Mind Action Series Mathematics Grade 12 Pdf Download

Unformatted text preview: ALL RIGHTS RESERVED ©COPYRIGHT BY THE AUTHOR The whole or any part of this publication may not be reproduced or transmitted in any form or by any means without permission in writing from the publisher. This includes electronic or mechanical, including photocopying, recording, or any information storage and retrieval system. Every effort has been made to obtain copyright of all printed aspects of this publication. However, if material requiring copyright has unwittingly been used, the copyrighter is requested to bring the matter to the attention of the publisher so that the due acknowledgement can be made by the author. Maths Textbook Grade 12 NCAPS ISBN 13: 978-1-86921-482-1 Product Code: MAT 106 Authors: M.D Phillips J. Basson C. Botha First Edition: April 2013 Approved for the National Catalogue PUBLISHERS ALLCOPY PUBLISHERS P.O. Box 963 Sanlamhof, 7532 Tel: (021) 945 4111, Fax: (021) 945 4118 Email: [email protected] Website: i MATHS GRADE 12 NCAPS TEXTBOOK INFORMATION The AUTHORS MARK DAVID PHILLIPS B.SC, H Dip Ed, B.Ed (cum laude) University of the Witwatersrand. The author has over twenty years of teaching experience in both state and private schools. He has led the Mathematics Department in two schools and is currently teaching at The King's School West Rand. The author has presented teacher training and learner seminars for various educational institutions including Excelearn, Isabelo, Kutlwanong Centre for Maths, Science and Technology, Learning Channel, Oracle and UJ. Mark is currently a TV presenter for "Matrics Uploaded" on SABC 1. He has previously published highly successful Mathematics textbooks through Allcopy Publishers and is a qualified assessor. Mark has travelled extensively in the United Kingdom, Europe and America, gaining invaluable international educational experience. The textbook contains methods used in top schools all over the world. JURGENS BASSON BA (Mathematics and Psychology) HED Randse Afrikaanse Universiteit (UJ) Jurgens Basson has over twenty years of teaching experience in secondary and tertiary education. He is the founder of The Maths Centre, a private study centre which offers extra maths and science tuition to more than 500 students in two different areas in Gauteng. In 2003, he established the RAU/Oracle School of Maths. Over 2000 previously disadvantaged students have had the opportunity to benefit from this programme. With the introduction of the new curriculum in 2006, Jurgens saw the need to train and empower educators to implement the new curriculum. He obtained a sponsorship from Oracle to do educator development workshops in five of the nine provinces to date. Jurgens is a highly sought after educational consultant with a passion for Maths and the new curriculum. In 2010, he served on the panel to finalise the CAPS curriculum for Maths. CONRAD BOTHA Maths Hons, PGCE, B.Sc (Mathematics and Psychology) Rand Afrikaans University. Conrad Botha has been an educator in both state and private schools for the past eight years. He is currently Head of Department at The King's School West Rand and has also been extensively involved in teaching Maths at The Maths Centre (Grades 4-12). The institution tutors both primary and high school learners in Maths, Science as well as conducts examination training sessions for Grade 11 and 12 learners. The