Cambridge A Level Mathematics 9709 — 2012 May/June Paper 1 · Variant 3

*6682689949* 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 2012 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 4 printed pages. JC12 06_9709_13/RP © UCLES 2012 [Turn over

2 1 (i) Prove the identity tan2θ −sin2θ ≡tan2θ sin2θ. [3] (ii) Use this result to explain why tan θ > sin θ for 0◦< θ < 90◦. [1] 2 Relative to an origin O, the position vectors of the points A, B and C are given by −−→ OA = 2 −1 4 !, −−→ OB = 4 2 −2 ! and −−→ OC = 1 3 p !. Find (i) the unit vector in the direction of −−→ AB, [3] (ii) the value of the constant p for which angle BOC = 90◦. [2] 3 The first three terms in the expansion of (1 −2x)2(1 + ax)6, in ascending powers of x, are 1 −x + bx2. Find the values of the constants a and b. [6] 4 (i) Solve the equation sin 2x + 3 cos 2x = 0 for 0◦≤x ≤360◦. [5] (ii) How many solutions has the equation sin 2x + 3 cos 2x = 0 for 0◦≤x ≤1080◦? [1] 5 O x y A 1 B (6, 1) x = – 2 8 y2 The diagram shows part of the curve x = 8 y2 −2, crossing the y-axis at the point A. The point B (6, 1) lies on the curve. The shaded region is bounded by the curve, the y-axis and the line y = 1. Find the exact volume obtained when this shaded region is rotated through 360◦about the y-axis. [6] 6 The first term of an arithmetic progression is 12 and the sum of the first 9 terms is 135. (i) Find the common difference of the progression. [2] The first term, the ninth term and the nth term of this arithmetic progression are the first term, the second term and the third term respectively of a geometric progression. (ii) Find the common ratio of the geometric progression and the value of n. [5] © UCLES 2012 9709/13/M/J/12

3 7 The curve y = 10 2x + 1 −2 intersects the x-axis at A. The tangent to the curve at A intersects the y-axis at C. (i) Show that the equation of AC is 5y + 4x = 8. [5] (ii) Find the distance AC. [2] 8 O B X A r In the diagram, AB is an arc of a circle with centre O and radius r. The line XB is a tangent to the circle at B and A is the mid-point of OX. (i) Show that angle AOB = 1 3π radians. [2] Express each of the following in terms of r, π and √3: (ii) the perimeter of the shaded region, [3] (iii) the area of the shaded region. [2] 9 A curve is such that d2y dx2 = −4x. The curve has a maximum point at (2, 12). (i) Find the equation of the curve. [6] A point P moves along the curve in such a way that the x-coordinate is increasing at 0.05 units per second. (ii) Find the rate at which the y-coordinate is changing when x = 3, stating whether the y-coordinate is increasing or decreasing. [2] 10 The equation of a line is 2y + x = k, where k is a constant, and the equation of a curve is xy = 6. (i) In the case where k = 8, the line intersects the curve at the points A and B. Find the equation of the perpendicular bisector of the line AB. [6] (ii) Find the set of values of k for which the line 2y + x = k intersects the curve xy = 6 at two distinct points. [3] © UCLES 2012 9709/13/M/J/12 [Turn over

4 11 The function f is such that f(x) = 8 −(x −2)2, for x ∈>. (i) Find the coordinates and the nature of the stationary point on the curve y = f(x). [3] The function g is such that g(x) = 8 −(x −2)2, for k ≤x ≤4, where k is a constant. (ii) State the smallest value of k for which g has an inverse. [1] For this value of k, (iii) find an expression for g−1(x), [3] (iv) sketch, on the same diagram, the graphs of y = g(x) and y = g−1(x). [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. 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. © UCLES 2012 9709/13/M/J/12

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2012 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 2012 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 2012 9709 13 © University of Cambridge International Examinations 2012 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 2012 9709 13 © University of Cambridge International Examinations 2012 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 2012 9709 13 © University of Cambridge International Examinations 2012 1 θ θ θ θ 2 2 2 2 sin tan sin tan = − (i) 2 2 2 s c s − → 2 2 2 2 c c s s − = 2 2 2 ) 1( c c s − → 2 2s t (ii) RHS > 0 → tan² θ > sin ²θ QED tan θ > sin θ if θ acute. M1 M1 A1 [3] B1 [1] Use of s ÷ c = t Use of s² + c² = 1 All ok Realises RHS > 0 2           − = 4 1 2 OA ,           − = 2 2 4 OB ,           = p OC 3 1 . (i)           − = 6 3 2 AB Modulus = ( ) 36 9 4 + + Unit Vector =           −6 3 2 7 1 (ii) p OC OB 2 6 4 . − + = = 0 → p = 5 B1 M1 A1 [3] M1A1 [2] co. Correct method for modulus co for his vector AB. Dot product = 0. co 3 6 2 ) 1( ) 2 1( ax x + − Coeff of x in 6) 1( ax + = 6ax Coeff of x² in 6) 1( ax + = 15a²x² B1 B1 6C1 needs removing (here or later) 6C2 needs removing (here or later) Multiplies by (1 − 4x + 4x²) 2 terms in x 6a − 4 = −1 → a = ½ M1 A1 Needs to consider 2 terms in equation Co 3 terms in x² 15a² − 24a + 4 = b → b = −4¼ M1 A1 [6] Needs to consider 3 terms in equation 4 0 2 cos 3 2 sin = + x x (i) → tan 2x = −3 2x = 180 – 71.6 or 360 – 71.6 x = 54.2º or 144.2º Also 234.2º and 324.2º (ii) 12 answers. M1 M1 A1A1 A1 [5] B1 [1] Uses tan2x = k and works with “2x”. Finds “2x” before ÷ 2 co. co (both of these need 2nd M) for 180º + his answer(s) for 3 times the number of solns to (i).

