Y420/01 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 Y420/01 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. Y420/01 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. Y420/01 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. Y420/01 Mark Scheme June 2023 9 Question Answer Marks AO Guidance 1 (a) (i) a – ib B1 1.2 [1] 1 (a) (ii) iz = –b +ai so Re(iz) = –b M1 A1 1.1 1.1 [2] 1 (b) (i) DR 5 + √3i 2 −√3i = (5 + √3 i) (2 + √3 i) (2 −√3 i) (2 + √3 i) M1 1.1a = 7 + 7√3i 7 = 1 + √3i A1 1.1 an intermediate step must be seen before final answer [2] 1 (b) (ii) ห1 + √3iห= 2 arg( 1 + √3 i) = 𝜋 3 B1ft 1.1 for either their modulus or argument soi so 𝑤= 2(cos గ ଷ+ i sin గ ଷ) B1 1.1 cao. Allow 60° [2] Y420/01 Mark Scheme June 2023 10 Question Answer Marks AO Guidance 2 DR Normal to plane is i + 3j + 2k B1 1.1a soi Angle between 3i + 2j + k and i + 3j + 2k is  cos 𝜃= 3 × (−1) + 2 × 3 + 1 × 2 √3ଶ+ 2ଶ+ 1ଶඥ(−1)ଶ+ 3ଶ+ 2ଶ= 5 √14√14 M1 1.1 use of scalar product formula with their normal seen   = 69.07… A1 1.1 or 1.2… rads or arccos ହ ଵସ 90 − their 𝜃 so angle with plane is 20.9° M1 A1 1.1 2.2a or గ ଶ− their 𝜃 or 0.365 rads Alternative method 1 Normal to plane is i + 3j + 2k B1 soi Angle between 3i + 2j + k and the plane is  sin𝜃= 3 × (−1) + 2 × 3 + 1 × 2 √3ଶ+ 2ଶ+ 1ଶඥ(−1)ଶ+ 3ଶ+ 2ଶ= 5 √14√14 M2 = 20.9° A2 or 0.365 rads Alternative method 2 Normal to plane is i + 3j + 2k B1 soi Angle between 3i + 2j + k and i + 3j + 2k is  sin𝜃= อ൭ 3 2 1 ൱× ൭ −1 3 2 ൱อ √3ଶ+ 2ଶ+ 1ଶඥ(−1)ଶ+ 3ଶ+ 2ଶ= อ൭ 1 −7 11 ൱อ √14√14 M1 complete method with vector product seen   = 69.07… A1 or 1.2… rads or arcsin √ଵ଻ଵ ଵସ 90 − their 𝜃 so angle with plane is 20.9° M1 A1 or గ ଶ− their 𝜃 or 0.365 rads [5] Y420/01 Mark Scheme June 2023 11 Question Answer Marks AO Guidance 3 (a) 1 𝑟(𝑟+ 2) = 𝐴 𝑟+ 𝐵 (𝑟+ 2) 1 = 𝐴(𝑟+ 2) + 𝐵𝑟 M1 1.1 𝑟= 0 ⇒𝐴= ଵ ଶ, 𝑟= −2 ⇒𝐵= − ଵ ଶ A1 1.1 ෍ 1 𝑟(𝑟+ 2) ௡ ௥ୀଵ = 1 2 ෍൬1 𝑟− 1 𝑟+ 2൰ ௡ ௥ୀଵ = 𝑘൤1 −1 3 + 1 2 −1 4 + 1 3 −1 5 +. . . + 1 𝑛− 1 𝑛+ 2൨ M1 1.1 enough terms to show consistent cancellation in their series = 1 2 ൤1 + 1 2 − 1 𝑛+ 1 − 1 𝑛+ 2൨ A1 1.1 = 3 4 − 2𝑛+ 3 2(𝑛+ 1)(𝑛+ 2) A1 1.1 [5] 3 (b) 3 4 B1 2.2a [1] Y420/01 Mark