BMOS MENTORING SCHEME (Senior Level) February 2011 (Sheet 5) Solutions
|
|
- Coleen McDowell
- 6 years ago
- Views:
Transcription
1 MOS MENTORING SCHEME (Senior Level) February 011 (Sheet 5) Solutions These solutions represent one way of solving the questions. They are not the only way, and just because your solution is different does not mean that it is worse or wrong. You will get most from the solutions if you have at least attempted the questions. Only then will you appreciate the difficulties that a problem posed and how they were addressed by the solution. Relevant theorems have been identified where possible, and references to further information on the Web have been given. If you are still unclear, ask your mentor. A. Rzym & J. Cranch c UKMT Let f(n) denote the average number of circles in the bowl having started with n lengths of string. Starting with n lengths of string and having selected the first end, the second end is selected from one of the n 1 remaining ends. There are then two possibilities: (a) In one case (i.e. with probability 1/(n 1)) the second end was the other end of the same string. In this case, we have created a circle and have n 1 strings left (yielding f(n 1) circles on average), i.e. a total of 1 + f(n 1) circles on average. (b) In the remaining n cases (i.e. with probability (n )/(n 1)), the second end is of a string distinct from the first. On joining, we are left with n 1 strings, i.e. a total of f(n 1) circles on average. Probability weighting these possibilities (and given that f(1)=1) gives and hence f(n) = 1 n (1 + f(n 1)) + n 1 n 1 f(n 1) = 1 + f(n 1) n 1 f(6) = 1/11 + 1/9 + 1/7 + 1/5 + 1/ If α, β are the roots of x Ax = 0 then αβ = and α + β = A. Defining S k = α k + β k, we have S 0 =, S 1 = A and S k = (α k + β k ) = (α k 1 + β k 1 )(α + β) αβ(α k + β k ) = AS k 1 + S k S k = A + (1) S k 1 S k 1 /S k Using (1) for k = n, and substituting (1) for k = n 1 we obtain S n S n 1 = A +. A + S n /S n 3 Repeating the process and observing that S 1 /S 0 = A/ gives the desired result. Supported by the Man Group plc Charitable Trust
2 3. (a) Repeatedly squaring the given equation we obtain x + x 1 + x x 1 = (x + x 1) + (x x 1) + x x + 1 = (x 1) = 1 x 1 = 1. We had hoped that squaring the original equation would give us a finite number of possible values for x (even if not all are actually solutions), but this has not happened. Instead, we will now consider whether the argument above can guide us to start with an identity and derive the required equation. For x 1 we have 1 x = 1 x and therefore (1 x) = 1 x x + x x + 1 =. Providing x 1/, x 1 exists and so (x + x 1) + (x x 1) + x x + 1 =. Since 1/ x 1, x + x 1 0 and x x 1 0, therefore the above equation implies ( x + ( x 1) + x x 1) + (x + x 1)(x x 1) =. Taking the square root of both sides yields the given equation. In other words, the given equation is satisfied if and only if 1/ x 1. (b) y squaring the given equation repeatedly we obtain x + x 1 + x x 1 = 1 (x + x 1) + (x x 1) + x x + 1 = 1 x + x x + 1 = 1/ 1 x = 1/ x. However there are no solutions to this equation, and therefore we conclude that there are no solutions to the given equation either. (c) y squaring the given equation repeatedly we obtain x + x 1 + x x 1 = (x + x 1) + (x x 1) + x x + 1 = 4 x x + 1 = x x = 3. Thus the only possible solution is x = 3/, and on substituting in the given equation we see that it is, indeed, a solution.
