Cambridge A Level Mathematics - Further 9231 — 2010 May/June Paper 1 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 May/June 2010 3 hours Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Centre number, candidate number and name on all the work you hand in. Write in dark blue or black pen. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. 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 5 printed pages and 3 blank pages. © UCLES 2010 [Turn over *1900609380*

2 1 The variables x and y are such that y = −1 when x = 1 and x2 + y2 + dy dx 3 = 29. Find the values of dy dx and d2y dx2 when x = 1. [5] 2 The curve C has polar equation r = a1 −e−θ, where a is a positive constant and 0 ≤θ < 2π. (i) Draw a sketch of C. [3] (ii) Show that the area of the region bounded by C and the lines θ = ln 2 and θ = ln 4 is 1 2a2ln 2 −13 32. [4] 3 At any point (x, y) on the curve C, dx dt = t √(t2 + 4) and dy dt = −t √(4 −t2), where the parameter t is such that 0 ≤t ≤2. Show that the length of C is 4 √2. [3] Given that y = 0 when t = 2, determine the area of the surface generated when C is rotated through one complete revolution about the x-axis, leaving your answer in an exact form. [4] 4 The sum SN is defined by SN = N ∑ n=1 n5. Using the identity n + 1 2 6 −n −1 2 6 ≡6n5 + 5n3 + 3 8n, find SN in terms of N. [You need not simplify your result.] [4] Hence find lim N→∞N−λSN, for each of the two cases (i) λ = 6, (ii) λ > 6. [3] 5 Let In = ã e 1 x(ln x)n dx, where n ≥1. Show that In+1 = 1 2e2 −1 2(n + 1)In. [3] Hence prove by induction that, for all positive integers n, In is of the form Ane2 + Bn, where An and Bn are rational numbers. [6] © UCLES 2010 9231/11/M/J/10

3 6 The equation x3 + x −1 = 0 has roots α, β, γ. Use the relation x = √y to show that the equation y3 + 2y2 + y −1 = 0 has roots α2, β2, γ 2. [2] Let Sn = αn + βn + γ n. (i) Write down the value of S2 and show that S4 = 2. [3] (ii) Find the values of S6 and S8. [4] 7 The lines l1 and l2 have vector equations r = 4i −2j + λ(2i + j −4k) and r = 4i −5j + 2k + µ(i −j −k) respectively. (i) Show that l1 and l2 intersect. [3] (ii) Find the perpendicular distance from the point P whose position vector is 3i −5j + 6k to the plane containing l1 and l2. [3] (iii) Find the perpendicular distance from P to l1. [4] 8 The matrix A is given by A =  4 1 −1 −4 −1 4 0 −1 5 . Given that one eigenvector of A is 1 −2 −1 !, find the corresponding eigenvalue. [2] Given also that another eigenvalue of A is 4, find a corresponding eigenvector. [2] Given further that 1 −4 −1 ! is an eigenvector of A, with corresponding eigenvalue 1, find matrices P and Q, together with a diagonal matrix D, such that A5 = PDQ. [6] © UCLES 2010 9231/11/M/J/10 [Turn over

4 9 (i) Write down the five fifth roots of unity. [2] (ii) Hence find all the roots of the equation ß5 + 16 + (16 √3)i = 0, giving answers in the form reiqπ, where r > 0 and q is a rational number. Show these roots on an Argand diagram. [4] Let w be a root of the equation in part (ii). (iii) Show that 4 ∑ k=0 w 2 k = 3 + i √3 2 −w . [3] (iv) Identify the root for which |2 −w| is least. [2] 10 Find the set of values of a for which the system of equations x + 4y + 12ß = 5, 2x + ay + 12ß = a −1, 3x + 12y + 2aß = 10, has a unique solution. [4] Show that the system does not have any solution in the case a = 18. [2] Given that a = 8, show that the number of solutions is infinite and find the solution for which x + y + ß = 1. [5] © UCLES 2010 9231/11/M/J/10