Scheme June 2023 12 Question Answer Marks AO Guidance 4 (a) (i) f ′(𝑥) = (1 + 2𝑥)ିଵ ଶ B1 1.1 f ″(𝑥) = −(1 + 2𝑥)ିଷ ଶ B1 1.1 [2] 4 (a) (ii) f(0) = 1, f ᇱ(0) = 1, f ᇱᇱ(0) = −1 1 + 𝑥−1 2 𝑥ଶ M1 A1 [2] 1.1 1.1 Their f(0), f’(0) and f’’(0) evaluated and substituted into Maclaurin Must come from correct expressions for f ′(𝑥), f ′′(𝑥); cannot come from binomial expansion. Ignore subsequent terms. 4 (b) ට1 + 2 × ଵ ଼ or 1 + ଵ ଼− ଵ ଶቀ ଵ ଼ቁ ଶ M1 3.1a using 𝑥= ଵ ଼ in their expansion  √5 ≈ ଵସଷ ଺ସ A1 2.2a AG [2] Y420/01 Mark Scheme June 2023 13 Question Answer Marks AO Guidance 5 (a) DR Let 𝑧= 𝑟(cos 𝜃+ i sin𝜃) or 𝑧= 𝑟e௜ఏ 𝑧଺= 64(cos 𝜋+ i sin𝜋) or 64e୧గ M1 1.1 condone −𝜋 for 𝜋 r = 2 B1 1.1 soi z = 2ei/6, 2ei/2, 2e5i/6, 2e–i/6, 2e–i/2, 2e–5i/6 A1 2.5 1 correct A1 1.1 all correct or 2e7i/6, 2e3i/2, 2e11i/6 [4] 5 (b) M1 A1 A1 1.1 1.1 1.1 six roots lying on approximate circle centre O forming an approximate regular hexagon root at 2i or −2i oe indicated. If part (a) contains incorrect roots do not award final A1. [3] 2 i Y420/01 Mark Scheme June 2023 14 Question Answer Marks AO Guidance 6 (a) DR MN = ቀ0 1 2 0ቁ or ቀ0 1 1 0ቁቀ2 0 0 1ቁ= ቀ0 1 2 0ቁ NM = ቀ0 2 1 0ቁ or ቀ2 0 0 1ቁቀ0 1 1 0ቁ= ቀ0 2 1 0ቁ M1 A1* 1.1a 1.1 both MN and NM calculated both correct so not commutative A1 2.2a dep A1* [3] 6 (b) (i) M is reflection in y = x B1 [1] 1.1 6 (b) (ii) N is stretch parallel to the x-axis scale factor 2 M1 A1 1.1 1.1 [2] 6 (c) order of transformations matters B1 2.2a must refer to transformations not just matrices [1] 6 (d) ቀ2 0 0 1ቁቀ𝑥 𝑦ቁ= ൬2𝑥 𝑦൰ or ቀ2 0 0 1ቁቀ 𝑥 𝑚𝑥+ 𝑐ቁ= ቀ 2𝑥 𝑚𝑥+ 𝑐ቁ M1 2.1 Allow SC1 for correct geometrical argument, e.g. stretch in x-direction leaves y-coordinates unchanged y-coordinate unchanged so lines parallel to x-axis invariant A1 2.2a www. Must refer to invariant lines. [2] Y420/01 Mark Scheme June 2023 15 Question Answer Marks AO Guidance 7 (a) A: r = a , = 0 B1 1.1 or = 2𝜋 B: r = a , = /2 B1 1.1 C: r = 3a , = 3/2 B1 1.1 [3] 7 (b) sin𝜃> 1 2 M1 3.1a Allow sin𝜃= ଵ ଶ (or < ଵ ଶ or ≤ ଵ ଶ or ≥ ଵ ଶ) 1 6 𝜋, 5 6 𝜋 A1 1.1 both 1 6 𝜋< 𝜃< 5 6 𝜋 A1 1.1 condone  for < [3] Y420/01 Mark Scheme June 2023 16 Question Answer Marks AO