Cambridge IGCSE Mathematics (with coursework) 0581 — 2014 May/June Paper 4 · Variant 3

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 fl uid. DO NOT WRITE IN ANY BARCODES. Answer all questions. If working is needed for any question it must be shown below that question. Electronic calculators should be used. If the degree of accuracy is not specifi ed in the question, and if the answer is not exact, give the answer to three signifi cant fi gures. Give answers in degrees to one decimal place. For π, use either your calculator value or 3.142. 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 of the marks for this paper is 130. MATHEMATICS 0581/43 Paper 4 (Extended) May/June 2014 2 hours 30 minutes Candidates answer on the Question Paper. Additional Materials: Electronic calculator Geometrical instruments Tracing paper (optional) This document consists of 16 printed pages. [Turn over IB14 06_0581_43/FP © UCLES 2014 *3494048905* Cambridge International Examinations Cambridge International General Certifi cate of Secondary Education PAPA CAMBRIDGE

2 0581/43/M/J/14 © UCLES 2014 1 In July, a supermarket sold 45 981 bottles of fruit juice. (a) The cost of a bottle of fruit juice was $1.35 . Calculate the amount received from the sale of the 45 981 bottles. Give your answer correct to the nearest hundred dollars. Answer(a) $ … [2] (b) The number of bottles sold in July was 17% more than the number sold in January. Calculate the number of bottles sold in January. Answer(b) … [3] (c) There were 3 different fl avours of fruit juice. The number of bottles sold in each fl avour was in the ratio apple : orange : cherry = 3 : 4 : 2. The total number of bottles sold was 45 981. Calculate the number of bottles of orange juice sold. Answer(c) … [2] (d) One bottle contains 1.5 litres of fruit juice. Calculate the number of 330 ml glasses that can be fi lled completely from one bottle. Answer(d) … [3] (e) 9 5 of the 45 981 bottles are recycled. Calculate the number of bottles that are recycled. Answer(e) … [2] __________________________________________________________________________________________

3 0581/43/M/J/14 © UCLES 2014 [Turn over 2 4 3 2 1 10 20 30 Amount ($x) 40 50 60 0 Frequency density A survey asked 90 people how much money they gave to charity in one month. The histogram shows the results of the survey. (a) Complete the frequency table for the six columns in the histogram. Amount ($x) 0 < x Y 10 Frequency 4 [5] (b) Use your frequency table to calculate an estimate of the mean amount these 90 people gave to charity. Answer(b) $ … [4] __________________________________________________________________________________________

4 0581/43/M/J/14 © UCLES 2014 3 (a) P X Q R 17 cm 12 cm NOT TO SCALE The diagram shows triangle PQR with PQ = 12 cm and PR = 17 cm. The area of triangle PQR is 97 cm2 and angle QPR is acute. (i) Calculate angle QPR. Answer(a)(i) Angle QPR = … [3] (ii) The midpoint of PQ is X. Use the cosine rule to calculate the length of XR. Answer(a)(ii) XR = … cm [4]

5 0581/43/M/J/14 © UCLES 2014 [Turn over (b) 37° 42° a cm 9.4 cm NOT TO SCALE Calculate the value of a. Answer(b) a = … [4] (c) sin x = cos 40°, 0° Y x Y 180° Find the two values of x. Answer(c) x = … or x = … [2] __________________________________________________________________________________________

6 0581/43/M/J/14 © UCLES 2014 4 The table shows some values for the function y = 21 x + x , x ≠ 0. x –3 –2 –1 –0.5 0.5 1 2 3 4 y –2.89 –1.75 3.5 2 2.25 4.06 (a) Complete the table of values. [3] (b) On the grid, draw the graph of y = 21 x + x for –3 Y x Y – 0.5 and 0.5 Y x Y 4. y x 5 4 3 2 1 –1 –2 –3 0 –1 –2 –3 3 4 2 1 [5]

