Cambridge A Level Mathematics - Further 9231 — 2011 Oct/Nov Paper 1 · Variant 3

*8082949944* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/13 Paper 1 October/November 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 11_9231_13/2R © UCLES 2011 [Turn over

2 1 Verify that 1 n2 − 1 (n + 1)2 = 2n + 1 n2(n + 1)2 . [1] Let SN = N ∑ r=1 2r + 1 r2(r + 1)2 . Express SN in terms of N. [2] Let S = lim N→∞SN. Find the least value of N such that S −SN < 10−16. [3] 2 Prove by mathematical induction that, for all positive integers n, dn dxn 1 2x + 3 = (−1)n n!2n (2x + 3)n+1 . [6] 3 The equation x3 + 5x2 −3x −15 = 0 has roots α, β, γ. Find the value of α2 + β2 + γ 2. [3] Hence show that the matrix  1 α β α 1 γ β γ 1 is singular. [4] 4 A curve has parametric equations x = 2 sin 2t, y = 3 cos 2t, for 0 < t < 1 2π. For the point on the curve where t = 1 3π, find the value of (i) dy dx, [3] (ii) d2y dx2 . [4] 5 Use de Moivre’s theorem to express cos4θ in the form a cos 4θ + bcos 2θ + c, where a, b, c are constants to be found. [4] Hence evaluate ã 1 4π 0 cos4θ dθ, leaving your answer in terms of π. [3] © UCLES 2011 9231/13/O/N/11

3 6 Find the general solution of the differential equation d2x dt2 + 4dx dt + 4x = sin 2t. [6] Describe the behaviour of x as t →∞, justifying your answer. [2] 7 Show that d dt t(1 + t3)n = (3n + 1)(1 + t3)n −3n(1 + t3)n−1. [3] Let In = ã 1 0 (1 + t3)n dt. Using the above result, or otherwise, show that (3n + 1)In = 2n + 3nIn−1. [2] Hence evaluate I3. [4] 8 The curve C has polar equation r = 1 + sin θ for −1 2π ≤θ ≤1 2π. Draw a sketch of C. [2] The area of the region enclosed by the initial line, the half-line θ = 1 2π, and the part of C for which θ is positive, is denoted by A1. The area of the region enclosed by the initial line, and the part of C for which θ is negative, is denoted by A2. Find the ratio A1 : A2, giving your answer correct to 1 decimal place. [8] 9 Find a cartesian equation of the plane Π containing the lines r = 3i + k + s(2i + j −k) and r = 3i −7j + 10k + t(i −3j + 4k). [4] The line l passes through the point P with position vector 6i −2j + k and is parallel to the vector 2i + j −4k. Find (i) the position vector of the point where l meets Π, [3] (ii) the perpendicular distance from P to Π, [3] (iii) the acute angle between l and Π. [3] 10 A curve C has equation y = 5(x2 −x −2) x2 + 5x + 10 . Find the coordinates of the points of intersection of C with the axes. [2] Show that, for all real values of x, −1 ≤y ≤15. [4] Sketch C, stating the coordinates of any turning points and the equation of the horizontal asymptote. [7] [Question 11 is printed on the next page.] © UCLES 2011 9231/13/O/N/11 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The curve C has equation y = 1 3x 1 2(3 −x), for 0 ≤x ≤3. Find the mean value of y with respect to x over the interval 0 ≤x ≤3. [4] Show that ds dx = 1 2x −1 2 + x 1 2, where s denotes arc length, and find the arc length of C. [6] Find the area of the surface generated when C is rotated through 2π radians about the x-axis. [4] OR Find the eigenvalues and corresponding eigenvectors of the matrix A, where A =  1 1 2 0 2 2 −1 1 3 . [7] The linear transformation T : >3 →>3 is defined by x →Ax. Let e, f be two linearly independent eigenvectors of A, with corresponding eigenvalues λ and µ respectively, and let Π be the plane, through the origin, containing e and f. By considering the parametric equation of Π, show that all points of Π are mapped by T onto points of Π. [3] Find cartesian equations of three planes, each with the property that all points of the plane are mapped by T onto points of the same plane. [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/O/N/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the October/November 2011 question paper for the guidance of teachers 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the October/November 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 – October/November 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 – October/November 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 – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 1 Verifies result. 2 2 2 2 2 2 2 2 )1 ( 1 2 )1 ( 1 2 )1 ( 1 1 + + = + − + + = + − n n n n n n n n n n (AG) B1 1 Uses difference method SN       + − + +       − +       − = 2 2 2 2 2 2 )1 ( 1 1 … 3 1 2 1 2 1 1 1 N N M1 to sum. 2)1 ( 1 1 + − = N A1 2 Considers difference between sum and sum to infinity. 16 2 16 10 )1 ( 1 10 − − < + ⇒ < − N S S N M1 8 10 )1 ( > + ⇒N A1 Solves inequality. 8 10 = ⇒ N least A1 3 [6] 2 States proposition. 1 ) 3 2 ( 2! )1 ( 3 2 1 : + + − =       + n n n n n n x n x x d d P Proves base case.       + 3 2 1 x x d d 2 ) 3 2 )( 1 ( 2 × + − = − x M1 2) 3 2 ( 2 !1 )1 ( + × − = x 1P ⇒ is true. A1 States inductive hypothesis. Assume Pk is true. i.e. 1 ) 3 2 ( 2! )1 ( 3 2 1 + + − =       + k k k k k x k x x d d B1 Shows 1 + ⇒ k k P P . 2 1 1 1 ) 3 2 ( 2! )1 ( 2 )1 ( 3 2 1 + + + + + + − =       + k k k k k x k k x x d d M1 2 1 1 ) 3 2 ( 2 )! 1 ( )1 ( + + + + + − = k k k x k A1 1 P P + ⇒ ∴ k k States conclusion. Since 1P is true and 1 P P + ⇒ k k , hence by the principle of mathematical induction n P is true . + ∈ ∀ Z n A1 6 [6]

