Y540/01 Mark Scheme June 2023 5 5. 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. Y540/01 Mark Scheme June 2023 6 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. Y540/01 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. 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 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 2023 8 Question Answer Marks Guidance 1 DR ( ) ( ) 50 50 50 2 2 3 1 1 1 2 2 16 16 16 1 50 51 101 50 51 6 4 686800 1625625 938825 r r r r − = − =    −  = − = −    M1 M1 A1 Separating and using correct formulae for 3 2 and r r   Substituting must be seen Including all notation correct (Sigmas do not need limits) Alternative method: ( ) ( ) ( ) 50 50 2 2 3 1 1 1 2 2 2 2 16 16 16 1 ( 1)(2 1) ( 1) 6 4 8 1 ( 1) (2 1) ( 1) 3 4 ( 1) 32 61 3 12 50 51 32 61 50 3 50 12 938825 n r r r r n n n n n n n n n n n n n n − = − = + + − +   = + + − +     + = + −  = +  − = −    M1 M1 A1 Separating and using the correct formulae Substituting anywhere in the algebra [3] Y540/01 Mark Scheme June 2023 9 Question Answer Marks Guidance 2 (a) DR ( ) ( ) 4 3 4 3 2 4 2 1 4 1 ...... 0 4 6 4 1 ....... 3 2 0 w w w w w w w w − + − + =  − + − + +  + + = M1 M1 A1 Substitute z = w – 1 Expanding with at least (w1)4 seen Convincingly shown AG. Must include = 0 on last line. Brackets must be fully expanded in working or evidence of collection of like terms A0 if variable used is not w. Alternative method: ( ) ( ) 4 2 4 3 4 3 2 1 3 1 ...... 0 4 ....... 4 9 10 6 0 0 z z z z z z z z + + + + =  + +  + + + + = = M1 A1 Substitute w = z + 1 into end result Alternative method using symmetry of roots 4, 9. 10, 6 ( 1) 4 4 4 0 ( 1)( 1) 3 6 9 12 6 3           = − = = − = + = + = −+ = + + = + + = − + =         ( )( )( ) ( )( )( )( ) For 1 1 1 0 and 1 1 1 1 2        + + + = + + + + =  B1 B1 B1 For sums from original equation and finding the sum of the new roots For showing convincingly the sum of new roots in pairs For the last two [3] (b) 2 1, 2 w = −− M1 Solving quadratic equation in w2 (or using their variable) i, 2i w  =  M1 Square rooting their w2, including ±, as long as their w2 not both non- negative and real. z= i 1, 2i 1  −  − A1 cao Answers with no working is 0 [3] Y540/01 Mark Scheme June 2023 10 Question Answer Marks Guidance 3 (a) ( ) ( ) 2 2 1 1 3 3 12 3 2 2 r = − + = = B1 AG so must show use of 2 2 z a b = + 3 arctan 3 6 5 6 6        − = −        = − + = B1 AG. Or 𝜃= 𝜋−arctan (√3 3 ), may be indicated on a diagram, but clear reasoning must be shown (eg.finding complementary angle, or use of Pythagoras’ theorem and then arcsin or arccos) Alternative method: 5 i 6 5 5 3e 3 cos isin 6 6 3 1 3i 3 3 i 2 2 2          = +               − = − + =       M1 A1 AG Clearly shown [2] (b) ( ) 1 1 5 5 5 2 9 3 3 3 r   = = =     B1 For 𝑟= √3 oe (including 1.73....) ( ) 1 5 2 5 12 for 0,1,2,3,4 5 6 30 r r r       = + = + =     M1 For their 5 6 𝜋+ 2𝜋𝑛 from (a) divided by 5 (either in terms of n, or for at least two values of n). 