Cambridge A Level Mathematics - Further 9231 — 2016 Oct/Nov Paper 1 · Variant 1

*1030774021* Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 2016 3 hours Additional Materials: List of Formulae (MF10) READ THESE INSTRUCTIONS FIRST An answer booklet is provided inside this question paper. You should follow the instructions on the front cover of the answer booklet. If you need additional answer paper ask the invigilator for a continuation booklet. 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 and 1 insert. JC16 11_9231_11/2R © UCLES 2016 [Turn over

2 1 Use the method of differences to find n Ð r=1 1 2r2 −1. [4] Deduce the value of ∞ Ð r=1 1 2r2 −1. [1] 2 Find the cubic equation with roots !, " and ' such that ! + " + ' = 3, !2 + "2 + '2 = 1, !3 + "3 + '3 = −30, giving your answer in the form x3 + px2 + qx + r = 0, where p, q and r are integers to be found. [6] 3 Find a matrix A whose eigenvalues are −1, 1, 2 and for which corresponding eigenvectors are ` 1 0 0 a , ` 1 1 0 a , ` 0 1 1 a , respectively. [7] 4 Using factorials, show that @ n r −1 A + @ n r A = @n + 1 r A . [2] Hence prove by mathematical induction that a + xn = @ n 0 A an + @n 1 A an−1x + à + @ n r A an−rxr + à + @n n A xn for every positive integer n. [4] 5 The linear transformation T : >4 →>4 is represented by the matrix A, where A = € 1 3 5 7 2 8 7 9 3 13 9 11 6 24 21 27  . Find (i) the rank of A, [3] (ii) a basis for the range space of T, [1] (iii) a basis for the null space of T. [4] © UCLES 2016 9231/11/O/N/16

3 6 Find the general solution of the differential equation d2x dt2 + 7dx dt + 10x = 116 sin 2t. [8] State an approximate solution for large positive values of t. [1] 7 The curve C has equation y = e−2x. Find, giving your answers correct to 3 significant figures, (i) the mean value of dy dx over the interval 0 ≤x ≤2, [2] (ii) the coordinates of the centroid of the region bounded by C, x = 0, x = 2 and y = 0. [9] 8 A curve C has equation x2 + 4xy −y2 + 20 = 0. Show that, at stationary points on C, x = −2y. [3] Find the coordinates of the stationary points on C, and determine their nature by considering the value of d2y dx2 at the stationary points. [8] 9 Evaluate Ó 1 20 0 x sinx dx. [2] Given that In = Ó 1 20 0 xn sin x dx, prove that, for n > 1, In = n1 20n−1 −nn −1In−2. [4] By first using the substitution x = cos−1u, find the value of Ó 1 0 cos−1u3 du, giving your answer in an exact form. [5] 10 Let z = cos 1 + i sin 1. Show that zn + 1 zn = 2 cos n1 and zn −1 zn = 2i sin n1. [2] By considering @ z −1 z A4 @ z + 1 z A2 , show that sin41 cos21 = 1 32cos 61 −2 cos 41 −cos 21 + 2. [7] Hence find the exact value of Ó 1 40 0 sin41 cos21 d1. [3] [Question 11 is printed on the next page.] © UCLES 2016 9231/11/O/N/16 [Turn over

4 11 Answer only one of the following two alternatives. EITHER The lines l1 and l2 have equations r = 6i −3j + s3i −4j −2k and r = 2i −j −4k + ti −3j −k respectively. The point P on l1 and the point Q on l2 are such that PQ is perpendicular to both l1 and l2. Show that the position vector of P is 3i + j + 2k and find the position vector of Q. [7] Find, in the form r = a + ,b + -c, an equation of the plane which passes through P and is perpendicular to l1. [3] The plane meets the plane r = pi + qj in the line l3. Find a vector equation of l3. [4] OR A curve C has parametric equations x = 1 −3t2, y = t1 −3t2, for 0 ≤t ≤1 ï3. Show that @dx dt A2 + @dy dt A2 = 1 + 9t22. [2] Hence find (i) the arc length of C, [2] (ii) the surface area generated when C is rotated through 20 radians about the x-axis. [3] Use the fact that t = y x to find a cartesian equation of C. Hence show that the polar equation of C is r = sec 11 −3 tan21, and state the domain of 1. [4] Find the area of the region enclosed between C and the initial line. [3] 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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 2016 9231/11/O/N/16