3 4. We will describe a sum of non-negative powers of without regard to order and where no power can be used more than twice as an a-partition, and denote the number of a-partitions of a positive integer n by a(n). We seek the value of a(011). We will describe a sum of positive powers of without regard to order and where no power can be used more than twice as a b-partition, and denote the number of b-partitions of a positive integer n by b(n). For every b-partition of an even integer, k, we can halve each term to give an a-partition of the integer k and vice-versa. This correspondence implies that b(k) = a(k). For every a-partition of an odd number, k + 1, there will be exactly one 0 in the sum. Removing this gives a b-partition of k and vice-versa. This correspondence implies a(k + 1) = b(k) = a(k). Every a-partition of an even integer, k, contains either no powers of 0 (in which case it is also a b-partition of k) or two powers of 0 (in which case, by dropping these, we have a b-partition of k. We conclude that a(k) = b(k) + b(k ) = a(k) + a(k 1). To summarise: a(k + 1) = a(k), a(k) = a(k) + a(k 1). Observing that a(1) = 1, a() = we then have a(011) = a(1005) = a(50) = a(51) + a(50) = a(15) + a(14) = 3a(6) + a(61) = 3a(31) + 4a(30) = 7a(15) + 4a(14) = 11a(7) + 4a(6) = 15a(3) + 4a() = 15a(1) + 4a() = Denote the centre of the circle of radius r by O. Denote the area of a triangle P QR by P QR. y the law of sines, Also, we have a sin A = b sin = c sin C = R AC = AO + OC + COA = 1 cr + 1 ar + 1 br = 1 r(a + b + c) = rs ut we also have AC = 1 bc sin A, which from the law of sines above can be written as AC = 1 bc sin A = 1 ( a ) bc = abc R 4R Equating these two expressions for AC we have AC = abc 4R = rs abc = 4srR 3
4 6. Defining p = d/a, q = e/b, r = f/c, dividing the inequality to be proved by abc then cubing it, we see that the inequality to be proved follows providing that we can prove (1 + p 3 )(1 + q 3 )(1 + r 3 ) (1 + pqr) (p 3 + q 3 + r 3 ) + (p 3 q 3 + q 3 r 3 + r 3 p 3 ) + (pqr) pqr + 3(pqr) + (pqr) 3. Our proof therefore proceeds as follows: Applying AM-GM to the collection {p 3, q 3, r 3 } we have p 3 + q 3 + r p 3 q 3 r 3 = 3pqr. () Applying AM-GM to the collection {p 3 q 3, q 3 r 3, r 3 p 3 } gives p 3 q 3 + q 3 r 3 + r 3 p p 6 q 6 r 6 = 3(pqr). (3) Using () and (3) we have 1 + (p 3 + q 3 + r 3 ) + (p 3 q 3 + q 3 r 3 + r 3 p 3 ) + (pqr) pqr + 3(pqr) + (pqr) 3 from which the result follows. (1 + p 3 )(1 + q 3 )(1 + r 3 ) (1 + pqr) 3 7. Let F denote the function mapping the quadruple q = (a, b, c, d) onto the quadruple ( a b, b c, c d, d a ). We will refer to the maximum value of the quadruple q as meaning the maximum value of the four integers a, b, c, d. We will prove that a quadruple q cannot exist such that repeated application of F does not eventually yield (0, 0, 0, 0) by assuming the converse and deriving a contradiction. Suppose such a q does, in fact, exist. Then all elements of F (q) are positive and therefore a q must exist with all elements positive. If q = (a, b, c, d) and a, b, c, d 0 then the maximum value of q is finite and non-negative. y assumption, not all of a, b, c, d are zero, and therefore the maximum value of q is positive. Furthermore, if repeated application of F on q does not yield (0, 0, 0, 0) it must be the case that after a certain number iterations, the maximum value of the quadruple is positive and does not decrease with further applications of F. Thus there exists q = (a, b, c, d) with the property that the maximum of q and F (q) = ( a b, b c, c d, d a ) are equal. Without loss of generality we can take a to be the maximum value of q. If a b takes the (same) maximum value, then b = 0. If b c takes the maximum value, a, then b = a and c = 0 etc. Thus one of the elements of q which takes the maximum value of q must be followed by a zero. Without loss of generality assume that these are a and b respectively. Thus we now have q = (a, 0, c, d), with a > 0, c, d 0 and a equal to the maximum value of q. Then q and F (F (q)) = ( a c, c c d, c d d a, d a a ) have the same maximum value. There are then four cases to consider: 4
5 If the first element of F (F (q)) is equal to the maximum value of q (a), this implies c = 0, i.e. the quadruple q is of the form (a, 0, 0, d); If the second element of F (F (q)) is equal to the maximum value of q (a), then c = a, d = 0, i.e. the quadruple q is of the form (a, 0, a, 0); If the third element of F (F (q)) is equal to the maximum value of q (a), then either d = 0, c = 0 or d = a, c = 0, i.e. the quadruple is of the form q = (a, 0, 0, d) or (a, 0, 0, a); If the fourth element of F (F (q)) is equal to the maximum value of q (a), then d = a, i.e. the quadruple q is of the form q = (a, 0, c, a). We now dispose of these possibilities. For example, if q = (a, 0, 0, d) then F (q) = (a, 0, d, d a ) F (F (q)) = (a, d, d d a, d a a ) F (F (F (q))) = ( a d, d d d a, d d a d a a, a d a a ). Each of the final terms can only equal a if d is one of 0, a. ut if q = (a, 0, 0, 0) then F (F (F (F (q)))) = (0, 0, 0, 0). Similarly, if q = (a, 0, 0, a) then F (F (F (q))) = (0, 0, 0, 0). The other cases can be similarly disposed of, so we conclude that there is no q for which repeated application of F does not lead to (0, 0, 0, 0). References Vieta s Fundamental Theorem of Algebra s formulas Law of sines: of sines AM-GM inequality: of arithmetic and geometric means 5
MeritPath.com. Problems and Solutions, INMO-2011
