Cambridge IGCSE Mathematics - Additional 0606 — 2015 Oct/Nov Paper 2 · Variant 3

This document consists of 16 printed pages. DC (KN/CGW) 98325/2 © UCLES 2015 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 1 2 9 6 1 1 8 2 2 1 * ADDITIONAL MATHEMATICS 0606/23 Paper 2 October/November 2015 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 0606/23/O/N/15 © UCLES 2015 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 0606/23/O/N/15 © UCLES 2015 [Turn over 1 Find the equation of the tangent to the curve y x x x 3 5 7 3 2 = + - - at the point where x = 2. [5]

4 0606/23/O/N/15 © UCLES 2015 2 Find the values of k for which the line y x k 2 2 = + + cuts the curve ( ) y x k x 2 2 8 2 = + + + in two distinct points. [6]

5 0606/23/O/N/15 © UCLES 2015 [Turn over 3 (a) Given that y x x 2 2 3 = - , find x y d d . [3] (b) Given that y x x 4 6 = + , show that ( ) x y x k x 4 6 1 d d = + + and state the value of k. [3]

6 0606/23/O/N/15 © UCLES 2015 4 Solve the following simultaneous equations, giving your answers for both x and y in the form a b 3 + , where a and b are integers. x y 2 9 + = x y 3 2 5 + = [5]

7 0606/23/O/N/15 © UCLES 2015 [Turn over 5 The roots of the equation x ax bx c 0 3 2 + + + = are 1, 3 and 3. Show that c 9 – = and find the value of a and of b. [4]

8 0606/23/O/N/15 © UCLES 2015 6 Solve the following equation. ( ) log log x 29 15 3 2 2 x 2 - = + [5]

9 0606/23/O/N/15 © UCLES 2015 [Turn over 7 The velocity, v ms 1 - , of a particle travelling in a straight line, t seconds after passing through a fixed point O, is given by ( ) v t 2 10 2 = + . (i) Find the acceleration of the particle when t = 3. [3] (ii) Explain why the particle never comes to rest. [1] (iii) Find an expression for the displacement of the particle from O after time t s. [3] (iv) Find the distance travelled by the particle between t = 3 and t = 8. [2]

10 0606/23/O/N/15 © UCLES 2015 8 (i) Prove that sec cosec sec cosec x x x x 2 2 2 2 + = . [4] (ii) Hence, or otherwise, solve sec cosec tan x x x 4 2 2 2 + = for x 90 270 1 1 c c. [4]

11 0606/23/O/N/15 © UCLES 2015 [Turn over 9 Given that ( )x x x 3 12 2 f 2 = + + , (i) find values of a, b and c such that ( ) ( ) x a x b c f 2 = + + , [3] (ii) state the minimum value of f(x) and the value of x at which it occurs, [2] (iii) solve y 1 0 f = c m , giving each answer for y correct to 2 decimal places. [3]

12 0606/23/O/N/15 © UCLES 2015 10 (i) Given that e x kx d d e x x 2 2 2 2 = - - ^ h , state the value of k. [1] (ii) Using your result from part (i), find x x 3 e d x 2 2 - y . [2] (iii) Hence find the area enclosed by the curve y x 3 e x 2 2 = - , the x-axis and the lines x 1 = and x 2 = . [2]

13 0606/23/O/N/15 © UCLES 2015 [Turn over (iv) Find the coordinates of the stationary points on the curve y x 3 e x 2 2 = - . [4]

14 0606/23/O/N/15 © UCLES 2015 11 The trees in a certain forest are dying because of an unknown virus. The number of trees, N, surviving t years after the onset of the virus is shown in the table below. t 1 2 3 4 5 6 N 2000 1300 890 590 395 260 The relationship between N and t is thought to be of the form N Ab t = - . (i) Transform this relationship into straight line form. [1] (ii) Using the given data, draw this straight line on the grid below. [3]

15 0606/23/O/N/15 © UCLES 2015 [Turn over (iii) Use your graph to estimate the value of A and of b. [3] If the trees continue to die in the same way, find (iv) the number of trees surviving after 10 years, [1] (v) the number of years taken until there are only 10 trees surviving. [2] Question 12 is printed on the next page.

16 0606/23/O/N/15 © UCLES 2015 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 plane that can travel at 250 kmh–1 in still air sets off on a bearing of 070°. A wind with speed w kmh–1 from the south blows the plane off course so that the plane actually travels on a bearing of 060°. Find, in kmh–1, the resultant speed V of the plane and the windspeed w. [5]

® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME for the October/November 2015 series 0606 ADDITIONAL MATHEMATICS 0606/23 Paper 2 , 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 October/November 2015 series for most Cambridge IGCSE®, Cambridge International A and AS Level components and some Cambridge O Level components.

