Y420/01 Mark Scheme June 2022 3 2. Subject-specific Marking Instructions for A Level Mathematics B (MEI) 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. Y420/01 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. 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. Y420/01 Mark Scheme June 2022 5 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 reasoning”, or the command words “Show” and “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 2022 6 Question Answer Marks AOs Guidance 1 (a) (𝑟+ 1)3 −𝑟3 = 𝑟3 + 3𝑟2 + 3𝑟+ 1 −𝑟3 = 3𝑟2 + 3𝑟+ 1 B1 1.1 Soi by M1A1 ∑(3𝑟2 + 3𝑟+ 1) 𝑛 𝑟=1 = ∑[(𝑟+ 1)3 −𝑟3] 𝑛 𝑟=1 = 23 −13 + 33 −23 + ⋯+ 𝑛3 −(𝑛−1)3 + (𝑛+ 1)3 −𝑛3 M1 2.5 Cannot use standard summation formulae for final 2 marks = (𝑛+ 1)3 −1 A1 2.2a isw [3] 1 (b) 3 ∑𝑟(𝑟+ 1) 𝑛 𝑟=1 + 𝑛= (𝑛+ 1)3 −1 M1* 2.5 ∑ 1 𝑛 𝑟=1 = 𝑛used and starting to rearrange ⇒∑𝑟(𝑟+ 1) 𝑛 𝑟=1 = 1 3 [(𝑛+ 1)3 −1 −𝑛] A1 3.1a = 1 3 (𝑛+ 1)[(𝑛+ 1)2 −1] M1dep * 1.1 Factorising with n or n + 1 correctly = 1 3 𝑛(𝑛+ 1)(𝑛+ 2) A1 2.2a Allow SC2 for correct solution using standard summation formulae. [4] 2 DR 2 2 3 3 1 1 d d 4 5 ( 2) 1 k k x x x x x = − + − +      M1 3.1a completing the square, (ignore limits) = [arctan( 𝑥−2)]3 𝑘 A1 1.1 [arctan( 𝑥−2)] or [arctan 𝑢] if 𝑢= 𝑥−2 (ignore limits) lim 𝑘→∞[arctan( 𝑘−2)] = 𝜋 2 = arctan( 𝑘−2) − 𝜋 4or 𝜋 2 − 𝜋 4 E1 B1 2.4 1.1 Clear limit argument ⇒integral = 𝜋 4 B1 [5] 2.2a Y420/01 Mark Scheme June 2022 7 Question Answer Marks AOs Guidance 3 DR 3 cosh 𝑥= 2 sinh2 𝑥= 2(cosh2 𝑥−1) M1 3.1a sinh2 𝑥= cosh2 𝑥−1 used ⇒2 cosh2 𝑥−3 cosh 𝑥−2 = 0 A1 1.1 ⇒(2 cosh 𝑥+ 1)(cosh 𝑥−2) = 0 M1 1.1 Solve their quadratic ⇒cosh 𝑥= −1 2 or 2 A1 1.1 Don't need to see –½ if factorisation or quadratic formula shown ⇒𝑥= ln(2 + √3) A1 1.1 or −ln(2 + √3) A1 1.1 or ln(2 −√3) [6] 4 (a) | 𝑚 2 1 0 1 −2 2 0 3 | = 3𝑚−2 × 4 + 1 × (−2) = 3𝑚−10 M1 A1 3.1a 1.1 finding determinant so 3𝑚−10 = 0 ⇒𝑚= 10 3 A1 1.1 Alternative solution ( 𝑚 0 2 ) . ( 2 1 0 ) × ( 1 −2 3 ) = ( 𝑚 0 2 ) . ( 3 −6 −5 ) = 3𝑚−10 [= 