Cambridge IGCSE Mathematics - Additional 0606 — 2023 Oct/Nov Paper 1 · Variant 3

This document has 16 pages. [Turn over Cambridge IGCSE™ DC (CE/SW) 317872/2 © UCLES 2023 ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2023 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 [ ]. * 9 4 3 5 3 7 8 1 2 6 *

2 0606/13/O/N/23 © UCLES 2023 Mathematical Formulae 1. ALGEBRA Quadratic Equation For the equation ax bx c 0 2 + + = , x a b b ac 2 4 2 ! = - - Binomial Theorem ( ) a b a a b a b a b n n n r b 1 2 n n n n n r r n 1 2 2 f f + = + + + + + + - - - e e e o o o where n is a positive integer and ( )! ! ! n r n r r n = - e o Arithmetic series ( ) u a n d 1 n = + - ( ) { ( ) } S n a l n a n d 2 1 2 1 2 1 n = + = + - Geometric series u ar n n 1 = - ( ) ( ) S r a r r 1 1 1 n n ! = - - ( ) S r a r 1 1 1 = - 3 2. TRIGONOMETRY Identities sin cos A A 1 2 2 + = sec tan A A 1 2 2 = + ec cos cot A A 1 2 2 = + Formulae for ∆ABC sin sin sin A a B b C c = = cos a b c bc A 2 2 2 2 = + - sin bc A 2 1 T =

3 0606/13/O/N/23 © UCLES 2023 [Turn over 1 (a) On the axes, sketch the graphs of y x 2 5 = + and y x 4 3 = - , stating the intercepts with the coordinate axes. [3] y x O (b) Solve the inequality . x x 4 3 2 5 1 - + [3]

4 0606/13/O/N/23 © UCLES 2023 2 The perpendicular bisector of the line joining the points ,3 3 2 - e o and ,6 3 7 - b l passes through the point (2, k). Find the value of k. [4]

5 0606/13/O/N/23 © UCLES 2023 [Turn over 3 On the axes, draw the graph of sin y x 2 3 1 = - for ° °. x 360 360 G G - [4] y x 0 – 360° – 270° – 180° – 90° 1 – 1 – 2 – 3 – 4 2 3 4 90° 180° 270° 360°

6 0606/13/O/N/23 © UCLES 2023 4 The polynomial P is given by ( )x ax bx x 3 2 P 3 2 = + + + , where a and b are integers. ( )x P has a factor of . x 2 1 + ( )x P has a remainder of 6 - when divided by . x 1 + (a) Find the values of a and b. [5] (b) Show that the equation ( )x 0 P = has only one real root. [3]

7 0606/13/O/N/23 © UCLES 2023 [Turn over 5 (a) A 5-character password is to be formed from the following 10 characters. Letters A B C X Y Z Symbols * $ # & No character can be used more than once in any 5-character password. (i) Find the number of passwords that can be formed. [1] (ii) Find the number of passwords that can be formed if the password has to contain at least one symbol. [2] (iii) Find the number of passwords that can be formed if the password has to start with two letters and end with two symbols. [2] (b) A team of 8 people is to be chosen from 5 doctors, 4 teachers and 6 police officers. Find how many possible teams have the same number of doctors as teachers. [5]

8 0606/13/O/N/23 © UCLES 2023 6 The polynomial ( )x q is given by ( ) ( )( ) . x x x 3 1 2 1 3 q 2 =- - + (a) Find the x-coordinates of the stationary points on the curve ( ) y x q = . [4] (b) On the axes, sketch the graph of ( ) y x q = stating the intercepts with the coordinate axes. [3] y x O

9 0606/13/O/N/23 © UCLES 2023 [Turn over (c) Find the values of k such that ( ) q x k = has exactly one solution. [3] 7 Solve the equation . x x 6 2 1 0 3 1 3 1 - - = - Give your answers in exact form. [4]

10 0606/13/O/N/23 © UCLES 2023 8 The first three terms, in descending powers of x, in the expansion of x x 2 4 1 n 2 - e o can be written in the form x ax bx 256 c 16 13 + + , where n, a, b and c are integers. Find the values of n, a, b and c. [6]

11 0606/13/O/N/23 © UCLES 2023 [Turn over 9 Given that ( ) ( ) y x x 1 5 2 2 3 1 = - + , show that x y d d can be written in the form ( ) ( ) ( ) x x Ax B 3 5 2 1 3 3 2 + - - + , where A and B are integers. [5]

12 0606/13/O/N/23 © UCLES 2023 10 In this question, all lengths are in centimetres and all angles are in radians. C O A 20 B i The diagram shows the sector, OAB, of a circle with centre O and radius 20. The perimeter of this sector is 65. The lines CA and CB are both tangents to the circle at the points A and B, so that the triangle ABC is isosceles, with AC CB = . The angle AOB is equal to i. Find the area of the shaded region. [9]

13 0606/13/O/N/23 © UCLES 2023 [Turn over Additional working space for question 10.

