Cambridge A Level Mathematics - Further 9231 — 2009 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 2009 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 2009 [Turn over *1484530378*

2 1 The equation x4 −x3 −1 = 0 has roots α, β, γ, δ. By using the substitution y = x3, or by any other method, find the exact value of α6 + β6 + γ 6 + δ 6. [5] 2 Verify that, for all positive values of n, 1 (n + 2)(2n + 3) − 1 (n + 3)(2n + 5) = 4n + 9 (n + 2)(n + 3)(2n + 3)(2n + 5). [2] For the series N ∑ n=1 4n + 9 (n + 2)(n + 3)(2n + 3)(2n + 5), find (i) the sum to N terms, [3] (ii) the sum to infinity. [1] 3 The equation of a curve is y = λx2, where λ > 0. The region bounded by the curve, the x-axis and the line x = a, where a > 0, is denoted by R. The y-coordinate of the centroid of R is a. Show that λ = 10 3a. [6] 4 A curve has equation y = 1 3x3 + 1. The length of the arc of the curve joining the point where x = 0 to the point where x = 1 is denoted by s. Show that s = ã 1 0 √(1 + x4) dx. [2] The surface area generated when this arc is rotated through one complete revolution about the x-axis is denoted by S. Show that S = 1 9π(18s + 2 √2 −1). [4] [Do not attempt to evaluate s or S.] 5 Draw a sketch of the curve C whose polar equation is r = θ, for 0 ≤θ ≤1 2π. [2] On the same diagram draw the line θ = α, where 0 < α < 1 2π. [1] The region bounded by C and the line θ = 1 2π is denoted by R. Find the exact value of α for which the line θ = α divides R into two regions of equal area. [4] © UCLES 2009 9231/01/M/J/09

3 6 A curve has equation (x + y)(x2 + y2) = 1. Find the values of dy dx and d2y dx2 at the point (0, 1). [7] 7 Let In = ã 1 0 tne−t dt, where n ≥0. Show that, for all n ≥1, In = nIn−1 −e−1. [3] Hence prove by induction that, for all positive integers n, In < n!. [5] 8 Find the general solution of the differential equation 4d2y dx2 + 4dy dx + 65y = 65x2 + 8x + 73. [6] Show that, whatever the initial conditions, y x2 →1 as x →∞. [2] 9 The matrix A = 3 1 4 1 5 −1 2 1 5 ! has eigenvalues 1, 5, 7. Find a set of corresponding eigenvectors. [5] Find a matrix P and a diagonal matrix D such that An = PDP−1. [3] [The evaluation of P−1 is not required.] Determine the set of values of the real constant k such that knAn tends to the zero matrix as n →∞. [3] 10 The curve C has equation y = x2 x + λ , where λ is a non-zero constant. Obtain the equation of each of the asymptotes of C. [3] In separate diagrams, sketch C for the cases λ > 0 and λ < 0. In both cases the coordinates of the turning points must be indicated. [8] © UCLES 2009 9231/01/M/J/09 [Turn over

4 11 The line l1 is parallel to the vector 4j −k and passes through the point A whose position vector is 2i + j + 4k. The variable line l2 is parallel to the vector i −(2 sin t)j, where 0 ≤t < 2π, and passes through the point B whose position vector is i + 2j + 4k. The points P and Q are on l1 and l2, respectively, and PQ is perpendicular to both l1 and l2. (i) Find the length of PQ in terms of t. [5] (ii) Hence find the values of t for which l1 and l2 intersect. [2] (iii) For the case t = 1 4π, find the perpendicular distance from A to the plane BPQ, giving your answer correct to 3 decimal places. [5] © UCLES 2009 9231/01/M/J/09

