Cambridge IGCSE Mathematics - International 0607 — 2013 Oct/Nov Paper 4 · Variant 1

This document consists of 18 printed pages and 2 blank pages. IB13 11_0607_04/2RP © UCLES 2013 [Turn over *3769158093* For Examiner's Use UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/04 Paper 4 (Extended) October/November 2013 2 hours 15 minutes Candidates answer on the Question Paper. Additional Materials: Geometrical Instruments Graphics 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. Do not use staples, paper clips, highlighters, glue or correction fluid. You may use a pencil for any diagrams or graphs. DO NOT WRITE IN ANY BARCODES. Answer all the questions. Unless instructed otherwise, give your answers exactly or correct to three significant figures as appropriate. Answers in degrees should be given to one decimal place. For π, use your calculator value. You must show all the relevant working to gain full marks and you will be given marks for correct methods, including sketches, even if your answer is incorrect. 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 120. www.XtremePapers.com

2 © UCLES 2013 0607/04/O/N/13 Formula List For the equation ax2 + bx + c = 0 x = 2 _ ± 4 2 _ b b ac a Curved surface area, A, of cylinder of radius r, height h. A = 2πrh Curved surface area, A, of cone of radius r, sloping edge l. A = πrl Curved surface area, A, of sphere of radius r. A = 4πr2 Volume, V, of pyramid, base area A, height h. V= 1 3 Ah Volume, V, of cylinder of radius r, height h. V = πr2h Volume, V, of cone of radius r, height h. V = 1 3 πr2h Volume, V, of sphere of radius r. V = 4 3 πr3 = = sin sin sin a b c A B C a2 = b2 + c2 – 2bc cos A Area = 1 2 bc sin A A a B C b c

3 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use Answer all the questions. 1 Manuel buys a car for $8000. (a) Each year the value of the car decreases by 8% of its value at the start of the year. (i) Calculate the value of the car after 5 years. Answer(a)(i) $ [2] (ii) Calculate how many more years it takes for the value of the car to be less than $4000. Answer(a)(ii) [2] (b) Manuel has a journey of 235 km. The journey takes 3 h 15 min and the car uses 19.7 litres of fuel. (i) Calculate the average speed of the journey in kilometres per hour. Answer(b)(i) km/h [2] (ii) Find the rate at which the car uses fuel. Give your answer in litres per 100 km. Answer(b)(ii) l/100 km [1]

4 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 2 W T y x 5 4 3 2 1 –1 –2 –3 –4 –5 0 –1 1 2 3 4 5 –2 –3 –4 –5 (a) (i) Reflect triangle T in the x-axis. Label the image U. [2] (ii) Rotate triangle U clockwise through 90° about (0, 0). Label the image V. [2] (iii) Describe fully the single transformation that maps triangle T onto triangle V. [2] (b) Describe fully the single transformation that maps triangle T onto triangle W. [3]

5 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use 3 North N W A 346 km 271 km 493 km NOT TO SCALE The diagram shows the straight line distances between the cities Auckland (A), Napier (N) and Wellington (W) in New Zealand. (a) The bearing of W from A is 179°. Calculate the bearing of N from A. Answer(a) [4] (b) A map shows the three cities. The scale of the map is 1 : 10 000 000. Calculate the area of triangle ANW on the map. Give your answer in square centimetres. Answer(b) cm2 [3]

6 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 4 O C NOT TO SCALE P 14 cm 12 cm The diagram shows a hollow cone of height 12 cm and sloping edge, OP, 14 cm. C is the centre of the base of the cone. (a) Calculate (i) the radius of the base of the cone, Answer(a)(i) cm [3] (ii) the volume of the cone. Answer(a)(ii) cm3 [2]

7 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use (b) The cone is cut along the sloping edge OP and opened out to make a sector of a circle. O A B NOT TO SCALE (i) Calculate the area of the sector and show that it rounds to 317 cm2, correct to 3 significant figures. [2] (ii) Calculate the reflex angle AOB. Answer(b)(ii) [3]