7 0581/43/M/J/14 © UCLES 2014 [Turn over (c) Use your graph to solve the equation 21 x + x – 3 = 0 . Answer(c) x = … or x = … or x = … [3] (d) Use your graph to solve the equation 21 x + x = 1 – x. Answer(d) x = … [3] (e) By drawing a suitable tangent, fi nd an estimate of the gradient of the curve at the point where x = 2. Answer(e) … [3] (f) Using algebra, show that you can use the graph at y = 0 to fi nd 1 3 - . Answer(f) [3] __________________________________________________________________________________________

8 0581/43/M/J/14 © UCLES 2014 5 (a) 5 4 3 2 1 0 1 2 3 4 5 6 7 8 x y A B (i) Write down the position vector of A. Answer(a)(i) f p [1] (ii) Find ì ì , the magnitude of . Answer(a)(ii) … [2] (b) p q O Q S P R NOT TO SCALE O is the origin, = p and = q. OP is extended to R so that OP = PR. OQ is extended to S so that OQ = QS. (i) Write down in terms of p and q. Answer(b)(i) = … [1] (ii) PS and RQ intersect at M and RM = 2MQ. Use vectors to fi nd the ratio PM : PS, showing all your working. Answer(b)(ii) PM : PS = … : … [4] __________________________________________________________________________________________

9 0581/43/M/J/14 © UCLES 2014 [Turn over 6 In this question, give all your answers as fractions. N A T I O N The letters of the word NATION are printed on 6 cards. (a) A card is chosen at random. Write down the probability that (i) it has the letter T printed on it, Answer(a)(i) … [1] (ii) it does not have the letter N printed on it, Answer(a)(ii) … [1] (iii) the letter printed on it has no lines of symmetry. Answer(a)(iii) … [1] (b) Lara chooses a card at random, replaces it, then chooses a card again. Calculate the probability that only one of the cards she chooses has the letter N printed on it. Answer(b) … [3] (c) Jacob chooses a card at random and does not replace it. He continues until he chooses a card with the letter N printed on it. Find the probability that this happens when he chooses the 4th card. Answer(c) … [3] __________________________________________________________________________________________

10 0581/43/M/J/14 © UCLES 2014 7 (a) Y X A B F E D C x° x° 32° t ° p° q° NOT TO SCALE ABCDEF is a hexagon. AB is parallel to ED and BC is parallel to FE. YFE and YABX are straight lines. Angle CBX = 32° and angle EFA = 90°. Calculate the value of (i) p, Answer(a)(i) p = … [1] (ii) q, Answer(a)(ii) q = … [2] (iii) t, Answer(a)(iii) t = … [1] (iv) x. Answer(a)(iv) x = … [3]

11 0581/43/M/J/14 © UCLES 2014 [Turn over (b) 63° y° x° R Q S T U P NOT TO SCALE P, Q, R and S are points on a circle and PS = SQ. PR is a diameter and TPU is the tangent to the circle at P. Angle SPT = 63°. Find the value of (i) x, Answer(b)(i) x = … [2] (ii) y. Answer(b)(ii) y = … [2] __________________________________________________________________________________________

12 0581/43/M/J/14 © UCLES 2014 8 (a) (i) Show that the equation 4 7 x + + 2 2 3 x - = 1 can be simplifi ed to 2x2 + 3x – 6 = 0 . Answer(a)(i) [3] (ii) Solve the equation 2x2 + 3x – 6 = 0 . Show all your working and give your answers correct to 2 decimal places. Answer(a)(ii) x = … or x = … [4] (b) The total surface area of a cone with radius x and slant height 3x is equal to the area of a circle with radius r. Show that r = 2x. [The curved surface area, A, of a cone with radius r and slant height l is A = πrl.] Answer(b) [4] __________________________________________________________________________________________

