Y540/01 Mark Scheme June 2022 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. Y540/01 Mark Scheme June 2022 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. Y540/01 Mark Scheme June 2022 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. Y540/01 Mark Scheme June 2022 6 Question Answer Marks AO Guidance 1 (a) DR 2ln3 2ln3 e e cosh(2ln3) 2 1 1 41 9 2 9 9 − + =   = + =     M1 A1 1.1 2.1 Correct use of definition of cosh 𝑥𝑥 must be seen AG, must see either eln 9 and eln1 9 or 32 and 3−2 or 1 1 9 2 9   +     [2] (b) DR ( ) [ ] ( ) ( ) 2ln3 2 0 2ln3 0 3 sinh d cosh cosh(2ln3) cosh 0 41 1 9 32 cm 9 V x x x π π π π π = = = −   = −     = ∫ oe M1 A1 M1 A1 3.3 1.1 3.4 1.1 oe, intention to integrate 𝑦𝑦2. Condone missing 𝜋𝜋, ignore limits. For + cosh 𝑥𝑥. Ignore any reference to c Substituting correct limits and subtracting Ignore units [4] Y540/01 Mark Scheme June 2022 7 Question Answer Marks AO Guidance 2 (a) DetA =3 2 2 1 8 × −−× = B1 1.1 [1] (b) 3 2 1 1 2 8     −   B1 1.1 ft their Det A [1] (c) 1 1 A 2 x y − −     =         M1 1.1 Sight of their A-1 multiplied by 1 2 −       1 5 , 8 8 x y ⇒ = = A1 1.1 ft their 𝐀𝐀−1. Could be given as a vector. [2] (d) 3 2 1 1 2 2     −   oe B1 2.2a ft their 𝐀𝐀−1. Accept 4𝐀𝐀−1. [1] (e) ( ) DC = p B1 1.1 Must be a matrix; do not award for just p. 0 4 2 CD = 0 0 0 0 2 p p           M1 A1 1.1 1.1 for 3 × 3 matrix with at least one correct row or (non-0) column. SC B2 for correct matrices, but CD and DC interchanged or unspecified SC B1 for one correct matrix. [3] (f) Commutativity B1 1.2 Or the “commutative property”. Accept “commutative”. Allow also “non commutative” [1] Y540/01 Mark Scheme June 2022 8 Question Answer Marks AO Guidance 3 (a) DR 2 36 1 3 i 4 2 2 z z ± − = ⇒ = ± or ( ) 2 2 2 1 9 2 5 2 4 2 1 9 1 3 1 3 i i 2 4 2 2 2 2 z z z z z z z   − = −⇒ − + = −       − = − ⇒ − = ± ⇒ = ±     M1 A1 1.1 1.1 Using correct quadratic formula with correct substitutions Or completing the square (Root of negative number must be seen) Final answer [2] (b) (i) B1 B1 B1 1.1 1.1 2.2a Correct line Im(𝑧𝑧) = 1 Circle centre 2 Circle intersect on imaginary axis at ±i [3] (ii) Major segment (area below line and inside circle) shaded. B1 1.1 ft their circle and their horizontal line from (ii) [1] Y540/01 Mark Scheme June 2022 9 Question Answer Marks AO Guidance 3 (c) (i) DR 2 2 3 3 3 3 2 i 2 2 2 2 z     − = − ± = − + ±         M1 2.1 ft For calculating modulus of their |𝑧𝑧−2| for at least one of their values of z. 9 10 5 2 2   = < =       A1 2.2a AG. Must see clear statement of inequality (eg as shown or by values). Both roots If shown that |𝑧𝑧−2|2 < 5, must explain that |𝑧𝑧−2| < √5 follows from |𝑧𝑧−2| ≥0. [2] (ii) 1 3 i 2 2 − because Im 1 3 3 i 1 2 2 2   − = − <     B1 2.2a Or 1 3 5 1 3 5 2 i i 2i 2 2 2 2 2 2 − = < − − = ft their roots as long as their roots are conjugate pairs with imaginary part of magnitude greater than one. Could be explained in terms of loci (ie “because (the point representing) 1 2 − 3 2 i is closer to O than it is to (the point representing) 2i”) but must be consistent with their diagram after (d). Or “below the perpendicular bisector” or “below axis” [1] isw after acceptable answer (d) M1 A1 3.1a 1.1 For a complex conjugate pair. Approximate positions inside their circle above and below the line [2] Y540/01 Mark Scheme June 2022 10 Question Answer Marks AO Guidance 4 Direction of y-axis is ൭ 0 1 0 ൱ B1 3.1a Correct direction vector representation of the y-axis. 1 0 2 3 . 1 2 3 0 3    =      −    M1 1.1 Correct use of dot product with 1 2 3 3         −   and their direction vector for y-axis.soi ( ) ( ) 2 2 0 2 3 cos 1 1 2 3 3 2 3 4 or 30 6 θ π θ = × + + − = ⇒ = M1 A1 1.1 1.1 Correct use of dot product with their vectors to find cosine of angle soi . Condone eg.൫√3൯ 2 in place of ൫−√3൯ 2. Or 0 3 cos or 30 2 6 π ϕ θ π ϕ = − ⇒ = − = Accept 0.524c Mark the final answer SC B2 right answer only www [4] Y540/01 Mark Scheme June 2022 11 Question Answer Marks AO Guidance 5 (a) AG ( ) ( ) ( ) ( ) ( ) 2 2 3 2 4 6 3 4 2 2 2 3 1 3 2 3 9 9 xy r r r r xy x y r x y x y x y   = −     ⇒ = − = − ⇒ = − ⇒ + = − M1 A1 A1 1.1 1.1 Fully shown (AG). Use of double angle formula and 𝑦𝑦= 𝑟𝑟sin 𝜃𝜃, 𝑥𝑥= 𝑟𝑟cos 𝜃𝜃 oe Any correct form of the equation with 𝜃𝜃 eliminated. Fully shown [3] (b) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) 3 4 2 2 3 4 2 2 4 4 3 4 2 2 9 9 9 1 9 y x y x x y x y x y x y x y ⇒ + = − ⇒ + = − − = − − ⇒ + = − M1 A1 3.1a 2.4 Interchange of 𝑥𝑥 and 𝑦𝑦 Showing/explaining clearly that