Cambridge A Level Mathematics - Further 9231 — 2016 May/June Paper 1 · Variant 3

*8948221148* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 2016 3 hours Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST An answer booklet is provided inside this question paper. You should follow the instructions on the front cover of the answer booklet. If you need additional answer paper ask the invigilator for a continuation booklet. 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 4 printed pages and 1 insert. JC16 06_9231_13/2R © UCLES 2016 [Turn over

2 1 Verify that 1 3r + 13r + 4 = 1 3 @ 1 3r + 1 − 1 3r + 4 A . [1] Let SN denote N Ð r=1 1 3r + 13r + 4 and let S denote ∞ Ð r=1 1 3r + 13r + 4. Find the least value of N such that S −SN < 1 10 000. [5] 2 It is given that a diagonal of a polygon is a line joining two non-adjacent vertices. Prove, by mathematical induction, that an n-sided polygon has 1 2nn −3 diagonals, where n ≥3. [6] 3 Find the two values of the constant k for which the equations kx + y + z = 2, x + ky + z = −1, x + y + kz = −1, have no unique solution. [4] Show that, for one of these values of k, the equations have no solution, and solve the equations for the other value of k. [3] 4 The curve C has equation y = −ln1 −x2 for −1 2 ≤x ≤1 2. Show that 1 + @dy dx A2 = P1 + x2 1 −x2 Q2 . [2] Show further that 1 + x2 1 −x2 may be expressed in the form P 1 + x + Q 1 −x + R, where P, Q and R are constants to be determined. [2] Find the exact arc length of C. [4] 5 The curve C has equation y = x + 2 x2 −9. Show that dy dx < 0 at all points on C. [3] State the equations of the asymptotes of C. [2] Sketch C, showing the coordinates of any points of intersection with the coordinate axes. [3] 6 Let In = Ó 2 0 xn4 −x2 1 2 dx, for n ≥1. By considering d dx 4 xn4 −x2 3 2 5 , show that n + 3In+1 = 4nIn−1, where n ≥2. [4] Find the value of I1 and deduce the exact value of I3. [4] © UCLES 2016 9231/13/M/J/16

3 7 A curve has polar equation r = 1 1 −cos 1, for 0 < 1 < 20. Find, in the form y2 = fx, the cartesian equation of the curve. [3] Hence sketch the curve, and shade the region whose area is given by 1 2Ô 3 20 1 20 1 1 −cos 12 d1. [3] Using the cartesian equation of the curve, find the area of this region. [3] 8 The cubic equation z3 −z2 −z −5 = 0 has roots !, " and '. Show that the value of !3 + "3 + '3 is 19. [4] Find the value of !4 + "4 + '4. [2] Show that the cubic equation with roots ! −1 ! , " −1 " and ' −1 ' may be found using the substitution z = 1 1 −x, and find this equation, giving your answer in the form px3 + qx2 + rx + s = 0, where p, q, r and s are constants to be determined. [4] 9 Use de Moivre’s theorem to show that cos41 = 1 8cos 41 + 4 cos 21 + 3. [4] Find the corresponding expression for sin41 in terms of cos 41 and cos 21. [4] Hence find the exact value of Ó 1 80 0 cos41 + sin41 d1. [3] 10 Given that y is a function of x and that x = eu, show that xdy dx = dy du and x2 d2y dx2 = d2y du2 −dy du. [3] Given also that x2 d2y dx2 + 3xdy dx + 17y = 34 ln x + 21, deduce that d2y du2 + 2 dy du + 17y = 34u + 21. [1] Find y in terms of x given that y = 0 and dy dx = −1 when x = 1. [9] [Question 11 is printed on the next page.] © UCLES 2016 9231/13/M/J/16 [Turn over