5 12 Answer only one of the following two alternatives. EITHER By considering n−1 ∑ k=0 (1 + i tan θ)k, show that n−1 ∑ k=0 cos kθ seck θ = cot θ sin nθ secn θ, provided θ is not an integer multiple of 1 2π. [7] Hence or otherwise show that n−1 ∑ k=0 2k cos1 3kπ = 2n √3 sin1 3nπ. [2] Given that 0 < x < 1, show that n−1 ∑ k=0 cos(k cos−1x) xk = sin(n cos−1x) xn−1 √(1 −x2). [4] OR The linear transformations T1 : >4 →>4 and T2 : >4 →>4 are represented by the matrices M1 and M2, respectively, where M1 =  1 1 1 2 1 4 7 8 1 7 11 13 1 2 5 5 , M2 =  2 0 −1 −1 5 1 −3 −3 3 −1 −1 −1 13 −1 −6 −6 . (i) Find a basis for R1, the range space of T1. [4] (ii) Find a basis for K2, the null space of T2, and hence show that K2 is a subspace of R1. [5] The set of vectors which belong to R1 but do not belong to K2 is denoted by W. (iii) State whether W is a vector space, justifying your answer. [1] The linear transformation T3 : >4 →>4 is the result of applying T1 and then T2, in that order. (iv) Find the dimension of the null space of T3. [3] © UCLES 2009 9231/01/M/J/09

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/09

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2009 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 May/June 2009 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 2009 9231 01 © UCLES 2009 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 2009 9231 01 © UCLES 2009 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 2009 9231 01 © UCLES 2009 1 ( ) 3 4 3 / 1 1 y y y x + = ⇒ = M1A1 0 1 3 3 2 3 4 = − − − − ⇒ y y y y A1 ( ) 7 3 2 1 6 = − × − = ∑α M1A1 or 0 1 2 2 3 4 2 = + − − ⇒ = y y y x y has roots 2 2 2 , , γ β α M1 ( )( ) ( )( ) ( ) 2 2 2 2 2 2 4 2 6 3 γ β α β α α α α α ∑ ∑ ∑ ∑ ∑ ∑ + − = M1A2 = 5 – (–2) + 0 = 7 A1 or For last 2 marks z = y2 ⇒ z4 – 7z3 + z2 – 3z + 1 = 0 M1 ( ) 7 1 7 6 = − − = ∑α A1 or Use of SN+4 = SN + SN+3 M1 S–1 = 1 0 = 0 S2 = 12 – 2 × 0 = 1 (both) M1A1 S3 = 0 + 1 = 1 S4 = 1 + 4 = 5 (both) A1 S5 = 5 + 1 = 6 S6 = 6 + 1 = 7 (both) A1 2 Verifies displayed result M1A1 (i) SN = 1/15 – 1 / (N + 3)(2N + 5) M1A1A1 (ii) S∞ = 1 / 15 A1 ft Note: Must see working for preliminary result Either (2n2 + 11n + 15) – (2n2 + 7n + 6) (oe) Or (n + 3)(2n + 5) – (n + 2) (2n + 3) A1 A1 } in numerator

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 3 ∫ ∫ = ydx dx y y 2 2 1 M1 3 / 3 0 a ydx a λ = ∫ B1 ∫ = a a dx y 0 5 2 2 5 / λ M1A1 10 / 3 ... 2 a y λ = = A1 a a y 3 10 = ⇒ = λ (AG) A1 Finding x is not a misread and scores M0 4 dy / dx = x2 and range of x is [0,1] ( ) dx x s 2 / 1 1 0 4 1 ∫ + = ⇒ (AG) M1A1 ( )( ) dx x x S 2 / 1 1 0 4 3 1 1 3 / 2 ∫ + + = π (AEF) (completely correct) M1 ( ) ( ) dx x x s π S 2 / 1 4 1 0 3 1 3 / 2 2 + + = ∫ π A1 ( ) ( )( ) 2 / 3 4 2 / 1 4 3 1 6 / 1 1 x dx x x + = + ∫ B1 ( )( )1 2 2 18 9 / − + = s S π (AG) cwo A1

