Cambridge A Level Mathematics - Further 9231 — 2011 May/June Paper 1 · Variant 3

*3313671834* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/13 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_13/2R © UCLES 2011 [Turn over

2 1 Find 22 + 42 + . . . + (2n)2. [2] Hence find 12 −22 + 32 −42 + . . . −(2n)2, simplifying your answer. [3] 2 Let A = 2 3 0 1 . Prove by mathematical induction that, for every positive integer n, An = 2n 3(2n −1) 0 1 . [5] 3 Find a cubic equation with roots α, β and γ, given that α + β + γ = −6, α2 + β2 + γ 2 = 38, αβγ = 30. [3] Hence find the numerical values of the roots. [3] 4 The curve C has equation 2xy2 + 3x2y = 1. Show that, at the point A (−1, 1) on C, dy dx = −4. [3] Find the value of d2y dx2 at A. [5] 5 Let In = ã 1 4π 0 tannx dx, where n ≥0. Use the fact that tan2x = sec2x −1 to show that, for n ≥2, In = 1 n −1 −In−2. [4] Show that I8 = 1 7 −1 5 + 1 3 −1 + 1 4π. [4] 6 The curves C1 and C2 have polar equations C1: r = a, C2: r = 2a cos 2θ, for 0 ≤θ ≤1 4π, where a is a positive constant. Sketch C1 and C2 on the same diagram. [3] The curves C1 and C2 intersect at the point with polar coordinates (a, β). State the value of β. [1] Show that the area of the region bounded by the initial line, the arc of C1 from θ = 0 to θ = β, and the arc of C2 from θ = β to θ = 1 4π is a21 6π −1 8 √3. [4] © UCLES 2011 9231/13/M/J/11

3 7 A curve C has parametric equations x = et cos t, y = et sin t, for 0 ≤t ≤π. Find the arc length of C. [4] Find the area of the surface generated when C is rotated through 2π radians about the x-axis. [7] 8 Find the general solution of the differential equation d2x dt2 + 2dx dt + 5x = 10 sin t. [6] Find the particular solution, given that x = 5 and dx dt = 2 when t = 0. [4] State an approximate solution for large positive values of t. [1] 9 The curve C with equation y = ax2 + bx + c x −1 , where a, b and c are constants, has two asymptotes. It is given that y = 2x −5 is one of these asymptotes. (i) State the equation of the other asymptote. [1] (ii) Find the value of a and show that b = −7. [3] (iii) Given also that C has a turning point when x = 2, find the value of c. [3] (iv) Find the set of values of k for which the line y = k does not intersect C. [4] 10 The lines l1 and l2 have equations l1: r = 6i + 5j + 4k + λ(i + j + k) and l2: r = 6i + 5j + 4k + µ(4i + 6j + k). Find a cartesian equation of the plane Π containing l1 and l2. [4] Find the position vector of the foot of the perpendicular from the point with position vector i + 10j + 3k to Π. [4] The line l3 has equation r = i + 10j + 3k + ν(2i −3j + k). Find the shortest distance between l1 and l3. [5] [Question 11 is printed on the next page.] © UCLES 2011 9231/13/M/J/11 [Turn over

