen Permanent values all correct C 4 9 D 5 11 11 9 15 12 11 Or from D: A 5 11 B 3 7 C 2 2 D 1 0 E 4 9 15 11 7 2 9 Shortest path: A – B – C – D B1 1.1 A – B – C – D (not in reverse) [3] 2 (b) BE = 2 ✓ CD = 2 ✓ AE = 3 ✓ AB = 4 BC = 5 ✓ CE = 8 DE = 9 AD = 15 M1 A1 1.1 1.1 Using Kruskal, list of (at least the first 5) arcs (and weights) in increasing order of weight, o.e. Allow BE, CD, AE, BC listed in order, with no others BE and CD may be interchanged Not choosing arc AB in list AB indicated in a different way from BE etc Allow AB completely missing from list Minimum spanning tree: BE, CD, AE, BC B1 1.1 Choosing arcs BE, CD, AE, BC (only), allow A-E-B-C-D From a correct tree seen or arcs listed (separately from Kruskal working) (need not show arc weights) For reference: [3] Y534/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 3 (a) (i) First pass 10 18 7 23 54 31 62 (82) M1 1.1 Bubble increasing Result at end of first pass starts 10 18 7 … Second pass 10 7 18 23 31 54 (62 82) Third pass 7 10 18 23 31 (54 62 82) A1 1.1 Results correct at end of first, second and third passes (cao) Fourth pass 7 10 18 23 31 54 62 82 A1 1.1 A final fourth pass in which nothing changes Allow at most one misread for max M1 A0 A1 Decreasing: first pass starts 23 18 10 …  M1 A0 A0 [3] 3 (a) (ii) First = 5 M1 1.1 5 swaps in first pass (allow FT provided M1 gained in (a)(i)) Second = 2, third = 1 Fourth pass = 0 A1 1.1 2 swaps in second pass and 1 swap in third pass (cao) Ignore swaps for fourth pass (and any others that are shown) Swaps must be written in figures (not tallies) [2] 3 (b) (i) First pass 10 23 18 7 62 54 31 82 M1 1.1 Shuttle increasing Result at end of first pass starts 10 23 Second pass 10 18 23 (7 62 54 31 82) Third pass 7 10 18 23 (62 54 31 82) Fourth pass 7 10 18 23 62 (54 31 82) A1 1.1 Results correct at end of second, third and fourth passes (cao) Fifth pass 7 10 18 23 54 62 (31 82) Sixth pass 7 10 18 23 31 54 62 (82) Seventh pass 7 10 18 23 31 54 62 82 A1 1.1 Results correct at end of fifth, sixth and seventh passes (cao) Seven passes used in total Allow at most one misread for max M1 A1 A0 Decreasing: first pass starts 23 10 with 0 swaps  M1 A0 A0 [3] Y534/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (b) (ii) First = 1, second = 1, third = 3, fourth = 0 M1 1.1 Number of swaps correct for first four passes (allow FT provided M1 gained in (b)(i) and consistently increasing, or decreasing) Fifth = 1, sixth = 2, seventh = 0 A1 1.1 All correct, with exactly 7 passes used Swaps must be written in figures (not tallies [2] 3 (c) E.g. Both use 8 swaps but shuttle sort uses fewer comparisons so is more efficient B1 3.1a Fewer comparisons or smaller total (from valid reasoning and increasing) Number of comparisons need not be evaluated (or be correct) (bubble = 7+6+5+4 = 22, shuttle = 1+2+3+1+2+3+1=13) but shuttle sort must have fewer comparisons than bubble sort (their) 13 < (their) 22 or (their) 21 < (their) 30 Not using ratios [1] 3 (d) E.g. Takes (approximately) 9 times as long B1 2.2a 32 or 9 seen in context, (3000 1000) 2 = 9 (so run time is) 9 times as long Allow ‘increase by 9’ (BOD scale factor), allow 9 ‘exponential’  B0 unless 9 (o.e.) also seen [1] Y534/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) (i) In complete graph K4 each vertex has degree 3 B1 2.4 Complete (or K4) degree 3, allow odd Diagrams are not sufficient explanation In (a simply connected) Eulerian graph (with 4 vertices) each vertex has even degree so there is a contradiction B1 2.4 Eulerian (or G) no odd (or all even), allow degree 2 o.e Allow ‘some even’ or (at most) two odd (i.e. semi-Eulerian) Diagrams are not sufficient explanation Allow K4 has odd vertex degrees so not Eulerian for B1 B1 [2] 4 (a) (ii) Number of arcs in graph G is 4 B1 2.2a 4 G is Eulerian so vertex degrees are even G is connected so vertex degrees are not 0 G is simple so no vertex degree > 3 Hence each vertex of G has degree 2 B1 2.4 Explaining the significance of G being (Eulerian,) simple (and connected) e.g. stating that each degree < 4, not just ‘no loops’ and hence each vertex has degree 2 Diagrams are not sufficient explanation [2] 4 (a) (iii) Each vertex of G is directly connected to (only) 2 of the others Use vertices that are not directly connected to form the two sets. B1 1.1 Showing that G is bipartite (no FT) This could be shown in a diagram provided the two sets are identified e.g. by drawing a ring round the vertices or from labels (set 1 and set 2 o.e.) If no indication of the sets, a diagram or with K2,2 stated [1] Y534/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 4 (b) (i) The leading diagonal has all zero entries B1 2.2a No arcs for each vertex to itself Allow ‘A – A = 0 etc.’ o.e. (with B, C, D implied from etc.) Diagrams are not sufficient explanation [1] 4 (b) (ii) The sum of the entries in the matrix is 10 B1 2.2a 1+ 1+ … = 10 or 2 + 2 + 4 + 2 = 10, or similar Allow 10 (without calculation) [1] 4 (b) (iii) A D B C B1 1.1 This graph o.e. with 10 directed arcs For reference: To A B C D A 0 1 0 1 From B 0 0 2 0 C 2 1 0 1 D 0 1 1 0 [1] 4 (b) (iv) e.g. C – A – B – C – A – D – C – D – B – C – B M1 1.1 Trail starts at C and ends at B (not reversed) and includes every letter at least once A1 1.1 A valid trail (using each directed arc exactly once, no FT) that passes through A, B and D (exactly) twice and passes through C three times 2 A’s, 3 B’s, 4 C’s, 2 D’s [2] Y534/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 5 (a) 2.5 2.5 B(2) C(2.5) D(2) E(2.5) 2.5 2.5 A(0.5) F(1) G(1.5) J(0.5) 0 0.5 0.5 1.5 3 3 4.5 5 H(2) I(1.5) 2.5 3.5 B1 3.4 Forward pass seen (first value in bold at each vertex) Allow 0 and/or 5 missing but otherwise correct 5 (hours) B1 1.1 cao 5 stated (not implied from diagram) [2] 5 (b) See answer given in part (a) B1 3.4 Backward pass seen (second value in bold at each vertex) Allow 0 and/or 5 missing but otherwise correct A, B, C, D, E (in any order) B1 1.1 cao [2] Y534/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 5 (c) M1 ft 1.1 Any two correct, or from their forward and backward passes Working need not be seen F = 3 – 0.5 – 1 = 1.5 (hours) G = 4.5 – 1.5 – 1.5 = 1.5 (hours) J = 5 – 3 – 0.5 = 1.5 (hours) A1 1.1 F, G, J = 1.5 (cao) Working need not be seen H = 3.5 – 0.5 – 2 = 1 (hour) I = 5 – 2.5 – 1.5 = 1 (hour) A1 1.1 H, I = 1 (cao) Working need not be seen [3] 5 (d) (i) B C D E A F G L J I H B1 B1 3.3 3.5c C, E, G, I correct Activity L is after C, E, G and I (directions may be implied) Activity L is before J (directions may be implied) [2] 5 (d) (ii) Time to start of L = 5 hours So L + J take 2.5 hours B1 3.4 Appropriate working seen, with evidence of what the values represent e.g A + B + C + L + J = 7.5 or A + D + E + L + J = 7.5 o.e. use of critical path No FT Duration of L = 2 hours B1 2.2a 2 (cao) [2] Y534/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 6 (a) Delete W B1 2.2a From column headings in a reduced