8 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 5 200 students each record the number of hours, h, they spend on homework in one week. The cumulative frequency curve shows the results. 200 180 160 140 120 100 80 60 40 20 5 10 15 20 Time spent on homework (hours) 25 30 35 0 h Cumulative frequency (a) Find (i) the median, Answer(a)(i) h [1] (ii) the lower quartile, Answer(a)(ii) h [1] (iii) the inter-quartile range, Answer(a)(iii) h [1] (iv) the 90th percentile, Answer(a)(iv) h [1] (v) the number of students who spend more than 10 hours on homework. Answer(a)(v) [2]

9 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use (b) (i) Use the cumulative frequency curve to complete the frequency table. Time spent on homework h hours 0 < h Ğ 10 10 < h Ğ 15 15 < h Ğ 20 20 < h Ğ 25 25 < h Ğ 35 Frequency 20 20 50 [2] (ii) Calculate an estimate of the mean number of hours spent on homework. Answer(b)(ii) h [2] (iii) The data is used to draw a histogram. Complete the frequency density table. (Do not draw the histogram.) Time spent on homework h hours 0 < h Ğ 15 15 < h Ğ 20 20 < h Ğ 25 25 < h Ğ 35 Frequency density 10 [3]

10 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 6 y x 10 –10 –8 8 0 f(x) = ) 2 ( ) 3 2 ( + − x x (a) On the diagram, sketch the graph of y = f(x). [3] (b) Write down the value of f(0). Answer(b) [1] (c) Solve the equation f(x) = 0. Answer(c) [1] (d) Write down the equations of the asymptotes. Answer(d) [2] (e) Find the range of f(x) for the domain 0 Ğ x Ğ 8 . Answer(e) [2]

11 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use (f) g(x) = 3 – x (i) On the diagram, sketch the graph of y = g(x). [1] (ii) Solve the equation f(x) = g(x). Answer(f)(ii) x = or x = [2] (iii) Show that the equation f(x) = g(x) can be re-arranged into x2 + x – 9 = 0 . [3] (iv) The exact solutions of the equation x2 + x – 9 = 0 are 2 1 k ± − . Find the value of k. Answer(f)(iv) k = [2]

12 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 7 y x 0 A B NOT TO SCALE A (1, 6) is joined to B (5, 2) by the line AB. (a) Calculate the length of the line AB. Answer(a) [3] (b) Find the equation of the straight line that passes through A and B. Answer(b) [3] (c) (i) Find the equation of the line which is perpendicular to AB and passes through the origin. Answer(c)(i) [2] (ii) Find the co-ordinates of the point of intersection of the line in part (c)(i) and the line AB. Answer(c)(ii) ( , ) [1]

13 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use 8 Find the nth term of each of the following sequences. (a) 21, 17, 13, 9, 5, ……… Answer(a) [2] (b) 3, 6, 12, 24, 48, ……… Answer(b) [2] (c) 4 1 , 5 4 , 6 9 , 7 16 , 8 25 , ……… Answer(c) [2] (d) 0, 6, 24, 60, 120, ……… Answer(d) [4]

14 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 9 If the weather is fine, the probability that Alex goes to the beach is 10 9 . If the weather is not fine, the probability that Alex goes to the beach is 10 3 . The probability that the weather will be fine is 6 5 . (a) Complete the tree diagram. Yes Weather is fine Alex goes to the beach No Yes No No Yes … … … … … … [3] (b) Find the probability that Alex goes to the beach. Answer(b) [3] (c) Which combination of these events has a probability of 12 1 ? Answer(c) [1]