5 11 Answer only one of the following two alternatives. EITHER The variables ß and x are related by the differential equation 3ß2 d2ß dx2 + 6ß2 dß dx + 6ßdß dx 2 + 5ß3 = 5x + 2. Use the substitution y = ß3 to show that y and x are related by the differential equation d2y dx2 + 2dy dx + 5y = 5x + 2. [3] Given that ß = 1 and dß dx = −2 3 when x = 0, find ß in terms of x. [9] Deduce that, for large positive values of x, ß ≈x 1 3. [2] OR The curve C has equation y = x(x + 1) (x −1)2 . (i) Obtain the equations of the asymptotes of C. [3] (ii) Show that there is exactly one point of intersection of C with the asymptotes and find its coordinates. [2] (iii) Find dy dx and hence (a) find the coordinates of any stationary points of C, (b) state the set of values of x for which the gradient of C is negative. [6] (iv) Draw a sketch of C. [3] © UCLES 2010 9231/11/M/J/10

8 BLANK PAGE 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. University of 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. 9231/11/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/11 Paper 11, maximum raw mark 100 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 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: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 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: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 1 29 ) ( 1 1 3 = ′ + + y ⇒ 27 ) ( 3 = ′y ⇒ 3 = ′y B1 y y x ′ + 2 2 , 0 ) (3 2 = ′′ ′ + y y M1A1, A1 2 – 6 + 27 y ′′ = 0 ⇒ 27 4 = ′y A1 [5] 2 (i) Sketch of C: Approximately correct shape and location for π θ 2 0 < ≤ B1 Shows initial line to be tangential to C at the pole B1 Asymptotic approach to circle r = a B1 [3] (ii) ∫ − −+ − = ln4 ln2 2 2 ) 2 1( ) 2 / ( θ θ θ d e e a A M1A1 4 ln 2 ln 2 2 ] ) 2 / 1( 2 )[ 2 / ( θ θ θ − −− + = e e a A1 = ... = ) 32 / 13 2 )(ln 2 / ( 2 − a (AG) A1 [4] 3 t t t t t t t s 2 2 8 ) 4 ( ) 4 ( d d 2 2 2 = = − + + = B1 [ ] 2 4 2 d 2 2 2 0 2 2 0 = = = ∫ t t t s (AG) M1A1 [3] y = (1 / 3)(4 – t2)3 / 2 B1 t t t S d ) 4 ( 2 2 3 / 2 2 0 2 / 3 2 ∫ − + = π M1 [ ] 2 0 2 / 5 2) 4 )( 15 / 2 4 ( ... t − − = = π A1 15 / 2 128 π = A1 [4] 4 (N + 1 / 2)6 – 1 / 64 = 6SN + (5 / 4)N2(N + 1)2 + 3N(N + 1) / 16 M1A1A1 M1 for application of difference method: A1 for LHS correct: A1 for RHS correct SN = (1 / 6)(N + 1 / 2)6 – (5 / 24)N2(N + 1)2 – (1 / 32)N(N + 1) – 1 / 384 Or       + − + − − + )1 ( 16 3 4 )1 ( 