Problems and Solutions, INMO-011 1. Let,, be points on the sides,, respectively of a triangle such that and. Prove that is equilateral. Solution 1: c ka kc b kb a Let ;. Note that +, and hence. Similarly,
More informationThe CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Wednesday, April 15, 2015
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 015 Euclid Contest Wednesday, April 15, 015 (in North America and South America) Thursday, April 16, 015 (outside of North America
More informationFIITJEE ALGEBRA-2 Pre RMO
FIITJEE ALGEBRA- Pre RMO A. AP, GP 1. Consider the sequence 1,,,, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7,... and evaluate its 016 th term.. Integers 1,,,..., n, where n >, are written on a board. Two numbers
More informationProofs Not Based On POMI
s Not Based On POMI James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 1, 018 Outline Non POMI Based s Some Contradiction s Triangle
More information2007 Hypatia Contest
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 007 Hypatia Contest Wednesday, April 18, 007 Solutions c
More informationThe Law of Averages. MARK FLANAGAN School of Electrical, Electronic and Communications Engineering University College Dublin
The Law of Averages MARK FLANAGAN School of Electrical, Electronic and Communications Engineering University College Dublin Basic Principle of Inequalities: For any real number x, we have 3 x 2 0, with
More informationProofs Not Based On POMI
s Not Based On POMI James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 12, 2018 Outline 1 Non POMI Based s 2 Some Contradiction s 3
More informationThe CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca Euclid Contest. Tuesday, April 12, 2016
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 016 Euclid Contest Tuesday, April 1, 016 (in North America and South America) Wednesday, April 13, 016 (outside of North America
More informationComplete Syllabus of Class XI & XII
Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-0005 Ph.: 0-7656 Fa : 0-767 MM : 0 Sample Paper : Campus Recruitment Test Time : ½ Hr. Mathematics (Engineering) Complete Syllabus of Class XI & XII
More informationMATHEMATICS. Candidates exhibited the following as some of their strengths:
MATHEMATICS 1. STANDARD OF THE PAPER The standard of the paper compared favourably with that of previous years. Candidates performance this year was slightly better than that of previous years. 2. SUMMARY
More informationCh 3.2: Direct proofs
Math 299 Lectures 8 and 9: Chapter 3 0. Ch3.1 A trivial proof and a vacuous proof (Reading assignment) 1. Ch3.2 Direct proofs 2. Ch3.3 Proof by contrapositive 3. Ch3.4 Proof by cases 4. Ch3.5 Proof evaluations
More informationMATH 135 Fall 2006 Proofs, Part IV
MATH 135 Fall 006 s, Part IV We ve spent a couple of days looking at one particular technique of proof: induction. Let s look at a few more. Direct Here we start with what we re given and proceed in a
More informationClassroom. The Arithmetic Mean Geometric Mean Harmonic Mean: Inequalities and a Spectrum of Applications
Classroom In this section of Resonance, we invite readers to pose questions likely to be raised in a classroom situation. We may suggest strategies for dealing with them, or invite responses, or both.
More informationOlympiad Correspondence Problems. Set 3
(solutions follow) 1998-1999 Olympiad Correspondence Problems Set 3 13. The following construction and proof was proposed for trisecting a given angle with ruler and Criticizecompasses the arguments. Construction.
More informationYear 1: Fall. Year 1: Spring. HSB Topics - 2 Year Cycle
Year 1: Fall Pigeonhole 1 Pigeonhole 2 Induction 1 Induction 2 Inequalities 1 (AM-GM) Geometry 1 - Triangle Area Ratio Theorem (TART) Contest (Math Battle) Geometry 2 - Inscribed Quadrilaterals, Ptolemy
More informationSixth Term Examination Papers 9465 MATHEMATICS 1 THURSDAY 8 JUNE 2017
Sixth Term Examination Papers 9465 MATHEMATICS 1 THURSDAY 8 JUNE 217 INSTRUCTIONS TO CANDIDATES AND INFORMATION FOR CANDIDATES six six Calculators are not permitted. Please wait to be told you may begin
More informationRoots and Coefficients of a Quadratic Equation Summary
Roots and Coefficients of a Quadratic Equation Summary For a quadratic equation with roots α and β: Sum of roots = α + β = and Product of roots = αβ = Symmetrical functions of α and β include: x = and
More information2009 Euclid Contest. Solutions
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 009 Euclid Contest Tuesday, April 7, 009 Solutions c 009
More informationInternational GCSE Mathematics Formulae sheet Higher Tier. In any triangle ABC. Sine Rule = = Cosine Rule a 2 = b 2 + c 2 2bccos A
Arithmetic series Sum to n terms, S n = n 2 The quadratic equation International GCSE Mathematics Formulae sheet Higher Tier [2a + (n 1)d] Area The solutions of ax 2 + bx + c = 0 where a ¹ 0 are given
More informationYear 11 Mathematics: Specialist Course Outline
MATHEMATICS LEARNING AREA Year 11 Mathematics: Specialist Course Outline Text: Mathematics Specialist Units 1 and 2 A.J. Unit/time Topic/syllabus entry Resources Assessment 1 Preliminary work. 2 Representing
More information2016 Notes from the Marking Centre - Mathematics
2016 Notes from the Marking Centre - Mathematics Question 11 (a) This part was generally done well. Most candidates indicated either the radius or the centre. Common sketching a circle with the correct
More information11 is the same as their sum, find the value of S.
