Cambridge IGCSE Mathematics - Additional 0606 — 2017 May/June Paper 1 · Variant 1

This document consists of 12 printed pages. DC (RW/FD) 129466/3 © UCLES 2017 [Turn over * 9 2 7 2 0 8 0 0 1 6 * ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2017 2 hours Candidates answer on the Question Paper. Additional Materials: Electronic calculator READ THESE INSTRUCTIONS FIRST 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 an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. DO NOT WRITE IN ANY BARCODES. 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 80. Cambridge International Examinations Cambridge International General Certificate of Secondary Education

2 0606/11/M/J/17 © UCLES 2017 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)n = an + ( n 1)an–1 b + ( n 2)an–2 b2 + … + ( n r)an–r br + … + bn, where n is a positive integer and ( n r) = n! (n – r)!r! 2. TRIGONOMETRY Identities sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A Formulae for ∆ABC a sin A = b sin B = c sin C a2 = b2 + c2 – 2bc cos A ∆ = 1 2 bc sin A

3 0606/11/M/J/17 © UCLES 2017 [Turn over 1 The line y kx 5 = - , where k is a positive constant, is a tangent to the curve y x x 4 2 = + at the point A. (i) Find the exact value of k. [3] (ii) Find the gradient of the normal to the curve at the point A, giving your answer in the form a b 5 + , where a and b are constants. [2]

4 0606/11/M/J/17 © UCLES 2017 2 It is given that x x ax bx 48 p 3 2 = + + - ^ h . When x p^ h is divided by x 3 - the remainder is 6. Given that 1 0 p = l^ h , find the value of a and of b. [5] 3 (a) Simplify x y x y 8 10 3 6 ' - 3 , giving your answer in the form x y a b, where a and b are integers. [2] (b) (i) Show that t t 4 2 5 2 2 1 2 3 - + - ^ ^ h h can be written in the form t qt r 2 p - + ^ ^ h h, where p, q and r are constants to be found. [3] (ii) Hence solve the equation t t 4 2 5 2 0 2 1 2 3 - + - = ^ ^ h h . [1]

5 0606/11/M/J/17 © UCLES 2017 [Turn over 4 (a) It is given that x 3 5 f e x 4 = + - ^ h for x R ! . (i) State the range of f. [1] (ii) Find f 1 - and state its domain. [4] (b) It is given that x x 5 g 2 = + ^ h and ln x x h = ^ h for x 0 2 . Solve x 2 hg = ^ h . [3]

6 0606/11/M/J/17 © UCLES 2017 5 (a) O A M C B b a c The diagram shows a figure OABC, where a OA = , b OB = and c OC = . The lines AC and OB intersect at the point M where M is the midpoint of the line AC. (i) Find, in terms of a and c, the vector OM . [2] (ii) Given that : : OM MB 2 3 = , find b in terms of a and c. [2]

7 0606/11/M/J/17 © UCLES 2017 [Turn over (b) Vectors i and j are unit vectors parallel to the x-axis and y-axis respectively. The vector p has a magnitude of 39 units and has the same direction as i j 10 24 - + . (i) Find p in terms of i and j. [2] (ii) Find the vector q such that p q 2 + is parallel to the positive y-axis and has a magnitude of 12 units. [3] (iii) Hence show that q k 5 = , where k is an integer to be found. [2]

8 0606/11/M/J/17 © UCLES 2017 6 O A D B C 8 cm 12 cm The diagram shows a circle, centre O, radius 12 cm. The points A and B lie on the circumference of the circle and form a rectangle with the points C and D. The length of AD is 8 cm and the area of the minor sector AOB is 150 cm2. (i) Show that angle AOB is 2.08 radians, correct to 2 decimal places. [2] (ii) Find the area of the shaded region ADCB. [6] (iii) Find the perimeter of the shaded region ADCB. [3]

9 0606/11/M/J/17 © UCLES 2017 [Turn over 7 Show that the curve y x 3 8 2 3 5 = + ^ h has only one stationary point. Find the coordinates of this stationary point and determine its nature. [8]

10 0606/11/M/J/17 © UCLES 2017 8 (i) On the axes below sketch the graphs of y x 2 5 = - and y x x 9 80 16 2 = - . [5] –1 1 0 –10 –5 5 10 y 2 3 4 5 6 x (ii) Solve x 2 5 4 - = . [3] (iii) Hence show that the graphs of y x 2 5 = - and y x x 9 80 16 2 = - intersect at the points where y 4 = . [1] (iv) Hence find the values of x for which x x x 9 2 5 80 16 2 G - - . [2]

