Cambridge A Level Mathematics - Further 9231 — 2008 May/June Paper 1 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/01 Paper 1 May/June 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 5 printed pages and 3 blank pages. © UCLES 2008 [Turn over *8437663496*

2 1 The finite region enclosed by the line y = kx, where k is a positive constant, the x-axis for 0 ≤x ≤h, and the line x = h is rotated through 1 complete revolution about the x-axis. Prove by integration that the centroid of the resulting cone is at a distance 3 4h from the origin O. [4] [The volume of a cone of height h and base radius r is 1 3πr2h.] 2 Given that un = ln
1 + xn+1 1 + xn , where x > −1, find N ∑ n=1 un in terms of N and x. [3] Find the sum to infinity of the series u1 + u2 + u3 + . . . when (i) −1 < x < 1, [1] (ii) x = 1. [1] 3 Show that if λ is an eigenvalue of the square matrix A with e as a corresponding eigenvector, and µ is an eigenvalue of the square matrix B for which e is also a corresponding eigenvector, then λ + µ is an eigenvalue of the matrix A + B with e as a corresponding eigenvector. [2] The matrix A =  3 −1 0 −4 −6 −6 5 11 10  has  1 −1 1  as an eigenvector. Find the corresponding eigenvalue. [1] The other two eigenvalues of A are 1 and 2, with corresponding eigenvectors  1 2 −3  and  1 1 −2  respectively. The matrix B has eigenvalues 2, 3, 1 with corresponding eigenvectors  1 −1 1 ,  1 2 −3 ,  1 1 −2  respectively. Find a matrix P and a diagonal matrix D such that (A + B)4 = PDP−1. [3] [You are not required to evaluate P−1.] © UCLES 2008 9231/01/M/J/08

3 4 The curves C1 and C2 have polar equations r = θ + 2 and r = θ2 respectively, where 0 ≤θ ≤π. (i) Find the polar coordinates of the point of intersection of C1 and C2. [2] (ii) Sketch C1 and C2 on the same diagram. [2] (iii) Find the area bounded by C1, C2 and the line θ = 0. [3] 5 The equation x3 + x −1 = 0 has roots α, β, γ . Show that the equation with roots α3, β3, γ 3 is y3 −3y2 + 4y −1 = 0. [4] Hence find the value of α6 + β6 + γ 6. [3] 6 The curve C is defined parametrically by x = 4t −t2 and y = 1 −e−t, where 0 ≤t < 2. Show that at all points of C, d2y dx2 = (t −1)e−t 4(2 −t)3 . [4] Show that the mean value of d2y dx2 with respect to x over the interval 0 ≤x ≤7 4 is 4e−1 2 −3 21 . [4] 7 Prove by induction that n ∑ r=1 3r5 + r3 = 1 2n3(n + 1)3, for all n ≥1. [5] Use this result together with the List of Formulae (MF10) to prove that n ∑ r=1 r5 = 1 12n2(n + 1)2Q(n), where Q(n) is a quadratic function of n which is to be determined. [3] © UCLES 2008 9231/01/M/J/08 [Turn over

4 8 (i) Given that In =  1 2π 0 tn sin t dt, show that, for n ≥2, In = n
π 2 n−1 −n(n −1)In−2. [5] (ii) A curve C in the x-y plane is defined parametrically in terms of t. It is given that dx dt = t4(1 −cos 2t) and dy dt = t4 sin 2t. Find the length of the arc of C from the point where t = 0 to the point where t = 1 2π. [5] 9 The curve C has equation y = x2 −2x + λ x + 1 , where λ is a constant. Show that the equations of the asymptotes of C are independent of λ. [3] Find the value of λ for which the x-axis is a tangent to C, and sketch C in this case. [4] Sketch C in the case λ = −4, giving the exact coordinates of the points of intersection of C with the x-axis. [3] 10 By considering N ∑ n=1 2n−1, where  = eiθ, show that N ∑ n=1 cos(2n −1)θ = sin(2Nθ) 2 sin θ , where sin θ ≠0. [6] Deduce that N ∑ n=1 (2n −1) sin(2n −1)π N  = −N cosec π N . [4] 11 Show that, with a suitable value of the constant α, the substitution y = xαw reduces the differential equation 2x2 d2y dx2 + (3x2 + 8x)dy dx + (x2 + 6x + 4)y = f(x) to 2d2w dx2 + 3dw dx + w = f(x). [5] Find the general solution for y in the case where f(x) = 6 sin 2x + 7 cos 2x. [6] © UCLES 2008 9231/01/M/J/08

