Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 1 · Variant 1

*6593575770* 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 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 11_9709_11/2R © UCLES 2012 [Turn over

2 1 The first term of an arithmetic progression is 61 and the second term is 57. The sum of the first n terms is n. Find the value of the positive integer n. [4] 2 A curve is such that dy dx = −8 x3 −1 and the point (2, 4) lies on the curve. Find the equation of the curve. [4] 3 An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the sea. At midday the radius of the patch of oil is 50 m and is increasing at a rate of 3 metres per hour. Find the rate at which the area of the oil is increasing at midday. [4] 4 (i) Find the first 3 terms in the expansion of (2x −x2)6 in ascending powers of x. [3] (ii) Hence find the coefficient of x8 in the expansion of (2 + x)(2x −x2)6. [2] 5 A curve has equation y = 2x + 1 (x −1)2 . Verify that the curve has a stationary point at x = 2 and determine its nature. [5] 6 O B A D C r q rad The diagram shows a sector OAB of a circle with centre O and radius r. Angle AOB is θ radians. The point C on OA is such that BC is perpendicular to OA. The point D is on BC and the circular arc AD has centre C. (i) Find AC in terms of r and θ. [1] (ii) Find the perimeter of the shaded region ABD when θ = 1 3π and r = 4, giving your answer as an exact value. [6] 7 (i) Solve the equation 2 cos2θ = 3 sin θ, for 0◦≤θ ≤360◦. [4] (ii) The smallest positive solution of the equation 2 cos2(nθ) = 3 sin(nθ), where n is a positive integer, is 10◦. State the value of n and hence find the largest solution of this equation in the interval 0◦≤θ ≤360◦. [3] © UCLES 2012 9709/11/O/N/12

3 8 x y O a 1 2 y x 2 = 2 – 1 3 = 2 – 1 y x The diagram shows the curve y2 = 2x −1 and the straight line 3y = 2x −1. The curve and straight line intersect at x = 1 2 and x = a, where a is a constant. (i) Show that a = 5. [2] (ii) Find, showing all necessary working, the area of the shaded region. [6] 9 The position vectors of points A and B relative to an origin O are a and b respectively. The position vectors of points C and D relative to O are 3a and 2b respectively. It is given that a = 2 1 2 ! and b = 4 0 6 ! . (i) Find the unit vector in the direction of −−→ CD. [3] (ii) The point E is the mid-point of CD. Find angle EOD. [6] 10 The function f is defined by f(x) = 4x2 −24x + 11, for x ∈>. (i) Express f(x) in the form a(x −b)2 + c and hence state the coordinates of the vertex of the graph of y = f(x). [4] The function g is defined by g(x) = 4x2 −24x + 11, for x ≤1. (ii) State the range of g. [2] (iii) Find an expression for g−1(x) and state the domain of g−1. [4] [Question 11 is printed on the next page.] © UCLES 2012 9709/11/O/N/12 [Turn over

4 11 O x y A (1, 2) C B E y x = (6 + 2) 1 3 The diagram shows the curve y = (6x + 2) 1 3 and the point A (1, 2) which lies on the curve. The tangent to the curve at A cuts the y-axis at B and the normal to the curve at A cuts the x-axis at C. (i) Find the equation of the tangent AB and the equation of the normal AC. [5] (ii) Find the distance BC. [3] (iii) Find the coordinates of the point of intersection, E, of OA and BC, and determine whether E is the mid-point of OA. [4] 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 9709/11/O/N/12

CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the October/November 2012 series 9709 MATHEMATICS 9709/11 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 2012 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.

Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 11 © 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 Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 11 © 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 – October/November 2012 9709 11 © Cambridge International Examinations 2012 1

122    14 

122    14 2  31 0  31 M1 A1 DM1 A1 [4] Attempt sum formula with a = 61, d = –4 Equated to n cao Attempt to solve. Accept div. by n cao 2   
   Sub (2, 4) ⟶ 5 M1A1 DM1A1 [4] Attempt integration. cao Dependent on c present 3   →   2        used   3 soi 300 (or 942) B1 M1 B1 A1 [4] 4 (i) 2   64  192  240 (ii)  2   coeff of  = 2 240  192 288 B1B1B1 [3] M1 A1 [2] cao Looks at exactly 2 terms 5   2  2  1  Sub x = 2 →   2  2 0 ⇒ stat value at x = 2 

6  1  (and sub x = 2) (At x = 2, 

6) > 0 ⟹ Minimum B2,1,0 B1 M1 A1 [5] –1 each error in 2, –2,   1  AG Reasonable attempt to diff form   1
Correct 

and ‘minimum’ is required Or other valid method for last 2 marks 6 (i)     cos ! (ii) arc "   arc #    $%&'     4  4 cos    "# 4 sin    $%&'  2√3  2 Perimeter =    2√3  2 B1 [1] B1 M1A1 M1A1 A1 [6] Allow   $%&'  for M1. Allow 3.14 Allow 1.46 cao Accept √12 7 (i) 21  sin! 3 sin ! 2 sin !  1sin !  2 0 ! 30° or 150° (ii)     3 $%&' 3! 720  $%&'150 870 ! 290° M1 M1 A1A1 [4] B1 M1 A1 [3] Use   . 1 Attempt to solve cao ft provided n is an integer Allow full list up to at least 870 cao

Page 5 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 11 © Cambridge International Examinations 2012 8 (i)  3 ⇒  3 0 ⇒ 3 or 0    or 5 ⇒a 5 AG (ii) 0   

1
2 2 , 4    
   5 4    05, 4       6     75 Subtract areas at some stage   oe M1 A1 [2] B1B1B1 M1 M1 A1 [6] OR form equation in x and attempt solution OR sub x=5 each eq (M1) → y = 3(twice) (A1) (5,3) subst only once scores 0/2 Or ∆ = ½(5 – ½)3 Apply limits ½ and 5 for, at least, curve Dependent on some integration cao 9/4 with no working scores 0/6, but 9 − 27/4 = 9/4 scores 1/6 (M1 subtraction) 9 (i) # 888889 3 : 2 1 2 ;  2 : 4 0 6 ; : 2 3 6 ; Unit vector =  : 2 3 6 ; (ii) <= 888889 : 6 3 6 ;  : 1 1½ 3 ; : 7 1½ 9 ; <= 888889. <# 8888889 = 56 + 0 + 108 = 164 │<= 888889│= √132.25 (= 11.5); │<# 8888889│= √208 164 = √132.25  √208  cos ! ! 8.6° cao B1 M1A1 [3] M1A1 M1 M1 M1 A1 [6] Allow M1A1 for  : 2 3 6 ; or 
     etc or equivalent method Use of     AA Correct method for moduli All connected correctly. Dependent on <= 888889, =<, 8888889 <#, 88888889 #< 8888889 used 10 (i) 4  3  25 Vertex is (3, –25) (ii) range is ( g(x) ) ≥ –9 Allow > (iii)   3     25   3 B   C  25 g  3    √  25 Domain is x ≥ –9 B1B1B1 B1 [4] B1B1 [2] M1 DM1 A1 B1 [4] Or E 4, F 3,  25 ft to their (b, c). Accept if not ‘hence’ B1 for ≥ , B1 for –9 Accept e.g. [−9,∞] Attempt to square root both sides cao ft from their (ii)

Page 6 Mark Scheme Syllabus Paper GCE AS/A LEVEL – October/November 2012 9709 11 © Cambridge International Examinations 2012 11 (i)  
6 
  6  2
 Equation of tangent is   2 G  1 Equation of normal is   2      1 Both eqns correct with G   cao (ii) B = (0, 1½); C = (2, 0) BC = H2  1
 = 2½ (iii) BC:   1        0 or       2 Intersection (E) :      1   2   ;    Mid-point of OA = (½, 1) →E not mid-point B1B1 M1 M1 A1 [5] B1 M1A1 [3] M1 M1 A1 B1 [4] Independent Where m = numerical   Including use of GG 1 SC 1/3 Blatant tangent/normal reversal Both cao ft from their B and C or       1   cao Dependent on correct x values or y values for both E and the mid-point of OA