Cambridge IGCSE Mathematics - Additional 0606 — 2020 May/June Paper 1 · Variant 1

Cambridge IGCSE™ This document has 16 pages. Blank pages are indicated. DC (LEG/SW) 196733/3 R © UCLES 2020 [Turn over * 4 4 4 0 0 2 6 5 8 1 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2020 2 hours You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You should use a calculator where appropriate. ● You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator. ● Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question. INFORMATION ● The total mark for this paper is 80. ● The number of marks for each question or part question is shown in brackets [ ].

2 0606/11/M/J/20 © UCLES 2020 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) … … a b a n a b n a b n r a b b 1 2 n n n n n r r n 1 2 2 + = + + + + + + - - - J L KK J L KK J L KK N P OO N P OO N P OO where n is a positive integer and ( )! ! ! n r n r r n = - J L KK N P OO Arithmetic series u a n d S n a l n a n d 1 2 1 2 1 2 1 n n − − = + = + = + ^ ^ ^ h h h # - Geometric series u ar S r a r r S r a r 1 1 1 1 1 n n n n 1 1 ! − − − = = = 3 − ^ ^ ^ h h h 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC sin sin sin A a B b C c = = a2 = b2 + c2 – 2bc cos A ∆ = 2 1 bc sin A

3 0606/11/M/J/20 © UCLES 2020 [Turn over 1 The diagram shows the graph of a cubic curve f( ) y x = . y x 5 0 5 – 1 – 2 ( ) y x f = (a) Find an expression for f( )x . [2] (b) Solve f( )x 0 G . [2]

4 0606/11/M/J/20 © UCLES 2020 2 (a) Write down the period of cos x 2 3 1 - . [1] (b) On the axes below, sketch the graph of cos y x 2 3 1 = - for ° ° x 360 360 G G - . [3] y x – 360° – 180° 180° 360° – 1 0 1 – 2 2 – 3 3

5 0606/11/M/J/20 © UCLES 2020 [Turn over 3 The radius, r cm, of a circle is increasing at the rate of 5 cms–1. Find, in terms of r, the rate at which the area of the circle is increasing when r 3 = . [4]

6 0606/11/M/J/20 © UCLES 2020 4 DO NOT USE A CALCULATOR IN THIS QUESTION. Find the positive solution of the equation x x 5 4 7 4 2 7 1 0 2 + + - - = ` ` j j , giving your answer in the form a b 7 + , where a and b are fractions in their simplest form. [5]

7 0606/11/M/J/20 © UCLES 2020 [Turn over 5 Find the equation of the tangent to the curve ln y x x 2 3 1 2 = + - ` j at the point where x 1 = . Give your answer in the form y mx c = + , where m and c are constants correct to 3 decimal places. [6]

8 0606/11/M/J/20 © UCLES 2020 6 The line y x 5 6 = + meets the curve xy 8 = at the points A and B. (a) Find the coordinates of A and of B. [3] (b) Find the coordinates of the point where the perpendicular bisector of the line AB meets the line y x = . [5]

9 0606/11/M/J/20 © UCLES 2020 [Turn over 7 12 cm rad i 9.6 cm C D A B M O The diagram shows an isosceles triangle OAB such that OA OB = and angle AOB i = radians. The points C and D lie on OA and OB respectively. CD is an arc of length 9.6 cm of the circle, centre O, radius 12 cm. The arc CD touches the line AB at the point M. (a) Find the value of i. [1] (b) Find the total area of the shaded regions. [4] (c) Find the total perimeter of the shaded regions. [3]

10 0606/11/M/J/20 © UCLES 2020 8 (a) Show that x x 2 3 3 2 3 3 - + + can be written as x x 4 9 12 2 - . [2] (b) Hence find d x x x 4 9 12 2 - y , giving your answer as a single logarithm and an arbitrary constant. [3]

11 0606/11/M/J/20 © UCLES 2020 [Turn over (c) Given that a d ln x x x 4 9 12 5 5 2 2 - = y , where a 2 2 , find the exact value of a. [4]

12 0606/11/M/J/20 © UCLES 2020 9 (a) An arithmetic progression has a second term of 14 - and a sum to 21 terms of 84. Find the first term and the 21st term of this progression. [5]

13 0606/11/M/J/20 © UCLES 2020 [Turn over (b) A geometric progression has a second term of p 27 2 and a fifth term of p5. The common ratio, r, is such that r 0 1 1 1 . (i) Find r in terms of p. [2] (ii) Hence find, in terms of p, the sum to infinity of the progression. [3] (iii) Given that the sum to infinity is 81, find the value of p. [2]

14 0606/11/M/J/20 © UCLES 2020 10 (a) (i) Show that sec sec cot 1 1 1 1 2 2 i i i - - + = . [3] (ii) Hence solve sec sec x x 1 1 1 1 2 2 6 - - + = for ° x 90 90° 1 1 - . [5]

15 0606/11/M/J/20 © UCLES 2020 [Turn over (b) Solve cosec y 3 2 r + = b l for y 0 2 G G r radians, giving your answers in terms of r. [4] Question 11 is printed on the next page.

