Cambridge A Level Mathematics (US) 9280 — 2014 Oct/Nov Paper 1 · Variant 1

*0524221283* Cambridge International Examinations Cambridge International Advanced Subsidiary and Advanced Level MATHEMATICS (US) 9280/11 Paper 1 Pure Mathematics 1 (P1) October/November 2014 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. JC14 11_9280_11/FP © UCLES 2014 [Turn over

2 1 In the expansion of 2 + ax7, the coefficient of x is equal to the coefficient of x2. Find the value of the non-zero constant a. [3] 2 Find the value of x satisfying the equation sin−1x −1 = tan−13. [3] 3 Solve the equation 13 sin21 2 + cos 1 + cos 1 = 2 for 0Å ≤1 ≤180Å. [4] 4 The line 4x + ky = 20 passes through the points A 8, −4 and B b, 2b, where k and b are constants. (i) Find the values of k and b. [4] (ii) Find the coordinates of the mid-point of AB. [1] 5 Find the set of values of k for which the line y = 2x −k meets the curve y = x2 + kx −2 at two distinct points. [5] 6 Relative to an origin O, the position vector of A is 3i + 2j −k and the position vector of B is 7i −3j + k. (i) Show that angle OAB is a right angle. [4] (ii) Find the area of triangle OAB. [3] 7 (i) A geometric progression has first term a (a ≠0), common ratio r and sum to infinity S. A second geometric progression has first term a, common ratio 2r and sum to infinity 3S. Find the value of r. [3] (ii) An arithmetic progression has first term 7. The nth term is 84 and the (3n)th term is 245. Find the value of n. [4] © UCLES 2014 9280/11/O/N/14

3 8 ! rad O A B C D 4 cm In the diagram, AB is an arc of a circle with center O and radius 4 cm. Angle AOB is ! radians. The point D on OB is such that AD is perpendicular to OB. The arc DC, with center O, meets OA at C. (i) Find an expression in terms of ! for the perimeter of the shaded region ABDC. [4] (ii) For the case where ! = 1 60, find the area of the shaded region ABDC, giving your answer in the form k0, where k is a constant to be determined. [4] 9 The function f is defined for x > 0 and is such that f ′x = 2x −2 x2 . The curve y = fx passes through the point P 2, 6. (i) Find the equation of the normal to the curve at P. [3] (ii) Find the equation of the curve. [4] (iii) Find the x-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum. [4] 10 (i) Express x2 −2x −15 in the form x + a2 + b. [2] The function f is defined for p ≤x ≤q, where p and q are positive constants, by f : x →x2 −2x −15. The range of f is given by c ≤fx ≤d, where c and d are constants. (ii) State the smallest possible value of c. [1] For the case where c = 9 and d = 65, (iii) find p and q, [4] (iv) find an expression for f −1x. [3] [Question 11 is printed on the next page.] © UCLES 2014 9280/11/O/N/14 [Turn over

4 11 x y O P 0, 1 Q 2, 3 ! y = 4x + 1 1 2 y = 1 2x2 + 1 The diagram shows parts of the curves y = 4x + 1 1 2 and y = 1 2x2 + 1 intersecting at points P 0, 1 and Q 2, 3. The angle between the tangents to the two curves at Q is !. (i) Find !, giving your answer in degrees correct to 3 significant figures. [6] (ii) Find by integration the area of the shaded region. [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. 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 2014 9280/11/O/N/14

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International Advanced Subsidiary and Advanced Level MARK SCHEME for the October/November 2014 series 9280 MATHEMATICS 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 October/November 2014 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 – October/November 2014 9280 11 © Cambridge International Examinations 2014 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 – October/November 2014 9280 11 © Cambridge International Examinations 2014 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 – October/November 2014 9280 11 © Cambridge International Examinations 2014 1 7C1 × 26 × a (=) 7C2 × 25 × a2 soi 3 2 2 21 2 7 5 6 =       × × = a oe B2, 1, 0 B1 [3] Treat the same error in each expression as a single error 2 tan–1(3) = 1.249 or 71.565º sin 1.25 or sin71.6 or 0.949 soi (x =) 1.95 cao, accept 1 + 10 3 oe M1 M1 A1 [3] Attempt at tan–13 or right angle triangle with attempt at hypotenuse = √10 Attempt at sin tan–13 Answer only B3 3 θ θ θ θ cos 2 4 cos cos 2 sin 13 2 2 + = + + 4 1 2 2 2 sin 4 sin 1 sin 13 = → = − + θ θ θ or 4 3 2 2 2 cos 4 cos cos 13 13 = → = + − θ θ θ 30º, 150º M1 M1 A1A1 [4] Attempt to multiply by θ cos 2 + Use of s2 + c2 appropriately SC both answers correct in radians, A1 only Ft on 180 – their first value of
4 (i) 3 20 4 32 = ⇒ = − k k 20 2 3 4 = × + b b 2 = b M1A1 M1 A1 [4] Sub (8, −4) Sub (b, 2b), [alt: ( ) ( ) k b b / 4 8 / 4 2 − = − + 20 2 4 = + bk b M1 both M1 solving A1, A1 ] (ii) Mid-point = (5, 0) B1 [1] Ft on their b 5 ) 0 )( 2 ( ) 2 ( 2 = − + − + k k x x ) 0 )( 2 ( 4 ) 2 ( 2 > − − − k k soi ) 0 )( 6 )( 2 ( > − − k k 2 < k or 6 > k (condone Y, [) Allow {–∞, 2}U{6, ∞} etc. M1 M1 DM1 A2 [5] Equate and move terms to one side of equ. Apply ac b 4 2 − (>0). Allow [ at this stage. Attempt to factorise or solve or find 2 solns. SCA1 for 2, 6 seen with wrong inequalities 6 (i) AB or BA = ±[(7i − 3j + k) – (3i + 2j – k)]= ±(4i − 5j + 2k) (AO.AB) = ±(12 – 10 – 2) [allow as column if total given] = 0 hence OAB = 90° M1A1 DM1 A1 [4] May be seen in part (ii) OR 59 , 14 , 45 2 2 2 = = = OB AO AB Hence 2 2 2 OB AO AB = + Hence OAB = 90º (ii) 14 1 4 9 = + + = OA , 45 4 25 16 = + + = AB ( ) 5. 12 45 14 Area 2 1 = = ∆ B1 M1A1 [3] At least one magnitude correct in (i) or (ii) Accept 12.6, 2 70 3         oe