0] ⇒𝑚= 10 3 M1 A1 A1 3.1a 1.1 1.1 [3] 4 (b) det 𝑴= 4𝑘+ 3 B1 1.2 det( 𝑵𝑴) = det 𝑵× det 𝑴 M1 3.1a Soi ⇒(3𝑘+ 1)(4𝑘+ 3) = 2 M1 1.1 Equating to 2 𝑘= −1 or −1 12 A1 1.1 Alternative solution det(𝑁𝑀) = (𝑎𝑘−3𝑏)(𝑐+ 4𝑑) −(𝑎+ 4𝑏)(𝑐𝑘−3𝑑) (𝑎𝑘−3𝑏)(𝑐+ 4𝑑) −(𝑎+ 4𝑏)(𝑐𝑘−3𝑑) = 2 (3𝑘+ 1)(4𝑘+ 3) = 2 𝑘= −1 or −1 12 B1 M1 M1 A1 1.2 3.1a 1.1 1.1 [4] Y420/01 Mark Scheme June 2022 8 Question Answer Marks AOs Guidance 5 (a) M1 A1 1.1 1.1 symmetrical loop about the initial line correct shape with cusp at O [2] 5 (b) 2 2 2 0 1 (1 cos ) d 2 A a    = −  M1 1.1a correct integral and limits, condone missing 𝑑𝜃 limits can be soi by later work may see ∫𝑎2(1 −cos 𝜃)2 𝑑𝜃 𝜋 0 ( ) 2 2 0 1 1 1 2cos [1 cos2 ] d 2 2 a     = − + +  M1 M1 1.1 3.1a Expanding correctly substituting for cos2 = 1 4 𝑎2 [3𝜃−4 sin 𝜃+ 1 2 sin 2 𝜃] 0 2𝜋 B1 1.1 𝑘[3𝜃−4 sin 𝜃+ 1 2 sin 2 𝜃] = 3 2 𝜋𝑎2 A1cao 1.1 [5] 6 ( 2 0 −1 1) = ( 2 0 1 −2 1) so true when 𝑛= 1 B1 2.1 [Assume true for 𝑛= 𝑘] ( 2 0 −1 1) 𝑘+1 = ( 2𝑘 0 1 −2𝑘 1) ( 2 0 −1 1) M1 2.1 = ( 2𝑘+1 0 2 −2𝑘+1 −1 1) M1 2.1 Intermediate step seen = ( 2𝑘+1 0 1 −2𝑘+1 1) [so true for 𝑛= 𝑘+ 1] A1 2.3 As true for 𝑛= 1, and if true for 𝑛= 𝑘 then true for 𝑛= 𝑘+ 1, true for all n B1cao 2.4 Must receive all 4 previous marks for this to be awarded [5] O Y420/01 Mark Scheme June 2022 9 Question Answer Marks AOs Guidance 7 𝑥+ 1 (𝑥−1)(𝑥2 + 1) = 𝐴 𝑥−1 + 𝐵𝑥+ 𝐶 𝑥2 + 1 M1 2.1 correct partial fractions ⇒𝑥+ 1 = 𝐴(𝑥2 + 1) + (𝐵𝑥+ 𝐶)(𝑥−1) 𝑥= 1 ⇒2 = 2𝐴⇒𝐴= 1 A1 1.1 coefficient of 𝑥2:    0 = 𝐴+ 𝐵⇒𝐵= −1 A1 2.1 constants: 1 = 𝐴−𝐶⇒𝐶= 0 A1 2.1 ( ) 3 3 2 2 2 2 1 1 d d ( 1)( 1) 1 1 x x x x x x x x + = − − + − +      = [ln( 𝑥−1) −1 2 𝑙𝑛( 𝑥2 + 1)] 2 3 B1ft M1 A1ft 2.1 2.1 1.1 ln( 𝑥−1) 𝑘ln( 𝑥2 + 1) or 2 1 1 d 2 u x u u = +  𝑘= − 1 2 or − 1 2 ln 𝑢 = ln 2 −1 2 ln 1 0 + 1 2 ln 5 [−ln 1] = ln 2 − 1 2 ln 10 5 M1 1.1 Combining two of their logarithm terms correctly = ln 2 −1 2 ln 2 = 1 2 ln 2 A1 2.2a AG [9] Y420/01 Mark Scheme June 2022 10 Question Answer Marks AOs Guidance 8 (a) M1 A1 B1 B1 [4] 1.1 1.1 1.1 1.1 half line