experience he obtained at The Maths Centre resulted in him gaining valuable knowledge and teaching methodology, particularly in the implementation of the new curriculum. Conrad has also done editing work for the Learning Channel. The CONCEPT The purpose of this publication is to cover all Maths topics in the new curriculum (CAPS) using a user-friendly, modern approach. The examples presented in each chapter of the textbook cover all of the main concepts in each topic. There is a logical, to-the-point, progression from one example to the next and the exercises reinforce the concepts inherent in the particular topic dealt with. At the end of each chapter there is a mixed revision exercise. This exercise is for revising all of the concepts dealt with in the chapter. There is also a "Some Challenges" exercise which provides invaluable extension and problem solving for top learners. At the end of the book, short answers to all of the exercises have been provided. ENDORSEMENT "This textbook has always been a life-saver for me. I just don't have the time to waste trying to create my own lessons using other books and then still trying to get to assess my learners in the way that we are supposed to. This textbook has helped me to get through the content as quickly and effectively as possible leaving more time for me to assess my learners. I can also have a life outside of school and not get so wrapped up with so much work. I like the assessment tasks in the Teacher Guide. I have used them to meet the requirements in the policy documents." Hester Jansen Van Vuuren, Educator ii MATHEMATICS TEXTBOOK GRADE 12 NCAPS CONTENTS 1 CHAPTER 1 SEQUENCES AND SERIES CHAPTER 2 INVERSE FUNCTIONS 38 CHAPTER 3 FINANCIAL MATHEMATICS 62 CHAPTER 4 TRIGONOMETRY 110 CHAPTER 5 ALGEBRA 147 CHAPTER 6 CALCULUS 160 CHAPTER 7 ANALYTICAL GEOMETRY 217 CHAPTER 8 EUCLIDEAN GEOMETRY 233 CHAPTER 9 STATISTICS 288 CHAPTER 10 THE COUNTING PRINCIPLE AND PROBABILITY 305 328 SOLUTIONS TO EXERCISES iii iv CHAPTER 1 – SEQUENCES AND SERIES In Grade 10 and 11 you learnt about linear and quadratic number patterns. Linear number patterns have a constant difference between consecutive terms while quadratic number patterns have a constant second difference. REVISION OF QUADRATIC NUMBER PATTERNS In Grade 11 we dealt with quadratic number patterns having a general term of the form Tn = an 2 + bn + c . EXAMPLE Consider the following number pattern: 2 ; 3 ; 6 ;11; ... (a) Determine the nth term (general term) and hence the value of the 42nd term. (b) Determine which term will equal 1091. Solutions (a) a+b+c 3a + b 2 3 1 2a 6 3 2 2a = 2 ∴a = 1 11 5 2 3a + b = 1 ∴ 3(1) + b = 1 ∴ b = −2 a+b+c = 2 ∴ (1) + (−2) = 2 ∴c = 3 ∴ Tn = n 2 − 2n + 3 ∴ T42 = (42)2 − 2(42) + 3 = 1683 (b) Tn = 1091 ∴ n 2 − 2n + 3 = 1091 ∴ n 2 − 2n − 1088 = 0 ∴ (n − 34)(n + 32) = 0 ∴ n = 34 or n = −32 But n ≠ −32 ∴ n = 34 The 34th term will equal 1091 REVISION EXERCISE 1. Determine the general term for each number pattern below. (a) (d) 2; 6 ; 14 ; 26 ; ... −1; 0 ; 3 ; 8 ; ... 4 ; 9;16 ; 25 ; ... (c) −3 ; − 6; − 11; − 18 ; ... (f) (b) (e) 1 1; 3 ; 6 ;10 ; ... 10 ; 6 ; 3 ;1; ... 2. 3. 