table Ignore if any rows also deleted In each row, value in col W > value in col Y So col W is dominated by col Y or col Y dominates col W Delete Z B1 2.2a From column headings in a reduced table Ignore if any rows also deleted In each row, value in col Z > value in X So col Z is dominated by col X or col X dominates col Z X Y A 0 2 B 2 -3 C 4 -4 D -1 5 Alternative method A B C D X 0 -2 -4 1 Y -2 3 4 -5 Or transposed [2] For reference: W X Y Z A 4 0 2 1 B 0 2 -3 4 C 1 4 -4 5 D 6 -1 5 0 Y534/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 6 (b) X has probability p and Y has prob 1 – p A: 0p + 2(1 – p) = 2 – 2p B: 2p – 3(1 – p) = 5p – 3 C: 4p – 4(1 – p) = 8p – 4 D: –p + 5(1 – p) = 5 – 6p M1 ft 3.1a Definition of p may be implied for M marks Using their reduced table (or correct) and their (single) variable At least two correct expressions (need not be simplified) seen or implied from graph Using values from original table means that these are losses for Casey, so will need the minimax (lowest point of upper boundary) Alternative method (change signs) X has probability p and Y has prob 1 – p A: 0p – 2(1 – p) = –2 + 2p B: –2p + 3(1 – p) = –5p + 3 C: –4p + 4(1 – p) = –8p + 4 D: p – 5(1 – p) = –5 + 6p M1 ft Definition of p may be implied for M marks Using their reduced table (or correct) and their (single) variable At least two correct expressions (need not be simplified) seen or implied from graph Using negatives of values from original table means that these are gains for Casey, so will need the maximin (highest point of lower boundary) 8p – 4 = 5 – 6p M1 ft 3.1a Solving for (their) p (graphically or algebraically or implied) p = 9 14 A1 1.1 p = 0.643 (or better) (0.6428571…) (cao) Optimal mixed strategy: choose randomly between X and Y, choosing X with prob 9 14 and Y with prob 5 14 A1 3.2a Correct probabilities described in context involving both X and Y Or correct ratios o.e. e.g randomly choose X 1.8 times as often as Y [4] Y534/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 6 (b) cont If W and Z are chosen X has probability p and Y has prob 1 – p A: 4p + (1 – p) = 1 + 3p B: 0 + 4(1 – p) = 4 –4p C: p + 5(1 – p) = 5 –4p D: 6p + 0(1 – p) = 6p 5 – 4p = 6p (= 1 2) M1 ft M1 ft A0 A0 Or the negatives of these for the alternative method Definition of p may be implied for M marks Using their reduced table (or correct) and their (single) variable At least two correct expressions (need not be simplified) seen or implied from graph Or 4p – 5 = -6p for the alternative method Y534/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 7 (a) B1 B1 B1 1.1 1.1 1.1 Boundary line y = 3x –30, at least from (10, 0) to (12, 6) Boundary line y = 15 – x, at least from (6, 9) to (12, 3) Boundary line y = 1 3x – 2, at least from (6, 0) to (12, 2) Ignore any extra lines x y P = 4x + y 6 0 24 10.5 1.5 43.5 11.25 3.75 48.75 0 15 15 M1 1.1 Checking any vertex (other than the origin) or using a sliding profit line with gradient –4 (approx.) seen, or implied from answer P = 48.75 A1 1.1 cao x = 11.25, y = 3.75 A1 1.1 cao [6] Y534/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 7 (b) (i) Additional constraint y  kx 3.75  11.25 = 1 3 M1 3.1a 1 3 or (their y)  (their x) Solution is unchanged for k  1 3 A1ft 1.1  (their) 1 3 Accept with a lower bound of 0, 0 < k  (their) 1 3, provided their upper bound is positive, but not with a lower bound other than 0 [2] 7 (b) (ii) For k > 1 3, optimal solution is at intersection of y = kx and x + y = 15  x + kx = 15 M1 3.1a Evidence of finding point where y = kx meets x + y = 15 May be implied from any of x, y, P correct P = 15( 4+𝑘 1+𝑘) A1 1.1 cao (or equivalent) x = 15( 1 1+𝑘) A1 1.1 cao (or equivalent) without wrong working y = 15( 𝑘 1+𝑘) A1 1.1 cao (or equivalent) without wrong working [4] 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. OCR is an exempt charity. OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals. OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these resources. Though we make every effort to check our resources, there may be contradictions between published support and the specification, so it is important that you always use information in the latest specification. We indicate any specification changes within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy between the specification and a resource, please contact us. Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more information using our Expression of Interest form. Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.

Y535/01 Mark Scheme June 2023 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. Y535/01 Mark Scheme June 2023 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. 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. Y535/01 Mark Scheme June 2023 5 • 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. Y535/01 Mark Scheme June 2023 6 Question Answer Marks AO Guidance 1 (a) 205 = 7  29 + 2 i.e. q = 29, r = 2 B1 1.1 [1] 1 (b) From (a), since 7 does not divide exactly into 205 (and 7 is prime), 7 must divide 8666 M1 2.4 Use of Euclid’s Lemma No need to note that this is because r  0 If 7 | (205×8066), then, by Euclid’s Lemma … A1 2.2a Allow description of the “Euclid’s Lemma” condition instead A complete justification with assumption and conclusion (hcf(7, 205) = 1) [2] Question Answer Marks AO Guidance 2 (a) 𝑢𝑛 + 5 = = 3(n + 5)2 + 3(n + 5) + 7 = 3n2 + 33n + 97= M1 3.1a Expanding and considering at least one term mod 10 or evaluating 𝑢𝑛 from n=1 to n=10 = 3n2 + 3n + 7 (mod 10)  𝑢𝑛 (mod 10) A1 1.1 Correct conclusion from correct algebraic work [2] 2 (b) The sequence is periodic (with period 5) B1 1.1 Allow repeating/cyclic Any shorter periodicity would have to be a factor of 5. Since the sequence is not constant, there are no smaller possibilities. B1 2.4 Allow statement only that sequence is non-constant or sight of {3, 5, (3, 7, 7, …)}. [2] Y535/01 Mark Scheme June 2023 7 Question Answer Marks AO Guidance 3 2 3 2 2 + − =   y xy x z B1 1.1 1 3 2 2 + − =   x y x y z B1 1.1 Setting both first partial derivatives equal to zero and attempt to eliminate one variable M1 1.1a Either directly via 2 2 2 3 y y x − = or 2 2 1 3 x x y − = OR indirectly via x x y y xy 1 3 2 3 2 − = − =  y = 2x Either 0 2 3 3 = + −y y or 0 1 3 4 3 = + −x x i.e. (y – 1)2(y + 2) = 0 or (2x – 1)2(x + 1) = 0 M1 1.1 Any cubic equation in one variable (x, y, z) = ( 2 1 , 1, 4 3 ) A1 1.1 First SP correct BC www = (−1, −2, −6) A1 1.1 Second SP correct BC www SC1 for both pairs of (x, y) correct with z’s missing or z incorrect NB Extra SP A1A0 [6] Y535/01 Mark Scheme June 2023 8 Question Answer Marks AO Guidance 4 (a) b and d are the direction vectors of l … B1 1.2 … so b = md for some scalar m or b  d=0 B1 2.2a or statement that one of b, d is a multiple of the other or that they are parallel a and c are (the position vectors of) points on l … B1 1.2 a – c = µb or a – c = γd or (a – c)  b = 0 or (a – c)  d = 0 B1 2.4 