Guidance 8 when n = 1, 8n – 3n = 5 div by 5 B1 2.1 [Assume true when n = k] so 8k = 3k + 5m M1 2.1 or 3k = 8k – 5m or 8k −3k = 5m or 8k −3k div 5 8k+1 – 3k+1 = 8(3k + 5m) – 33k M1 2.1 assumption used = 53k + 40m = 5(3k + 8m) div by 5 A1 2.2a successful completion 3k = 8k – 5m used  5(8௞+ 3𝑚) div 5 www 8k −3k = 5m used 5(8𝑚+ 3௞) div 5 www so if true for n = k then true for n = k+1 As true for n = 1, true for all n. A1 2.4 must receive all previous marks for this to be awarded Alternative method when n = 1, 8n – 3n = 5 div by 5 B1 [Assume true when n = k] so 𝑢௞= 8k−3k = 5m M1 𝑢௞ାଵ−𝑢௞= 8௞ାଵ−3௞ାଵ−(8௞−3௞) M1 considering difference between 𝑢௞ାଵ and 𝑢௞ = 2൫8௞−3௞൯+ 5(8௞) = 2(5𝑚) + 5൫8௞൯= 5൫2𝑚+ 8௞൯ div by 5 A1 so if true for n = k then true for n = k+1 As true for n = 1, true for all n. A1 must receive all previous marks for this to be awarded [5] Y420/01 Mark Scheme June 2023 17 Question Answer Marks AO Guidance 9 𝑘න sinଶ𝑛𝑡d𝑡 ଶగ ௡ ଴ = 1 2 𝑘න (1 −cos 2𝑛𝑡)d𝑡 ଶగ ௡ ଴ M1* A1 3.1a 1.1 use of double angle formula for sinଶ𝑛𝑡 = 1 2 𝑘ቈ𝑡−1 2𝑛sin2𝑛𝑡቉ 0 2𝜋 𝑛 M1dep 1.1 ൤𝑡−1 2𝑛sin 2𝑛𝑡൨ = 1 2 𝑎ଶ2𝜋 𝑛= 𝜋𝑎ଶ 𝑛 A1 1.1 www mean value of I2 = 2 2 2 2 a a n n   M1 1.1 for dividing by ଶగ ௡  RMS value = ௔ √ଶ A1 2.2a www AG Alternative method 1 ቀ2𝜋 𝑛ቁ 𝑘න sinଶ𝑛𝑡d𝑡 ଶగ ௡ ଴ = 𝑛 2𝜋𝑘න sinଶ𝑛𝑡d𝑡 ଶగ ௡ ଴ M1 1.1 for dividing their integral by ଶగ ௡ = 1 2 × 𝑛 2𝜋𝑘න (1 −cos2𝑛𝑡)d𝑡 2𝜋 𝑛 0 M1* A1 3.1a 1.1 use of double angle formula for sinଶ𝑛𝑡 = 𝑛 4𝜋𝑘ቈ𝑡−1 2𝑛sin2𝑛𝑡቉ 0 2𝜋 𝑛 M1dep 1.1 ൤𝑡−1 2𝑛sin 2𝑛𝑡൨ = 𝑎ଶ𝑛 4𝜋൬2𝜋 𝑛൰= 𝑎ଶ 2 A1 1.1 www  RMS value = ௔ √ଶ A1 2.2a www AG [6] Y420/01 Mark Scheme June 2023 18 Question Answer Marks AO Guidance 10 (a) 𝛼+ 𝛽+ 𝛼+ 𝛽= 4 [⇒𝛼+ 𝛽= 2] B1 3.1a 𝛼𝛽+ (𝛼+ 𝛽)𝛼+ (𝛼+ 𝛽)𝛽= 7 B1 1.1 or 𝛼𝛽+ 2𝛼+ 2𝛽= 7 or 𝛼𝛽+ 2(𝛼+ 𝛽) = 7 ⇒𝛼ଶ+ 𝛽ଶ+ 3𝛼𝛽= 7 𝑥+ ଷ ௫= 2 or 3𝑥(2 −𝑥) + (2 −𝑥)ଶ+ 𝑥ଶ= 7 M1 3.1a substitution which could lead to a quadratic in 𝛼 or 𝛽 or 𝑥 ⇒𝑥ଶ– 2𝑥+ 3 = 0 A1 1.1 could be in 𝛼 or 𝛽 ⇒𝑥= 2 ± √−8 2 = 1 ± i √2 M1 1.1 a method to solve their quadratic so roots are 1 + i√2, 1 −i√2 and 2 A1 2.2a [6] 10 (b) [𝑐=] −6 B1 2.2a [1] Y420/01 Mark Scheme June 2023 19 Question Answer Marks AO Guidance 11 d𝑦 d𝑥−2𝑦sinh 𝑥 cosh 𝑥= 1 B1 3.1a ୢ௬ ୢ௫−2𝑦tanh 𝑥= 1 