Cambridge IGCSE Mathematics - Additional 0606 — 2016 Feb/March Paper 2 · Variant 2

This document consists of 16 printed pages. DC (NF/FD) 106874/3 © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 5 7 6 4 7 8 3 2 7 3 * ADDITIONAL MATHEMATICS 0606/22 Paper 2 February/March 2016 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.

2 0606/22/F/M/16 © UCLES 2016 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax2 + bx + c = 0, x b b ac a = − − 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 0606/22/F/M/16 © UCLES 2016 [Turn over 1 Two variables x and y are such that y x 9 5 = - for x 9 2 . (i) Find an expression for x y d d . [2] (ii) Hence, find the approximate change in y as x increases from 13 to 13 + h, where h is small. [2]

4 0606/22/F/M/16 © UCLES 2016 2 The sets A, B and C are such that , , , , , . C A B C A B B B C A B 2 12 14 19 n n n n + + , , Q 1 = = = = = ^ ^ ^ ^ h h h h Complete the Venn diagram to show the sets A, B and C and hence state A B C n + + l l ^ h.  A B C n + + l l ^ h = … [4]

5 0606/22/F/M/16 © UCLES 2016 [Turn over 3 Find the equation of the curve which passes through the point (1, 7) and for which x y x x 9 3 d d 2 4 = - . [4]

6 0606/22/F/M/16 © UCLES 2016 4 (a) cos x a bx c f + = ^ h has a period of 60°, an amplitude of 10 and is such that f (0) = 14. State the values of a, b and c. [2] (b) Sketch the graph of sin y x 3 4 2 = - for ° x 0 180° G G on the axes below. [3] 45° O y x 90° 135° 180°

7 0606/22/F/M/16 © UCLES 2016 [Turn over 5 (i) Find, in ascending powers of x, the first 3 terms of the expansion of kx 3 7 + ^ h , where k is a constant. Give each term in its simplest form. [3] (ii) Given that, in the expansion of kx 3 7 + ^ h , the coefficient of x2 is twice the coefficient of x, find the value of k. [2]

8 0606/22/F/M/16 © UCLES 2016 6 Do not use a calculator in this question. x 1 3 3 + 6 2 3 + 5 3 - The diagram shows two parallelograms that are similar. The base and height, in centimetres, of each parallelogram is shown. Given that x, the height of the smaller parallelogram, is p q 6 3 + , find the value of each of the integers p and q. [5]

9 0606/22/F/M/16 © UCLES 2016 [Turn over 7 (a) Given that 4 2 6 0 8 4 A = - J L KK N P OO and 6 7 1 2 2 1 B = - J L KK N P OO , find A − 3B. [2] (b) Given that 2 4 0 1 C = - J L KK N P OO and 4 3 3 5 D = - - J L KK N P OO , find (i) the inverse matrix C−1, [2] (ii) the matrix X such that XD−1 = C. [3]

10 0606/22/F/M/16 © UCLES 2016 8 The line 2y = x + 2 meets the curve x xy y 3 12 2 2 + - = at the points A and B. (i) Find the coordinates of the points A and B. [5] (ii) Given that the point C has coordinates (0, 6), show that the triangle ABC is right-angled. [2]

11 0606/22/F/M/16 © UCLES 2016 [Turn over 9 P 1 m x m Q T R rad S i PQRS is a quadrilateral with PS parallel to QR. The perimeter of PQRS is 3 m. The length of PQ is 1 m and the length of PS is x m. The point T is on QR such that ST is parallel to PQ. Angle SRT is q radians. (i) Find an expression for x in terms of q. [3] (ii) Show that the area, A m2, of PQRS is given by cosec A 1 2 i = - . [2] (iii) Hence find the exact value of q when A 1 3 3 = - c m m2. [2]

12 0606/22/F/M/16 © UCLES 2016 10 (a) The vectors p and q are such that p = 11i − 24j and q = 2i + aj. (i) Find the value of each of the constants a and b such that p + 2q = (a + b)i − 20j. [3] (ii) Using the values of a and b found in part (i), find the unit vector in the direction p + 2q. [2]