4 11 Answer only one of the following two alternatives. EITHER It is given that 1 and 4 are eigenvalues of the matrix A, where A = ` 1 −3 −3 −8 6 −3 8 −2 7 a . Find eigenvectors corresponding to each of these eigenvalues. [3] Given further that ` 0 1 −1 a is an eigenvector of A, find the corresponding eigenvalue. [2] Write down matrices P and D such that P−1AP = D, where D is a diagonal matrix, and find P−1. [5] Write down a matrix C such that C2 = D, and deduce a matrix B such that B2 = A. [4] OR The position vectors of the points A, B, C, D are a = 2i + ,j −3k, b = 6i + 3j −2k, c = i + 2j −k, d = i + 7j + 4k respectively. It is given that the shortest distance between the lines AB and CD is 3. (i) Show that ,2 + , −20 = 0. [7] (ii) The planes p1 and p2 are the planes through A, B and D corresponding to the two values of , satisfying the equation in part (i). Find the acute angle between p1 and p2. [7] 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 2016 9231/13/M/J/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 10 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 2016 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 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 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 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.

Page 3 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 The following abbreviations may be used in a mark scheme or used on the scripts: AEF Any Equivalent Form (of answer is equally acceptable) AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) BOD Benefit of Doubt (allowed when the validity of a solution may not be absolutely clear) 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 MR Misread PA Premature Approximation (resulting in basically correct work that is insufficiently accurate) SOS See Other Solution (the candidate makes a better attempt at the same question) 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.

