Cambridge A Level Mathematics - Further 9231 — 2013 May/June Paper 1 · Variant 2

*2496540878* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Level FURTHER MATHEMATICS 9231/12 Paper 1 May/June 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 06_9231_12/FP © UCLES 2013 [Turn over

2 1 Find the area of the region enclosed by the curve with polar equation r = 21 + cos 1, for 0 ≤1 < 20. [4] 2 Prove by mathematical induction that 52n −1 is divisible by 8 for every positive integer n. [5] 3 The cubic equation x3 −2x2 −3x + 4 = 0 has roots !, ", '. Given that c = ! + " + ', state the value of c. [1] Use the substitution y = c −x to find a cubic equation whose roots are ! + ", " + ', ' + !. [3] Find a cubic equation whose roots are 1 ! + " , 1 " + ' , 1 ' + !. [2] Hence evaluate 1 ! + "2 + 1 " + '2 + 1 ' + !2 . [2] 4 Let In = Ô 1 0 1 1 + x2n dx. Prove that, for every positive integer n, 2nIn+1 = 2−n + 2n −1In. 5 Given that I1 = 1 40, find the exact value of I3. [3] 5 Use the method of differences to show that N Ð r=1 1 2r + 12r + 3 = 1 6 − 1 22N + 3. [5] Deduce that 2N Ð r=N+1 1 2r + 12r + 3 < 1 8N . [4] 6 The matrix A is given by A = ` 4 −5 3 3 −4 3 1 −1 2 a . Show that e = ` 1 1 1 a is an eigenvector of A and state the corresponding eigenvalue. [2] Find the other two eigenvalues of A. [4] The matrix B is given by B = ` −1 4 0 −1 3 1 1 −1 3 a . Show that e is an eigenvector of B and deduce an eigenvector of the matrix AB, stating the corresponding eigenvalue. [3] © UCLES 2013 9231/12/M/J/13

3 7 By considering the binomial expansion of @ Ï −1 Ï A6 , where Ï = cos 1 + i sin 1, express sin61 in the form 1 32p + q cos 21 + r cos 41 + s cos 61, where p, q, r and s are integers to be determined. [6] Hence find the exact value of Ó 1 40 0 sin61 d1. [4] 8 The linear transformations T1 : >4 →>4 and T2 : >4 →>4 are represented by the matrices M1 and M2 respectively, where M1 = € 1 −2 3 5 3 −4 17 33 5 −9 20 36 4 −7 16 29  and M2 = € 1 −2 0 −3 2 −1 0 0 4 −7 1 −9 6 −10 0 −14  . The null spaces of T1 and T2 are denoted by K1 and K2 respectively. Find a basis for K1 and a basis for K2. [6] It is given that a = € 1 2 3 4  . The vectors x1 and x2 are such that M1x1 = M1a and M2x2 = M2a. Given that x1 −x2 = € p 5 7 q  , find p and q. [4] 9 Find x in terms of t given that 4d2x dt2 + 4dx dt + x = 6e−2t, and that, when t = 0, x = 5 3 and dx dt = 7 6. [9] State lim t→∞x. [1] [Questions 10 and 11 are printed on the next page.] © UCLES 2013 9231/12/M/J/13 [Turn over

