Cambridge A Level Mathematics - Further 9231 — 2008 Oct/Nov Paper 1 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/01 Paper 1 October/November 2008 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 4 printed pages. © UCLES 2008 [Turn over *3668489426*

2 1 The curve C is defined parametrically by x = t4 −4 ln t, y = 4t2. Show that the length of the arc of C from the point where t = 2 to the point where t = 4 is 240 + 4 ln 2. [5] 2 Let y = ex. Find the mean value of y with respect to x over the interval 0 ≤x ≤2. [2] Show that the mean value of x with respect to y over the interval 1 ≤y ≤e2 is e2 + 1 e2 −1. [4] 3 The curve C has polar equation r =
1 2π −θ 2, where 0 ≤θ ≤1 2π. Draw a sketch of C. [3] Find the area of the region bounded by C and the initial line, leaving your answer in terms of π. [3] 4 The matrix A has λ as an eigenvalue with e as a corresponding eigenvector. Show that e is an eigenvector of A2 and state the corresponding eigenvalue. [3] Given that one eigenvalue of A is 3, find an eigenvalue of the matrix A4 + 3A2 + 2I, justifying your answer. [3] 5 The curve C has equation x2 −xy −2y2 = 4. Show that, at the point A (2, 0) on C, dy dx = 2. [2] Find the value of d2y dx2 at A. [5] 6 The matrix A is defined by A = ⎛⎜⎜⎝ 1 −1 −2 −3 −2 1 7 2 −3 3 6 α 7 −6 −17 −17 ⎞⎟⎟⎠. (i) Show that if α = 9 then the rank of A is 2, and find a basis for the null space of A in this case. [5] (ii) Find the rank of A when α ≠9. [2] © UCLES 2008 9231/01/O/N/08

3 7 Let In =  1 0 1
1 + x4n dx. By considering d dx x
1 + x4n , show that 4nIn+1 = 1 2n + (4n −1)In. [4] Given that I1 = 0.866 97, correct to 5 decimal places, find I3. [4] 8 Find y in terms of t, given that 5d2y dt2 + 6dy dt + 5y = 15 + 12t + 5t2, and that y = dy dt = 0 when t = 0. [9] 9 Use induction to prove that N ∑ n=1 4n + 1 n(n + 1)(2n −1)(2n + 1) = 1 − 1 (N + 1)(2N + 1). [6] Show that 2N ∑ n=N+1 4n + 1 n(n + 1)(2n −1)(2n + 1) < 3 8N2. [4] 10 Use de Moivre’s theorem to express cos 8θ as a polynomial in cos θ. [5] Hence (i) express cos 8θ as a polynomial in sin θ, [2] (ii) find the exact value of 4x4 −8x3 + 5x2 −x, where x = cos2
1 8π. [3] 11 The plane Π1 has equation r = i + 2j + k + θ(2j −k) + φ(3i + 2j −2k). Find a vector normal to Π1 and hence show that the equation of Π1 can be written as 2x + 3y + 6 = 14. [4] The line l has equation r = 3i + 8j + 2k + t(4i + 6j + 5k). The point on l where t = λ is denoted by P. Find the set of values of λ for which the perpendicular distance of P from Π1 is not greater than 4. [4] The plane Π2 contains l and the point with position vector i + 2j + k. Find the acute angle between Π1 and Π2. [4] © UCLES 2008 9231/01/O/N/08 [Turn over