Answers: (998-99 HKMO Final Events) Created by: Mr. Francis Hung Last updated: July 08 Individual Events I P 4 I a 8 I a 6 I4 a I5 a IS a Q 8 b 0 b 7 b b spare b 770 R c c c c 0 c 57 S 0 d 000 d 990 d
More information1. Suppose that a, b, c and d are four different integers. Explain why. (a b)(a c)(a d)(b c)(b d)(c d) a 2 + ab b = 2018.
New Zealand Mathematical Olympiad Committee Camp Selection Problems 2018 Solutions Due: 28th September 2018 1. Suppose that a, b, c and d are four different integers. Explain why must be a multiple of
More informationSequences. 1. Number sequences. 2. Arithmetic sequences. Consider the illustrated pattern of circles:
Sequences 1. Number sequences Consider the illustrated pattern of circles: The first layer has just one blue ball. The second layer has three pink balls. The third layer has five black balls. The fourth
More information= 10 such triples. If it is 5, there is = 1 such triple. Therefore, there are a total of = 46 such triples.
. Two externally tangent unit circles are constructed inside square ABCD, one tangent to AB and AD, the other to BC and CD. Compute the length of AB. Answer: + Solution: Observe that the diagonal of the
More informationBaltic Way 2003 Riga, November 2, 2003
altic Way 2003 Riga, November 2, 2003 Problems and solutions. Let Q + be the set of positive rational numbers. Find all functions f : Q + Q + which for all x Q + fulfil () f ( x ) = f (x) (2) ( + x ) f
More informationYou must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Write your name here Surname Other names Pearson Edexcel International GCSE Mathematics A Paper 3HR Centre Number Wednesday 14 May 2014 Morning Time: 2 hours Candidate Number Higher Tier Paper Reference
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 2 NOVEMBER 2011
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY NOVEMBER 011 Mark Scheme: Each part of Question 1 is worth four marks which are awarded solely for the correct
More informationWeb Solutions for How to Read and Do Proofs
Web Solutions for How to Read and Do Proofs An Introduction to Mathematical Thought Processes Sixth Edition Daniel Solow Department of Operations Weatherhead School of Management Case Western Reserve University
More information3301/1H. MATHEMATICS (SPECIFICATION A) 3301/1H Higher Tier Paper 1 Non-Calculator. General Certificate of Secondary Education November 2005
Surname Other Names Leave blank Centre Number Candidate Number Candidate Signature General Certificate of Secondary Education November 2005 MATHEMATICS (SPECIFICATION A) 330/H Higher Tier Paper Non-Calculator
More informationHere is a link to the formula booklet:
IB MATH SL2 SUMMER ASSIGNMENT review of topics from year 1. We will be quizzing on this when you return to school. This review is optional but you will earn bonus points if you complete it. Questions?
More informationNAME: Date: HOMEWORK: C1. Question Obtained. Total/100 A 80 B 70 C 60 D 50 E 40 U 39
NAME: Date: HOMEWORK: C1 Question Obtained 1 2 3 4 5 6 7 8 9 10 Total/100 A 80 B 70 C 60 D 50 E 40 U 39 1. Figure 2 y A(1, 7) B(20, 7) D(8, 2) O x C(p, q) The points A(1, 7), B(20, 7) and C(p, q) form
More informationMathematics Revision Guides Vectors Page 1 of 19 Author: Mark Kudlowski M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier VECTORS
Mathematics Revision Guides Vectors Page of 9 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Higher Tier VECTORS Version:.4 Date: 05-0-05 Mathematics Revision Guides Vectors Page of 9 VECTORS
More informationTime: 1 hour 30 minutes
Paper Reference (complete below) Centre No. Surname Initial(s) Candidate No. Signature Paper Reference(s) 6663 Edexcel GCE Pure Mathematics C Advanced Subsidiary Specimen Paper Time: hour 30 minutes Examiner
More informationSMT 2011 General Test and Solutions February 19, F (x) + F = 1 + x. 2 = 3. Now take x = 2 2 F ( 1) = F ( 1) = 3 2 F (2)
SMT 0 General Test and Solutions February 9, 0 Let F () be a real-valued function defined for all real 0, such that ( ) F () + F = + Find F () Answer: Setting =, we find that F () + F ( ) = Now take =,
More informationBaltic Way 2008 Gdańsk, November 8, 2008
Baltic Way 008 Gdańsk, November 8, 008 Problems and solutions Problem 1. Determine all polynomials p(x) with real coefficients such that p((x + 1) ) = (p(x) + 1) and p(0) = 0. Answer: p(x) = x. Solution:
More information2017 Canadian Team Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 017 Canadian Team Mathematics Contest April 017 Solutions 017 University of Waterloo 017 CTMC Solutions Page Individual Problems
More informationPossible C2 questions from past papers P1 P3
Possible C2 questions from past papers P1 P3 Source of the original question is given in brackets, e.g. [P1 January 2001 Question 1]; a question which has been edited is indicated with an asterisk, e.g.