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 5 Sketch with correct shape, location and orientation B1 Shows tangency to the initial line at the pole B1 Ignore extra in diagram Draws, in the same diagram, a straight line passing through the origin and with positive gradient (distinct half-line and not a construction line) B1 ( )∫ = 2 / 0 3 2 48 / 2 / 1 π d π θ θ M1A1 ( )∫ = α d 0 3 2 6 / 2 / 1 α θ θ A1 3 / 4 3 3 2. 96 / 6 / − = ⇒ = π α π α (aef) A1 or for previous 4 marks: ( ) ( ) ∫ ∫ = α π α d d 0 2 / 2 2 2 / 1 2 / 1 θ θ θ θ M1 ] ) 2 / )[( 6 / 1( 6 / 3 3 3 α π α − = A1A1 ( ) 3 / 1 3 3 16 16 / − = ⇒ = ⇒ π α π α , or equivalent A1

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 6 (1 + y1)(x2 + y2) + (x + y)(2x + 2yy1) = 0 B1B1 Puts x = 0, y = 1 to obtain y1 = –1/3 B1 (x2 + y2)y2 + 2(1 + y1)(2x + 2yy1) + (x + y)(2 + 2 1 2y + 2yy2) = 0 B1B1B1 Puts x = 0, y = 1, y1 = –1/3 to obtain y2 = –4/9 (cwo) B1 or x3 + xy2 + x2y +y3 = 1 (expanding) Differentiating 3x2 + y2 + 2xyy′ + 2xy + x2y′ + 3y2y′ = 0 B1 B1 Sub (0,1) 1 + 3y′ = 0 ⇒ y′ = – 3 1 B1 Differentiating again 6x + 2yy′ + 2yy′ + 2x(yy″ + (y′)2) + 2y + 2xy′ + x2y″ + 2xy′ + 6y(y′)2 + 3y2y″ = 0 B1 B1 B1 Sub (0,1) 9 4 0 3 3 4 − = ′ ⇒ = ′′ + y y B1 For those who write ( ) ( ) 2 2 2 2 3 2 2 3 d d y xy x y xy x x y + + + + − = 4th, 5th and 6th marks are awarded as follows: )1 B ( )1 B ( d d )1 B ( d d 2v x v u x u v −

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 7 [ ] ∫ − − − + − = 1 0 1 1 0 dt e t n e t I t n t n n M1A1 1 1 − −− = e nI I n n (AG) A1 Hk : Ik < k! for some k B1 Hk :⇒ Ik + 1 < (k + 1)k! – e–1 < (k + 1)! M1A1 I1 = 1 – 2e–1 B1 (Do not allow I0 unless I1 is deduced from it) Completes induction argument A1 This requires at least: ‘The assumption is true and it is true for all positive integers’ for the final mark, which is only awarded if all previous marks have been gained in the proof part. 8 Roots of AQE are –1/2 ± 4i (allow 1 sign/coefficient error) M1 CF = e–x/2 (A sin 4x + B cos 4x) A1 PI of the form ax2 + bx + c M1 PI = x2 + 1 M1A1 General solution: y = e–x/2(A sin 4x + B cos 4x) + x2 + 1 A1 CF/x2 → 0∀A and B ⇒ CF/x2 → 0 M1 ⇒ y/x2 →1 as x → ∞ (AG, CWO) A1 Allow e–kx → 0 as x → ∞ provided K > 0 for M1 [Use of D operator for PI M1 Obtaining PI M1A1]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 9 Exhibits A – λ I in a numerical form, where λ = 1, 5 or 7 M1 Uses this form in some way to obtain1 eigenvector M1 Eigenvectors: any non-zero scaling of         − − 7 6 17 ,         − 1 2 1 ,         1 0 1 A1A1A1 P =         − − − 1 1 7 0 2 6 1 1 17 Do not allow         0 0 0 B1 ft D =         n n 7 0 0 0 5 0 0 0 1 M1A1 1 7 0 0 0 5 0 0 0 −         = P P A n n n n n n n k k k k M1A1 –1/7 < k < 1/7 A1 For first 2 marks Forms [ 1 vector product M1 Evaluates [ 1 vector product M1