interchange of 𝑥𝑥 and 𝑦𝑦 leaves the equation unchanged. Alternative method sin 2 ൬1 4 𝜋𝜋+ 𝛼𝛼൰= sin 2 ൬1 4 𝜋𝜋−𝛼𝛼൰ or sin 2 ቀ 5 4 𝜋𝜋+ 𝛼𝛼ቁ= sin 2 ቀ 5 4 𝜋𝜋−𝛼𝛼ቁ M1 Showing that sin 2𝜃𝜃 is symmetrical about 𝜃𝜃= 1 4 𝜋𝜋 (or 𝜃𝜃= 5 4 𝜋𝜋). May be done graphically or using trig identities. So 𝜃𝜃= 1 4 𝜋𝜋 (and 𝜃𝜃= 5 4 𝜋𝜋 are lines of symmetry). So 𝑦𝑦= 𝑥𝑥 is a line of symmetry. A1 so 𝑟𝑟= 3(1 −sin 2𝜃𝜃) has lines of symmetry at 𝜃𝜃= 1 4 𝜋𝜋 and 𝜃𝜃= 5 4 𝜋𝜋 and the line 𝑦𝑦= 𝑥𝑥 has polar equation 𝜃𝜃= 1 4 𝜋𝜋 and 𝜃𝜃= 5 4 𝜋𝜋. Any mention of 𝜃𝜃= 5 4 𝜋𝜋 is unnecessary since it is implied by the 𝜃𝜃= 1 4 𝜋𝜋 case. [2] Y540/01 Mark Scheme June 2022 12 Question Answer Marks AO Guidance (c) DR ( ) ( ) ( ) 5 4 2 4 5 4 2 4 5 4 4 5 4 4 1 9 1 sin 2 d 2 1 9 1 2sin 2 sin 2 d 2 1 1 9 1 2sin 2 1 cos4 d 2 2 9 3 1 cos2 sin 4 2 2 8 9 3 5 27 0 0 2 2 4 4 4 A π π π π π π π π θ θ θ θ θ θ θ θ θ θ θ π π π = − = − +   = − + −       = + −         = − + + =         ∫ ∫ ∫ Other loop is the same B1 M1 M1* M1dep M1 A1 2.2a 1.1 1.1 1.1 3.1a 2.2a For identifying appropriate limits correctly 2 1 Use of r d 2 A θ = ∫ Expand correctly and use of ( ) 2 1 sin 2 1 cos4 2 θ θ = − Ignore limits Integrate Substituting correct limits and subtracting Answer for each loop which must be stated Alternative method for last 2 marks Using limits 0 to 2π ( ) 2 0 9 3 1 cos2 sin 4 2 2 8 9 27 3 0 0 for both loops 2 2 27 for one loop by symmetry in 4 y x π θ θ θ π π π   = + −     = + + = ⇒ = M1 A1 Substituting correct limits and subtracting Answer for each loop which must be stated [6] Y540/01 Mark Scheme June 2022 13 Question Answer Marks AO Guidance 6 Base case: d cosh cosh sinh d (2 1 1)cosh sinh y y x x x x x x x x x = ⇒ = + = × − + So true for n = 1 B1 2.5 Evidence of correct differentiation using product rule and a substitution n = 1 into RHS Assume result holds for n = k ( ) ( ) ( ) ( ) ( ) 2 1 2 1 2 2 2 1 2 1 2( 1) 1 2( 1) 1 d sinh 2 1 cosh d d sinh cosh 2 1 sinh d d cosh cosh sinh 2 1 cosh d sinh 2 1 cosh d i.e. sinh 2( 1) 1 cosh d k k k k k k k k y x x k x x y x x x k x x y x x x x k x x x x k x y x x k x x − − + + + − + − = + − ⇒ = + + − ⇒ = + + + − = + + = + + − So if true for n = k then also true for n = k + 1 But it is true for n= 1 and so is true generally M1* M1dep M1dep A1* A1dep 