15 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use 10 D C B A X 6x° 3x° x° 8.55 cm 2.78 cm 9.23 cm NOT TO SCALE A, B, C and D lie on the circumference of a circle. AC and BD intersect at X. (a) Angle CDX = x°, angle DCX = 3x° and angle CXD = 6x°. Show that angle ABX = 54°. [3] (b) (i) Complete the statement Triangles CDX and BAX are [1] (ii) AB = 9.23 cm, DC = 8.55 cm and XC = 2.78 cm. Calculate the length of BX. Answer(b)(ii) cm [2] (iii) Find the value of BAX CDX triangle of Area triangle of Area . Give your answer correct to 2 decimal places. Answer(b)(iii) [2]

16 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 11 (a) 0 y x NOT TO SCALE The sketch shows the graph of y = logax . On the same diagram, sketch the graph of y = 2logax . [2] (b) 3log x = log16 – 2log x Find the value of x. Answer(b) x = [3] (c) Solve the equation 5 y = 100 . Give your answer correct to 4 significant figures. Answer(c) y = [3]

17 © UCLES 2013 0607/04/O/N/13 [Turn over For Examiner's Use 12 5x 2x NOT TO SCALE The diagram shows a rectangle with length 5x and width 2x. One of the shorter sides is joined to a semicircle with radius x. (a) Find a formula, in terms of x and π, for the total area, A, of the shape. Answer(a) A = [2] (b) Make x the subject of your formula in part (a). Answer(b) x = [3] (c) Find the value of x when A = 200. Answer(c) x = [1]

18 © UCLES 2013 0607/04/O/N/13 For Examiner's Use 13 (a) (i) Factorise. 2x2 – x – 1 Answer(a)(i) [2] (ii) Write as a single fraction in its simplest form. 1 4 1 2 1 2 − + − − x x x Answer(a)(ii) [3] (b) Simplify. qt pt q p q p 5 5 25 2 2 − − + − Answer(b) [4]

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20 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. University of 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. © UCLES 2013 0607/04/O/N/13 BLANK PAGE

CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2013 series 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/04 Paper 4 (Extended), maximum raw mark 120 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 2013 series for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components. www.XtremePapers.com

Page 2 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0607 04 © Cambridge International Examinations 2013 1 (a) (i) 5272.65 (allow 5270, 5272 to 5273) 2 M1 for 5 92 .0 8000 × oe (ii) 4 (allow 3.31, 3.312 to 3.313) nfww 2 M1 for 8000 × 0.92n = 4000 oe or SC1 for 9 or 8.31 or 8.312 to 8.313 (b) (i) 72.3 (72.30 to 72.31) 2 M1 for 235 ÷ 3.25 oe (ii) 8.38 (8.382 to 8.383) 1 2 (a) (i) Triangle at (1, –1), (4, –1), (4, –2) 2 SC1 for reflection in y-axis (ii) Triangle at (–1, –1), (–1, –4 ), ( –2, –4) 2 FT FT SC case only SC1 for anti-clockwise rotation of 90° about (0, 0) (iii) Reflection y = – x oe B1FT B1FT FT the transformation FT full description B’s independent but both marks lost if more than one transformation stated (b) Enlargement (or reduction) (0, 2) [factor] 0.5 B1 B1 B1 B’s independent but all 3 marks lost if more than one transformation stated No ratios 3 (a) 147 nfww 4 B3 for [A =] 31.9 to 32.1 nfww or M2 for [cos angle A =] 493 346 2 271 493 346 2 2 2 × × − + oe or M1 for correct implicit expression with angle A B1 FT 179 – their angle A (b) 4.52 (4.519 to 4.520) 3 M2 for 0.5 × 4.93 × 3.46 × sin (their A) oe e.g. 0.5 × 493 × 346 × sin (their A) ÷ 1002 or use of Hero’s formula or M1 for scale correctly applied or correct use of C absin 5.0 or correct use of Hero’s formula figs 4519 to 4520 imply M1

