Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 1 · Variant 1

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/11 Paper 1 Pure Mathematics 1 (P1) October/November 2009 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. © UCLES 2009 [Turn over *8401882611*

2 1 Solve the equation 3 tan(2x + 15◦) = 4 for 0◦≤x ≤180◦. [4] 2 The equation of a curve is y = 3 cos 2x. The equation of a line is x + 2y = π. On the same diagram, sketch the curve and the line for 0 ≤x ≤π. [4] 3 (i) Find the first 3 terms in the expansion of (2 −x)6 in ascending powers of x. [3] (ii) Given that the coefficient of x2 in the expansion of (1 + 2x + ax2)(2 −x)6 is 48, find the value of the constant a. [3] 4 The equation of a curve is y = x4 + 4x + 9. (i) Find the coordinates of the stationary point on the curve and determine its nature. [4] (ii) Find the area of the region enclosed by the curve, the x-axis and the lines x = 0 and x = 1. [3] 5 A B C D O 6 cm The diagram shows a semicircle ABC with centre O and radius 6 cm. The point B is such that angle BOA is 90◦and BD is an arc of a circle with centre A. Find (i) the length of the arc BD, [4] (ii) the area of the shaded region. [3] 6 A curve is such that dy dx = k −2x, where k is a constant. (i) Given that the tangents to the curve at the points where x = 2 and x = 3 are perpendicular, find the value of k. [4] (ii) Given also that the curve passes through the point (4, 9), find the equation of the curve. [3] © UCLES 2009 9709/11/O/N/09

3 7 The equation of a curve is y = 12 x2 + 3. (i) Obtain an expression for dy dx. [2] (ii) Find the equation of the normal to the curve at the point P (1, 3). [3] (iii) A point is moving along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.012 units per second. Find the rate of change of the y-coordinate as the point passes through P. [2] 8 The first term of an arithmetic progression is 8 and the common difference is d, where d ≠0. The first term, the fifth term and the eighth term of this arithmetic progression are the first term, the second term and the third term, respectively, of a geometric progression whose common ratio is r. (i) Write down two equations connecting d and r. Hence show that r = 3 4 and find the value of d. [6] (ii) Find the sum to infinity of the geometric progression. [2] (iii) Find the sum of the first 8 terms of the arithmetic progression. [2] 9 Relative to an origin O, the position vectors of the points A, B and C are given by −−→ OA = 2 3 −6 ! , −−→ OB = 0 −6 8 ! and −−→ OC = −2 5 −2 ! . (i) Find angle AOB. [4] (ii) Find the vector which is in the same direction as −−→ AC and has magnitude 30. [3] (iii) Find the value of the constant p for which −−→ OA + p −−→ OB is perpendicular to −−→ OC. [3] 10 Functions f and g are defined by f : x →2x + 1, x ∈>, x > 0, g : x →2x −1 x + 3 , x ∈>, x ≠−3. (i) Solve the equation gf(x) = x. [3] (ii) Express f−1(x) and g−1(x) in terms of x. [4] (iii) Show that the equation g−1(x) = x has no solutions. [3] (iv) Sketch in a single diagram the graphs of y = f(x) and y = f−1(x), making clear the relationship between the graphs. [3] © UCLES 2009 9709/11/O/N/09

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. 9709/11/O/N/09

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2009 question paper for the guidance of teachers 9709 MATHEMATICS 9709/11 Paper 11, 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. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2009 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 A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 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 A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 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 A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 1 3 tan(2x + 15°) = 4 tan(2x + 15°) = 1⅓ Sets the bracket to tan–1(1⅓) 2x + 15 = 53.13º or 233.13º → x = 19.1º or 109.1º M1 M1 A1 A1√ [4] Removes the “3” first by division. Looks up tan–11⅓, then uses bracket co. √ for (90 + 1st answer) and no other answers in the range. 2 B1 B1 B1 B1 [4] 1 complete oscillation 0 → π Range from −3 to 3 All correct (V shape B0) Line correct. 3 (i) (2 – x)6 64 − 192x + 240x² (ii) (1 + 2x + ax2)(2 – x)6 Coeff of x² = 240 − 384 + 64a Equates to 48 → a = 3 3 × B1 [3] M1 M1 A1 [3] co Allow 26. Considers at least 2 terms in x². Considers exactly 3 terms + solution co 4 y = x4 + 4x + 9 (i) Differential = 4x3 + 4 Sets to 0 + solution → (−1, 6) 2nd differential = 12x² Positive, → Minimum (ii) A =       + + x x x 9 2 5 2 5 Limits from 0 to 1 → 11.2 B1 M1 A1 B1 [4] B1 M1 A1 [3] co Differentiates and sets to 0. co. Statement only. co. Value at “1” − value at “0” in integral of y. 5 r = 6 cm (i) AB = √(62 + 62) = √72 Angle BAD = ¼π or 45º Arc length = √72 × ¼ π = 6.66(7) (ii) Sector area = ½r2θ = ½ × 72 × ¼π Area of triangle = ½ × 6 × 6 Shaded area = 10.3 or 9π − 18. B1 B1 M1 A1 [4] M1 B1 A1 [3] Use of Pythagoras – or trig (8.5 ok) In degrees or radians Use of s=rθ with θ in rads only – or correct with degrees. Use of r = 6 M0. Use of ½r2θ with θ in rad, and r ≠ 6. co co