IF is e∫ିమ౩౟౤౞ೣ ౙ౥౩౞ೣୢ௫ M1 2.1 or 𝑒∫ିଶ୲ୟ୬୦௫ௗ௫. Integral must come from an attempt to get ୢ௬ ୢ௫ on its own = eିଶ୪୬ୡ୭ୱ୦௫= ൫𝑒୪୬ୱୣୡ୦௫൯ ଶ= sechଶ𝑥 A1 2.2a d d𝑥(𝑦sechଶ𝑥) = sechଶ𝑥 M1 2.1 or multiplying through by their IF 𝑦sechଶ𝑥= ∫sechଶ𝑥d𝑥+ 𝑐= tanh 𝑥+ 𝑐 A1 1.1 substituting 𝑥= 0, 𝑦= 1 gives 𝑐= 1 M1 1.1 substituting 𝑥= 1 and 𝑦= 1 to lead to a value for 𝑐 𝑦= coshଶ𝑥(tanh 𝑥+ 1) A1 2.2a oe, e.g. cosh 𝑥(sinh 𝑥+ cosh 𝑥) or e௫cosh 𝑥 [7] Y420/01 Mark Scheme June 2023 20 Question Answer Marks AO Guidance 12 Considering ቀ𝑧− ଵ ௭ቁ ହ M1 3.1a 32i sinହ𝜃 B1 1.1 seen at any stage = 𝑧ହ−5𝑧ଷ+ 10𝑧−10 1 𝑧+ 5 1 𝑧ଷ−1 𝑧ହ M1 1.1 expansion of ቀ𝑧− ଵ ௭ቁ ହ = 𝑧ହ−1 𝑧ହ−5 ൬𝑧ଷ−1 𝑧ଷ൰+ 10 ൬𝑧−1 𝑧൰ A1 2.1 correct expansion with 𝑧௡, ଵ ௭೙ terms paired and factorised Using 𝑧௡− ଵ ௭೙= 2i sin𝑛𝜃 with sum of terms M1 2.1 FT their expansion = 2i sin 5𝜃−10i sin 3𝜃+ 20i sin𝜃 A1 1.1 i must appear in each term sinହ𝜃= ଵ ଵ଺sin 5𝜃− ହ ଵ଺sin 3𝜃+ ହ ଼sin𝜃 A1 2.2a www Alternative method 1 Considering ൫e୧ఏ−eି୧ఏ൯ ହ M1 32i sinହ𝜃 B1 = eହ୧ఏ−5eଷ୧ఏ+ 10e୧ఏ−10eି୧ఏ+ 5eିଷ୧ఏ−eିହ୧ఏ M1 expansion of ൫e୧ఏ−eି୧ఏ൯ ହ = eହ୧ఏ−eିହ୧ఏ−5൫eଷ୧ఏ−eିଷ୧ఏ൯+ 10(e୧ఏ−eି୧ఏ) A1 correct expansion with e୧ఏ, eି୧ఏ terms paired and factorised Using e୧ఏ−eି୧ఏ= 2i sin𝑛𝜃with sum of terms M1 FT their expansion = 2i sin 5𝜃−10i sin 3𝜃+ 20i sin𝜃 A1 i must appear in each term sinହ𝜃= ଵ ଵ଺sin 5𝜃− ହ ଵ଺sin 3𝜃+ ହ ଼sin𝜃 A1 www Alternative method 2 Equating Im components of (cos 𝜃+ i sin 𝜃)ହ using binomial expansion and de Moivre’s theorem sin 5𝜃= 5 cosସ𝜃sin 𝜃−10 cosଶ𝜃sinଷ𝜃+ sinହ𝜃 B1 = 5(1 −sinଶ𝜃)ଶsin θ −10(1 −sinଶ𝜃) sinଷ𝜃+ sinହ𝜃 M1* correct expression for sin 5𝜃 and substituting in cosଶ𝜃= 1 − sinଶ𝜃 sin 5𝜃= 16 sinହ𝜃−20 sinଷ𝜃+ 5 sin 𝜃 A1 or 16 sinହ𝜃= sin 5𝜃+ 20 sinଷ𝜃−5 sin 𝜃 or any correct rearrangement Equating Im components of (cos 𝜃+ i sin 𝜃)ଷ using binomial expansion and de Moivre’s theorem sin3𝜃= 3 cosଶ𝜃sin 𝜃−sinଷ𝜃 M1* correct expression for sin 3𝜃 in terms of sin 𝜃 and cos 𝜃 sinଷ𝜃= 3 4 sin 𝜃−1 4 sin 3𝜃 A1 or sin 3𝜃= 3 sin 𝜃−4 sinଷ𝜃 or any correct rearrangement 16 sinହ𝜃= sin 5𝜃+ 20 ቀ ଷ ସsin 𝜃− ଵ ସsin 3𝜃ቁ−5 sin 𝜃 M1dep substituting expression for sinଷ𝜃 into sinହ𝜃. sinହ𝜃= ଵ ଵ଺sin 5𝜃− ହ ଵ଺sin 3𝜃+ ହ ଼sin𝜃 A1 www [7] Y420/01 