Cambridge A Level Mathematics - Further 9231 — 2017 May/June Paper 1 · Variant 2

*7565775426* Cambridge International Examinations Cambridge International Advanced Level CANDIDATE NAME CENTRE NUMBER CANDIDATE NUMBER FURTHER MATHEMATICS 9231/12 Paper 1 May/June 2017 3 hours Candidates answer on the Question Paper. Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST Write your Centre number, candidate number and name in the spaces at the top of this page. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. The use of a calculator is expected, where appropriate. Results obtained solely from a graphic calculator, without supporting working or reasoning, will not receive credit. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 27 printed pages and 1 blank page. JC17 06_9231_12/RP © UCLES 2017 [Turn over

2 BLANK PAGE © UCLES 2017 9231/12/M/J/17

3 1 It is given that n Ð r=1 ur = n22n + 3, where n is a positive integer. (i) Find 2n Ð r=n+1 ur. [2] … … … … … … … … … … (ii) Find ur. [3] … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

4 2 Prove, by mathematical induction, that 5n + 3 is divisible by 4 for all non-negative integers n. [5] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

5 3 A curve C has equation tan y = x, for x > 0. (i) Use implicit differentiation to show that d2y dx2 = −2x @dy dx A2 . [3] … … … … … … … … … … … … … (ii) Hence find the value of d2y dx2 at the point 1, 1 40 on C. [2] … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

6 4 (i) Find the value of k for which the set of linear equations x + 3y + kz = 4, 4x −2y −10z = −5, x + y + 2z = 1, has no unique solution. [3] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

7 (ii) For this value of k, find the set of possible solutions, giving your answer in the form ` x y z a = a + tb, where a and b are vectors and t is a scalar. [3] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

8 5 The matrix A, given by A = ` 1 2 −2 6 4 −6 6 5 −7 a , has eigenvalues 1, −1 and −2. (i) Find a set of corresponding eigenvectors. [4] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

9 … … … … … … … (ii) The matrix B is given by B = A −2I, where I is the 3 × 3 identity matrix. Write down the eigenvalues of B, and state a set of corresponding eigenvectors. [2] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

10 6 Let In = Ó 1 20 0 xn sin x dx. (i) Prove that, for n ≥2, In + nn −1In−2 = n1 20n−1. [4] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

11 (ii) Calculate the exact value of I1 and deduce the exact value of I3. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

12 7 By finding a cubic equation whose roots are !, " and ', solve the set of simultaneous equations ! + " + ' = −1, !2 + "2 + '2 = 29, 1 ! + 1 " + 1 ' = −1. [8] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

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14 8 (i) Let z = cos 1 + i sin 1. Show that z −1 z = 2i sin 1 and hence express 16 sin5 1 in the form sin 51 + p sin 31 + q sin 1, where p and q are integers to be determined. [6] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

15 (ii) Hence find the exact value of Ó 1 30 0 16 sin51 d1. [3] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

16 9 The curve C has equation y = x2 −3x + 6 1 −x . (i) Find the equations of the asymptotes of C. [3] … … … … … … … … (ii) Find the coordinates of the turning points of C. [3] … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

17 (iii) Find the coordinates of any intersections with the coordinate axes. [2] … … … … … … … … … (iv) Sketch C. [3] © UCLES 2017 9231/12/M/J/17 [Turn over

18 10 It is given that x = t 1 2, where x > 0 and t > 0, and y is a function of x. (i) Show that dy dx = 2t 1 2 dy dt and d2y dx2 = 2dy dt + 4td2y dt2 . [3] … … … … … … … … … … … (ii) Hence show that the differential equation d2y dx2 − @ 8x + 1 x A dy dx + 12x2y = 4x2e−x2 * reduces to the differential equation d2y dt2 −4dy dt + 3y = e−t. [1] … … … … … … … © UCLES 2017 9231/12/M/J/17

19 (iii) Find the general solution of *, giving y in terms of x. [7] … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

20 11 The curve C has polar equation r = a1 + sin 1 for −0 < 1 ≤0, where a is a positive constant. (i) Sketch C. [2] (ii) Find the area of the region enclosed by C. [4] … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