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 10 One asymptote is x = –λ B1 y = x – λ +λ 2/(x + λ ) M1 Other asymptote is y = x – λ A1 or for previous 2 marks: mx + c = x2/(x + λ )⇒(m – 1)x2 + (c + λ m)x + λ c = 0 m – 1 = 0, c + λ m = 0 (both) M1 m = 1, c = –λ (both) A1 Differentiates and y′ = 0 M1 Identifies turning point at (–2λ , –4λ ) and (0, 0) A1A1 (Award D1 if correct turning points stated with no working) First graph: Axes and both asymptotes B1 Branches B1B1 Deduct at most 1, overall, for bad forms at infinity Second graph: Axes and both asymptotes + 1 branch B1 Other branch B1 Deduct at most 1, overall, for bad forms at infinity Choose which is 1st graph and which is 2nd to the benefit of candidates, i.e. best = 1st For differentiating a special case M1A0A0 Graphs of special cases B2 (max) B1 (max) If stated in text, turning points need not be labelled on graph

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 11 (i) (i – (2sint)j) × (4j – k) = (2sint)i + j + 4k M1A1 PQ = (i – j).[(2sint)i + j + 4k]/ 17 sin 4 2 + t dM1A1 PQ = 17 sin 4 / sin 2 1 2 + − t t A1 Condone disappearing modulus sign – Deduct 1 mark if no modulus sign at all (ii) 6 / 5 ,6 / 2 / 1 sin 0 π π = ⇒ = ⇒ = t t PQ or 0.524, 2.62 M1A1 (both) (iii) Obtains some vector ⊥ plane BPQ, e.g., ( ) ( ) k j i j i k j i 3 4 2 4 2 4 2 − + = − × + + M1A1 ( ) ( ) ( ) 57 / . 3 4 2 4 : j i k j i − − + = BPQ A p M1A1 ( ) 219 .0 57 / 1 2 4 = − = A1 or For (i)         − +         = = 1 4 0 4 1 2 1 λ r l         − +         = = 0 sin 2 1 4 2 1 2 t r µ l         − − − = ⇒ λ µ λ µ t PQ sin 2 4 1 1 0 1 4 0 =         − ⋅ PQ 4 sin 8 17 = + ⇒ t µ λ 0 0 sin 2 1 =         − ⋅ t PQ ( ) t t t sin 2 1 sin 4 1 sin 8 2 + = + + ⇒ µ λ M1 Whence ( ) ( )t t 2 sin 4 17 sin 2 1 4 + − = λ and ( )t t 2 sin 4 17 sin 2 17 + + = µ A1 (both)

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 For PQ ( ) ( )t t t x 2 sin 4 17 sin 2 1 sin 2 + − = ( )t t y 2 sin 4 17 sin 2 1 + − = ( ) t t z 2 sin 4 17 sin 2 1 4 + − = dM1A1 (λ and µ of the form t t b a 2 sin 4 17 sin + + ) ( ) ( ) 2 2 2 2 2 2 2 4 1 sin 4 sin 4 17 sin 2 1 + + + − = + + = t t t z y x PQ t t 2 sin 4 17 sin 2 1 + − = A1 For (iii) Eliminates λ and µ from parametric equation of plane BPQ         +         − +         =         4 1 2 0 2 1 4 2 1 µ λ z y x M1A1 to obtain cartesian equation of plane BPQ ( )1 2 4 3 4 2 4 − = − + z y x * A1 Uses distance of point from a line formula with * and (2, 1, 4) M1 d = ( ) 57 1 2 4 9 16 32 4 2 4 12 4 2 8 − = + + + − − + = 0.219 cao A1