4 5 6 Page 3 4 (a) (i) (ii) (b) (i) (ii) 5 (a) (i) (ii) (iii) (iv) (v) (b) (i) (ii) (iii) 6 (a) (b) (c) (d) 3 7.21 (7.21 653 (653. 317.1 to 3 185 (185. 20 16 9 29 180 60, 50 20.125 (or 2.67 (2.66 12 5 – 1.5 oe 1.5 oe x = –2, y = IGCSE – © Ca 1..) or 2 1 2 to 653.5… 17.2… 3 to 185.5) r 20.1 or 20.1 66 to 2.667) = 2 Mark Sch – October/N ambridge Inte 3 …) or π 208 12 to 20.13) oe heme November 2 ernational Ex 2013 xaminations 3 M r 2FT F M 2 M 3 M or im 3 1 1 1 1 2 M or 1, 1 2FT 1 1FT 1FT F 1 M im F F 3 M hy A x A y- 1 D 1 D 1, 1 Syllab 060 2013 M2 for 2 14 2 2 14 12 = + FT their (a)(i) M1 for ( 3 1 th π M1 for (thei π M2 for 14 ( ( π their r M1 for π the mplicit statem 14 360 2 × ×π x M1 for 20 ind r SC1 for an FT their (b)(i 10 M1 for at leas mplied FT their (b)(i FT their (b)(i M1 for reason yperbola sha A1 for asymp = – 2 and y = A1 for x-inter -intersection Do not allow Do not allow bus 07 2 12 − or M 2 4 oe ) 1( )) )( ( 2 i a heir ) 14 ))( )( ( i a ir 360 ) 4 ) )( ( 2 × i b o 2) 14 ( ) )( ( π i b eir oe ment e.g. 317 or 317 = dicated e.g. o nswer of 20 i) only if answ st 3 mid-valu i) i) nable rectang ape ptotes approx = 2 (soi) rsection posit n negative co-ordinates co-ordinates Paper 04 M1 for ) 12 oe or correct 2. 317 to 1.7 on y-axis wers add to ues seen or gular ximately tive and s s

Page 4 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0607 04 © Cambridge International Examinations 2013 (e) –1.5  f(x)  1.3 oe 2 Strict inequality at either end or both ends scores only 1 Allow in words but “between – 1.5 and 1.3” scores only 1 B1 for – 1.5 and 1.3 seen or for f(x)  – 1.5 or for f(x)  1.3 (f) (i) Reasonable x y − = 3 added to sketch 1 (ii) –3.54 (–3.541…), 2.54 (2.541…) 1, 1 (iii) ) 3 )( 2 ( 3 2 x x x − + = − x x x x x 2 6 3 )] 3 )( 2 [( 2 − + − = − + 0 9 2 = − + x x M1 B1 E1 Allow ) 2 ( ) 2 ( 3 3 2 − − − = − x x x x or ) 3 ( 2 ) 3 ( 3 2 x x x x − + − = − Allow 2 6 x x − + No errors or omissions (iv) 37 2 M1 for ) 9 )( 1( 4 1 4 2 2 − − = −ac b seen or ( ) 9 4 1 2 2 1 = − + x or better 7 (a) 5.66 (5.656 to 5.657) or 2 4 3 M2 for 2 2 ) 2 6 ( )1 5 ( − + − oe or better or M1 for 5 – 1 and 6 – 2 (or 2 – 6) soi (b) 7 = + y x oe 3 M1 for gradient = 1 5 6 2 − − oe M1 for using (1, 6) or (5, 2) in y = mx + c oe (c) (i) y = x 2 FT M1 for gradient = ) ( 1 b in gradient their − (ii) (3.5, 3.5) oe cao 1 8 (a) 25 – 4n oe 2 M1 for answer of c n + −4 (b) 1 2 3 − × n oe 2 M1 for 3 × q 2 seen and with no other terms (c) 3 2 + n n oe 2 B1 for fraction with either numerator or denominator correct (d) n n − 3 oe 4 M3 for comparing sequence with values of n3 or d cn bn an + + + 2 3 with 4 values of n substituted correctly oe or M2 for attempting cubic expression oe or listing values of n3 or M1 for reaching equal third differences