14 0606/13/O/N/23 © UCLES 2023 11 A O B Z Y X In the triangle OAB, OA a = and OB b = . The straight line XYZ is such that: • OX b 5 4 = • AY AB 3 1 = • AZ a n = , where n is a constant • YZ XY m = , where m is a constant. (a) Show that . XY a b 3 2 15 7 = - [3]

15 0606/13/O/N/23 © UCLES 2023 [Turn over (b) Find YZ in terms of m, a and b. [1] (c) Find YZ in terms of n, a and b. [2] (d) Hence find the values of m and n, [3] Question 12 is printed on the next page.

16 0606/13/O/N/23 © UCLES 2023 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 Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. 12 Solve the equation r , cosec x 3 3 2 3 4 2 - = e o for . x 0 3 1 G r Give your answers in terms of r. [5]

This document consists of 13 printed pages. © UCLES 2023 [Turn over Cambridge IGCSE™ ADDITIONAL MATHEMATICS 0606/13 Paper 1 October/November 2023 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 International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the October/November 2023 series for most Cambridge IGCSE, Cambridge International A and AS Level components, and some Cambridge O Level components.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 2 of 13 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.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 3 of 13 Mathematics-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, non-integer 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 or sign 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 A or B 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

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 4 of 13 Question Answer Marks Guidance 1(a) 3 B1 for a V shaped graph with a vertex on the positive x-axis. B1 for 0.75 and 3 marked correctly and dependent on first B1 B1 for a straight line passing through –2.5 and 5 marked correctly, axis with a gradient such that there are two points of intersection. The second point of intersection may be implied. 1(b) 4 2 5 4 3 so − +   x x x B1 1 4 3 so 3 2 5 + − + − x x x B1 nfww 1 4 3 −   x B1 Dependent on both B1 SC2 for the values 1 3 − and 4 without any or with wrong inequality signs nfww Alternative 2 3 11 4 0 or = 0 − −  x x (M1) For squaring each side of the inequality and forming a 3-term quadratic. Allow multiples. 1 3 − , 4 (A1) Critical values 1 4 3 −   x (A1) 2 Mid-point 3 5 , 2 6   −     B1 Do not allow if unsimplified Gradient = 1 3 − B1 Allow unsimplified 5 3 3 2 6 2   + =  −     k oe M1 For attempt at perpendicular bisector Must be with their perpendicular gradient and their mid-point 2 3 = k A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 5 of 13 Question Answer Marks Guidance 3 4 B1 for a correct shape, starting in approximately correct places between 2 and 3 − − and finishing in approximately correct places between 0 and 1, having an amplitude of 2 and crossing the x- axis only once, on the positive x- axis. B1 for a correct shape and (0, −1) B1 for a correct shape and a max and a min in approximately correct places. ( ) ( ) 270 ,1 and 270 , 3 − − B1 for a correct shape crosses at ( ) 90 ,0 4(a) 1 3 P : 2 0 2 8 4 2   − − + − + =     a b M1 Allow one arithmetic error. Must be equated to 0 soi Allow unsimplified P( 1): 3 2 6 − −+ −+ = − a b M1 Allow unsimplified Must be equated to−6 soi −a + 2b + 4 = 0 oe −a + b + 5 = 0 oe A1 For both allow unsimplified a = 6, b = 1 2 M1 dep on at least one previous M1 for attempt to solve their simultaneous equations to find at least one of their unknowns. A1 for both. 4(b) 2 ) 2 1 3 ( )( 2 + − + x x x 2 M1 for correct attempt to obtain 2 terms of the quadratic for their P(x). Must divide by (2x + 1) A1 for correct quadratic 2 (3 ) 2 − + x x 2 For 3 2, the dicriminant is 0 1 so only one real root of oe 2 − +  − x x B1 Must have a valid attempt to evaluate the discriminant. 5(a)(i) 30 240 B1 5(a)(ii) 720 B1 the number of passwords with no symbols. Not part of a product 29 520 B1 5(a)(iii) (6 5) 6(4 3) = 2160 oe 2 B1 for (6 5) and (4 3) soi