Page 4 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 1 ( ) ( ) ( )( ) ( )( ) 3 4 3 1 1 1 1 1 1 3 3 1 3 4 3 3 1 3 4 3 1 3 4 r r r r r r r r   + − +   − = =       + + + + + +     AG 1 1 1 1 1 1 1 1 1 1 3 4 7 7 10 3 1 3 4 3 4 3 4 N S N N N             = − + − +…+ − = −             + + +             1 12 S ⇒ = ⇒ S – SN = ( ) 1 1 3 3 4 10000 N < + 10000 3 4 3 N ⇒ + > ⇒N > 1109 7 9 . Thus least N is 1110. B1 [1] M1 A1 A1 M1 A1 [5] 2 With 3 n = , ( ) 1 3 0 2 n n − = A triangle has no diagonals ⇒ H3 is true. Assume Hk is true: A k-gon has ( ) 1 3 2 k k − diagonals for some integer 3 ≥ Adding an extra vertex, a further (k – 1) diagonals can be drawn. ( ) ( )( ) 2 1 2 1 3 2 2 3 1 2 2 2 k k k k k k k k + − − + − − + −= = ( )( ) 1 1 1 3 2 k k = + + − (So Hk ⇒ Hk+1) ⇒Hn is true for all integers 3 . n
M1 A1 B1 M1 A1 A1 [6] 3 1 1 1 1 0 1 1 k k k = ⇒ 3 3 2 0 k k − + = ( ) ( ) 2 1 2 0 k k − + = ⇒ k = 1, –2 1: 2 and 1 inconsistent . k x y z x y z = ⇒ + + = + + = − ⇒ 2: 1 k x z = − ⇒ = − Hence (x,y,z) = ( [–1 + t], t, t ) M1A1 M1A1 [4] B1 M1 A1 [3]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 4 2 d 2 d 1 y x x x = − ( ) ( ) ( ) ( ) 2 2 2 2 2 4 2 2 2 2 2 2 2 1 d 4 1 2 4 1 1 dx 1 1 1 x y x x x x x x x + − + +   + = + = =     − − − (AG) 2 2 1 1 1 1 1 1 1 x x x x + ≡ + − + − − 1 2 0 1 1 2 1 d 1 1 s x x x   = + −   + −   ∫ (oe) 1 2 0 1 2 ln 1 x x x   +   = −     −     2ln3 1 = − (AEF) B1 B1 [2] M1A1 [2] M1 M1A1 A1 [4] 5 ( ) ( ) ( ) 2 2 2 9 2 2 d d 9 x x x y x x − − + = − ( ) ( ) ( ) 2 2 2 2 2 2 2 5 4 9 d 0 d 9 9 x x x y x x x − + − − − − = = ⇒ < − − Asymptotes: 3 ; 0 x y = ± = Sketch: Axes and asymptotes ; Outside branches Middle branch, showing ( ) 2 0, and 2,0 9   − −     . (Deduct at most 1 mark for poor forms at infinity.) B1 M1A1 [3] B1B1 [2] B1B1 B1 [3]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 6 ( ) ( ) ( ) 3 1 3 2 1 2 1 2 2 2 2 d 4 3 4 4 d n n n x x x x nx x x + −     − = − − + −       ( ) ( ) ( ) 1 1 1 2 1 2 2 2 2 3 4 4 4 n n x x nx x x + − − − + − − ( ) 2 3 2 2 1 1 1 0 4 0 3 4 n n n n x x I nI nI + − +   ⇒ − = = − + −       ( ) 1 1 3 4 n n n I nI + − ⇒ + = (AG) ( ) 2 3 2 2 1 0 1 8 4 3 3 I x   = − − =       3 1 3 64 5 8 15 I I I = ⇒ = B1 M1 M1 A1 [4] M1A1 M1A1 [4] 7 2 2 2 2 2 2 1 1 1 x y x y x x x y + = ⇒ + − = − + 2 1 2 y x ⇒ = + or equivalent RHS Sketch: Parabola symmetrical about x-axis. Intercepts at ( ) 1 , 0 and 0 , 1 2   − ±     . Shading correct area Recognises ( ) 3 2 2 2 1 1 2 1 cos d π π θ θ − ∫ as the area of the region between parabola and y-axis. ( ) 0 1 2 1 2 2 1 2 d x x − = × + ∫ ( ) 0 3 2 1 2 1 2 2 1 2 3 3 x −   = + =     M1A1 A1 [3] B1 B1 B1 [3] M1 M1A1 [4]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 8 Let Sn denote αn + βn +γn. ( ) 2 2 2 1 2 1 2 1 3 S S αβ = − ∑ = − × − = S3 = S2 + S1 +15 = 3 + 1 + 15 = 19 (AG) Alt method: 1, α ∑ = Σαβ= -1, and αβγ = 5 M1, A1 S3 = S1 3- 3Σα Σαβ + 3αβγ =19 M1, A1 4 3 2 1 5 19 3 5 27 S S S S = + + = + + = 1 1 1 1 1 1 1 z x x z z z z x − = = − ⇒ = − ⇒ = − Hence ( ) ( ) ( ) 3 2 1 1 1 5 0 1 1 1 x x x − − − = − − − is the required equation. ( ) ( ) ( ) 2 3 1 1 1 5 1 0 x x x ⇒− − − − − − = 3 2 5 16 18 6 0 x x x ⇒…⇒ − + − = M1A1 M1A1 [4] M1A1 [2] B1 M1 M1A1 [4] 9 4 4 2 