2.1 1.1 3.1a 2.2a 2.4 Differentiate using the product rule Differentiate second time using the product rule Must be equated to correct derivative form Dependent on previous A1 and the B1 [6] Y540/01 Mark Scheme June 2022 14 Question Answer Marks AO Guidance 7 (a) ( ) ( ) ( ) 2 2 2 2 18 9 9 x Ax x B x Cx D x + ≡ + + + + + B1 1.1 Correct multiplying out of fractions ( ) e.g. 0 9 18 2 1 10 10 19 1 10 10 19 10 19 1 3i 9 3 i 9 0 10 20 1 19 0 x B B x A B C D x A B C D B D D x D C C A A = ⇒ = ⇒ = = ⇒ + + + = = −⇒− + − + = ⇒ + = ⇒ = − = ⇒− + = ⇒ = ⇒ + −= ⇒ = M1 A1 1.1 1.1 Any substitutions to get a set of (at least) four simultaneous equations solvable for 𝐴𝐴, 𝐵𝐵, 𝐶𝐶 and 𝐷𝐷. Or equating coefficients which gives 𝐴𝐴+ 𝐶𝐶= 0, 𝐵𝐵+ 𝐷𝐷= 1, 9𝐴𝐴= 0, 9𝐵𝐵= 18. Any two coefficients correct. i.e. A = 0, B = 2, C = 0, D = −1 A1 1.1 All four coefficients correct. [4] SC B1 after M0 if one or more coefficients are correct. (b) DR ( ) 1 2 2 2 2 3 1 1 1 2 1 2 1 d tan 9 3 3 2 1 d 9 2 2 1 lim tan tan 1 3 3 3 2 2 1 lim lim tan 3 12 3 3 2 1 0 3 12 3 2 2 3 12 k k k x x c x x x x x x k k k k π π π π − ∞ − − →∞ − →∞ →∞   − = − − +   +     ⇒ −   +         = − − − −               = − + −     = − + − × = − ∫ ∫ M1 A1 M1 A1 A1 A1 1.1 1.1 1.1 2.1 2.1 1.1 integration including a tan-1 term ft their part (a) Use of limiting process on their integrated function. Ignore notation for limits for lim 𝑘𝑘→∞ቀ 1 𝑘𝑘ቁ= 0, or as 𝑘𝑘→∞, 1 𝑘𝑘→0, A0 for eg. 1 ∞= 0. lim 𝑘𝑘→∞ቀtan−1 𝑘𝑘 𝑐𝑐ቁ= 1 2 𝜋𝜋, or as 𝑘𝑘→∞, tan−1 𝑘𝑘 𝑐𝑐→ 1 2 𝜋𝜋, A0 for eg.tan−1∞ = 1 2 𝜋𝜋. In both cases must see some evidence of the limiting process. [6] Y540/01 Mark Scheme June 2022 15 Question Answer Marks AO Guidance 8 (a) (i) ( ) 2 2 . 2 2 2 2 d d 2 78 2 78 d d d d 2 156 78 2 d d d d 80 78 d d d d 80 78 d d x x x y t t x x x x k t t x x k t t x x k t t = − + −   = − + − + −     ⇒ = − + ⇒ + = M1 A1 3.3 1.1 Differentiate d d x t and substitute dy dt AG convincingly shown after second substitution. [2] (ii) 2 80 0 0, 80 λ λ λ + = ⇒ = − M1 1.1 Attempt to solve auxiliary equation for their DE Complementary function is 80 e t x A B − = + A1 2.2a Trial function is x = at (+ b) M1 3.1a Correct (or recovered) trial function for their CF. 2 2 80 d d , 0 d d 39 80 78 40 39 GS is e 40 t x x a t t a k a k x A