4 10 The curve C has equation y = 2x2 −3x −2 x2 −2x + 1 . State the equations of the asymptotes of C. [2] Show that y ≤25 12 at all points of C. [4] Find the coordinates of any stationary points of C. [3] Sketch C, stating the coordinates of any intersections of C with the coordinate axes and the asymptotes. [4] 11 Answer only one of the following two alternatives. EITHER The curve C has equation y = 2 sec x, for 0 ≤x ≤1 40. Show that the arc length s of C is given by s = Ó 1 40 0 2 sec2x −1 dx. 4 Find the exact value of s. [2] The surface area generated when C is rotated through 20 radians about the x-axis is denoted by S. Show that (i) S = 40 Ó 1 40 0 2 sec3x −sec x dx, [3] (ii) d dxsec x tan x = 2 sec3x −sec x. [3] Hence find the exact value of S. [2] OR The points A, B, C and D have coordinates as follows: A 2, 1, −2, B 4, 1, −1, C 3, −2, −1 and D 3, 6, 2. The plane 1 passes through the points A, B and C. Find a cartesian equation of 1. [4] Find the area of triangle ABC and hence, or otherwise, find the volume of the tetrahedron ABCD. [6] [The volume of a tetrahedron is 1 3 × area of base × perpendicular height.] The plane 2 passes through the points A, B and D. Find the acute angle between 1 and 2. [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 2013 9231/12/M/J/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Level MARK SCHEME for the May/June 2013 series 9231 FURTHER MATHEMATICS 9231/12 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 May/June 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 – May/June 2013 9231 12 © 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 – May/June 2013 9231 12 © 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 – May/June 2013 9231 12 © Cambridge International Examinations 2013 Qu No Commentary Solution Marks Part Mark Total 1 Use of ∫ 2 2 1 r dθ A = ∫ + + π θ θ 2 0 2 ) cos cos 2 1( 4 2 1 dθ M1 Use of double angle formula and attempt to integrate. = ( ) ∫ + + π θ θ 2 0 2 cos cos 4 3 dθ M1 Integrates correctly. = π θ θ θ 2 0 2 2 sin sin 4 3     + + A1 Finds value. = π 6 (CWO) Accept 18.8 A1 4 [4] 2 Proves base case. Pn: 1 52 − n is divisible by 8. ⇒ × = = − 8 3 24 1 52 P1 is true B1 B1 States inductive hypothesis. Assume Pk is true: λ 8 1 52 = − k for some k. 1 5 5. 24 1 5. 25 1 5 2 2 2 2 2 − + = − = − + k k k k = λ 8 5.8 3 2 + × k M1 Proves inductive step. ∴Pk ⇒Pk+1 A1 States conclusion. (Since P1 is true and Pk ⇒Pk+1)∴ Pn is for every positive integer n (by PMI). A1 5 [5] 3 Uses ∑ − = a b α . 2 = c B1 1 Uses substitution ( γ β α − = + c etc.) ⇒ y c x x c y − = ⇒ − = 0 4 ) 2 ( 3 ) 2 ( 2 ) 2 ( 2 3 = + − − − − − y y y …(their c) M1 M1 to obtain required cubic equation. 0 2 4 2 3 = + + − ⇒ y y y A1 3 Obtains equation whose roots are reciprocals of those in previous cubic equation. Uses 1 − = y z to obtain 0 1 4 2 2 3 = + − + z z z M1A1 2 Uses ( ) ∑ ∑ ∑ − = αβ α α 2 2 2 ∑ = − −       = + 4 1 4 ) 2 ( 2 2 1 ) ( 1 2 2 β α M1A1 2 [8]

Page 5 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 4 Integrates by parts. ∫ + = 1 0 2 ) 1( 1 n n x I dx ∫ + − + +       + = 1 0 2 ) 1 ( 2 1 0 2 2. ) 1( ) 1( x x n x x n n dx M1A1 Rearranges. ∫ − + + + = + − − 1 0 2 ) 1 ( 2 )1 1( ) 1( 2 2 x x n n n dx M1A1 Obtains result. n n n I n nI )1 2 ( 2 2 1 − + = ∴ − + . (AG) A1 5 Uses reduction formula to find 2I . π π 8 1 4 1 4 1 2 1 2 2 2 + = ⇒ + = I I M1A1 Uses reduction formula to find 3I . π π 32 3 4 1 8 3 4 3 4 1 4 3 3 + = ⇒ + + = I I A1 3 [8] 5 Finds partial fractions. ( )( )   + − + = + + 3 2 1 1 2 1 2 1 3 2 1 2` 1 r r r r ∑ = + + N r r r 1 )3 2 )( 1 2 ( 1 M1A1 Expresses terms as differences.       + − + + +       − = 3 2 1 1 2 1 2 1 ... 5 1 3 1 2 1 N N M1A1 Shows cancellation. ) 3 2 ( 2 1 6 1 + − = N (AG) A1 5 Uses ∑ ∑ ∑ − = + N N N N 1 2 1 2 1 . ∑ +       + − −       + − = N N N N 2 1 ) 3 2 ( 2 1 6 1 )3 4 ( 2 1 6 1 M1 Applies result       + − + = 3 4 1 3 2 1 2 1 N N A1 and simplifies. ) 3 4 )( 3 2 ( + + = N N N M1 Deduces inequality. < N N N N 8 1 4. 2 = (AG) A1 4 [9]

