Cambridge A Level Mathematics 9709 — 2010 May/June 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) May/June 2010 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. © UCLES 2010 [Turn over *9210562645*

2 1 The acute angle x radians is such that tan x = k, where k is a positive constant. Express, in terms of k, (i) tan(π −x), [1] (ii) tan(1 2π −x), [1] (iii) sin x. [2] 2 (i) Find the first 3 terms in the expansion of 2x −3 x 5 in descending powers of x. [3] (ii) Hence find the coefficient of x in the expansion of 1 + 2 x2 2x −3 x 5 . [2] 3 The ninth term of an arithmetic progression is 22 and the sum of the first 4 terms is 49. (i) Find the first term of the progression and the common difference. [4] The nth term of the progression is 46. (ii) Find the value of n. [2] 4 x y O y = 5 y x x = 6 – 2 The diagram shows the curve y = 6x −x2 and the line y = 5. Find the area of the shaded region. [6] 5 The function f is such that f(x) = 2 sin2x −3 cos2x for 0 ≤x ≤π. (i) Express f(x) in the form a + b cos2x, stating the values of a and b. [2] (ii) State the greatest and least values of f(x). [2] (iii) Solve the equation f(x) + 1 = 0. [3] 6 A curve is such that dy dx = 3x 1 2 −6 and the point (9, 2) lies on the curve. (i) Find the equation of the curve. [4] (ii) Find the x-coordinate of the stationary point on the curve and determine the nature of the stationary point. [3] © UCLES 2010 9709/11/M/J/10

3 7 O C A B x y y = 2 – 18 2 + 3 x The diagram shows part of the curve y = 2 − 18 2x + 3, which crosses the x-axis at A and the y-axis at B. The normal to the curve at A crosses the y-axis at C. (i) Show that the equation of the line AC is 9x + 4y = 27. [6] (ii) Find the length of BC. [2] 8 x y O C B (15, 22) A (3, –2) The diagram shows a triangle ABC in which A is (3, −2) and B is (15, 22). The gradients of AB, AC and BC are 2m, −2m and m respectively, where m is a positive constant. (i) Find the gradient of AB and deduce the value of m. [2] (ii) Find the coordinates of C. [4] The perpendicular bisector of AB meets BC at D. (iii) Find the coordinates of D. [4] © UCLES 2010 9709/11/M/J/10 [Turn over

4 9 The function f is defined by f : x →2x2 −12x + 7 for x ∈>. (i) Express f(x) in the form a(x −b)2 −c. [3] (ii) State the range of f. [1] (iii) Find the set of values of x for which f(x) < 21. [3] The function g is defined by g : x →2x + k for x ∈>. (iv) Find the value of the constant k for which the equation gf(x) = 0 has two equal roots. [4] 10 O A B C The diagram shows the parallelogram OABC. Given that −−→ OA = i + 3j + 3k and −−→ OC = 3i −j + k, find (i) the unit vector in the direction of −−→ OB, [3] (ii) the acute angle between the diagonals of the parallelogram, [5] (iii) the perimeter of the parallelogram, correct to 1 decimal place. [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 2010 9709/11/M/J/10

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Advanced Subsidiary Level and GCE Advanced Level MARK SCHEME for the May/June 2010 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 May/June 2010 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 2010 9709 11 © UCLES 2010 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 2010 9709 11 © UCLES 2010 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 2010 9709 11 © UCLES 2010 1 tan x = k (i) ) tan( x − π = −k B1 [1] co. www Mark final answers (ii)       −x 2 tan π = k 1 B1 [1] co. www (iii) sin x = 2 1 k k + from 90° triangle. M1 A1 [2] Any valid method – 90° triangle or formulae. 2 5 3 2       −x x (i) x x x 720 240 32 3 5 + − 3 × B1 [3] co. SC B2 for other 3 terms (i.e. ascending) (ii)      + 2 2 1 x ( x x x 720 240 32 3 5 + − ) Coeff of x (1 × 720) + (2 × –240) → 240 M1 A1√ [2] Looks at exactly 2 terms. co from his answer to (i). 3 9th term = 22, S4 = 49 (i) 22 8 = + d a 49 ) 3 2 ( 2 = + d a Soln of sim eqns → d = 1.5, a = 10 (ii) 46 )1 ( = − + d n a Substitutes for a and d → n = 25 B1 B1 M1 A1 [4] M1 A1 [2] co co Solution of two linear sim eqns. co Correct formula needed and attempt to solve. co. 4 2 6 x x y − = Meets y = 5 when x = 1 or x = 5. B1 co Integral = 3 3 1 2 3 x x − M1 A1 attempt to integrate. co. Their limits (1 to 5) used → 30⅔ Area of rectangle = 20 Shaded area = 10⅔ (integral of 5 6 2 − −x x B1, M1, A1 DM1 as above, then “−5x” B1√ A1) DM1 B1√ A1 [6] value at top limit − value at lower co to his x values co

