Cambridge A Level Mathematics - Further 9231 — 2012 Oct/Nov Paper 1 · Variant 3

*1588720257* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 October/November 2012 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. JC12 11_9231_13/RP © UCLES 2012 [Turn over

2 1 Show that 2n ∑ r=n+1 r2 = 1 6n(2n + 1)(7n + 1). [4] 2 Find the set of values of a for which the system of equations ax + y + 2ß = 0, 3x −2y = 4, 3x −4y −6aß = 14, has a unique solution. [4] 3 Let SN = 1 2! + 2 3! + 3 4! + . . . + N (N + 1)!. Prove by mathematical induction that, for all positive integers N, SN = 1 − 1 (N + 1)!. [5] 4 The points A, B and C have position vectors i + 2j + 2k, 2i + 4j + 5k and 2i + 3j + 4k respectively. Find −−→ AB × −−→ AC. [3] Deduce, in either order, the exact value of (i) the area of the triangle ABC, (ii) the perpendicular distance from C to AB. [3] 5 The curve C has polar equation r = 1 + 2 cos θ. Sketch the curve for −2 3π ≤θ < 2 3π. [2] Find the area bounded by C and the half-lines θ = −1 3π, θ = 1 3π. [4] 6 The curve C has parametric equations x = t2, y = 1 4t4 −ln t, for 1 ≤t ≤2. Find the area of the surface generated when C is rotated through 2π radians about the y-axis. [7] 7 A cubic equation has roots α, β and γ such that α + β + γ = 4, α2 + β2 + γ 2 = 14, α3 + β3 + γ 3 = 34. Find the value of αβ + βγ + γ α. [2] Show that the cubic equation is x3 −4x2 + x + 6 = 0, and solve this equation. [6] © UCLES 2012 9231/13/O/N/12

3 8 Let ß = cos θ + i sin θ. Show that 1 + ß = 2 cos 1 2θ(cos 1 2θ + i sin 1 2θ). [3] By considering (1 + ß)n, where n is a positive integer, deduce the sum of the series n 1 sin θ + n 2 sin 2θ + . . . + n n sin nθ. [6] 9 The curve C has equation y = x2 −3x + 3 x −2 . Find the equations of the asymptotes of C. [3] Show that there are no points on C for which −1 < y < 3. [4] Find the coordinates of the turning points of C. [3] Sketch C. [2] 10 The curve C has equation x3 + y3 = 3xy, for x > 0 and y > 0. Find a relationship between x and y when dy dx = 0. [4] Find the exact coordinates of the turning point of C, and determine the nature of this turning point. [8] 11 Show that ã x1 −x2 1 2 dx = −1 31 −x2 3 2 + c, where c is a constant. [1] Given that In = ã 1 0 xn1 −x2 1 2 dx, prove that, for n ≥2, (n + 2)In = (n −1)In−2. [5] Use the substitution x = sin u to show that ã 1 0 1 −x2 1 2 dx = 1 4π. [5] Find I4. [2] © UCLES 2012 9231/13/O/N/12 [Turn over

4 12 Answer only one of the following two alternatives. EITHER The vector e is an eigenvector of each of the n × n matrices A and B, with corresponding eigenvalues λ and µ respectively. Prove that e is an eigenvector of the matrix AB with eigenvalue λµ. [2] It is given that the matrix A, where A =  3 2 2 −2 −2 −2 1 2 2 , has eigenvectors 0 1 −1 ! and 1 0 −1 !. Find the corresponding eigenvalues. [2] Given that 2 is also an eigenvalue