Mark Scheme June 2023 21 Question Answer Marks AO Guidance 13 (a) (i) M1 A1 A1 1.1 1.1 1.1 circle, centre O radius √5 shaded inside (oe). Candidates may shade the region that is not required, but should clearly indicate that what they have shaded is not required. [3] 13 (a) (ii) M1 A1 A1 1.1 1.1 1.1 (–2, 4) and (2, 6) identified perpendicular bisector of (–2, 4) and (2, 6) shaded on RHS of line (oe). Candidates may shade the region that is not required, but should clearly indicate that what they have shaded is not required. [3] Y420/01 Mark Scheme June 2023 22 Question Answer Marks AO Guidance 13 (b) gradient = –2 M1 1.1 passing through (0, 5) B1 3.1a equation y = –2x + 5 A1 1.1 oe. Allow inequality. Could be obtained from diagram in (a). circle is x2 + y2 = 5 B1 3.1a allow inequality x2 + (5 – 2x)2 = 5 M1 2.1 or ቀ ହ ଶ− ௬ ଶቁ ଶ + 𝑦ଶ= 5. Must be an equation. 5x2 – 20x + 20 = 0 M1 1.1 simplifying to a three-term quadratic equation x = 2 [only] A1* 2.2a or 𝑦= 1 [only] [unique solution is] z = 2 + i A1dep 3.2a Alternative method (𝑥+ 2)ଶ+ (𝑦−4)ଶ= (𝑥−2)ଶ+ (𝑦−6)ଶ M1 squaring both sides of equations or inequality M1 expanding all four sets of brackets y = –2x + 5 A1 oe. Allow inequality. circle is x2 + y2 = 5 B1 allow inequality x2 + (5 – 2x)2 = 5 M1 or ቀ ହ ଶ− ௬ ଶቁ ଶ + 𝑦ଶ= 5. Must be an equation. 5x2 – 20x + 20 = 0 M1 simplifying to a three term quadratic equation x = 2 [only] A1* or 𝑦= 1 [only] [unique solution is] z = 2 + i A1dep [8] Y420/01 Mark Scheme June 2023 23 Question Answer Marks AO Guidance 14 (a) อ 𝑘 0 −1 −1 𝑘 2 2𝑘 2 3 อ= 𝑘(3𝑘−4) −1(−2 −2𝑘ଶ) M1 M1 1.1 1.1 considering correct determinant finding determinant using any row or column = 5k2 – 4k +2 A1 1.1 discriminant = 16 – 40 = – 24 < 0 M1 3.1a oe, e.g. completing the square, considering quadratic formula or showing no real roots so determinant is never zero planes always meet at a point A1 2.2a www. Both statements required. [5] 14 (b) 𝐌ିଵ= 1 5𝑘ଶ−4𝑘+ 2 ൭ 3𝑘−4 −2 𝑘 4𝑘+ 3 5𝑘 −2𝑘+ 1 −2 −2𝑘ଶ −2𝑘 𝑘ଶ ൱ M1 A1 M1 M1 1.1 1.1 1.1 3.1a at least 5 cofactors correct (need not be in matrix) all cofactors correct cofactor matrix transposed Multiplying by ଵ ୲୦ୣ୧୰ୢୣ୲ୣ୰୫୧୬ୟ୬୲ A1 1.1 correct inverse 𝐌ିଵ൭ 2 1 0 ൱= 1 5𝑘ଶ−4𝑘+ 2 ൭ 6𝑘−10 13𝑘+ 6 −4𝑘ଶ−2𝑘−4 ൱ M1 1.1 finding 𝐌ିଵ൭ 2 1 0 ൱ ቆ 6𝑘−10 5𝑘ଶ−4𝑘+ 2 , 13𝑘+ 6 5𝑘ଶ−4𝑘+ 2 , −4𝑘ଶ−2𝑘−4 5𝑘ଶ−4𝑘+ 2 ቇ A2,1,0 1.1, 1.1 Award A1 for one of x, y or z coordinates correct or all three correct FT their determinant [8] Y420/01 Mark Scheme June 2023 24 Question Answer Marks AO Guidance 15 DR 1 + 2𝑥−𝑥ଶ= −(𝑥ଶ−2𝑥) + 1 = −(𝑥−1)ଶ+ 2 M1 A1 3.1a 1.1 completing the square ൤arcsin 𝑥−1 𝑘 ൨ ଵ ଶ M1 2.1 = ൤arcsin 𝑥−1 √2 ൨ ଵ ଶ A1 2.1 = arcsin ൬1 √2 ൰−arcsin 0 = 𝜋 4 A1 2.2a substitution or evaluation of both limits must be seen Alternative method 1 + 2𝑥−𝑥ଶ= −(𝑥ଶ−2𝑥) + 1 = −(𝑥−1)ଶ+ 2 M1 A1 completing the square Let 𝑢= 𝑥−1 න 1 √2 −𝑢2 d𝑢 1 0 M1 complete substitution including limits = ൤arcsin 𝑢 √2 ൨ ଴ ଵ A1 = arcsin ൬1 √2 ൰−arcsin 0 = 𝜋 4 A1 substitution or evaluation of both limits must be seen [5] Y420/01 Mark Scheme June 2023 25 Question Answer Marks AO Guidance 16 Distance from point to plane = |ଶ×ସାଵ×ଵା଴×ଶ| √ଶమାଵమାଶమ M1* 3.1a = 3 units A1 1.1 Line is 𝐫= 3𝐢+ 𝐣−5𝐤+ 𝝀(2𝐢+ 𝑏𝐣+ 3𝐤) M1 3.1a 𝑨𝑷 ሬሬሬሬሬሬ⃗= 𝐢+ 5𝐤 A1 2.1 or −𝑨𝑷 ሬሬሬሬሬሬ⃗. Could be seen as part of distance calculation. 𝑨𝑷 ሬሬሬሬሬሬ⃗× 𝒅= (𝐢+ 5𝐤) × (2𝐢+ 𝑏𝐣+ 3𝐤) = −5𝑏𝐢+ 7𝐣+ 𝑏𝐤 M1 A1 2.1 2.1 or 𝒅× 𝑨𝑷 ሬሬሬሬሬሬ⃗ dist from P to line |−5𝑏𝐢+ 7𝐣+ 𝑏𝐤| |𝒅| = √26𝑏ଶ+ 49 √13 + 𝑏ଶ M1* A1 3.1a 1.1 so ଶ଺௕మାସଽ ଵଷା௕మ= 9 26𝑏ଶ+ 49 = 117 + 9𝑏ଶ M1dep 2.1 equating the two distances 𝑏= 2 A1 3.2a Alternative method 1 Distance from point to plane = |ଶ×ସାଵ×ଵା଴×ଶ| √ଶమାଵమାଶమ M1* = 3 units A1 Line is 𝐫= 3𝐢+ 𝐣−5𝐤+ 𝜆(2𝐢+ 𝑏𝐣+ 3𝐤) M1 (2𝜆−1)𝐢+ 𝑏𝜆𝐣+ (3𝜆−5)𝐤 A1 vector from P to a point on the line ൭ 2𝜆−1 𝑏𝜆 3𝜆−5 ൱∙൭ 2 𝑏 3 ൱= 0 M1* λ = 17 13 + 𝑏ଶ A1 dist from P to line ඨቀ2 ቀ ଵ଻ ଵଷା௕మቁ−1ቁ ଶ + ൬𝑏ቀ ଵ଻ ଵଷା௕మቁ൰ ଶ + ቀ3 ቀ ଵ଻ ଵଷା௕మቁ−5ቁ ଶ M1 substituting 𝜆 into ห𝑨𝑷 ሬሬሬሬሬሬ⃗ห = √26𝑏ସ+ 387𝑏ଶ+ 637 13 + 𝑏ଶ A1 simplified So ଶ଺௕రାଷ଼଻௕మା଺ଷ଻ (ଵଷା௕మ)మ = 9 M1dep equating the two distances 17𝑏ସ+ 153𝑏ଶ−884 = 0 𝑏= 2 A1 Y420/01 Mark Scheme June 2023 26 Question Answer Marks AO Guidance Alternative method 2 Distance from point to plane = |𝟐×𝟒ା𝟏×𝟏ା𝟎×𝟐| ඥ𝟐𝟐ା𝟏𝟐ା𝟐𝟐 M1* = 3 units A1 Line is 𝐫= 3𝐢+ 𝐣−5𝐤+ 𝜆(2𝐢+ 𝑏𝐣+ 3𝐤) M1 (2𝜆−1)𝐢+ 𝑏𝜆𝐣+ (3𝜆−5)𝐤 A1 vector from P to a point on