Cambridge A Level Mathematics - Further 9231 — 2011 May/June Paper 1 · Variant 2

*8728606253* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 May/June 2011 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. JC11 06_9231_12/FP © UCLES 2011 [Turn over

2 1 Express 1 (2r + 1)(2r + 3) in partial fractions and hence use the method of differences to find n ∑ r=1 1 (2r + 1)(2r + 3). [4] Deduce the value of ∞ ∑ r=1 1 (2r + 1)(2r + 3). [1] 2 The roots of the equation x3 + px2 + qx + r = 0 are β k , β, kβ, where p, q, r, k and β are non-zero real constants. Show that β = −q p. [4] Deduce that rp3 = q3. [2] 3 The linear transformation T : >4 →>4 is represented by the matrix M =  1 3 −2 4 5 15 −9 19 −2 −6 3 −7 3 9 −5 11 . (i) Find the rank of M. [3] (ii) Obtain a basis for the null space of T. [3] 4 It is given that f(n) = 33n + 6n−1. (i) Show that f(n + 1) + f(n) = 28(33n) + 7(6n−1). [2] (ii) Hence, or otherwise, prove by mathematical induction that f(n) is divisible by 7 for every positive integer n. [4] 5 The curve C has polar equation r = 2 cos 2θ. Sketch the curve for 0 ≤θ < 2π. [4] Find the exact area of one loop of the curve. [4] 6 The line l1 passes through the point with position vector 8i + 8j −7k and is parallel to the vector 4i + 3j. The line l2 passes through the point with position vector 7i −2j + 4k and is parallel to the vector 4i −k. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. In either order, (i) show that PQ = 13, (ii) find the position vectors of P and Q. [9] © UCLES 2011 9231/12/M/J/11

3 7 The variables x and y are related by the differential equation y2 d2y dx2 + 2y2 dy dx + 2ydy dx 2 −5y3 = 8e−x. Given that v = y3, show that d2v dx2 + 2dv dx −15v = 24e−x. [4] Hence find the general solution for y in terms of x. [7] 8 Find the eigenvalues and corresponding eigenvectors of the matrix A =  4 −1 1 −1 0 −3 1 −3 0 . [8] Find a non-singular matrix P and a diagonal matrix D such that A5 = PDP−1. [3] 9 The curve C has equation y = x 3 2. Find the coordinates of the centroid of the region bounded by C, the lines x = 1, x = 4 and the x-axis. [7] Show that the length of the arc of C from the point where x = 5 to the point where x = 28 is 139. [5] 10 Let In = ã 1 2π 0 cosnx dx, where n ≥0. Show that, for all n ≥2, In = n −1 n In−2. [4] A curve has parametric equations x = a sin3t and y = a cos3t, where a is a constant and 0 ≤t ≤1 2π. Show that the mean value m of y over the interval 0 ≤x ≤a is given by m = 3a ã 1 2π 0 (cos4t −cos6t) dt. [4] Find the exact value of m, in terms of a. [4] [Question 11 is printed on the next page.] © UCLES 2011 9231/12/M/J/11 [Turn over

