Cambridge IGCSE Mathematics - Additional 0606 — 2016 Oct/Nov Paper 2 · Variant 3

This document consists of 16 printed pages. DC (LK/SG) 116670/2 © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education * 3 6 6 6 4 5 3 2 4 9 * ADDITIONAL MATHEMATICS 0606/23 Paper 2 October/November 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 0606/23/O/N/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 0606/23/O/N/16 © UCLES 2016 [Turn over 1 Without using a calculator, show that k 5 3 5 3 3 2 + + = - where k is an integer to be found. [3] 2 Solve the equation e e 6 x x 3 = . [3]

4 0606/23/O/N/16 © UCLES 2016 3 (i) Show that d d cos sin cos x x x x 1 1 1 + = + c m . [4] (ii) y x 3 2 1 0 1 2 y = 1 1 + cos x The diagram shows part of the graph of cos y x 1 1 = + . Use the result from part (i) to find the area enclosed by the graph and the lines , x x 0 2 = = and y 0 = . [2]

5 0606/23/O/N/16 © UCLES 2016 [Turn over 4 The cubic given by ( ) p x x ax bx 24 3 2 = + + - is divisible by x 2 - . When ( ) p x is divided by x 1 - the remainder is 20 - . (i) Form a pair of equations in a and b and solve them to find the value of a and of b. [4] (ii) Factorise ( ) p x completely and hence solve ( ) p x 0 = . [4]

6 0606/23/O/N/16 © UCLES 2016 5 In this question all lengths are in centimetres. A B C 60° In the triangle ABC shown above, AC 3 1 = + , BC 3 1 = - and angle ° ACB 60 = . (i) Without using a calculator, show that the length of AB 6 = . [3] (ii) Show that angle ° CAB 15 = . [2] (iii) Without using a calculator, find the area of triangle ABC. [2]

7 0606/23/O/N/16 © UCLES 2016 [Turn over 6 A curve has equation tan y x 7 = + . Find (i) the equation of the tangent to the curve at the point where x 4 r = , [4] (ii) the values of x between 0 and r radians for which d d x y y = . [4]

8 0606/23/O/N/16 © UCLES 2016 7 In this question all lengths are in metres. h 0.5h + 2 r A conical tent is to be made with height h , base radius r and slant height 0.5h + 2, as shown in the diagram. (i) Show that . r h h 2 4 0 75 2 2 = + - . [2]

9 0606/23/O/N/16 © UCLES 2016 [Turn over The volume of the tent, V, is given by r h 3 1 2 r . (ii) Given that h can vary find, correct to 2 decimal places, the value of h which gives a stationary value of V. [5] (iii) Determine the nature of this stationary value. [2]

10 0606/23/O/N/16 © UCLES 2016 8 10 cm 6 cm A C T B O The points A, B and C lie on a circle centre O, radius 6 cm. The tangents to the circle at A and C meet at the point T. The length of OT is 10 cm. Find (i) the angle TOA in radians, [2]

11 0606/23/O/N/16 © UCLES 2016 [Turn over (ii) the area of the region TABCT, [6] (iii) the perimeter of the region TABCT. [2]

12 0606/23/O/N/16 © UCLES 2016 9 In this question i is a unit vector due east and j is a unit vector due north. Units of length and velocity are metres and metres per second respectively. The initial position vectors of particles A and B, relative to a fixed point O, are i +5j and qi – 15j respectively. A and B start moving at the same time. A moves with velocity pi – 3j and B moves with velocity 3i – j. (i) Given that A travels with a speed of 5 ms–1, find the value of the positive constant p. [1] (ii) Find the direction of motion of B as a bearing correct to the nearest degree. [2] (iii) State the position vector of A after t seconds . [1] (iv) State the position vector of B after t seconds. [1]

13 0606/23/O/N/16 © UCLES 2016 [Turn over (v) Find the time taken until A and B meet. [2] (vi) Find the position vector of the point where A and B meet. [1] (vii) Find the value of the constant q. [1]

14 0606/23/O/N/16 © UCLES 2016 10 The functions f and g are defined for x 1 2 by ( ) f ln x x 2 = + , ( ) g e x 2 3 x = + . (i) Find ( ) fg x . [1] (ii) Find ( ) ff x . [1] (iii) Find ( ) g x 1 - . [2]

15 0606/23/O/N/16 © UCLES 2016 [Turn over (iv) Solve the equation ( ) f x 4 = . [1] (v) Solve the equation ( ) gf x 20 = . [4] Question 11 is printed on the next page.