16 0606/11/M/J/20 © UCLES 2020 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. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which itself is a department of the University of Cambridge. 11 A curve is such that d d cos x y x 5 2 2 2 = . This curve has a gradient of 4 3 at the point , 12 4 5 r r - b l. Find the equation of this curve. [8]

This document consists of 9 printed pages. © UCLES 2020 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2020 MARK SCHEME Maximum Mark: 80 Published Students did not sit exam papers in the June 2020 series due to the Covid-19 global pandemic. This mark scheme is published to support teachers and students and should be read together with the question paper. It shows the requirements of the exam. The answer column of the mark scheme shows the proposed basis on which Examiners would award marks for this exam. Where appropriate, this column also provides the most likely acceptable alternative responses expected from students. Examiners usually review the mark scheme after they have seen student responses and update the mark scheme if appropriate. In the June series, Examiners were unable to consider the acceptability of alternative responses, as there were no student responses to consider. Mark schemes should usually be read together with the Principal Examiner Report for Teachers. However, because students did not sit exam papers, there is no Principal Examiner Report for Teachers for the June 2020 series. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the June 2020 series for most Cambridge IGCSE™ and Cambridge International A & AS Level components, and some Cambridge O Level components.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 2 of 9 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptors for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with: • the specific content of the mark scheme or the generic level descriptors for the question • the specific skills defined in the mark scheme or in the generic level descriptors for the question • the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively: • marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate • marks are awarded when candidates clearly demonstrate what they know and can do • marks are not deducted for errors • marks are not deducted for omissions • answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 3 of 9 Maths-Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be 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 in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 4 of 9 Question Answer Marks Partial Marks 1(a) ( )( )( ) 1 2 1 5 2 y x x x = − + + − B1 For 1 2 − B1 For ( )( )( ) 2 1 5 x x x + + − 1(b) –2 ⩽ x ⩽ –1 B1 x ⩾ 5 B1 2(a) 1080° B1 2(b) B1 For correct shape and symmetry about the y-axis B1 For correct x-intercepts B1 For correct y-intercept 3 d 5 d r t = B1 d 2π d A r r = B1 d d d d d d A A r t r t = × leading to d 10π d A r t = M1 Use of the chain rule, may be implied by 5 6π × d 30π d A t = A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 5 of 9 Question Answer Marks Partial Marks 4 ( ) ( ) ( )( ) ( ) 2 4 2 7 4 2 7 4 5 4 7 1 2 5 4 7 x − − + − − + − = + M1 For correct use of quadratic formula, allow inclusion of ± until final answer ( ) ( ) 4 2 7 16 28 16 7 20 16 7 2 5 4 7 x − − + + − + + = + ( ) ( ) 4 2 7 8 2 5 4 7 x − − + = + M1 For attempt to simplify discriminant, must see attempt at expansion and subsequent simplification ( ) 4 2 7 2 5 4 7 x + = + or ( ) 2 7 5 4 7 x + = + A1 For either ( ) 2 7 5 4 7 5 4 7 5 4 7 x + − = × − + 10 5 7 8 7 28 25 112 x + − − = − M1 For attempt to rationalise, must see attempt at expansion and subsequent simplification 6 7 29 29 x = + A1 5 ( ) ( ) ( ) 2 2 2 6 2 ln 3 1 d 3 1 d 2 x x x y x x x + − − − = + B1 B1 for 2 6 3 1 x x − M1 For attempt to differentiate a quotient or an equivalent product, must have correct order of terms and correct sign A1 When ln 2 1, 3 x y = = or ( ) 0.231 0 B1 When d 1, 0.92298 d y x x = = , allow 0.923 B1 0.923 0.692 y x = − B1 6(a) ( ) 5 6 8 x x + = 2 5 6 8 0 x x + − = M1 For attempt to equate and obtain a 3- term quadratic in either x or y 4 , 10 5       A1 Allow A1 if only x-coordinates or only y-coordinates are given (–2, –4) A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 6 of 9 Question Answer Marks Partial Marks 6(b) Midpoint 3 , 3 5   −     B1 Gradient 5 B1 1 3 3 5 5 y x   − = − +     M1 Attempt at perp bisector using their midpoint and perp gradient 1 3 3 5 5 x x   − = − +     M1 For use of y x = and attempt to solve 12 12 , 5 5       A1 7(a) 0.8 B1 7(b) Sector area = ( ) 2 112 0.8 2 57.6 B1 Allow unsimplified tan0.4 12 AM = 12tan0.4 AM = 5.074 M1 Attempt at AM using their 2 θ Allow unsimplified Area of triangle ( ) 1 5.074 2 2 12 2 = × × × 60.88 M1 Area of triangle using their AM, allow unsimplified Shaded area 3.28 A1 7(c) sin0.4 AM OA = 5.074 sin0.4 OA = 13.03 M1 Attempt to find OA using their 2 θ and their AM Perimeter ( ) ( ) 2 1.03 9.6 2 5.074 = + + M1 Allow if using their 2 θ and their CM Perimeter = 21.8 A1 8(a) ( ) ( ) 2 3 2 3 3 2 3 4 9 x x x + + − − M1 Must see for M1 2 12 4 9 x x − A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 7 of 9 Question Answer Marks Partial Marks 8(b) 3 3 d 2 3 2 3 x x x + − +  B2 ( ) ( ) 3 3 ln 2 3 ln 2 3 2 2 x x = − + + B1 for each correct term, having made use of (a) ( ) 2 3 ln 4 9 2 x c − + or ( )( ) ( ) 3 ln 2 3 2 3 2 x x c − + + or ( ) 3 2 2 ln 4 9 x c − + B1 8(c) ( ) 3 3 3 2 2 2 2 ln 4 9 ln7 ln5 a − − = M1 For correct application of limits, allow equivalent forms 2 4 9 35 a − = A1 For a correct method of dealing with logarithms and eliminating them 11 a = M1 For solving a quadratic equation, dep on first M mark A1 9(a) Second term: 14 a d + = − B1 Sum: 4 10 a d = + B1 2 d = B1 16 a = − B1 Last term = 24 B1 Ft on their d and their a 9(b)(i) 2 27 ar p = 4 5 ar p = B1 For both equations 3 p r = B1 9(b)(ii) 81 a p = M1 M1 for attempt to find a in terms of p A1 81 1 3 p S p ∞= − or 243 3 p p − B1 Follow through on their a and their r