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 3 Uses ∑ − = 5 α ∑ − = 3 αβ B1 ∑ ∑ ∑ − = αβ α α 2 ) ( 2 2 ∑ = − × − − = 31 )3 ( 2 ) 5 ( 2 2 α M1A1 3 Evaluates determinant. Det           1 1 1 γ β γ α β α = αβγ γ β α 2 ) ( 1 2 2 2 + + + − M1A1 15 ) 15 ( = − − = αβγ 15 2 31 1 × + − ⇒ M1 Shows it is zero. = 0 ⇒ matrix is singular. A1 4 [7] 4 Finds first derivative. t t t x y x d y d 2 tan 2 3 2 cos 4 2 sin 6 − = − = =

M1A1 Evaluates. When 3 π = t , 2 3 3 = x d y d A1 3 (ii) Finds second derivative. t t x d td x d dy td d x d y d 2 sec 4 1 2 sec 3 2 2 2 × − = ×       = M1A1 t2 sec 4 3 3 − = A1 Evaluates. When 3 π = t , 6 8 4 3 2 2 = × = x d y d A1 4 [7] Alternatively (i) Finds cartesian equation and differentiates implicitly. 2 3 3 3 3 2 4 9 4 9 1 3 2 2 2 = − × − = − = ′ ⇒ =       +       y x y y x M1A1 A1 3 (ii) Differentiates again. [ ] 6 3 1 2 3 2 1 0 ) ( 9 2 2 1 2 = ′′ ⇒ ′′ = + ⇒ = ′′ + ′ + y y y y y M1A2 A1 4 [7]

