Cambridge A Level Mathematics - Further 9231 — 2009 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 2009 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 2009 [Turn over *0307464767*

2 1 Given that y = x2 sin x, (i) show that the mean value of dy dx with respect to x over the interval 0 ≤x ≤1 2π is 1 2π, [2] (ii) find the mean value of d2y dx2 with respect to x over the interval 0 ≤x ≤1 2π. [2] 2 Relative to an origin O, the points A, B, C have position vectors i, j + k, i + j + θk, respectively. The shortest distance between the lines AB and OC is 1 √2. Find the value of θ. [6] 3 The curve C has equation y = x2 −5x + 4 x + 1 . (i) Obtain the coordinates of the points of intersection of C with the axes. [2] (ii) Obtain the equation of each of the asymptotes of C. [3] (iii) Draw a sketch of C. [3] 4 It is given that x = t + sin t, y = t2 + 2 cos t, where −π < t < π. Find dy dx in terms of t. [2] Show that d2y dx2 = 2t sin t (1 + cos t)3. [4] Show that dy dx increases with x over the given interval of t. [2] 5 The equation x3 + 5x + 3 = 0 has roots α, β, γ . Use the substitution x = −3 y to find a cubic equation in y and show that the roots of this equation are βγ, γ α, αβ. [4] Find the exact values of β2γ 2 + γ 2α2 + α2β2 and β3γ 3 + γ 3α3 + α3β3. [5] © UCLES 2009 9231/01/O/N/09

3 6 Show that d dxxn−1 √(4 −x2) = 4(n −1)xn−2 √(4 −x2) − nxn √(4 −x2). [3] Let In = ä 1 0 xn √(4 −x2) dx, where n ≥0. Prove that nIn = 4(n −1)In−2 −√3, for n ≥2. [2] Given that I0 = 1 6π, find I4, leaving your answer in an exact form. [4] 7 Use de Moivre’s theorem to express sin6 θ in the form a + b cos 2θ + c cos 4θ + d cos 6θ, where a, b, c, d are constants to be found. [5] Hence evaluate ã 1 4π 0 sin6 2x dx, leaving your answer in terms of π. [4] 8 (a) The curve C1 has equation y = −ln(cos x). Show that the length of the arc of C1 from the point where x = 0 to the point where x = 1 3π is ln(2 + √3). [5] (b) The curve C2 has equation y = 2 √(x + 3). The arc of C2 joining the point where x = 0 to the point where x = 1 is rotated through one complete revolution about the x-axis. Show that the area of the surface generated is 8 3π(5 √5 −8). [5] © UCLES 2009 9231/01/O/N/09 [Turn over

4 9 Show that if y depends on x and x = eu then x2 d2y dx2 = d2y du2 −dy du. [4] Given that y satisfies the differential equation x2 d2y dx2 + 5xdy dx + 3y = 30x2, use the substitution x = eu to show that d2y du2 + 4dy du + 3y = 30e2u. [2] Hence find the general solution for y in terms of x. [5] 10 The curve C has polar equation r = a sin 3θ, where 0 ≤θ ≤1 3π. (i) Show that the area of the region enclosed by C is 1 12πa2. [3] (ii) Show that, at the point of C at maximum distance from the initial line, tan 3θ + 3 tan θ = 0. [3] (iii) Use the formula tan 3θ = 3 tan θ −tan3 θ 1 −3 tan2θ to find this maximum distance. [4] (iv) Draw a sketch of C. [2] © UCLES 2009 9231/01/O/N/09

5 11 Answer only one of the following two alternatives. EITHER Prove by induction that N ∑ n=1 n3 = 1 4N2(N + 1)2. [5] Use this result, together with the formula for N ∑ n=1 n2, to show that N ∑ n=1 (20n3 + 36n2) = N(N + 1)(N + 3)(5N + 2). [3] Let SN = N ∑ n=1 (20n3 + 36n2 + µn). Find the value of the constant µ such that SN is of the form N2(N + 1)(aN + b), where the constants a and b are to be determined. [3] Show that, for this value of µ, 5 + 22 N < N−4SN < 5 + 23 N , for all N ≥18. [3] OR One of the eigenvalues of the matrix A = 1 −4 6 2 −4 2 −3 4 a ! is −2. Find the value of a. [3] Another eigenvalue of A is −5. Find eigenvectors e1 and e2 corresponding to the eigenvalues −2 and −5 respectively. [3] The linear space spanned by e1 and e2 is denoted by V. (i) Prove that, for any vector x belonging to V, the vector Ax also belongs to V. [4] (ii) Find a non-zero vector which is perpendicular to every vector in V, and determine whether it is an eigenvector of A. [4] © UCLES 2009 9231/01/O/N/09

