Cambridge A Level Mathematics - Further 9231 — 2013 Oct/Nov Paper 1 · Variant 1

*0581709578* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 2013 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. JC13 11_9231_11/RP © UCLES 2013 [Turn over

2 1 The curve C has polar equation r = 2e1, for 1 60 ≤1 ≤1 20. Find (i) the area of the region bounded by the half-lines 1 = 1 60, 1 = 1 20 and C, [2] (ii) the length of C. [3] 2 The cubic equation x3 −px −q = 0, where p and q are constants, has roots !, ", '. Show that (i) !2 + "2 + '2 = 2p, [1] (ii) !3 + "3 + '3 = 3q, [2] (iii) 6!5 + "5 + '5 = 5!3 + "3 + '3!2 + "2 + '2. [3] 3 It is given that Sn = n Ð r=1 ur = 2n2 + n. Write down the values of S1, S2, S3, S4. Express ur in terms of r, justifying your answer. [4] Find 2n Ð r=n+1 ur. 3 4 It is given that In = Ô 1 0 xn 1 + 2x dx. Show that, for n ≥1, 2n + 1In = ï3 −nIn−1. 3 Show that I3 = 2 35ï3 + 1. 4 5 It is given that y = 1 + x2 ln1 + x. Find d3y dx3 . [3] Prove by mathematical induction that, for every integer n ≥3, dny dxn = −1n−1 2n −3! 1 + xn−2 . 5 © UCLES 2013 9231/11/O/N/13

3 6 The linear transformation T : >4 →>4 is represented by the matrix M, where M = € 1 −3 −1 2 4 −10 0 2 1 −1 3 −4 5 −12 1 1  . Find, in either order, the rank of M and a basis for the null space K of T. [6] Evaluate M € 1 −2 −3 −4  , and hence show that every solution of Mx = € 2 16 10 22  has the form x = € 1 −2 −3 −4  + ,e1 + -e2, where , and - are real numbers and e1, e2 is a basis for K. [3] 7 The square matrix A has , as an eigenvalue with e as a corresponding eigenvector. Show that e is an eigenvector of A2 and state the corresponding eigenvalue. [3] Find the eigenvalues of the matrix B, where B = ` 1 3 0 2 0 2 1 1 2 a . 4 Find the eigenvalues of B4 + 2B2 + 3I, where I is the 3 × 3 identity matrix. [3] 8 The plane 1 has equation r = ` 2 3 −1 a + s ` 1 0 1 a + t ` 1 −1 −2 a . Find a cartesian equation of 1. [3] The plane 2 has equation 2x −y + Ï = 10. Find the acute angle between 1 and 2. [2] Find an equation of the line of intersection of 1 and 2, giving your answer in the form r = a + ,b. [5] 9 The curve C has parametric equations x = t2, y = t −1 3t3, for 0 ≤t ≤1. Find the surface area generated when C is rotated through 20 radians about the x-axis. [6] Find the coordinates of the centroid of the region bounded by C, the x-axis and the line x = 1. [6] © UCLES 2013 9231/11/O/N/13 [Turn over