B kt − = = ⇒ = ⇒ = ⇒ = + + A1 1.1 For GS Y540/01 Mark Scheme June 2022 16 Question Answer Marks AO Guidance Alternative method for first 4 marks: 𝑥𝑥̇ + 80𝑥𝑥= 78𝑘𝑘𝑘𝑘+ 𝑐𝑐 e∫80d𝑡𝑡(𝑥𝑥̇ + 80𝑥𝑥) = (78𝑘𝑘𝑘𝑘+ 𝑐𝑐)e∫80d𝑡𝑡 M1 Integrating 𝑥𝑥̈ + 80𝑥𝑥̇ = 78𝑘𝑘 wrt 𝑡𝑡 and then using integrating factor d d𝑡𝑡(e80𝑡𝑡𝑥𝑥) = (78𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 e80𝑡𝑡𝑥𝑥= න(78𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡d𝑡𝑡+ 𝑑𝑑 M1 Integrating wrt 𝑡𝑡 e80𝑡𝑡𝑥𝑥= (78𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 80 −න78𝑘𝑘 80 e80𝑡𝑡d𝑡𝑡 + 𝑑𝑑 M1 Integrating by parts (78kt + c may be separated) e80𝑡𝑡𝑥𝑥= (78𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 80 −39𝑘𝑘 3200 e80𝑡𝑡+ 𝑑𝑑 𝑥𝑥= 39 40 𝑘𝑘𝑘𝑘+ 𝑐𝑐 80 −39𝑘𝑘 3200 + 𝑑𝑑e−80𝑡𝑡 = 𝐴𝐴+ 𝐵𝐵e−80𝑡𝑡+ 39 40 𝑘𝑘𝑘𝑘 A1 GS of form 𝑥𝑥= 𝐴𝐴+ 𝐵𝐵e−80𝑡𝑡+ 39 40 𝑘𝑘𝑘𝑘 Y540/01 Mark Scheme June 2022 17 Question Answer Marks AO Guidance When t = 0, x = 0 ⇒ Α + Β = 0 M1 3.3 Using t = 0 and x = 0 in their GS to find an equation in A and B. ( ) 80 80 d 39 80 e d 40 d When 0, 0, 0 0 0 d 39 80 40 1 1 , 3200 3200 Particular solution: 1 e 3120 3200 t t x B k t x t x y k t k B k B k A k k x t − − = − + = = = ⇒ = + + ⇒ = − + ⇒ = − = = − + M1 A1 3.1a 3.3 Differentiating and using 𝑡𝑡= 0, 𝑥𝑥= 0, (𝑦𝑦= 0), 𝑥𝑥̇ = 𝑘𝑘 to find another equation in (𝐴𝐴 and) 𝐵𝐵. Could also use 𝑥𝑥̇ + 𝑦𝑦̇ = 𝑘𝑘 and x + y = kt. 3120 39 N.B. 3200 40 = Alternative method for this M mark: 78𝑦𝑦= 𝑥𝑥̇ + 2𝑥𝑥−𝑘𝑘 = −78𝐵𝐵e−80𝑡𝑡−1 40 𝑘𝑘+ 2𝐴𝐴+ 78 40 𝑘𝑘𝑘𝑘 When 𝑡𝑡= 0, 𝑥𝑥= 0, 𝑦𝑦= 0 0 = 2𝐴𝐴−78𝐵𝐵−1 40 𝑘𝑘 M1 Finding GS for 𝑦𝑦 and using 𝑥𝑥= 0, 𝑦𝑦= 0 to find another equation for 𝐴𝐴 and 𝐵𝐵 [7] (iii) ( ) 4000 4000 156000 When 50, 3200 because e 1 so 1 e 0 i.e. 48.75 292.5 for 6 i.e. 250 so fails safety food standards k t x x k k x − − = > < − > > > > > B1 3.4 Use of t = 50 must be seen to give a term in k Must reference k > 6. Condone the idea of ignoring exponential term as negligible Sight of 292.5 >250 earns the mark [1] Y540/01 Mark Scheme June 2022 18 Question Answer Marks AO Guidance (b) ( ) ( ) ( ) 41 1677 40.95... 