1 17 29 41 53 i i i i i 6 30 30 30 30 3e , 3e , 3e , 3e , 3e z       = A1 Allow √3e 1 30(5+12𝑛)𝜋i for n = 0, 1, 2, 3, 4. Accept only r = ( ) 1 2 3 or 3 For last two marks, If M0 then SC B1 for all five roots [3] Y540/01 Mark Scheme June 2023 11 Question Answer Marks Guidance 4 (a) 0 1 0 1 1 0 1 0 1 0 0 1 −      = =      −      BA i.e. a reflection in the x-axis M1 A1 Multiplication in correct order Or reflection in the line y = 0 Answer with no working B2 Alternative method: A represents a rotation anti-clockwise of 900 B represents a reflection in the line y = x Taken one after the other gives a reflection in the x- axis M1 A1 [2] (b) TA is a rotation 90 degrees (anti-clockwise about O) B1 (423 has remainder 3 when divided by 4) so A423 = A3 M1 For A4 = I or TA repeated four times is a 360 degree rotation. Condone clockwise instead of anticlockwise for TA so notes that 423 is divisible by 4 with remainder 3 so A423 = A3 So 423 3 0 1 1 0   = =   −   A A A1 As A3 represents a 270 degrees rotation anti-clockwise (or 90 degrees clockwise) or by direct calculation of A3. [3] (c) (i) 1 Det 6 C = M1 1 3 × 1 2 or 1 6 seen. Or area of ellipse = πab = 𝜋× 2 × 3 1 Area of R' = 36 6 6     = A1 cao [2] Y540/01 Mark Scheme June 2023 12 Question Answer Marks Guidance 4 (c) (ii) . M1 A1 A1 A1 Correct shape, (approximately elliptical, possibly identified by scales, closed, radius on horizontal axis > radius on vertical axis ) Intercepts labelled at x = ± 3 and y = ± 2(Give A1only rather than A2 if only positive intercepts are labelled on both) Give A1 if x and y axes interchanged. Correctly shaded or labelled R’ and everything else correct. [4] (0,2) (3,0) (0,−2) (−3, 0) Y540/01 Mark Scheme June 2023 13 Question Answer Marks Guidance 5 (a) ( ) 2 Auxiliary equation: 2 5 0 1 2i e cos2 sin 2 x n n n y A x B x − + =  =   = + oe B1 B1 Correct roots of auxiliary equation. ft complex k only. Or 𝑦= 𝑅e𝑥cos(2𝑥+ 𝜑) or 𝑦= 𝑅e𝑥sin(2𝑥+ 𝜑) Or 𝑦= 𝐴e(2i+1)𝑥+ 𝐵e(−2i+1)𝑥 Final equation must be y = f(x) [2] (b) 2 Trial function: y ax bx c = + + B1 Allow any extraneous terms in the trial function (eg.dx3) as long as d shown to be zero, , , 0 a b c  2 2 2 2 ' 2 , '' 2 2 2(2 ) 5( ) 4 5 5 ( 4 5 ) 5 2 2 4 5 y ax b y a a ax b ax bx c x x ax x a b c b a x x  = + =  − + + + +  −  + − + + − +  − M1 Differentiates their trial function to find 𝑑𝑦 𝑑𝑥 and 𝑑2𝑦 𝑑𝑥2 and substitutes 2 1, 0, 5 a b c  = − = = A1 ( ) 2 2 GS: e cos2 sin 2 5 x y A x B x x  = + − + A1ft ft their particular integral, and their CF from (a) (dependent on CF containing exactly two arbitrary constants) [4] Y540/01 Mark Scheme June 2023 14 Question Answer Marks Guidance 6 DR ( ) 0.5 0 1 Mean value = 20 20tanh(1.44 ) d 0.5 t t −  B1 For using the definition of the mean value of p wrt t, correct limits and 1/0.5. ( ) ( ) ( ) 0.5 1.44 1.44 0 0.72 0.72 40 40 ln e e 1.44 40 20 ln e e ln 2 1.44 t t t − −   = − +       = − + −     M1 A1 For ∫tanh 1.44𝑡d𝑡= 𝑘ln | e1.44𝑡+ e−1.44𝑡|(+𝑐) k can = 1 Or ∫tanh 1.44𝑡d𝑡= 𝑘ln | cosh 1.44𝑡|(+𝑐) May see integration by substitution, eg.