Cambridge IGCSE Mathematics - International 0607 — 2010 Oct/Nov Paper 6 · Variant 1

This document consists of 10 printed pages and 2 blank pages. IB10 11_0607_06/4RP © UCLES 2010 [Turn over *6357012477* UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/06 Paper 6 (Extended) October/November 2010 1 hour 30 minutes Candidates answer on the Question Paper. Additional Materials: 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. Answer both parts A and B. You must show all relevant working to gain full marks for correct methods, including sketches. In this paper you will also be assessed on your ability to provide full reasons and communicate your mathematics clearly and precisely. At the end of the examination, fasten all your work securely together. The total number of marks for this paper is 40. www.XtremePapers.com

2 © UCLES 2010 0607/06/O/N/10 For Examiner's Use Answer both parts A and B. A INVESTIGATION THE FIBONACCI SEQUENCE 24 marks You are advised to spend no more than 55 minutes on this part. The Fibonacci sequence is a sequence of numbers that is found in many real-life situations. The Fibonacci sequence begins 1 1 2 3 5 ……. where, apart from the first two terms, each term is the sum of the previous two terms. For example 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5 and so on. 1 Complete the table for the first 15 Fibonacci numbers. You must show your working. Term position 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Fibonacci number 1 1 2 3 5 8 13 21 34 55 89 144 233

3 © UCLES 2010 0607/06/O/N/10 [Turn over For Examiner's Use 2 (a) The table shows Fibonacci numbers that are multiples of 2. Complete the table. Term position 3 9 Fibonacci number 2 8 Notice that: 2 is the third term in the Fibonacci sequence, 8 is the sixth term in the Fibonacci sequence, and so on. Every third term in the Fibonacci sequence is a multiple of 2. (b) The next two tables show other patterns. Complete the tables and the statements that follow. (i) Term position 4 8 12 Fibonacci number 3 3 is the term in the Fibonacci sequence. Every term in the Fibonacci sequence is a multiple of 3. (ii) Term position 20 Fibonacci number 5 55 6 765 5 is the term in the Fibonacci sequence. Every term in the Fibonacci sequence is a multiple of . (c) Complete the following statement. Every term in the Fibonacci sequence is a multiple of 8.

4 © UCLES 2010 0607/06/O/N/10 For Examiner's Use 3 A Golden Rectangle is a rectangle with width and length that are consecutive Fibonacci numbers. 2 3 5 etc. 3 5 8 When a Golden Rectangle is divided into the least number of squares, the length of the side of each square is a Fibonacci number. 3 2 1 1 1 The diagram above shows the 2 by 3 Golden Rectangle. The least number of squares it can be divided into is three. These squares have sides 1, 1 and 2. 5 3 The diagram above shows the 3 by 5 Golden Rectangle. The least number of squares it can be divided into is four. These squares have sides 1, 1, 2 and 3.

5 © UCLES 2010 0607/06/O/N/10 [Turn over For Examiner's Use (a) On the grid below, draw the 5 by 8 Golden Rectangle. Show how this can be divided into the least number of squares. These squares have sides 1, 1, 2, 3 and 5. (b) On the grid below, draw the 8 by 13 Golden Rectangle. Show how this can be divided into the least number of squares.

6 © UCLES 2010 0607/06/O/N/10 For Examiner's Use (c) (i) Complete the table to show the least number of squares in each of the first six Golden Rectangles. Size of rectangle 1 by 1 1 by 2 2 by 3 3 by 5 5 by 8 8 by 13 Least number of squares 1 4 (ii) Write down the least number of squares there are in the 21 by 34 Golden Rectangle. (iii) When the least number of squares is 11, write down the width and the length of this Golden Rectangle. and (d) When the width and the length of a Golden Rectangle are the (nO1)th and the nth terms of the Fibonacci sequence, write down the least number of squares in terms of n.