Page 2 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 23 © Cambridge International Examinations 2015 Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied www without wrong working 1 3 2 2 3 5 7 d 3 6 5 d y x x x y x x x = + − − = + − d 2 19 d y x x = → = 3 = y eqn of tangent: 3 19 ( 19 35) 2 y y x x − = → = − − M1 A1 A1FT B1 A1FT Differentiate on their d d y x 2 2 2 2 2 ( 2) 8 x k x k x + + = + + + 2 2 6 ( 0) x kx k + + − = ) 6 ( 2 4 4 2 2 k k ac b − × − = − 2 8 48 ( 0) k k + − > ( 12)( 4) ( 0) k k + − > 12 − < k or 4 > k M1 A1 M1 DM1 A1 A1 eliminate y or x correct quadratic use discriminant attempt to solve 3 term quadratic 12 k = − and 4 k = 3 (a) 2 2 3 2 4 2 2 2 2 d (2 )3 ( 2 ) 6 d (2 ) (2 ) y x x x x x x x x x   − − − − = =   − −   M1 A2,1,0 For quotient rule (or product rule on correct y) (b) 5.0 5.0 ) 6 4 ( 4 ) 6 4 ( 2 1 d d + + × + × = − x x x x y 0.5 6( 1) 6 (4 6) x k x + = → = + M1 A1 A1 product rule 4 13 ) 3 4 ( = − x ( ) ( )( ) 13 4 3 4 3 4 3 4 3 x + = − + = + 3 2 1− = y M1 A1 M1 A1 A1 eliminate y or x simplified rationalisation

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 23 © Cambridge International Examinations 2015 5 0 )1 )( 3 )( 3 ( = − − − x x x 3 2 7 15 9 0 7 15 9 x x x a b c − + − = = − = = − M1 A1 A1 A1 AG for c 6 2 2 log 2 log 2 log x x = 2 2 2 log log 2 x x = 8 log 3 2 = 2 8 29 15 ( 0) (8 5)( 3) ( 0) x x x x − + = → − − = 8 5 = x or 3 = x B1 B1 B1 M1 A1 obtain quadratic and attempt to solve 7 (i) 3) 2 ( 20 + − = t a 3 0.16 t a = → = − m/s2 M1 A1 A1FT
  2
 oe 20 k = − (ii) 2) 2 ( 10 + t is never zero. B1 (iii) 5 2 10 + + − = t s M1 A1 A1 integrate 2 k t + 10 k = − +5 (iv) 8 3 10 1 2 2 1 s t   = − = −+   +   = M1 A1 insert limits and subtract

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 23 © Cambridge International Examinations 2015 8 (i) 2 2 2 2 2 2 2 2 2 2 2 2 1 1 sec cosec cos sin sin cos sin cos 1 sin cos sec cosec x x x x x x x x x x x x + = + + = = = B1 B1 B1 B1 add fractions use of 2 2 sin cos 1 x x + = fully correct solution (ii) 2 2 2 2 2 1 sin 4 cos sin cos 4sin 1 1 sin 2 135 , 225 x x x x x x x = → = = ± = o o M1 A1 A1, A1 correct simplified equation 9 (i) 2 2 f( ) 3 12 2 3( 2) 10 x x x x = + + = + − a = 3 b = 2 c = –10 B1 B1 B1 (ii) minimum f ( ) 10 x = − at 2 − = x B1FT B1FT (iii) 1 1 10 f 0 ( ) 2 3 y y     = → = ± −         5.74, 0.26 y = − − M1 A1, A1 obtain explicit expression for 1 y or y

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 23 © Cambridge International Examinations 2015 10 (i) 2 2 2 2 d (e ) 2 e d x x x x − − = − B1 k = –2 (ii) 2 2 3e 2 x c − − + M1 A1FT 2 2e x D − 2 3 − = D or 3 k (iii) 2 2 2 1 3e 3 3 e 2 2 2 x −   − = − +       2.58 M1 A1 insert limits on their (ii) and subtract (iv) 2 2 3 e x y x − = 2 2 2 2 d 3 ( 2 e ) 3e d x x y x x x − − = − + d 0 d y x = → 707 .0 2 1 ± = ± = x 1.5 3 e 9.51 2 y = ± = ± M1 A1 A1 A1 product rule both
or a pair both  11 (i) log log log N A t b = − B1 (ii) t 1 2 3 4 5 6 log N 3.30 3.11 2.95 2.77 2.60 2.41 ln N 7.60 7.17 6.79 6.38 5.98 5.56 M1 M1 A1 find logs of N plot log N or ln N against t or –t straight line passing through five points (iii) 2.415 3.3 gradient log 1.5 5 b b − = − = → = intercept = log A = 3.47 2950 A → = DM1 DM1 A1 set gradient = –log b and solve set intercept = log A and solve both values correct (iv) 10 2950 10 51 1.5 t N = → = = B1 (v) log295 10 1.5 295 log1.5 14years t N t = → = →= = M1 A1 substitute N = 10, their A, b into given or transformed equation 1 2 3 4 5 6 7 1 2 3 4 x y

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2015 0606 23 © Cambridge International Examinations 2015 12 0 250cos20 cos30 , , 250sin 20 sin30 p r w V v v v w V       = = =             o o o o 0 cos30 250cos20 sin30 250sin 20 r p w v v v V w V = +       = +             o o o o 250cos20 cos30 271km/hr V = = o o sin30 250sin 20 50.1km/hr w V = − = o o OR triangle with sides 250 V
opposite angles ° 60 ° 110 ° 10 250 sine rule: sin10 sin60 50.1km/hr 250 sin110 sin60 271km/hr w w V V = = = = o o o o B1 M1 A1 M1 A1 B1 M1 A1 M1 A1 equate  components and solve equate  components and solve apply to correct triangle and solve apply to correct triangle and solve