4 10 The curve C has equation y = px2 + 4x + 1 x + 1 , where p is a positive constant and p ≠3. (i) Obtain the equations of the asymptotes of C. [3] (ii) Find the value of p for which the x-axis is a tangent to C, and sketch C in this case. [4] (iii) For the case p = 1, show that C has no turning points, and sketch C, giving the exact coordinates of the points of intersection of C with the x-axis. [5] 11 Answer only one of the following two alternatives. EITHER State the fifth roots of unity in the form cos 1 + i sin 1, where −0 < 1 ≤0. [2] Simplify x −cos 2 50 + i sin 2 50 x −cos 2 50 −i sin 2 50 . 2 Hence find the real factors of x5 −1. 2 Express the six roots of the equation x6 −x3 + 1 = 0 as three conjugate pairs, in the form cos 1 ± i sin 1. [6] Hence find the real factors of x6 −x3 + 1. 2 OR Given that y2 d2y dx2 −6y2 dy dx + 2y @dy dx A2 + 3y3 = 25e−2x and that v = y3, show that d2v dx2 −6dv dx + 9v = 75e−2x. 4 Find the particular solution for y in terms of x, given that when x = 0, y = 2 and dy dx = 1. [10] 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 2013 9231/11/O/N/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2013 series 9231 FURTHER MATHEMATICS 9231/11 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 2013 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 2013 9231 11 © Cambridge International Examinations 2013 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 2013 9231 11 © Cambridge International Examinations 2013 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 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 1 (i) (ii) Uses area formula. Obtains result. Uses arc length formula. Obtains result. [ ] ∫ = = 2 6 2 6 2 2 e d e 4 2 1 Area π π π π θ θ θ 3 e e π π − = ( = 20.3) Arc length = ∫ ∫ = + 2 6 2 6 2 2 d e 2 2 d e 4 e 4 π π π π θ θ θ θ θ [ ] ( ) 8.83 e e 2 2 e 2 2 6 2 2 6 =         − = = π π π π θ M1 A1 M1A1 A1 2 3 [5] 2 (i) (ii) (iii) Finds S2. Finds S3. Finds S5. ( ) ( ) γα βγ αβ γ β α γ β α + + − + + = + + 2 2 2 2 2 ( ) p p 2 2 0 = − − = (AG) q q q p 3 3 0 3 3 3 3 = + = + ∑ = + + α γ β α (AG) ∑ ∑ + = + + 2 3 5 5 5 α α γ β α q p pq p q q p 5 2. 3. = + = 2 3 5 5 pq 30 6 α α α ∑ ∑ = = ∑ ⇒ (AG) B1 M1A1 M1 A1 A1 1 2 3 [6] 3 Writes first four sums. Deduces first four terms, conjectures and justifies result. Obtains required sum. S1…S4 ~ 3, 10, 21, 36 u1…u4 ~ 3, 7, 11, 15 ⇒ ur = 4r ‒ 1 since ( ) { } n n n n Sn + = − + = 2 2 1 4 6 2 as given. Or ( ) ( )1 1 2 2 2 2 1 − − − − + = − = − r r r r S S u r r r = 4r – 1 ( ) ( ) ( )       − + − − + = − ∑+ n n n n n n r n n 2 1 .4 2 2 1 2 2 .4 1 4 2 1 ( ) n n n n n n + = + − + = 2 2 2 6 2 2 8 Or Sum of AP ( ) n n n n n + = − + + = 2 6 1 8 3 4 2 B1 B1B1 B1 B1B1 B1 M1A1 A1 4 3 [7]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 4 Integrates by parts. Obtains result. Alternatively: Finds I0 (or I1). Uses Red. Form. Finds I2 and I3. ( ) ( ) ( ) x x x nx x x I n n n d 2 1 2 1 2 1 1 0 2 1 1 1 0 2 1 ∫ + + −     + − ( ) ( ) x x x n x x x n n n d 2 1 2 d 2 1 3 1 0 1 0 2 1 2 1 1 ∫ ∫ + − + − = − ( ) (AG) 3 1 2 1 − − = + ⇒ n n nI I n ( ) ( ) ( ) 2 1 1 2 1 2 1 2 1 . 