13 0581/43/M/J/14 © UCLES 2014 [Turn over 9 f(x) = 4 – 3x g(x) = 3–x (a) Find f(2x) in terms of x. Answer(a) f(2x) = … [1] (b) Find ff(x) in its simplest form. Answer(b) ff(x) = … [2] (c) Work out gg(–1). Give your answer as a fraction. Answer(c) … [3] (d) Find f –1(x), the inverse of f(x). Answer(d) f –1(x) = … [2] (e) Solve the equation gf(x) = 1. Answer(e) x = … [3] __________________________________________________________________________________________

14 0581/43/M/J/14 © UCLES 2014 10 (a) 8 cm r cm NOT TO SCALE The three sides of an equilateral triangle are tangents to a circle of radius r cm. The sides of the triangle are 8 cm long. Calculate the value of r. Show that it rounds to 2.3, correct to 1 decimal place. Answer(a) [3] (b) 8 cm 12 cm NOT TO SCALE The diagram shows a box in the shape of a triangular prism of height 12 cm. The cross section is an equilateral triangle of side 8 cm. Calculate the volume of the box. Answer(b) … cm3 [4]

15 0581/43/M/J/14 © UCLES 2014 [Turn over (c) The box contains biscuits. Each biscuit is a cylinder of radius 2.3 centimetres and height 4 millimetres. Calculate (i) the largest number of biscuits that can be placed in the box, Answer(c)(i) … [3] (ii) the volume of one biscuit in cubic centimetres, Answer(c)(ii) … cm3 [2] (iii) the percentage of the volume of the box not fi lled with biscuits. Answer(c)(iii) … % [3] __________________________________________________________________________________________ Question 11 is printed on the next page.

16 0581/43/M/J/14 © UCLES 2014 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. 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 Diagram 1 Diagram 2 Diagram 3 The fi rst three diagrams in a sequence are shown above. Diagram 1 shows an equilateral triangle with sides of length 1 unit. In Diagram 2, there are 4 triangles with sides of length 2 1 unit. In Diagram 3, there are 16 triangles with sides of length 4 1 unit. (a) Complete this table for Diagrams 4, 5, 6 and n. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Diagram n Length of side 1 2 1 4 1 Length of side as a power of 2 20 2–1 2–2 [6] (b) (i) Complete this table for the number of the smallest triangles in Diagrams 4, 5 and 6. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of smallest triangles 1 4 16 Number of smallest triangles as a power of 2 20 22 24 [2] (ii) Find the number of the smallest triangles in Diagram n, giving your answer as a power of 2. Answer(b)(ii) … [1] (c) Calculate the number of the smallest triangles in the diagram where the smallest triangles have sides of length 128 1 unit. Answer(c) … [2]

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the May/June 2014 series 0581 MATHEMATICS 0581/43 Paper 4 (Extended), maximum raw mark 130 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 2014 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. PAPA CAMBRIDGE

Page 2 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Abbreviations cao correct answer only dep dependent FT follow through after error isw ignore subsequent working oe or equivalent SC Special Case nfww not from wrong working soi seen or implied Qu Answers Mark Part Marks 1 (a) 62100[.00] Final answer 2 B1 for 62 074[. 35] or 62 070 (b) 39300 3 M2 for 45 981÷ 1.17 oe or M1 for 45 981 associated with 117 [%] (c) 20436 2 M1 for 45 981÷ (3+4+2) or 45 981 × 4 (d) 4 3 M2 for 330 1000 5.1 × oe or M1 for figs 4545... or 455 (e) 25545 2 M1 for 45 981 × 9 5 2 (a) 10 < x ≤ 25 25 < x ≤ 30 30 < x ≤ 35 35 < x ≤ 50 50 < x ≤ 60 13 33 19 [4] 15 6 2 3 5 correct B1 for 3 or 4 correct or SC1 for all correct but in the form 10 to 25 or 10 − 25 B2 for 4 correct or B1 for 3 correct (b) 25.1[0] or 25.13 to 25.14 nfww 4 M1 for mid-values soi, condone one error or omission 5 17.5 27.5 32.5 42.5 55 soi and M1 for ∑fx for any x in intervals including boundaries, but all fs must be integers, condone one further error or omission and M1 dep for ∑fx ÷ 90 Dep on 2nd M mark earned