from 10 at 45 to Real axis implied by angle shown or meeting the Imaginary axis at 10 circle centre 3 + 6𝑖 circle meeting half line on Imaginary axis 8 (b) DR one point is 10i B1 1.1 𝑘2 = (3 −0)2 + (6 −10)2 M1* 3.1a ⇒𝑘= 5 A1 1.1 soi line 𝑥+ 𝑦= 10 (𝑥−3)2 + (10 −𝑥−6)2 = 25 M1de p* 3.1a solving 𝑥+ 𝑦= 10 and circle equation simultaneously ⇒2𝑥2 −14𝑥= 0 M1 1.1 Rearranging into a quadratic = 0 ⇒𝑥= 7,   𝑦= 3 other point is 7 + 3𝑖 A1 A1 1.1 3.2a Could see solutions as √58(cos 0.405 + 𝑖sin 0.405)or √58𝑒0.405𝑖 Alternative solution one point is 10i B1 eqn of perp from (3, 6) to chord is 𝑦= 𝑥+ 3 M1 oe, e.g. midpoint of chord has equal x and y displacements from (3, 6) solving with 𝑥+ 𝑦= 10 midpoint of chord is (3 1 2 ,   6 1 2) M1 A1 or by inspection oe other end of chord is (2 × 3 1 2 −0,   2 × 6 1 2 −10) M1  other point of intersection is (7, 3) A1 this represents 7 + 3𝑖 A1 Could see solutions as √58(cos 0.405 + 𝑖sin 0.405)or √58𝑒0.405𝑖 [7] Y420/01 Mark Scheme June 2022 11 Question Answer Marks AOs Guidance 9 (a) sinh 𝑘> −1 ⇒𝑘> sinh−1( −1) = ln(−1 + √2) M1 A1 2.1 2.2a May be in exponential form AG [2] 9 (b) (i) 𝑓′(𝑥) = cosh 𝑥 1 + sinh 𝑥 M1 A1 1.1 1.1 chain rule [2] 9 (b) (ii) 𝑓″(𝑥) = (1 + sinh 𝑥) sinh 𝑥−cosh2 𝑥 (1 + sinh 𝑥)2 M1 1.1 quotient or product rule 𝑓″(𝑥) = sinh 𝑥−1 (1 + sinh 𝑥)2 M1 A1 3.1a 1.1 𝑐𝑜𝑠ℎ2 𝑥−𝑠𝑖𝑛ℎ2 𝑥= 1 used [3] 9 (c) 𝑓(0) = 0,  𝑓′(0) = 1,  𝑓″(0) = −1 B1ft 1.1 soi 𝑓(𝑥) = 𝑥−1 2 𝑥2 M1 A1cao 1.1 1.1 Maclaurin expansion attempted, must see their values substituted in [3] 9 (d) ln( 1 + sinh( 0.1)) −0.095 ln( 1 + sinh( 0.1)) × 100 M1 1.1 = 0.48% A1 1.1 Allow –0.48% [2] Y420/01 Mark Scheme June 2022 12 Question Answer Marks AOs Guidance 10 (a) 𝛼× 2 𝛼× 𝛽× 3𝛽= 6 4 M1 3.1a ⇒𝛽= −1 2 A1 1.1 𝛼+ 2 𝛼+ 𝛽+ 3𝛽= −16 4 M1 3.1a ⇒𝛼2 + 2𝛼+ 2 = 0 A1 1.1 𝛼= −2 ± √4 −8 2 = −1 + 𝑖,   −1 −𝑖 M1 1.1 ` Solving their quadratic to find 𝛼 so roots are −1 + 𝑖,  −1 −𝑖,  − 1 2 ,  − 3 2 A1 3.2a [6] 10 (b) (−1 + 𝑖) (−1 −𝑖) + (−1 + 𝑖) (−1 2) + (−1 + 𝑖) (−3 2) +(−1 −𝑖) (−1 2) + (−1 −𝑖) (−3 