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. 9231/01/O/N/09

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 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, 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 October/November 2009 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 – October/November 2009 9231 01 © UCLES 2009 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 – October/November 2009 9231 01 © UCLES 2009 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 – October/November 2009 9231 01 © UCLES 2009 1 (i) MV of y1 over 0 Y x Y π/2 = 2/π [x2 sinx] 2 / 0 π M1 = ... = π/2 (AG, CWO) A1 (ii) MV of y2 over 0 Y x Y π/2 = 2/π [2x sinx + x2 cosx] 2 / 0 π = 2 M1A1 2 Evaluates some relevant vector product, e.g., (i – j – k) × (i + j + θk) = (1 – θ)i – (1 + θ)j + 2k M1A1 [ ] 6 2 / . 2 ) (1 ) (1 p 2 + + + = − − θ θ θ i k j i M1 6 2 )/ (1 2 + = − θ θ A1 Puts 2 / 1 = p to obtain a horizontal equation such as 2θ2 + 6 = 2(θ – 1)2 M1 ⇒…⇒ θ = –1 A1 3 (i) (1,0), (4,0) B1 (0,4) B1 (ii) One asymptote is x = –1 B1 y = x – 6 + 10/(x + 1) M1 Other asymptote: y = x – 6 A1 (iii) Sketch: Axes and asymptotes B1 Upper branch: Correct location and orientation B1 Lower branch correctly located and orientated B1 4 dx/dt = 1 + cost, dy/dt = 2t – 2sint dy/dx = (2t – 2sint)/(1 + cost) M1A1 Any correct result for d(dy/dx)/dt in terms of t B1 d2y/dx2 = [(2 – 2cost)(1 + cost) + sint(2t – 2sint)]/(l + cost)3 (AEF) M1A1 d2y/dx2 = 2tsint/(l + cost)3 (AG) A1 Consideration of sign of d2y/dx2 M1 –π < t < 0: (–)(–)/+ > 0; 0 < t < π : (+)(+)/+ > 0 ⇒ d2y/dx2 > 0, ∀ non-zero t ∈ (–π,π) A1 5 Shows y = –3/x ⇒ y3 – 5y2 – 9 = 0 B1 y = –3/x ⇒ y = αβγ/x M1 ⇒ y = βγ, γα, αβ when x = α, β, γ, respectively M1A1 OR for previous 3 marks: βγ + γα + αβ = 5 B1 α2βγ + αβ2γ + αβγ2 = αβγ(α + β + γ) = αβγ × 0 = 0 B1 βγγααβ = (αβγ)2 = 9 B1 Σα2β2 = 25 – (2 × 0) = 25 M1A1 Σα3β3 – 5 Σα2β2 – 27 = 0 M1A1 ⇒…⇒ Σα3β3 = 152 A1

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2009 9231 01 © UCLES 2009 6 [ ] ( ) 2 2 2 2 1 4 / 4 1 4 x x x n n n x n x x dx d − − − − − − = − B1 Shows that this preliminary result implies first displayed result (AG) M1A1 [ ] ( ) n n n nI I n x x − − = − − − 2 1 0 2 1 1 4 4 M1 ⇒…⇒ second displayed result (AG) A1 ,2 / 3 3 / 6 / 2 0 − = ⇒ = π π I I ,3 12 4 2 4 − = I I (all) B1B1 4 / 3 7 4 − = π I M1A1 OR for last 4 rnarks: 3 6 / .1.4 2 2 − = π I M1 2 / 3 3 / 2 − = ⇒ π I A1 ( ) 3 2 / 3 3 / 3.4 4 4 − − = ⇒ π I A1 4 / 3 7 4 − = ⇒ π I A1 7 ( ) ( ) θ θ i z z z e / 1 sin 64 6 6 = − − = M1 ( ) ( ) ( ) 20 / 1 15 / 1 6 / 1 2 2 4 4 6 6 + + − + + + − = z z z z z z M1A1 ( ) ( ) ( ) θ θ θ θ 6 cos 32 / 1 4 cos 16 / 3 2 cos 32 / 15 16 / 5 sin 6 − + − = M1A1 ( ) ( ) ( ) x x x x 12 cos 32 / 1 8 cos 16 / 3 4 cos 32 / 15 16 / 5 2 sin 6 − + − = M1 Any one of ∫ = /4 0 0 cos π dx kx for k = 1, 2, 3 B1 Further B1 if all 3 are written down or implied B1 ∫ = /4 0 6 /64 5 2 sin π π dx x or 4 / a π (CWO) A1 OR for last 4 marks ( ) ( ) ( ) u d u u u du u I ∫ ∫ − + − = = /2 0 /2 0 6 cos6 6cos4 15cos2 10 1/64 sin 1/2 π π M1 ( )[ ] /2 0 6 / 6 sin 2 / 4 sin 3 2 / 2 sin 15 10 64 / 1 π u u u u − + − = M1A1 64 / 5π = (CWO) A1 8 (a) y = tanx B1 x y sec 1 2 1 = + M1 ∫ = π/3 dx x s 0 sec (AEF) A1 ( ) [ ] ( ) 3 2 ln ... tan sec ln /3 0 + = = + = π x x (AG) M1A1 (b) ( ) ( ) ( ) dx x x S 2 / 1 2 / 1 1 0 3 / 1 1 3 4 + + + = ∫ π (AEF) M1A1 ( ) dx x 2 / 1 1 0 4 4 ... ∫ + = = π A1 ( )( ) [ ] 1 0 2 / 3 4 3 / 8 + = x π A1 ( )[ ] 8 5 5 3 / 8 − = π (AG) A1