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 6 of 13 Question Answer Marks Guidance 5(b) 1 of each and 6 police officers = 20 B1 For 4 5 6 1 1 6   C C C must be evaluated, could be implied by a correct total 2 of each and 4 police officers = 900 B1 For 4 5 6 2 2 4   C C C must be evaluated, could be implied by a correct total 3 of each and 2 police officers = 600 B1 For 4 5 6 3 3 2   C C C must be evaluated, could be implied by a correct total 4 of each and no police officers = 5 B1 For 4 5 4 4  C C must be evaluated, could be implied by a correct total Total = 1525 B1 6(a) 2 1 q'( ) (2(2 1)( 3) 2( 3) )oe 3 = − − + + + x x x x 2 M1 for attempt to differentiate, allow one arithmetic slip. A1 – allow unsimplified. 2 2 q'( ) (3 11 6) 0 3   = − + + =     x x x 2 3 and 3 = − = − x x 2 Dep M1 for equating their q(x)to zero and attempt to solve their 3- term quadratic to get two solutions for x = … A1 for both x values correct nfww 6(b) 3 B1 for correct cubic shape with maximum point in correct quadrant. B1 for correct cubic shape touching at (−3, 0) and passing through (0.5, 0), intercepts must be marked. B1 for correct cubic shape passing through (0, 3) intercept must be marked. 6(c) 0  k B1 Condone 0  y 2 343 , or 4.23 3 81 = − = = x y y M1 For finding the value of y at their max point. If incorrect must see substitution of their 2 3 = − x nfww 343or 4.23 81   k k A1 Condone 343or 4.23 81   y y

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 7 of 13 Question Answer Marks Guidance 7 2 1 1 3 3 6 2 0   − − =       x x or 1 2 3 6 2 0 where oe m m m x − − = = B1 1 1 3 3 2 1 = , = oe 3 2 x x − M1 For attempt to solve 3-term quadratic equation in the form 2 6 2 0   = m m and obtain 1 3 =… or ... = x m from correct work only 8 1 = , = 27 8 − x x 2 Dep M1 for dealing with the power of 1 3 correctly at least once. A1 for both 8 ( ) 2 16 2 256 soi 8 x x n = = B1 ( ) 7 2 13 8 1 2 oe 1 4 leading to 256 x ax x a    − =       = − 2 M1 for attempt at 2nd term with their n to find a, need to see one step to evaluate. Allow a sign error in simplifying but not missing in 1 4 −x ( ) 2 6 2 8 1 2 2 4 leading to 112 c x bx x b    − =       = 2 M1 for attempt at 3rd term with their n to find b, need to see one step to evaluate. Allow a sign error but not missing in 2 1 4   −     x 10 = c B1 Can be seen by observation.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 8 of 13 Question Answer Marks Guidance 9 2 1 2 3 3 4 5 (5 2) ( 1) 2(5 2) ( 1) d 3 d ( 1) − +  − − + −   =   −   x x x x y x x or 2 1 2 3 3 3 d 5 (5 2) ( 1) (5 2) d 3 2 ( 1) y x x x x x − − −   = = +  − + +     − − 3 B1 for 2 3 5 (5 2) 3 − + x M1 for attempt at differentiation of a quotient or product. A1 all other terms correct. 2 3 3 (5 2) (5 5 30 12) 3( 1) − + −− − − x x x x M1 Dep M1 for attempt to simplify by factorising 2 3 (5 2) or ( 1) − + − x x nfww to the given form, allow one arithmetic slip and/or one sign slip. 2 3 3 (25 17) 3( 1) (5 2) − + − + x x x A1 Must be in correct form.