4 11 Answer only one of the following two alternatives. EITHER Use de Moivre’s theorem to prove that tan 3θ = 3 tan θ −tan3θ 1 −3 tan2θ . [6] State the exact values of θ, between 0 and π, that satisfy tan 3θ = 1. [2] Express each root of the equation t3 −3t2 −3t + 1 = 0 in the form tan(kπ), where k is a positive rational number. [3] For each of these values of k, find the exact value of tan(kπ). [3] OR The curve C has equation y = x2 + λx −6λ 2 x + 3 , where λ is a constant such that λ ≠1 and λ ≠−3 2. (i) Find dy dx and deduce that if C has two stationary points then −3 2 < λ < 1. [5] (ii) Find the equations of the asymptotes of C. [3] (iii) Draw a sketch of C for the case 0 < λ < 1. [3] (iv) Draw a sketch of C for the case λ > 3. [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 2011 9231/12/M/J/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/12 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 May/June 2011 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 2011 9231 12 © University of Cambridge International Examinations 2011 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 2011 9231 12 © University of Cambridge International Examinations 2011 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 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 1 Any method including cover-up rule.       + − + = + + 3 2 1 1 2 1 2 1 ) 3 2 )( 1 2 ( 1 r r r r B1 Expresses all terms as differences. n S             + − +       − +       − = 3 2 1 1 2 1 … 7 1 5 1 5 1 3 1 2 1 n n M1A1 Finds sum. ) 3 2 ( 2 1 6 1 + − = n (acf) A1 4 6 1 = ∞ S (B0M1A1√ A0A1√ if signs reversed.) A1 1 [5] 2 Sum of roots. p k k k − = + + 2 β β β B1 Sum of products in pairs. q k k = + + 2 2 2 β β β B1 Factorises. p k k k − =       + + ⇒ 1 2 β and q k k k =       + + 1 2 2 β M1 p q − = ⇒β (AG) A1 4 Product of roots. r − = 3 β B1 3 3 3 3 q rp r p q = ⇒ − = − ⇒ (AG) B1 2 [6] 3 (i) Reduces matrix to echelon form.             − − − − − − 11 5 9 3 7 3 6 2 19 9 15 5 4 2 3 1 . →             − − 0 0 0 0 0 0 0 0 1 1 0 0 4 2 3 1 M1A1 Uses rank = dimension – nullity r(A) = 4 – 2 = 2 A1 3 (ii) Obtains a set of equations. 0 0 4 2 3 = − = + − + t z t z y x B1 Finds basis vectors. Basis is                        −            − 0 0 1 3 , 1 1 0 2 (OE) M1A1 3 [6]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 4 (i) Establishes initial result. n n n n n n 6 3 6 3 1) f( ) f( 3 3 1 3 + + + = + + + − ) 6 1( 6 ) 27 1( 3 1 3 + + + = − n n ) 6 ( 7 ) 3 ( 28 1 3 − + = n n (AG) M1 A1 2 (ii) States inductive hypothesis. Proves base case. Shows Pk ⇒ Pk+1. States conclusion. Hk: f(k) = 7λ ⇒ × = = + 7 4 28 6 3 0 3 H1 is true f(k + 1) + f(k) = f(k +1) + 7λ = ) 6 ( 7 ) 3 ( 28 1 3 − + k k = µ 7 ) ( 7 )1 ( f λ µ − = + ⇒ k ∴Hk⇒Hk+1 (Hence by the principle of mathematical induction Hn is) true for all positive integers n. B1 B1 M1 A1 4 [6] 5 Right-hand loop Left-hand loop Deduct 1 mark for extra loops (r < 0). Position and through pole and (2, 0). Position and through pole and (2, π). B1B1 B1B1 4 Uses A = ∫ 2 2 1 r dθ A = ∫ θ θ d cos 4 2 1 2 dθ M1 Uses double angle formula. ∫ + = )1 4 cos ( θ dθ (LNR) M1 Integrates. = [ ] θ θ + 4 sin (LNR) A1 Inserts any appropriate limits which legitimately give the result. e.g. [ ]4 4 4 sin π π θ θ − + 2 π = A1 4 [8]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 6 (i) Uses vector product to find vector perpendicular to both lines. n = = −1 0 4 0 3 4 k j i –3i + 4j – 12j