21 (iii) Show that the length of the arc of C from the pole to the point furthest from the pole is given by s = ï2a Ô 1 20 −1 20 1 + sin 1 d1. [3] … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

22 (iv) Show that the substitution u = 1 + sin 1 reduces this integral for s to ï2aÔ 2 0 1 2 −u du. Hence evaluate s. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

23 12 Answer only one of the following two alternatives. EITHER The curve C has equation y = 1 2 ex + e−x for 0 ≤x ≤4. (i) The region R is bounded by C, the x-axis, the y-axis and the line x = 4. Find, in terms of e, the coordinates of the centroid of the region R. [10] … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

24 … … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

25 (ii) Show that ds dx = 1 2 ex + e−x, where s denotes the arc length of C, and find the surface area generated when C is rotated through 20 radians about the x-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

26 OR The position vectors of the points A, B, C, D are i + j + 3k, 3i −j + 5k, 3i −j + k, 5i −5j + !k, respectively, where ! is a positive integer. It is given that the shortest distance between the line AB and the line CD is equal to 2ï2. (i) Show that the possible values of ! are 3 and 5. [7] … … … … … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17

27 … … … … … … … (ii) Using ! = 3, find the shortest distance of the point D from the line AC, giving your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … © UCLES 2017 9231/12/M/J/17 [Turn over

28 (iii) Using ! = 3, find the acute angle between the planes ABC and ABD, giving your answer in degrees. [4] … … … … … … … … … … … … … … … … … … … … … … Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. © UCLES 2017 9231/12/M/J/17

® IGCSE is a registered trademark. This document consists of 15 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 May/June 2017 MARK SCHEME Maximum Mark: 100 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the May/June 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 15 Mark Scheme Notes Marks are of the following three types: M Method mark, awarded for a valid method applied to the problem. Method marks are not 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. Correct application of a formula without the formula being quoted obviously earns the M mark and in some cases an M mark can be implied from a correct answer. 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). B Mark for a correct result or statement independent of method marks. • When a part of a question has two or more “method” steps, the M marks are generally independent unless the scheme specifically says otherwise; and similarly when there are several B marks allocated. The notation DM or DB (or dep*) is used to indicate that a particular M or B mark is dependent on an earlier M or B (asterisked) mark in the scheme. When two or more steps are run together by the candidate, the earlier marks are implied and full credit is given. • The symbol FT implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A or B marks are given for correct work only. A and B marks are not given for fortuitously “correct” answers or results obtained from incorrect working. • Note: B2 or A2 means that the candidate can earn 2 or 0. B2/1/0 means that the candidate can earn anything from 0 to 2. The marks indicated in the scheme may not be subdivided. If there is genuine doubt whether a candidate has earned a mark, allow the candidate the benefit of the doubt. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored. • Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. • For a numerical answer, allow the A or B mark if a value is obtained which is correct to 3 s.f., or which would be correct to 3 s.f. if rounded (1 d.p. in the case of an angle). As stated above, an A or B mark is not given if a correct numerical answer arises fortuitously from incorrect working. For Mechanics questions, allow A or B marks for correct answers which arise from taking g equal to 9.8 or 9.81 instead of 10.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 3 of 15 The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) CAO Correct Answer Only (emphasising that no “follow through” from a previous error is allowed) CWO Correct Working Only – often written by a ‘fortuitous’ answer ISW Ignore Subsequent Working SOI Seen or implied SR Special Ruling (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the light of a particular circumstance) Penalties MR –1 A penalty of MR –1 is deducted from A or B marks when the data of a question or part question are genuinely misread and the object and difficulty of the question remain unaltered. In this case all A and B marks then become “follow through” marks. MR is not applied when the candidate misreads his own figures – this is regarded as an error in accuracy. An MR –2 penalty may be applied in particular cases if agreed at the coordination meeting. PA –1 This is deducted from A or B marks in the case of premature approximation. The PA –1 penalty is usually discussed at the meeting.