5 12 Answer only one of the following two alternatives. EITHER The position vectors of the points A, B, C, D are 7i + 4j −k, 3i + 5j −2k, 2i + 6j + 3k, 2i + 7j + λk respectively. It is given that the shortest distance between the line AB and the line CD is 3. (i) Show that λ 2 −5λ + 4 = 0. [7] (ii) Find the acute angle between the planes through A, B, D corresponding to the values of λ satisfying the equation in part (i). [7] OR The linear transformation T : 4 →4 is represented by the matrix ⎛⎜⎜⎝ 1 2 −1 −1 1 3 −1 0 1 0 3 1 0 3 −4 −1 ⎞⎟⎟⎠. The range space of T is denoted by V. (i) Determine the dimension of V. [3] (ii) Show that the vectors ⎛⎜⎜⎝ 1 1 1 0 ⎞⎟⎟⎠, ⎛⎜⎜⎝ 2 3 0 3 ⎞⎟⎟⎠, ⎛⎜⎜⎝ −1 −1 3 −4 ⎞⎟⎟⎠are linearly independent. [4] (iii) Write down a basis of V. [1] The set of elements of 4 which do not belong to V is denoted by W. (iv) State, with a reason, whether W is a vector space. [1] (v) Show that if the vector ⎛⎜⎜⎝ x y  t ⎞⎟⎟⎠belongs to W then y − −t ≠0. [5] © UCLES 2008 9231/01/M/J/08

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/M/J/08

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 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 May/June 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 LEVEL – May/June 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 LEVEL – May/June 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 LEVEL – May/June 2008 9231 01 © UCLES 2008 1 dx h rx x dx xy 2 2 ) / ( h h 0 0 ∫ = ∫ M1 /4 ... 2 2h r = = (Q) A1 /4 3 / h V Q x = = M1A1 OR dx x ρπK x h ρπK 3 2 3 2 (1/3) h 0∫ = M1M1 [ ] h x x h 0 4 3 /4 (1/3) = A1 /4 3h x = (AG) A1 2 ) ln(1 ) ln(1 1 x x u n n + − + = + or for ln{Product of fractions} B1 ( ) ( ) [ ] x x S u N N N n n + + = = + = ∑ 1 / 1 ln 1 1 (AEF) Cancels → result M1A1 (i) ) ln(1 x S + − = ∞ OR       + x 1 1 ln A1 (ii) 0 = ∞ S B1 3 Ae = λ e and Be = µe ⇒ Ae + Be = λ e + µe M1 ⇒ (A + B)e = ( λ + µ)e A1 λ = 4 B1           − − − = 1 3 2 1 2 1 1 1 1 P B1           = 1296 0 0 0 0 81 256 0 0 D M1A1 (ft)

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 4 (i) θ = 2, r = 4 B1B1 Ignore extra values Accept as written in MS – co-ords not required (ii) Graphs: correct location, orientation and concavity required B1B1 Separate diagrams 1/2, B1 Shapes correct B1 Intersection correct (iii) A1 = (1/2) ( ) θ θ d 2 2 + ∫2 0 (LNR) M1 = … = 28/3 A2 = (1/2) θ θ 4d 2 0∫ = … = 16/5 (LR) A1 M1 for 1 correct integral representation plus A1 if both correct Area = A1 – A2 = 92/15 (6.13) A1 S.C. –92/15 M1 A0 A1 Alternative layout: ( ) θ θ θ d 2 2 1 4 2 − + ∫2 0 M1 (LNR) = 2 0 5 2 3 5 4 2 3 2 1       − + + θ θ θ θ A1 (LR) = 92/15 A1 5 Uses substitution y = x3 M1 Obtains y + y1/3 – 1 = 0 A1 y = (1 – y)3 A1 ⇒…⇒ y3 – 3y2 + 4y – 1 = 0 (AG) A1 ( ) 3 3 2 3 6 2 γ β α α ∑ ∑ ∑ − = B1 = 9 – 8 = 1 M1A1 OR put y = z1/2 to obtain z3 – z2 + 10z – 1 = 0 M1A1 ∑ 6 α = –coefficient of z2, = 1 A1

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 6 y1 = e–t/(4 – 2t) (Tolerate sign error) M1 y2 = any correct form in t Omission of dx dt M0 M1A1 y2 = (t – 1)e–t/4(2 – t)3 (AG) A1 Mean value = (4/7) x y d 2 7/4 0∫ M1 (Limits may be given as t = 0 to t = 1/2) [ ] ( ) 1/4 1/3 d 1/2 7/4 1 2 − = = ∫ − e y x y 0 4 / 7 0 (AEF) M1(LNR) A1(LR) = (1/21) (4e–1/2 – 3) (AG) A1 7 Verifies H1 to be true B1 Hk : ∑= k r 1 (3r5 + r3) = (1/2)k3(k + 1)3 B1 Hk ⇒ ∑ + = 1 1 k r (3r5 + r3) = (1/2)k3(k + 1)3 + 3(k + 1)5 + (k + 1)3 M1 = … = (1/2)(k + 1)3(k + 2)3 A1 Thus Hk ⇒ Hk+1 and concludes A1 ∑= n r 1 3 r5 + (1/4)n2(n + 1)2 = (1/2)n3(n + 1)3 M1 ⇒…⇒∑= n r 1 r5 = (1/12)n2(n + 1)2(2n2 + 2n – 1) M1A1