4 11 Answer only one of the following two alternatives. EITHER A 3 × 3 matrix A has eigenvalues −1, 1, 2, with corresponding eigenvectors 0 1 −1 !, −1 0 1 !, 1 1 0 !, respectively. Find (i) the matrix A, (ii) A2n, where n is a positive integer. [14] OR Determine the rank of the matrix A =  1 −1 −1 1 2 −1 −4 3 3 −3 −2 2 5 −4 −6 5 . [3] Show that if Ax = p  1 2 3 5 + q  −1 −1 −3 −4 + r  −1 −4 −2 −6 , where p, q and r are given real numbers, then x =  p + λ q + λ r + λ λ , where λ is real. [4] Find the values of p, q and r such that p  1 2 3 5 + q  −1 −1 −3 −4 + r  −1 −4 −2 −6 =  3 7 8 15 . [3] Find the solution x =  α β γ δ of the equation Ax =  3 7 8 15 for which α2 + β2 + γ 2 + δ 2 = 11 4 . [4] 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 9231/13/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 9231/13 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 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 13 © 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 13 © 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 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 1 Finds four times sum of first n squares. Subtracts eight times sum of first n squares from sum of first 2n squares. Simplifies. 6 )1 2 )( 1 ( 4 ) 2 ( ... 4 2 2 2 2 + + = + + + n n n n 2 2 2 2 2 ) 2 ( ... 4 3 2 1 n − + − + − 6 )1 2 )( 1 ( 8 6 )1 4 )( 1 2 ( 2 + + − + + = n n n n n n ( ) )1 2 ( 4 4 1 4 3 )1 2 ( + − = − − + + = n n n n n n Or 6 )1 2 )( 1 ( 4 2 )1 ( 4 6 )1 2 )( 1 ( 4 + + − + + − + + n n n n n n n n n n n − − = 2 2 M1A1 M1A1 A1 (M1A1) (A1) 2 3 [5] 2 States proposition. Shows base case is true. Proves inductive step. States conclusion. Let Pn be the proposition: A = ⇒       1 0 3 2 An =       − 1 0 )1 2 ( 3 2 n n A1 = ⇒       − × =       1 0 )1 2 ( 3 2 1 0 3 2 1 P1 is true. Assume Pk is true for some integer k. Ak+1 =       −       1 0 )1 2 ( 3 2 1 0 3 2 k k       − =       + − = + + + 1 0 )1 2 ( 3 2 1 0 3 )1 2 ( 2.3 2 1 1 1 k k k k Since P1 is true and Pk ⇒ Pk+1, hence by PMI Pn is true ∀ positive integers n. B1 B1 M1 A1 A1 5 [5] 3 Uses ( ) ∑ ∑ ∑ + = αβ α α 2 2 2 States equation with required roots. Factorises Gives values of α, β, γ. ∑ ∑ − = ⇒ + = 1 2 38 36 αβ αβ 0 30 6 2 3 = − − + ∴ t t t is the required equation. 0 ) 5 )( 3 )( 2 ( = + + − ⇒ t t t Hence α, β, and γ are 2, –3 and –5 (in any order). N.B. Answers written down with no working get B1. M1A1 A1 M1A1 A1 3 3 [6]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 4 Differentiates with respect to x. Substitutes (–1, 1) Differentiates again. Substitutes (–1, 1) and 4 − = ′y 0 3 6 4 2 2 2 = ′ + + ′ + y x xy y xy y 4 0 3 6 4 2 − = ′ ⇒ = ′ + − ′ − y y y (AG) 0 3 6 6 6 4 ) 4 4 ( 4 2 = ′′ + ′ + ′ + + ′′ + ′ ′ + + ′ y x y x y x y y xy y y x y y y 0 3 24 24 6 4 80 16 = ′′ + + + + ′′ − − − y y 42 − = ′′ ⇒y B1B1 B1 B1B1 B1 M1 A1 3 5 [8] 5 Uses 1 sec tan 2 2 − = x x Integrates Obtains reduction formula. Evaluates I0 Uses reduction formula. ∫ ∫ − = = − 4 0 2 2 4 0 d )1 (sec tan d tan π π x x x x x I n n n =∫ − − − − −       − = − 4 0 2 4 0 1 2 2 2 1 tan d sec tan π π n n n n I n x I x x x Or [ ]4 0 1 tan π x I n n − = ∫ − − − − − 4 0 2 2 3 d tan sec tan ) 2 ( π n n I x x x x n ∫ − − − + − + = 4 0 2 2 2 d ) tan 1( tan ) 2 ( 1 π n n I x x x n 2 1 1 − − − = nI n (AG) ∫ = = 4 0 0 4 d 1 π π x I [ ] ∫ − = − = − = 4 0 4 0 2 2 4 1 tan d )1 (sec π π π x x x x I 0 2 1 I I − = 0 4 1 3 1 I I + − = 0 6 1 3 1 5 1 I I − + − = 4 1 3 1 5 1 7 1 8 π + − + − = I (AG) M1 M1A1 (M1) (A1) A1 B1 (B1) M1A1 A1 4 4 [8]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 5 Alternative for first part: x x n x x n n 2 2 1 sec tan )1 ( ) (tan d d − − − = ) tan 1( tan )1 ( 2 2 x x n n + − = − x n x n n n tan )1 ( tan )1 ( 2 − + − = − Integrating with respect to x, between 0 and 4 π [ ] )1 ( )1 ( tan 2 4 0 1 − + − = − − n I n x n n π n n I n I n )1 ( )1 ( 1 2 − + − = ⇒ − 2 1 1 − − − = ⇒ n n I n I M1 A1 M1 A1 4 6 Sketches each curve on same diagram. States the value of β. Adds 12 1 of area of circle to sector of C2 from 2 to 6 π θ π θ = = . Uses double angle formula and integrates. Obtains printed result. Sketch of C1 (relevant part only required). Sketch of C2 (generous on tangency features). 6 π β = . ∫ + 4 6 2 2 2 d 2 cos 4 2 1 12 1 π π θ θ π a a ∫ + + = 4 6 2 2 d )1 4 (cos 12 1 π π θ θ π a a = 4 6 2 2 4 4 sin 12 1 π π θ θ π     + + a a =       −8 3 6 2 π a (AG) B1 B2 B1 B1M1 M1 A1 3 1 4 [8]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 7 Differentiates Uses arc length formula Uses surface area formula and obtains correct integral. Integrates by parts twice. Sees original again. Obtains surface area. ) sin (cos e t t x t − = & ) cos (sin e t t y t + = & t 2 e 2 ) cos sin 2 1 cos sin 2 1( e = + + − = t t t t s t & ∫ = π 0 d e 2 t s t )1 e( 2 − = π (= 31.3) ∫ ∫ = = π π π π 0 0 2 d sin e 2 2 d e 2 sin e 2 t t t t S t t t Let ∫ = t t I t d sin e2 ∫ + − = t t t t t d cos e 2 cos e 2 2 ∫ − + − t t t t t t t d sin e 4 sin e 2 cos e 2 2 2 t t I t t cos e sin e 2 5 2 2 − = ) cos sin 2 ( 5 e2 t t I t − = ⇒ ( ) π π 0 2 cos sin 2 5 e 2 2       − = t t S t ( )1 e 5 2 2 2 + = π π (= 953) (N.B. If 953 written down with no working award B1 in place of the final 5 marks.) B1 B1 M1 A1 M1A1 M1 A1 M1 A1 A1 4 7 Alternative method for integrating∫ t t t d sin e2 . { } { }t t t t t d e Im d e e Im i) 2 ( i 2 ∫ ∫ + = ⋅       − + =       + = + i) 2 )( sin i (cos 5 e Im i 2 e Im 2 i) 2 ( t t t t ) cos sin 2 ( e 5 1 2 t t t − = M1 A1M1 A1 [11]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 8 Forms and solves AQE. States CF States form for PI. Substitutes in equation. Obtains values for p and q by comparing coefficients. States GS. Uses initial conditions to evaluate constants. States particular solution. Gives req. approximate solution. 2 1 0 5 2 2 ± − = ⇒ = + + m m m i CF ) 2 sin 2 cos ( e t B t A t + − (OE) PI t q t p x sin cos + = ⇒ t q t p x cos sin + − = & ⇒ t q t p x sin cos − − = && t t q t p t q t p t q t p sin 10 sin 5 cos 5 cos 2 sin 2 sin cos = + + + − − − 0 2 4 = + q p and 10 4 2 = + − q p ⇒ 1 − = p , 2 = q GS t t t B t A x t cos sin 2 ) 2 sin 2 cos ( e − + + = − (OE) 6 5 0 = ⇒ = = A x t t t t B t A t B t A x t t sin cos 2 ) 2 cos 2 2 sin 2 ( e ) 2 sin 2 cos ( e + + + − + + − = − − & 3 2 2 6 2 = ⇒ + + − = B B t t t t x t cos sin 2 ) 2 sin 3 2 cos 6 ( e − + + = − (OE) As ∞ → t t t x cos sin 2 − ≈ (The final mark is independent of A and B). M1 A1 M1 M1 A1 A1 B1 M1 A1 A1 B1 6 4 1 [11] 9 (i) (ii) States vertical asymptote. States the value of a. Divides. Compares coefficients to obtain b. x = 1 2 = a )1 ( − + + + + + = x c b a b a ax y 7 5 2 − = ⇒ − = + b b (AG) Or 1 5 2 − + − = x a x y 1 5 7 2 2 − + + − = x a x x Equate coefficients to obtain 2 = a , 7 − = b B1 B1 M1 A1 (M1) (B1A1) 1 3