𝑢= e1.44𝑡+ e−1.44𝑡or 𝑢= cosh 1.44𝑡. If so, award this mark for 𝑘ln |𝑢| seen For fully correct integration of tanh 1.44𝑡. Either for ∫tanh 1.44𝑡d𝑡= 1 1.44 ln | e1.44𝑡+ e−1.44𝑡|(+𝑐) or ∫tanh 1.44𝑡d𝑡= 1 1.44 ln | cosh 1.44𝑡|(+𝑐) or ∫tanh 1.44𝑡d𝑡= 1 1.44 ln |𝑢| (+𝑐) with u as above. Condone missing modulus. 40 2.5412 20 ln 13.35(W) 1.44 2   = − =     A1 cao, with clear working. [4] Y540/01 Mark Scheme June 2023 15 Question Answer Marks Guidance 7 (a) (i) e.g. sign of d dt y is +ve because the force is in the same direction of motion B1 Convincingly shown [1] (ii) 2 2 2 2 2 1 d 5cosh 0 3 5 2 d 3 d 6(m ) d x t x s t − = −− = −  = − M1 A1 Substitutes t = 0 and d 5 d x t Or 6 towards O [2] (b) (i) ( ) Maclaurin: f (0) f '(0) ... d When 0, 6 f (0) 6, f '(0) 5 d 6 5 .... x t x t x t x t = + +   = = = = = = = −      = − B1 First two terms ofMaclaurin’s formula, possibly in generalised form, must be seen AG Convincingly shown. [1] (ii) 3rd term of Maclaurin is 2 f ''(0) 2! t With (from (a)(ii)) f ''(0) 6 = − So 3rd term is ( ) 2 2 6 3 2! t t − = − B1 AG Convincingly shown [1] Y540/01 Mark Scheme June 2023 16 Question Answer Marks Guidance 7 (b) (iii) 2 3 2 3 d d 1 d 5cosh 5( 1)sinh 0.5 d d 2 d x x x t t t t t t  + + − + = M1 A1 For ( ) d cosh sinh d t t t = and product rule attempted Fully correct differentiation 3 3 1 When 0, 5 0 2.5 6 f '''(0) 2 f '''(0) 3 1 4th term = f '''(0) 3! 2 t t t = + + − =  =  = M1 A1 Substitute t = 0 AG. Allow embedded answer. [4] (c) (i) 2 3 1 6 5 3 2 When 0.25, 6 1.25 0.1875 0.0078 4.570... Distance travelled 6 4.570 1.430... x t t t t x = − − +  = = − − −   −  So suitable as value close B1 Substitutes t = 0.25 to obtain an approximation and correct conclusion [1] (ii) e.g. more terms may be required Higher terms may be large The candidate calculates the term in t4 (11t4/12) and indicates that this term is large for values of t > 1. B1 Allow any correct explanation that explains/implies that for t > 1 some of the higher power terms are large and so non-negligible. f(10) = 156 is too large is not enough [1] Y540/01 Mark Scheme June 2023 17 Question Answer Marks Guidance 8 (a) 2 1 2 , 1 2 3 PQ PR         = − =         − −     B1 Both correct or other way round ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 1 2 . 