Y436/01 Mark Scheme June 2022 3 2. Subject-specific Marking Instructions for AS/A Level Further 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. Y436/01 Mark Scheme June 2022 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. Y436/01 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. Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance 1 (a) (i) B1 B1 B1 [3] 1.1b 1.1b 1.1b Shape relative to axes, points on axes. Shape relative to axes, points on axes. Shape relative to axes, points on axes. Y436/01 Mark Scheme June 2022 7 Question Answer Marks AOs Guidance (ii) Closed curve B1 [1] 1.2 Bounded also allowed. ‘Maximum’ B0 ‘Continuous’ B0 (iii) When a = 1 the curve is the points (x, y) which satisfy 𝑥2 + 𝑦2 + 2𝑥𝑦= 1 ⇔(𝑥+ 𝑦)2 = 1 ⇔𝑥+ 𝑦= 1 or 𝑥+ 𝑦= −1 ⇔𝑦= 1 −𝑥 or 𝑦= −1 −𝑥 Therefore the two straight lines are y = - x + 1 and y = -x – 1. M1 A1 [2] 1.1b 2.2a Working must be shown for M1A1, otherwise M0A0. (b) (i) 2 2 2 1 x y axy + + = Setting cos and sin x r y r   = = gives 2 2 2 sin cos 1 r ar   + = . Therefore a polar form for the curve is 𝑟2 = 1 1 + 𝑎sin 2𝜃 (So 𝑟= (±)√ 1 1 + 𝑎sin 2𝜃) . M1 M1 A1 [3] 3.1a 1.1b 2.2a Allow correct formula for r2. Allow formula lwith +/- in front of the brackets. Allow formula with 2sin cos  rather than sin2for A1. Note entire curve is given for 0 2     (taking values of θ for which 1 sin2 0 a  +  Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance (ii) When a = 2 the area is ( ) ( ) 2 2 2 0 0 1 1 1 2 2 1 2sin(2 ) 3 3 ln 7 4 3 ln 7 4 3 12 12 r d d      = = + − − = +   M1 A1 [2] 1.1a 1.1b Need integration formula with correct limits. Answer given is equivalent to ( ) 3 ln 2 3 6 + . Note that exact form is required. √3 6 ln ( 2 −√3 7 −4√3 ) is also equivalent to the answer. Y436/01 Mark Scheme June 2022 9 (c) Solving 2 2 2 1 y mx x y axy = + + = as a pair of simultaneous equations gives solution pairs (x, y) of 2 2 2 2 2 2 2 2 2 1 2 1 , 2 1 2 1 and 2 1 2 1 , 2 1 2 1 m am m m am m am m am m am m m am m am m am   + + + +     + + + +     + + + +   − −   + + + +   These two points only exist if 2 2 1 0 m am + + , in which case they are distinct. 2 2 2 2 2 2 1 0 ( ) 1 0 ( ) 1 m am m a a m a a + +   + + −   +  − If 1 1 a −  then this is true for all values of m. If a = 1 this is true only when m a −, so when 1 m − If 1 a = −this is true only when m a −, so when 1 m  M1 M1 M1 A1 B1 B1 B1 1.1a 2.5 3.1a 3.1a 1.1a 1.1a 1.1a Sub mx as y to get 𝑥2 + (𝑚𝑥)2 + 2𝑎𝑥(𝑚𝑥) = 1 or 𝑥2 + 𝑚2𝑥2 + 2𝑎𝑚𝑥2 = 1 (**) or better Only x-coordinates required. 𝑥2(𝑚2 + 2𝑎𝑚+ 1) = 1 or better is needed for this M1 Commenting on existence of the points (soi). Allow use of discriminant For finding conditions for existence of the points (implies previous M1) Putting discriminant > 0 for (**) as a quadratic in x is another way to get to 𝑚2 + 2𝑎𝑚+ 1 > 0 Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance If a < −1 or a > 1 then require that 2 2 1 or 1 m a a m a a +  − + − − i.e 2 2 1 or 1 m a a m a a −+ − −− − M1 A1 [9] 3.2a 3.2a For case descriptor and at least one inequality (allow reasonable FT (e.g. sign errors or minor coefficient errors) from earlier work. For both inequalities. Note that if this final case appears with descriptor “a ≤ −1 or a ≥ 1” then the previous two B1 marks are implied. Must be strict inequalities for M1A1 but for non-strict award M1A0. Y436/01 Mark Scheme June 2022 11 Question Answer Marks AOs Guidance 2 (a) (i) 59 191 309 441 559 691 809 941 No need to state program example given on the right B1 [1] 2.5 for i in range(0,1000): if i*i%1000==481: print(i) Need all the values for the mark (ii) 481 is a quadratic residue modulo 1000 because there is an integer x such that 2 481 (mod 1000) x  . For example x = 59. B1 [1] 2.4 Need an actual example or reference to working in 2(a)(i) to get this mark. (iii) Quadratic residues modulo 11 are 2 2 2 2 2 1 1 (mod 11) 4 2 (mod 11) 9 3 (mod 11) 5 4 (mod 11) 3 5 (mod 11)      (squaring 6, 7, 8, 9, 10 give values already in this list) M1 A1 [2] 2.5 1.2 Accept appropriate code instead, must be stated SC1 for 1, 3, 4, 5 and 9. Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance (iv) Quadratic residues modulo 13 are 2 2 2 2 2 2 1 1 (mod 13) 4 2 (mod 13) 9 3 (mod 13) 3 4 (mod 13) 12 5 (mod 13) 10 6 (mod 13)       (squaring 7, 8, 9, 10, 11, 12 give values already in this list) M1 A1 [2] 2.5 1.2 Accept appropriate code instead SC1 for 1, 3, 4, 9, 10 and 12. (v) 2 2 2 2 ( ) 2 ( ( 2 )) n m n mn m m n n m − = − + = + − So 2 ( ) n m − and 2 m differ by a multiple of n. Therefore 2 2 ( ) (mod ) m n m n  − M1 A1 [2] 1.1a 2.1 Multiplying out the brackets Explanation or see n(n – 2m), and consideration modulo n. Allow e.g. 𝑛2 −2𝑚𝑛+ 𝑚2 = 0 ± 0 + 𝑚2 (mod 𝑛) Y436/01 Mark Scheme June 2022 13 Question Answer Marks AOs Guidance (vi) By part (v) 2 2 ( ) mod m p m p  − 2 2 2 2 2 2 1 ( 1) mod p 2 ( 2) mod p ... 1 1 mod p 2 2 p p p p  −  − − +              Therefore the number of quadratic residues mod p is at most 1 2 p − . Suppose n, m are such that 1 1 2 p m n −    and that 2 2 mod m n p = . Then p divides n2 – m2 = (n – m)(n + m) and so p divides either n – m or n + m (since p is prime). But 0 n m n m p  −  +  and so n – m = 0 Therefore m = n. Therefore each distinct integer between 1 and 1 2 p − has a distinct square mod p and so there are 1 2 p − quadratic residues mod p. M1 A1 M1 A1 [4] 3.1a 2.1 3.1a 2.1 Use of previous result with n replaced by p. Explanation that pairing up p – 1 values gives 1 2 p − pairs. m may be replaced by a specific value. Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance (b) (i) If 91 were prime then 90 2 1mod 91  by Fermat’s little theorem. However 90 2 64mod 91  M1 A1 [2] 2.2a 3.2a soi. Need to see idea that 290 – 1 is not a multiple of 91 implies 91 is not prime. Can use values of a other than 2 here. Y436/01 Mark Scheme June 2022 15 Question Answer Marks AOs Guidance (ii) 561 by the program on the right Appropriate structure program Loops with correct range dependent on m, n. Check for common divisors and divisors with if statement (and tracking greatest one found in case of coprime function) Fully correct program. M1 M1 M1 M1 M1 A1 [6] 3.3 3.3 3.4 2.1 2.1 2.3 Pseudo code accepted, condone lack of syntax, give reasonable BOD on possible transcription errors def isprime(n): prime=1 for i in range(2,n): if n%i==0: prime=0 return prime def coprime(m,n): coprime=1 for i in range(2,m+1): if n%i==0 and m%i==0: coprime=i return coprime for i in range(500,600): candidate = 1 if isprime(i)==1: candidate = 0 if isprime(i)==0: for j in range(1,i): if coprime(j,i)==1 and ((j**(i-1))%i)!=1: candidate = 0 if candidate == 1: print(i) M1 for prime test M1 for coprime test M1 for test over range (can be implied by any stated output in this range). M1 for rejecting primes M1 for testing equation for appropriate values. A1 for value 561. Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance 3 (a) (i) B1 B1 B1 [3] 1.2 1.1a 1.1a For sketch of 1 y x = + only needed for 0 x  . For region d 0 1 d y y x x    + For region d 0 1 d y y x x    + Need general shape correct (decreasing positive gradient and includes (0,1) and other coordinates approximately correct). Must be correct in upper right quadrant. Ignore anything in other quadrants. Only award if general shape correct (see above) Only award if general shape correct (see above) (ii) Solution is ( ) ( ) 2 2 2 2 2 2 1 y b x b x b = − + − + B1 [1] 1.2 oe Y436/01 Mark Scheme June 2022 17 Question Answer Marks AOs Guidance (iii) With y as in (ii) ( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 2 1 b x b dy dx b x b x b − + − = − + − + So 2 2 1 0 2 dy b x dx b − =  = − This has a solution with 0 x  if and only if 1 2 b   When 2 2 1 2 b x b − = − in the solution in (ii) 2 1 2 y b = − M1 A1 A1 B1 [4] 1.1a 1.1b 3.2a 1.1b Could also use 2 1 y x = + direct from (*). Numerator = 0 or equivalent. For x in terms of b from 0 dy dx = FT using their x Allow anything equivalent to this y and FT using their x (b) (i) Around a = 0. B1 [1] 1.1b Allow 0.5 0.5 a −   (ii) Around a = 0.6 B1 [1] 1.1b Allow 0.5 1 a   (has to be different to answer to (b)(i) to be awarded). Y436/01 Mark Scheme June 2022 (iii) = B1 B1 [2] 1.1b 1.1b Needs to have a maximum and negative second derivative throughout. Must extend to at least x = 4.5. No turning point and negative second derivative throughout. Must extend to at least x = 4.5. Y436/01 Mark Scheme June 2022 19 Question Answer Marks AOs Guidance (iv) One curve has a turning point (local max between 0 and 5), the other doesn’t. B1 [1] 1.2 Also allow ‘not strictly increasing’ or ‘intersects the x -axis’ or ‘stationary point’. Not ‘asymptote’. (c) (i) Insert 0 and 2 resp into cells A1 and B1. Using cells I1 and I2 for the values of h and a respectively other formulae then could be Cell C1: ==I$1*((B1^I$2)/(A1+1)-1/B1) Cell A2: =A1+I$1 Cell B2: =B1+C1 And copy down. B1 B1 B1 [3] 1.1a 3.1a 2.1 Give reasonable BOD on possible transcription errors and consider correct answers to 3(c)(ii), 3(c)(iii) as evidence of correct formulae in the spreadsheet. Cols for x and y and initialised x and y values. Allows for a and h to be varied (doesn’t need cell reference, can be implied) Formulae for xn+1 and yn+1. (ii) Spreadsheet gives that y is approximately 2.222487439 (to no. of dec places shown) when x = 3. B1 [1] 1.1b Allow any value between 2.2 and 2.23 inclusive. Y436/01 Mark Scheme June 2022 Question Answer Marks AOs Guidance (iii) Using the spreadsheet with h = 0.01 and a = -0.2 gives (layout as in c(i), rows 81 to 86 shown below) This suggest that the local maximum has x- coordinate between 0.82 and 0.84 and so would be a value which rounds to 0.8 to one decimal place. Smaller values of h (and use of other numerical methods confirms this) B1 M1 A1 [3] 1.1a 3.1a 2.2b For stating existence of increasing then decreasing y values (as x increases). For analysis with values of h less than or equal to 0.05. For correct answer of 0.8. Or stating existence of positive then negative values of dy dx (as x increases). Note that h = 0.01 to be sure of accuracy of correct answer (but not required to state this to award A1). Need to get in touch? 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