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 13 © University of Cambridge International Examinations 2012 5 2 8 2 − = y x ; at x = 0, y = 2 → 4 32 64 2 4 2 + − = y y x Integral of x² = y y y 4 1 32 3 64 1 3 + − − − − − Uses limits 1 to 2 → 6⅔π B1 B1B1B1 M1 A1 [6] co All co. Uses 1 to 2 or 2 to 1. co. 6 (i) Uses Sn 135 ) 8 24 ( 2 9 = + d → d = ¾ (ii) 9th term of AP = 12 + 8×¾ = 18 GP 1st tern 12, 2nd term 18 Common ratio = r = 18 ÷ 12 = 1½ 3rd term of GP = ar² = 27 nth term of AP is 12 + (n − 1)¾ 12 + (n − 1)¾ = 27 → n = 21 M1 A1 [2] B1 M1 M1 M1A1 [5] Uses correct formula co on “d” Uses “ar” Uses ar² or “ar” × r Links AP with GP. co 7 2 1 2 10 − + = x y (i) = x y d d 2)1 2 ( 10 + − x × 2 At A, y = 0, → x = 2 m at x = 2, is 5 4 − Eqn of tangent is ) 2 ( 5 4 − − = x y → 5y + 4x = 8 (ii) C (0, 1.6) d = ( ) 2 2 2 6.1 + = 2.56 B1 B1 B1 M1 A1 [5] M1 A1 [2] Without the “×2”. For the “×2”. For x = 2 Must be using differential as m co – answer given. Correct method – needs . co 8 (i) OBX = 90º, cos θ = r r 2 → θ = ⅓ π. M1 A1 [2] Needs 90º + cos (or Pyth + sin or tan) co ag (ii) Arc length AB = ⅓ rπ BX = rtan(⅓π) = r 3 P = r + (⅓ rπ + r 3 ) B1 B1 B1 [3] r + sum of other two (iii) Area = π 2 6 1 2 2 1 3 r r − B1 B1 [2] on tan(⅓π). co

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 13 © University of Cambridge International Examinations 2012 9 2 2 d d x y = x 4 − (i) x y d d = c x + − 2 2 x y d d = 0 when x = 2, → c = 8 x x y 8 3 2 3 + − = (+C) Subs (2, 12) → C = 3 4 (iii) t y d d = x y d d × t x d d = −10 × 0.05 → decreasing at 0.5 units per second B1 B1 B1 B1 M1 A1 [6] M1 A1 [2] For −2x² c = 8 For each term – on “c”– ignore (+C) Uses (2, 12) to find C. Must use. Enough to see product of gradient and rate. bod over notation. 10 k x y = + 2 6 = xy (i) 8 2 = + x y → 6 ) 2 8 ( = − y y 0 6 8 2 2 = + −y y or x² − 8x + 12 = 0 →(6, 1) and (2, 3) Midpoint M (4, 2) m = −½ Perpendicular m = 2 → ) 4 ( 2 2 − = − x y (ii) 6 ) 2 ( = − y y k → 0 6 2 2 = + −ky y or x² − kx + 12 = 0 Uses b² − 4ac (0) → k² > 48 → k < − 48 and k > 48 M1 DM1A1 M1 M1 A1 [6] M1 A1 A1 [3] Complete elimination of x (or y) DM1 soln of quadratic. co for their 2 points Uses m1m2 = −1 to find perp. gradient co unsimplified Any use of b² − 4ac on a quadratic = 0 For √48 on its own All correct.

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 13 © University of Cambridge International Examinations 2012 11 f(x) = 2) 2 ( 8 − −x , (i) Stationary point at x = 2 y – coordinate = 8 Nature Maximum (or y 4 4 2 + + − = x x y −2x + 4 = 0 → (2, 8) Max) B1 B1 B1 [3] co co co independent of first two marks (ii) k = 2 B1 [1] on “x-value” (iii) y = 2) 2 ( 8 − −x → 8 ) 2 ( 2 = + − y x → (x − 2) = ± y − 8 → g –1 = 2 + x − 8 M1 M1 A1 [3] Attempt to make x the subject Order of operations correct Must be f(x). (iv) B1 B1 B1 [3] B1 arc 1st quad (no tp, no axes) B1 Evidence of symmetry about y = x B1 all correct as shown left