Cambridge A Level Mathematics 9709 — 2015 May/June Paper 1 · Variant 2

*2057450266* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS 9709/12 Paper 1 Pure Mathematics 1 (P1) May/June 2015 1 hour 45 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulae (MF9) 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. 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 an electronic calculator is expected, where appropriate. 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. The total number of marks for this paper is 75. Questions carrying smaller numbers of marks are printed earlier in the paper, and questions carrying larger numbers of marks later in the paper. This document consists of 3 printed pages and 1 blank page. JC15 06_9709_12/FP © UCLES 2015 [Turn over

2 1 The function f is such that f ′x = 5 −2x2 and 3, 5 is a point on the curve y = fx. Find fx. [3] 2 21 rad O A B X Y r In the diagram, AYB is a semicircle with AB as diameter and OAXB is a sector of a circle with centre O and radius r. Angle AOB = 21 radians. Find an expression, in terms of r and 1, for the area of the shaded region. [4] 3 (i) Find the coefficients of x2 and x3 in the expansion of 2 −x6. [3] (ii) Find the coefficient of x3 in the expansion of 3x + 12 −x6. [2] 4 Variables u, x and y are such that u = 2xy −x and x + 3y = 12. Express u in terms of x and hence find the stationary value of u. [5] 5 (i) Prove the identity sin 1 −cos 1 sin 1 + cos 1  tan 1 −1 tan 1 + 1. [1] (ii) Hence solve the equation sin 1 −cos 1 sin 1 + cos 1 = tan 1 6 , for 0Å ≤1 ≤180Å. [4] 6 A tourist attraction in a city centre is a big vertical wheel on which passengers can ride. The wheel turns in such a way that the height, h m, of a passenger above the ground is given by the formula h = 601 −coskt. In this formula, k is a constant, t is the time in minutes that has elapsed since the passenger started the ride at ground level and kt is measured in radians. (i) Find the greatest height of the passenger above the ground. [1] One complete revolution of the wheel takes 30 minutes. (ii) Show that k = 1 150. [2] (iii) Find the time for which the passenger is above a height of 90 m. [3] © UCLES 2015 9709/12/M/J/15

3 7 The point C lies on the perpendicular bisector of the line joining the points A 4, 6 and B 10, 2. C also lies on the line parallel to AB through 3, 11. (i) Find the equation of the perpendicular bisector of AB. [4] (ii) Calculate the coordinates of C. [3] 8 (a) The first, second and last terms in an arithmetic progression are 56, 53 and −22 respectively. Find the sum of all the terms in the progression. [4] (b) The first, second and third terms of a geometric progression are 2k + 6, 2k and k + 2 respectively, where k is a positive constant. (i) Find the value of k. [3] (ii) Find the sum to infinity of the progression. [2] 9 Relative to an origin O, the position vectors of points A and B are given by −−→ OA = 2i + 4j + 4k and −−→ OB = 3i + j + 4k. (i) Use a vector method to find angle AOB. [4] The point C is such that −−→ AB = −−→ BC. (ii) Find the unit vector in the direction of −−→ OC. [4] (iii) Show that triangle OAC is isosceles. [1] 10 The equation of a curve is y = 4 2x −1. (i) Find, showing all necessary working, the volume obtained when the region bounded by the curve, the x-axis and the lines x = 1 and x = 2 is rotated through 360Å about the x-axis. [4] (ii) Given that the line 2y = x + c is a normal to the curve, find the possible values of the constant c. [6] 11 The function f is defined by f : x →2x2 −6x + 5 for x ∈>. (i) Find the set of values of p for which the equation fx = p has no real roots. [3] The function g is defined by g : x →2x2 −6x + 5 for 0 ≤x ≤4. (ii) Express gx in the form ax + b2 + c, where a, b and c are constants. [3] (iii) Find the range of g. [2] The function h is defined by h : x →2x2 −6x + 5 for k ≤x ≤4, where k is a constant. (iv) State the smallest value of k for which h has an inverse. [1] (v) For this value of k, find an expression for h−1x. [3] © UCLES 2015 9709/12/M/J/15

4 BLANK PAGE 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 2015 9709/12/M/J/15

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the May/June 2015 series 9709 MATHEMATICS 9709/12 Paper 1, maximum raw mark 75 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 2015 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 AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 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 AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 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 Cambridge International AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 1 2 2 5 ) ( f x x − = ′ and (3, 5) f(x) = 5x − 3 2 3 x (+c) Uses (3, 5) → c = 8 B1 M1 A1 [3] For integral Uses the point in an integral co 2 Radius of semicircle = 1 2 AB = rsinθ Area of semicircle = 1 2 πr²sin²θ = A1 Shaded area = semicircle – segment = A1 − 1 2 r²2θ + 1 2 r²sin2θ B1 B1 B1B1 [4] aef Uses 1 2 πr² with r = f(θ) B1 ( –sector ), B1 for + (triangle) 3 (i) 6) 2 ( x − Coeff of x² is 240 Coeff of x³ is − 20 × 8 = −160 B1 B2,1 [3] co B1 for +160 (ii) 6) 2 )( 1 3 ( x x − + Product needs exactly 2 terms → 720 – 160 = 560 M1 A1 [2] 3 × their 240 + their -160 for candidate’s answers. 4 ) ( 2 x y x u − = and 12 3 = + y x ,       − − = x x x u 3 12 2 2 8 3 8 x x = − 16 3 d 8 d x u x = − = 0 when x = 1 1 2 → ( y = 3 1 2 ) → u = 6 M1 A1 M1 A1 A1 [5] Expresses u in terms of x Differentiate candidate’s quadratic, sets to 0 + attempt to find x, or other valid method Complete method that leads to u Co 5 (i) θ θ θ θ cos sin cos sin + − . Divides top and bottom by cos θ → 1 1 t t − + B1 [1] Answer given. (ii) θ θ θ θ θ tan 6 1 cos sin cos sin = + − → 1 1 6 t t t −= + → 0 6 5 2 = + −t t → t = 2 or t = 3 → θ = 63.4º or 71.6º B1 M1 A1 A1 [4] Using the identity. Forms a 3 term quadratic with terms all on same side. co co

