Cambridge A Level Mathematics 9709 — 2011 May/June Paper 1 · Variant 3

*5299987921* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/13 Paper 1 Pure Mathematics 1 (P1) May/June 2011 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 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 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. JC11 06_9709_13/2R © UCLES 2011 [Turn over

2 1 The coefficient of x3 in the expansion of (a + x)5 + (1 −2x)6, where a is positive, is 90. Find the value of a. [5] 2 Find the set of values of m for which the line y = mx + 4 intersects the curve y = 3x2 −4x + 7 at two distinct points. [5] 3 The line x a + y b = 1, where a and b are positive constants, meets the x-axis at P and the y-axis at Q. Given that PQ = √(45) and that the gradient of the line PQ is −1 2, find the values of a and b. [5] 4 (a) Differentiate 2x3 + 5 x with respect to x. [3] (b) Find ã (3x −2)5 dx and hence find the value of ã 1 0 (3x −2)5 dx. [4] 5 O A B C E F G D P R i j k Q In the diagram, OABCDEFG is a rectangular block in which OA = OD = 6 cm and AB = 12 cm. The unit vectors i, j and k are parallel to −−→ OA, −−→ OC and −−−→ OD respectively. The point P is the mid-point of DG, Q is the centre of the square face CBFG and R lies on AB such that AR = 4 cm. (i) Express each of the vectors −−→ PQ and −−→ RQ in terms of i, j and k. [3] (ii) Use a scalar product to find angle RQP. [4] 6 (a) A geometric progression has a third term of 20 and a sum to infinity which is three times the first term. Find the first term. [4] (b) An arithmetic progression is such that the eighth term is three times the third term. Show that the sum of the first eight terms is four times the sum of the first four terms. [4] © UCLES 2011 9709/13/M/J/11

3 7 O B X A 6 cm 1 3p In the diagram, AB is an arc of a circle, centre O and radius 6 cm, and angle AOB = 1 3π radians. The line AX is a tangent to the circle at A, and OBX is a straight line. (i) Show that the exact length of AX is 6 √3 cm. [1] Find, in terms of π and √3, (ii) the area of the shaded region, [3] (iii) the perimeter of the shaded region. [4] 8 (i) Prove the identity  1 sin θ − 1 tan θ  2 ≡1 −cos θ 1 + cos θ . [3] (ii) Hence solve the equation  1 sin θ − 1 tan θ  2 = 2 5, for 0◦≤θ ≤360◦. [4] 9 A curve is such that dy dx = 2 √x −1 and P (9, 5) is a point on the curve. (i) Find the equation of the curve. [4] (ii) Find the coordinates of the stationary point on the curve. [3] (iii) Find an expression for d2y dx2 and determine the nature of the stationary point. [2] (iv) The normal to the curve at P makes an angle of tan−1k with the positive x-axis. Find the value of k. [2] 10 Functions f and g are defined by f : x →3x −4, x ∈>, g : x →2(x −1)3 + 8, x > 1. (i) Evaluate fg(2). [2] (ii) Sketch in a single diagram the graphs of y = f(x) and y = f −1(x), making clear the relationship between the graphs. [3] (iii) Obtain an expression for g′(x) and use your answer to explain why g has an inverse. [3] (iv) Express each of f −1(x) and g−1(x) in terms of x. [4] © UCLES 2011 9709/13/M/J/11

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. 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. 9709/13/M/J/11

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2011 question paper for the guidance of teachers 9709 MATHEMATICS 9709/13 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 must be read in conjunction with the question papers and the report on the examination. • Cambridge will not enter into discussions or correspondence in connection with these mark schemes. Cambridge is publishing the mark schemes for the May/June 2011 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 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: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 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: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 1 ( ) ( )6 5 2 1 x x a − + + Coeff of x3 in 1st = 10 × a2 Coeff of x3 in 2nd = 20 × (–2)3 → 10a2 – 160 = 90 → a = 5 B1 B1 + B1 M1 A1 [5] co co Forming an equation for a + solution co (condone ±) 2 y = mx + 4 y = 3x2 – 4x + 7 Equate → 3x2 – (4 + m)x + 3 = 0 Uses b2 – 4ac → (4 + m)2 – 36 Solution of quadratic m = 2 or –10 Set of values m > 2 or m < –10 M1 M1 DM1 A1 A1 [5] Eliminates y (or x) completely Any use of b2 – 4ac Method shown. Correct end-values co 3 1 = + b y a x P (a, 0) and Q(0, b) Distance → ( ) 45 2 2 = + b a Gradients → 2 1 − = − b a Solution of sim eqns → a = 6, b = 3 M1 A1 M1 A1 A1 [5] M1 even if sign(s) incorrect. Correct values a and b (both) 4 (a) x x x x y 5 2 5 2 2 3 + = + = d/dx = 2 5 4 x x − or 2 5 4 − −x x (b) ( ) ( ) 6 2 3 d 2 3 6 5 − = − ∫ x x x ÷ 3 (+c) ( ) ( )         − = − ∫ 18 2 3 d 2 3 6 1 0 5 x x x Limits used correctly → –3½ M1 A1 + A1 [3] B1 B1 M1 A1 [4] Knows to divide numerator by x co B1 without “÷ 3”. B1 for “÷ 3”. (ignore (+c)) Uses limits after integration. co 5 (i) PQ = 3i + 6j – 3k RQ = –3i + 8j + 3k (ii) PQ . RQ = –9 + 48 – 9 = 30 = √54 √82 cos RQP → RQP = 63.2° B2,1 B1 [3] M1 M1 M1 A1 [4] Allow B2,1 for either one, B1 for the other. Use of x1x2 + y1y2 + z1z2 Correct use of modulus All linked correctly co