2) + (−1 2) (−3 2) = 𝑎 4 M1 1.1 Allow in terms of 𝛼 and 𝛽, no missing terms ⇒𝑎= 27 A1 1.1 (−1 + 𝑖) (−1 −𝑖) (−1 2) + (−1 + 𝑖) (−1 −𝑖) (−3 2) +(−1 + 𝑖) (−1 2) (−3 2) + (−1 −𝑖) (−1 2) (−3 2) = −𝑏 4 M1 1.1 Allow in terms of 𝛼 and 𝛽, no missing terms ⇒𝑏= 22 A1 1.1 Alternative solution (𝑥+ 1 + 𝑖)(𝑥+ 1 −𝑖)(2𝑥+ 1)(2𝑥+ 3) = 0 M1 ⇒(𝑥2 + 2𝑥+ 2)(4𝑥2 + 8𝑥+ 3) = 0 M1 ⇒4𝑥4 + 16𝑥3 + 27𝑥2 + 22𝑥+ 6 = 0 A1 ⇒𝑎= 27,   𝑏= 22 A1 Alternative solution Y420/01 Mark Scheme June 2022 13 Question Answer Marks AOs Guidance 𝑓(−3 2) = 81 4 −54 + 9𝑎 4 −3𝑏 2 + 6 = 0 𝑓(−1 2) = 1 4 −2 + 𝑎 4 −𝑏 2 + 6 = 0 𝑎= 27 𝑏= 22 M1 M1 A1 A1 [4] Y420/01 Mark Scheme June 2022 14 Question Answer Marks AOs Guidance 11 (a) (i) M1 A1 [2] 1.1 1.1 on a circle centre O form an approximate equilateral triangle B and C must be labelled 11 (a) (ii) 𝑧1 + 𝑧2 + 𝑧3 = 𝑧1(1 + 𝑒2 𝑖𝜋/3 + 𝑒4 𝑖𝜋/3 ) = 𝑧1 1 −𝑒2 𝑖𝜋 1 −𝑒2 𝑖𝜋/3 = 0 M1 A1 2.1 2.2a sum of GP used or other correct method Alternative solution 𝑧1 + 𝑧2 + 𝑧3 = 𝑧1 [1 + cos 2𝜋 3 + cos 4𝜋 3 + 𝑖(sin 2𝜋 3 + sin 4𝜋 3 )] M1 = 𝑧1 [1 −1 2 −1 2 + 𝑖(√3 2 −√3 2 )] = 0 A1 [2] 11 (b) 𝑧3 = 8 (cos 1 2 𝜋+ 𝑖sin 1 2 𝜋) = 8𝑒𝑖𝜋/2 |𝑧| = 2 B1 B1 3.1a 1.1 Exponential or modulus argument form soi Soi 𝑧= 2𝑒𝑖𝜋/6,   2𝑒5𝑖𝜋/6,   2𝑒3𝑖𝜋/2 = √3 + 𝑖,  −√3 + 𝑖,  −2𝑖 B1B1 1.1 B1B0 if two out of three roots given [4] Im z2 z1 Re z3 A B C O Y420/01 Mark Scheme June 2022 15 Question Answer Marks AOs Guidance 12 d𝑦 d𝑥− 𝑥 4 −𝑥2 𝑦= 1 4 −𝑥2 B1 2.1 IF 2 d 4 e x x x −  − M1 2.1 Integral must come from an attempt to get 𝑑𝑦 𝑑𝑥 on its own = e 1 2 ln(4−𝑥2) M1 2.1 For integrating = √4 −𝑥2 A1 2.2a d d𝑥(√4 −𝑥2 𝑦) = 1 √4 −𝑥2 M1 2.1 Multiplying both sides by their IF 2 2 1 4 d 4 x y x x − = −   = arcsin 𝑥 2 + 𝑐 A1 1.1 𝑦= arcsin (1 2 𝑥) + 𝑐 √4 −𝑥2 M1 2.2a Rearranging into the form y=, equation must come from an attempt at integration having used IF and include c when 𝑥= 0,   𝑦= 1 ⇒𝑐= 2 M1 2.1 Substituting in x = 0, y = 1 to lead to a value of c 𝑦= arcsin (1 2 𝑥) + 2 √4 −𝑥2 A1 2.2a [9] Y420/01 Mark Scheme June 2022 16 Question Answer Marks AOs Guidance 13 (a) AB ⃗⃗⃗⃗⃗ = 6𝒊+ 4𝒋−2𝒌 B1 1.1 Any multiple soi angle between line and normal is  where𝑐𝑜𝑠𝜃= (3𝒊+2𝒋−𝒌).