4 12 Answer only one of the following two alternatives. EITHER The curve C has equation y = (x −2)(x −a) (x −1)(x −3), where a is a constant not equal to 1, 2 or 3. (i) Write down the equations of the asymptotes of C. [2] (ii) Show that C meets the asymptote parallel to the x-axis at the point where x = 2a −3 a −2 . [2] (iii) Show that the x-coordinates of any stationary points on C satisfy (a −2)x2 + (6 −4a)x + (5a −6) = 0, and hence find the set of values of a for which C has stationary points. [6] (iv) Sketch the graph of C for (a) a > 3, (b) 2 < a < 3. [4] OR The roots of the equation x4 −5x2 + 2x −1 = 0 are α, β, γ , δ. Let Sn = αn + βn + γ n + δn. (i) Show that Sn+4 −5Sn+2 + 2Sn+1 −Sn = 0. [2] (ii) Find the values of S2 and S4. [3] (iii) Find the value of S3 and hence find the value of S6. [6] (iv) Hence find the value of α2(β4 + γ 4 + δ4) + β2(γ 4 + δ4 + α4) + γ 2(δ4 + α4 + β4) + δ2(α4 + β4 + γ 4). [3] 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. © UCLES 2008 9231/01/O/N/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2008 question paper 9231 FURTHER MATHEMATICS 9231/01 Paper 1, 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. All Examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. 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 October/November 2008 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 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 GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 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 GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 1 x& = 4t3 – 4/t, y& =8t B1 ( ) [ ] ∫ + − = 4 0 1/2 2 2 3 64 / 4 4 dt t t t s M1A1 ( ) [ ] t t t t t / 4 4 64 / 4 4 3 2 / 1 2 2 3 + = + − B1 2 ln 4 240 ] ln 4 [ 4 2 4 + = + = t t s A1 2 MV (y wrt x) over [0, 2] = (1/2) ∫ 2 0 e dx x = (1/2) 2 0] e[ x = (e2 – 1)/2 (=3.19) M1A1 1 e ln 2 e 1 2 − ∫ dy y M1 [ ] 1 e ln 2 e 1 2 − − = y y y M1A1 (for integration of ln y) – can be earned independently     − − −     − − = 1 e 1 1 e e e 2 2 2 2 2 (oew) 1 e 1 e 2 2 − + = A1 (AG)

Page 5 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 3 Approximately correct curve passing through the pole, O, and the point A(π2/4,0). B1 Negative gradient at A B1 Correct form at O. B1 Area ( ) ( ) ∫ − = /2 0 4 2 / 2 / 1 π dθ θ π M1 ( ) [ ] 2 / 0 5 2 / ) 10 / 1( π θ π − − = A1 320 / 5 π = A1 4 Ae = λe B1 A2e = A(A)e = A(λe) = λ(Ae) = λ2e ⇒ eigenvalue is λ2 M1A1 Ae = 3e for some e ⇒ (A4 + 3A2 + 2I)e = 81e + 27e + 2e = 110e M1M1 ⇒ an eigenvalue is 110 A1 OR 3 is an eigenvalue of A ∴ 32 = 9 is an eigenvalue of A2 and 34 = 81 is an eigenvalue of A4 (either of these) M1 eigenvalue of A4 + 3A2 + 2I = 81 + 3×9 + 2 (Adding [2 terms) M1 = 110 A1

Page 6 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 5 2x – xy1 – y – 4yy1 = 0 B1 ⇒…y1(2) = 2 (AG) B1 2 – xy2 – y1 – y1 – 2 1 4y – 4yy2 = 0 M1A2 2 – 2y2(2) – 2 – 2 – 16 = 0 ⇒ y2(2) = –9 M1A1 OR ( ) x y y x dx dy dx dy x y y x + − = ⇒ + = − 4 2 4 2 ( )( ) ( )( ) ( ) 2 1 2 2 4 4 1 2 2 4 1 y x y y x y y x dx y d + + − − − + = ⇒ M1 Use of quotient rule A1 [ term of numerator= A1 All correct 4 9 4 0 8 × − × = M1 Substitution of values 9 − = A1 6 (i) Reduction of A to echelon form, e.g.,             − − − − − − 0 0 0 0 9 0 0 0 4 3 1 0 3 2 1 1 α M1A1 ⇒ = 9 α last 2 rows consist entirely of zeros ( ) 2 = ⇒ A r A1 A basis for the null space of A is , 1 0 4 1 , 0 1 3 5                         −             or equivalent M1A1 (ii) 0 9 ≠ − α M1 ( ) 3 = A r A1