4 2 1 1 1 4 6 z z z z z z       + = + + + +             Taking cos isin z θ θ = + , ( ) 4 2cos 2cos4 8cos2 6 θ θ θ = + + ( ) ( ) 4 1 1 cos 2cos4 8cos2 6 cos4 4cos2 3 16 8 θ θ θ θ θ ⇒ = + + = + + (AG) 4 4 2 4 2 1 1 1 4 6 z z z z z z       − = + − + +             ( ) 4 2isin 2cos4 8cos2 6 θ θ θ ⇒ = − + ( ) ( ) 4 1 1 sin 2cos4 8cos2 6 cos4 4cos2 3 16 8 θ θ θ θ θ ⇒ = − + = − + ( ) ( ) 1 1 8 8 4 4 0 0 1 cos sin d cos4 3 d 4 π π θ θ θ θ θ + = + ∫ ∫ 1 8 0 1 sin 4 3 4 4 π θ θ   = +     = 1 3 16 32π + (oe) M1 M1A1 A1 [4] M1 M1A1 A1 [4] M1 M1A1 [3]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 10 d d d d 1 d d d d d d y y u y y x x x x u x u x u = × = × = (AG) 2 2 2 2 2 d d d d 1 d d d d d d y y y x x x x x x x u x     = =         2 2 1 d 1 d d d d d d d y y u x u x u u x x     = − +         2 2 2 2 d d 1 d d . d d d d y y y y x u x u u u = − + = − (AG) Substituting in differential equation: 2 2 2 2 d d d d d 3 17 34 21 2 17 34 21 d d d d d y y y y y y u y u u u u u u − + + = + ⇒ + + = + (AG) ( ) 2 2 17 0 1 4i e cos4 sin 4 u c m m m y A u B u − + + = ⇒ = −± ⇒ = + and 0 2 17 17 34 21 p y cu d y c y c cu d u = + ⇒ = = ⇒ + + ≡ ′ + ′ ′ ⇒ 2 , 1 2 1 p c d y u = = ⇒ = + [ ] [ ] ( ) 1 cos 4ln sin 4lnx 2lnx 1 y A x B x ⇒ = + + + (or in terms of u) 0 when 1 0 1 1 y x A A = = ⇒ = + ⇒ = − or u = 0, y= 0 etc [ ] [ ] ( ) [ ] [ ] 2 d 1 1 4 4 2 cos 4ln sin 4lnx sin 4ln . cos 4ln . d y A x B A x B x x x x x x x   = − + + − + +     d 1 when 1 1 1 4 2 1 d y x B B x = − = ⇒−= + + ⇒ = − or all in terms of u Hence [ ] [ ] ( ) 1 2lnx 1 cos 4ln sin 4lnx y x x = + − + (oe) B1 M1 A1 [3] B1 [1] M1A1 M1 A1 A1 B1 M1 A1 A1 [9]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 11 (e) 1 : λ = 24 1 0 3 3 24 ~ 1 8 5 3 24 1         − − =         − − − −     i j k (Or by equations.) 15 1 4 : 3 3 3 15 ~ 1 8 2 3 30 2 λ         = − − − =         − − − −     i j k 1 3 3 0 0 8 6 3 1 9 9 8 2 7 1 9 λ − −           − − = ⇒ =           − − −      P = 1 1 0 1 1 1 1 2 1         − − −   D = 1 0 0 0 4 0 0 0 9           Det P =1 Adj P = 1 1 1 0 1 1 1 1 0     − −     −   ⇒ P–1 = 1 1 1 0 1 1 1 1 0     − −     −   . (Or by row operations) C = 1 0 0 0 2 0 0 0 3           P–1AP = C2 ⇒ A = PC2P–1 = PCP–1. PCP–1 (i.e. B = PCP–1) PC = 1 2 0 1 2 3 1 4 3         − − −   1 1 1 2 2 1 2 0 3 − −     ⇒ = − −       B (Intermediate step required.) (Note: B may not be unique.) M1A1 A1 [3] M1A1 [2] B1B1 B1 M1A1 [5] B1 M1 M1A1 [4]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – May/June 2016 9231 13 © Cambridge International Examinations 2016 Qu Solution Marks 11 (o) (i) (ii) Direction perpendicular to AB and CD: 2 0 1 1 4 4 3 1 4 λ λ −     = =     − −   i j k n 5 4 6 DB     = −     −  


 . Hence ( ) 2 5 2 4 . 4 6 4 3 2 1 6 1 6 λ λ −       −       − −    = − + + or equivalent ( ) ( ) 2 2 5 2 9 4 36 λ λ λ ⇒ − = − + 2 20 0 λ λ ⇒…⇒ + − = (AG) ( )( ) 5 4 0 5, 4 λ λ λ + − = ⇒ = − 2 4 4 3 λ     = ⇒ =     −   a ⇒ Normal to ABD = 10 4 1 1 29 1 3 7 11 −     − = −     −   i j k 2 5 5 3 λ     = −⇒ = −     −   a ⇒ Normal to ABD = 44 4 8 1 29 1 12 7 56     = −     −   i j k 2 2 2 2 2 2 440 841 616 1017 cos 1062 5913 10 29 11 44 29 56 θ − + + = = + + + + 66.1 θ ⇒ = ° M1A1 M1A1 M1M1 A1 [7] B1 M1A1 A1 M1A1 A1 [7]