7 © UCLES 2010 0607/06/O/N/10 [Turn over For Examiner's Use (e) In this part, numbers that are two positions apart in the Fibonacci sequence are used for the width and length of a rectangle. For example, 1 by 2, 1 by 3, 2 by 5, 3 by 8 and so on. 1 1 2 3 5 8 ... 1 Fibonacci number Term position 2 3 4 5 6 ... Write down, in words, the connection between the term positions of the two Fibonacci numbers used for the width and length of a rectangle and the least number of squares in the rectangle.

8 © UCLES 2010 0607/06/O/N/10 For Examiner's Use B MODELLING THE SOLAR SYSTEM 16 marks You are advised to spend no more than 35 minutes on this part. Logarithms to base 10 are written as log. 1 The table below shows information about seven planets in the Solar System. Planet Distance from the Sun (S km) Time to orbit the Sun (T days) log S log T Mercury 5.79 x 107 88 7.8 1.9 Venus 1.08 x 108 225 8.0 2.4 Earth 1.50 x 108 365 8.2 2.6 Mars 2.28 x 108 687 Jupiter 7.78 x 108 4 330 Saturn 1.43 x 109 10 800 Pluto 5.91 x 109 90 800 9.8 5.0 Complete the table of values for log S and log T. Give each value correct to 2 significant figures. 2 (a) On the grid opposite, plot the seven points (log S, log T). (b) Plot the mean point (8.6, 3.2) and use this to draw a line of best fit. (Do this by eye. Do not use your calculator.) 3 The time taken for the planet Uranus to orbit the Sun is 30 685 days. Use your graph to estimate the distance of Uranus from the Sun. Give your answer correct to 2 significant figures. km 4 Let x = log S and y = log T. The equation of the line of best fit is y = mx + c. Use your calculator to find the values of m and c. Give each answer correct to 2 significant figures. m = c = 5 A model for this is log T = mlog S + c. The distance of the planet Neptune from the Sun is 4.50 x 109 km. Use the model to find the time taken for Neptune to orbit the Sun. Give your answer in standard form correct to 2 significant figures. days

9 © UCLES 2010 0607/06/O/N/10 [Turn over For Examiner's Use 7 8 9 10 5 4 3 2 1 0 log S log T Question 6 is on the next page.

10 © UCLES 2010 0607/06/O/N/10 For Examiner's Use 6 Writing c as log k, the model can be written as log T = m log S + log k. (a) Show that T = kSm. (b) Find the value of k. k = (c) Write the model T = kSm using your values of k and m. Use the data for Earth to test this model.

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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS International General Certificate of Secondary Education MARK SCHEME for the October/November 2010 question paper for the guidance of teachers 0607 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/06 Paper 6 (Extended), maximum raw mark 40 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 must be read in conjunction with the question papers and the report on the examination. • CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses. www.XtremePapers.com

Page 2 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0607 06 © UCLES 2010 A INVESTIGATION THE FIBONACCI SEQUENCE 1 Term position ... 12 13 14 15 Fibonacci number ... 144 233 377 610 2 C1 1 1ft C1 for showing working ft for 610 – 233 + ‘their 377’ 2 (a) Term position 3 6 9 12 Fibonacci number 2 8 34 144 2 1 for both in row 1 1 for both in row 2 (b) (i) Term position 4 8 12 16 Fibonacci number 3 21 144 987 3 is the 4th term... Every 4th term... 5 1 2ft for all 3 in row 2 –1 eeoo ft from Q1 for 987 – ‘their 377’ + ‘their 610’ (ii) Term position 5 10 15 20 Fibonacci number 5 55 610 6765 5 is the 5th term... Every 5th term in the... is a multiple of 5 5 2 for all 3 in row 1 –1eeoo 1ft 1 1 for both entries ft from Q1 for ‘their 610’ (c) Every 6th term in the... 1