or (a – c) is a multiple of (or parallel to) b (or d) SC1 (a +  b – c ) d = 0 [4] 4 (b) (a – c)  d = 0 since d, a – c parallel (from (a)) M1 1.1  a  d = c  d A1 1.1  a . (c  d) = a . (a  d) = 0 since a  d is perpendicular to a A1 2.4 Correct answer, fully justified ALT. Substitution for r and use of the Distributive property of the VP: (a  d) + ( b  d) – (c  d) = 0 M1 But b  d = 0 since b, d parallel  a  d = c  d A1  a . (c  d) = a . (a  d) = 0 since a  d is perpendicular to a A1 Correct answer, fully justified [3] Y535/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 5 (a) 23 7119 = 9 + (1  23) + (1  232) + (7  233) M1 1.1 Clear use of base-23 column-values = 85 73010 A1 1.1 [2] 5 (b) 7n + 11  9 (mod 23)  7n + 11  32 (mod 23) OR 7n  −2 (mod 23)  7n  21 (mod 23) M1 1.1 Use of modular arithmetic to gain a multiple of 7 on the RHS  n  3 (mod 23) or n = 23k + 3 (k  ℤ) A1 1.1 Any correct, complete statement Accept n  −20 (mod 23) hcf(7, 23) = 1 for division to be valid B1 2.4 Statement that hcf(7, 23) = 1 for division to be valid or equivalent [3] 5 (c) i 3N – 7M = 30a + 3b – 7a – 49b = 23(a – 2b) B1 1.1 Must be written explicitly as a multiple of 23, here or later on If N = 23k, then 7M = 23(3k – a + 2b) M1 2.1 Proof attempted in one direction (“proof” includes attempt to obtain a multiple of 23) If M = 23k, then 3N = 23(7k + a – 2b) M1 2.1 Proof attempted in other direction (“proof” includes attempt to obtain a multiple of 23) Both correctly shown multiples of 23 with explanation that hcf(7, 23) = 1 and hcf(3, 23) = 1 A1 2.4 At least one justification must be noted [4] 5 (c) ii 711 965 → 71 231 → 7130 → 713 → 92 (→ 23) M1 1.1 Implementation of procedure, at least first step correct Process may be stopped at either 92 or 23, giving divisibility by 23 A1 1.1 Conclusion must be noted Accept process stopped at 713 if 23 is shown to be a factor Y535/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance [2] Question Answer Marks AO Guidance 6 (a) P6 = 2×𝑃5 6 + 5 𝑃5 used with P5 = 266.67 M1 1.1 Must consider a range of possible values of P6 (ignoring </ at endpoints) so that P6 = 88.91 to 2d.p. A1 1.1 Shown carefully (showing that 𝑃6 is monotonic in the interval is not required) [1] SC1 88.91 (by substituting 266.67 soi) 6 (b) i Number of bacteria initially increases, then declines B1 2.2a Must be in context (refers to the number of bacteria) [1] 6 (b) ii On the 8th day B1 3.4 (∵ the number of bacteria falls below 10 000) [1] 6 (c) i f(n) = n B1 3.3 Other possibilities for f may also be fine. At least three values need to approximately fit [1] 6 (c) ii Suggest n = 100 (accept n = 99) B1 3.1b FT their f(n) even from B0 in part (c) provided >80 [1] 6 (c) iii Using x x 99 100 2 08 . 10 + = to create a quadratic eqn. in x = P99 M1 3.2a i.e. 2x2 – 1008x + 9900 = 0 which gives P99 = 10.02 … A1 3.5b It is not necessary to verify that P98 < 10 (working back another step gives P98 = 9.98) [2] Y535/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 6 (d)       + + = + n n n P n n P P 1 2 INT 1 (+1) B1 3.5c [1] Question Answer Marks AO Guidance 7 (a) x = 11 B1 3.1a Since, e.g., x = 2  16 = 32  11 (mod 21) OR Calculating the powers of 2, mod 21: 2, 4, 8, 16, 11 … B1 2.4 Correct working. ALT G consists of all n, 0 < n < 21, with n co-prime to 21 [2] 7 (b) Element 2 4 5 8 10 13 16 17 19 20 x Order 6 3 6 2 6 2 3 6 6 2 6 B1 1.1 At least 3 correct B1 B1 1.1 1.1 At least 6 correct All correct [2] 7 (c) i Subgroup of order 3 is {1, 4, 16} B1 1.1 [1] 7 (c) ii Subgroups of order 6 are {1, 2, 4, 8, 16, 11} B1 1.1 Accept x for 11 {1, 5, 4, 20, 16, 17} B1 1.1 {1, 19, 4, 13, 16, 10} B1 1.1 Withhold final B1 if extras appear [3] 7 (d) The order of the elements of the subgroup must divide 4 B1 2.5 Or a group with identity + 3 self-inverse elements Y535/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance A cyclic group of order 4 does not exist since G has no element of order 4 B1 2.1 Or the only possible order of the elements (other than the identity) is two Only subgroup is {1, 8, 13, 20} M1 3.1a Checking for Closure: 8  13 = 20, 8  20 = 13, 13  20 = 8 A1 1.1 Visibly checked [4] 7 (e) G has no element of order 12, hence not cyclic B1 2.3 Correct answer with valid stated reason; no further justification required. (NB This is easily seen from part (c) (iii), where each element appears in a 6-subgroup so cannot possibly generate the whole group). [1] Question Answer Marks AO Guidance 8 A single, continuous curve drawn in the x-z plane B1 1.1 Accept a closed curve Curve lies entirely in −1  z  1 B1 1.1 A single point with z = −1 at (0, −1) B1 1.1 Two maxima at (1, 1) B1 1.1 Maxima do not need to be stationary points As x →  , z → −1 B1 1.1 An adequately, and essentially completely correct, solution curve drawn B1 1.1 SC1 one of the following in correct location without labels: Curve lies entirely in −1  z  1 A single point with z = −1 at (0, −1) Two maxima at (1, 1) Y535/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance [6] APPENDIX Exemplar answers for question 8 Exemplar 1 Y535/01 Mark Scheme June 2023 14 A single, continuous curve drawn in the x-z plane B1 Curve lies entirely in −1  z  1 B1 A single point with z = −1 at (0, −1) B1 Two maxima at (1, 1) B1 As x →  , z → −1 B1 An adequately, and essentially completely correct, solution curve drawn B1 Y535/01 Mark Scheme June 2023 15 Exemplar 2 A single, continuous curve drawn in the x-z plane B1 Curve lies entirely in −1  z  1 B1 A single point with z = −1 at (0, −1) B1 Two maxima at (1, 1) B1 As x →  , z does not approach −1 B0 A partially incorrect solution curve drawn B0 Y535/01 Mark Scheme June 2023 16 Exemplar 3 A single, continuous curve drawn in the x-z plane B1 Curve does not lie entirely in −1  z  1 B0 A single point with z = −1 at (0, −1) B1 No maxima at (1, 1) B0 As x →  , z does not approach −1 B0 An incorrect solution curve drawn B0 Y535/01 Mark Scheme June 2023 17 Exemplar 4 A single, continuous curve drawn in the x-z plane B1 SC1 one of the following in correct location without labels: Curve lies entirely in −1  z  1 A single point with z = −1 at (0, −1) Two maxima at (1, 1) B1 As x →  , z does not approach −1 B0 An incorrect, solution curve drawn B0 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. OCR is an exempt charity. OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals. OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these resources. Though we make every effort to check our resources, there may be contradictions between published support and the specification, so it is important that you always use information in the latest specification. We indicate any specification changes within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy between the specification and a resource, please contact us. Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more information using our Expression of Interest form. Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.