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total (iii) (iv) Differentiates and uses given value of x to obtain c. Forms quadratic in x. Uses discriminant. Obtains required result. 0 )1 ( ) 5 ( 2 2 = − − − = ′ x c y When x = 2 then c = 7 Let k x x x y = − + − = )1 ( 7 7 2 2 0 7 ) 7 ( 2 2 = + + + − ⇒ k x k x No real roots 0 ) 7 ( 8 ) 7 ( 2 < + − + ⇒ k k 1 7 0 )1 )( 7 ( 0 7 6 2 < < − ⇒ < − + ⇒ < − + ⇒ k k k k k M1A1 A1 B1 M1 A1 A1 3 4 [11] 10 Uses vector product to find normal to plane. Uses r.n = constant. Obtains cartesian equation of plane. 1 6 4 1 1 1 k j i = –5i + 3j + 2k Equation of plane: 5x – 3y – 2z = constant 30 – 15 – 8 = 7 5x – 3y – 2z = 7 Alternatively:           +           +           =           1 6 4 1 1 1 4 5 6 µ λ z y x µ λ 4 6 + + = x µ λ 6 5 + + = y µ λ + + = 4 z Eliminates λ and µ. Obtains 5x – 3y – 2z = 7 M1A1 M1 A1 (M1) (A1) (M1) (A1) 4