Page 6 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 6 Shows e is an eigenvector of A and gives eigenvalue. Ae = 2e ⇒ e is an eigenvector with eigenvalue 2. M1A1 2 Finds characteristic equation. 0 2 2 2 3 = + − − λ λ λ M1A1 Factorises. 0 )1 )( 2 ( 2 = − − ⇒ λ λ A1 States other eigenvalues. Other eigenvalues are –1 and 1. A1 4 Repeats for B. Be = 3e ⇒ e is an eigenvector with eigenvalue 3 B1 States result for AB. ABe = A.3e = 3Ae = 3.2e = 6e AB has eigenvector e with eigenvalue 6 M1A1 3 [9] 7 Expands and groups. Use of 1 − −z z and 1 − + z z . Correctly. ( ) ( ) ( ) ( ) 20 15 6 2 2 4 4 6 6 6 1 − + + + − + = − − − − − z z z z z z z z M1A1 M1 Obtains result. ( ) 20 2 cos 30 4 cos 12 6 cos 2 sin i2 6 − + − = θ θ θ θ ( ) θ θ θ θ 6 cos 4 cos 6 2 cos 15 10 32 1 sin 6 − + − = (Allow p = 10 , q = –15 , r = 6 , s = –1) A1A1 A1 6 Integrates correctly. 4 0 192 6 sin 64 4 sin 3 64 2 sin 15 16 5 π θ θ θ θ     − + − M1A1 Inserts limits and evaluates. 48 11 64 5 192 1 64 15 64 5 − = + − π π (SC: 1f power of 2 consistently wrong ¾ for 2nd part.) M1A1 4 [10]

Page 7 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 8 Reduces matrices to echelon form.             − − → →             − − − − − − −             − → →             − − − − 0 0 0 0 1 1 0 0 2 0 1 0 3 0 2 1 14 0 10 6 9 1 7 4 0 0 1 2 3 0 2 1 0 0 0 0 2 1 0 0 9 4 1 0 5 3 2 1 29 16 7 4 36 20 9 5 33 17 4 3 5 3 2 1 ... ... M1 A1 A1 Finds basis for each null space.                         − ⇒    = + = + + = + + − 1 2 1 1 0 2 0 9 4 0 5 3 2 t z t z y t z y x                         − ⇒    = + = + = − − 1 1 2 1 0 0 2 0 3 2 t z t y t y x M1A1 A1 6 Writes x1 and x2 appropriately.             − +             = 1 2 1 1 4 3 2 1 1 λ x and             − +             = 1 1 2 1 4 3 2 1 2 µ x B1√ Finds difference ⇒             + − − − − = − µ λ µ λ µ λ µ λ 2 2 2 1 x x ⇒ 5 2 = −µ λ and 7 2 = −µ λ 3 = ⇒λ and 1 − = µ M1 A1 and solves. ) 4 7 5 4 ( x x 2 1 − = − T 4 = ⇒p and q = –4 A1 4 [10]

Page 8 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 9 Finds complementary function. 2 1 0 )1 2 ( 0 1 4 4 2 2 − = ⇒ = + ⇒ = + + m m m m C.F.: 2 2 e e t t Bt A x − − + = M1 A1 Finds particular integral. P.I.: t t t k x k x k x 2 2 2 e 4 e 2 e − − − = ⇒ − = ⇒ = && & 3 2 6 8 16 = ⇒ = + − k k k k t x 2 e 3 2 − = ⇒ M1 A1 Adds for general solution. G.S.: t t t Bt A x 2 2 2 e 3 2 e e − − − + + = A1 Uses initial conditions to find constants. 1 3 2 3 5 3 5 ) 0 ( = ⇒ + = ⇒ = A A x t t t t Bt B x 2 2 2 2 e 3 4 e 2 1 e e 2 1 − − − − − − + − = & 3 3 4 2 1 6 7 6 7 ) 0 ( = ⇒ − + − = ⇒ = B B x& B1 M1 A1 Gives particular solution. t t t t x 2 2 2 e 3 2 e 3 e − − − + + = A1 9 States limit. ∞ → t lim x = 0 B1 1 [10]