More informationThe Geometric Mean and the AM-GM Inequality
The Geometric Mean and the AM-GM Inequality John Treuer February 27, 2017 1 Introduction: The arithmetic mean of n numbers, better known as the average of n numbers is an example of a mathematical concept
More informationIndividual Round CHMMC November 20, 2016
Individual Round CHMMC 20 November 20, 20 Problem. We say that d k d k d d 0 represents the number n in base 2 if each d i is either 0 or, and n d k ( 2) k + d k ( 2) k + + d ( 2) + d 0. For example, 0
More informationHANOI OPEN MATHEMATICAL COMPETITON PROBLEMS
HANOI MATHEMATICAL SOCIETY NGUYEN VAN MAU HANOI OPEN MATHEMATICAL COMPETITON PROBLEMS HANOI - 2013 Contents 1 Hanoi Open Mathematical Competition 3 1.1 Hanoi Open Mathematical Competition 2006... 3 1.1.1
More information2010 Euclid Contest. Solutions
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Euclid Contest Wednesday, April 7, 00 Solutions 00 Centre
More informationCircles, Mixed Exercise 6
Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5
More information*P59022A0228* International GCSE Mathematics Formulae sheet Higher Tier DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA DO NOT WRITE IN THIS AREA
Arithmetic series Sum to n terms, S n = n 2 The quadratic equation International GCSE Mathematics Formulae sheet Higher Tier [2a + (n 1)d] Area The solutions of ax 2 + bx + c = 0 where a ¹ 0 are given
More informationINTER-SCHOOL MATHEMATICAL COMPETITION 1978
INTER-SCHOOL MATHEMATICAL COMPETITION 978 The problems of the Inter-school Mathematical Competition 978 are reproduced below. The Competition Subcommittee expresses its regret that the original version
More informationKENDRIYA VIDYALAYA GACHIBOWLI, GPRA CAMPUS, HYD 32
KENDRIYA VIDYALAYA GACHIBOWLI, GPRA CAMPUS, HYD 32 SAMPLE PAPER 05 (2018-19) SUBJECT: MATHEMATICS(041) BLUE PRINT : CLASS X Unit Chapter VSA (1 mark) SA I (2 marks) SA II (3 marks) LA (4 marks) Total Unit
More information2011 Olympiad Solutions
011 Olympiad Problem 1 Let A 0, A 1, A,..., A n be nonnegative numbers such that Prove that A 0 A 1 A A n. A i 1 n A n. Note: x means the greatest integer that is less than or equal to x.) Solution: We
More informationSenior Math Circles February 18, 2009 Conics III
University of Waterloo Faculty of Mathematics Senior Math Circles February 18, 2009 Conics III Centre for Education in Mathematics and Computing Eccentricity of Conics Fix a point F called the focus, a
More informationYou must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Write your name here Surname Other names Pearson Edexcel International GCSE Mathematics A Paper 4HR Centre Number Tuesday 16 January 2018 Morning Time: 2 hours Candidate Number Higher Tier Paper Reference
More information2010 Fermat Contest (Grade 11)
Canadian Mathematics Competition An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 010 Fermat Contest (Grade 11) Thursday, February 5, 010
More informationSTEP Support Programme. Hints and Partial Solutions for Assignment 5
STEP Support Programme Hints and Partial Solutions for Assignment 5 Warm-up 1 (i) As always, a diagram might be useful: cos θ b c and sin θ a c, so (using Pythagoras Theorem), cos 2 θ + sin 2 θ b2 + a
More informationMathematics A Paper 3HR
P45864A 2016 Pearson Education Ltd. 1/1/1/1/ Write your name here Surname Pearson Edexcel International GCSE Mathematics A Paper 3HR Thursday 26 May 2016 Morning Time: 2 hours Centre Number Other names
More informationx n+1 = ( x n + ) converges, then it converges to α. [2]
1 A Level - Mathematics P 3 ITERATION ( With references and answers) [ Numerical Solution of Equation] Q1. The equation x 3 - x 2 6 = 0 has one real root, denoted by α. i) Find by calculation the pair
More informationMathematics A Level 1/2 Paper 2H
