Cambridge A Level Mathematics - Further 9231 — 2010 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 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 4 printed pages. © UCLES 2010 [Turn over *5014820134*

2 1 The curve C has equation y = 1 4e2x + e−2x. Show that the length of the arc of C from the point where x = 0 to the point where x = 1 2 is e2 −1 4e . [4] 2 Use the method of differences to find SN, where SN = N ∑ n=1 1 n(n + 2). [4] Deduce the value of lim N→∞SN. [1] 3 A finite region R in the x-y plane is bounded by the curve with equation y = √x −1 √x, the x-axis between x = 1 and x = 4, and the line x = 4. Find the exact value of the y-coordinate of the centroid of R. [5] 4 Prove by mathematical induction that, for all non-negative integers n, 72n+1 + 5n+3 is divisible by 44. [5] 5 Let In = ã 1 0 (1 −x)n sin x dx for n ≥0. Show that In+2 = 1 −(n + 1)(n + 2)In. [4] Hence find the value of I6, correct to 4 decimal places. [4] 6 The linear transformation T : >4 →>4 is represented by the matrix A, where A =  1 2 −1 α 2 3 −1 0 2 1 2 −2 0 1 −3 −2 . Given that the dimension of the range space of T is 4, show that α ≠1. [3] It is now given that α = 1. Show that the vectors  1 2 2 0 ,  2 3 1 1  and  −1 −1 2 −3  form a basis for the range space of T. [2] Given also that the vector  p 1 1 q is in the range space of T, find a condition satisfied by p and q. [3] © UCLES 2010 9231/01/O/N/10

3 7 The roots of the equation x3 + 4x −1 = 0 are α, β and γ. Use the substitution y = 1 1 + x to show that the equation 6y3 −7y2 + 3y −1 = 0 has roots 1 α + 1, 1 β + 1 and 1 γ + 1. [2] For the cases n = 1 and n = 2, find the value of 1 (α + 1)n + 1 (β + 1)n + 1 (γ + 1)n . [2] Deduce the value of 1 (α + 1)3 + 1 (β + 1)3 + 1 (γ + 1)3 . [2] Hence show that (β + 1)(γ + 1) (α + 1)2 + (γ + 1)(α + 1) (β + 1)2 + (α + 1)(β + 1) (γ + 1)2 = 73 36. [3] 8 The curves C1 and C2 have polar equations given by C1 : r = 3 sin θ, 0 ≤θ < π, C2 : r = 1 + sin θ, −π < θ ≤π. (i) Find the polar coordinates of the points, other than the pole, where C1 and C2 meet. [2] (ii) In a single diagram, draw sketch graphs of C1 and C2. [3] (iii) Show that the area of the region which is inside C1 but outside C2 is π. [5] 9 Find the eigenvalues and corresponding eigenvectors of the matrix A =  3 −1 0 −1 2 −1 0 −1 3 . [7] Find a non-singular matrix M and a diagonal matrix D such that (A −2I)3 = MDM−1, where I is the 3 × 3 identity matrix. [3] 10 By using de Moivre’s theorem to express sin 5θ and cos 5θ in terms of sin θ and cos θ, show that tan 5θ = 5t −10t3 + t5 1 −10t2 + 5t4, where t = tan θ. [5] Show that the roots of the equation x4 −10x2 + 5 = 0 are tan1 5nπ for n = 1, 2, 3, 4. [2] By considering the product of the roots of this equation, find the exact value of tan1 5π tan2 5π. [3] © UCLES 2010 9231/01/O/N/10 [Turn over