13 0606/22/F/M/16 © UCLES 2016 [Turn over (b) O B C A a b The points A and B have position vectors a and b with respect to an origin O. The point C lies on AB and is such that AB : AC is 1 : l. Find an expression for OC in terms of a, b and l. [3] (c) The points S and T have position vectors s and t with respect to an origin O. The points O, S and T do not lie in a straight line. Given that the vector 2s + mt is parallel to the vector (m + 3)s + 9t, where m is a positive constant, find the value of m. [3]

14 0606/22/F/M/16 © UCLES 2016 11 A curve has equation y x x 1 2 = + . (i) Find the coordinates of the stationary points of the curve. [5]

15 0606/22/F/M/16 © UCLES 2016 [Turn over (ii) Show that x y x px qx 1 d d 2 2 2 3 3 = + + ^ h , where p and q are integers to be found, and determine the nature of the stationary points of the curve. [5] Question 12 is printed on the next page.

16 0606/22/F/M/16 © UCLES 2016 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. 12 A particle P is projected from the origin O so that it moves in a straight line. At time t seconds after projection, the velocity of the particle, v ms−1, is given by v t t 9 63 90 2 = - + . (i) Show that P first comes to instantaneous rest when t = 2. [2] (ii) Find the acceleration of P when t = 3.5. [2] (iii) Find an expression for the displacement of P from O at time t seconds. [3] (iv) Find the distance travelled by P (a) in the first 2 seconds, [2] (b) in the first 3 seconds. [2]

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the March 2016 series 0606 ADDITIONAL MATHEMATICS 0606/22 Paper 22, maximum raw mark 80 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 March 2016 series for most Cambridge IGCSE® and Cambridge International A and AS Level components.

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Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2016 0606 22 © Cambridge International Examinations 2016 Question Answer Marks Guidance 4 (a) a = 10 b = 6 c = 4 or 10cos6 4 x + B2,1,0 for B1 allow correct FT of c from a e.g. their c = 14 – their a (b) B3,2,1,0 Correct shape; two cycles; both maximum at 1 and minimum at –5; starting at (0, –2) and ending at (180, –2) 5 (i) 2 2 2187 5103 5103 kx k x + + B3 1 for each term; ignore extra terms (ii) 2 2(5103 ) 5103 k k = M1 must not include x, x2 k = 2 A1 A0 if k = 0 also given as a solution 6 5 3 1 3 3 6 2 3 x − = + + oe soi M1 4 14 3 ( ) 6 2 3 x −+ = + oe M1 ( ) 4 14 3 6 2 3 6 2 3 6 2 3 x −+ − = × + − M1 p = −27, q = 23 isw A1 + A1 allow 27 23 3 ( ) 6 x − + = 45 90 135 180 O -2 1 -5 y x ° ° ° °

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2016 0606 22 © Cambridge International Examinations 2016 Question Answer Marks Guidance 7 (a) 4 6 8 2 0 4     −   − 18 3 6 21 6 3     −   M1 for attempt to multiply and subtract 14 3 2 23 6 1 −     −   A1 (b) (i) 1 0 1 4 2 2   −   − −   oe B1 + B1 1 mark for 1 2   −     and 1 mark for 1 0 4 2     − −   k (ii) Valid method M1 XD–1D = CD 8 6 13 7 − −       A2,1,0 −1 each error If M0 then SC1 for DC = 4 3 14 5     − −   8 (i) Eliminate x (or y) M1 ( ) ( ) 2 2 3 2 2 2 2 12 y y y y − + − − = 2 2 2 2 3 12 2 2 x x x x + +     + − =         2 13 26 0 y y − = or 2 13 13 0 4 x − = oe A1 13y(y – 2) or x2 = 4 M1 x = −2, x = 2 y = 0 y = 2 isw A1 + A1FT or for (−2, 0) or (2, 2) from correct working FT their x or y values to find their y or x values; or A1 for (−2, 0) and (2, 2) (ii) their 1 2 AB m = or their 2 BC m = − soi M1 may be unsimplified or Pythagoras’ theorem correctly applied to their (0, −2), their (2, 2) and (0, 6) use of ( ) ( ) 1 AB BC m m × = − and conclusion A1 or use of h2 = a2 + b2 and conclusion

