Cambridge A Level Mathematics - Further 9231 — 2010 May/June Paper 1 · Variant 3

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 May/June 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 5 printed pages and 3 blank pages. © UCLES 2010 [Turn over *4278730524*

2 1 Given that 5 is an eigenvalue of the matrix A =  5 −3 0 1 2 1 −1 3 4 , find a corresponding eigenvector. [2] Hence find an eigenvalue and a corresponding eigenvector of the matrix A + A2. [2] 2 By considering the identity cos[(2n −1)α] −cos[(2n + 1)α] ≡2 sin α sin 2nα, show that if α is not an integer multiple of π then N ∑ n=1 sin(2nα) = 1 2 cot α −1 2 cosec α cos[(2N + 1)α]. [4] Deduce that the infinite series ∞ ∑ n=1 sin2 3nπ does not converge. [1] 3 The sequence x1, x2, x3, . . . is such that x1 = 3 and xn+1 = 2x2 n + 4xn −2 2xn + 3 for n = 1, 2, 3, . . . . Prove by induction that xn > 2 for all n. [6] 4 The parametric equations of a curve are x = cos t + t sin t, y = sin t −t cos t. The arc of the curve joining the point where t = 0 to the point where t = 1 2π is rotated about the x-axis through one complete revolution. Find the area of the surface generated, leaving your result in terms of π. [7] 5 Use de Moivre’s theorem to show that sin 5θ = 16 sin5θ −20 sin3θ + 5 sin θ. [4] Hence find all the roots of the equation 32x5 −40x3 + 10x + 1 = 0 in the form sin(qπ), where q is a positive rational number. [4] © UCLES 2010 9231/13/M/J/10

3 6 The curve C has equation y = x2 −3x −7 x + 1 . (i) Obtain the equations of the asymptotes of C. [3] (ii) Show that dy dx > 1 at all points of C. [2] (iii) Draw a sketch of C. [3] 7 It is given that x = t2e−t2 and y = te−t2. (i) Show that dy dx = 1 −2t2 2t −2t3 . [3] (ii) Find d2y dx2 in terms of t. [5] 8 Obtain the general solution of the differential equation d2y dx2 + 5dy dx + 4y = 10 sin 3x −20 cos 3x. [5] Show that, for large positive x and independently of the initial conditions, y ≈R sin(3x + φ), where the constants R and φ, such that R > 0 and 0 < φ < 2π, are to be determined correct to 2 decimal places. [4] 9 Let In = ã 1 2π 0 sinnθ dθ, where n is a non-negative integer. Show that In+2 = n + 1 n + 2 In. [4] The region R of the x-y plane is bounded by the x-axis, the line x = π 2m and the curve whose equation is y = sin4mx, where m > 0. Find the y-coordinate of the centroid of R. [6] © UCLES 2010 9231/13/M/J/10 [Turn over

4 10 The equation x4 + x3 + cx2 + 4x −2 = 0, where c is a constant, has roots α, β, γ, δ. (i) Use the substitution y = 1 x to find an equation which has roots 1 α , 1 β , 1 γ , 1 δ . [2] (ii) Find, in terms of c, the values of α2 + β2 + γ 2 + δ 2 and 1 α2 + 1 β2 + 1 γ 2 + 1 δ 2. [3] (iii) Hence find α −1 α 2 + β −1 β 2 + γ −1 γ 2 + δ −1 δ 2 in terms of c. [2] (iv) Deduce that when c = −3 the roots of the given equation are not all real. [3] 11 The curve C has polar equation r = a 1 + θ , where a is a positive constant and 0 ≤θ ≤1 2π. (i) Show that r decreases as θ increases. [2] (ii) The point P of C is further from the initial line than any other point of C. Show that, at P, tan θ = 1 + θ, and verify that this equation has a root between 1.1 and 1.2. [4] (iii) Draw a sketch of C. [3] (iv) Find the area of the region bounded by the initial line, the line θ = 1 2π and C, leaving your answer in terms of π and a. [3] © UCLES 2010 9231/13/M/J/10