Y414/01 Mark Scheme June 2023 5 11. Annotations Annotation 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Y414/01 Mark Scheme June 2023 6 Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result 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. BP Blank Page Seen Highlighting

Y414/01 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 Y414/01 Mark Scheme June 2023 8 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. 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. Y414/01 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. Y414/01 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. Y414/01 Mark Scheme June 2023 11 Question Answer Mark AO Guidance 1 (a) (i) 1.30258 × 10‒9 or 0.000 000 001 302 58 B1 1.1 [1] 1 (a) (ii) Because the values in L22 and M22 are stored to a higher precision than they are displayed B1 1.1 Accept ‘greater accuracy’ but do not allow e.g. ‘the spreadsheet stores values to a different degree of accuracy than it displays’ And they are not equal B1 1.1 Accept “they are both not equal to 1” [2] 1 (b) (i) 1.051… soi B1 1.1 ‒0.000 271 (096 376) to 3 or more sf B1 1.1 Condone 0.000 271 [2] 1 (b) (ii) It will be different because (although the numbers and operations are the same) the order of operations is different B1 1.2 Must be a written explanation [1] 2 (𝑥−5)(𝑥−8) (3−5)(3−8) × 9.26 + (𝑥−3)(𝑥−8) (5−3)(5−8) × 19.3 + (𝑥−3)(𝑥−5) (8−3)(8−5) × 37.96 M1 A1 1.1 1.1 Allow sign errors and one substitution error All substitutions correct, may be implied by later work 𝑦= 0.24𝑥2 + 3.1𝑥−2.2 A1 1.1 2 terms correct A1 1.1 All correct in the required form in terms of x, with y = seen cao [4] 1.1 3 (a) 43.2 2.145+2.175 or 43.2 2.135+2.165 M1 1.1a For either expression correct, may be implied by 10 or 10.047 10 < P ≤ 10.047 A1 1.1 10.047 to at least 5sf (10.0465116279=432 43 ) Allow < or ≤ for either, or interval notation / words. ISW [2] 3 (b) 43.2 2.175−2.135 or 43.2 2.165−2.145 M1 1.1a For either expression correct, may be implied by 1080 or 2160 1080 < Q < 2160 A1 1.1 Allow < or ≤ for either, or interval notation / words. ISW [2] 3 (c) There is such a big difference because the denominator of Q involves the subtraction of nearly equal numbers B1 2.4 allow eg division by subtraction of nearly equal numbers [1] Y414/01 Mark Scheme June 2023 12 Question Answer Mark AO Guidance 4 (a) 0.002 × 0.576 M1 1.1 Use of g(1.802) ≈ g(1.8) + 0.002× g′(1.8), may see 0.001152 error ≈ 0.00115 A1 2.2b Mark the final answer, must be rounded correctly to 3sf SCB1 for 0.00115 without working Alternative Method: g(1.8) ‒ g(1.802) calculated (±) 0.00115 M1 A1 May be implied by sight of ‒0.00115088631 (from using given expression for g(x) from later part of question) Mark the final answer, must be rounded correctly to 3sf [2] 4 (b) 𝑥0 =1.8 𝑥1 =1.80082266922 𝑥2 =1.80129614714 M1 1.1 𝑥1, 𝑥2 seen to 5 or more dp (ignore labelling) 𝑥3 =1.8015686022 𝑥4 =1.80172536548 𝑥5 =1.80181555739 𝑥6 =1.80186744644 𝑥7 =1.80189729856 𝑥8 =1.80191447248 𝑥9 =1.80192435258 𝑥10 =1.80193003654 𝑥11 =1.80193330648 𝑥12 =1.80193518765 𝑥13 =1.80193626988 (𝑥14 =1.80193689247) M1 1.1 𝑥3, 𝑥4 or any 2 further correct iterations seen to 6 or more dp (ignore labelling) 1.80194 cao A1 2.2a Correct answer to 5dp, supported by at least 4 correct iterations. Must be seen separately from their list of iterations as a clear final answer. [3] 4 (c) Because the condition ‒1 < g′(0.445) < 1 is not satisfied B1 2.4 Allow eg because g′(0.445) > 1, must reference the point (x ≈)0.445 or 𝛽 (not just ‘because the gradient is more than 1’) [1] Y414/01 Mark Scheme June 2023 13 Question Answer Mark AO Guidance 4 (d) 𝑥𝑛+1 = (1 −(−0.258))𝑥𝑛+ (−0.258)g(𝑥𝑛) B1 1.1 Sight of the correct relaxed iteration formula with given values. May see 𝑔(𝑥𝑛) = √𝑥𝑛2 + 2𝑥𝑛−1 3 , may be seen in working. 𝑥0 =0.445 𝑥1 =0.44504176118 𝑥2 =0.445041867581 𝑥3 =0.445041867912 𝑥4 =0.445041867913 𝑥5 =0.445041867913 M1 1.1 𝑥2, 𝑥3 seen to 6 or more dp (ignore labelling) 0.44504187 cao A1 2.2a Correct answer to 8dp, supported by at least 3 correct iterations. Must be seen separately from their list of iterations as a clear final answer. [3] 5 (a) 1.30258554+1.28983372 2 M1 1.1 1.29620963 A1 1.1 Must be correctly rounded to 8dp. SCB1 for correct answer without working. [2] 5 (b) 2×1.30258554+1.28983372 3 or 2×1.29948881+their 1.29620963 3 M1 3.1a Or equivalent correct working [S2 =] 1.29833493… A1 1.1 Accept rounded correctly to 6dp or more SCB1 for correct answer without working. [S4 =] 1.29839575 A1 1.1 Accept rounded correctly to 6dp or more SCB1 for correct answer without working. [3] 5 (c) 1.298 (is secure) because S4 and S2 agree to 3dp B1 2.2b Allow ‘by comparison of S4 with S2’ or ‘1.2984 (is probable) because S4 is (more accurate / better) than S2’ but not referencing T/M values. [1] 5 (d) 16×1.29839575−1.29833493 15 M1 3.1a May be implied by 1.29839980467 Allow partial extrapolation for this mark only 1.298399 to 1.2984 A1 1.1 1.29840 by comparison with S4 A1 2.2b www, allow 1.298400 or 1.2984000 Accept ‘because extrapolation increases accuracy’ Y414/01 Mark Scheme June 2023 14 Question Answer Mark AO Guidance [3] 6 (a) t W ΔW ΔW² ΔW³ 0 35.90 ‒1.99 1 33.91 0.46 ‒1.53 ‒0.06 2 32.38 0.40 ‒1.13 ‒0.06 3 31.25 0.34 ‒0.79 4 30.46 M1 A1 1.1 1.1 Differences found, with at least three correct in first column All correct [2] 6 (b) The third differences are equal B1 2.4 Allow third differences are constant or fourth differences are zero [1] 6 (c) 35.9 + 𝑡× −1.99 + 𝑡(𝑡−1) 2! × 0.46 + 𝑡(𝑡−1)(𝑡−2) 3! × −0.06 M1 A1 3.3 1.1 Allow sign errors in substitution or one numerical error At least 2 terms correctly substituted 𝑊= −0.01𝑡3 + 0.26𝑡2 −2.24𝑡+ 35.9 A1 1.1 3 terms correct A1 3.3 All correct; A0 if different variable used or “W =” omitted [4] 6 (d) 𝑡= 6, 𝑊= 29.66 (which is ≈29.8) so model is a good fit B1ft 3.4 Ft 𝑊|𝑡=6 in their equation from (c) [1] 6 (e) In the long run model predicts weight decreases forever oe B1 3.5b Allow ‘modelled weight will eventually be negative’ [1] 7 (a) 16.17𝑥2 −4.34𝑥−1.11 seen M1* 3.1a Allow one slip in differentiation 𝑥− f(𝑥) their f′(𝑥) used M1 dep* 1.1 Need to see at least 3 iterates, including their 𝑥1 and 𝑥2 May be implied by correct iterates. Y414/01 Mark Scheme June 2023 15 Question Answer Mark AO Guidance (𝑥0 =‒0.5) 𝑥1 = ‒0.458598726 𝑥2 = ‒0.454582156 𝑥3 = ‒0.454545458 𝑥4 = ‒0.454545455 (𝑥5 = ‒0.454545455) M1 1.1 𝑥3, 𝑥4 seen to 6 or more dp (ignore labelling) ‒0.4545455 A1 3.2a Correct answer to 7dp, supported by at least 4 correct iterations. Must be seen separately from their list of iterations as a clear final answer. [4] 7 (b) (i) = B10 ‒ B9 B1 2.2a [1] 7 (b) (ii) = C12/C11 B1 2.2a [1] 7 (c) Ratio of differences is converging to 0.5 B1 2.2b Accept ‘approaching 0.5’ but not ‘is constant’ or ‘=0.5’ So convergence appears to be 1st order B1 2.2b Unusual because Newton-Raphson method usually has 2nd order convergence