Page 7 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 7 ( ) ( ) ( ) 1 4 4 4 4 1 4 1 ] 1 [ − − − − + − + = + n n n x nx x x x D M1 ( )( ) ( ) 1 4 4 1 4 1 4 1 − − − + + + − = n n x n x n A1 ( ) [ ] ( ) 1 1 0 4 4 4 1 1 + − + − = + ⇒ n n n nI I n x x& M1 ( ) n n n I n nI 1 4 2 4 1 − + = ⇒ − + (AG) A1 1 2 2 3 3 2 / 1 4 , 7 4 / 1 8 I I I I + = + = B1B1 ( ) 710 .0 or 7096 .0 32 / 21 64 / 9 1 3 ≈ + = I I M1A1 OR n = 1 7752275 .0 86697 .0 3 2 1 4 2 2 = ⇒ × + = I I n = 2 7095740625 .0 7752275 .0 7 4 1 8 3 3 = ⇒ × + = I I 710 .0 or 7096 .0 3 = ∴I (No penalty for correct 5 dp value.) M1 Use of formula A1 Gets I2 A1ft Subs value for I2 in I3 formula A1 obtains I3 correct (cao)

Page 8 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 8 AQE has roots –3/5 ± (4/5)i M1 CF: e–3t/5 [A cos(4t/5) + B sin(4t/5)] A1 PI = at2 + bt + c ⇒ 10a + 6(2at + b) + 5(at2 + bt + c) ≡ 5t2 + 12t + 15 M1 5a = 5, 12a + 5b = 12, 10a + 6b + 5c = 15 A1 ⇒ a = 1, b = 0, c = 1 A1 GS: y = e–3t/5 [A cos(4t/5) + B sin(4t/5)] + t2 + 1 A1 y(0) = 0 ⇒ 0 = A + 1 ⇒ A = –1 B1 y& = –(3/5 e–3t/5 [A cos(4t/5) + B sin(4t/5)] + e–3t/5 [(–4A/5) sin(4t/5) + (4B/5) cos(4t/5)] + 2t ⇒y& (0) = 0 ⇒ –3A/5 + 4B/5 = 0 M1 ⇒ B = –3/4 ⇒ y = –(1/4)e–3t/5 [(4 cos(4t/5) + 3 sin(4t/5)] + t2 + 1 A1 or –1.25 cos(0.8t – 0.64) or 1.25 cos(0.8t + 2.50) etc.

Page 9 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 9 Set up ( )( )( ) ( )( )1 2 1 1 1 1 2 1 2 1 1 4 : 1 + + − = + − + + ∑ = k k n n n n n H k n k B1 for some positive integer k ( )( )( ) ( )( ) ( )( )( )( ) 3 2 1 2 2 1 5 4 1 2 1 1 1 1 2 1 2 1 1 4 1 1 + + + + + + + + − = + − + + ⇒∑ + = k k k k k k k n n n n n H k n k M1 ( )( )( )( ) 3 2 1 2 2 1 1 3 2 1 2 + + + + + + − = k k k k k k A1 ( )( ) 3 2 2 1 1 ... + + − = = k k A1 Verifies H1 is true. B1 Correct completion of induction argument A1 ( )( )( ) ( )( ) ( )( )1 4 1 2 1 1 2 1 1 ... 1 2 1 2 1 1 4 2 1 + + − + + = = + − + + ∑ + = N N N N n n n n n N N n M1A1 ( )( )( ) 2 8 3 4. 2. 3 1 4 1 2 1 3 N N N N N N N N N = < + + + = M1A1 OR N N N N N N N 1 2 2 3 7 14 8 3 1 7 14 8 3 + + + = + + + = Since N [=1 0 7 14 1 > + + N N ∑< ∴ 2 8 3 N