(𝒊−2𝒋) √14√5 = −1 √14√5 M1 A1 3.1a 1.1 𝜃= 96.9°, so angle with plane is 6.9 A1 1.1 7 or better 0.1198 rad [4] 13 (b) eqn of AB is [𝒓=]4𝒊−𝒌+ 𝜆(3𝒊+ 2𝒋−𝒌) B1ft 2.1 Soi Note may use another position vector as long as it is correct Substituting 𝑥= 4 + 3𝜆,   𝑦= 2𝜆,   𝑧= −1 −𝜆: M1 1.1 4 + 3𝜆−4𝜆= 5 ⇒𝜆= −1 meets 1 at (1, 2, 0) (say C) A1 2.2a substituting into 2 M1 3.1a or solving with 2 2 × 1 + 3 × (−2) −0 = −4, so C lies on 2 A1 2.2a [5] Alternative method eqn of AB is [𝒓=]4𝒊−𝒌+ 𝜆(3𝒊+ 2𝒋−𝒌) Substituting 𝑥= 4 + 3𝜆,   𝑦= 2𝜆,   𝑧= −1 −𝜆: 4 + 3𝜆−4𝜆= 5 ⇒𝜆= −1 Substituting 𝑥= 4 + 3𝜆,   𝑦= 2𝜆,   𝑧= −1 −𝜆: 2(4 + 3𝜆) + 3(2𝜆) −(−1 −𝜆) = −4 ⇒𝜆= −1 Finding equal 𝜆 for both planes (1,-2,0) B1 M1 M1 A1 A1 soi [5] 13 (c) (i) (𝒊−2𝒋) × (2𝒊+ 3𝒋−𝒌) = 2𝒊+ 𝒋+ 7𝒌 B1 [1] 1.1 13 (c) (ii) √54 B1 3.1a soi √5 × √14 sin 𝜃= √54 M1 1.1 ⇒𝜃= 61.4° A1 1.1 1.07 rad [3] Y420/01 Mark Scheme June 2022 17 Question Answer Marks AOs Guidance 13 (c) (iii) 2𝒊+ 𝒋+ 7𝒌 is direction vector of line of intersection B1ft 3.1a Soi by using in both numerator and denominator in correct method for d ( 3 2 −1 ) × ( 2 1 7 ) = ( 15 −23 −1 ) M1 1.1 𝑑= √152 + (−23)2 + (−1)2 √22 + 12 + 72 M1 1.1 = 3.74 A1cao 1.1 [4] Y420/01 Mark Scheme June 2022 18 Question Answer Marks AOs Guidance 14 (a) (3 −𝑒2𝑖𝜃) (3 −𝑒−2𝑖𝜃) = 9 −3(𝑒2𝑖𝜃+ 𝑒−2𝑖𝜃) + 1 = M1 1.1 For expanding correctly = 10 −6 cos 2 𝜃 A1 1.1 [2] 14 (b) let 𝑆= sin 𝜃+ 1 3 sin 3 𝜃+ 1 9 sin 5 𝜃+ 1 27 sin 7 𝜃+. .. and 𝐶= cos 𝜃+ 1 3 cos 3 𝜃+ 1 9 cos 5 𝜃+ 1 27 cos 7 𝜃+. .. 𝐶+ 𝑖𝑆= e𝑖𝜃+ 1 3 e3 𝑖𝜃+ 1 9 e5 𝑖𝜃+ 1 27 e7 𝑖𝜃+ . .. M1 2.1 At least 2 terms of C + iS soi by correct GP formula = 𝑒𝑖𝜃 1 −1 3 e2𝑖𝜃 M1 A1 2.1 2.2a sum to infinity of GP formula for their series (which must be geometric) oe = 3 e𝑖𝜃 3 −e2 𝑖𝜃 = 3 e𝑖𝜃( 3 −e−2 𝑖𝜃) 10 −6 cos 2 𝜃 M1* 3.1a multiply numerator and denominator by 3 −e−2𝑖𝜃 = 9(cos 𝜃+ 𝑖sin 𝜃) −3(cos 𝜃−𝑖sin 𝜃) 10 −6 cos 2 