B1 2.3 [3] 8 (a) 1.4439304−1.3258177 0.8 M1 1.1 0.147640875 A1 1.1 Accept rounded correctly to 6dp or more. SCB1 for correct answer without working. [2] 8 (b) 1.4439304−0.9638087 1.6 M1 1.1 0.3000760625 (at least 6dp) A1 1.1 Accept rounded correctly to 6dp or more. SCB1 for correct answer without working. [2] Y414/01 Mark Scheme June 2023 16 Question Answer Mark AO Guidance 8 (c) Answer to part (b) likely to be closer to true value since central difference method is (generally) a 2nd order method, whereas forward difference method is (generally) 1st order method B1 2.4 Allow answer to part (b) probably closer to true value since central difference approximation takes values either side of 2 but forward difference (uses a step in the positive x-direction only) just takes a value to the right oe Explanation must mention both methods [1] 8 (d) 0.2355276 + (‒ 0.0007017) × 0.25 0.75 M1 A1 3.1a 2.1 allow slip in substitution 0.25 or any rounded version of 0.2484204 0.2352937 to 0.2352962 A1 1.1 allow 0.235, 0.2353, 0.23529 or 0.235294 because extrapolation improves accuracy significantly oe A1 3.2a www Allow 0.24 is certain by comparison of extrapolated value with 0.2355276 if M0 allow SC2 for 0.2355276 + (‒ 0.0007017) × 0.2484204 ≈ 0.235 as final answer; allow rounded versions of difference and ratio [4] Need to get in touch? 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Y410/01 Mark Scheme June 2023 3 5. Annotations Annotation 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Y410/01 Mark Scheme June 2023 4 Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result 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. BP Blank Page Seen Highlighting

Y410/01 Mark Scheme June 2023 5 6. 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 Y410/01 Mark Scheme June 2023 6 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. Y410/01 Mark Scheme June 2023 7 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. Y410/01 Mark Scheme June 2023 8 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. Y410/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 1 (a) 1 0 0 1 −   =     M B1 1.1 must be 2 by 2 [1] 1 (b) 2 1 0 0 1   =     M B1 1.1 Condone from 1 0 0 1   =   −   M [1] 1 (c) [M2 is the identity matrix] It represents the combination of two reflections, which is the identity transformation. B1 2.4 M2 must be an identity matrix (condone 33) oe, e.g. R is self-inverse [1] Y410/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 2 DR  +  = k,  = 2k B1 1.1a 2 2        + + = M1 2.3 for combining fractions correctly 2 ( ) 2     + − = M1 3.1a 2 + 2 = ( + )2 −2 (soi) 2 4 1 2 2 2 k k k k − = = − A1 1.1 must be simplified (accept 𝑘−4 2 ) Alternative solution 2 8 2  − = k k k x B1 by formula or completing the square 2 2 2 2 8 8 8 8     + − − − + = + − − + − k k k k k k k k k k k k ( ) ( ) 2 2 2 2 2 2 8 8 ( 8 )( 8 ) + − + − − = − − + − k k k k k k k k k k k k M1 combining fractions 2 2 2 2 2( 8 ) 8 + − = − + k k k k k k A1 expanding correctly 2 2(2 8 ) 1 2 8 2 − = = − k k k k A1 must be simplified (accept 𝑘−4 2 ) [4] Y410/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 DR 3 27 63 f 24 15 9 9 0 2 4 4  = − + − = −+ =     [so 3/2 is a root] B1 1.1 must see some substitution, [so 3 f 0 2  =     alone is B0]  2z – 3 is a factor M1 2.2a or z − 3/2 f(z) = (2z – 3)(z2 – 2z + 5) M1 A1 1.1 1.1 attempt to factorise (oe, e.g. long division) or (z – 3/2)(2z2 – 4z + 10) 2 16 2 z  − = M1 1.1 or by completing the square or using sum and prod of roots [z = a  ib, 2a = 2, a2 + b2 = 5  a = 1, b = 2] 3 1 2i, 1 2i, 2 z = + − B1 2.2a [3/2 may be stated as a root earlier] If no working shown then award no marks Alternative solution Other roots are  and  where 3 7 3 3 3 15 , 8, 2 2 2 2 2 2       + + = + + = = so  = 5,  +  = 2 M1 A1 symmetric property of roots used (condone 7, 15, 16 or sign errors) – must have at least 2 of the 3  2 − 2 + 5 = 0 A1  2 16 2   − = M1 or by completing the square or using sum and prod of roots [z = a ib, 2a = 2, a2 + b2 = 5  a = 1, b = 2] 3 1 2i, 1 2i, 2 z = + − B1 [6] Y410/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 1 1 1 ( ) 1 n n n r r r ar b a r b = = = + = +    M1 1.1a splitting sum 1 ( 1) 2 an n bn + + A1 A1 1.2 1.1 1 ( 1) 2 an n + … …+ bn 2 1 ( 1) 2 an n bn n + + = 1 1 2 2 1, 0 a a b = + = M1 3.1a equating coefficients (or substituting two values for n) So a = 2 b = –1 A1 A1 1.1 1.1 Alternative solution 1 ( ) = +  n r ar b is an AP with 1st term a+ b, common diff a M1 so 1 ( ) [2( ) ( 1) ] 2 = + = + + −  n r n ar b a b n a M1A1 use of sum formula (need not be simplified) 2 1 1 [2 2 ] 2 2 2 = + + − = + + n a b na a na nb n a 1 1 2 2 1, 0 a a b = + = So a = 2 b = –1 M1 A1 A1 equating coeffs or substituting two values for n Alternative solution substitute n = 1: a + b = 1 substitute n = 2: 3a + 2b = 4 solving simultaneously: 2a + 2b = 2 so a = 2 b = −1 M1 A1 A1 As the question requires finding 1 ( ) = +  n r ar b in terms of a, b and n, allow a maximum of 3 marks for this method [6] Y410/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 5 B1 B1 B1 B1 1.1 1.1 1.1 1.1 z*: reflection of z in Re axis 1 + z: 1 unit to right of z 1/z: arg = − arg z (must be < 45), ½ unit from O iz: rotation of z 90anticlockwise about O Give mark if intention is clear. [4] Y410/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (a) 2 3 4 3 2 4 3 5 4 , , 2 1 3 2 4 3       = = =       − − − − − −       M M M B2 1.1 Allow B1 if M2 correct [2] 6 (b) 1 1 n n n n n +   =   − −+   M B1 [1] 1.1 oe – allow correct unsimplified expressions allow if correct expression is seen in part (c) 6 (c) 1 2 1 1 1 1 1: 1 0 1 1 1   +     = = =       − − −+       n M so true B1 2.1 Assume true for n = k, so 1 1 k k k k k +   =   − − +   M 2.1 1 1 2 1 1 1 0 + +    =    − − + −    k k k k k M M1 1.1 or M  Mk 2 1 1 + +   =   −− −   k k k k ( 1) 1 1 ( 1) ( 1) 1 + + +   ==   − + − + +   k k k k So true for n = k + 1 A1 A1 or using target expression As true for n = 1, and if true for n = k then true for n = k + 1, true for all n. B1 2.2a dep first three marks gained. Must have if … then… (oe) [5] Y410/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) DR (3 + i)5=(3)5 + 5(3)4i + 10(3)3i2 + 10(3)2i3 + 53i4 + i5 M1 A1 1.1a 1.1 Binomial theorem showing correct pattern and coeffs (could use 5C0, etc correct expression (must evaluate nCr s) = –163 + 16 i [so a = –163 and b = 16] A1 1.1 Alternative method (3 + i)2= (2 + 23i) B1 or 3 + 23i − 1 (3 + i)3= 8i or (3 + i)4 = −8 + 83i B1 Either seen (3 + i)5= –163 + 16i so a = –163 and b = 16 B1 If without working, award no marks [3] 7 (b) (i) 3 + i| = 2 B1 1.1 modulus = 2 arg(3 + i) = /6 B1 1.1 arg = /6 or 30 So z = 2(cos /6 + i sin /6) B1ft 2.5 ft their modulus and argument [3] 7 (b) (ii) z5 = 25 (cos 5/6 +i sin 5/6) = 32 (cos 5/6 +i sin 5/6) B1ft 1.1 (their modulus)5 B1ft 1.1 5  their argument [2] If fully correct (condone 150) by converting −163 + 