Page 10 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 10 Write , cosθ = c , sinθ = s ( ) θ n cn cos = ( ) 8 6 2 4 4 2 6 8 8 8 8 8 28 70 28 i s s c s c s c c c is c s c + − + − = ⇒ + = + M1A1 ( ) ( ) ( ) ( ) 8 6 4 2 6 4 2 2 4 2 4 2 6 8 8 4 6 4 1 3 3 1 28 2 1 70 1 28 c c c c c c c c c c c c c c c + − + − + − + − − + − + − − = ⇒ M1A1 1 32 160 256 128 2 4 6 8 8 + − + − = ⇒ c c c c c (*) A1 (i) θ π θ − → 2 / in (*) leads to: 1 32 160 256 128 2 4 6 8 8 + − + − = s s s s c M1A1 (ii) From (*), ( ) 1 cos 1 5 8 4 32 8 / cos 2 3 4 2 − = = + − + − ⇒ = π π x x x x x M1M1 16 / 1 5 8 4 2 3 4 − = − + − ⇒ x x x x A1 11 (2j – k) × (3i + 2j – 2k) = –2i – 3j – 6j (oew) M1A1 14 6 3 2 : 1 = + + Π z y x (AG) M1A1 Perpendicular distance, p, of P from l in terms of 1 parameter, e.g., ( ) ( ) ( ) ( ) 14 5 2 6 6 8 3 4 3 2 7 / 1 − + + + + + = λ λ λ p M1 λ 8 4 + = A1 p Y 4 ⇒ –1 Y λ Y 0 M1A1 (3i + 8j + 2k) – (i + 2j + k) = 2i + 6j + k (2i + 6j + k) × (4i + 6j + 5k) = 24i – 6j – 12k M1A1 ( )( ) 21 / 1 21 7 / 2 4 . 6 3 2 cos = − − + + = k j i k j i α M1 ° = 4. 77 α A1

Page 11 Mark Scheme Syllabus Paper GCE A/AS LEVEL – October/November 2008 9231 01 © UCLES 2008 12 EITHER (i) x = 1, x = 3 (both) B1 y = 1 B1 (ii) Solves (x – 2)(x – a)/(x – 1)(x – 3) = 1 to obtain x = ξ where ξ = (2a – 3)/(a – 2) M1A1 (iii) y1 = 0 ⇒ (x – 2)(x – a)(2x – 4) = (x – 1)(x – 3)(2x – 2 – a) M1 ⇒ (–4 – 4 – 2a)x2 + (4a + 8 + 4a)x – 8a = (–8 – 2 – a)x2 + (6 + 8 + 4a)x – 6 – 3a A1 ⇒ (a – 2)x2 + (6 – 4a)x + (5a – 6) = 0 (AG) A1 (6 – 4a)2 [ 4(a – 2)(5a – 6) M1 ⇒ a2– 4a + 3 Y 0 ⇒ (a – l)(a – 3) Y 0 M1 ⇒ 1 < a < 3 (a ≠ 2 given) A1 (iv) (a) Axes and asymptotes B1 Branches (all) B1 (b) Middle branch with maximum value in the range 0 < y < 1 B1 Outside branches with correctly placed minimum point B1 OR (i) α a root of given equation ⇒ α4 – 5α2 + 2α – 1 = 0 ⇒ αn + 4 – 5αn + 2 + 2αn + 1 – αn = 0 M1 Summing over α, Ä, Ö, Ç, leads to Sn + 4 – 5Sn + 2 + 2Sn + 1 – Sn = 0 A1 (ii) S2 = 10 B1 S4 = 5S2 – 2S1 + 4 = 50 – 0 + 4 = 54 M1A1 (iii) S–1 = 2 from e.g., y4 – 2y3+5y2 – 1 = 0 M1A1 S3 = 5S1 – 2S0 + S–1 = –6 M1A1 OR 2S3 = 3S1S2 – S 3 1 + ∑αβγ 6 M1A1 = 3 × 10 × 0 – 0 + 6 × (–2) M1 ⇒ S3 = –6 A1 S6 = 5S4 – 2S3 + S2 = 292 M1A1 (iv) ∑ 4 2β α = S2S4 – S6 = 540 – 292 = 248 M2A1