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 15 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International Advanced Level FURTHER MATHEMATICS 9231/11 Paper 1 October/November 2016 MARK SCHEME Maximum Mark: 100 Published 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 2016 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 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 Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 The following abbreviations may be used in a mark scheme or used on the scripts: AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid) 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 SOI Seen or implied 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 Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 1 ( )( ) 1 1 1 1 2 1 2 1 2 2 1 2 1 r r r r   = −   − + − +   ( ) ( ) 2 1 1 1 1 1 1 1 1 1 1 1 1 OE 2 3 3 5 2 1 2 1 2 2 1 2 1 n r n n n r =           = − + − +…+ − = −           − + +         −   ∑ ( ) 2 1 1 1 1 1 1 2 2 1 2 1 2 2 1 r n n n r ∞ =   − = ⇒ =   + +   − ∑ M1A1 M1A1 B1 [4] [1] 2 2 9 1 4 αβ αβ ∑ = −⇒∑ = Use of, e.g.: ( ) 3 2 3 α αβγ α α αβ ∑ − = ∑ ∑ −∑ or ( ) 3 3 3 3 α α α αβ αβγ ∑ = ∑ + ∑ ∑ − Correct substitution in formula 7 αβγ ⇒ = − Required cubic equation is 3 2 3 4 7 0 x x x − + + = must see final equation ALT METHOD: S3 – 3S2+4S1 + 3r = 0 M1 3r = 30 +3 × 1 – 4 × 3 A1 r=7 A1 M1A1 M1 A1 A1 A1 [6]

Page 5 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 3 1 1 0 0 1 1 0 0 1     =       P , 1 0 0 0 1 0 0 0 2 −     =       D soi 1 1 − − = ⇒ = P AP D A PDP soi 1 1 1 1 0 1 1 0 0 1 − −     = −       P 1 1 0 0 1 2 0 0 2 −     =       PD or 1 1 1 1 0 1 1 0 0 2 − − −     = −       DP 1 2 2 0 1 1 0 0 2 − −     =       A . ALT METHOD: Av1 = – v1, Av2 = v2, Av3 = 2v3 M1, A1 Multiply out, solve each equation M1, A1, A1, A1 1 2 2 0 1 1 0 0 2 − −     =       A B1√ B1B1 M1 M1A1 M1 A1 [7]

Page 6 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 4 1 n n r r    +    −    ( ) ( ) ( ) ( ) ( ) ! ! ! 1 1 1 ! 1 ! ! ! 1 ! ! 1 n n n r n r r n r r n r n r r   = + = +   − −+ − − − −+   ( ) ( ) ( ) ( ) ( ) 1 1 ! ! 1 1 ! ! 1 ! 1 ! n n n r n r r r n r r n r r n r   + +   + −+ = = =       − − −+ −+     ( ) 1 1 1 1 H is true. 0 1 a x a x a x   + = + = + ⇒     Assume Hk is true,i.e. ( ) 1 0 1 k k k k r r k k k k k a x a a x a x x r k − −     + = + +…+ +…+         Multiplying by ( ) a x + , the coefficient of 1 k r r a x −+ is: 1 1 k k k r r r +      + =      −      ⇒ Hk+1 is true. Hence Hn is true for all positive integers. M1 A1 B1 B1 M1 A1 [2] [4]

Page 7 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 5 (i) 1 3 5 7 2 8 7 9 3 13 9 11 6 24 21 27             ⇒…⇒ 1 3 5 7 0 2 3 5 0 0 0 0 0 0 0 0     − −         . r(A) = 4 – 2 = 2 M1A1 A1 [3] (ii) Basis for range space is 1 3 2 8 , 3 13 6 24                                 oe B1 [1] (iii) 3 5 7 0 x y z t + + + = 2 3 5 0 y z t − − = Solving 29 19 5 3 A basis for null space is , 0 2 2 0   − −              ⇒                  oe B1 M1 A1A1 [4]