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 15 Question Answer Marks Guidance 1(i) ( ) ( ) ( ) 2 2 2 1 2 4 3 2 3 n n ru n n n n + = + − + ∑ M1 Method mark for using S2n – Sn 3 2 14 9 n n = + A1 Total: 2 1(ii) ( ) ( ) ( ) 2 2 2 3 1 2 1 ru r r r r = + − − + M1A1 Method mark for using Sr – Sr–1 OE 2 6 1 r = − A1 SR: CAO B1 without wrong working Total: 3 2 Let Pn be the proposition that 5n+3 is divisible by 4 50 +3 =4 ⇒ P0 is true (allow P1) B1 Some explanation of what Pk being true means Assume that Pk is true for some non-negative integer k. B1 or e.g. 5k +3 = 4α for 2nd B1 ( ) 1 5 3 5 4 3 3 k α + + = − + M1 Alt method: Use f(k+1) – f(k) M1 A1 ( ) 20 12 4 5 3 α α = − = − (or shows that ( ) 1 5 3 5.5 5.3 4.3 5 5 3 4.3 k k k + + = + − = + − ) A1 P0 is true and Pk ⇒Pk+1, hence Pn is true for all non-negative integers n. A1 Total: 5

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 15 Question Answer Marks Guidance 3(i) ( ) 2 d 1 1 d y x x + = or 2 sec 1 dy y dx = ⇒ ( ) 2 d 1 1 d y x x + = or 2 cos dy y dx = M1 Using implicit differentiation ⇒ ( ) 2 2 2 d d 2 1 0 d d y y x x x x + + = ⇒ 2 2 2 d 2 d y dy x dx x   = −     (AG) M1 A1 M1 for good attempt at product rule Alt method: ( ) 2 2 2 2 d d d d cos cos sin 2 d d d d y y y y y y y x x x x x   = ⇒ = − = −     M1 A1 M1 for good attempt at implicit differentiation Total: 3 3(ii) ( ) ( ) ( ) 2 1 4 2 ' 1 cos ' 1 y y π = ⇒ = B1 ( ) 1 2 " 1 y ⇒ = − B1 FT Total: 2 4(i) 1 3 2 1 5 0 1 1 2 k     − − =       M1 M1 Using the determinant Alt method: Uses row operations 8 k ⇒ = A1 Total: 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 15 Question Answer Marks Guidance 4(ii) 3 8 4 x y z + + = (1) 4 2 10 5 x y z − − = − (2) M1 From (1) and (2) obtain 3 y x = − or other correct expression 2 1 x y z + + = (3) Substitute x t = and 3 y t = − in (3) to obtain z. A1 FT Alternative: Find REF for augmented matrix and form equations (M1) Correctly (A1) 1 2 0 1 0 3 1 x y t z           = + −                or 1 2 3 2 1 3 0 1 t −         + −             (OE) A1 Total: 3 5(i) Eigenvectors are i + 2j + 2k, i + k and j + k (OE) for λ = 1, –1 and –2 respectively. M1A1 (Award M1A1 for any one correct and A1 for each of the other two.) A1A1 Total: 4 5(ii) Eigenvalues for B are –1, –3 and –4. B1 Eigenvectors are the same as for A respectively. B1 FT Total: 2

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 15 Question Answer Marks Guidance 6(i) 1 2 2 cos cos d 0 0 n n nI x x nx x x π π −   = − +   ∫ M1A1 Uses integration by parts with u n x = ( ) 1 2 2 2 0 sin 1 sin d 0 0 n n nx x n n x x x π π − −   = + − −   ∫ A1 ( ) ( ) ( ) ( ) 1 1 1 1 2 2 2 2 1 1 n n n n n n n n I I n n I n π π − − − − = − − ⇒ + − = (AG) A1 Total: 4 6(ii) [ ] 2 2 2 1 sin d cos cos d 0 0 0 I x x x x x x x π π π = = − + ∫ ∫ M1 [ ] 2 sin 1 0 x π = = A1 ( ) 2 2 3 3 2 4 3 3 3 2 1 6 n I π π = ⇒ = −× × = − B1 FT Total: 3

