Cambridge A Level Mathematics (US) 9280 — 2015 May/June Paper 1 · Variant 1

*2619244666* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS (US) 9280/11 Paper 1 Pure Mathematics 1 (P1) May/June 2015 1 hour 45 minutes Additional Materials: Answer Booklet/Paper Graph Paper List of Formulas (MF9) (US) READ THESE INSTRUCTIONS FIRST If you have been given an Answer Booklet, follow the instructions on the front cover of the Booklet. Write your Center number, candidate number, and name on 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 4 printed pages. JC15 06_9280_11/FP © UCLES 2015 [Turn over

2 1 Given that 1 is an obtuse angle measured in radians and that sin 1 = k, find, in terms of k, an expression for (i) cos 1, [1] (ii) tan 1, [2] (iii) sin1 + 0. [1] 2 x y O Q X −2, 0 P p, 0 y = 2x2 The diagram shows the curve y = 2x2 and the points X −2, 0 and P p, 0. The point Q lies on the curve and PQ is parallel to the y-axis. (i) Express the area, A, of triangle XPQ in terms of p. [2] The point P moves along the x-axis at a constant rate of 0.02 units per second and Q moves along the curve so that PQ remains parallel to the y-axis. (ii) Find the rate at which A is increasing when p = 2. [3] 3 (i) Find the first three terms, in ascending powers of x, in the expansion of (a) 1 −x6, [2] (b) 1 + 2x6. [2] (ii) Hence find the coefficient of x2 in the expansion of 1 −x1 + 2x6. [3] 4 Relative to the origin O, the position vectors of points A and B are given by −−→ OA = ` 3 0 −4 a and −−→ OB = ` 6 −3 2 a . (i) Find the cosine of angle AOB. [3] The position vector of C is given by −−→ OC = ` k −2k 2k −3 a . (ii) Given that AB and OC have the same length, find the possible values of k. [4] © UCLES 2015 9280/11/M/J/15

3 5 A piece of wire of length 24 cm is bent to form the perimeter of a sector of a circle of radius r cm. (i) Show that the area of the sector, A cm2, is given by A = 12r −r2. [3] (ii) Express A in the form a −r −b2, where a and b are constants. [2] (iii) Given that r can vary, state the greatest value of A and find the corresponding angle of the sector. [2] 6 The line with slope −2 passing through the point P 3t, 2t intersects the x-axis at A and the y-axis at B. (i) Find the area of triangle AOB in terms of t. [3] The line through P perpendicular to AB intersects the x-axis at C. (ii) Show that the mid-point of PC lies on the line y = x. [4] 7 (a) The third and fourth terms of a geometric progression are 1 3 and 2 9 respectively. Find the sum to infinity of the progression. [4] (b) A circle is divided into 5 sectors in such a way that the angles of the sectors are in arithmetic progression. Given that the angle of the largest sector is 4 times the angle of the smallest sector, find the angle of the largest sector. [4] 8 The function f : x →5 + 3 cos1 2x is defined for 0 ≤x ≤20. (i) Solve the equation fx = 7, giving your answer correct to 2 decimal places. [3] (ii) Sketch the graph of y = fx. [2] (iii) Explain why f has an inverse. [1] (iv) Obtain an expression for f −1x. [3] 9 The equation of a curve is y = x3 + px2, where p is a positive constant. (i) Show that the origin is a stationary point on the curve and find the coordinates of the other stationary point in terms of p. [4] (ii) Find the nature of each of the stationary points. [3] Another curve has equation y = x3 + px2 + px. (iii) Find the set of values of p for which this curve has no stationary points. [3] [Question 10 is printed on the next page.] © UCLES 2015 9280/11/M/J/15 [Turn over

4 10 x y O B x = 4 y = 8 3x + 4 P Q A 0, 4 The diagram shows part of the curve y = 8 3x + 4. The curve intersects the y-axis at A 0, 4. The normal to the curve at A intersects the line x = 4 at the point B. (i) Find the coordinates of B. [5] (ii) Show, with all necessary working, that the areas of the regions marked P and Q are equal. [6] 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 9280/11/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 9280 MATHEMATICS (US) 9280/11 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 9280 11 © 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 9280 11 © 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 9280 11 © Cambridge International Examinations 2015 1 θ is obtuse , sin θ = k (i) cos θ = −√(1 – k²) B1 [1] cao (ii) tan θ = θ θ cos sin used → tan θ = − ) 1( 2 k k − aef M1 A1 [2] Used, attempt at cosine seen in (i) Ft for their cosine as a function of k only, from part (i) (iii) sin (θ + π) = −k B1 [1] cao 2 y = 2x² , X(−2, 0) and P(p, 0) (i) A = 1 2 × (2 + p) × 2p² (= 2p² + p³) M1 A1 [2] Attempt at base and height in terms of p and use of 2 bh (ii) p A d d = 4p + 3p² t p p A t A d d d d d d × = = 0.02 × 20 = 0.4 or t p p t p p t A d d 3 d d 4 d d 2 + = B1 M1 A1 [3] cao any correct method, cao 3 6 2 ) 2 1( ) 1( x x + − . (i) (a) 6) 1( x − = 1 – 6x + 15x² B2,1 [2] −1 each error (b) 6) 2 1( x + = 1 + 12x + 60x² B2,1 [2] −1 each error SC B1 only, in each part, for all 3 correct descending powers SC only one penalty for omission of the ‘1’ in each expansion (ii) Product of (a) and (b) with >1 term → 60 – 72 + 15 = 3 M1 DM1A1 [3] Must be 2 or more products M1 exactly 3 products. cao, condone 3x2