4 11 It is given that x ≠0 and xd2y dx2 + 2dy dx + 4xy = 8x2 + 16. Show that if ß = xy then d2ß dx2 + 4ß = 8x2 + 16. [3] Find y in terms of x, given that y = 0 and dy dx = −2 when x = 1 2π. [9] 12 Answer only one of the following two alternatives. EITHER The curve C has equation y = x2 + 2λx x2 −2x + λ , where λ is a constant and λ ≠−1. (i) Show that C has at most two stationary points. [3] (ii) Show that if C has exactly two stationary points then λ > −5 4. [2] (iii) Find the set of values of λ such that C has two vertical asymptotes. [2] (iv) Find the x-coordinates of the points of intersection of C with (a) the x-axis, (b) the horizontal asymptote. [3] (v) Sketch C in each of the cases (a) λ < −2, (b) λ > 2. [4] OR The plane Π1 has equation r = 2i + j + 4k + λ(2i + 3j + 4k) + µ(−i + k). Obtain a cartesian equation of Π1 in the form px + qy + rß = d. [4] The plane Π2 has equation r.(i −4j + 5k) = 12. Find a vector equation of the line of intersection of Π1 and Π2. [3] The line l passes through the point A with position vector ai + (2a + 1)j −3k and is parallel to 3ci −3j + ck, where a and c are positive constants. Given that the perpendicular distance from A to Π1 is 15 √6 and that the acute angle between l and Π1 is sin−1 2 √6, find the values of a and c. [7] 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 2010 9231/01/O/N/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2010 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 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 – October/November 2010 9231 01 © 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 – October/November 2010 9231 01 © 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 – October/November 2010 9231 01 © UCLES 2010 1 2 2 2 2 2 2 2 ) e e( 4 1 )) e e( 2 1 ( 1 d d 1 x x x x x y − − + = − + =       + M1A1 expression simplified Length = ( ) ( ) [ ]2 1 0 2 2 2 2 2 1 0 e e 4 1 d e e 2 1 x x x x x − − − = + ∫ M1 integrate = ( ) ( ) e 4 1 e e e 4 1 e e 4 1 2 0 0 1 1 − = − − − − AG A1 cao [4] 2 nth term is       + − 2 1 1 2 1 n n M1A1                     − +       − +       − − +       + − − +       + − = 3 1 1 1 4 1 2 1 ... 1 2 1 1 1 1 1 2 1 1 2 1 N N N N N N S N M1 sum of terms =     + − + − 1 1 2 1 2 3 2 1 N N A1 after cancellation [4] Limit = ¾ B1√ [1] 3 Area = 3 / 8 2 3 2 d 4 1 4 1 2 1 2 3 2 1 2 1 =         − =         − ∫ − x x x x x B1 A x x x A x x x y 4 1 2 4 1 ln 2 2 2 1 d ) 1 2 ( 2 1       + − = + − = ∫ M1 use of A x y∫ d 2 1 2 M1 integrate A1 correct Final answer:       + 4 3 2 ln 8 3 or       + 2 3 4 ln 16 3 or 32 9 2 ln 8 3 + etc (ACF) A1 [5]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 01 © UCLES 2010 4 n = 0: 71 + 53 = 132 which is divisible by 44 B1 Assume 72k + 1 + 5k + 3 is divisible by 44 B1 Consider 72(k + 1) + 1 + 5(k + 1) + 3 = 7272k + 1 + 5.5k + 3 M1 (k + 1) th term = 49(72k + 1 + 5k + 3) – 44.5k + 3 M1 in appropriate form