Write your name here Surname Other names Pearson Edexcel International GCSE Mathematics A Level 1/2 Paper 2H Specimen Paper Time: 2 hours Centre Number Candidate Number Higher Tier Paper Reference 4MA1/2H
More informationPURE MATHEMATICS AM 27
AM Syllabus (014): Pure Mathematics AM SYLLABUS (014) PURE MATHEMATICS AM 7 SYLLABUS 1 AM Syllabus (014): Pure Mathematics Pure Mathematics AM 7 Syllabus (Available in September) Paper I(3hrs)+Paper II(3hrs)
More informationPLC Papers. Created For:
PLC Papers Created For: Algebra and proof 2 Grade 8 Objective: Use algebra to construct proofs Question 1 a) If n is a positive integer explain why the expression 2n + 1 is always an odd number. b) Use
More informationFurther Mathematics Summer work booklet
Further Mathematics Summer work booklet Further Mathematics tasks 1 Skills You Should Have Below is the list of the skills you should be confident with before starting the A-Level Further Maths course:
More informationYou must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Write your name here Surname Other names Pearson Edexcel International GCSE Mathematics A Paper 3HR Thursday 26 May 2016 Morning Time: 2 hours Centre Number Candidate Number Higher Tier Paper Reference
More informationNational Quali cations Date of birth Scottish candidate number
N5FOR OFFICIAL USE X747/75/01 FRIDAY, 5 MAY 1:00 PM :00 PM National Quali cations 017 Mark Mathematics Paper 1 (Non-Calculator) *X7477501* Fill in these boxes and read what is printed below. Full name
More information36th United States of America Mathematical Olympiad
36th United States of America Mathematical Olympiad 1. Let n be a positive integer. Define a sequence by setting a 1 = n and, for each k > 1, letting a k be the unique integer in the range 0 a k k 1 for
More informationCHAPTER 1 NUMBER SYSTEMS. 1.1 Introduction
N UMBER S YSTEMS NUMBER SYSTEMS CHAPTER. Introduction In your earlier classes, you have learnt about the number line and how to represent various types of numbers on it (see Fig..). Fig.. : The number
More information2015 Canadian Intermediate Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 015 Canadian Intermediate Mathematics Contest Wednesday, November 5, 015 (in North America and South America) Thursday, November
More informationIYGB. Special Paper U. Time: 3 hours 30 minutes. Created by T. Madas. Created by T. Madas
IYGB Special Paper U Time: 3 hours 30 minutes Candidates may NOT use any calculator Information for Candidates This practice paper follows the Advanced Level Mathematics Core Syllabus Booklets of Mathematical
More informationBREVET BLANC N 1 JANUARY 2012
Exercise. (5 pts) duration: h Presentation and explanations (4 points) Numerical Activities Consider the figure opposite, which is made up of two squares.. a) Calculate the area A of the white part. b)
More informationB293A. MATHEMATICS B (MEI) Paper 3 Section A (Higher Tier) GENERAL CERTIFICATE OF SECONDARY EDUCATION. Tuesday 12 January 2010 Morning WARNING
H GENERAL CERTIFICATE OF SECONDARY EDUCATION MATHEMATICS B (MEI) Paper 3 Section A (Higher Tier) B293A *OCE/14183* Candidates answer on the Question Paper OCR Supplied Materials: None Other Materials Required:
More informationCOT 2104 Homework Assignment 1 (Answers)
1) Classify true or false COT 2104 Homework Assignment 1 (Answers) a) 4 2 + 2 and 7 < 50. False because one of the two statements is false. b) 4 = 2 + 2 7 < 50. True because both statements are true. c)
More informationb UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW = (8 2) + ( 2 6) = 100
Circles 6F a U(, 8), V(7, 7) and W(, ) UV = ( x x ) ( y y ) = (7 ) (7 8) = 8 VW = ( 7) ( 7) = 64 UW = ( ) ( 8) = 8 Use Pythagoras' theorem to show UV UW = VW 8 8 = 64 = VW Therefore, UVW is a right-angled
More informationCandidates are expected to have available a calculator. Only division by (x + a) or (x a) will be required.