Page 13 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 12 EITHER ( ) ( ) θ θ θ θ k i k i k k sin cos sec tan 1 + = + M1A1 ( ) θ θ θ tan /]1 ) tan 1 [( tan 1 1 0 i i i n k n k − + = + ∑ − = M1A1 ( ) θ θ θ θ θ θ θ cot cos sec cot cot sin sec n i n n n − + = A1 θ θ θ θ θ cot sin sec cos sec 1 0 n k n k n k = ⇒∑ − = (AG) M1A1 Put θ = 3 / π to obtain second result (AG) M1A1 x x / 1 sec cos = ⇒ = θ θ , 2 1 / cot x x − = θ M1A1 Uses these results to obtain final result (AG) M1A1 First 7 marks ( ) ( ) ( ) { } θ θ θ θ θ θ tan tan 1 1 sec sin cos tan 1 1 0 1 0 i i k i k i n k n n k − + − = + = + ∑ ∑ − − M1A1 M1A1 ( )       + − = ∴∑ − θ θ θ θ θ θ tan sec sin cos 1 Re sec cos 1 0 i n i n k n n k M1A1 = θ θ θ cot sec sin n n (AG) A1

Page 14 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 12 OR (i) Reduces M1 to col echelon form by elementary col operations or T 1 M to row echelon form by elementary row operations             − − → 0 0 0 0 1 1 0 0 7 0 3 0 1 0 0 3 1 T M hence                         −                         1 1 0 0 , 7 0 3 0 , 1 0 0 3 M1 Any correct echelon form A1 Selects 3 li cols from reduced M1 (or equiv from reduced T 1 M ) M1 (Allow M1A1M1 for any other valid method) Any correct basis of R1, e.g.                         −                         1 1 0 0 , 1 6 3 0 , 1 1 1 1 or                         −                         − 1 1 0 0 , 7 0 3 0 , 1 0 0 3 A1 or for (i) Reduces M1 to row echelon form by elementary row operations             0 0 0 0 1 2 0 0 2 2 1 0 2 1 1 1 M1 Any correct row echelon form A1 No working for echelon matrix here, or in (ii) gets B1 rr(M1) = 3 ⇒ any 3 li columns form a basis of M1 May be implied by answer M1 Any correct basis of R1, e.g.,                                                 5 11 7 1 , 2 7 4 1 , 1 1 1 1 A1 or detM1 = 0 (calculator) M1 3 1 ≤ ⇒M r A1 First 3 columns of M1 are clearly linearly independent M1 Hence basis A1

Page 15 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2009 9231 01 © UCLES 2009 (ii) Reduces M2 to echelon form by elementary row operations             − − − − 0 0 0 0 0 0 0 0 1 1 2 0 1 1 0 2 M1 Any correct row echelon form A1 Valid method to find basis of K2 M1 (Allow M1A1M1 for any other valid method) Any correct basis of K2, e.g.,                                     0 2 1 1 , 2 0 1 1 or                         −             1 1 0 0 , 1 1 1 1 A1 Shows each basis element of K2 is in R1 (AG) B1 First 4 marks d c 2c b a b a 0 0 0 0 0 0 0 0 1 1 1 0 13 3 5 2 d c b a 6 1 3 1 6 1 3 1 1 1 1 0 13 3 5 2 2 − + +             − − →             − − − − − − − − − − = T M M1A1 Basis is                         −             1 1 0 0 , 0 2 1 1 M1A1 (iii) Any valid argument, e.g., W does not contain zero vector so W not a vector space B1 (iv) For any vector x, M2M1x = M2(α b1 + β b2 + γ b3), where b1, b2, b3 are any 3 l.i. basis vectors of R1, 2 of which must be basis vectors of K2 M1 Hence if b1 and b2 are basis vectors of K2, then M2M1x = γ M2b3 A1 Hence as dim(range of T3) = 1, then the dimension of the null space of T3 = 4 – 1 = 3 A1 or M2M1 =             − − − − − − − − − − − − 90 90 45 0 20 20 10 0 36 36 18 0 14 14 7 0 B1 Nullity = 4 – r(M2M1) = 3 M1A1