Page 3 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Qu Answers Mark Part Marks 3 (a) (i) 72[.0] or 71.98 to 71.99 nfww 3 M2 for [sin P = ] 17 12 97 2 1 × × oe or M1 for implicit version (ii) 16.2 or 16.18 to 16.19 nfww 4 M2 for ) 72 ( cos 17 6 2 17 6 2 2 their × × × − + or M1 for implicit form and A1 for [XR2 =] 261.8 to 262 (b) 7.61 or 7.612... nfww 4 M3 for [a =] 9.4 × sin 37 ÷ cos 42 oe or [a = ] 9.4sin37/sin(90–42) or M2 for [a =] their height ÷ cos 42 oe or ) 42 90 sin( 4.9 37 sin − = a oe or M1 for their height ÷ a = cos 42 or for [their height = ] 9.4 × sin 37 oe or B1 for 48° correctly used or seen in correct position on diagram (c) 50 130 1 1 4 (a) 0, 4.5, 3.11[1…] 3 B1, B1, B1 (b) Complete correct curve with minimum below y = 2 -2 -1 1 2 3 -2 -1 1 2 3 4 x y 5 B3 FT for 9 points correctly plotted B2 FT for 7 or 8 points correctly plotted or B1 FT 5 or 6 points correctly plotted and B1 indep two separate branches not touching or cutting y-axis (c) − 0.5 to − 0.6 0.6 to 0.7 2.8 to 2.9 1 1 1 if 0 SC1 for y = 3 indicated (d) Correct line or no line and − 0.7 to − 0.6 nfww 3 Must check line - not if wrong line B2 for y = 1 − x ruled correctly or SC1 for ruled line with either gradient –1 or y-intercept 1 but not line y = 1 or correct freehand line

Page 4 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Qu Answers Mark Part Marks (e) tangent ruled at x = 2 and 0.62 to 0.8 3 Accept integer/integer provided in range B1 for correct tangent drawn and M1 for change in y / change in x dep on any tangent or close attempt at tangent at any point Must see correct or implied calculation from a drawn tangent (f) x x − = 2 1 or 1 + x3 = 0 3 1 x − = or 1 3 − = x 3 1 − = x M1 M1 A1 dep M1 dep M2 5 (a) (i)       4 2 1 (ii) 5.83 to 5.831 2 M1 for 32 + 52 seen (b) (i) − 2p + q oe 1 accept unsimplified (ii) = PS − p +2q or SP = p – 2q MS = – 3 2 p+ 3 4 q seen or SM = 3 2 p – 3 4 q seen or = RM 3 2 ( − 2p + q) soi or MR = 3 2 ( 2p – q) soi or MQ = 3 1 (– 2p + q) soi or QM = 3 1 (2p – q) soi = PM p + RM or p – MR or – p + q + QM or – p + q – MQ [ = – 3 1 p + 3 2 q ] 1 : 3 nfww B1 B1 M1 A1 Any correct route for PM eg RM PR + After 0 scored, SC1 for 1 : 3

Page 5 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Qu Answers Mark Part Marks 6 (a) (i) 6 1 1 (ii) 6 4 oe 1 (iii) 6 2 oe 1 (b) 36 16 oe 3 M2 6 2 6 4 6 4 6 2 × + × only oe or M1 for one of 6 4 6 2 × or 6 2 6 4 × soi by 9 2 (c) 360 48 oe 3 M2 for 3 2 4 2 5 3 6 4 × × × only oe or M1 for denominators 6, 5, 4, 3 soi in product of four fractions 7 (a) (i) 148 1 (ii) 122 2 B1 for 58 seen at A or 32 seen at Y (iii) 148 1 (iv) 106 nfww 3 B1 for [sum of interior angles =] 720 and M1 for 2 1 {(their 720) − (p+q+t+90)} (b) (i) 63 2 B1 for angle RPS = 27 or 90 at P or at S seen or stated (ii) 54 2 B1 for their x or 63 or letter x at Q seen or state