1 2 3 cos 2 2 2 1 1 3 6 12 11 6 8 22 2 sin 1 2 22 11 11 11 12 11         −       − −    = + − + −  + + − =        = − = = =         M1 A1 Correct use of scalar product including correct method for magnitudes and dot product AG All working must be using exact forms Alternative for M1 A1 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 1 2 3 sin 2 2 2 1 1 3 2 4 1 1 4 6 4 2 22 11 12 11 132 22        −        − −    = + − + −  + + −      = = = = M1 A1 Correct use of vector product to find |sinθ| including correct method for magnitudes, and correct method for ( 2 −2 −2 ) × ( 1 1 −3 ) soi. Condone sinθ instead of |sinθ| AG. [3] Y540/01 Mark Scheme June 2023 18 Question Answer Marks Guidance 8 (a) Alternative method: 11, 11, 12 12 11 11 3 Cosine rule cos = 11 2 11 12 3 8 2 sin 1 22 11 11 11 QR RP PQ   = = = + −  =    = − = = Or: 11, 11, 12 QR RP PQ = = = Drop perpendicular from R to QP at M 2 1 11 12 8 2 8 2 sin 22 11 11 RM    = − =      = = B1 M1 A1 B1 M1 A1 Sides of triangle Cosine rule Sides of triangle Recognises an isosceles triangle so uses a median line [3] Y540/01 Mark Scheme June 2023 19 Question Answer Marks Guidance 8 (b) 2 1 2 2 1 4 1 2 3 1 PQ PR          = −  =         − −     2 Sub for a point 5 2 5 x y z d d x y z  + + =  =  + + = B1 M1 A1 BC or any other relevant vector product. Can be awarded if seen in (a) Attempt to use r.n = d to form linear equation. ft their vector product. Substitutes coordinates of P, Q or R to find d. (multiples accepted) [3] (c) 5 2 3 . 1 5 1 1 7 2 6 1 1 D    −     −   = =      M1 A1 Uses the formula given in formula book, or any other complete method for the shortest distance, ft their 𝑃𝑄 ⃗⃗⃗⃗⃗ × 𝑃𝑅 ⃗⃗⃗⃗⃗ . Correct shortest distance. 1 1 sin 3 2 1 1 2 7 2 3 11 22 3 2 11 6 14 3 V PQ PR D    =       =      = M1 A1 Uses 1 2 |𝑃𝑄 ⃗⃗⃗⃗⃗ ||𝑃𝑅 ⃗⃗⃗⃗⃗ | sin 𝜃oe, multiplied by their D/3, or 1 2 |( 2 −2 −2 ) × ( 1 1 −3 )| multiplied by their D/3, but must indicate that 1 2 |( 2 −2 −2 ) × ( 1 1 −3 )| is the area of the base. AG. [4] Y540/01 Mark Scheme June 2023 20 Question Answer Marks Guidance 8 (d) ( ) 2 2 2 2 2 cos 0 sin 5 2 2 0 1 0 3 ... sin 0 cos 1 ... 5cos sin 2 2 sin 5cos 2 2 1 cos 25cos 20 2 cos 8 26cos 20 2 cos 7 0 2 7 2 cos , 2 26                        =           − −       − =  = − − = − +  − + =  = M1 M1 A1 For rotation matrix multiplied by 𝑂𝑅 ⃗⃗⃗⃗⃗ or 𝑂𝑆 ⃗⃗⃗⃗⃗ . For a correct step to form quadratic equation in sin 𝜙 or cos 𝜙 only. For reference: 2 26sin 4 2sin 17 0   + − = Solves quadratic equation in sin 𝜙 or cos 𝜙. (Exact answers required) 2 17 2 sin 5cos 2 2 , 2 26   = − = − M1 Uses their sin 𝜙 or cos 𝜙 with 5 cos 𝜙−sin 𝜙= 2√2to find cos 𝜙 or sin 𝜙 respectively, (even if only one root) cos 0 sin 1 cos 0 1 0 3 3 sin 0 cos 0 sin 7 2 2 26 2 3 , 3 2 17 2 2 26 2 2 7 2 17 2 i.e. ' ,3, or ,3, 2 2 26 26 R               =         − −                              −             = −             A1 or fromsin 𝜙= √1 −cos2 𝜙 or cos 𝜙= √1 −sin2 𝜙 (condone inclusion of ±), or repeats previous