2 1 2 1 d d − − + + + =   + x x nx x x x x n n n ( ) ( ) n n n I x x x n x x + + + =     + ⇒ − 2 1 2 1 2 1 1 1 0 2 1 ( ) (AG) 3 1 2 1 − − = + ⇒ n n nI I n [ ] 1 3 2 1 1 0 0 − = + = x I ( ) 3 1 1 3 3 3 1 1 = ⇒ − − = I I ( )1 3 35 2 15 2 5 3 3 2 + = ⇒ − = ⇒ I I (AG) M1A1 A1 (M1) (A1) (A1) B1 M1 A1A1 3 4 [7] 5 (i) (ii) Differentiates once, twice and three times. Proves base case. States inductive hypothesis. Differentiates Proves inductive step and states conclusion. ( ) ( ) ( ) x x x y + + + + = ′ 1 1 ln 1 2 ( ) 3 1 ln 2 + + = ′′ x y x y + = ′′′ 1 2 (Allow B1 if constant term in previous line incorrect.) ( ) true. is H 1 2 1 !0.2. 1 d d 3 2 3 3 ⇒ + = + − = x x x y ( ) ( ) ( ) . some for 1 ! 3 .2. 1 d d : H 2 .1 k x k x y k k k k k − − + − − = ( ) ( ) ( )( )( ) ( ) 1 1 1 1 1 2 1 ! 3 2. 1 d d − − − + + + − − − − = k k k k x k k x y ( ) ( ) ( ) true is H 1 ! 2 .2. 1 1 1 + − ⇒ + − − = k k k x k Hence by PMI Hn is true for all integers [=3 B1 B1 B1 B1 B1 M1 A1 A1 3 5 [8]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 6 Reduces to echelon form. Obtains rank and basis for null space. Evaluates matrix product. Finds general solution of equations.             − − − → →             − − − − − − 0 0 0 0 0 0 0 0 3 2 1 0 2 1 3 1 1 1 12 5 4 3 1 1 2 0 10 4 2 1 3 1 K r(M) = 4 ‒ 2 = 2    = − + = + − − 0 3 2 0 2 3 t z y t z y x λ µ λ µ λ µ 5 7 , 2 3 , , − = − = = = ⇒ x y z t Basis is : 1 0 3 7 , 0 1 2 5                                     − − (OE) e.g.             − 2 3 0 1 or             5 7 1 0 M             =             − − − 22 10 16 2 4 3 2 1 x K 4 3 2 1 ∈             − − − − x (AG) 4 3 2 1 2 1 e e µ λ + +             − − − = M1A1 A1 M1 A1 A1 B1 M1 A1 6 3 9 7 Proves first result. Obtains eigenvalues of B. Obtains eigenvalues of related matrix. Ae = λe ⇒ A2e = AAe = A λe = λ Ae = λ2e ⇒ result. (e ≠ 0 ⇒ λ2 is an eigenvalue of A2.) ( )( )( ) 0 2 4 1 = + − − λ λ λ ⇒ λ = ‒2, 1, 4 . eigenvalue an is 6 3 1 2 1 2 4 ⇒ = + × + 6e e e e ( ) ( ) . eigenvalue an is 27 27 3 2 2 2 2 4 ⇒ = + − × + − e e e e . eigenvalue an is 291 291 3 4 2 4 2 4 ⇒ = + × + e e e e B1 M1A1 M1A1 A1A1 M1A1 A1 3 4 3 10

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 8 Finds normal to Π1 . Finds Cartesian equation. Finds angle between normals, using scalar product. Finds direction of line of intersection, using vector product. Finds point common to both planes. States vector equation. k j i k j i − + = − − 3 2 1 1 1 0 1 Equation of Π1 :