5 ) 2 1 ( ) 2 1 ( 6 1 2 2 6 6 N N N N N A1 [4] (i) For λ = 6, S} = 1/6 B2 (ii) For λ > 6, S} = 0 B1 [3]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 5 D [(x2 / 2)(lnx)n] = x(lnx)n + (nx / 2)(lnx)n – 1 M1 ⇒ 1 1 2 ) 2 / ( ] ) )(ln 2 / [( − + = n n e n I n I x x A1 ⇒ ... ⇒ In = e2 / 2 – (n / 2)In – 1 for n [ 2 ⇒ In + 1 = e2 / 2 – (n + 1)2In – 1 for n [ 1 A1 [3] OR ∫ ∫ − − = e 1 1 e 1 1 2 ) )(ln 2 / ( ] ) )(ln 2 / [( ) (ln dx x nx x x dx x x n e n n M1A1 ⇒ In = e2 / 2 – (n / 2)In – 1 ⇒ In + 1 = nI n e 2 )1 ( 2 2 + − A1 Hk : Ik = Ak e2 + Bk , where Ak and Bk are rational B1 Hk ⇒ Ik + 1 = e2 / 2 – (k + 1)(Ak e2 + Bk) / 2, = Ak + 1e2 + Bk + 1, where Ak + 1 = 1 / 2 – (k + 1)Ak / 2, Bk + 1 = – (k + 1)Bk / 2, M1 ⇒ Ak + 1 and Bk + 1 are rational A1 I1 = e2 / 4 + 1 / 4 ⇒ A1 = 1 / 4, B1 = 1 / 4 ⇒ H1 is true M1A1 Completion of induction argument A1 [6] 6 Obtains an equation in y not involving radicals, e.g., y(y + 1)2 = 1 M1 ⇒ ... ⇒ y3 + 2y2 + y – 1 = 0 (AG) A1 [2] (i) S2 = –2 B1 S4 = 4 – 2 = 2 M1A1 [3] (ii) S6 = –2S4 – S2 + 3 = 1 M1A1 OR 2 2 − = Σα , 1 2 2 = Σ β α , 1 2 2 = βγ α 2 2 2 2 2 2 3 2 6 3 3 ) ( γ β α β α α α + Σ Σ − Σ = S M1 = (–2)3 – 3 × (–2) × 1 + 3 = –8 + 6 + 3 = 1 A1 S8 = –2S6 – S4 + S2 = –6 M1A1 [4]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 7 (i) Solves any 2 of the equations: 4 + 2λ = 4 + µ , –2 + λ = –5 – µ , –4λ = 2 – µ to obtain λ = –1, µ = –2 M1A1 Checks consistency with the third equation A1 [3] (ii) P = | (i – 2j + 2k).(5i + 2j + 3k) / 38 | M1A1 =7 / 38 = 1.14 A1 [3] OR n = –5i – 2j – 3k M1 Plane is 5x + 2y + 3z = 16 A1 38 7 3 2 5 16 18 10 15 2 2 2 = + + − + − = P A1 OR Plane is 5x + 2y + 3z = 16 (as above) M1A1 Sub. general pt on perpendicular 38 7 3 6 2 5 5 3 − = ⇒         + + − + t t t t ⇒ P =         t t t 3 2 5 = 1.14 A1 (iii) (i – 2j + 2k) × (2i + j – 4k) = 6i + 8j + 5k OR (i + 3j – 6k) × (2i + j – 4k) = –6i – 8j – 5k, etc. B1 d = |6i + 8j + 5k| / 21 = 21 / 125 = 2.44 M1A1A1 [4] OR Let Q be the foot of the perpendicular from P to l, and A be the known point on l1 AQ = 21 8 21 ) 4 2 ( ). 2 2 ( = − + + − k j i k j i M1A1 AP2 = 12 + (–2)2 + 22 = 9 B1 PQ2 = 9 – 21 5 5 21 125 21 64 = ⇒ = PQ A1 OR 0 4 1 2 4 6 3 2 1 4 6 3 2 1 6 5 3 4 2 2 4 =         − ⋅        − − + + ∴         − − + + =         − −         − + − + = t t t t t t t t t PQ M1 ⇒ t = – 21 29 A1 44 .2 10 34 37 21 1 10 34 37 21 1 2 2 2 = + + = ⇒         − − = PQ PQ M1A1