5 12 Answer only one of the following two alternatives. EITHER The line l1 passes through the point A whose position vector is 3i + j + 2k and is parallel to the vector i + j. The line l2 passes through the point B whose position vector is −i −k and is parallel to the vector j + 2k. The point P is on l1 and the point Q is on l2 and PQ is perpendicular to both l1 and l2. (i) Find the length of PQ. [4] (ii) Find the position vector of Q. [5] (iii) Show that the perpendicular distance from Q to the plane containing AB and the line l1 is √3. [4] OR The linear transformation T : >4 →>4 is represented by the matrix M =  1 1 5 7 3 9 17 25 1 7 7 11 3 6 16 23 . (i) In either order, (a) show that the dimension of R, the range space of T, is equal to 2, (b) obtain a basis for R. [5] (ii) Show that the vector  1 −15 −17 −6 belongs to R. [3] (iii) It is given that {e1, e2} is a basis for the null space of T, where e1 =  14 1 −3 0 and e2 =  19 2 0 −3 . Show that, for all λ and µ, x =  4 −3 0 0 + λe1 + µe2 is a solution of Mx =  1 −15 −17 −6  (∗) . [3] (iv) Hence find a solution of (∗) of the form  α 0 γ δ . [2] © UCLES 2010 9231/13/M/J/10

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. 9231/13/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2010 question paper for the guidance of teachers 9231 FURTHER MATHEMATICS 9231/13 Paper 13, 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 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 – May/June 2010 9231 13 © 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 – May/June 2010 9231 13 © 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 – May/June 2010 9231 13 © UCLES 2010 1 Relevant working from A – 5I =           − − − − 1 3 1 1 3 1 0 3 0 or the equivalent in equations M1 to obtain an eigenvector of the form           −1 0 1 A1 [2] An eigenvalue of A + A2 is 5 + 25 = 30 B1 Corresponding eigenvector, as above B1 [2] No penalty for not hence methods. Accept λ = 6 with           − − 1 1 1 and λ = 20 with           −1 1 3 Accept linear scaling of eigenvectors. 2 ( ) ( ) [ ] α α α α 1 2 cos cos 2 sin sin 2 1 + − = ∑= N n N n M1A1 ⇒ displayed result (AG) M1A1 [4] ( ) 3 / 1 2 cos π + N oscillates finitely as ( ) 3 / 2 sin 1 π n n n ∑ ∞ = ⇒ ∞ → does not converge (CWO) B1 Require 3 π α = , ‘oscillate’ or values of ( ) 3 1 2 cos π + N given as 2 1 or –1 [1] 3 2 : > k k x H for some k B1 ( ) ( ) 3 2 / 8 2 2 2 1 + − = − + k k k x x x M1A1 1 1 2 2 0 8 2 + + ⇒ > ⇒ > − ⇒ k k k k H x x H A1 1 1 2 3 H x ⇒ > = is true B1 CWO Completion of the induction argument A1 [6]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 Alternatively for lines 2 and 3: ( ) 3 2 3 2 1 2 1 1 + − + = + k k k x x x M1A1 1 7 3 2 + ⇒ > + ⇒ k k k H x H A1 OR ( ) 3 2 2 1 + + = − + k k k k x x x x M1A1 2 2 1 > ⇒ > + k k x x A1 OR ( ) 3 2 2 1 + = − − + k k k k x x x x M1A1 2 2 1 > > ⇒ > + k k k x x x A1 Minimum conclusion is ‘Hence true for n [ 1’. 4 t t dt dy t t dt dx sin / , cos / = = (both) B1 ( ) ( ) t dt dy dt dx = + 2 2 / / B1 ( )dt t t t t S cos sin 2 2 2 / 0 − ∫ = π π (AEF) M1A1 t t t dt t t sin cos sin + − = ∫ B1 t t t t t dt t t sin 2 cos 2 sin cos 2 2 − + = ∫ B1 ( )[ ] 2 12 2 / π π − = S Accept forms such as 2 / 6 3 π π − , etc. A1 [7] OR for lines 4 and 5 ( ) ( ) { } [ ] 2 / 0 2 sin 2 sin cos cos 2 π π ∫ ∫ − − + − dt t t t t dt t t t (LNR) M1 [ ] 2 / 0 2 sin 2 cos 2 sin sin cos 2 π π t t t t t t t t + − − + − = (LNR) A1