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 10 Contd. Finds equation of perpendicular to plane through given point. Finds value of parameter at point in plane. Obtains foot of perpendicular. Alternatively: Form sufficient equations, using orthogonality. Two will suffice if foot of perpendicular is expressed using parametric equation of plane. Finds direction of common perpendicular. Forms vector between known points on l1 and l3. Finds shortest distance by projection. Equation of perpendicular: r = i + 10j + 3k + t ( 5i – 3j – 2k ) 1 7 ) 2 3 ( 2 ) 3 10 ( 3 ) 5 1( 5 = ⇒ = − − − − + t t t t Foot of perpendicular is 6i + 7j + k. Let foot of perpendicular be ai + bj + ck and using othogonality: 14 0 1 1 1 3 10 1 = + + ⇒ =           ⋅           − − − c b a c b a 67 6 4 0 1 6 4 3 10 1 = + + ⇒ =           ⋅           − − − c b a c b a ai + bj + ck lies in plane of l1 and l2 : 7 2 3 5 = − − c b a k j i + + ⇒ 7 6 k j i k j i 5 4 1 3 2 1 1 1 − + = −           − =           −           1 5 5 3 10 1 4 5 6 ) 54 .1 ( 42 10 5 1 4 1 5 5 25 1 16 1 = =           − ⋅           − + + M1 M1 A1 A1 (M1A1) (M1A1) M1A1 M1A1 A1 4 5 [13]

Page 11 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 10 Contd. Alternative for last part: Let P be on l1 and Q be on l3.           + + + = λ λ λ 4 5 6 p and           + − + = v v v 3 3 10 2 1 q           − − − − − + − − = ⇒ v v v PQ 3 1 3 5 2 5 λ λ λ Uses orthogonality conditions: 3 1 0 3 1 0 1 1 1 − = ⇒ = − − ⇒ =           ⋅ ⇒ λ λ PQ 7 13 0 14 26 0 1 3 2 = ⇒ = + − ⇒ =           − ⋅ v v PQ           − − = ⇒ 25 5 20 21 1 PQ 42 21 5 5 1 4 21 5 2 2 2 = + + = ⇒PQ (M1) (M1) (A1) (A1) (A1) (5)

Page 12 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 (i) (ii) EITHER Writes P and D. (Note: Columns can be in any order, but must match.) Finds Det P. Finds inverse of P. (Adj ÷ Det) Finds expression for A. Evaluates A. Finds expression for A2n Evaluates. P =           − − 0 1 1 1 0 1 1 1 0 D =          − 2 0 0 0 1 0 0 0 1 Det P = 2 P–1 =           − − − 1 1 1 1 1 1 1 1 1 2 1 (No working 1/3) Row operations M1A1A1 ( 3 errors). A = PDP–1 A =           − − 0 1 1 2 0 1 2 1 0 .           − − − 1 1 1 1 1 1 1 1 1 2 1 =           − 0 1 1 5.1 5.0 5.1 5.0 5.0 5.1 A2n = PD2nP–1 =           − − −                     − − 1 1 1 1 1 1 1 1 1 2 0 0 0 1 0 0 0 1 0 1 1 1 0 1 1 1 0 2 1 2n =           − − −           − − 1 1 1 1 1 1 1 1 1 0 1 1 2 0 1 2 1 0 2 1 2 2 n n =           − + − − − + 2 0 0 1 2 1 2 1 2 1 2 1 2 1 2 2 1 2 2 2 2 2 2 n n n n n n B1B1 B1 M1A1 M1 M1A1 A1 M1 A1 M1A1 A1 9 5 [14]