the line ඥ(2𝜆−1)ଶ+ (𝑏𝜆)ଶ+ (3𝜆−5)ଶ M1* finding magnitude of this vector ඥ(13 + 𝑏ଶ)𝜆ଶ−34𝜆+ 26 A1 expanding and simplifying ඥ(13 + 𝑏ଶ)𝜆ଶ−34𝜆+ 26 = 3 M1dep setting distances equal (13 + 𝑏ଶ)𝜆ଶ−34𝜆+ 17 = 0 A1 correct quadratic equation = 0. So (−34)ଶ−4(13 + 𝑏ଶ)(17) = 0 M1 setting discriminant equal to 0 𝑏= 2 A1 Alternative method 3 Distance from point to plane = |ଶ×ସାଵ×ଵା଴×ଶ| √ଶమାଵమାଶమ M1* = 3 units A1 Line is 𝐫= 3𝐢+ 𝐣−5𝐤+ 𝜆(2𝐢+ 𝑏𝐣+ 3𝐤) M1 𝑨𝑷 ሬሬሬሬሬሬ⃗= 𝐢+ 5𝐤 A1 or −𝑨𝑷 ሬሬሬሬሬሬ⃗. Could be seen as part of distance calculation. ൭ 1 0 5 ൱∙൭ 2 𝑏 3 ൱= อ൭ 1 0 5 ൱ออ൭ 2 𝑏 3 ൱อcos 𝜃 M1 scalar product formula with 𝑨𝑷 ሬሬሬሬሬሬ⃗∙𝒅 to find a value for cos 𝜃 1 × 2 + 5 × 3 = ඥ1ଶ+ 5ଶඥ2ଶ+ 𝑏ଶ+ 3ଶcos 𝜃 M1 evaluating scalar product and magnitudes cos 𝜃= 17 √26√13 + 𝑏ଶ A1 expression for cos𝜃 √26 cos 𝜃= 17 √13 + 𝑏ଶ M1* expression for the distance from (3, 1, -5) to the foot of the perpendicular to the line 26 −ቆ 17 ඥ13 + 𝑏2ቇ 2 = 32 M1dep using their value correctly with Pythagoras oe to lead to a value for 𝑏 𝑏= 2 A1 [10] Y420/01 Mark Scheme June 2023 27 Question Answer Marks AO Guidance 17 (a) (i) dଶ𝑥 d𝑡ଶ= 𝑘d𝑥 d𝑡−𝑎d𝑦 d𝑡 M1* 3.1a differentiate ୢ௫ ୢ௧ wrt t = 𝑘 ୢ௫ ୢ௧−𝑎(𝑘𝑦−𝑏𝑥) M1dep 1.1 substitution for ୢ௬ ୢ௧ = 𝑘 ୢ௫ ୢ௧+ 𝑎𝑏𝑥−𝑘ቀ𝑘𝑥− ୢ௫ ୢ௧ቁ M1dep 3.1a substitution for 𝑦 dଶ𝑥 d𝑡ଶ−2𝑘d𝑥 d𝑡+ (𝑘ଶ−𝑛ଶ)𝑥= 0 A1 2.1 correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛ଶ. AE 𝑚ଶ−2𝑘𝑚+ (𝑘ଶ−𝑛ଶ) = 0 ⇒𝑚= 𝑘± 𝑛 M1 2.1 giving and solving their AE Hence GS is 𝑥= 𝐴𝑒(௞ା௡)௧+ 𝐵𝑒(௞ି௡)௧ A1 2.3 AG www Alternative method 𝑦= 𝑘 𝑎𝑥−1 𝑎 d𝑥 d𝑡⇒d𝑦 d𝑡= 𝑘 𝑎 d𝑥 d𝑡−1 𝑎 dଶ𝑥 d𝑡ଶ M1* differentiate 𝑦 wrt t 𝑘 𝑎 d𝑥 d𝑡−1 𝑎 dଶ𝑥 d𝑡ଶ= 𝑘𝑦−𝑏𝑥 M1dep substitution for ୢ௬ ୢ௧ 𝑘 𝑎 d𝑥 d𝑡−1 𝑎 dଶ𝑥 d𝑡ଶ= 𝑘൬𝑘 𝑎𝑥−1 𝑎 d𝑥 d𝑡൰−𝑏𝑥 M1dep substitution for 𝑦 dଶ𝑥 d𝑡ଶ−2𝑘d𝑥 d𝑡+ (𝑘ଶ−𝑛ଶ)𝑥= 0 A1 correct differential equation = 0. Could see 𝑎𝑏 instead of 𝑛ଶ. AE 𝑚ଶ−2𝑘𝑚+ (𝑘ଶ−𝑛ଶ) = 0 ⇒𝑚= 𝑘± 𝑛 M1 giving and solving their AE Hence GS is 𝑥= 𝐴e(௞ା௡)௧+ 𝐵e(௞ି௡)௧ A1 AG www [6] 17 (a) (ii) ୢ௫ ୢ௧= 𝐴(𝑘+ 𝑛)e(௞ା௡)௧+ 𝐵(𝑘−𝑛)e(௞ି௡)௧ M1 2.1 differentiation of 𝑥 must be seen 𝑦= ௡ ௔൫−𝐴e(௞ା௡)௧+ 𝐵e(௞ି௡)௧൯ A1 2.2a may not be factorised but must be simplified [2] Y420/01 Mark Scheme June 2023 28 Question Answer Marks AO Guidance 17 (b) (i) 𝑛= √0.004 = 0.02 ⇒𝑘+ 𝑛= 0.035 and 𝑘−𝑛 = −0.005 M1 1.1 all three values established 𝑥଴= 𝐴+ 𝐵, 𝑦଴= −1 2 (𝐴−𝐵) ⇒𝐴= 1 2 (𝑥଴−2𝑦଴), 𝐵= 1 2 (𝑥଴+ 2𝑦଴) M1 3.3 a method to find both A and B (ft their y) from expressions for 𝑥଴and 𝑦଴ 𝑥= 1 2 (𝑥଴−2𝑦଴)e଴.