Page 9 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 10 States asymptotes. Vertical: 1 = x and Horizontal: 2 = y B1B1 2 Obtains quadratic form in x. 2 3 2 2 2 2 − − = + − x x y yx yx 0 ) 2 ( )3 2 ( ) 2 ( 2 = + + − − − ⇒ y x y x y M1A1 Uses 0 4 2 ≥ − AC B for real roots. For real x 0 ) 2 )( 2 ( 4 ) 3 2 ( 2 ≥ + − − − y y y 12 25 25 12 ≤ ⇒ ≤ ⇒ y y . M1 A1 4 Finds condition for 0 = ′y . ⇒ = ′ 0 y 0 ) 2 2 )( 2 3 2 ( )3 4 )( 1 2 ( 2 2 = − − − − − + − x x x x x x M1 Solves 0 )1 )( 7 ( 0 7 8 2 = − − ⇒ = + − ⇒ x x x x 7 = ⇒x , (since 1 = x is vertical asymptote). A1 Obtains stationary point. Stationary point is       12 25 , 7 A1 3 Sketch showing: Axes and asymptotes (-0.5,0) , (2,0) , (0,–2) and (4,2) Left hand branch. Right hand branch. B1 B1 B1 B1 4 [13]

Page 10 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 11 E Differentiation. x x y x y tan sec 2 sec 2 = ′ ⇒ = M1 Use of x x 2 2 tan 1 sec + = . ( ) )1 (sec sec 4 1 1 2 2 2 − + = ′ + x x y 1 sec 4 sec 4 2 4 + − = x x ( ) 2 2 1 sec 2 − = x M1 A1 Substitute in arc length formula. ∫ − = 4 0 2 d )1 sec 2 ( π x x s A1 4 Integrate. [ ]4 0 tan 2 π x x − = M1 Substitute limits.    − = π 4 1 2 A1 2 (i) Use surface area formula ∫ − = 4 0 2 d )1 sec 2 ( sec 2 2 π π x x x S M1A1 Obtain correct form. ( ) ∫ − = 4 0 3 d sec sec 2 4 π π x x x (AG) A1 3 (ii) Differentiates ( ) x x x x x x 3 2 sec tan sec tan sec d d + = ( ) x x x 3 2 sec 1 sec sec + − = M1A1 and obtains printed result. x x sec sec 2 3 − = (AG) A1 3 Uses result to deduce surface area. [ ] 4 0 tan sec 4 π π x x S = 2 4π = M1 2 [14] A1

Page 11 Mark Scheme Syllabus Paper GCE A LEVEL – May/June 2013 9231 12 © Cambridge International Examinations 2013 11 O Vector perpendicular to Π1         − − = − 6 1 3 1 3 1 1 0 2 k j i M1A1 Obtains cartesian equation 17 ) 6 ( ) 2 ( )1 ( 1 2 3 = − × − + − × + × 17 6 3 = − − ⇒ z y x M1 A1 4 Obtains area of triangle ABC. 46 2 1 6 1 3 2 1 2 2 2 = + + (=3.39) M1 A1 Obtains length of perpendicular from D to triangle ABC. 46 26 6 1 3 17 12 6 9 2 2 2 = + + − − − M1A1 Uses × 3 1 Base area× Height. Either 3 13 46 26 46 2 1 3 1 = × × M1A1 or triple scalar product method. Or e.g. 3 13 6 26 6 1 3 4 5 1 6 1 = =         − − ⋅         6 Obtains normal to ABD.         − = 10 7 5 1 0 2 4 5 1 k j i M1A1 Uses scalar product         − ⋅         − − = + + + + 10 7 5 6 1 3 cos 10 7 5 6 1 3 2 2 2 2 2 2 θ M1 to find angle between normals and hence angle between Π1 and Π2. = ⇒ = ⇒ θ θ 174 46 68 cos 40.5º A1 4 [14]