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 5 Binomial expansion and groups. 6 ) ( 4 ) ( ) ( 2 2 4 4 4 1 + + + + = + − − − z z z z z z , M1A1 where ) sin (cos θ θ i z + = . Uses de M’s Thm. 6 2 cos 8 4 cos 2 ) cos 2 ( 4 + + = θ θ θ M1 Simplifies 8 3 2 cos 2 1 4 cos 8 1 cos 4 + + = θ θ θ A1 4 Integrates result ∫ ∫       + + = 4 0 4 0 4 8 3 2 cos 2 1 4 cos 8 1 cos π π θ θ θ θ d d M1 correctly. 4 0 8 3 4 2 sin 32 4 sin π θ θ θ     + + = A1 Evaluates. 32 3 4 1 π + = A1 3 [7] 6 Forms auxiliary equation and factorises. ⇒ m = –2 m2 + 4m + 4 = 0 ⇒ (m + 2)2 = 0 M1 States CF. CF: -2t -2t e Bt e A + A1 States form of PI PI: t q t p x 2 cos 2 sin + = M1 and differentiates twice. t q t p x 2 sin 2 2 cos 2 − =
; t q t p x 2 cos 4 2 sin 4 − − =

Compares coefficients and solves. 0 1 8 = = − ⇒ 8p ; q 0 8 1 = − = ⇒ p ; q M1A1 States GS. GS: t e Bt e A x t t 2 cos 8 1 2 2 − + = − − A1 6 Reason. As t → ∞ e–2t and te–2t → 0 B1 Behaviour. Hence x oscillates. ( Accept t x 2 cos 8 1 − ≈ .) B1 2 [8]

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 7 Differentiates. { } n n n 3 t t n t t t(1 td d ) 1( ) 1( 3 ) 3 1 3 3 + + + = + − B1 Rearranges. n n t t t n ) 1( ) 1 )( 1 1( 3 3 1 3 3 + + + − + = − M1 1 3 3 ) 1( 3 ) 1 )( 1 3 ( − + − + + = n n t n t n (AG) A1 3 Integrates wrt t. [ ] 1 1 0 3 3 ) 1( )1 3 ( − + + = + n n nI t t I n M1 Obtains reduction formula. 1 3 2 )1 3 ( − + = + n n n nI I n (AG) A1 2 Evaluates I1 (or I0 ) directly. Uses reduction formula. ∫ + = 1 0 3 1 ) 1( t I dt [ ] 1 0 4 25 .0 t t + = = 1.25 B1M1 Obtains I2. 14 23 25 .1 6 4 7 2 2 = ⇒ × + = I I A1 Obtains I3. 140 319 14 23 9 8 10 3 3 = ⇒ × + = I I ( = 2.28) A1 4 [9] 8 Sketches graph. Arc above initial line. Arc below initial line. B1 B1 2 Uses ∫ θ d r 2 2 1 ∫ + 2) sin 1( 2 1 θ dθ = ) sin sin 2 1( 2 1 2 ∫ + + θ θ dθ M1 Uses double angle formula. = ∫       − + θ θ 2 cos 2 1 sin 2 2 3 2 1 dθ M1 Integrates. = c +     − − θ θ θ 2 sin 4 1 cos 2 2 3 2 1 M1A1 Inserts limits. A1 = 1 8 3 2 sin 4 1 cos 2 2 3 2 1 2 0 + =           − − π θ θ θ π M1A1 A2 = 1 8 3 2 sin 4 1 cos 2 2 3 2 1 0 2 − =           − − − π θ θ θ π A1 2. 12 1 8 3 1 8 3 =       − ÷       + = π π n (1d.p.) A1 8 [10]