Page 13 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total (i) (ii) EITHER (Alternative) Uses Ae = λe (3 times) Forms 3 linear equations (3 times) Solves one set of equations. Solves other two sets. Writes A. Writes P and D. Finds inverse of P. Finds A2n.           − =           −           1 1 0 1 1 0 j h g f e d c b a 1 1 0 = − − = − = − j h f e c b          − =          −           1 0 1 1 0 1 j h g f e d c b a 1 0 1 = + − = + − − = + − j g f d c a           =                     0 2 2 0 1 1 j h g f e d c b a 0 2 2 = + = + = + h g e d b a A =           − 0 1 1 5.1 5.0 5.1 5.0 5.0 5.1 P =           − − 0 1 1 1 0 1 1 1 0 D =          − 2 0 0 0 1 0 0 0 1 P–1 =           − − − 1 1 1 1 1 1 1 1 1 2 1 A2n = PD2nP–1 =           − − −                     − − 1 1 1 1 1 1 1 1 1 2 0 0 0 1 0 0 0 1 0 1 1 1 0 1 1 1 0 2 1 2n =           − − −           − − 1 1 1 1 1 1 1 1 1 0 1 1 2 0 1 2 1 0 2 1 2 2 n n =           − + − − − + 2 0 0 1 2 1 2 1 2 1 2 1 2 1 2 2 1 2 2 2 2 2 2 n n n n n n M1A1 M1 A1 B1B1 B1 M1A1 M1 A1 M1A1 A1 4 5 5 [14]

Page 14 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 OR Reduces matrix to echelon form. Obtains rank. Use system of equations, or any other method (see below). Finds basis of null space. Obtains general solution. Finds values of p, q and r, e.g. by solving a set of equations. Award B2 (all correct) or B1 (two correct), with no working. Solves for 4 11 2 2 2 2 = + + + δ γ β α .             − − − − − − − − 5 6 4 5 2 2 3 3 3 4 1 2 1 1 1 1             − − − − → 0 0 0 0 1 1 0 0 1 2 1 0 1 1 1 1 ... r(A) = 4 – 1 = 3 0 0 2 0 = − = + − = + − − t z t z y t z y x λ = ⇒t , λ = z , λ = y , λ = x ∴ Basis of null space is                         1 1 1 1 =                         − 0 r q p Ax 0             +             = ⇒ 1 1 1 1 0 λ r q p x (AG) 8 2 3 3 7 4 2 3 = − − = − − = − − r q p r q p r q p p = 1, q = –1 r = –1 0 4 1 2 4 4 11 2 2 2 2 2 = + − ⇒ = + + + λ λ δ γ β α             − − = ⇒ = ⇒ =       − ⇒ 25 .0 75 .0 75 .0 25 .1 4 1 0 2 1 2 2 x λ λ M1A1 A1 M1 A1 M1A1 M1 A1, A1 M1A1 M1A1 3 4 3 4 [14]

Page 15 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – May/June 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Mark Total 11 Contd. Alternative methods for 2nd part. Writes             = 4 3 2 1 x x x x x and forms equations from Ax =             − − − − +             − − − − +             6 2 4 1 4 3 1 1 5 3 2 1 r q p r q p x x x x − − = + − − 4 3 2 1 r q p x x x x 4 2 3 4 2 4 3 2 1 − − = + − − r q p x x x x 2 3 3 2 2 3 3 4 3 2 1 − − = + − − r q p x x x x 6 4 5 5 6 4 5 4 3 2 1 − − = + − − Obtains, for example, p x x + = 4 1 q x x + = 4 2 r x x + = 4 3 Sets λ = 4 x to obtain:             + + + = λ λ λ λ r q p x Mark similarly if equations obtained from reduced augmented matrix. Those who work in reverse direction and merely verify the result get M1A1 i.e. 2/4. M1A1 M1 A1