଴ଷହ௧+ 1 2 (𝑥଴+ 2𝑦଴)eି଴.଴଴ହ௧ A1 2.2a AG dependent upon both M1s Alternative method 𝑛= √0.004 = 0.02 ⇒𝑘+ 𝑛= 0.035, 𝑘−𝑛 = −0.005 M1 all three values established ୢ௫ ୢ௧= 0.015𝑥଴−0.04𝑦଴, ୢ௫ ୢ௧= 0.035𝐴−0.005𝐵 ⇒𝐴= 1 2 (𝑥଴−2𝑦଴), 𝐵= 1 2 (𝑥଴+ 2𝑦଴) M1 a method to find both A and B from two expressions for ୢ௫ ୢ௧ 𝑥= 1 2 (𝑥଴−2𝑦଴)e଴.଴ଷହ௧+ 1 2 (𝑥଴+ 2𝑦଴)eି଴.଴଴ହ௧ A1 AG dependent upon both M1s [3] 17 (b) (ii) 𝑦= 1 2 ൬−1 2 (𝑥଴−2𝑦଴)e଴.଴ଷହ௧+ 1 2 (𝑥଴+ 2𝑦଴)eି଴.଴଴ହ௧൰ = 1 4 (𝑥଴+ 2𝑦଴)eି଴.଴଴ହ௧−1 4 (𝑥଴−2𝑦଴)e଴.଴ଷହ௧ B1 2.2a AG first step or equivalent substitution must be seen [1] 17 (c) (i) 𝑥= −50e଴.଼଻ହ+ 550eି଴.ଵଶହ, 𝑦= 275eି଴.ଵଶହ+ 25e଴.଼଻ହ M1 1.1 for either may be implied by awrt 365 or 303 𝑥= 365, 𝑦= 302 A1 2.2b for both correct allow 𝑦= 303 [2] Y420/01 Mark Scheme June 2023 29 Question Answer Marks AO Guidance 17 (c) (ii) DR −50e଴.଴ଷହ௧+ 550eି଴.଴଴ହ௧= 275eି଴.଴଴ହ௧+ 25e଴.଴ଷହ௧ M1 3.1b equating 𝒙 and 𝒚 with 𝒙𝟎, 𝒚𝟎 substituted 275eି଴.଴଴ହ௧= 75e଴.଴ଷହ௧ M1 2.1 collecting like terms 0.04𝑡= ln ൬11 3 ൰ M1 3.1a oe. Taking logs for a single term in 𝒕. 𝑡= 32.5, so numbers equal after about 32 or 33 years A1 2.2a [4] 17 (c) (iii) x does become zero, as 550eି଴.଴଴ହ௧→0 as t increases, and −50e଴.଴ଷହ୲is always negative M1 A1 3.4 3.4 M1 for correct conclusion with some explanation A1 for complete explanation (e.g. x = 0 at t = 59.94...) y is the sum of two positive terms, so is never zero B1 3.5a reason must be given, e.g. 𝑦= 0 when 𝑡= 25 ln(−11) which is undefined or e଴.଴ସ௧= −11 has no solution. Implied by M1A1 without further working seen unless incorrect conclusion for 𝑦. [3] 17 (d) (i) 𝑥଴= 2𝑦଴ B1 3.5b oe. Subscripts must be seen. [1] 17 (d) (ii) 𝑥= 𝐶eି଴.଴଴ହ௧, 𝑦= ଵ ଶ𝐶eି଴.଴଴ହ௧ M1 3.3 oe; e.g. C is 𝟏 𝟐(𝒙𝟎+ 𝟐𝒚𝟎) or 𝟐𝒚𝟎or 𝒙𝟎 so both population numbers tend to zero A1 2.4 oe, e.g. both species disappear. Must be supported by explanation. [2] Need to get in touch? 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