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 15 Question Answer Marks Guidance 7 2 1 29 14 αβ αβ = − ⇒ = − ∑ ∑ M1A1 14 1 14 αβ αβγ αβγ αβγ − = = −⇒ = ∑ M1A1 FT 3 2 14 14 0 x x x ⇒ + − − = A1 ( )( ) 2 1 14 x x ⇒ + − M1A1 Attempt to factorise cubic ⇒Solution is –1, in 14 ± any order. Accept ±3.74 (awrt) SR B1 for correct roots without working A1 Total: 8 8(i) ( ) ( ) 1 cos isin cos isin 2isin z z θ θ θ θ θ − − = + − − − − = B1 ( ) ( ) ( ) ( ) 5 3 5 5 3 1 1 1 1 5 10 z z z z z z z z − = − − − + − M1A1 5 32sin i=2i sin5 10isin3 20isin θ θ θ θ ⇒ − + M1A1 Grouping not required at this stage. 5 16sin sin5 5sin3 10sin θ θ θ θ ⇒ = − + A1 M1 for grouping and applying initial result Total: 6

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 9 of 15 Question Answer Marks Guidance 8(ii) 1 1 5 cos5 5cos3 3 3 5 3 16sin d 10cos 0 0 θ θ π π θ θ θ   = − + −   ∫ M1A1FT 5 5 53 1 1 10 3 5 3 30 5 10     = − − − −−+ − =     A1 Total: 3 9(i) 1 x = B1 2 y x = − M1A1 Total: 3 9(ii) ( ) 2 ' 1 4 1 0 y x − = −+ − = M1 1,3 x ⇒ = − A1 Turning points are ( ) 1,5 − and ( ) 3, 3 − A1 Total: 3 9(iii) (0 , 6) B1 2 0 3 6 y x x = ⇒ − + ; 9 24 ∆= − ⇒ No intersection with x-axis B1 Total: 2

923 © U Q 31/12 CLES 2017 Question 9(iv) 10(i) d d y x = 2 2 d d y x 10(ii) Subs 1 2 d d d 2 d d d y t y t t x t = × = d d d d d d y y t t x x   = × =     stitute in (*): d 2 d y t d ⇒ C A y t (AG) 1 1 2 2 2 2 d d 2 d d y y t t t t −  = +   2 2 d d 4 16 d d y y y t t t t t + − 2 2 d d 4 3 d d y y y t t − + = Cambridge Inter Answer 1 2 2 d 2 2 4 d y y t t t  = +   d 2 12 4 d y y ty t t − + = t e− (AG) rnational A Leve PUBLISHED Page 10 of 15 2 2 d d y t (AG) 4 t te− l – Mark Schem Mar B1 B Total: M Total: Total: me rks 1 FT Asymptote B1B1 Each branc 3 B1 M1A1 3 B1 1 Guidanc es correct. ch. May/June 2 ce 2017

923 © U Q 31/12 CLES 2017 Question 10(iii) CF: PI: GS: 11(i) ( )( ) 1 3 m m − − = 3 t t y Ae Be = + ' t y ke y − = ⇒ = − 4 3 t t ke ke k − − + + 1 8 k ⇒ − 3 t t y Ae Be = + + 2 2 3 x x y Ae Be = + C A 0 " t t ke y ke − − − ⇒ = t t ke e − − = 1 8 t e− 2 1 8 x e− + Cambridge Inter Answer t rnational A Leve PUBLISHED Page 11 of 15 l – Mark Schem Mar A1 A1 Total: Total: me rks M1 A1 M1 M1 A1 1 FT 1 FT 7 B1 Sketch of a B1 Correct ori 2 Guidanc a cardioid ientation and labe May/June 2 ce elled 2017