16i then allow SC B2 (but see below) 7 (b) (iii) 32(cos 5/6 + i sin 5/6) [= 32(–3/2 + ½ i)] B1 Condone no intermediate working for B2 provided 7(b)(ii) correct and from using de Moivre = –163 + 16 i as before B1 Or 322 = (−163)2 + 162 oe B1, tan−1(−16/163) = 5/6 as in 2nd quadrant B1 (but do not allow if (b)(ii) done by converting). In this case, if de Moivre used here: 32 = 25 or 532 = 2 SCB1, and   5 = 5 /6 or 5 / 5 =  SCB1 NB if (b)(ii) is not fully correct, award no marks for part (iii) [2] Y410/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 8 (a) 2 1 3 3 1 2 [ 0] 4 3 7 − − = − M1 1.1a finding determinant of correct matrix det = 0  planes do not meet at a point A1 1.1 [2] 8 (b) Method 1: use 2 equations to find e.g. z and y in terms of x, and substitute in 3rd equation e.g. (1) + (2): 5x + z = 5, M1 2.1 eliminating one variable  z = 5 – 5x, y = 13x – 12 M1 2.1 finding e.g. y and z in terms of x substitute in (3): –4x + 39x – 36 + 35 – 35x = k M1 2.1 substituting in 3rd equation  k = −1 A1 2.2a Method 2: eliminate one variable from 2 pairs of eqns, and compare eqns in 2 remaining v’bles, e.g. from (1) and (2): 5x + z = 5 (4) M1 eliminating one variable using two equations from (1) and (3): 10x + 2z = 9 − k (5) M1 eliminating same variable using two other equations  9 − k = 10 M1 eliminating both v’bles using (4) and (5) equations (coeffs in (4) and (5) must be correct)  k = −1 A1 Method 3: find linear combination by inspection (3) = (1) − 2(2) M3 oe  k = 3 − 22 = −1 A1 for k = −1 unsupported, allow SCB3 Method 4: substitute a value of x, y or z , e.g. substituting x = 0 y + 3z = 3, −y − 2z = 2 M1 any value for x, y or z substituted  z = 5, y = −12 A1 solve 2 equations for y, z substitute in (3): −36 + 35 = k  k = −1 M1 A1 substitute values into third equation [4] Y410/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 9 1 1 1 k x kx x y k y x ky + − + −     =     +     M1 1.1 If a specific value for k used, allow max of 3 M marks (SC) y = mx  x + ky = m(kx + x – y) M1 3.1a or x + ky = m(kx + x − y) + c  x + kmx = m(kx + x – mx) A1 2.1  x + k(mx + c) = m(kx + x − mx − c) + c  1 + km = km + m – m2  1 + km = km + m – m2 and kc = c − mc  m2 – m + 1 = 0 A1 1.1 soi discriminant = (−1)2 – 4 = −3 < 0 M1 3.1a oe or by solving to get 1 3 i 2 2 =  m so no real roots, and no invariant lines A1 3.2a without wrong working [6] If invariant point (instead of line) only first M1 is available Y410/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 Normal to x − z = 3 is vector i − k B1 3.1a may be implied by i − k seen (or column vector) 2 ( ) (2 ) cos45 2 5 a a − + − = + i k . i j k M1 1.1a allow 1 slip  2 1 3 2 2 5 a = + A1 1.1 must have 1/2 for cos 45  2 5 3 a + =  a = 2 A1 1.1 Plane equation is 2x + ay – z = k M1 1.1 or with their a (3, –1, 1) lies in plane  6 – a – 1 = k M1 3.1a substituting (3, −1, 1) into plane equation (k = 5 − their a)  k = 3 so plane equation is 2x + 2y – z = 3 A1 3.2a [7] Y410/01 Mark Scheme June 2023 19 APPENDIX Exemplar responses for Qxx Response Mark Need to get in touch? 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Y411/01 Mark Scheme June 2023 6 2. Subject-specific Marking Instructions for A Level Mathematics B (MEI) 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, 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 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 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. Y411/01 Mark Scheme June 2023 7 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. 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. This rule should be applied to each case. When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. Follow through should be used so that only one mark is lost for each distinct accuracy error, except for errors due to premature approximation which should be penalised only once in the examination. There is no penalty for using a wrong value for g. E marks will be lost except when results agree to the accuracy required in the question. g 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. Marks designated as cao may be awarded as long as there are no other errors. E marks are lost unless, by chance, the given results are established by equivalent working. ‘Fresh starts’ will not affect an earlier decision about a misread. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a graphical calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers (provided, of course, that there is nothing in the wording of the question specifying that analytical methods are required). 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. Y411/01 Mark Scheme June 2023 8 Question Answer Marks AOs Guidance 1 (a) Let the driving force on the car be D N 5000 20 D = ( = 250) B1 3.4 soi 𝐷= 𝑘(20)2 M1 3.3 Using driving force = resistance e.g. 5000 = 𝑘(20)2 × 20 implies B1 and M1 3 5000 5 8 20 k  = = A1 1.1 AG Must be correctly obtained [3] (b) Power is 5 8 (28)3 = 13720 (W) B1 1.1 Accept 13 700 www Accept 14 000 from correct method [1] (c) Let the mass of the car be m kg 13000 20 = 5 8 (20)2 + 𝑚𝑔sin 2° 650 = 250 + 𝑚𝑔sin 2° B1 M1 1.1 1.1 Correct weight component 𝑚𝑔sin 2° (B0 for 𝑊sin 2° unless 𝑊= 𝑚𝑔 used later) Resolving parallel to the plane – attempt at driving force (using 13000), resistance, and attempt to resolve weight (condone missing g) Driving force 13000 is M0 Mass is 1170 (kg) A1 1.1 1169.53911… [3] Y411/01 Mark Scheme June 2023 9 Question Answer Marks AOs Guidance 2 (a) Let the speed of the ball just before and just after hitting the ground be u and v m s-1. Let the coefficient of restitution be e. 𝑢2 = 02 + 2𝑔× 12.8 or 1 2 2 12.8 =  mu mg 𝑢2 = 128 5 𝑔= 250.88, 𝑢= 15.84 M1 3.3 Attempt at equation for u 02 = 𝑣2 −2𝑔× 5 or 1 2 2 5 =  mv mg 𝑣2 = 10𝑔= 98, 𝑣= 9.90 M1 3.3 Attempt at equation for v OR 5 = 12.8𝑒2 M2 𝑒= 𝑣 𝑢= 5 8 (= 0.625) A1 1.1 Accept 0.62 to 0.63 from correct method (e.g. 9.9/15.8 = 0.627) A0 for −5/8 (as final answer) [3] (b) ‘Perfectly elastic’ means no energy loss B1 1.2 Or Speed is unchanged by the impact Just 𝑒= 1 is not sufficient ( ) 12.8 12.8 5 5  − + =  m mg g E M1 3.3 WEP Allow sign errors M1 normally implies B1 OR 1 2 𝑚𝑣1 2 = 𝑚𝑔× 12.8 −12.8𝐸 , 1 2 𝑚𝑣2 2 = 𝑚𝑔× 5 + 5𝐸 𝑣2 = 𝑣1 M1 B1 WEP in two stages ⇒17.8𝐸= 7.8𝑚𝑔⇒𝐸= 39 89 𝑚𝑔 A1 2.2a AG Must be convincingly shown. Note that (39/89)g = 1911/445 [3] (c) Model A. Speed just after second bounce is 𝑤= 𝑒𝑣= 5 8 × √98 = 6.187, 𝑤2 = 125 32 𝑔= 38.28125 M1 3.1b Attempt to find w or w2 OR Maximum height is 5𝑒2 = 5 × ( 5 8) 2 or 5 × 5 12.8 M1 Maximum height is 125 64 (= 1.953125) m A1 1.1 Accept 1.9 to 2.0 from correct method Model B. Let maximum height be h m after second bounce. ( ) 39 89 5 5 − + = mg h mgh mg M1 1.1 WEP ( ) 128 250 125 89 89 64 1.953125  =  = = h h (which is the same) A1 2.2a Both A1’s can only be awarded if values agree to at least 3 sf [4] Y411/01 Mark Scheme June 2023 10 Question Answer Marks AOs Guidance 3 (a)  2 3 M T L   = G M1 1.1 Allow units (kg, m, s)  1 3 2 M L T − −  = G A1 1.1 Accept L3/(MT2) isw Do not allow units [2] (b) [v] = 1 LT− B1 1.1 Correct dimensions for v Allow units 1 LT− ( ) ( ) ( )( )( ) 1 3 2 1 2 M L T LT L M M    − − − = M1 1.1 Setting up an equation in M, L and T using given equation and their [G] Condone (M + M), 2M etc Allow units M0 for [𝑚1𝑚2(𝑚1 + 𝑚2)] = M or M2 [ −𝛼+ 3 = 0 ⟹ ] 3 = B1 1.1 cao 3 1    + + = 1 2  − − = − M1 1.1a Setting up equations using L and T FT their dimensions equation. Allow one error 5 = − and 3 = − A1 1.1 cao [5] (c) ( ) 3 8.64 6.13 ( 2.8  ) or ( ) 3 6.13 8.64 ( 0.357  ) M1 3.4 For ( 6.13 8.64) ± their 𝛾 M0 if their 𝛾= 0 So the stars approach 2.8 times faster when closer together. A1 2.2b cao M1A0 for 2.8 obtained from 𝛾= 3 [2] Y411/01 Mark Scheme June 2023 11 Question Answer Marks AOs Guidance 4 (a) Let the velocity of B after collision be u m s-1 in the direction AB . 