Cambridge A Level Mathematics 9709 — 2013 May/June Paper 1 · Variant 2

*2740852128* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS General Certificate of Education Advanced Subsidiary Level and Advanced Level MATHEMATICS 9709/12 Paper 1 Pure Mathematics 1 (P1) May/June 2013 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. JC13 06_9709_12/FP © UCLES 2013 [Turn over

2 1 A curve is such that dy dx = 6 x2 and 2, 9 is a point on the curve. Find the equation of the curve. [3] 2 Find the coefficient of x2 in the expansion of (i) @ 2x −1 2x A6 , [2] (ii) 1 + x2 @ 2x −1 2x A6 . [3] 3 The straight line y = mx + 14 is a tangent to the curve y = 12 x + 2 at the point P. Find the value of the constant m and the coordinates of P. [5] 4 A O D C B X 10 cm The diagram shows a square ABCD of side 10 cm. The mid-point of AD is O and BXC is an arc of a circle with centre O. (i) Show that angle BOC is 0.9273 radians, correct to 4 decimal places. [2] (ii) Find the perimeter of the shaded region. [3] (iii) Find the area of the shaded region. [2] 5 It is given that a = sin 1 −3 cos 1 and b = 3 sin 1 + cos 1, where 0Å ≤1 ≤360Å. (i) Show that a2 + b2 has a constant value for all values of 1. [3] (ii) Find the values of 1 for which 2a = b. [4] © UCLES 2013 9709/12/M/J/13

3 6 Relative to an origin O, the position vectors of points A and B are given by −−→ OA = i −2j + 2k and −−→ OB = 3i + pj + qk, where p and q are constants. (i) State the values of p and q for which −−→ OA is parallel to −−→ OB. [2] (ii) In the case where q = 2p, find the value of p for which angle BOA is 90Å. [2] (iii) In the case where p = 1 and q = 8, find the unit vector in the direction of −−→ AB. [3] 7 The point R is the reflection of the point −1, 3 in the line 3y + 2x = 33. Find by calculation the coordinates of R. [7] 8 The volume of a solid circular cylinder of radius r cm is 2500 cm3. (i) Show that the total surface area, S cm2, of the cylinder is given by S = 20r2 + 5000 r . 2 (ii) Given that r can vary, find the stationary value of S. [4] (iii) Determine the nature of this stationary value. [2] 9 A function f is defined by fx = 5 1 −3x, for x ≥1. (i) Find an expression for f ′x. [2] (ii) Determine, with a reason, whether f is an increasing function, a decreasing function or neither. [1] (iii) Find an expression for f −1x, and state the domain and range of f −1. [5] 10 (a) The first and last terms of an arithmetic progression are 12 and 48 respectively. The sum of the first four terms is 57. Find the number of terms in the progression. [4] (b) The third term of a geometric progression is four times the first term. The sum of the first six terms is k times the first term. Find the possible values of k. [4] [Question 11 is printed on the next page.] © UCLES 2013 9709/12/M/J/13 [Turn over

4 11 x y O A C B y x = (1 + 4 ) Ö The diagram shows the curve y = 1 + 4x, which intersects the x-axis at A and the y-axis at B. The normal to the curve at B meets the x-axis at C. Find (i) the equation of BC, [5] (ii) the area of the shaded region. [5] 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 2013 9709/12/M/J/13

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2013 series 9709 MATHEMATICS 9709/12 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 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2012 9709 12 © 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 12 © 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 Syllabus Paper GCE AS/A LEVEL – May/June 2013 9709 12 © Cambridge International Examinations 2013 1 2 6 x y = dx d 1 6 − − = x y + c Uses (2, 9) → c = 12 1 6 − − = x y + 12 B1 M1 A1 [3] Integration only – unsimplified Uses (2, 9) in an integral 2 6 2 1 2       − x x (i) Coeff of x² = 15×16×(−½)² = 60 (ii) Constant term is 20 × 8x³ ×(−1 ÷ 8x³) ×(1 + x²) needs to consider 2 terms → 60 − 20 = 40 B1 B1 [2] B1 M1 A1 [3] B1 for 2/3 parts. B1 B1 unsimplified Needs to consider the constant term 3 2 12 14 + = + x mx → 0 12 12 2 = − + x mx Uses ac b 4 2 = → m = −3 0 12 12 3 2 = − + − x x → P (2, 8) [Or m = −12x−2 M1 Sub M1 x = 2 A1] [→ m = −3 and y = 8 M1 A1] M1 M1 A1 DM1 A1 [5] Eliminates x (or y) Any use of discriminant Any valid method. 4 (i) BOC = 2tan−1½ = 0.9273 (ii) OB = √(10² + 5²) or 11.2 = r Arc BXC = √125 × 0.9273 → Perimeter = 20.4 cm (iii) Area = ½r²θ − ½.10.10 → 7.96 cm². M1 A1 [2] B1 M1 A1 [3] M1 A1 [2] Correct trigonometry. (ans given) Use of trig (or Pyth) for the OB = √125. Use of s = rθ with θ in rads , r ≠10 Correct formula used with rads, r ≠ 10. Allow 7.95 or 7.96 5 θ θ cos 3 sin − = a , θ θ cos sin 3 + = b (i) a ² + b² = ( ) 6 9 ( ) 6 9 2 2 2 2 sc c s sc c s + + + − + 10c² + 10s² = 10 (ii) c s c s + = − 3 6 2 → s = −7c → tanθ = −7 → 98.1° and 278.1° B1 M1 A1 [3] M1 A1 A1 A1 [4] Correct squaring Use of s² + c² =1 to get constant. (can get 2/3 for missing 6sc) Collecting and t = s÷c For 180⁰ + first answer, providing no extra answers in the range.