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 8 (i) [ ] cdt t n c t sdt t n π n n 1 /2 0 − ∫ + − = ∫ 2 / 0 2 / 0 π π M1A1(LR) [ ] 2 /2 0 1 1) ( − − − − = n π n I n n s nt M1A1 2 1 1) ( /2) ( − − − − = ⇒ n n n I n n n I π (AG) A1 (ii) L = length of arc = dt y x 2 2 & & + ∫ 2 / 0 π with integrand expressed in terms of t M1 s t y x 4 2 2 2 = + & & B1 4 4 2 2 I dt s t L = = ∫ A1 2 π 2 − = I A1 48 24π π3 + − = L (AEF), e.g., ( ) 2) 24(π π/2 8 − − = 3 L Accept 3.61 A1 9 One asymptote is 1 − = x (Allow 1 − → x B1 3 − →x y ) 1) 3)/( (λ 3 + + + − = x x y M1 3 − = ⇒ x y A1 1 λ = B1 Axes plus both asymptotes drawn B1 ft DEF: Here Z denotes ‘correct shape, orientation and approximately correct location’ RH branch with Z B1 ft LH branch with Z B1 4 λ − = : RH branch with Z B1 LH branch with Z B1 Intersections with x-axis (1 + 5 , 0), (1 – 5 , 0) B1

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 10 ( ) ( ) 2 1 2 1 1 2 1 / z z z z N N n n − − = + = − ∑ (Sum of G.P.) B1 ( ) ( ) [ ] 2 1 2 1 / z z z S N N − − ℜ = + (Real Part) M1 ( )( ) ( ) [ ] θ z z z S N N 2cos2 2 1 2 1 2 − − − ℜ = + / (AEF with a real denominator) M1 ( ) ( ) ( ) ( ) θ θ N θ N 2cos2 2 / 1 2 cos 1 2 cos ... − + − − = = (AEF in θ and N, only.) M1A1 Multiplying out numerator θ Nθ/2sin sin2 ... = = (AG) A1 ( ) ( ) θ θ Nθ θ Nθ N θ n n N n cot csc (1/2)sin2 csc cos2 1 2 sin 1 2 1 − = − − −∑= (AEF) M1A1(LHS) A1(RHS) Put N θ π/ = to obtain required result (AG) A1 11 w x α α w x 2 w x y w αx w x y α α α α α 2 1 1 2 2 1 1 1 1) ( , − − − − + + = + = α B1B1 (Must see general results with α ) Obtain any DE of the form ) ( ) , ( ) , ( ) , ( 1 2 x f w α x R w α x Q w α x P = + + M1 Sets 2 − = α and obtains required w x − DE (AG) M1A1 oew Complementary function /2 x x Be Ae − −+ = M1A1 Puts x Q x P cos2 sin2 + to obtain 2 linear equations in P and Q M1A1 Solves to obtain 1 0, − = = Q P A1 [ ] x Be Ae x y x x cos2 /2 2 − + = − − − A1 Allow [ ] x Be Ae y x x x cos2 /2 2 − + = − −

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 12 EITHER (i) k j i 4 12) (4λ 2) (λ − − + − = × CD AB M1A1 [ ] 30 3λ 4 12) (4λ 2) (λ ). 4 2 5 ( − = − − + − + + − k j i k j i M1A1 3 16 12) (4λ 2) (λ / 30 3λ 2 2 = + − + − − M1A1 0 4 5λ λ 2 = + − (AG) A1 (ii) 4 1, λ − B1 k j i n 7 13 5 : 1 λ 1 − + = = M1A1 cao k j i n 7 25 8 : 4 λ 2 − + = = M1A1 cao Acute angle between planes ° = = − 12.1 / . cos 2 1 2 1 1 n n n n M1A1 cao [For lines 2 and 3: ( ) ( ) 3 4 4λ 12 λ 2 16 8λ 24 5λ 10 2 2 2 = + − + − − + − − M1A1 2 2 17λ 100λ 164 10) (λ + − = − M1A1] 12 OR (i) Transform given matrix to an echelon form, e.g.,             − − 0 0 0 0 1 1 0 0 1 0 1 0 1 1 2 1 . M1A1 3 ) ( = V dim OR BY EQUIVALENT METHOD If written down with no working 1/3 B1 (ii) 0, α α α 4 3 1 1 α 3 0 3 2 α 0 1 1 1 α 3 2 1 3 2 1 = = = ⇒ =             − − − +             +             0 shown ⇒ linear independence M2A2 (iii)                         − − −                         4 3 1 1 , 3 0 3 2 , 0 1 1 1 B1

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2008 9231 01 © UCLES 2008 (iv) W not a vector space since W does not contain the zero vector, or equivalent or not closed wrt addition B1 (v) Reduces             + + − + + − − − →             − − − t z y z y x x y x t z y x 0 0 0 2 3 4 0 0 0 1 0 1 2 1 4 3 0 3 0 1 1 3 1 1 2 1 M1A1 0 iff = − − ∈             ≡ ⇒ t z y V t z y x b or equivalent method M1A1 so that 0 ≠ − − ⇒ ∈ t z y W b A1 Alternative: suppose             − − − +             +             =             4 3 1 1 γ 3 0 3 2 β 0 1 1 1 α t z y x (i.e. in V) 0 ... ... ... ... ... = = − − ⇒ = = = = t z y t z y x Hence 0 ≠ − − ⇒ ∈             t z y W t z y x