9 1 1 Page 5 9 (a) (b) (c) 10 (a) (b) (i) (ii) (iii) 11 (a) 5 10 9 , 6 1 , 6 5 placed 60 48 oe ( Fine weath beach 6 3 + + x x x = 18 angles in t similar 3[.00] or 2 0.86 IGCSE – © Ca 10 7 , 10 3 , 10 1 , ( 8.0 , 20 16 etc her but Alex 180 = x or 10 the same seg 2.990 to 3.00 Mark Sch – October/N ambridge Inte 0 7 oe all cor c.) does not go 0x = 180 ment oe 02 heme November 2 ernational Ex rrectly to the 2013 xaminations 3 B 3 is M or 1 B1 B1 B1 A or 1 N 2 M s. 2 M to or ⎜ ⎝ ⎛ 2 M A y- gr Syllab 060 2013 B1 for each p sw any cance M2 for 10 9 6 5 × r M1 for one Allow angles r same chord No alternative M1 for 23 .9 55 .8 = .f = 1.08 or 1 M1 for 23 .9 55 .8 ⎜ ⎝ ⎛ o 0.859 or 1. r 5.0 78 .2 5.0 × × theirB ⎟ ⎠ ⎞ ⎝ ⎛ … 2008 . 11 . 61476 .9 M1 for shape A1 for throug -values appro raph bus 07 air correctly elling or conv 10 3 6 1 0 9 × + e of the prod subtended b d es BX 78 .2 = oe a 1.079 to 1.08 2 3 5⎟ ⎠ ⎞ oe (impl 16 to 1.17) sin 23 .9 54 sin 55 .8 8 × × BX gh (1, 0) and ox. double th Paper 04 placed verting ucts by itself by the same a allow 80 lied by 0.857 54 n 4 positive hose on log x f arc 7 x

Page 6 Mark Scheme Syllabus Paper IGCSE – October/November 2013 0607 04 © Cambridge International Examinations 2013 (b) ) 16 log( ) log( 5 = x or 16 5 = x or ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = 2 3 16 log log x x or 2 3 16 x x = or appropriate sketch 1.74 (1.741…) or 5 16 or 8.0 2 oe M2 B1 M1 for using a rule of logarithms once correctly (c) ylog5 = log100 or 100 log 5 = y or 5 log 100 log or sketch 2.861 M1 B2 e.g. for sketch x y 5 = with y = 100 B1 for 2.86 or 2.8613 to 2.8614 12 (a) 2 2 1 2 10 x x π + oe final answer 2 B1 for 2 10x or 2 2 1 x π seen (b) ) 10 ( 2 1 2 π + = x A or ) 20 ( 2 2 π + = x A π 2 1 10 2 + = A x or π + 20 2A π 2 1 10 + A or π + 20 2A final answer 3 M1 for correctly taking x2 as a factor from two terms, one containing π M1 for correct division by other factor which has two terms and no x in it M1 for correct square root to give x (c) 4.16 (4.157 to 4.158) cao B1 13 (a) (i) )1 )( 1 2 ( − + x x 2 SC1 for )1 )( 1 ( − + bx ax where ab = 2 or b – a = – 1 or for answer 1 , 2 1 = − = x x but only from factors (ii) )1 )( 1 2 ( 5 8 − + + x x x oe final answer 3 B2 for 8x + 5 seen or M1 for )1 2 ( 4 1 2 − − + − x x x or better seen e.g. 1 + 4(2x + 1) B1 for denominator )1 )( 1 2 ( − + x x oe in final answer (b) t q p − − 1 5 oe nfww final answer 4 B1 for ) 5 )( 5 ( q p q p − + B2 for ) 1 )( 5 ( t q p − + or B1 for ) 5 ( 5 q p t q p + − + or ) 1( 5 ) 1( t q t p − + −