Revision Checklist Unit C2: Core Mathematics 2 Unit description Algebra and functions; coordinate geometry in the (x, y) plane; sequences and series; trigonometry; exponentials and logarithms; differentiation;
More informationDO NOT OPEN THIS TEST BOOKLET UNTIL YOU ARE ASKED TO DO SO
DO NOT OPEN THIS TEST BOOKLET UNTIL YOU ARE ASKED TO DO SO T.B.C. : P-AQNA-L-ZNGU Serial No.- TEST BOOKLET MATHEMATICS Test Booklet Series Time Allowed : Two Hours and Thirty Minutes Maximum Marks : 00
More information8.5 Taylor Polynomials and Taylor Series
8.5. TAYLOR POLYNOMIALS AND TAYLOR SERIES 50 8.5 Taylor Polynomials and Taylor Series Motivating Questions In this section, we strive to understand the ideas generated by the following important questions:
More informationSydney University Mathematical Society Problems Competition Solutions.
Sydney University Mathematical Society Problems Competition 005 Solutions 1 Suppose that we look at the set X n of strings of 0 s and 1 s of length n Given a string ɛ = (ɛ 1,, ɛ n ) X n, we are allowed
More information1. Number a. Using a calculator or otherwise 1 3 1 5 i. 3 1 4 18 5 ii. 0.1014 5.47 1.5 5.47 0.6 5.1 b. Bus tour tickets . Algebra a. Write as a single fraction 3 4 11 3 4 1 b. 1 5 c. Factorize completely
More informationYEAR 9 SCHEME OF WORK - EXTENSION
YEAR 9 SCHEME OF WORK - EXTENSION Autumn Term 1 Powers and roots Spring Term 1 Multiplicative reasoning Summer Term 1 Graphical solutions Quadratics Non-linear graphs Trigonometry Half Term: Assessment
More informationYou must have: Ruler graduated in centimetres and millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Tracing paper may be used.
Write your name here Surname Other names Edexcel Certificate Edexcel International GCSE Mathematics A Paper 3H Friday 11 May 2012 Afternoon Time: 2 hours Centre Number Candidate Number Higher Tier Paper
More information184/09 MATHEMATICS HIGHER TIER PAPER 1. P.M. MONDAY, 4 June (2 Hours) CALCULATORS ARE NOT TO BE USED FOR THIS PAPER
Candidate Name Centre Number Candidate Number WELSH JOINT EDUCATION COMMITTEE General Certificate of Secondary Education CYD-BWYLLGOR ADDYSG CYMRU Tystysgrif Gyffredinol Addysg Uwchradd 184/09 MATHEMATICS
More informationHMMT February 2018 February 10, 2018
HMMT February 018 February 10, 018 Algebra and Number Theory 1. For some real number c, the graphs of the equation y = x 0 + x + 18 and the line y = x + c intersect at exactly one point. What is c? 18
More informationSome practice questions for CIMC.
1. Determine the value of {2008 Cayley #2} Some practice questions for CIMC. 25 16 25 16. 25 16 25 16 = 9 5 4 = 3 1 = 3 2. In the diagram, P T and QS are straight lines intersecting at R such that QP =
More informationVerulam School Mathematics. Year 9 Revision Material (with answers) Page 1
Verulam School Mathematics Year 9 Revision Material (with answers) Page 1 Q1. (a) Simplify a 2 a 4 Answer... (b) Simplify b 9 b 3 Answer... (c) Simplify c 5 c c 5 Answer... (Total 3 marks) Q2. (a) Expand
More informationMath Day at the Beach 2016
Multiple Choice Write your name and school and mark your answers on the answer sheet. You have 30 minutes to work on these problems. No calculator is allowed. 1. What is the median of the following five
More informationHANOI OPEN MATHEMATICS COMPETITON PROBLEMS
HANOI MATHEMATICAL SOCIETY NGUYEN VAN MAU HANOI OPEN MATHEMATICS COMPETITON PROBLEMS 2006-2013 HANOI - 2013 Contents 1 Hanoi Open Mathematics Competition 3 1.1 Hanoi Open Mathematics Competition 2006...
More informationIn Z: x + 3 = 2 3x = 2 x = 1 No solution In Q: 3x = 2 x 2 = 2. x = 2 No solution. In R: x 2 = 2 x = 0 x = ± 2 No solution Z Q.
THE UNIVERSITY OF NEW SOUTH WALES SCHOOL OF MATHEMATICS AND STATISTICS MATH 1141 HIGHER MATHEMATICS 1A ALGEBRA. Section 1: - Complex Numbers. 1. The Number Systems. Let us begin by trying to solve various
More informationCN#5 Objectives 5/11/ I will be able to describe the effect on perimeter and area when one or more dimensions of a figure are changed.
CN#5 Objectives I will be able to describe the effect on perimeter and area when one or more dimensions of a figure are changed. When the dimensions of a figure are changed proportionally, the figure will
More information2003 Solutions Pascal Contest (Grade 9)
Canadian Mathematics Competition An activity of The Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario 00 Solutions Pascal Contest (Grade 9) for The CENTRE for
More informationThe Bridge to A level. Diagnosis Worked Solutions
The Bridge to A level Diagnosis Worked Solutions 1 1 Solving quadratic equations Solve x 2 + 6x + 8 = 0 (x + 2)(x + 4) = 0 x = 2 or 4 Solve the equation y 2 7y + 12 = 0 Hence solve the equation x 4 7x
More informationCO-ORDINATE GEOMETRY. 1. Find the points on the y axis whose distances from the points (6, 7) and (4,-3) are in the. ratio 1:2.