M1A1 Finds BA and its scalar product with unit perpendicular vector. BA = i + 10j – 11j perp.dist. = 13 169 169 12 4 3 11 12 10 4 1 3 2 2 2 = = + + × + × + × − = 13 (AG) (No penalty for sign errors made in n, which lead to correct result.) M1A1 4 (ii) p =           − + + 7 3 8 4 8 λ λ q =           − − + µ µ 4 2 4 7 PQ =           − − − + − − µ λ µ λ 11 3 10 4 4 1 =           − − 12 4 3 t B1 M1A1 1 2 1 − = − = − = ⇒ µ λ t M1 p = 2j – 7k q = 3i – 2j + 5k (Award B1B1B1 if t assumed to be ±1 i.e. 3/5) A1 5 [9] 6 (ii) Alternative Solution: Finds two parameter representation for PQ p =           − + + 7 3 8 4 8 λ λ q =           − − + µ µ 4 2 4 7 Uses scalar product between PQ and direction vector of at least one line and equates to zero PQ =           − − − + − − µ λ µ λ 11 3 10 4 4 1 B1 34 25 16 = − λ µ M1A1 15 16 17 = − λ µ A1 Solves simultaneously. 1 − = µ 2 − = λ M1A1 Obtains position vectors for P and Q p = 2j – 7k q = 3i – 2j + 5k A1 7 (i) Obtains length of PQ. 13 12 ) 4 ( 3 2 2 2 = + − + (AG) M1A1 2 [9]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 7 Differentiates twice. 2 2 2 3 ) ( 6 3 3 y y y y v y y v y v ′ + ′′ = ′′ ⇒ ′ = ′ ⇒ = B1B1 Substitutes. x v v y y y y v − = − ′ + ′ + ′ − ′′ e 8 5 3 2 ) ( 2 ) ( 2 3 1 2 2 M1 Obtains v-x equation. x v x x v - 2 2 e 24 15 d dv 2 d d = − + ⇒ (AG) A1 4 Solves AQE = ⇒ = − + m m m 0 15 2 2 5 − , 3 M1 Finds CF. CF: x x B A 3 5 e e + − A1 Differentiates form for PI. PI: x k v -e = x x k v k v − − = ′′ ⇒ − = ′ ⇒ e e M1 Substitutes. x x x x k k k − − − − = − − e 24 e 15 e 2 e M1 Obtains PI 2 3 e 24 e 16 − = ⇒ = − ⇒ − − k k x x A1 Obtains GS GS: x x x B A v − − − + = e 2 3 e e 3 5 A1 Obtains y in terms of x (from a complete, if incorrect, GS). 3 1 3 5 e 2 3 e e   − + = − − x x x B A y A1√ 7 [11] 8 Finds characteristic equation and solves. Det(A – λI) = 0 0 30 11 4 2 3 = + − − ⇒ λ λ λ = ⇒λ –3, 2, 5 (Any one) (All three) M1A1 A1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 3 − = λ e1 =           =           = − − − 1 1 0 20 20 0 3 3 1 1 1 7 t k j i M1A1 2 = λ e2 =           − =           − = − − − − 1 1 1 5 5 5 3 2 1 1 1 2 t k j i A1 5 = λ e3 =           − =           − = − − − − − 1 1 2 4 4 8 3 5 1 1 1 1 t k j i A1 8 Forms P from eigenvectors and D with fifth powers of eigenvalues on leading diagonal. Note: Columns of P and D can be permuted, but must match. P =           − − 1 1 1 1 1 1 2 1 0 D =          − 3125 0 0 0 32 0 0 0 243 B1√ M1 A1√ 3 [11]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 9 Uses formulae for coords of centroid. x x x x x d d 4 1 2 3 4 1 2 5 ∫ ∫ = ∫ ∫ = 4 1 2 3 4 1 3 d d 2 1 x x x x y M1M1 Numerators. 7 254 7 2 4 1 2 7 =         x 8 255 8 1 4 1 4 =     x A1A1 Denominator. 5 62 5 2 4 1 2 5 =         x B1 Obtains values. (Accept rational values.) 217 635 = x (= 2.93) 496 1275 = y (= 2.57) M1A1 7 Differentiates y wrt x and squares. 4 9 ) ( 2 3 2 2 3 x y x y x y = ′ ⇒ = ′ ⇒ = B1 Subs in arc length formula (LR). x x s ∫ + = 28 5 d 4 9 1 B1 Integrates. 