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 6 (a) ar2 = 20 a r a 3 1 = − Soln of equations → (r = ⅔) a = 45 (b) a + 7d = 3(a +2d) → 2a = d S8 = 4(2a + 7d) = 32d or 64a S4 = 2(2a + 3d) = 8d or 16a B1 B1 M1 A1 [4] M1 A1 M1 A1 [4] co co Complete method to find a. co Use of a + (n – 1)d co correct use of Sn formula once. ag 7 (i) AX = 6tan 3 π = 6√3 (ii) Area of triangle = ½ × 6 × 6√3 Area of sector = ½ 62 × 3 π Area shaded = 18√3 – 6π (iii) Arc AB = 6 × 3 π = 2π OX = 6 ÷ cos 3 π = 12, BX = 6 Perimeter = 6√3 + 2π + 6 B1 [1] M1 M1 A1 [3] M1 B1 M1 A1 [4] ag Use of ½bh Use of ½r2θ co Use of rθ Use of trig to find (OX and then) BX. 8 (i) θ θ θ θ cos 1 cos 1 tan 1 sin 1 2 + − ≡       − ( ) θ θ θ θ θ 2 2 2 sin cos 1 sin cos sin 1 − =       − ( )( ) θ θ θ θ θ cos 1 cos 1 cos 1 cos 1 cos 1 2 + − = − − − = (ii) 5 2 tan 1 sin 1 2 =       − θ θ 5 2 cos 1 cos 1 = + − θ θ 7 3 cos = θ θ = 64.6° or 295.4° M1 M1 A1 [3] M1 A1 A1 A1 √ [4] Use of tan = sin/cos Use of sin2 + cos2 = 1. All correct. (NB ag. – ensure cancelling has been done) Uses part (i) to obtain an eqn in cosθ co co. √ for 360 – “1st answer”.

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 9 1 2 d d − = x x y P(9, 5) (i) ( ) c x x y + − = 4 Uses (9, 5) in an integrated expression → c = 2 (ii) 0 d d = x y → x = 4, y = 6 (iii) 2 3 2 2 d d − − = x x y → –ve → Max (iv) 3 1 d d − = x y Perpendicular m = 3 tanθ = 3 Angle is tan-13 k = 3 B1 B1 M1 A1 [4] M1 A1 A1 [3] B1 B1√ [2] M1 A1 [2] Ignore + c. Substitution of point after integration. co. Attempt to solve dy/dx = 0. x correct. y correct. co. √ for correct deduction. Use of m1m2 = –1 Needs k = 3

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2011 9709 13 © University of Cambridge International Examinations 2011 10 4 3 : f − x x a ( ) 8 1 2 : g 3 + − x x a (i) fg(2) = f(10) = 26 (ii) (iii) g′(x) = 6(x – 1)2 g′(x) > → no turning points → g is 1 : 1, g has an inverse. (iv) ( ) 3 4 f 1 + = − x x Attempt at making x Order correct. – 8, ÷ 2, 3 , + 1 ( ) 1 2 8 g 3 1 + − = − x x M1 A1 [2] B1 B1 B1 [3] B1 B1√ B1√ [3] B1 M1 M1 A1 [4] Must use g first, then f. co y = f(x) correct in 1st, 4th quadrants. y = f –1(x) correct in 1st, 2nd quadrants. y = x marked, or quoted. co allow only for incorrect “6” following from incorrect “6” co May change x and y first. Must all be correct, but allow for + 8, – 1 co as function of x, not y. f(x) f –1(x)