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 6 x y d d = k – 2x (i) At x = 2, m = (k − 4) x = 3 m = (k − 6) (k − 4)(k − 6) = −1 → k = 5 (ii) y = kx – x2 (+ c) Substitutes (4, 9) → c = 5 M1 M1 DM1A1 [4] B1√ M1 A1 [3] Obtains either gradient as f(k). Uses m1m2 – −1 with gradients f(k) Soln of quadratic = 0. co (watch for fortuitous answers) For integration without c Realises need to substitute for x and y co (nb If k = 5 is fortuitous, loses last A1) 7 3 12 2 + = x y (i) x y d d = –12(x2 + 3)–2 × 2x (ii) At x = 1, m = – 2 3 m of normal = ⅔ Eqn of normal y – 3 = 3 2 (x – 1) (iii) 012 .0 d d d d d d 2 3 × − = × = t x x y t y → −0.018 B1 B1 [2] M1 M1 A1 [3] M1 A1√ [2] Without the “×2x”. For “×2x”. Accept unsimplified answer Uses m1m2 – −1 …algebraic ok. Correct form of equation. co unsimplified Correct link between differentials co to his x y d d . (Omission of x in part (i) causes fortuitous results in (ii) and (iii).) 8 (i) 8 + 4d = 8r 8 + 7d = 8r2 Eliminates one of the variables → 4r2 – 7r + 3 = 0 Solution → r = ¾ → d = −½ (ii) S∞ = r a − 1 → 32 (iii) S8 = 4(16 + 7d) = 50 B1 B1 M1 DM1 A1 A1 [6] M1 A1 [2] M1 A1 [2] co – but allow if a in place of 8. co – but allow if a in place of 8. Complete elimination of either r or d. Correct method of solution. nb answer for r given. co (assumes r = ¾, give B1B1 for equations, B1 for d) Correct formula used. Correct formula used. 64 + 28d ok co

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 9 . 2 5 2 , 8 6 0 , 6 3 2         − − =         − =         − = OC OB OA (i) Scalar product = −18 − 48 −66 = abcos θ a= 7 and b= 10 → Angle AOB = 160.5º (ii)          − = − = 4 2 4 a c AC Modulus = 6 Vector = 5 ×          − 4 2 4 or          − 20 10 20 (iii) 0 2 5 2 8 6 6 3 2 =         − −         + − − . p p → p = 2 1 M1 M1 M1 A1 [4] B1 M1 A1 [3] B1 M1 A1 [3] Use of x1x2 + y1y2 + z1z2 Linking everything correctly Correct modulus of either a or b. co co. allow ±. For modulus and multiplying by “5” co For OB p OA + as single vector. Scalar product = 0. Co (beware fortuitous answers)

Page 7 Mark Scheme: Teachers’ version Syllabus Paper GCE A/AS LEVEL – October/November 2009 9709 11 © UCLES 2009 10 f : x a 2x + 1, x∈o, x > 0 g : x 3 1 2 + − x x a , x∈o, x ≠ –3. (i) gf(x) = 3 1 2 1 )1 2 ( 2 + + − + x x = 4 2 1 4 + + x x Equates to x → 2x2 = 1 → x = ½√2 (ii) f –1(x) = ½(x − 1) To find g –1(x) , make x the subject Order must be correct → g –1(x) = 2 3 1 − − − x x or x x − + 2 3 1 (iii) x x − + 2 3 1 = x → x2 + x + 1 = 0 Looks at b2 – 4ac → negative → no roots. (iv) M1 M1 A1 [3] B1 M1 M1 A1 [4] M1 M1 A1 [3] B1 B1 B1 [3] Must be gf , needs x replacing twice. Forms quadratic + solution Co. condone ±. Co Attempt at x as the subject. Order correct. Allow for sign errors. Co – must be f(x). Forms quadratic equation. Looks at discriminant or attempts to solve and finds √(negative). Co Correct y = 2x + 1 on graph from (0, 1) Correct y = ½(x − 1) on graph from (1, 0) (if −ve x plotted, B1 s.c. for both) Shows or states or implies that f, f –1 are reflections in y = x.