16 0606/23/O/N/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. 11 It is given that A p q 2 3 = J L KK N P OO and that A A I 5 2 2 - = , where I is the identity matrix. (i) Find a relationship connecting the constants p and q. [4] (ii) Given that p and q are positive and that det A p 3 =- , find the value of p and of q. [4]

® IGCSE is the registered trademark of Cambridge International Examinations. This document consists of 6 printed pages. © UCLES 2016 [Turn over Cambridge International Examinations Cambridge International General Certificate of Secondary Education ADDITIONAL MATHEMATICS 0606/23 Paper 2 October/November 2016 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 October/November 2016 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 2016 0606 23 © UCLES 2016 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 Question Answer Mark Part Marks 1 ( ) ( ) ( ) ( ) 5 3 3 5 3 5 3 5 3 + − × + − = 5 3 15 15 9 5 3 + − − − 2 15 4 2 − = = 15 2 − M1 A1 A1 rationalise with ( ) 5 3 − numerator (3 or 4 terms) denominator and completion 2 lne3x = ln6ex 3x = ln6ex 3x = ln6 + lnex 3x = ln6 + x x = 1 2 ln6 or ln 6 or 0.896 M1 M1 A1 one law of indices/logs second law of indices/logs www oe in base 10 3 (i) ( ) ( ) 2 1+cos cos +sin sin d sin d 1+cos 1+cos x x x x x x x x  =     = ( ) 2 2 2 cos +cos +sin 1+cos x x x x ( ) 2 1+cos 1+cos x x = M1 A1 B1 A1 Quotient Rule (or Product Rule from (sinx)(1 + cosx)-1) correct unsimplified use of 2 2 sin +cos 1 x x = oe completion (ii) 2 2 0 0 1 sin d 1+cos 1+cos x x x x     =         ∫ awrt 1.56 M1 A1 correct integrand

Page 3 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 23 © UCLES 2016 Question Answer Mark Part Marks 4 (i) ( ) ( ) p 2 0 8 4 2 24 0 4 2 16 a b a b = → + + − = → + = ( ) ( ) p 1 20 1 24 20 3 a b a b = − →+ + − = − → + = 5 and 2 a b = = − B1 B1 M1 A1 solve their linear equations for a or b (ii) ( ) ( ) 3 2 2 p 5 2 24 2 ( 7 12) x x x x x x x = + − − = − + + ( )( )( ) 2 3 4 x x x = − + + ( ) p 0 2, 3, 4. x x = → = − − M1 A1 M1 A1 find quadratic factor correct quadratic factor soi factorise quadratic factor and write as product of 3 linear factors if 0 scored, SC2 for roots only 5 (i) ( ) ( ) ( )( ) 2 2 2 3 1 3 1 2 3 1 3 1 cos60 AB = + + − − + − 3 1 2 3 3 1 2 3 2 = + + + + − − 6 = M1 A1 A1 use cosine rule at least 7 terms correct completion AG (ii) sin sin60 3 1 6 A = − ( ) 3 1 sin60 6 2 sin 4 6 A − − = = oe or 0.259 or 0.2588… M1 A1 sine rule (or cosine rule) correct explicit expression for sinA AG (iii) Area = ( )( ) 1 3 1 3 1 sin60 2 + − = 3 2 M1 A1 correct substitution into 1 sin 2 ab C 6 (i) 2 d sec d y x x = 2 d sec 2 4 d 4 y x x π π = → = = 8 y = Equation of tangent 8 2 4 y x π − = − (4 – 2y = π – 16, y = 2x + 6.429…, π 4 = 0.7853… ) B1 B1 B1 B1 evaluated