Page 5 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9280 11 © Cambridge International Examinations 2014 7 (i) r a S − =1 , r a S 2 1 3 − = 1 – r = 3 – 6r 5 2 = r B1 M1 A1 [3] At least r a S 2 1 3 − = Eliminate S (ii) 84 )1 ( 7 = − + d n and/or 245 )1 3 ( 7 = − + d n [ 77 )1 ( = − d n , 238 )1 3 ( = − d n , 161 2 = nd ] 238 77 1 3 1 = − − n n (must be from the correct un formula) 23 = n (d = 22 77 = 3.5) B1 B1 M1 A1 [4] At least one of these equations seen Two different seen – unsimplified ok Or other attempt to elim d. E.g. sub n d 2 161 = (if n is eliminated d must be found) 8 (i) α 4 Arc = AB α α) cos 4 ( Arc = DC α cos 4 4 ) (or − = DB AC Perimeter α α α α cos 8 8 4 cos 4 − + + = B1 B1 B1 B1 [4] (ii)        = = 3 2 6 cos 4 π OD Shaded area = 2 1 2 6 4 π   × ×   ( ) 2 1 2 6 2 3 π   − ×     3 π B1 B1B1 B1 [4] Or 3 1 = k 9 (i) 7 2 2 7 2 1 normal of gradient 4 ) 2 ( f − = → = − = ′ ) 2 ( 6 7 2 − − = − x y AEF B1M1 A1 [3] Ft from their ( ) 2 f ′ (ii) ) ( 2 ) ( 2 c x x x f + + = 1 1 4 6 = ⇒ + + = c c B1B1 M1A1 [4] Sub (2, 6) – dependent on c being present (iii) 0 2 2 0 2 2 3 2 = − ⇒ = − x x x 1 = x ( ) 3 4 2 f x x + = ′′ or any valid method ( ) 6 1 f = ′′ OR > 0 hence minimum M1 A1 M1 A1 [4] Put ( ) 0 f = ′ x and attempt to solve Not necessary for last A mark as x > 0 given Dependent on everything correct

Page 6 Mark Scheme Syllabus Paper Cambridge International AS/A Level – October/November 2014 9280 11 © Cambridge International Examinations 2014 10 (i) 16 )1 ( 2 − − x B1B1 [2] (ii) –16 B1 [1] Ft from (i) (iii) 9 Y (x – 1)2 – 16 Y=65 OR 4 ,6 9 15 2 2 − → = − −x x 25 Y=(x – 1)2 Y=81 8 , 10 65 15 2 2 − → = − −x x 5 Y=x – 1 Y=9 p = 6 6 Y=x Y=10 q = 10 M1 M1 A1 A1 [4] OR x2 – 2x – 24 [ 0, x2 – 2x – 80 Y 0, (x – 6)(x + 4) [=0 (x – 10)(x + 8) Y 0 x [ 6 x Y 10 SC B2, B2 for trial/improvement (iv) 16 )1 ( 2 − − = y x [interchange x/y] 16 ) ( 1 + ± = − x y 16 1 ) ( f 1 + + = − x x M1 M1 A1 [3] OR 16 )1 ( 2 + = − y x 16 ) ( 1 + ± + = y x 16 1 ) ( f 1 + + = − x x 11 (i) For 2 1)1 4 ( + = x y , ] 4 [ )1 4 ( d d 2 1 2 1 × + =           − x x y When x = 2, gradient 3 2 1 = m For 1 2 2 1 + = x y , 2 gradient d d 2 = → = m x x y 1 1 2 1 tan tan m m − − − = α 7. 29 69 . 33 43 . 63 = − = α cao B1B1 B1 B1 M1 A1 [6] Ft from their derivative above (ii) ] 4 [ 3 / 2 ) 1 4 ( d ) 1 4 ( 2 3 2 1 ÷ ∫ + = +               x x x ( ) x x x x + = + ∫ 3 6 1 2 2 1 d 1 ∫ − = + 2 0 6 1 ]1 27 [ d )1 4 ( 2 1 x x , ( ) [ ] 2 d 1 6 8 2 0 2 2 1 + = + ∫ x x 3 10 3 13 − 1 B1B1 B1 M1 M1 A1 [6] Apply limits 0 → 2 to at least the 1st integral Subtract the integrals (at some stage)