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9709 12 © Cambridge International Examinations 2013 6 , 2 2 k j i + − = OA k j i q p OB + + = 3 (i) p = −6, q = 6 (ii) dot product = 0 → 3 − 2 p + 4p = 0 → p = −1.5 (iii) AB = b − a = 2i + 3j + 6k Unit vector = (2i + 3j + 6k) ÷ 7 B1 B1 [2] M1 A1 [2] B1 M1 A1 [3] Use of x1x2 + y1y2 + z1z2 = 0 not for b – a. M1 for division by modulus. on B1. 7 . 33 2 3 = + x y Gradient of line = −⅔ Gradient of perpendicular = 3/2 Eqn of perp )1 ( 3 2 3 + = − x y Sim Eqns → (3, 9) (−1, 3) → (3, 9) → (7, 15) B1 M1 M1 M1 A1 M1 A1 [7] Use of m1m2 = −1 with gradient of line Correct form of perpendicular eqn. Sim eqns. Vectors or other method. 8 (i) π π 250 2 = h r → 2 250 r h = → 2 2 2 r rh S π π + = → r r S π π 500 2 2 + = (ii) r d dS 2 500 4 r r π π − = = 0 when r³ = 125 → r = 5 → S= 150π (iii) 2 2 d d r S 3 1000 4 r π π + = This is positive → Minimum M1 M1 [2] B1 B1 M1 A1 [4] M1 A1 [2] Makes h the subject. h r 2 π must be right Ans given – check all formulae.. B1 for each term Sets differential to 0 + attempt at soln Any valid method. 2nd differential must be correct – no need for numerical answer or correct r. 9 f(x) = x 3 1 5 − , x ≥ 1 (i) f ′(x) = 2) 3 1( 5 x − − × −3 (ii) 15 > 0 and (1 − 3x)² >0, f′(x) > 0 → increasing (iii) y = x 3 1 5 − → 3x = y 5 1− → f −1(x) = x x 3 5 − or ⅓ – x 3 5 Range is ≥ 1 Domain is − 2.5 ≤ x < 0 B1 B1 [2] B1 [1] M1 A1 B1 B1 B1 [5] B1 without × −3. B1 for ×−3, even if first B mark is incorrect providing ( )² in denominator. Attempt to make x the subject. Must be in terms of x. must be ≥ condone <

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – May/June 2013 9709 12 © Cambridge International Examinations 2013 10 (a) 57 = 2(24 + 3d) → d = 1.5 48 = 12 + (n − 1)1.5 → n = 25 (b) ar² = 4a r = ±2 ka r r a = − − 1 )1 ( 6 → k = 63 or k = − 21 M1 A1 M1 A1 [4] B1 B1 B1 B1 [4] Use of correct Sn formula. Use of correct Tn formula. (allow for r = 2) 11 x y 4 1+ = (i) 2 1 ) 4 1( 2 1 − + = x y dx d × 4 = 2 at B (0, 1) Gradient of normal = −½ Equation y − 1 = −½ x (ii) At A x = −¼ ∫ + dx x 4 1 = 2 3 2 3 ) 4 1( x + ÷ 4 Limits −¼ to 0 → 6 1 Area BOC = ½ × 2 × 1 = 1 → Shaded area = 6 7 B1 B1 M1 M1 A1 [5] B1 B1 B1 B1 B1 [5] B1 Without “×4”. B1 for “×4” even if first B mark lost. Use of m1m2=−1 Correct method for eqn. B1 Without the “÷4”. For “÷4” even if first B mark lost. For 1 + his “1/6”.