 3    12 6 11 1 3 2 cos − − = θ ° = ⇒ = 7. 75 66 2 θ or 1.32 rad. k j i k j i 7 3 2 1 1 2 1 3 1 − − = − − Point on both planes is e.g. (6,2,0) ( ) k j i j i r 7 3 2 2 6 − − + + = t (OE) M1A1 A1 M1 A1 M1A1 M1A1 A1 3 2 5 [10]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 9 Obtains area of surface of revolution. Finds coordinates of centroid, using relevant formulae. 2 1 , 2 t y t x − = = & & ( ) 2 2 4 2 2 2 1 2 1 4 d d t t t t t s + = + − + =       ⇒ ( ) t t t t s y d 1 3 1 2 d 2 2 1 0 3 +      − = ∫ ∫ π π ∫       − + =       − + = 1 0 1 0 6 4 2 5 3 18 6 2 2 d 3 1 3 2 2 t t t t t t t π π π 9 11 = or 3.84 15 8 5 3 2 3 2 d 3 2 2 d d d 1 0 1 0 5 3 4 2 =       − =       − =∫ ∫ t t t t t t t x y 105 32 7 . 3 2 5 2 d 3 2 2 d d d 1 0 1 0 7 5 6 4 =       − =       − = ∫ ∫ t t t t t t t x xy 72 11 72 9 4 d 9 1 3 2 d d d 2 1 1 0 1 0 8 6 4 7 5 3 2 =       + − =       + − = ∫ ∫ t t t t t t t t t x y Centroid is       192 55 , 7 4 Or (0.571,0.286) M1A1 M1A1 M1 A1 B1 M1A1 M1A1 A1 6 6 [12]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 10 (i) (ii) (iii) Vertical asymptote. Oblique asymptote. Obtains value of p. Sketches curve. Proves required result. Sketches graph. Asymptotes : x = –1 ( )( ) p px y x p p px y − + = ⇒ + − + − + = − 4 1 3 4 1 ⇒ = 4 p x-axis is a tangent Correct location of turning points and asymptotes. Each branch. ( ) ( ) 2 1 1 2 1 1 2 3 1 − − + + = ′ ⇒ + − + = ⇒ = x y x x y p ([ 1) Intersections on x-axis at ( ) 0 , 3 2 ± − Each branch. B1 M1A1 B1 B1 B1B1 M1A1 B1 B1B1 3 4 5 [12] 11E Obtains all fifth roots. Simplifies expression. Obtains factors. Solves quadratic in x3. Expresses them in polar form. Finds factors. . 2 , 1 , 0 , 5 2 sin i 5 2 cos ± ± = + = k k k z π π 1 5 2 cos 2 2 + − x x π ( )1 1 5 4 cos 2 1 5 2 cos 2 2 2 −       + −       + − x x x π π 3 sin i 3 cos 2 3 i 2 1 3 π π ± = ± = x or 3 13 sin i 3 13 cos or 3 7 sin i 3 7 cos π π π π ± ± 9 13 sin i 9 13 cos , 9 7 sin i 9 7 cos , 9 sin i 9 cos π π π π π π ± ± ± = x       + −       + − 1 9 7 cos 2 1 9 cos 2 2 2 π π x x x       + − 1 9 13 cos 2 2 π x (ACF) B1B1 M1A1 M1A1 M1A1 A1 A1 M1A1 M1A1 2 2 2 6 2 [14]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – October/November 2013 9231 11 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 11O Uses substitution to obtain v–x equation. Finds CF. Finds PI. Uses initial conditions to find constants. Writes solution of y–x equation explicitly. 2 2 2 2 2 3 d d 3 d d 6 d d 3 x y y x y y v x y y v y v +       = ′′ ⇒ = ′ ⇒ = x v x v x v 2 2 2 e 25 3 d d 2 d d 3 1 − = + − 2x 2 2 e 75 9 d d 6 d d − = + − ⇒ v x v x v (AG) 3 0 9 6 2 = ⇒ = + − m m m x x Bx A v 3 3 e e + = x x x k v k v k v 2 2 2 e 4 e 2 e − − − = ′′ ⇒ − = ′ ⇒ = 3 75 9 12 4 = ⇒ = + + k k k k k x x x Bx A v 2 3 3 e 3 e e − + + = 5 3 8 8 , 2 , 0 = ⇒ + = ⇒ = = = A A v y x x x x x B Bx v 2 3 3 3 e 6 e e 3 e 15 − − + + = ′ 12 1 , 2 , 0 = ′ ⇒ = ′ = = v y y x 3 6 15 12 = ⇒ − + = B B x x x x v y 2 3 3 3 e 3 e 3 e 5 − + + = = { }3 1 2 3 3 e 3 e 3 e 5 x x x x y − + + = B1B1 M1 A1 M1 A1 M1 A1 A1 B1 M1A1 A1 A1 4 10 [14]