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 8 of 9 Question Answer Marks Partial Marks 9(b)(iii) 81 81 1 3 p p = − or 243 81 3 p p = − M1 For attempt to solve using their answer to (ii) as far as p = … 3 4 p = A1 10(a)(i) ( ) ( ) 2 sec 1 sec 1 sec 1 θ θ θ + − − − M1 For dealing with the fractions 2 2 tan θ M1 For use of the correct identity 2 2cot θ A1 A1 for given answer, must see 2 8 tan θ first 10(a)(ii) 2 2cot 2 6 x = 1 tan 2 3 x = ± M1 M1 for use of (i) and attempt to simplify A1 M1 M1 for attempt to solve, may be implied by one correct solution o o o o 2 150 , 30 , 30 , 150 x = − − o o o o 75 , 15 , 15 , 75 x = − − A2 A1 for each pair of correct solutions 10(b) π 1 sin 3 2 y   + =     M1 For dealing with cosec and an attempt to solve π 5π 13π , 3 6 6 y + = M1 M1 for a complete method of solution, may be implied by a correct solution π 2 y = A1 11π 6 y = A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2020 © UCLES 2020 Page 9 of 9 Question Answer Marks Partial Marks 11 ( ) d 5 sin 2 d 2 y x c x = + M1 M1 for sin 2 k x A1 Condone omission of c 3 5 π sin 4 2 6 c   = − +     M1 Dep on first M1 for attempt to find c 2 c = A1 ( ) 5 cos2 2 4 y x x d = − + + M1 M1 for attempt to integrate their d d y x A1 Condone omission of d 5 5 π π cos 4 4 6 6 d π   = − − − +     M1 Dep on previous M1 for attempt to find d 17 5 3 12 8 d π = + 5 17 5 3 cos2 2 4 12 8 y x x π = − + + + or 5 cos2 2 5.53 4 y x x = − + + A1 Must have the equation for A1