Page 8 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 6 ( )( ) 2 7 10 0 2 5 0 2 or 5 m m m m m + + = ⇒ + + = ⇒ = − − CF: 2 5 e e t t A B − − + PI: ¨ sin 2 cos2 2 cos2 2 sin 2 4 sin 2 4 cos2 x p t q t x p t q t x p t q t = + ⇒ = − ⇒ = − −  Substituting: 14 6 0 p q + = 6 14 116 p q − = Solving: p = 3 , q = –7 ⇒ 3sin 2 7cos2 x t t = − GS: 2 5 e e 3sin 2 7cos2 t t x A B t t − − = + + − For large positive values of t x 3sin 2 7cos2 t t ≈ − . M1 A1 M1A1 M1 M1A1 A1 B1 [8] [1]

Page 9 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 7 (i) M.V. = ( ) 2 2 4 0 e 1 0.491 2 0 2 x e − −   −  = = − − M1A1 [2] (ii) Coordinates are 2 2 0 2 2 0 e d e d x x x x x x − − = ∫ ∫ and 2 4 0 2 2 0 e d 1 2 e d x x x y x − − = ∫ ∫ ( ) 2 2 2 2 4 0 0 1 1 e d e 1 e 2 2 x x x − − −   = − = −     ∫ (= 0.4908…) (N.B. As in (i)) ( ) 2 2 2 2 2 2 2 2 4 2 0 0 0 0 xe e e e 5e 1 e d d 0.2271.. 2 2 2 4 4 4 x x x x x x x x x − − − − − −     = − + = − − = − + =         ∫ ∫ 2 2 4 8 4 0 0 1 1 e 1 1 e e d ( 0.12495 ) 2 2 4 2 4 4 x x x − − −     = − = − = …         ∫ (not straight from calculator) 0.463 x = , 0.255 y = (Final answers in algebraic form SRA1) M1M1 B1 M1A1 M1A1 A1A1 [9]

Page 10 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 8 ( ) 2 4 ' 2 ' 0 x xy y yy + + − = ( ) * ' 0 2 4 0 2 y x y x y = ⇒ + = ⇒ = − (AG) At stationary points 2 2 2 2 4 8 20 0 5 20 2 y y y y y − − + = ⇒ = ⇒ = ± Coordinates of stationary points are (4 , –2) and (–4,2) From ( ) * : ( ) 2 ' ' 0 x xy y yy + + − = Differentiating: ( ) [ ] ( ) 2 1 2 ' 0 xy y y yy y + + + ′′ ′ ′′ ′ − + = (or by quotient rule) At (4 , –2) with y′ = 0 : 1 1 8 2 0 maximum. 10 y y y ′′ ′′ ′′ + + = ⇒ = − ⇒ At (–4 , 2) with y′ = 0 : 1 1 8 2 0 minimum. 10 y y y ′′ ′′ ′′ − − = ⇒ = ⇒ M1A1 A1 M1A1 A1 M1A1 M1A1 A1 [3] [8]

Page 11 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 9 [ ] [ ] 2 2 2 2 0 0 0 0 sin d cos cos d 0 sin 1 x x x x x x x x π π π π = − + = + = ∫ ∫ 2 1 2 0 0 cos cos d n n nI x x nx x x π π −   = − +   ∫ ( ) 2 1 2 2 0 0 cos sin 1 sin d n n n x x nx x n n x x x π π − −   = − + − −   ∫ ( ) 1 2 1 2 n n n n n I π − −   = − −     1 d cos cos sin d u x u u x x x − = ⇒ = ⇒ = − ; 0 ,1 / 2 , 0 u x π = ⇒ = ( ) 1 0 /2 3 1 3 3 3 0 /2 0 cos d sin d sin d u u x x x x x x I π π −   = − = =   ∫ ∫ ∫ 2 2 3 1 3 3 3 2 I 6 2 4 I π π   = −× × = −     (OE) M1A1 M1A1 A1 A1 B1B1 B1 M1A1 [2] [4] [5]