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 8         − −         − −         − − − − 3 6 3 1 2 1 5 1 0 4 1 4 1 1 4 M1 ⇒ eigenvalue = 3 A1 [2] Eigenvector corresponding to 4 is         − − 4 4 1 M1A1 [2] D = diag(1 243 1024) B1 P =         − − − − − − 4 1 1 4 2 4 1 1 1 (f.t.) A1 Q = P–1 B1 =         − − − − 3 / 1 0 3 / 1 0 2 / 1 2 3 / 1 2 / 1 3 / 2 ft on P M1A2√ M1 for any valid method: A2 if completely correct A1 if exactly 1 error: A0 if > 1 errors [6] 9 (i) exp (2πki/5), k = 0, 1, 2, 3, 4 (AEF) M1A1 [2] M1 for 1 correct fifth root of unity A1 for exactly 5 distinct, correct roots (ii) z5 = 32 exp (–2πi/3) M1 zk = 2 exp (–2πi/15 + 2πki/5) A1 Roots equally spaced on circle |z| = 2; correctly placed M1A1 [4] (iii) ] 2 / 1[ ] ) 2 / ( 1[ ) 2 / ( 5 4 0 w w w k k − − = ∑ = M1 = [ ] 2 / 1 ) 3 16 16 )( 32 / 1( 1 w i − − − − A1 = ... = (3 + i3 )/(2 – w) (AG) A1 [3] (iv) Deduces from diagram in (ii) that minimum of |2 – w| occurs when w = 2e–2πi/15 or i e 15 28 2 π M1A1 [2] OR Evaluates 5 possible values of |2 – w| M1 Identifies minimum of |2 – w| correctly A1

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 10 A =         a a 2 12 3 12 2 12 4 1 → ... → =         − − − 36 2 0 0 12 8 0 12 4 1 a a M1A1 ⇒ r(A) = 3 provided a ≠ 18 and a ≠ 8 A1 ⇒ unique solution for all values of a except a = 18 and a = 8 A1 [4] OR for first 3 marks: det A = 0 ⇒ ... ⇒ a2 – 26a + 144 = 0 (M1A1) ⇒ a = 8 or 18 (A1) a = 18 ⇒ 0z = –5 which is impossible for any finite z, or equivalent contradiction M1A1 [2] When a = 8 system reduces to 2 equations: x + 4y = 2 and z = 1/4 M1 All solutions then of form: x = λ, y = (2 – λ)/4, z = 1/4 where λ is real A1 May be parametrised in any equivalent way ⇒ an infinite number of solutions A1 λ + (2 – λ)/4 + 1/4 = 1 M1 ⇒ λ = 1/3 ⇒ x = 1/3, y = 5/12, z = 1/4 A1 [5] 11 EITHER y′ = 3z2z′ B1 y′′ = 6z(z′)2 + 3z2z′′ B1 Obtains given y, x DE (AG) B1 [3] CF: y = e–x [Acos2x + Bsin2x] M1A1 PI: y = x M1A1 GS: y = e–x[Acos2x + Bsin2x] + x A1 When x = 0, y = 1, y′ = –2 (both) B1 Finds A, B : A = 1, B = –1 (both) M1A1 z = [e–x(cos2x – sin2x) + x]1/3 A1 [9] OR for previous 4 marks: z = [e–x(Acos2x + Bsin2x) + x]1/3 (B1) When x = 0, z = 1, z′ = –2/3 (both) (B1) Finds A, B : A = 1, B = –1 (both) (M1A1) For large positive x : e–x (cos2x – sin2x) ≈ 0 M1 ⇒ z ≈ x1/3 (AG) A1 [2]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 11 © UCLES 2010 11 OR (i) x = 1 B1 y = 1 + O(1/x) as |x| → ∞ M1 Second asymptote is y = 1 A1 [3] (ii) x(x + 1) / (x – 1)2 = 1 ⇒ x = 1/3, y = 1 M1A1 [2] (iii) (a) dy / dx = 0 ⇒ [(2x + 1)(x – 1)2 – 2x(x – 1)(x + 1)] / (x – 1)4 = 0 M1 ⇒ x = 1/3, y = –1/8 M1A1 (b) dy / dx = –(3x + 1) / (x – 1)3 M1 {x : x < –1/3} ∪ {x : x > 1} A1√A1 ft. [6] (iv) Sketch: Left-hand branch with approximately correct shape and location and passing through the origin and (–1, 0). B1 Intersection with y = 1 and location of minimum point consistent with results of (ii) and (iii) (cwo) B1 Right-hand branch with approximately correct forms at infinity B1 [3]