Page 5 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9709 11 © UCLES 2010 5 x x x 2 2 cos 3 sin 2 − a (i) x x 2 2 cos 3 ) cos 1( 2 − − M1 Uses 1 2 2 = + c s → x 2 cos 5 2 − (a = 2, b = −5) A1 co [2] (ii) Values are −3 and 2 B1√ B1√ [2] (iii) x 2 cos 5 2 − = −1 → cos2 x = 0.6 x = 0.685, 2.46 (accept 0.684) B1√ B1 B1√ [3] co √ for π − (first answer) SC B1 for both 39.2 and 140.8 6 x y d d = 6 3 − x (9, 2) (i) ) ( 6 3 2 3 2 3 c x x y + − = (9, 2) 2 = 54 − 54 + c → c = 2. B2,1 M1 A1 [4] Loses 1 for each error – ignore +c Uses (9, 2) with integration to find c. co. (ii) x y d d = 0 → x = 4 2 3 d d 2 1 2 2 − = x x y → +ve (or ¾) Minimum B1 M1 A1 [3] Ignore any y-value Any valid method. co. 7 3 2 18 2 + − = x y (i) A is (3, 0) B1 Anywhere – but not from given answer x y d d = 2 ) 3 2 ( 18 − + x × 2 B1 B1 B1 for 2 ) 3 2 ( 18 − + x , B1 for ×2 If x = 3, m = 9 4 . m of normal = − 4 9 M1 Use of m1m2 = −1 with m from dy/dx Equation of normal ) 3 ( 4 9 − − = x y M1 Correct method for normal → 27 9 4 = + x y A1 [6] co (answer was given) (ii) Normal meets y-axis at (0, 6¾) Curve meets y-axis at (0, −4) → BC = 10¾ M1 A1 [2] Needs to put x = 0 in both normal and curve. co

Page 6 Mark Scheme: Teachers’ version Syllabus Paper GCE AS/A LEVEL – May/June 2010 9709 11 © UCLES 2010 8 (i) y-step ÷ x-step = 2 → m = 1 (ii) Eqn of AC )3 ( 2 2 − − = + x y Eqn of BC ) 15 ( 22 − = − x y Sim eqns 4 2 = + x y , 7 + = x y → C (–1, 6) (iii) M is (9, 10) Perp gradient is −½ → , 29 2 = + x y 7 + = x y Sim eqns → D (5, 12) M1 A1 [2] M1 A1√ A1√ A1 [4] B1 M1 M1 A1 [4] Gradient = y-step ÷ x step used co Correct form of one of lines. √ to his m √ to his m co co Use of m1m2 = −1 Solve sim eqns for their BC & perp. bis co 9 (i) 7 12 2 2 + − x x = 11 ) 3 ( 2 2 − − x 3 × B1 [3] B1 for each value – accept if a, b, c not specifically quoted. (ii) Range of f [ − 11 B1√ [1] √ to his “c”. allow > or [. (iii) 7 12 2 2 + − x x < 21 → 14 12 2 2 − − x x or 32 ) 3 ( 2 2 < − x M1 3-term quadratic to 0 or 32 ) 3 ( 2 2 < − x → end-values of 7 or −1 → −1 < x < 7 A1 A1 [3] Correct end-values co (iv) gf(x) = k x x + + − ) 7 12 2 ( 2 2 = 0 Use of b2 − 4ac → 242 − 16(14 + k) → k = 22 M1 A1 M1 A1 [4] Puts f into g. co. Used correctly with quadratic co. 10 = OA i + 3j + 3k, OC = 3i − j + k. (i) = OB OA + OC = 4i + 2j + 4k B1 co Unit vector = 6 1 (4i + 2j + 4k) M1 A1√ Divides by the modulus. √ on OB . [3] (ii) AC = OC − OA = 2i − 4j − 2k B1 co AC . = OB 8 − 8 − 8 = −8 M1 Use of x1 x2 + y1 y2 + z1 z2 OB = 6; 24 = AC M1 Correct method for a modulus. −8 = 6 × 24 × cos θ M1 Connected correctly provided , OB AC used θ = 105.8° → 74.2° A1 co (accept acute or obtuse) [5] (iii) OA = 19 or OC = 11 Perimeter = 2(√ + √) → 15.4 B1 M1 A1 [3] Used as a length. co (accept 15.3)