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 11 © Cambridge International Examinations 2015 4 3 6 0 , 3 4 2 OA OB         = = −         −     uuur uuur ,           − − = 3 2 2 k k k OC (i) OA∙ OB = 18 – 8 = 10 Modulus of OA = 5, of OB = 7 Angle AOB = cos−1       35 10 aef → 35 10 or 7 2 M1 M1 A1 [3] Use of x1x2 + y1y2 + z1z2 All linked with modulus cao, (if angle given, no penalty), correct angle implies correct cosine (ii) = AB b – a =           − 6 3 3 k² + 4k² + (2k – 3)² = 9 + 9 + 36 → 9k² −12k −45( = 0) → k= 3 or k = −3 5 B1 M1 DM1 A1 [4] allow for a – b Correct use of moduli using their AB obtains 3 term quadratic. cao 5 (i) 24 = r + r + rθ → θ = r r 2 24 − A = 1 2 r²θ = 2 2 24 r r − = 12r – r². aef, ag M1 M1A1 [3] (May not use
) Attempt at s = rθ linked with 24 and r Uses A formula with θ as f(r). cao (ii) 2) 6 ( 36 ) ( − − = r A B1 B1 [2] cao (iii) Greatest value of A = 36 (r = 6) → θ = 2 B1 B1 [2] Ft on (ii). cao, may use calculus or the discriminant on 2 12 r r −

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 11 © Cambridge International Examinations 2015 6 (i) ) 8 2 )( 3 ( 2 2 t x y t x t y = + − − = − Set x to 0 → B(0, 8t) Set y to 0 → A(4t, 0) → Area = 16t² M1 M1 A1 [3] Unsimplified or equivalent forms Attempt at both A and B, then using cao (ii) m = 1 2 → 1 2 2 ( 3 )(2 ) y t x t y x t − = − = + Set y to 0 → C (−t, 0) Midpoint of CP is (t, t) This lies on the line y = x. B1 M1 A1 A1 [4] cao Unsimplified or equivalent forms co correctly shown. 7 (a) ar² = 3 1 , ar³ = 9 2 → r = 3 2 aef Substituting → a = 4 3 → S∞ = 3 1 4 3 = 2 1 4 aef M1 A1 M1 A1 [4] Any valid method, seen or implied. Could be answers only. Both a and r Correct formula with 1 < r , cao (b) d a a 4 4 + = → 3a = 4d 360 = S5 = ) 4 2 ( 2 5 d a + or 12.5a → a = 28.8º aef Largest = a + 4d or 4a = 115.2º aef B1 M1 A1 B1 [4] May be implied in ) 4 ( 2 / 5 360 a a + = Correct Sn formula or sum of 5 terms cao, may be implied (may use degrees or radians)

Page 7 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 11 © Cambridge International Examinations 2015 8 f : x⟼5 + 3cos 1 2 x       for 0 ø x ø 2π. (i) 5 + 3cos 1 2 x       = 7 cos 1 2 x       = 2 3 1 2 x = 0.84 x = 1.68 only, aef (in given range) B1 M1A1 [3] Makes cos 3 2 2 1 =       x Looks up cos−1 first, then ×2 (ii) B1 B1 [2] y always +ve, m always –ve. from (0, 8) to (2π, 2) (may be implied) (iii) No turning point on graph or 1:1 B1 [1] cao, independent of graph in (ii) (iv) y = 5 + 3cos 1 2 x       Order; −5, ÷3, cos−1, ×2 x = 2cos−1       − 3 5 x M1 M1 A1 [3] Tries to make x subject. Correct order of operations cao 2

Page 8 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 11 © Cambridge International Examinations 2015 9 2 3 px x y + = (i) x y d d = 3x² + 2px Sets to 0 → x = 0 or 3 2p − → (0, 0) or      − 27 4 , 3 2 3 p p B1 M1 A1 A1 [4] cao Sets differential to 0 cao cao, first A1 for any correct turning point or any correct pair of x values. 2nd A1 for 2 complete TPs (ii) 2 2 d d x y = 6x + 2p At (0, 0) → 2p +ve Minimum At      − 27 4 , 3 2 3 p p → −2p –ve Maximum M1 A1 A1 [3] Other methods include; clear demonstration of sign change of gradient, clear reference to the shape of the curve www (iii) px px x y + + = 2 3 → 3x² + 2px + p (= 0) Uses ac b 4 2 − → 4p² − 12p< 0 → 0 <p< 3 aef B1 M1 A1 [3] Any correct use of discriminant cao (condone ø)

Page 9 Mark Scheme Syllabus Paper Cambridge International AS/A Level – May/June 2015 9280 11 © Cambridge International Examinations 2015 10 4 3 8 + = x y (i) 2 3 ) 4 3 ( 4 + − = x dx dy × 3 aef → m(x=0) = − 2 3 Perpendicular m(x=0) = 3 2 Eqn of normal ) 0 ( 3 2 4 − = − x y Meets x = 4 at B       3 20 ,4 B1 B1 M1 M1 A1 [5] Without the “×3” For “×3” even if 1st B mark lost. Use of m1m2 = −1 after attempting to find x y d d (x=0) Unsimplified line equation cao (ii) 1 2 8 (3 4)) 8 d 3 (3 4) x x x + = ÷ + ∫ Limits from 0 to 4 → Area P = 3 32 Area Q = Trapezium – P Area of Trapezium = 3 64 4 3 20 4 2 1 = ×       + → Areas of P and Q are both 3 32 B1 B1 M1 A1 M1 A1 [6] Without “÷3”. For “÷3” Correct use of correct limits. cao Correct method for area of trapezium All correct.