Page 8 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 9 Finds normal to plane. k j i k j i 7 9 4 3 1 1 1 2 − − = − − M1A1 (i) Deduces equation. Π:: = − − z y x 7 9 constant Sub e.g. (1, –1, 2) ⇒ constant = – 4 Π:: 4 7 9 − = − − z y x M1A1 4 General point on line inserted in plane equation to find λ. l: λ λ λ 4 1 2 2 6 − = + − = + = z y x Sub in Π 4 28 7 9 18 2 6 − = + − − + + ⇒ λ λ λ M1 1 − = ⇒λ A1 Position vector of intersection is 4i – 3j + 5k. A1 3 (ii) Distance of point from plane formula or triple scalar product method. Either 49 81 1 4 7 18 6 + + + − + Or 49 81 1 ) 7 9 ).( 4 2 ( + + − − − + k j i k j i M1A1 131 21 = (=1.83) A1 3 (iii) Scalar product to find complement of angle. 21 ) 7 9 ).( 4 2 ( = − − − + k j i k j i M1 θ sin 49 81 1 16 1 4 + + + + = A1 0 6. 23 131 21 sin = ⇒ = ⇒ θ θ or 0.412 rad. A1 3 [13]

Page 9 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total 10 Intersections with axes. (–1,0) , (2,0) (0,–1) B1 B1 2 10 5 5 10 5 2 2 − − = + + x x y xy yx Rearranges as a quadratic equation. 0 )1 ( 10 ) 5 5 ( ) 5 ( 2 = + + + + − y x y x y Uses discriminant. For real x 0 4 2 ≥ −ac b ( ) 0 )1 ( 5 40 ) 5 5 ( 2 ≥ + − − + ⇒ y y y … M1A1 Solves inequality. 15 1 0 )1 )( 15 ( ≤ ≤ − ⇒ ≤ + − ⇒ y y y (AG) M1A1 4 Finds turning points. 0 1 = ⇒ − = x y 4 15 − = ⇒ = x y M1A1 A1 Turning points are (-4,15) and (0,–1) y = 5. B1 States asymptote. Axes and asymptote correct B1 Sketches graph. Graph correct. B1B1 7 [13] 11 EITHER Uses formula for mean value and integrates y wrt x, Mean value = 3 15 2 3 2 0 3 3 0 2 5 2 3 3 0 ÷         − = − ∫ x x x d y M1M1 A1 to obtain result. = 15 3 4 3 9 15 2 3 3 2 3 = ÷     × − × (= 0.462) A1 4 Uses 2) ( 1 y′ + x x y 2 1 2 1 − = ′       + − + = ⇒ x x x d s d 2 1 4 1 1 B1 and obtains result.       + =       + + = ⇒ x x x x x d s d 2 1 2 1 2 1 4 1 (AG) B1 Uses correct formula and integrates to find arc length. ∫         + =         + = − 3 0 3 0 2 3 2 1 2 1 3 2 2 2 1 2 1 x x x d x x s = 3 2 (= 3.46) M1A1 M1A1 6 Uses correct formula and integrates to obtain surface area. ∫         +         − = − 3 0 2 1 2 1 2 3 2 1 2 1 2 1 3 1 2 x d x x x x S π M1 ∫     − + =       − + = 3 0 3 0 3 2 2 9 1 3 1 3 1 3 2 1 x x x x d x x π π 3 = (OE) M1A1 M1 4 [14]

Page 10 Mark Scheme: Teachers’ version Syllabus Paper GCE A LEVEL – October/November 2011 9231 13 © University of Cambridge International Examinations 2011 Qu No Commentary Solution Marks Part Marks Total OR Forms characteristic equation. Det(A – λI) = 0 0 6 11 6 2 3 = − + − ⇒ λ λ λ M1A1 Solves. 3,2,1 = ⇒λ A1A1 Finds eigenvectors via equations or cross-products. e1 = –2j + k , e2 = i + j , e3 = 2i + 2j + k M1A1 A1 7 States equation of plane. r = se + tf A(se + tf) = sAe + tAf = (sλ)e + (tµ)f B1 M1A1 3 Cross-products of eigenvectors to obtain other plane equations. Either e1 × e2 = –i + j + 2k ⇒ x – y – 2z = 0 e1 × e3 = 2i – j – 2k ⇒ 2x – y – 2z = 0 e2 × e3 = i – j ⇒ x – y = 0 M1A1 A1 A1 4 [14]