41 e cosh 1677 sinh 1677 t a t b t − = < ⇒ + contains only negative exponentials when expanded M1 3.4 Turning function into exponentials. Could see eg e–0.0488t and e–81.95t. Condone error(s) in coefficients. Allow argument such as e-41t dominates so as t →∞, the exponential/hyperbolic parts → 0 so x → 20k < 240 < 250 since k < 12 so yes, food safety standards are met in the long run. A1 3.2a Argument must be complete and correct but could be based on sufficiently large values of t rather than formal limiting process. [2] SC B1 argument that 20k = 240 < 250 if 2nd term is assumed to tend to 0 (c) Pesticide is likely to be added periodically, eg, during the day; or depending on the weather/time of year; or only when it’s needed; or the amount of pesticide added changes as the amount of crop changes/grows or if it is subject to pest attack B1 3.5a The idea that regularly is not the same as continuously (and may not even mean “at a constant average rate”). Other sensible answers possible, but must be in context. [1] Y540/01 Mark Scheme June 2022 19 Question Answer Marks AO Guidance 9 Vertices of ABC satisfy z3 – 1 (= 0) B1 1.1 Or (𝑧𝑧−1) ቀ𝑧𝑧−e 2 3𝜋𝜋iቁቀ𝑧𝑧−e 4 3𝜋𝜋iቁ(= 0) Or all three values stated (Complex number represented by) M = −1 2 B1 3.1a Or one of 𝑀𝑀= 1 2 e𝜋𝜋i, 𝐿𝐿= 1 2 e 1 3𝜋𝜋i, 𝑁𝑁= 1 2 e−1 3𝜋𝜋i Vertices of LMN satisfy 8𝑧𝑧3 + 1 (= 0) M1 3.1a Attempt at polynomial relating to LMN. ቀ𝑧𝑧− 1 2 e𝜋𝜋iቁቀ𝑧𝑧− 1 2 e 1 3𝜋𝜋iቁቀ𝑧𝑧− 1 2 e−1 3𝜋𝜋iቁ suffices for this mark. (z3 – 1)(8𝑧𝑧3 + 1) (=0) M1 2.1 Attempt product of their two cubic factors. 8𝑧𝑧6 −7𝑧𝑧3 −1 = 0 A1 2.2a A0 without justification of 8z3 + 1 = 0. Must see = 0. [5] Alternative method ( ) 1 3 1 3 A 1 , B i , C i 2 2 2 2 1 3 1 3 1 L i , N i , M 4 4 4 4 2     − + − −                   + − −                 B1 B1 Finding A, B, C Finding L, M, N Quadratic satisfying B, C 2 1 0 z z + + = Quadratic satisfying L,N 2 4 2 1 0 z z − + = Quadratic satisfying A, M 2 2 1 0 z z − −= Eqn satisfying all 6 points ( )( )( ) 2 2 2 6 3 1 4 2 1 2 1 0 8 7 1 0 z z z z z z z z + + − + − − = ⇒ − −= M1 M1 A1 Combining to give a 6th degree polynomial Multiplying out some terms to give quadratics or cubics Must see = 0 Y540/01 Mark Scheme June 2022 20 Question Answer Marks AO Guidance Alternative methods By rotational symmetry Because ቀe 2 3𝜋𝜋iቁ 3 = 1 Calculation of vertices of L, M and N. Use of 1 + ω + ω2 = 0 (and ω3 = 1) to simplify eg (z – ½(ω + 1))(z – ½(ω2 + 1)) ))(z – ½(ω + ω2)) = (z + ½ω2))(z + ½ω) ))(z + ½) etc or