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 5 ( ) ( ) ( ) ( ) ( ) ( )5 4 3 2 2 3 4 5 5 5 10 10 5 is is c is c s c is c c is c + + + − + + = + M1 5 3 2 4 10 5 s s c s c + − = θ M1 ( ) ( ) 5 2 3 2 2 1 10 1 5 s s s s s + − − − = M1 s s s 5 20 16 3 5 + − = (AG) A1 OEW [4] 2 / 1 5 sin sin − = ⇒ = θ θ x M1 Roots are π q sin where q = 7/30, 11/30, 31/30, 35/30, 43/30 A3 A1 for 1 root: + A1 for 2 further roots: + A1 for completion [4] CWO CWO Alternative answers q = 30 11 , 30 23 , 30 35 , 30 47 , 30 59 or q = 30 7 , 30 19 , 30 31 , 30 43 , 30 55 6 (i) One asymptote is x = –1 B1 y = x – 4 – 3/(x + 1) M1 Require y = x + non-zero constant. Other asymptote is y = x – 4 A1 [3] Alternatively for last two marks: OR ( ) ( )1 / 7 3 2 + − − ≈ + x x x k x for large ( ) 7 3 1 2 2 − − ≈ + + + ⇒ x x k x k x x for large x M1 ⇒ − = ⇒ − = + ⇒ 4 3 1 k k other asymptote is y = x – 4 A1 OR ( ) ( ) ( ) ( ) 0 7 3 1 1 / 7 3 2 2 = + + + + + − ⇒ + = + − − c x c m x m c mx x x x M1 Put 0 3 ,0 1 = + + = − c m m to obtain other asymptote is y = x – 4 A1 (ii) Obtains any correct result for dy/dx M1 ( ) [ ] ( ) ( )2 2 2 1 / 3 1 1 / 3 1 / + + = + + + = ⇒ ⇒ x x x dx dy K A1 No comment required. [2] (iii) Axes and asymptotes correctly placed B1 Right-hand branch B1 Left-hand branch B1 Deduct 1 overall for bad form(s) at infinity [3]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 7 (i) ( ) 2 2 2 / 3 t e t t dt dx − − = , ( ) 2 2 1 / 2 t e t dt dy − − = (both, AEF) B1 Obtains displayed result (AG) M1A1 [3] (ii) Any correct result for ( ) dt dx dy d / / in terms of t (*) M1A1A1 M1 – Quotient Rule A1A1 for terms in numerator ( ) dx dt dx y d / / 2 2 × ∗ = expressed in terms of t M1 Simplify to, e.g., ( ) ( ) 3 2 3 4 2 1 4 2 1 2 t t e t t t − − + − A1 Two terms in numerator must be combined. [5] 8 Complementary function is x x Be Ae 4 − −+ M1A1 Particular integral of form x Q x P 3 cos 3 sin + , so that 10 15 5 = + − Q P and 20 5 15 − = −Q P M1 ,5 / 7 − = ⇒P 5 / 1 − = Q A1 General solution is x x Be Ae y x x 3 cos 2.0 3 sin 4.1 4 − − + = − − A1 [5] 0 4 → + − − x x Be Ae as +∞ → x , M1 7 1 tan = φ M1 , 41 .1 2 = = R ( ) 28 .3 7 / 1 arctan = + = π φ B1A1 Accept , 2 = R but must be positive. [4]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 9 [ ] ( ) , 1 2 2 1 c s n s c s D n n n + + − = + + where , / dx d D ≡ , sin x s = x c cos = M1 ( ) ( ) n n s n s n 1 2 2 + + + − = = + K A1 Integrates w.r.t. x over the range [ ] 2 / ,0 π to obtain ( ) ( ) n n I n I n 1 2 0 2 + + + − = + M1 Result (AG) A1 [4] OR [ ] ( ) dx s c n cs I n n n 2 2 2 / 0 2 / 0 1 1 − − ∫ − + − = π π M1 ( ) ( ) n n I n I n 1 1 2 − − − ⇒ − (n [=2) M1A1 ( ) [ ] ( ) ( ) [ ] n n n n I n n I I n n I 2 / 1 / 1 2 2 + + = ⇒ − = ⇒ + − for n [ 0 A1 OR Starts with 2 + nI and relates to nI directly: mark as above OR θ θ θ π d I n n cosec sin 2 2 2 / 0 + ∫ = [ ] ( ) θ θ θ θ θ θ π π d n n n cot cos sin 2 cot sin 1 2 / 0 2 / 0 2 + + + + − = ∫ (LNR) M1 ( ) ( ) θ θ θ π d n n sin 1 sin 2 0 2 2 / 0 − + + = ∫ (LR) M1 ( ) ( ) 2 2 2 + + − + = n n I n I n A1 ( ) ( ) n n I n n I 2 1 2 + + = ⇒ + A1