of A, find a corresponding eigenvector. [2] The matrix B, where B =  −1 2 2 2 2 2 −3 −6 −6 , has the same eigenvectors as A. Given that AB = C, find a non-singular matrix P and a diagonal matrix D such that P−1C2P = D. [8] OR Obtain the general solution of the differential equation d2x dt2 + 6dx dt + 13x = 75 cos 2t. [7] Given that x = 5 and dx dt = 0 when t = 0, find x in terms of t. [4] Show that, for large positive values of t and for any initial conditions, x ≈5 cos(2t −φ), where the constant φ is such that tan φ = 4 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 2012 9231/13/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2012 series 9231 FURTHER MATHEMATICS 9231/13 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 should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes for the October/November 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 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 – October/November 2012 9231 13 © Cambridge International Examinations 2012 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 – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 1 Use of: Use of: Obtains result. ∑ + n n 2 1 = ∑ ∑− n n 1 2 1 ∑ + + = n n n n r 1 2 6 )1 2 )( 1 ( 6 )1 2 )( 1 ( 6 )1 4 )( 1 2 ( 2 + + − + + n n n n n n = ( ) )1 7 )( 1 2 ( 6 1 1 2 8 )1 2 ( 6 1 + + = − − + + n n n n n n n (AG) M1 M1 A1 A1 4 [4] 2 Sets determinant ≠ 0. Factorises, or completes square. States result. 0 12 18 12 0 6 4 3 0 2 3 2 1 2 ≠ − + ⇒ ≠ − − − a a a a 0 ) 2 )( 1 2 ( 6 ≠ + − ⇒ a a 2 1 ≠ a or –2 (Or by row operations.) M1A1 M1 A1 4 [4] 3 Proposition. Proves base case. States inductive hypothesis. Proves inductive step. States conclusion. HN : )! 1 ( 1 1 + − = N S N ⇒ − = = = !2 1 1 2 1 !2 1 1 S H1 is true. Hk : Assume )! 1 ( 1 1 + − = k Sk is true. )! 2 ( )1 ( ) 2 ( )! 2 ( )! 2 ( 1 )! 1 ( 1 1 1 + + + + − + = + + + + − = ⇒ + k k k k k k k Sk )! 2 ( 1 1 1 + − = ⇒ + k Sk ∴ Hk ⇒ Hk+1. ∴ (By PMI Hn is) true for all positive integers N. B1 B1 M1 A1 A1 5 [5] 4 Finds vector product. Finds area of triangle. Finds length of perpendicular. AB =           3 2 1 AC =           2 1 1 AC AB × =           − = 1 1 1 2 1 1 3 2 1 k j i Area of triangle ABC = 3 2 1 1 1 1 2 1 =           − 14 3 3 2 1 3 2 1 2 1 2 2 2 = ⇒ = + + d d B1 M1A1 M1A1 A1 3 3 [6]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 5 Sketches graph. Uses area of sector formula. Uses double angle formula Integrates. Obtains result. Correct shape and orientation. Passing through (0,0) and (3,0) Area = ∫ + × 3 0 2 d ) cos 2 1( 2 1 2 π θ θ ∫ + + = 3 0 2 d ) cos 4 cos 4 1( π θ θ θ =∫ + + 3 0 ) 2 cos 2 cos 4 3 ( π θ θ dθ [ ] 3 0 2 sin sin 4 3 π θ θ θ + + =     + = 3 2 5 π B1 B1 M1 M1 A1 A1 2 4 [6] 6 Differentiates. Obtains 2 d d       t s Uses surface area formula about y-axis. t x 2 = & t t y 1 3 − = & 2 3 2 2 6 2 2 2 1 1 2 4 ) ( ) (       + = + − + = + t t t t t t y x & & ∫ = s x S d 2π t t t t t t t d ) ( 2 d 1 2 2 1 5 2 1 3 2 ∫ ∫ + =       + = π π 2 1 2 6 2 1 6 1 2     + = t t π π π 24 2 1 6 1 2 3 32 2 =           + −     + = B1 M1A1 M1A1 M1A1 7 [7] 7 ∑ = ⇒ + + = + + − + + 1 ) ( 2 ) ( 2 2 2 2 αβ γ β α γα βγ αβ γ β α Either Required equation is 0 4 2 3 = + + − c x x x ∑ ∑ = + + − ⇒ 0 3 4 4 2 3 c a α 6 18 4 34 56 3 = ⇒ = − − = ⇒ c c (AG) Or ) )( ( 3 2 2 2 3 3 3 γα βγ αβ γ β α γ β α αβγ γ β α − − − + + + + = − + + (or some other appropriate identity, e.g. ) 3 ) )( (3 ) ( 3 3 3 3 αβγ γα βγ αβ γ β α γ β α γ β α − + + + + + + + = + + 6 ... − = ⇒ ⇒ αβγ 0 6 4 2 3 = + + − ⇒ x x x (AG) ⇒ = − − + ⇒ 0 ) 3 )( 2 )( 1 ( x x x x = –1,2,3. M1A1 M1 M1 A1 (M1) (M1A1) A1 M1A1 2 6 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 8 Re-write. Obtains result. Use of Bin. Thm. Takes imaginary part and uses de M. Applies initial result. Equates imaginary parts to obtain result. θ θ θ 2 1 cos 2 1 sin i 2 2 1 cos 2 1 2 + = + z =       + θ θ θ 2 1 sin i 2 1 cos 2 1 cos 2 (AG) n n z n n z n z n z       + +       +       + = + ... 2 1 1 ) 1( 2 θ θ θ n n n n n z n sin ... 2 sin 2 sin 1 ) 1 Im(       + +       +       = + ∴ But θ θ 2 i e 2 1 cos 2 ) 1( n n n n z = + θ θ θ n n n n n sin ... 2 sin 2 sin 1       + +       +       ∴ θ θ 2 n sin 2 1 cos 2 n n = M1A1 A1 M1 M1 M1A1 B1 A1 3 6 [9] 9 States vertical asymptote. Divides and states oblique asymptote. Rearranges as quadratic in x. Uses discriminant to obtain condition stated. Differentiates, puts = 0 and obtains x-values. States stationary points. Sketch. Vertical asymptote is 2 = x . ⇒ oblique asymptote is 1 − = x y . 0 ) 2 3 ( )3 ( 3 3 2 2 2 = + + + − ⇒ + − = − y x y x x x y xy For real x, 0 4 2 ≥ − AC B 0 ) 2 3 ( 4 )3 ( 2 ≥ + − + ∴ y y 0 )) 1 )( 3 ( ... ≥ + − ⇒ ⇒ y y 1 − ≤ ⇒y or 3 ≥ y ∴ no points for 3 1 < < − y (AG) 3 or 1 0 ) 2 ( 1 2 = ⇒ = − − = ′ − x x y (Or uses y = –1 and 3 to obtain x-values.) Stationary points are (1,–1) and (3,3) One mark for each branch correctly placed. B1 M1 A1 B1 M1 A1 A1 M1 A1A1 B1B1 3 4 3 2 [12]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 10 Differentiates implicitly. Equates x y d d to zero. Obtains relationship. Substitutes for y. Obtains x and y. Differentiates Uses 0 = ′y . Identifies maximum. (Other watertight methods for showing a maximum accepted.) y x y y y x ′ + = ′ + 3 3 3 3 2 2 0 1 d d 2 2 = − − = ⇒ y x y x y 2 x y = ⇒ x y xy y xy y xy 2 2 3 2 3 3 = ⇒ = ⇒ = + ⇒ ( 0 ≠ y ) 3 1 3 4 2 2 2 = ⇒ = ⇒ = ⇒ x x x x and 3 2 2 = ⇒y ) 0 ( ≠ x y y x y y y y y x ′ + ′′ + ′ = ′ + ′′ + 3 3 3 ) ( 6 3 6 2 2 ) 1( 2 ) 3 3 ( 6 3 2 x x x y y x y x − = ′′ ⇒ − ′′ = ⇒ ⇒ − = − = ′′ ⇒ 2 2 1 2 y max B1B1 M1 A1 M1 A1A1 B1 B1B1 M1 A1 4 8 [12]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 11 Verifies result. Correct parts. Integrates by parts. Substitutes limits. Obtains reduction formula. Uses substitution correctly. Uses double angle formula. Integrates correctly. Uses reduction formula correctly. 