which is divisible by 44 A1 convincing argument [5] Alternative solution for final three marks: Consider (72k + 3 + 5k + 4) – (72k + 1 + 5k + 3) M1 = 48(72k + 1 + 5k + 3) – 44.5k + 3 M1 in appropriate form which is divisible by 44 A1 convincing argument 5 In + 2 = [–(1–x)n + 2 cosx] –∫ + − + x x x n n d cos ) 1 )( 2 ( 1 M1A1 ] d sin ) 1( ) sin ) 1 )[(( 2 ( )) 2 ( 1( 1 ∫ − + − + + + + = + x x x x x n n n n M1 integrate by parts again In + 2 = 1 – (n + 1)(n + 2) In AG A1 [4] I6 = 1 – 5 × 6I4; I4 = 1 – 4 × 3I2; I2 = 1 – 1 × 2I0 M1 ∫ − = = 1 0 0 1 cos 1 d sin x x I B1 I6 = 1 – 30(1 – 12(1 – 2I0)) = 0.0177 M1A1 [4] OR I0 = 1 – cos1 B1 I2 = 2cos1 – 1 M1 (use of RF) I4 = 13 – 24cos1 A1 I6 = 0.0177 A1 cao Accept decimal versions 6           − − − − − − − − 2 3 1 0 2 2 4 3 0 2 1 1 0 1 2 1 α α α →           − − − − − 6 6 0 0 0 2 4 1 0 0 2 1 1 0 1 2 1 α α α α M1A1 Dim = 4 ⇒ α ≠ 1 AG A1 [3] a + 2b – c = 0 2a + 3b – c = 0 Show a = b = c = 0 M1 attempt to solve 2a + b + 2c = 0 b – 3c = 0 Linearly independent and dim R(T) not 4: basis A1 [2] a + 2b – c = p 2a + 3b – c = 1 Attempt to find a, b, c in terms of q or p 2a + b + 2c = 1 b – 3c = q M1A1 6p + q = 3 A1 [3] Alternative solution: Use row operations as in (i) M1 Final column           − + − − 3 6 2 4 2 1 q p p p p A1 6p + q = 3 A1

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 01 © UCLES 2010 7 1 1 + = x y y y x − = ∴ 1 M1 use in given cubic equation Gives 6y3 – 7y2 + 3y – 1 = 0 AG A1 [2] n = 1: given expression = sum of roots = 7/6 B1 n = 2: ( ) ( ) ∑ ∑ ∑ = −       + = + 36 13 " " 2 1 1 1 1 2 2 αβ α α B1 [2] From cubic in y, 0 3 6 7 3 36 13 .7 1 1 6 3 = −       + −       + ∑α M1 216 / 73 1 1 3 =       + ∑α A1 [2] LHS = ∑       + + + + 3)1 ( 1 )( 1 )( 1 ( α α γ β M1 216 73 6 1 1 ×       = − M1 recognise product of roots = 73/36 AG A1 [3] 8 (i) 2 1 sin sin 3 sin 1 = ⇒ = + θ θ θ M1       6 , 2 3 π and       6 5 , 2 3 π A1 (both) [2] (ii) B1 circle B1 cardioid behaviour at origin B1 cardioid closed and symmetry [3] (iii) Subtract integrands M1 θ θ θ π π d ) sin 2 2 cos 4 3 ( 2 2 6 2 1 ∫ − − × M1 [ ] 2 6 cos 2 2 sin 2 3 π π θ θ θ + − = M1A1 = π AG A1 [5] Alternative: Area inside C1: 2 6 2 6 2 2 sin 2 1 2 9 d sin 9 2 1 2 π π π π θ θ θ θ ∫    − = × M1       + = 4 3 3 2 9 π A1

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 01 © UCLES 2010 Area inside C2: θ θ θ π π d ) 2 cos 1( 2 1 sin 2 1 2 1 2 2 6∫ − + + × 2 6 2 sin 4 1 cos 2 2 3 π π θ θ θ     − − = M1       + = 8 3 9 2 π (A1 if not earned earlier) Subtraction M1 Required area = π AG A1 [5] 9 [ ] 0 )) 3 ( (1 1 ) 3 )( 2 ( ) 3 ( = − − + − − − − λ λ λ λ M1 characteristic equation 0 ) 4 )( 1 )( 3 ( = − − − λ λ λ M1 factorise λ = 1, 3, 4 A1         =                 − − − − − − − 0 0 0 3 1 0 1 2 1 0 1 3 z y x λ λ λ Solve for λ = 1: (1, 2, 1) M1A1 Solve for λ = 3: (1, 0, –1) A1 Solve for λ = 4: (1, –1, 1) A1 [7] M =         − − 1 1 1 1 0 2 1 1 1 B1√ eigenvectors as columns (except         0 0 0 ) D =        − 8 0 0 0 1 0 0 0 1 M1A1√ ft on eigenvalues [3] 10 4 2 3 5 5 10 5 cos cs s c c + − = θ M1A1 use of de Moivre for (c + is)5 5 3 2 4 10 5 5 sin s s c s c + − = θ A1 4 2 3 5 5 10 1 5 10 5 tan t t t t t + − + − = θ AG M1A1 intermediate step needed [5] 5 0 5 tan π θ θ n = ⇒ = M1 Solutions 5 tan π n for n = 1, 2, 3, 4 A1 justify values of n [2] Roots ± 5 tan π , ± 5 2 tan π B1 Product of these roots = 5 M1 5 5 2 tan 5 tan = π π A1 [3]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 01 © UCLES 2010 11 z′ = y + xy′ B1 z′′ = 2y′ + xy′′ B1 Obtain result B1 [3] Auxiliary equation: m2 + 4 = 0 : m = ±2i M1 CF: x B x A 2 sin 2 cos + A1 PI: z = ax2 + bx + c Differentiate twice and substitute M1 a = 2, b = 0, c = 3 A1 GS: 3 2 2 sin 2 cos 2 + + + = x x B x A z A1√ their CF + their PI y = 0, x = 2 1 π : (z = 0) gives 3 2 2 + = π A B1 z′ = –2Asin2x + 2Bcos2x + 4x M1 y′ = –2, x = 2 π : (z′ = –π) gives 2 3π = B A1         + + +       + = 3 2 2 sin 2 3 2 cos 3 2 1 2 2 x x x x y π π A1 [9] 12 EITHER (i) y′ = 0 ⇒ (x2 – 2x + λ)(2x + 2λ) – (x2 + 2λx)(2x – 2) = 0 M1 ⇒ ... ⇒ (λ + 1)x2 – λx – λ2 = 0 A1 Hence at most 2 values of x and at most 2 stationary points A1 [3] (ii) For 2 real distinct roots, λ2 > 4(λ + 1)(– λ2) M1 use of discriminant 4 5 0 ) 4 5 ( 2 − > ∴ > + λ λ λ AG A1 [2] (iii) Vert. asymptotes when x2 – 2x + λ = 0 M1 b2 – 4ac > 0 ⇒ 4 – 4 λ > 0 For two vert. asymp. λ < 1 A1 [2] (iv) (a) y = 0 ⇒ x2 + 2λx = 0 M1 ⇒ x = 0 or –2λ A1 (both) (b) y = 1: x = 2 2 + λ λ B1 [3] (v) (a) λ < –2: no stat points: 2 vert. asymp B1 3 branches B1 completely correct shape (b) λ < 2: 2 stats points: no vert. asymp B1 max, min, horiz asymp B1 correct shape [4]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2010 9231 01 © UCLES 2010 OR Normal to plane: (2, 3, 4) × (–1, 0, 1) = (3, –6, 3) M1A1 r.(1, –2, 1) = d and point (2, 1, 4) M1 substitute point into plane eqn d = 4 x – 2y + z = 4 A1 [4] Alternative: x = 2 + 2λ – µ y = 1 + 2λ x + z = 6 + 6λ M1A1 z = 4 + 4λ + µ )1 ( 2 6 − + = + ∴ y z x M1 4 2 = + − ∴ z y x A1 x – 4y + 5z = 12 x – 2y + z = 4 Solve by eliminating one variable M1 Use parameter and express all 3 variables in terms of it M1 e.g. x = 3t – 4, y = 2t – 4, z = t r = (–4, –4, 0) + t (3, 2, 1) A1 or equivalent [3] Alternative: Direction of line =         =         − ×         − 1 2 3 5 4 1 1 2 1 t M1A1 Find any point on line e.g.         − − 0 4 4 ,         2 0 2 etc. ∴ r =         +         − − 1 2 3 0 4 4 t B1 Line l: r = (a, 2a + 1, –3) + α(3c, –3, c) Plane: x – 2y + z = 4 Distance A to plane: 6 15 6 4 3 )1 2 ( 2 = − − + − a a M1 3a + 9 =15 M1 correct use of modulus sign a = 2 A1 2 2 9 9 6 6 3 sin c c c c + + + + = θ M1A1 6 2 10 9 6 6 4 2 = + + ∴ c c M1 solve for c 6c2 – 12c = 0: c = 2 A1 (Penalise only once for negative values.) [7] }