11 0606/11/M/J/17 © UCLES 2017 [Turn over 9 (i) Show that tan sec x x 5 4 3 4 3 1 2 2 + = + J L KK J L KK N P OO N P OO . [1] (ii) Given that d d tan sec x x x 3 3 1 3 2 = J L KK J L KK c N P OO N P OO m , find sec x x 3 d 2J L KK N P OO y . [1] (iii) O y x y = 5 + 4 tan2 x 3 J L KK N P OO 2 The diagram shows part of the curve tan y x 5 4 3 2 = + J L KK N P OO. Using the results from parts (i) and (ii), find the exact area of the shaded region enclosed by the curve, the x-axis and the lines r x 2 = and r x = . [5] Question 10 is printed on the next page.

12 0606/11/M/J/17 © UCLES 2017 10 (a) Given that y x 4 1 e x 2 3 = + , find x y d d . [3] (b) Variables x, y and t are such that r r cos sin y x x 4 3 2 3 3 = + + + J L KK J L KK N P OO N P OO and t y 10 d d = . (i) Find the value of x y d d when r x 2 = . [3] (ii) Find the value of t x d d when r x 2 = . [2] 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 International Examinations Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cie.org.uk after the live examination series. 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.

® IGCSE is a registered trademark. This document consists of 8 printed pages. © UCLES 2017 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/11 Paper 1 May/June 2017 MARK SCHEME Maximum Mark: 80 Published 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 2017 series for most Cambridge IGCSE®, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 2 of 8 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 2017 © UCLES 2017 Page 3 of 8 Question Answer Marks Guidance 1(i) 2 5 4 − = + kx x x ( ) 2 4 5 0 + − + = x k x M1 equating line and curve equation and collecting terms to form an equation of the form 2 0 + + = ax bx c x terms must be gathered together, maybe implied by later work For a tangent ( ) 2 4 20 − = k DM1 correct use of discriminant 4 2 5 = + k A1 Accept 4 20 = + k Alternative Gradient of line = k Gradient of curve = d 2 4 d = + y x x Equating: 2 4 = + k x M1 substitution of 2 4 = + k x or 4 2 − = k x in 2 5 4 − = + kx x and simplify to a quadratic equation in k or x DM1 4 2 5 = + k A1 Accept 4 20 = + k 1(ii) Normal gradient = 1 4 2 5 4 2 5 4 2 5 − − × + − M1 use of negative reciprocal and attempt to rationalise using a form of 5 − a b or 20 − a or their equivalent from (i) = – 4 2 5 4 − − oe = 5 1 2 − A1 – 4 2 5 4 − − oe leading to 5 1 2 − 2 ( ) p 3 27 9 3 48 = + + − a b M1 attempt to find ( ) p 3 3 9 + = a b oe A1 ( ) 2 p 3 2 ′ = + + x x ax b ( ) p 1 3 2 ′ = + + a b M1 attempt to differentiate and find ( ) p 1 ′ must have 2 terms correct 2 3 + = − a b oe A1 12, 27 a b = = − A1 for both 3(a) 3 7 x y B2 B1 for each term

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 4 of 8 Question Answer Marks Guidance 3(b)(i) for ( ) ( ) ( ) 3 1 2 2 2 2 2 − = − − t t t soi M1 ( ) ( ) ( ) 1 2 2 4 5 2 − + − t t A1 ( ) ( ) 1 2 2 5 6 − − t t A1 3(b)(ii) 2 and 6 5 B1 FT on their ( ) ( ) 1 2 2 5 6 − − t t , must have 2 4(a)(i) f 5 > , ( ) f 5 > x B1 4(a)(ii) 4 5 e 3 − − = x y or 4 5 e 3 − − = y x B1 5 4 ln 3 −   − =     y x or 5 4 ln 3 −   − =     x y B1 leading to ( ) 1 1 5 f ln 4 3 − −   = −     x x or ( ) 1 1 3 f ln 4 5 −   =   −   x x or ( ) ( ) ( ) 1 1 f ln3 ln 5 4 − = − − x x or ( ) ( ) ( ) 1 1 f ln 5 ln3 4 − = − − − x x B1 Domain 5 > x B1 4(b) ( ) 2 ln 5 2 + = x B1 2 2 5 e + = x B1 1.55 = x or better or 2e 5 − B1 5(a)(i) ( ) 1 2 = + − uuuur uuur uuur uuur OM OC OA OC oe M1 may be implied by correct answer. ( ) 1 2 + a c A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 5 of 8 Question Answer Marks Guidance 5(a)(ii) 5 2 = uuuur OM b oe , 5 2 (their (i)) or ( ) 2 3 = − uuuur uuuur OM OM b M1 dealing with ratio correctly to relate b or uuur OB to uuuur OM = ( ) 5 4 + a c A1 5(b)(i) 10 24 26 − + = i j p 39 ( 10 24 26 = − + i j ) M1 magnitude of 10 24 − + i j and use with 39 p 15 36 = − + i j A1 5(b)(ii) If parallel to the y-axis, i component is zero M1 realising i component is zero so 2 12 + = p q j DM1 use of 12 q = 30 60 − i j A1 5(b)(iii) ( ) 2 2 30 1 2 = + − q or 900 5 × M1 attempt at magnitude of their q 30 5 = q A1 Answer Given: must have full and correct working 6(i) 2 1 12 150 2 θ × × = M1 use of sector area 2.083 θ = , so 2.08 θ = to 2dp A1 6(ii) Area of triangle AOB = 2 1 12 sin 2.08 2 × M1 correct method for area of triangle Area of segment = 2 1 150 12 sin 2.08 2 − × × A1 allow unsimplified, using 150 2.08, 2.083 or 72 θ = 2 sin1.04 12 = AB M1 correct trigonometric statement using 150 2.08, 2.083 or 72 θ = with attempt to obtain AB AB = awrt 20.7 A1 Shaded area = their AB 8 × −their segment area M1 execution of a correct ‘plan’(rectangle – segment) awrt 78.4 or 78.5 A1