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2009 9231 01 © UCLES 2009 9 du dy dx dy x = B1 Some relevant, correct intermediate result such as: ( ) ( ) 2 2 2 2 2 / / / du y d dx dy x dx y d x = + M1A1 du dy du y d dx y d x − = ⇒ 2 2 2 2 2 (AG) A1 Uses u x e = to obtain: u y du dy du y d 2 2 2 e 30 3 4 = + + (AG) M1A1 Complementary function u u B A 3 e e − −+ = M1A1 Particular integral u 2 e 2 = M1A1 General solution for y in the x-domain: 2 3 2 / / x x B x A y + + = A1 10 (i) Area ( )∫ = /3 0 2 2 3 sin 2 / π dθ θ a M1 ( ) ( ) ∫ − = /3 0 2 6 cos 1 4 / π dθ a θ A1 12 / ... 2 a π = = (AG) A1 (ii) Considers θ θ sin 3 sin a y = B1 0 cos 3 sin sin 3 cos 3 0 / = + ⇒ = θ θ θ θ θ d dy M1 0 tan 3 3 tan ... = + ⇒ ⇒ θ θ (AG) A1 (iii) Uses ( ) ( ) θ θ θ θ 2 3 tan 3 1 tan tan 3 3 tan / − − = to obtain 5 / 3 tan 2 = θ M1A1 Uses θ θ sin 3 sin = y with 5 / 3 tan = θ M1 Obtains y = 9a/16 (Accept 0.5625a, or 0.563a) A1 (iv) Closed loop entirely in the first quadrant with lower end at the pole and clearly tangential to the initial line at the pole B1 Symmetric about line 6 / π θ = with correct shape at (a, π /6) B1

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2009 9231 01 © UCLES 2009 11 EITHER ( ) ( ) 2 2 1 3 1 4 / 1 : + = = ∑= k k n S H k n k k for some k B1 ( ) ( ) ( ) 3 2 2 1 1 1 4 / 1 + + + = ⇒ + k k k S H k k M1 ( )( ) ( ) 2 2 2 1 4 / 1 ... + + = = k k so that 1 + ⇒ k k H H M1A1 Verifies H1 is true and completes induction argument A1 ( ) ( ) ( )( )1 2 1 6 1 5 36 20 2 2 1 2 3 + + + + = + ∑= N N N N N n n N n M1 ( )( )( ) 2 5 3 1 ... + + + = = N N N N (AG) M1A1 ( )( )( ) ( ) ( )1 2 / 2 5 3 1 + + + + + = N N N N N N SN µ M1 ( )( ) 2 / 6 17 5 1 2 µ + + + + = N N N N M1 Take 12 − = µ , then ( )( ) 17 5 1 2 + + = N N N SN so that a = 5, b = 17 A1 2 4 / 17 / 22 5 N N S N N + + = − , N / 22 5 + > , N ∀ [=1 M1, A1 N [=18 N N N / 1 / 17 17 2 < ⇒ > ⇒ N S N N / 23 5 4 + < ⇒ − (AG) A1 11 OR det (A + 2I) = ... = 2a + 16 M1A1 det (A + 2I) = 0 ⇒ a = –8 A1 Eigenvectors corresponding to –2 and –5: any non-zero scaling of           1 3 2 ,           −1 0 1 M1A1A1 (i) x ∈ V ⇒x of the form pe1 + qe2 B1 Ax = A(pe1 + qe2) = p(Ae1) + q(Ae2) M1A1 = –2pe1 – 5qe2 ∈ V A1 (ii) (2i + 3j + k) × (i – k) = –3i + 3j – 3k (AEF) M1A1 A(i – j + k) = 11i + 8j – 15k which is not a scaling of i – j + k M1 Hence i – j + k is not an eigenvector of A (CWO) A1 OR for the last 2 marks 3rd eigenvector is –4 with corresponding eigenvector           − − 4 1 4 B1 ∴ i – j + k is not an eigenvector B1