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 9 of 13 Question Answer Marks Guidance 10 40 20 65  + = *M1 1.25 = A1 1 2 sin 2 20   =     AB their 23.4 = AB or 1 2 AB = 11.7 2 Dep M1 for an attempt to find AB or 1 2 AB Either height of triangle tan 1 2 2   =     their ACB their AB Height of triangle = 8.44 Area of triangle = 98.8 3 DepM1 for a correct attempt to find the height of the triangle M1 for attempt to find the area of the triangle using their height and their AB A1 must be at least 3 significant figures. Or 1 2 cos 2 their AB their AC   =     14.4 AC = 1 Area of triangle sin 2 2    =        their AB their AC Area of triangle = 98.8 (3) DepM1 for a correct attempt to find CA M1 for attempt to find the area of the triangle using the sine rule with their CA. A1 must be at least 3 significant figures. ( ) 2 1 Area of the segment = 20 1.25 sin1.25 2 Area of the segment = 60.2   − B1 Area = 38.6 A1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 10 of 13 Question Answer Marks Guidance 10 Alternative 1 40 20 65  + = (*M1) 1.25 = (A1) tan oe soi 2 20   =     their AC AC =14.43 (2) DepM1 for a correct attempt to find the AC Area of triangle ACO = 1 20 14.43 144.3 2   = (2) M1 for a correct attempt to find the area of the triangle using their AC Area of the sector = 250 (B1) Area of half shaded region = (144.3 −125) 2 (M1) Dependent on a valid method for finding triangle ACO . Allow use of 144 Area = 38.6 (A1) Alternative 2 40 20 65  + = (*M1) 1.25 = (A1) 1 2 sin 2 20   =     AB their 23.4 = AB (2) Dep M1 for an attempt to find AB or 1 2 AB tan oe soi 2 20   =     their AC AC =14.43 (2) DepM1 for a correct attempt to find AC Shaded area 2 1 5 14.4 20 20 2 4 =  −   Area = 38.6 (3) M1 for area of Kite B1 for area of sector

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 11 of 13 Question Answer Marks Guidance 10 Alternative 3 40 20 65  + = (*M1) 1.25 = (A1) tan oe soi 2 20   =     their AC AC =14.43 (2) DepM1 for a correct attempt to find AC Area of triangle AOB ( ) 2 1 20 sin 2 189.[7969...] their  =   = (M1) Area of triangle ACB 1 sin 2 2 98.8 their AC their AB  =    = (M1) for a correct attempt to find the area of the triangle using their AC and their AB Area of the sector = 250 (B1) Area of half shaded region = area of triangle ACB + area of triangle AOB − area of sector = 189.8 + 98.8 – 250 (M1) Area = 38.6 (A1)

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 12 of 13 Question Answer Marks Guidance 10 Alternative 4 40 20 65  + = (*M1) 1.25 = (A1) 1 2 sin 2 20   =     AB their 23.4 = AB or 1 2 AB = 11.7 (2) Dep M1 for an attempt to find AB or 1 2 AB height of triangle tan 1 2 2   =     their ACB their AB Height of triangle = 8.44 (M1) DepM1 for a correct attempt to find the height of the triangle Height of triangle ABO 2 2 1 20 2 16.22 AB   = −    = (M1) for a correct attempt to find to find the height of the triangle Area of the sector = 250 (B1) Area of kite 1 23.4 (16.22 8.44) 2 288.5 =   + = (M1) Area = 288.5 −250 =38.5 (A1) 11(a) 1 oe soi 3 → → → = − + + XY OX AB a or 2 oe soi 3 → → → = − XY XB AB M1 4 1( ) oe soi 5 3 → = − + + − XY b a b a or 1 2 oe soi 5 3 → = + − XY b (a b) M1 For 1( ) 3  − b a or 4 5  b For 1 5 b or 2 3  − (a b) 2 7 cao 3 15 XY → = − a b A1 AG 11(b) 2 7 3 15  →   = −     YZ a b cao B1

0606/13 Cambridge IGCSE – Mark Scheme PUBLISHED October/November 2023 © UCLES 2023 Page 13 of 13 Question Answer Marks Guidance 11(c) 1 oe soi 3  → → = − YZ AB a M1 1( ) oe 3  → = − − YZ a b a A1 Allow unsimplified ISW from correct answer 11(d) 1 2 7 ( ) = soi 3 3 15    − − −     a b a a b M1 For equating their (b) and their (c) and attempt to equate coefficients of a or b at least once. 5 7 = A1 nfww 1 7 = A1 nfww 12 2 π 3 sin 3 3 2 2 π or tan 3 3 3   − =        − =      x x B1 Allow if is missing 3π 5π π, , , 3π 2 2 x = 4 M1dep on B1 for obtaining 2 π π 3 3 3 − = x or any valid value A1 for one correct solution A1 for a 2nd correct solution A1 for a 3rd and 4th correct solutions and no extras in the range