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 6 of 8 Question Answer Marks Guidance 6(iii) Arc AB = 25 or 24.96 B1 Perimeter = 25 + their AB + 16 M1 correct ‘plan’ ( arc + their AB + 2 8 × ) awrt 61.7 A1 7 differentiation to obtain answer in the form ( ) 2 2 3 3 8 + p x or ( ) 2 2 3 3 8 + qx x M1 ( ) 2 2 3 6 3 8 + x x B1 ( ) 2 2 3 d 5 6 3 8 d 3 = × + y x x x A1 all correct When d 0 d = y x only solution is 0 = x DM1 ( ) 2 2 3 3 8 0 + = qx x and attempt to solve 0 = x and 2 3 8 + x = 0 has no solutions A1 Stationary point at ( ) 0,32 A1 correct gradient method with substitution of x values either side of zero or equivalent valid method M1 correct conclusion from correct work using a correct d d y x A1 8(i) B5 B1 for shape of modulus function B1 for y intercept = 5 (for modulus graph only) B1 for x intercept = 2.5 at the V of a modulus graph B1 for shape of quadratic function for –1 ⩽ x ⩽ 6 B1 for intercepts at x = 0 and x = 5 for a quadratic graph 8(ii) 2 5 4 − = ± x B1 one correct answer 9 2 = x M1 solution of two different correct linear equations or solution of an equation obtained from squaring both sides or use of symmetry from first solution. 1 2 = x A1 second correct solution −1 1 2 3 4 5 6 −5 5 10 x y

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 7 of 8 Question Answer Marks Guidance 8(iii) 2 1 1 16 80 36 4 2 2     − + =         and 2 9 9 16 80 36 4 2 2     − + =         B1 verification using both x values or for forming and solving 2 16 80 36 0 − + = x x 8(iv) using their values from (ii) in an equality of the form a x b - - or < < a x b M1 1 9 2 2 x - - cao A1 9(i) 2 5 4 sec 1 3     + −         x leading to given answer B1 use of correct identity 9(ii) ( ) 3tan 3   +     x c B1 9(iii) attempt to integrate using (i) and/or (ii) M1 Area = π 2 π 2 4sec 1 d 3  +     ∫ x x A1 all correct π π 2 12tan 3    +         x x DM1 correct method for evaluation using limits in correct order π π π 12tan π 12tan 3 6 2     = + − +         A1 π 8 3 2 = + A1 10(a) differentiation of a quotient or equivalent product M1 correct differentiation of 3e x B1 ( ) ( ) 3 2 3 2 2 3e 4 1 8 e d d 4 1 + − = + x x x x y x x or ( ) 3 3 2 2 2 d 3e 8 e d 4 1 4 1 = − + + x x y x x x x A1 everything else correct including brackets where needed, allow unsimplified

0606/11 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2017 © UCLES 2017 Page 8 of 8 Question Answer Marks Guidance 10(b)(i) one term differentiated correctly M1 d π π 4sin 2 3cos d 3 3     = − + + +         y x x x A1 all correct When π 2 = x , d 5 d = − y x A1 10(b)(ii) d d d d d d × = y x y x t t d 5 10 d −× = x t oe M1 correct use of rates of change d 2 d = − y t A1 FT answer to (i)