8 0.5 1.6 0. 0. 5 3−  = u M1 3.3 COLM – all terms present but allow sign errors 3.2 = u A1 1.1 So coefficient of restitution 3.2 1 8 1.6 3 + = = A1 1.1 AG Must be correctly obtained [3] (b) Let B reach the lower section with speed v m s-1. 2 1 2 2 2 1 0.5 3.2 9.8 0.45 0 5 0.5 .     +  = v M1 3.3 Attempt at WEP: require two KE terms and a GPE term. M0 if 𝑣2 = 𝑢2 + 2𝑎𝑠 used ⇒𝑣2 = 19.06 , 𝑣= 4.36577 … A1 1.1 AG Must be correctly obtained [2] (c) Let the speed of C after collision be w m s-1. 0.7 0.5 = w v ( ⇒𝑤= 5 7 𝑣= 3.1184 ) M1 3.4 COLM KE of C is 1 2 × 0.7 × (3.1184)2 (= 3.40357 … ) J A1 1.1 C needs to gain 9.8 0.45 3.087 0.7  = J of GPE so C will collide with A next. A1 2.1 Argument must be clear. OR Assume C reaches the top section with speed x ms−1 1 2 × 0.7 × 3.11842 −0.7 × 9.8 × 0.45 = 1 2 × 0.7 × 𝑥2 A1 For KE of C 𝑥2 (= 0.9044) > 0 ( x = 0.951 ) So C will collide with A next A1 Or C can reach a height of 0.496 m ( > 0.45 ) [3] (d) OR OR [ If 𝑣B = 0, ] by COLM, 𝑣C > 𝑢B giving 𝑒> 1 (or increase of KE) which is impossible [contradicting 𝑒≤1 ] [ If 𝑣B = 0, since 𝑒≤1 ] 𝑣C ≤𝑢B and so 𝑚C𝑣C < 𝑚B𝑢B contradicting COLM 𝑚B𝑣B + 𝑚C𝑣C = 𝑚B𝑢B and 𝑣C −𝑣B ≤𝑢B ⟹(𝑚B + 𝑚C)𝑣B ≥(𝑚B −𝑚C)𝑢B ⟹ 𝑣B > 0 B2 3.5a Correct explanation; must mention momentum Give B1 for an explanation which includes at least one of the following (ignore incorrect statements) • This requires 𝑣C > 𝑢B • This requires 𝑒> 1 • If 𝑣C = 𝑢B then 𝑚C𝑣C < 𝑚B𝑢B [2] Y411/01 Mark Scheme June 2023 12 Question Answer Marks AOs Guidance 5 (a) Area of S ( ) 2 2 2 π 50 15 0 0 3 10 0  = = − − B1 1.1 Or ratio of masses (e.g. 25:9:1:15) used throughout 1 2 0 50 0 0 20 1500 0 900 0 40 0 0   −       = + +             x y M1 1.1 For (𝜋× 302)(20) or (𝜋× 102)(40) seen ⟹𝑥̅ = 12 A1 1.1 AG Needs convincing working 𝑦̅ = − 8 3 A1 1.1 Allow −2.67 [4] (b) arctan (12 ÷ 8 3) Angle is 77.47119…° M1 A1 FT 3.1b 1.1 tan 𝛼= 12 ÷ their |𝑦̅| (allow reciprocal) Just arctan(𝑥̅/𝑦̅) is not sufficient Accept 102.5°, −77.5° FT their |𝑦̅| [2] (c) 50 12  =  W F M1 1.1 Taking moments about O – both moments present 0.24  = F W A1 1.1 λ = 6/25 (or 0.24) [2] (d) Let the angle between A R and the horizontal be . 𝑅A cos 𝜃+ 𝐹sin 𝜃= 𝑅B cos 𝜃 0.6𝑅A + 0.8(0.24𝑊) = 0.6𝑅B Resolving horizontally. Must attempt to resolve all three forces. F may be their 0.24W 𝑅A sin 𝜃+ 𝑅B sin 𝜃= 𝑊+ 𝐹cos 𝜃 0.8𝑅A + 0.8𝑅B = 𝑊+ 0.6(0.24𝑊) Resolving vertically. Must include W and attempt to resolve the other three forces 60𝑅B sin 𝜃= 42𝑊 0.8 × 60𝑅B = 42𝑊 Moments about A. Both moments present. Must attempt to ‘resolve’ RB (or its distance) M1 M1 3.3 Any two of the above. See first lines and notes OR 18𝑊+ 60𝐹cos 𝜃= 60𝑅A sin 𝜃 18𝑊+ 0.6 × 60(0.24𝑊) = 0.8 × 60𝑅A M2 Moments about B. Must attempt to ‘resolve’ RA and F (or their distances) M1 1.1 Correct values for cos and sin used in at least one equation for which M1 has been earned. See second lines above. Allow cos(53.1°) etc Note Only sin needed in the third equation ⇒𝑅A = 0.555𝑊 ( 𝑅B = 0.875𝑊 ) A1 1.1 cao ( ) 0.24 16 0.555 37 0.432432   = = = W W A1 FT 2.2a FT their F and RA given as multiples of W, dependent on M3 [5] Y411/01 Mark Scheme June 2023 13 Question Answer Marks AOs Guidance 6 (a) Each support must exert ( ) 1 2 4 10 7 + = g g g N on the beam. B1 1.1 soi Taking moments about the left-most point: 𝑅A𝑥+ 𝑅B(𝑥+ 3.8) = 4𝑔× 1.95 + 10𝑔× 3 7𝑔𝑥+ 7𝑔(𝑥+ 3.8) = 4𝑔× 1.95 + 10𝑔× 3 M1 A1 3.3 1.1 Attempt to take moments about some point. All moments present. Allow one error Correct equation for x FT their 7g Correct equation www implies B1 ⇒𝑥= 0.8 A1 1.1 cao [4] (b) Moments about B, 3 5.3 sin60 10 2. L g T =  M1 3.1b Attempt at moments. No force acting at A; allow sin/cos switch. If moments taken about A, 𝑅A = 0 must be soi (e.g. by 𝑇𝐿sin 60° + 𝑅B = 10𝑔 ) Allow inequalities in (b) and (c) 49.10745 L T  = A1 1.1 AG Must be convincingly shown. [2] (c) Let the force at the supports be A R and B R N, and the friction at B be F N. B 0.4 = F R B1 3.4 Modelling friction. Must clearly be reaction at B 𝑇𝑆cos 60° = 𝐹  (⇒𝑇𝑆= 0.8𝑅B) M1 3.3 Resolving horizontally 𝑇𝑆sin 60° + 𝑅A + 𝑅B = 10𝑔 M1 1.1 Resolving vertically Moments about LH end, A B 1.5 5.3 3 10 + =  R R g M1 3.1b Attempt at moments. All moments present. Allow one error OR 3.8𝑅B = 1.5 × 10𝑔+ 1.5 × 𝑇𝑆sin 60° M2 Moments about A (⇒𝑅B = 53.2460 … 𝑅A = 7.864 …) 42.5968  = S T so beam will slide first (since 42.6 < 49.1) A1 2.2a cao [5] 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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Y412/01 Mark Scheme June 2023 3 4. Annotations Annotation 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 E Explanation mark 1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Y412/01 Mark Scheme June 2023 4 Other abbreviations in mark scheme Meaning E1 Mark for explaining a result or establishing a given result 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. BP Blank Page Seen Highlighting

Y412/01 Mark Scheme June 2023 5 5. 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 Y412/01 Mark Scheme June 2023 6 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. Y412/01 Mark Scheme June 2023 7 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. Y412/01 Mark Scheme June 2023 8 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. Y412/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 1 (a) 6 10 × 5 9 × 4 8 or for (6 3) ÷ (10 3 ) M1 1.1a M1 for correct denominator = 1 6 A1 1.1 AG correct working only. A0 for (𝟏𝟎 𝟑) × 𝟏 𝟏𝟎× 𝟏 𝟗× 𝟏 𝟖 [2] 1 (b) B1 B1 [2] 1.1 1.1 For heights. 