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 12 of 15 Question Answer Marks Guidance 11(ii) ( ) 2 2 1 2sin sin d 2 a A π θ θ θ π = + + −∫ M1 2 1 1 1 2sin cos2 d 2 2 2 a π θ θ θ π     = + + −     −   ∫ M1 2 2 3 1 3 2cos sin 2 2 2 4 2 a a π θ π θ θ π    = − − =   −     M1A1 Total: 4 11(iii) Show that when r=0, 2 π θ = − , when r=2a 2 π θ = and that d cos d r a θ θ = B1 ( ) 1 2 2 2 2 2 2 2 1 2 2 1 2sin sin cos d 2 1 sin d s a a a a π π π θ θ θ θ θ θ π = + + + = + − −∫ ∫ M1A1 Uses correct formula for arc length; AG Total: 3 11(iv) ( ) 2 2 d d 1 sin cos 1 1 2 u u u u u θ θ θ = + ⇒ = = − − = − M1 2 1 2 d 0 2 so s a u u = − ∫ AG A1 Including limits ( ) 1 2 2 2 2 2 4 0 a u a   = − − =     M1A1 Total: 4

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 13 of 15 Question Answer Marks Guidance 12E(i) ( ) ( ) 4 4 4 4 1 1 1 0 2 2 2 0 d e e d e e x x x x y x x e e − − −   = + = − = −   ∫ ∫ M1A1 ( ) 4 1 0 2 d d x x xy x x e e x − = + ∫ ∫ M1 ( ) ( ) ( ) 4 4 4 1 1 2 2 0 e 1 e 1 3e 2 5e x x x x − −   − − + = + −   A1 A1 ( ) 2 4 2 2 1 1 0 2 8 d e 2 e d x x y x x − = + + ∫ ∫ M1 ( ) 4 2 2 8 8 1 1 1 1 8 2 2 16 0 e 2 e 16 e x x x e − −   + + − = + −   A1 A1 Uses correct formulae for ݔ and ݕ 4 4 2 2 4 4 2 2 3e 2 5e 3e 5e e e e e x − − − −   + − + = =   − +   ; ( ) 8 8 4 4 1 e 16 e OE 8 e e y − −   + − =   −   M1 A1 Total: 10 12E(ii) ( ) ( ) ( ) ( ) 2 2 2 2 1 1 1 4 4 2 d 1 ' e 2 e e e e e d x x x x x x s y x − − − = + = + + = + = + (AG) B1 ( ) ( ) 2 2 2 4 4 1 2 e e d e 2 e d 0 0 4 2 x x x x S x x π π − − = + = + + ∫ ∫ M1 A1 2 2 8 8 4 e e e e 2 8 0 2 2 2 2 2 2 x x x π π − −     = + − = + −         A1 Total: 4

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 14 of 15 Question Answer Marks Guidance 12O(i) 2 2 2 2 2 4 2 2 1 AC AB CD α             = − = − = −             − −       uuur uuur uuur B1 May find ܣܦ ሬሬሬሬሬԦ, ܤܥ ሬሬሬሬሬԦ ݋ݎܤܦ ሬሬሬሬሬሬሬሬሬሬሬሬԦ instead of AC 5 5 1 1 1 3 ~ 3 2 4 1 2 2 AB CD α α α α α − −         × = − = − −         − − −     i j k uuur uuur M1A1 ( ) ( ) 2 2 2 2 5 2 3 2 2 2 2 2 2 2 16 38 8 5 3 4 α α α α α α −       − −       −    = ⇒ − + = − + − + M1A1 Substitutes their vectors into correct formula 2 2 2 16 38 8 8 15 0 α α α α ⇒ − + = ⇒ − + = A1 ( )( ) 3 5 0 3 or 5 α α α ⇒ − − = ⇒ = (AG) A1 Total: 7

9231/12 Cambridge International A Level – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 15 of 15 Question Answer Marks Guidance 12O(ii) 4 2 6 4 0 2 AD or CD             = − = −                   uuur uuur B1 Alt method: Let P be point on AC with parameter ࣅ M1 Distance of D from AC 56 3 1 4 6 0 4.32 1 1 1 1 1 1 = − = = + + − − i j k (or with CD) M1A1 Use DP.AC = 0 to find ࣅ (= -5/3) M1 Find length A1 Total: 3 12O(iii) ABC: 1 1 1 1 1 1 1 1 1 0 −     = − −= −     −   i j k n ABD: 2 3 1 1 1 2 2 3 0 1     = − =     − −   i j k n B1B1 3 2 0 2 14 cosθ θ −−+ = ⇒ =19.1° or 0.333 rads M1A1 Total: 4