method and multiplies out the matrices. For both, and no others. Accept given as 𝑂𝑅′ ⃗⃗⃗⃗⃗⃗⃗ . If M0M0A0M0A0, SCB1 for any R′with y-coordinate = 3, and no other y- coordinates. [5] Y540/01 Mark Scheme June 2023 21 Question Answer Marks Guidance 8 (d) Alternative method for first 3 marks: ( ) ( ) ( ) 2 5cos sin 2 2 5cos sin 26 cos( ) 2 2 1 1 5 where tan sin ,cos 5 26 26 26 sin( ) 26 2 2 3 2 cos cos cos( )cos sin( )sin 2 2 5 3 2 1 26 26 26 26 2 7 2 cos , 2 26                       − =  − = + = =  = + =  − =  = + − = + + + =     = M1 M1 A1 For rotation matrix multiplied by 𝑂𝑅 ⃗⃗⃗⃗⃗ or 𝑂𝑆 ⃗⃗⃗⃗⃗ . Expressing in the form 𝑅cos(𝜙+ 𝛼) or 𝑅sin(𝜙+ 𝛼) oe. Note that 5 cos 𝜙−sin 𝜙 = √26 sin (𝜙+ arctan (−1 5) + 𝜋) Solves for sin 𝜙 or cos 𝜙. For reference,√2 2 ≈0.707, − 17√2 26 ≈ −0.925 and 7√2 26 ≈0.381. [3] Y540/01 Mark Scheme June 2023 22 Question Answer Marks Guidance 9 (a) DR ( ) i -i i -i 4 4 e e 2isin e e 16sin       − =  − = B1 M1 Or eiθ+ e–iθ= 2cosθ May use z without definition oe, eg. (2i sin 𝜃)4 = 16 sin4 𝜃= (𝑒i𝜃−𝑒−i𝜃) 4. Award this mark for sinθto the power of four, and for (2i)4 = 16. Note that 16 may appear later. ( ) 4i 2i 2i 4i e 4e 6 4e e     − −  − + − + M1 Expanding (eiθ – e–iθ)4 with correct coefficients. ( ) ( ) ( ) 4i 2i 2i 4i 4i 4i 2i 2i 4 4 e 4e 6 4e e e e 4 e e 6 2cos4 8cos2 6 16sin 1 1 3 sin cos4 cos2 8 2 8 1 1 3 i.e. , , 8 2 8 A B C               − − − −  − + − + = + − + +  − + =  = − + = = − = M1 A1 Grouping terms and using eiθ+ e–iθ = 2 cosθ. cao, from fully correct reasoning. Allow A, B, C seen in the expression only. [5] Y540/01 Mark Scheme June 2023 23 Question Answer Marks Guidance 9 (b) DR ( ) ( ) 1 4 5 5 1 2 2 5 4 5 2 5 4 4 4 5 d 1 Let d 5 d 1 1 Let sin d 1 1 df f sin 4 8sin 2 12 4cos4 16cos2 12 d df df d d 1 1 . . 4cos4 16cos2 12 d d d d 5 1 df 1 1 3 32 cos4 cos2 32 sin 32 32 d 8 2 8 df Then u u x x x v v u u u x v v v v v v v u v v x x v u x x v v v u x v − − − =  = =  = = − −  = − +  = − + = = − +  −   = − + = = =     2 5 2 5 4 4 5 5 df d d 1 1 32 . . 32 d d d d 5 1 5 1 v u x x x v u x x x − = =   = − − B1 M1 A1 M1 M1 A1 Sight of ( ) 1 2 d sin 1 d 1 u u u − = − Uses chain rule Correct derivative, f’ Uses result from (a) Uses 1 5 4 4 1 5 sin sin x x −    =           AG Clearly shown [6] 9 (c)   ( ) ( ) ( ) 2 5 0 1 1 1 0 1 1 1 1 1 1 5 5 5 1 1 1 5 1 1 5 5 lim d lim f ( ) lim f ( ) f (0) 32 32 1 f (0) 0 f ( ) sin 4sin 8sin 2sin 12sin As 1 sin sin (1) 2 lim f ( ) sin 2 8sin 6 6 k k k k k k R x x k x k k k k k k k R      → → → − − − − − → = = = − − =           = − +                           → → =      = − + =  =  5 15 6 32 16    = M1 M1 A1 For use of part (b) and an upper limit of k < 1 Integral must be found in terms of k (which could be 1) For correct use of limits [3] Need to get in touch? 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