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2016 0606 22 © Cambridge International Examinations 2016 Question Answer Marks Guidance 9 (i) 1 tan RT θ = B1 or cot RT θ = 1 sin RS θ = B1 or cosec RS θ = 1 1 1 2tan 2sin x θ θ = − − oe or cot cosec 1 2 2 x θ θ = − − oe B1FT FT their RT and their RS, provided both are functions of trig ratios (ii) 1 cot 2 A x θ = + oe soi M1 correct completion to given answer cosec 1 2 A θ = − A1 (iii) 2 3 cosec 3 θ = oe M1 equivalent must be exact π 3 θ = cao A1 implies M1 10 (a) (i) (α + β)i – 20j = 15i + (2α − 24)j M1 implied by α + β = 15 or 2α − 24 = −20 α = 2 A1 β = 13 A1 (ii) ( )2 2 ( ) 20 their their α β + + − oe M1 15 – 2 2 0 5 i j oe A1FT FT their α + β provided non- zero (b) OC OA AB λ = +








 or (1 ) OC OB BA λ = + −





 B1 [ ] OC =


 a + λ( b – a) or [ ] OC =


 b + (1 − λ)( a – b) M1 [ ] OC =


 (1 − λ)a + λ b A1 (c) 2 3 9 µ µ = + M1 or multiplies one of the vectors by a general scale factor and finds a pair of simultaneous equations to solve Solves 2 3 18 0 µ µ + − = M1 or solves their correct equation to find their scale factor and attempts to use it to find µ µ = 3 A1 A0 if −6 not discarded

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2016 0606 22 © Cambridge International Examinations 2016 Question Answer Marks Guidance 11 (i) ( )( ) ( ) ( ) 2 2 2 1 1 (2 ) d d 1 x x x y x x + − = + oe M1* A1 Attempts to differentiate using the quotient rule correct; allow unsimplified 2 (1 ) 0 their x − = M1 dep* x = 1, x = −1 A1 from correct working only y = 0.5 , y = −0.5 oe A1 from correct working only or A1 for each of (1, 0.5), (−1, −0.5) oe from correct working; unsupported answers do not score (ii) ( ) ( ) ( )( ) 2 2 2 d 1 2 1 2 d x x x x + = + soi B1 ( ) 4 2 3 d 2 1 4 4 d x x x x x + + = + ( )( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 4 2 1 2 1 2 2 d 1 d 1 x their x their x x y x x x + − − − = + + M1 Applies quotient rule and factors out Correct completion to given answer ( ) 2 3 2 3 2 d 2 6 d 1 y x x x x − = + A1 When x = 1 ( ) 2 3 2 3 2 1 d 2(1) 6(1) d 1 1 x y their x = − = + oe < 0 therefore maximum B1FT Complete method including comparison to 0; FT their first or second derivative When x = −1 ( ) 2 3 2 3 2 1 d 2( 1) 6( 1) d ( 1) 1 x y their x =− − − − = − + oe > 0 therefore minimum B1FT Complete method including comparison to 0; FT their first or second derivative

Page 7 Mark Scheme Syllabus Paper Cambridge IGCSE – March 2016 0606 22 © Cambridge International Examinations 2016 Question Answer Marks Guidance 12 (i) 2 9 63 90 0 t t − + = ( )( ) 9 18 5 t t − − M1 showing that t = 2 is smaller value of t A1 must see evidence of solving e.g. t = 5 and t = 2 or factors (ii) d ( ) d v a t = attempted M1 18(3.5) – 63 = 0 cao A1 (iii) ( ) 2 9 63 90 d t t t − + ∫ M1 3 2 9 63 ( ) 90 3 2 t t s t = − + isw A2,1,0 −1 for each error or for +c left in (iv) (a) 3 2 9(2) 63(2) ( ) 90(2) 3 2 s = − + M1 or 2 3 2 0 9 63 90 3 2 t t t   − +     FT their (iii) 78 [m] A1 (b) 3 2 9(3) 63(3) ( ) 90(3) 67.5 3 2 s = − + = M1 FT their (iii) their 78 + 10.5 = 88.5 [m] A1FT