Page 12 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 10 ( ) cos isin and cos isin( ) cos isin n n z n n z n n n n θ θ θ θ θ θ − = + = − + − = − ⇒ 1 1 2cos and 2isin . n n n n z n z n z z θ θ + = − = (AG) 4 2 2 2 2 2 1 1 1 1 z z z z z z z z       − + = − −             (Or 1st bracket expanded) 2 4 2 4 1 1 2 2 z z z z    = − + − +       (and 2nd bracket expanded) 6 4 2 6 4 2 1 1 1 2 4 z z z z z z       = + − + − + +             (Expanded and grouped) ⇒ 4 2 64sin cos 2cos6 4cos4 2cos2 4 θ θ θ θ θ = − − + ⇒ ( ) 4 2 1 sin cos cos6 2cos4 cos2 2 32 θ θ θ θ θ = − − + (AG) (Uses initial result) 1 1 4 4 4 2 0 0 1 1 1 1 sin cos d sin 6 sin 4 sin 2 2 32 6 2 2 π π θ θ θ θ θ θ θ   = − − +     ∫ 1 1 1 3 4 0 32 6 2 2 192 π π −     = − − − + =         or 0.0283 (3sf) ALT METHOD: Expand and group each bracket separately and sub trig: M1, A1, A1 Multiply together two brackets and manipulate trig expressions M1,M1 LHS = RHS M1 A1 M1A1 M1 M1 M1A1 M1A1 A1 M1A1 A1 [2] [7] [3]

Page 13 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 11 E 2 1 3 1 1 3 4 2 5     − − = −     − −   i j k Obtains three equations. E.g. from OP PQ OQ + = JJJG JJJG JJJG 3 2 4 t s k − + = 3 4 2 t s k − + − = − 2 5 4 t s k −+ + = Solves: s = –1 , t = –1 , k = 1 p = 3i + j + 2k and 2 3 = + − q i j k (Both required) ALT METHOD: Find PQ in terms of s,t M1 Calculate scalar products direction vectors for l1 and l2 M1,A1,A1 Solve simultaneous equations M1 A1 p and q A1 22 3 4 2 19 2 1 5 5 −     − − = −     −   i j k 3 22 2 1 19 1 2 5 5 λ µ −           = + − + −                r M1A1 M1A1 M1A1 A1 B1 B1B1 [7] [3]

Page 14 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 Direction of l3 is 4 4 3 4 2 3 ~ 3 0 0 1 0 0 −       − − = −          i j k Normal to Π is 90 3 22 19 5 120 4 2 1 5 60 2         − = − = −         − − −     i j k Cartesian equation of Π is 3x – 4y – 2z = 1 (Since P lies in Π ) Vector equation of l3 is 1 4 1 3 0 0 λ −       = − +          r (OE) (Since z = 0 .) B1 M1 A1 B1 [4]

Page 15 Mark Scheme Syllabus Paper Cambridge International A Level – October/November 2016 9231 11 © UCLES 2016 11 O ( ) ( ) 2 2 2 2 2 4 2 4 2 36 1 18 81 1 18 81 1 9 AG x y t t t t t t + = + − + = + + = +   M1A1 [2] (i) ( ) 1 3 1 2 3 3 0 0 2 1 9 d 3 3 s t t t t   = + = + =   ∫ M1A1 [2] (ii) ( )( ) ( ) 1 1 3 3 2 2 3 5 0 0 2 1 3 1 9 d 2 6 27 d S t t t t t t t t π π = − + = + − ∫ ∫ 1 3 2 4 6 0 1 3 9 2 2 2 2 3 t t t π π   = + − =     Substitute y t x = in one of the parametric equations to obtain e.g. 2 2 1 3 y x x = − . Uses cos x r θ = and sin y r θ = ⇒ ( ) 2 2 cos 1 3tan sec 1 3tan r r θ θ θ θ = − ⇒ = − . States domain for θ : 0 6 π θ - - , ( ) 6 2 2 4 0 1 sec 1 6tan 9tan d 2 A π θ θ θ θ = − + ∫ 6 3 5 0 1 9 tan 2tan tan 2 5 π θ θ θ   = − +     4 3 45 = *M1 DM1A1 M1 M1A1 B1 M1 A1A1 [3] [4] [3]