28 5 2 3 4 9 1 27 8                + = x or ( ) 28 5 2 3 9 4 27 1     + x M1A1 Obtains printed result. = 27 7 16 or 27 7 8 3 3 3 4 − − = 139 (AG) A1 5 [12]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 10 Puts x x x n n cos . cos cos 1 − = and integrates by parts. = nI [ ] ∫ − − − + 2 0 2 2 2 0 1 sin cos )1 ( sin cos π π x x n x x n n dx M1A1 Uses x x 2 2 sin 1 cos − = and obtains reduction formula. ) cos (cos )1 ( 2 0 2 ∫ − − = − π x x n I n n n dx 2 1 − − = ⇒ n n I n n I ( 2 ≥ n ) (AG) M1 A1 4 Uses mean value formula. Mean value = a b x y b a − ∫ d M1 Transforms variable to t. = a t t t a t a ∫ 2 0 2 3 d cos sin 3. cos π M1A1 Uses t t 2 2 cos 1 sin − = to write in a form where reduction formula can be used. ∫ = 2 0 2 4 d sin cos 3 π t t t a ∫ − = 2 0 6 4 d ) cos (cos 3 π t t t a (AG) A1 4 Finds I0. Uses reduction formula to find I4 and I6. 2 0 π = I or I2 = π/4. B1 (Or equivalent.) π π 16 3 2 . 2 1 . 4 3 4 = = I , π π 32 5 . 16 3 . 6 5 6 = = I (e.g. 4 6 4 4 6 6 1 ) ( 6 5 I I I I I = − ⇒ = ) M1A1 Substitutes in integral to find mean value. Mean value = π 32 3a A1 4 [12]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 EITHER Uses de Moivre’s theorem and binomial theorem. θ θ θ θ 3 sin i 3 cos ) sin i (cos 3 + = + and θ θ θ θ θ θ 3 2 2 3 sin i sin cos 3 sin cos i3 cos − − + B1 M1A1 (If line 1 missing, or no reference made to cos3θ and sin3θ being real and imaginary parts.) Award B0M1A1M1M1A0 i.e.4/6 Equates real and imaginary parts. Uses A A A cos sin tan = θ θ θ θ 3 2 sin sin cos 3 3 sin − = θ θ θ θ 2 3 sin cos 3 cos 3 cos − = θ θ θ θ θ θ θ 2 3 3 2 sin cos 3 cos sin sin cos 3 3 tan − − = θ θ θ θ 2 3 tan 3 1 tan tan 3 3 tan − − = ∴ (*) (AG) M1 M1 A1 6 4 3 12 9 , 12 5 , 12 π π π π = B1 (one) B1 (all) 2 Put 1 3 tan = θ in (*) 0 1 3 3 2 3 = + − − ⇒ t t t M1 States roots of 1 3 tan = θ between 0 and π. Roots are 4 3 tan , 12 5 tan , 12 tan π π π A1 (one) A1 (all) 3 Obtains cubic equation and solves. 0 )1 4 )( 1 ( 2 = + − + t t t 3 2 , 1 ± − = ⇒t M1 Evaluates each root. 1 4 3 tan − = π 3 2 12 tan − = π 3 2 12 5 tan + = π A1 (one) A1 (all) 3 [14]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 12 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 (i) OR Differentiates y wrt x. Uses discriminant and factorises. Obtains result. (No equality.) 2 2 2 ) 3 ( ) 6 ( ) 2 )( 3 ( d d + − + − + + = x x x x x x y λ λ λ 2 2 2 ) 3 ( 6 3 6 + + + + = x x x λ λ 0 d d = x y has distinct roots if 0 2 3 0 24 12 36 2 2 > − − ⇒ > − − λ λ λ λ 0 ) 1 )( 2 3 ( > − + λ λ 1 2 3 < < − ⇒ λ (AG) M1 A1 M1 A1 A1 5 (ii) Uses division to obtain the form for recognition of oblique asymptote. 3 6 3 9 3 3 6 2 2 2 + − − + − + ≡ + − + x x x x x λ λ λ λ λ (Tolerate error on remainder term.) ⇒ asymptotes are: 3 − = x and 3 − + = λ x y . M1 B1 A1 3 (iii) For 0 < λ < 1 Axes and asymptotes. Upper branch with minimum below x-axis. Lower branch with maximum. B1 B1 B1 3 (iv) For 3 > λ Deduct 1 mark overall for wrong forms At infinity. Axes and asymptotes. (n.b. oblique asymptote has positive y intercept.) Left-hand branch. Right-hand branch. B1 B1 B1 3 [14]