2 1 2 2 1 2 2 3 2 ) 1( ) 2 ( ) 1( 2 3 3 1 ) 1( 3 1 d d x x x x c x x − = − × − × − =         + − − ∫ − = 1 0 2 1 2) 1( I x xn n dx = ∫ − − 1 0 2 1 2 1 ) 1( . x x xn dx ∫ − − +         − − = − − 1 0 2 3 2 2 1 0 2 3 2 1 ) 1( 3 1 . )1 ( ) 1( 3 1 . x x n x x n n dx ∫ − − − = − 1 0 2 1 2 2 2 ) 1 )( 1( 3 1 x x x n n dx n n I n I n 3 1 3 1 2 − − − = − 2 )1 ( ) 2 ( − − = + ⇒ n n I n I n (AG) u u x u x d cos d sin = ⇒ = Limits: 2 1 0 0 π = ⇒ = = ⇒ = u x u x ∫ ∫ = − 2 0 2 1 0 2 1 2 d cos d ) 1( π u u x x = u u d)1 2 (cos 2 1 2 0 ∫ + π = 4 2 2 sin 2 1 2 0 π π =     + u u (AG) 32 16 2 1 16 4 4 1 4 2 π π π π = × = ⇒ = × = ⇒ I I B1 M1 M1A1 M1 A1 M1 A1 M1 M1A1 M1A1 1 5 5 2 [13]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 12 EITHER Use of these results. Finds missing eigenvalues of A. Finds missing eigenvector of A. Calculates eigenvectors of B Uses initial result to find eigenvalues of C.(1 mark for one correct value, 2 marks for all three.) Finds P and D. e Ae λ = and e Be µ = e e Ae e A ABe λµ µλ µ µ = = = =           − − − 2 2 1 2 2 2 2 2 3           −1 1 0 =           0 0 0 0 = ⇒λ           − − − 2 2 1 2 2 2 2 2 3           −1 0 1 =           −1 0 1 1 = ⇒λ ⇒ = 2 λ           − = − − − 0 2 4 2 4 2 2 2 1 k j i ~           − 0 1 2           − − − − 6 6 3 2 2 2 2 2 1           −1 1 0 = 0 0 0 0 = ⇒           µ           − − − − 6 6 3 2 2 2 2 2 1           −1 0 1 = 3 3 0 3 − = ⇒          − µ           − − − − 6 6 3 2 2 2 2 2 1           − 0 1 2 = 2 0 2 4 − = ⇒          − µ ∴C has eigenvalues: 0 0 0 = × 3 ) 3 ( 1 − = − × 4 ) 2 ( 2 − = − ×           − − − = 0 1 1 1 0 1 2 1 0 P (OE)           = 16 0 0 0 9 0 0 0 0 D M1 A1 B1 B1 M1A1 B1 B1 B1 B2,1,0 B1 M1A1 2 2 2 8 [14]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2012 9231 13 © Cambridge International Examinations 2012 Qu No Commentary Solution Marks Part Marks Total 12 OR Finds complementary function. Finds particular integral. Finds general solution. Uses initial conditions.. Obtains solution. Obtains limit. 2 3 0 13 6 2 ± − = ⇒ = + + m m m i ) 2 sin 2 cos ( e x 3 t B t A t + = − t q t p x 2 sin 2 cos + = t q t p x 2 cos 2 2 sin 2 + − = & t q t p x 2 sin 4 2 cos 4 − − = && t t p q t q p 2 cos 75 2 sin ) 12 9 ( 2 cos ) 12 9 ( = − + + 3 = ⇒p 4 = q t t t B t A x 2 sin 4 2 cos 3 ) 2 sin 2 cos ( e-3t + + + = 5 = x when 2 3 5 0 = ⇒ + = ⇒ = A A t t t t B t A t B t A x t t 2 cos 8 2 sin 6 ) 2 cos 2 2 sin 2 ( e ) 2 sin 2 cos ( e 3 3 3 + − + − + + − = − − & 0 = x& when 1 2 8 6 0 0 − = ⇒ + + − = ⇒ = B B t t t t t x t 2 sin 4 2 cos 3 ) 2 sin 2 cos 2 ( e 3 + + − = − As ∞ → t , 0 e-3t → t t x 2 sin 4 2 cos 3 + ≈ ∴       − =       + ≈ ∴ − 3 4 tan 2 cos 5 2 sin 5 4 2 cos 5 3 5 1 t t t x (AG) M1 A1 M1 M1A1 A1 A1 B1 M1 A1 A1 M1 M1A1 7 4 3 [14]