3, 4, 5 & 6 labelled. B0 if tops joined. For labels to identify both axes and appropriate scale on the probability axis. 1 (c) DR E(𝑋) = 3 × 1 6 + 4 × 1 2 + 5 × 3 10 + 6 × 1 30 M1 1.1a Allow 0.5 + 2 + 1.5 + 0.2. Condone 1 error for M1. = 21 5 (= 4.2) A1 1.1 E(𝑋2) = 32 × 1 6 + 42 × 1 2 + 52 × 3 10 + 62 × 1 30 [= 91 5 ] M1 1.1 Allow 1.5 + 8 + 7.5 + 1.2. Condone 1 error for M1. Var(𝑋) = 91 5 −( 21 5 ) 2 M1 1.2 FT their E(X) and E(𝑋2). M0 if this leads to a non-positive answer. = 14 25 (=0.56) A1 1.1 cao SCB1 for 0.56 if M0M1 or M1M0 awarded [5] 1 (d) E(Y) = 2 × 21 5 + 3 = 57 5 (= 11.4) B1 2.2a FT their E(X) Var(𝑌) = 𝟐𝟐× 𝟏𝟒 𝟐𝟓 M1 3.5a For 4 × their Var(X) seen provided this is positive. = 56 25 (=2.24) A1 1.1 cao SCB1 for correct answer without working seen. [3] 0 1/6 1/3 1/2 3 4 5 6 Probability Value (£X) Y412/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 2 (a) Because we can assume that successive trials are independent, each trial has two possible outcomes with an equal probability of success (for each trial), and the number of trials up to (and including) the first success is being counted B1 2.4 B1 for ‘the number of trials up to (and including) the first success’ and one of the other three of these comments seen [1] 2 (b) 0.854 × 0.15 = 0.0783 B1 1.1 (0.078300…) [1] 2 (c) 0.8510 = 0.1969 B1 1.1 (0.196874….) [1] 2 (d) Mean = 20 3 [= 6.67 to 3 s. f. ] B1 3.1b Variance = 1−0.15 0.152 M1 1.1 M1 for 1−𝑝 𝑝2 used = 340 9 [=37.8 to 3 s.f.] A1 1.1 SCB1 for 37.8 without justification [3] Y412/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (a) B(20, 0.05) B1 3.3 For stating correct binomial distribution/calculation = 1 −0.7358 = 0.2642 B1 1.1 For 0.2642 Allow 0.264 [2] 3 (b) Var(X) [= 20 × 0.05 × 0.95] = 0.95 B1 1.1 [1] 3 (c) Var(Y) = 30 × 0.07 × 0.93 [= 1.953] M1 3.1b Var (X + Y) = 0.95 + 1.953 [=2.903] M1 1.1 FT their Var(X) + their Var(Y), provided their Var(Y) is identified. SD (X + Y) = 1.70 A1 1.1 cao (1.70381…) [3] 3 (d) e.g. Because there could be a fault that affected only some of the batches and it is more likely to be found if the sample comes from several batches B1 2.2b For a suitable comment that suggests that one batch may not be representative. [1] Y412/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) A binomial distribution (to model the number of successes) could be suitable as • There is a fixed number of parcels delivered each day. [n = 15000] • Each delivery has two possible outcomes (wrong address or not) • The deliveries could be assumed to be independent. • There is constant probability of delivery to a wrong address [p = 0.0005] B2,1,0 2.4 B1 for two of these conditions stated B2 for correct comments with at least three conditions stated, and at least one given in context with reference to binomial. SCB1 if first two B1 marks not awarded but B(15000, 0.0005) stated Because n = 15 000 is large and p = 0.0005 is small a Poisson distribution is also appropriate B1 2.4 Allow correctly worded explanations that Poisson could be suitable due to the mean being close to the variance [3] 4 (b) Poisson(𝟏𝟓𝟎𝟎𝟎× 𝟎. 𝟎𝟎𝟎𝟓) or Poisson(7.5) B1 3.3 Poisson soi. e.g. by 7.5 seen P(X = 5) = 0.1094 B1 1.1 BC (0.109374…) P(X ≥ 8) = 0.4754 B1 1.1 BC (0.475361…) [3] 4 (c) (i) Poisson (𝟓× 𝟕. 𝟓) or B(75000, 0.0005) oe seen B1 3.3 P(≥ 40) = awrt 0.363 B1 1.1 BC (0.362862… from Poisson or 0.362839… from binomial) [2] 4 (c) (ii) P(𝑋≥ 8)5 = 0.4753614 …5 M1 3.3 FT [their P(X ≥ 8)]5 Allow M1 for Y ~ B(5, their P(X ≥ 8)) and P(Y = 5) = 0.0243 A1 1.1 BC (0.024272…) cao SCB1 for correct answer without working shown. [2] Y412/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 5 (a) DR 1 20 21635 565 724 xy S = −   [= 1182] B1 1.1a Correct calculation seen. 2 1 20 17103 565 xx S = −  [= 1141.75] B1 1.1 For either xx S or yy S with correct calculation seen 2 1 20 29286 724 yy S = −  [= 3077.2] 1182 1141.75 3077.2 xy xx yy S r S S = =  M1 3.3 For general form including sq. root = 0.6306 A1 1.1 Allow 0.631 www [4] 5 (b) H0:  = 0, H1:  > 0 B1 3.3 For both hypotheses. Allow hypotheses in words provided these refer to the population correlation coefficient. where  is the population pmcc between x and y B1 2.5 For defining  in context. Must include population. For n = 20, the 5% 1-tailed critical value is 0.3783 B1 3.4 For correct critical value Since 0.6306 > 0.3783 so there is sufficient evidence to reject H0 M1 1.1 For suitable comparison and consistent conclusion regarding rejection of H0. FT their r (0 < r < 1) and their critical value. There is sufficient evidence at the 5% level to suggest that there is positive correlation between practice exam results A1 2.2b For non-assertive conclusion in context that refers to H1. Do not allow ‘relationship’ for ‘correlation’. No FT incorrect cv. [5] 5 (c) The test may not be valid since the population might not be bivariate Normal. B1 3.5b B0 for ‘the data might not be bivariate Normal’ [1] Y412/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (a) (i) 3 8 = 0.375 B1 [1] 1.1 6 (a) (ii) E(X) = 4.5 B1 1.1a BC Var(X) = 1 12 (82 −1) (⇒ SD(X) = 2.29..) B1 1.1 BC Need P(−0.1 < X < 9.1) M1 3.1b M1 for “4.5” + 2× “2.29” or “4.5” − 2× “2.29” = 𝟏, because this interval includes all the possible values. A1 1.1 AG For A1 needs to be fully correct. The given answer must be stated. [4] 6 (b) (i) Because P(score = n) × 80 = 𝟏 𝟖× 𝟖𝟎= 10 B1 2.2a [1] 6 (b) (ii) B9 = 13 B1 1.1 D5 contribution = (15−10)2 10 M1 3.4 = 2.5 A1 1.1 SCB1 for 2.5 if no method shown [3] 6 (b) (iii) DR H0: Uniform model is a good fit H1: Uniform model is not a good fit B1* 3.3 or H0: dice is unbiased H1: dice is biased X2 = 11.2 B1 1.1 FT 8.7 + “2.5” Refer to 2 7  M1 3.4 M1 for number of degrees of freedom = 7 Critical value at 5% level = 14.07 A1* 1.1 or 𝜒7 2(11.2) = 0.8699 11.2 < 14.07 M1* 1.1 M1 for suitable comparison. e.g. 0.8699 < 0.95 Do not reject H0. There is insufficient evidence to suggest that the uniform model is not a good fit. dep*A1 2.2b or Do not reject H0. There is insufficient evidence to suggest that the dice is biased. Condone Accept H0 Dep*A1 for non-assertive conclusion with reference to model or dice. FT their “11.2” Y412/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance [6] 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. Call us on 01223 553998 Alternatively, you can email us on support@ocr.org.uk For more information visit ocr.org.uk/qualifications/resource-finder ocr.org.uk Twitter/ocrexams /ocrexams /company/ocr /ocrexams OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge. For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR 2023 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA. Registered company number 3484466. OCR is an exempt charity. OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals. OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these resources. Though we make every effort to check our resources, there may be contradictions between published support and the specification, so it is important that you always use information in the latest specification. We indicate any specification changes within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy between the specification and a resource, please contact us. Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more information using our Expression of Interest form. Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.