Page 3 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0607 06 © UCLES 2010 3 (a) 5 by 8 rectangle drawn, divided into: one 5 by 5 square one 3 by 3 square one 2 by 2 square and two 1 by 1 squares 2 If not all correct 1 for any 2 squares shown excluding the two 1 by 1 squares (b) 8 by 13 rectangle drawn, divided into: one 8 by 8 square one 5 by 5 square one 3 by 3 square one 2 by 2 square and two 1 by 1 squares 2 If not all correct 1 for any 2 squares shown (c) (i) Size of rectangle 1 by 1 1 by 2 2 by 3 3 by 5 5 by 8 8 by 13 Least number of squares 1 2 3 4 5 6 1 1 for all 4 entries (ii) 8 1 (iii) 89 144 2 1 each (d) n – 1 1 oe e.g. n n n )1 ( − The least number of squares is: the same as the term number that comes between the position numbers of the width and the length OR the mean of the position numbers of the width and the length OR width (smallest ) position plus 1 or length (largest) position minus 1 OR e.g. for nth and (n + 2)th terms, answer of n + 1 oe 2 C1ft 1 identifying ‘term’ or ‘position’ number of width/length 1 method of calculation/showing connection C1ft sketches/working shown to identify/illustrate answer 1 for explaining least number of squares is sequential from 2 OR Identifying width/length as e.g. n and n + 2 ‘width’ + 1 scores 1 unless width is identified as shorter side, and same for ‘length’– 1 For C1 must show some understanding [Total: 26 + C2 = 28 scaled to 24]

Page 4 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0607 06 © UCLES 2010 B MODELLING THE SOLAR SYSTEM 1 8.4 2.8 8.9 3.6 9.2 4.0 3 2 for 5 or 4 correct 1 for 3 or 2 correct 0 for 1 or 0 correct Note: In Q 1, 3, 4, 5 a penalty of -1 once for not rounding to 2 sf 2 (a) 7 points plotted P2ft P1 ft for 4, 5 or 6 correct plots ft for 3 points in Q1 Condone inaccuracies of up to 1 mm in plotting (b) Mean (8.6, 3.2) plotted Line of best fit ruled through mean P1 L1 Between (7.6, 1.9) and (8, 1.9) and between (9.6, 5) and (10, 5) Condone inaccuracies of up to 1 mm in plotting and drawing 3 2.8 × 109 (km) / 3.2 × 109 (km) 3 C 1 for 4.5 seen (maybe on axis) 1ft for 9.45 / 9.5 oe ft from line of best fit 1ft for answer C opportunity for minimum of 4.5 on graph or 4.5 and 9.45/9.5 oe in working Note: In Q 1, 3, 4, 5 a penalty of -1 once for not rounding to 2 sf (anti-log value read from 4.5 and line of best fit) 4 (m =) 1.5 [1.3 – 1.7] (c =) –9.6 / –9.7 1 1ft C Maybe necessary to ft from m C opportunity if working shown for m and c Note: In Q 1, 3, 4, 5 a penalty of -1 once for not rounding to 2 sf (c = 3.2 – their m × 8.6) 5 7.6 × 104 (days) / 6.0 × 104 (days) 1ft C Maybe necessary to ft from m and c C opportunity if working shown Note: In Q 1, 3, 4, 5 a penalty of -1 once for not rounding to 2 sf (anti-log (their m × log(4.5 × 109) + their c))

Page 5 Mark Scheme: Teachers’ version Syllabus Paper IGCSE – October/November 2010 0607 06 © UCLES 2010 6 (a) log T = log Sm + log k log T= log kSm T = kSm (AG) M1 E1 ÷ by log = E0 (b) (k =) 2.0 × 10–10 / 2.5 × 10–10 1ft ft from their c (anti-log their c) (c) T = their k × (1.5 × 108)their m T ≈ 367 / 459 OR 365 = their k × Stheir m S ≈ 1.5 × 108 Comment that is appropriate to result of their test 1ft 1ft 1 C Substitution of their values ft from 6(b) and 4 and value of S or T from table Q1 C opportunity if working shown C1 1 for two C opportunities shown [Total: 20 scaled to 16]