UNIT- CO-ORDINATE GEOMETRY Mathematics is the tool specially suited for dealing with abstract concepts of any ind and there is no limit to its power in this field.. Find the points on the y axis whose
More informationThursday 11 June 2015 Afternoon
Oxford Cambridge and RSA H Thursday 11 June 2015 Afternoon GCSE METHODS IN MATHEMATICS B392/02 Methods in Mathematics 2 (Higher Tier) *4856252055* Candidates answer on the Question Paper. OCR supplied
More information19 Oliver records the distance from London to each of eight cities in the USA.
DO NOT WRITE IN THIS RE DO NOT WRITE IN THIS RE DO NOT WRITE IN THIS RE DO NOT WRITE IN THIS RE DO NOT WRITE IN THIS RE DO NOT WRITE IN THIS RE Title: Mock Set 3 Paper 2H Date: 1. Oliver records the distance
More informationExhaustion: From Eudoxus to Archimedes
Exhaustion: From Eudoxus to Archimedes Franz Lemmermeyer April 22, 2005 Abstract Disclaimer: Eventually, I plan to polish this and use my own diagrams; so far, most of it is lifted from the web. Exhaustion
More informationTHE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION. Sunday, March 14, 2004 Time: hours No aids or calculators permitted.
THE UNIVERSITY OF TORONTO UNDERGRADUATE MATHEMATICS COMPETITION Sunday, March 1, Time: 3 1 hours No aids or calculators permitted. 1. Prove that, for any complex numbers z and w, ( z + w z z + w w z +
More informationCandidate Number. General Certificate of Secondary Education Higher Tier June 2013
Centre Number Surname Candidate Number For Examiner s Use Other Names Candidate Signature Examiner s Initials General Certificate of Secondary Education Higher Tier June 2013 Pages 2 3 4 5 Mark Mathematics
More informationProblems and Solutions: INMO-2012
Problems and Solutions: INMO-2012 1. Let ABCD be a quadrilateral inscribed in a circle. Suppose AB = 2+ 2 and AB subtends 135 at the centre of the circle. Find the maximum possible area of ABCD. Solution:
More informationYear 8. Semester 2 Revisions
Semester Revisions Year 8 Semester Two Revisions 01 Students are advised to review previously completed tests, assignments and homework followed by this semester one revision booklet. Be sure to seek extra
More informationC.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series
C.7 Numerical series Pag. 147 Proof of the converging criteria for series Theorem 5.29 (Comparison test) Let and be positive-term series such that 0, for any k 0. i) If the series converges, then also
More informationSixth Form Entrance Mathematics
Sixth Form Entrance 2016 Mathematics 1 hour Attempt all questions if possible. Do not worry if there are topics you have never covered; do your best on whatever you can attempt. Questions are not necessarily
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mathematics SKE, Strand J STRAND J: TRANSFORMATIONS, VECTORS and MATRICES J4 Matrices Text Contents * * * * Section J4. Matrices: Addition and Subtraction J4.2 Matrices: Multiplication J4.3 Inverse Matrices:
More informationSome Basic Logic. Henry Liu, 25 October 2010
Some Basic Logic Henry Liu, 25 October 2010 In the solution to almost every olympiad style mathematical problem, a very important part is existence of accurate proofs. Therefore, the student should be
More informationBRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST,
BRITISH COLUMBIA SECONDARY SCHOOL MATHEMATICS CONTEST, 014 Solutions Junior Preliminary 1. Rearrange the sum as (014 + 01 + 010 + + ) (013 + 011 + 009 + + 1) = (014 013) + (01 011) + + ( 1) = 1 + 1 + +
More information2013 Canadian Senior Mathematics Contest
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2013 Canadian Senior Mathematics Contest Thursday, November 21, 2013 (in North America and South America) Friday, November 22, 2013
More informationMethods in Mathematics (Linked Pair Pilot)
Centre Number Surname Candidate Number For Examiner s Use Other Names Candidate Signature Examiner s Initials General Certificate of Secondary Education Higher Tier January 2013 Pages 3 4 5 Mark Methods
More informationDiagnostic Assessment Number and Quantitative Reasoning
Number and Quantitative Reasoning Select the best answer.. Which list contains the first four multiples of 3? A 3, 30, 300, 3000 B 3, 6, 9, 22 C 3, 4, 5, 6 D 3, 26, 39, 52 2. Which pair of numbers has
More information