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 dx mx dx mx y m m sin sin 4 2 / 0 8 2 / 0 2 1 ∫ ∫ = π π M1 (LR) let u = mx M1 du u du u y sin sin 4 2 / 0 8 2 / 0 2 1 ∫ ∫ = π π A1 2 0 π = I 4 2 1 2 π × = I      = × × = 16 3 2 2 1 4 3 4 π π I B1 2 2 1 4 3 6 5 6 π × × × = I      = × × × × = 768 105 2 2 1 4 3 6 5 8 7 8 π π I B1 96 35 4 8 2 1 = = I I y (or 0.365) B1 [6] OR for last 3 marks 48 35 4 8 = I I (oe) M1A1 96 35 = ∴y A1 10 (i) 0 1 4 2 / 1 2 3 4 = − − − − ⇒ = y cy y y y x M1A1 [2] (ii) c 2 1 2 − = ∑α M1A1 c + = ∑ − 4 2 α A1 (M1 is for use of ( ) ∑ ∑ ∑ − = αβ α α 2 2 2 in either part.) [3] (iii) ( ) 3 8 2 2 2 1 − − = − + = − = ∑ ∑ ∑ − − c S α α α α M1A1√ A1ft is for adding answers to (ii) correctly and subtracting 8. [2] (iv) 0 3 = ⇒ − = S c so that if all roots are real then 1 ± = α and similarly for δ γ β , , M1A1 CWO This is impossible since e.g., ,2 − = αβγδ or any other contradiction A1 CWO [3]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 11 (i) ( )2 1 / / θ θ + − = a d dr M1 Since ,0 > a then ,0 / < θ d dr ∀ points of C A1 [2] OR ( ) 0 > a constant and as ( ) 0 > θ increases, θ + 1 increases M1 θ + ∴1 a decreases A1 (ii) ( ) θ θ + = 1 / sin a y M1 ( ) 0 sin cos 1 0 / = − + ⇒ ⇒ = θ θ θ θ K d dy A1 θ θ + = ⇒ 1 tan (AG) A1 135 .0 1 tan − = − − θ θ when 372 .0 1 tan : 1.1 + = − − = θ θ θ when 2.1 = θ B1 [4] OR equivalent argument for B1 such as: ( ) ( ) 2.2 37 .2 2.1 tan ,1.2 96 .1 1.1 tan > ≈ < ≈ B1 or ( ) ( ) ( ) 14 .0 1.1 f 1 tan f − = ⇒ + − = θ θ θ ( ) 37 .0 2.1 f = (iii) Sketch: Approximately correct shape and placement for 0 Y θ Y π/2 and passing through ( ) 0 ,a and ( ), 2 / , 4.0 π a approximately, indicated in some way B1B1 Maximum in interval ( ) 2 / ,4 / π π B1 [3] (iv) ( ) ( ) θ θ π d a A 2 2 / 0 2 1 2 / − + = ∫ M1 ( ) ( ) [ ] 2 / 0 1 2 1 2 / π θ − + − = a A1 ( ) 2 2 / 2 + = = π πa K A1 Do not accept double minus signs or fractions in the denominator for the final mark. [3]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 12 EITHER (i) (i + j) × (j + 2k) = 2i – 2j + k M1A1 PQ = (4i + j + 3k).(2i – 2j + k)/3 = 9/3 = 3 M1A1 [4] (ii) (4 + λ)/2 = (1 + λ – µ)/(–2) = (3 – 2µ)/1 (AEF) OR (4 + λ) + (1 + λ – µ) = 0, (1 + λ – µ + 2(3 – 2µ) = 0, both M1A1 1 = ⇒ ⇒ µ K M1A1 Position vector of Q is –i + j + k A1 [5] Parts (i) and (ii) together:           − − + + = µ µ λ λ 2 3 1 4 PQ B1 0 2 5 0 0 1 1 . = − + ⇒ =           µ λ PQ M1A1 0 5 7 0 2 1 0 . = − + ⇒ =           µ λ PQ A1 λ = –2, µ = 1 M1A1 3 1 2 2 . = ⇒           − PQ PQ M1A1          − = 1 1 1 OQ A1 [9]