to find sum/product of roots. Y540/01 Mark Scheme June 2022 21 APPENDIX Mark scheme for Q8(a)(ii) if candidate solves for 𝑦𝑦, rather than 𝑥𝑥. 8 (a) (ii) 𝑦𝑦̈ = 2𝑥𝑥̇ −78𝑦𝑦̇ = 2(−2𝑥𝑥+ 78𝑦𝑦+ 𝑘𝑘) −78𝑦𝑦̇ 𝑦𝑦̈ = −2(𝑦𝑦̇ + 78𝑦𝑦) + 156𝑦𝑦+ 2𝑘𝑘−78𝑦𝑦̇ 𝑦𝑦̈ + 80𝑦𝑦̇ = 2𝑘𝑘 𝜆𝜆2 + 80𝜆𝜆= 0 M1 3.1a Auxiliary equation for their DE 𝜆𝜆= 0, −80 Complementary function is 𝑦𝑦= 𝐴𝐴+ 𝐵𝐵e−80𝑡𝑡 M1 3.3 Trial function is y= 𝑎𝑎𝑎𝑎 M1 2.2a Correct trial function for their CF 𝑦𝑦̇ = 𝑎𝑎, 𝑦𝑦̈ = 0 So 80𝑎𝑎= 2𝑘𝑘⇔𝑎𝑎= 1 40 𝑘𝑘 General solution is 𝑦𝑦= 𝐴𝐴+ 𝐵𝐵e−80𝑡𝑡+ 1 40 𝑘𝑘𝑘𝑘 Note the GS for 𝑥𝑥 in the main mark scheme achieves A1 but here A1 is not awarded until GS for 𝑥𝑥 found. Alternative method 𝑦𝑦̇ + 80𝑦𝑦= 2𝑘𝑘𝑘𝑘+ 𝑐𝑐 e∫80d𝑡𝑡(𝑦𝑦̇ + 80𝑦𝑦) = (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e∫80d𝑡𝑡 M1 Integrating 𝑦𝑦̈ + 78𝑦𝑦̇ = 2𝑘𝑘 wrt 𝑡𝑡 and then using integrating factor d d𝑡𝑡(e80𝑡𝑡𝑦𝑦) = (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 e80𝑡𝑡𝑦𝑦= න(2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡d𝑡𝑡+ 𝑑𝑑 M1 Integrating wrt 𝑡𝑡 e80𝑡𝑡𝑦𝑦= (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 80 −න2 80 𝑘𝑘e80𝑡𝑡d𝑡𝑡+ 𝑑𝑑 M1 Integrating by parts e80𝑡𝑡𝑦𝑦= (2𝑘𝑘𝑘𝑘+ 𝑐𝑐)e80𝑡𝑡 80 − 1 3200 𝑘𝑘e80𝑡𝑡+ 𝑑𝑑 𝑦𝑦= 1 40 𝑘𝑘𝑘𝑘+ 1 80 𝑐𝑐− 1 3200 𝑘𝑘+ 𝑑𝑑e−80𝑡𝑡 = 𝐴𝐴+ 𝐵𝐵e−80𝑡𝑡+ 1 40 𝑘𝑘𝑘𝑘 Y540/01 Mark Scheme June 2022 22 𝑦𝑦̇ = −80𝐵𝐵e−80𝑡𝑡+ 1 40 𝑘𝑘 𝑥𝑥= 1 2 (𝑦𝑦̇ + 78𝑦𝑦) = 39𝐴𝐴−𝐵𝐵e−80𝑡𝑡+ 39 40 𝑘𝑘𝑘𝑘+ 1 80 𝑘𝑘 A1 1.1 GS of form 𝑥𝑥= 𝐶𝐶+ 𝐷𝐷e−80𝑡𝑡+ 39 40 𝑘𝑘𝑘𝑘 When 𝑡𝑡= 0, 𝑥𝑥= 0, 𝑦𝑦= 0, 𝑦𝑦̇ = 0 + 0 = 0 0 = 𝐴𝐴+ 𝐵𝐵 0 = 39𝐴𝐴−𝐵𝐵+ 1 80 𝑘𝑘 0 = −80𝐵𝐵+ 1 40 𝑘𝑘 M1 M1 3.1a 1.1 Using two of 𝑥𝑥= 𝑦𝑦= 𝑦𝑦̇ = 0 to find simultaneous equations for 𝐴𝐴 and 𝐵𝐵 (M1 for each equation). 𝐵𝐵= 1 3200 𝑘𝑘 , 𝐴𝐴= − 1 3200 𝑘𝑘 Particular solution for 𝑥𝑥 is 𝑥𝑥= −39 3200 𝑘𝑘− 1 3200 𝑘𝑘e−80𝑡𝑡+ 39 40 𝑘𝑘𝑘𝑘+ 1 80 𝑘𝑘 = 1 3200 𝑘𝑘(1 −e−80𝑡𝑡+ 3120𝑡𝑡) A1 3.3 [7] Y540/01 Mark Scheme June 2022 23 Another alternative method for Q8(a)(ii) for first 4 marks 8 (a) (i) ( ) 80 80 80 80 80 80 80 80 80 80 80 d d D.E. is 80 78 where d d d I.F. e e 80e 78 e d d e 78 e d 39 39 e e e 40 40 d 39 39 e e d 40 40 80 t t t t t t t t t t t y x y k y t t y y k t y k t k k y A y A x k kt A A x B t − − − + = = ⇒ + = ⇒ = ⇒ = + ⇒ = + ⇒ = + ⇒ = − + M1 A1 M1 A1 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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