𝜃 M1de p* 2.1 e𝑖𝜃= cos 𝜃+ 𝑖sin 𝜃 used when denominator has been simplified to a real expression 𝑆= 9 sin 𝜃+ 3 sin 𝜃 10 −6 cos 2 𝜃= 6 sin 𝜃 5 −3 cos 2 𝜃 A1 2.2a AG [6] Y420/01 Mark Scheme June 2022 19 Question Answer Marks AOs Guidance 15 (a) (i) 𝑚𝑑2𝑥 d𝑡2 = −2𝑚𝑥⇒d2𝑥 d𝑡2 + 2𝑥= 0 B1 [1] 3.1b AG 15 (a) (ii) Simple harmonic motion B1 [1] 3.3 15 (a) (iii) 2𝜋 √2 = √2𝜋 (s) B1 [1] 1.2 Accept anything wrt 4.4 15 (a) (iv) General solution: 𝑥= 𝐴cos √2 𝑡+ 𝐵sin √2 𝑡 B1 1.1 𝑡= 0,   𝑥= 2 ⇒𝐴= 2 B1 3.3 Must come from correct GS d𝑥 d𝑡= −√2𝐴sin √2 𝑡+ √2𝐵cos √2 𝑡 M1 1.1 Must be differentiating a function in terms of cos and sin 𝑡= 0,  d𝑥 d𝑡= 1 ⇒𝐵= 1 √2 = √2 2 so 𝑥= 2 cos √2 𝑡+ √2 2 sin √2 𝑡 A1 [4] 3.3 15 (a) (v) Amplitude = √22 + (1 2 √2) 2 M1 3.4 = 3√2 2 (m) A1 1.1 Accept anything wrt 2.1 [2] 15 (b) (i) 𝑚 d2𝑥 d𝑡2 = −2𝑚𝑥−2𝑚 d𝑥 d𝑡 ⇒d2𝑥 d𝑡2 + 2 d𝑥 d𝑡+ 2𝑥= 0 B1 [1] 3.1b AG 15 (b) (ii) Underdamped as 22 −4 × 1 × 2 < 0 B1 3.5b Stating complex roots of auxillary equation is acceptable (−1 ± 𝑖) [1] Y420/01 Mark Scheme June 2022 20 Question Answer Marks AOs Guidance 15 (b) (iii) Auxiliary equation: 𝜆2 + 2𝜆+ 2 = 0 M1 1.2 𝜆= −2 ± √−4 2 = −1 ± 𝑖 A1 1.1 soi General solution: 𝑥= 𝑒−𝑡( 𝐴cos 𝑡+ 𝐵sin 𝑡) A1 1.1 [3] 15 (c) (i) Particular integral: 𝑥= 𝐶𝑐𝑜𝑠2 𝑡+ 𝐷𝑠𝑖𝑛2 𝑡 M1 2.1 −4𝐶+ 4𝐷+ 2𝐶= 2,  −4𝐷−4𝐶+ 2𝐷= 0 M1 3.3 ⇒𝐶= −0.2,   𝐷= 0.4 A1 2.2a 𝑥= 𝑒−𝑡( 𝐴cos 𝑡+ 𝐵sin 𝑡) −0.2 cos 2 𝑡+ 0.4 sin 2 𝑡 𝑡= 0,   𝑥= 2 ⇒𝐴= 2.2 B1 3.3 Has to come from a correct CF d𝑥 d𝑡= −𝑒−𝑡( 𝐴cos 𝑡+ 𝐵sin 𝑡) + 𝑒−𝑡( −𝐴sin 𝑡 + 𝐵cos 𝑡) +0.4 sin 2 𝑡+ 0.8 cos 2 𝑡 M1* 2.1 Must include use of product rule from x = CF + PI 𝑡= 0, d𝑥 d𝑡= 1 ⇒1 = −𝐴+ 𝐵+ 0.8 M1de p* 3.3 For substitution to lead to an equation in A and B ⇒𝐵= 2.4 𝑥= e−𝑡( 2.2 cos 𝑡+ 2.4 sin 𝑡) −0.2 cos 2 𝑡 + 0.4 sin 2 𝑡 A1cao 2.2a [7] 15 (c) (ii) As 𝑡→∞,   𝑥→−0.2 cos 2 𝑡+ 0.4 sin 2 𝑡 M1 3.5a Dependent on a function containing 𝑒−𝑘𝑡and cos 𝑝𝑡 and sin 𝑝𝑡 [this is SHM] with period 2𝜋 2 = 𝜋≈3.14 𝑠 A1 3.5a [2] Need to get in touch? 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