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 (iii) (i + j) × (4i + j + 3k) = 3i – 3j – 3k M1A1 ( )( ) [ ] 3 / 4 . 2 k i k j i + − − = p M1 3 = =K A1 [4] for final 2 marks 0 : = − − z y x π perpendicular distance 3 1 1 1 1 1 1 2 2 2 =       + + − − − = or any other method. M1 is for a complete strategy and A1 for .3 12 OR (i) (a)             → →             0 0 0 0 0 0 0 0 4 2 6 0 7 5 1 1 23 16 6 3 11 7 7 1 25 17 9 3 7 5 1 1 K OR Establishes equivalent result for MT OR Establishes 2 linearly independent relations for rows or columns M1A1 ( ) ( ) 2 dim 2 = ⇒ = ⇒ R r M A1 (b) Basis for RT is                                     6 7 9 1 , 3 1 3 1 M1A1 [5] (ii) Let             +             =             − − − 6 7 9 1 3 1 3 1 6 17 15 1 q p Solves any 2 of: p + q = 1, 3p + 9q = –15, p + 7q = –17, 3p + 6q = –6 to obtain p = 4, q = –3 M1A1 Checks consistency with other 2 equations A1 [3]

Page 13 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 (iii)             − − − =             − 6 17 15 1 0 0 3 4 M OR Works with equations (M written out fully.) (AG) B1 OEW e1 and e2 basis vectors for 0 ,0 2 1 = = ⇒ Me Me T M1 µ λ µ λ µ λ , , 6 17 15 1 0 0 6 17 15 1 0 0 3 4 2 1 ∀             − − − = + +             − − − = + +             − Me Me M A1 [3] (iv) Need –3 + λ + 2µ = 0 M1             − − = ⇒ = ⇒ = 3 3 0 37 x 1 1 µ λ A1 [2] Accept a parametric answer e.g.             − − +             − 1 2 0 3 0 9 0 46 t Augmented Matrix Method (i) (a)             − − − 6 23 16 6 3 17 11 7 7 1 15 25 17 9 3 1 7 5 1 1 ~             − − − 9 2 1 3 0 18 4 2 6 0 18 4 2 6 0 1 7 5 1 1 ~             − 0 0 0 0 0 0 0 0 0 0 9 2 1 3 0 1 7 5 1 1 M1A1 ( ) ( ) 2 dim r = = ⇒ R M A1 (b) Basis for R is                                     6 7 9 1 , 3 1 3 1 M1A1 [5]

Page 14 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2010 9231 13 © UCLES 2010 (ii) and (iii) 1 7 5 = + + + t z y x 9 2 3 − = + + t z y M1A1             − +             − +             − =             ⇒ 1 2 0 3 0 3 1 14 0 9 0 46 µ λ t z y x A1             − =             ⇒ = − = 0 0 3 4 0 ,3 t z y x µ λ B1             − − − =             − ∴ 6 17 15 1 0 0 3 4 M Me1 = Me2 = 0 M1             − − − =             + +             − ∴ 6 17 15 1 0 0 3 4 2 1 e e M µ λ A1 (iv) 0 = λ             − +             − =             1 2 0 3 0 9 0 46 µ t z y x M1A1 [8]