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 6 ) cos 1( 60 kt h − = (i) Max h when cos = −1 → 120 B1 [1] Co (ii) h = 0 and t = 30, or h =120 and t = 15 → cos30k = 1 or cos15k = –1 → 30k = 2π or 15k = π → k = 15 30 2 π π = M1 A1 [2] Substituting a correct pair of values into the equation. co ag (iii) 90 = 60(1 – cos kt) → coskt = 30 60 − = −0.5 → kt = 2 3 π or → kt = 4 3 π → Either t = 10 or 20 or both → t = 10 minutes B1 B1 B1 [3] co – but there must be evidence of correct subtraction. 7 A(4, 6), B (10, 2). (i) M = (7, 4) m of AB = −2 3 m of perpendicular = 2 3 → ) 7 ( 4 2 3 − = − x y B1 B1 M1 A1 [4] co co Use of m1m2 = −1 & their midpoint in the equation of a line. co (ii) Eqn of line parallel to AB through (3, 11) → 2 3 11 ( 3) y x − = − − Sim eqns → C (9, 7) M1 DM1A1 [3] Needs to use m of AB Must be using their correct lines. Co 8 (a) 1st, 2nd, nth are 56, 53 and −22 a = 56, d = −3 −22 = 56 + (n – 1)(−3) → n = 27 S27 = ( ) ( ) 3 26 112 2 27 − + → 459 M1 A1 M1 A1 [4] Uses correct un formula. co Needs positive integer n Co (b) 1st, 2nd, 3rd are 2k + 6, 2k and k + 2. (i) Either 2 2 2 6 2 k k k k + = + or uses a, r and eliminates → 0 12 10 2 2 = − − k k → k = 6 M1 DM1 A1 [3] Correct method for equation in k. Forms quad. or cubic equation with no brackets or fractions. Co

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 (ii) S∞ = 1 a r − with r = 2 2 6 k k+ or 2 2 k k + (= 2 3 ) → 54 M1 A1 [2] Needs attempt at a and r and S∞ Co 9 OA = 2i + 4j + 4k and OB =3i + j + 4k (i) OA .OB = 6 + 4 +16 = 26 36 = OA , 26 = OB Cos AOB = 26 6 26 → 31.8º M1 M1 M1 A1 [4] Must be numerical at some stage Product of 2 moduli All linked correctly co (ii) AB = b – a =           − 0 3 1 AB OC 2 4 4 2 +           = or AB +           4 1 3           − = 4 2 4 OC Unit vector ÷ modulus → 1 6 4 2 4     −       B1 M1 M1 A1 [4] Correct link ÷ by modulus. co (iii) OC = 6, OA = 6 B1 [1] co

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9709 12 © Cambridge International Examinations 2015 10 4 2 1 y x = −. B1 Correct without the ÷2 (i) 16 16 2 (2 1) 2 1 d x x x − − − = ∫ ÷ 2 Vol = π 8 2 1 x − −     with limits 1 and 2 → 16 3 π B1 M1 A1 [4] For the ÷2 even if first B1 is lost Use of limits in a changed expression. co (ii) m = 1 2 m of tangent = −2 d 4 2 d (2 1) 2 y x x − − = × Equating their d d y x to −2 → 2 3 = x or 2 1 − (y = 2 or – 2) → c = 5 2 or −7 2 M1 B1 B1 DM1 A1 A1 [6] Use of m1m2 = −1 Correct without the ×2 For the ×2 even if first B1 is lost co co 11 f : 5 6 2 2 + −x x x a (i) 0 5 6 2 2 = − + − p x x has no real roots Uses ) 5 ( 8 36 4 2 p ac b − − → − Sets to 0 → p < 1 2 M1 DM1 A1 [3] Sets to 0 with p on LHS. Uses discriminant. co – must be “<”, not “<”. (ii) ( ) 2 1 2 2 3 2 2 5 6 2 + − = + − x x x 3 × B1 [3] co (iii) Range of g 1 2 < g(x) < 13 h : 5 6 2 2 + −x x x a for k < x < 4 B1 B1 [2] on (ii) co from sub of x = 4 (iv) Smallest k = 2 3 B1 [1] on (ii) (v) h(x) = ( ) 2 1 2 2 3 2 + − x Order of operations ± 1 2 , ÷2, √, ± 2 3 → Inverse = ( ) 3 1 2 2 4 x + − M1 DM1 A1 [3] Using comp square form to try and get x as subject or y if transposed. Order must be correct co (without ±)