Page 6 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Qu Answers Mark Part Marks 8 (a) (i) ( )( ) ( ) 4 2 4 3 2 2 7 + = + − + × x x x 12 3 8 2 2 − − + x x x or better seen 0 6 3 2 2 = − + x x M1 B1 A1 Allow if bracket[s] omitted but recovers with no errors seen and brackets correctly expanded on both sides and no omission of brackets (ii) ( ) ( ) ( ) 6 2 4 3 2 − − or better p = − 3 and r = 2(2) 1.14 and − 2.64 cao B1 B1 B1B1 or 2 4 3       + x Must see r q p + or r q p − or both Or 4 3 − + or – 16 57 SC1 for 1.1 and − 2.6 final answer or 1.137 and – 2.637 final answer or 1.14 and − 2.64 seen in working or for -1.14 and 2.64 as final ans (b) 2 x × π + x x 3 × × π [ ] 2 4 x π =[ ] 2 r π 2x = r M2 M1 A1 or M1 for x x 3 × × π Dep on M2 with no errors seen 9 (a) 4 − 6x final answer 1 (b) 8 9 − x final answer 2 M1 for 4 − 3(4 − 3x) seen (c) 27 1 final answer 3 M2 for 3 3−soi by final answer 0.037037… to 3sf or better or M1 for [g( − 1) =] 3 soi (d) 3 4 x − oe final answer 2 M1 for a correct first step y x − = 4 3 oe or y x 3 4 − = or x y − = 3 4 3 (e) 3 4 or 3 1 1 or 1.33 or better 3 M2 for 3x − 4 = 0 or better or M1 for ) 3 4 ( 3 x − −

Page 7 Mark Scheme Syllabus Paper IGCSE – May/June 2014 0581 43 © Cambridge International Examinations 2014 Qu Answers Mark Part Marks 10 (a) [r =] 2.30[9...] 3 B2 for [r =] 2.31 or M2 for 30 tan 4 or M1 for 30 tan 4 = r (b) 333 or 332.5 to 332.6 4 M3 for 0.5 × 8 × 8 × sin 60 × 12 oe or M2 for 0.5 × 8 × 8 × sin 60 oe or M1 for their triangle area × 12 shown dep on ‘ 2 1 ’used within their area of triangle method (c) (i) 30 3 M2 for 12 ÷ 0.4 or 120 ÷ 4 or SC1 for figs 3 (ii) 6.65 or 6.647 to 6.648[...] 2 M1 for 4.0 3.2 2 × × π or SC1 for 4 3.2 2 × × π soi by 66.5 or 66.47 to 66.48[…] (iii) 40[.0] or 40.1 or 40.0 to 40.2 nfww 3 M2 for 100 ) ( ) )( ( ) )( ( 100 × × − b their ii c their i c their or 100 ) ( ) )( ( ) )( ( ) ( × × − b their ii c their i c their b their or M1 for 100 ) ( ) )( ( ) )( ( × × b their ii c their i c their or ) ( ) )( ( ) )( ( ) ( b their ii c their i c their b their × − 11 (a) 8 1 16 1 32 1 1 2 1 − n oe 3 2− 4 2− 5 2− n − 1 2 or )1 ( 2 − −n 2 2 1 1 B1 for 2 correct SC1 for n 2 1 oe (b) (i) 64 256 1024 6 2 8 2 10 2 1 1 (ii) )1 ( 2 2 − n or 22n-2 1 (c) 16 384 2 B1 for n = 8