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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