Page 4 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 23 © UCLES 2016 Question Answer Mark Part Marks (ii) 2 sec tan 7 x x = + 2 tan x −tan 6 0 x − = oe (tanx – 3)(tanx + 2) 0 = tan 3 or x = tan 2 x =− 1.25, 2.03 x = M1 M1 A1A1 use 2 2 sec 1 tan x x = + to obtain a 3 term quadratic in tanx solve three term quadratic for tanx extras in range lose final A1 7 (i) ( ) 2 2 2 0.5 2 r h h + = + oe 2 2 2 0.25 2 4 r h h h = + + − 2 2 2 4 0.75 r h h = + − M1 A1 correct expansion and r2 subject and completion www AG (ii) 2 1 π 3 V r h = = ( ) 2 3 π 2 4 0.75 3 h h h + − ( ) 2 d π 4 4 2.25 d 3 V h h h = + − 2 d 0 2.25 4 4 0 d v h h h = → − − = 2.49 h = only B1 M1 A1 M1 A1 any correct form in terms of h only differentiate V correct differentiation equate to 0 and solve 3 term quadratic cao (iii) ( ) 2 2 d π 4 4.5 when 2.49 3 d V h h h = − = (–7.545…) < 0 so maximum M1 A1 differentiate their 3 term d d V h and substitute their h draw correct conclusion www 8 (i) 6 cos 10 TOA = → 0.927 TOA = M1 A1 any method (ii) area of major sector = ( ) 2 1 6 2π 2 0.927 2 their − × (= 79.7) area of half kite = ( ) 2 2 1 6 10 6 2 − (=24) area of kite × 2 (=48) complete correct plan awrt 128 M2 M1 DM1 DM1 A1 or M1for 1 2 62(2 × their 0.927) DM1 for π×62 – 1 2 62(2 × their 0.927) any method their major sector + their kite (iii) arc length = 2 2 6 (2π 2 0.927) 2 10 6 their × −× + × − ) awrt 42.6 M1 A1 complete correct method

Page 5 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 23 © UCLES 2016 Question Answer Mark Part Marks 9 (i) 4 p = B1 (ii) tanα = ± 1 3 or ±3 or 18.4° or 71.6° seen 108 M1 A1 could use cos or sin (iii) 1 5 3 their p t    = +    −    Ar B1 (iv) 3 15 1 q t     = +     − −     Br B1 (v) 5 3 15 t t − = − − 10 t →= M1 A1 rA = rB and equate y/j and solve for t (vi) 41 25     −   only B1 (vii) 11 q = only B1 10 (i) ( ) ( ) fg ln 2e 3 2 x x = + + B1 isw (ii) ( ) ( ) ff ln ln 2 2 x x = + + B1 isw (iii) 2e 3 y x = + y 3 e 2 x − = ( ) -1 3 g ln 2 x x −   =     oe M1 A1 change x and y and make ey the subject (iv) 2e or 7.39 B1 (v) ( ) ( ) ln 2 gf 2e 3 20 x x + = + = ln 2 2e e 3 20 x + = 2 2 e 17 x = 2 17 2e x = or 1.15 B1 M1 M1 A1 gf correct and equation set up correctly one law of indices/logs second law of indices/logs www if 0 scored, SC2 for 17.3…

Page 6 Mark Scheme Syllabus Paper Cambridge IGCSE – October/November 2016 0606 23 © UCLES 2016 Question Answer Mark Part Marks 11 (i) 2 2 2 4 2 3 3 3 2 3 9 q q pq q q p p p p pq + +      = =      + +      A 2 5 2 4 10 2 or 9 15 2 8 pq pq pq − = → + − = + − = → = A A I B2,1,0 M1 A1 –1 each error equate top left or bottom right elements accept 8 8 , p q q p = = (ii) det 6 pq = − A 6 3 pq p − = − and solve 2 3 p → = q = 12 B1 M1 A1 A1 their det A = –3p and use their pq = k oe to solve for p or q FT from their pq = k