4. Consider the following number pattern: 1 ; 6 ; 15 ; 28 ; ... (a) Determine the general term and hence the value of the 20th term. (b) Determine which term will equal 3160. Consider the following number pattern: −4 ; − 10 ; − 18 ; − 28 ; ... (a) Determine the general term and hence the value of the 25th term. (b) Determine which term will equal −810 . Consider the following pattern that emerges when you add the terms of a linear number pattern. First the linear number pattern: 3 ; 7 ; 11 ; 15 ; ... Line 1: 3 Line 2: Line 3: 3 + 7 = 10 3 + 7 + 11 = 21 Line 4: 3 + 7 + 11 + 15 = 36 . . . (a) Determine the sum of the numbers in Line 5 and 6. (b) What type of number pattern is formed by the sum of the numbers in each line? (c) Hence or otherwise determine the sum of the numbers in Line n. ARITHMETIC SEQUENCES Consider the following linear number pattern: 7 ; 10 ; 13 ; 16 ; 19 ; ... We can rewrite this pattern using only the first term and the constant difference. T1 = 7 and the constant or common difference = T2 − T1 = T3 − T2 = T4 − T3 = ... = 3 T2 = 7 + 3 T3 = 7 + 3 + 3 = 7 + 2(3) T4 = 7 + 3 + 3 + 3 = 7 + 3(3) T5 = 7 + 3 + 3 + 3 + 3 = 7 + 4(3) T10 = 7 + 9(3) T100 = 7 + 99(3) ∴ Tn = 7 + ( n − 1) (3) Therefore the general term of the pattern is: Tn = 7 + (n − 1)(3) The general term can be simplified further as follows: Tn = 7 + (n − 1)(3) ∴ Tn = 7 + 3n − 3 ∴ Tn = 4 + 3n If the letter a is used for the first term and d for the constant difference of a linear pattern, then the pattern can be written as follows: T1 = a T2 = a + d T3 = (a + d ) + d = a + 2d T4 = (a + 2d ) + d = a + 3d 2 T5 = (a + 3d ) + d = a + 4d T6 = a + 5d T10 = a + 9d T100 = a + 99d ∴ Tn = a + (n − 1)d Therefore the general term of the pattern is: Tn = a + (n − 1)d Linear patterns are also called arithmetic sequences and have a general term of Tn = a + (n − 1)d where: a d n Tn represents the first term represents the constant or common difference represents the position of the term represents the nth term or general term (the value of the term in the nth position) EXAMPLE 1 Consider the arithmetic sequence 3 ; 5 ; 7 ; 9 ; ... (a) Determine a formula for the general term of the above sequence. (b) Find the value of the 50th term. Solution (a) T1 = 3 ∴a = 3 T2 − T1 = 5 − 3 = 2 T3 − T2 = 7 − 5 = 2 ∴d = 2 To find the general term Tn you have to substitute a = 3 and d = 2 into the general term for an arithmetic sequence, namely Tn = a + (n − 1)d ∴ Tn = 3 + (n − 1)(2) ∴ Tn = 3 + 2n − 2 (b) ∴ Tn = 2n + 1 Tn = 2n + 1 ∴ T50 = 2(50) + 1 = 101 EXAMPLE 2 Consider the following sequence −5 ; − 9 ; − 13 ; − 17 ;... (a) Show that the sequence is arithmetic. (b) Find the value of the 25th term of the sequence. Solutions (a) T2 − T1 = −9 − (−5) = −4 T2 − T1 = −13 − (−9) = −4 (b) Tn = a + (n − 1)d a = −5, d = −4 and n = 25 There is a constant difference of − 4 3 ∴ T25 = −5 + (25 − 1)( −4) = −101 EXAMPLE 3 Determine which term of the sequence −1 ; 2 ; 5 ; 8 ; ... is equal to 80. Solution T2 − T1 = 2 − (−1) = 3 and T3 − T2 = 5 − 2 = 3 ∴ d = 3 (the sequence is arithmetic) Write down what is given: a = −1, d = 3 and Tn = 80 (we know the actual term's value but not the position or n value) Tn = a + (n − 1)d (state the formula) ∴ 80 = −1 + ( n − 1)(3) (substitute a = −1, d = 3 and Tn = 80 ) ∴ 80 = −1 + 3n − 3 ∴ 84 = 3n ∴ n = 28 ∴ T28 = 80 EXAMPLE 4 x ; 4 x + 5 ; 10 x − 5 are the first three terms of an arithmetic sequence. Determine the value of x and hence the sequence. Solution Since the sequence is arithmetic, it is clear that d = T2 − T1 = (4 x + 5) − ( x) = 3x − 5 d = T3 − T2 = (10 x − 5) − (4 x + 5) = 10 x − 5 − 4 x − 5 = 6 x − 10 ∴ 3x + 5 = 6 x − 10 ∴− 3x = −15 ∴x = 5 ∴ T1 = x = 5 ∴ T2 = 4 x + 5 = 4(5) + 5 = 25 ∴ T3 = 10 x − 5 = 10(5) − 5 = 45 ∴ The sequence is 5 ; 25 ; 45 ; ........ EXERCISE 1 1. 2. Determine the general term of the following arithmetic sequences: (a) (c) −1 ;3 ; 7 ;.... (b) 4 ; − 2 ; −8 ; ...... 1 ; − 1 ; − 3 ; .... (d) 99;106;113;.... Determine the 38th term for each of the following arithmetic sequences: (a) (d) −4; − 8; − 12;.... 21 9 6; ; ; .... 4 2 (b) 2; − 1, 5; − 5;... (c) 99; 88; 77 ;.... (e) Tk = 3k − 4 (f) Tk = −2k + 5 4 3. 4. 5. 6. 7. (a) Which term of the arithmetic sequence −5; − 2;1;.... is equal to 94? (b) Which term of the arithmetic sequence 4 ; 2, 5 ; 1 ; ... is equal to −66, 5? (c) Find the number of terms in the sequence 12 ; 7 ; 2 ; ....; −203 . (d) Find the number of terms in the sequence −55; − 48; − 41;....; 85 Given the arithmetic sequence: −29; − 23; −17;.... (a) Determine the 31st term. (b) Determine which term is equal to 31. Given the arithmetic sequence: −13; − 9; −5;.... (a) Which term in the above sequence is 51? (b) Calculate the 51st term. p ; 2 p + 2 ; 5 p + 3 ; ... are the first three terms of an arithmetic sequence. (a) Calculate the value of p. (b) Determine the sequence. (c) Find the 49th term. (d) Which term of the sequence is 100 12 ? x + 3 ; 2 x + 6 ; 3 x + 9 ; ... are the first three terms of an arithmetic sequence. (a) Determine the 10th term in terms of x. (b) Determine the nth term in terms of x. GEOMETRIC SEQUENCES Consider the following number pattern: 6 ; 12 ; 24 ; 48 ; 96 ; ... This pattern is not linear (arithmetic) since there is no constant difference between the terms. In this pattern each successive term is obtained by multiplying the previous term by 2. Notice too that the following ratios between the terms are also equal to 2: T2 T3 T4 = = = ... = 2 T1 T2 T3 There is a constant ratio for this number pattern and this kind of number pattern is called an exponential or geometric sequence. We can rewrite this pattern using only the first term and the constant ratio: T1 = 6 T2 = 6 × 2 T3 = 6 × 2 × 2 = 6 × 2 2 T4 = 6 × 2 × 2 × 2 = 6 × 23 T5 = 6 × 2 4 T10 = 6 × 29 T100 = 6 × 299 ∴ Tn = 6 × 2n −1 Therefore the general term of the pattern is: Tn = 6 × 2 n −1 If the letter a is used for the first term and r for the constant ratio of an exponential number pattern, then the pattern can be written as follows: 5 T1 = a T2 = a × r = ar T3 = ( a × r ) × r = ar 2 T4 = ( a × r 2 ) × r = ar 3 T5 = ar 4 T10 = ar 9 T100 = ar 99 ∴ Tn = ar n −1 Therefore the general term of the pattern is: Tn = ar n −1 Exponential or geometric number patterns have a general term: Tn = ar n −1 where: a represents the first term d represents the constant or common ratio n represents the position of the term Tn represents the nth term or general term (the value of the term in the nth position) EXAMPLE 5 Consider the geometric sequence 2 ; 3 ; 4,5 ; 6, 25 ; ... (a) Determine a formula for the general term of the sequence. (b) Find the value of the 15th term. Solutions (a) T1 = a = 2 T2 3 = and T1 2 ∴r = T3 4,5 3 = = T2 3 2 3 2 Substitute a = 2 and r = (b) 3 Tn = 2 2 n −1 3 Tn = 2 2 n −1 3 into the general term Tn = ar n −1 2 15−1 3 T15 = 2 2 14 3 ∴ T15 = 2 2 ∴ T15 = 583,86 (rounded off to two decimal places) 6 EXAMPLE 6 Consider the following sequence −5 ; 10 ; − 20 ; 40 ;... (a) Show that the above sequence is geometric. (b) Determine the value of the 20th term of the above sequence. Solution (a) T −20 T2 10 = = −2 and 3 = = −2 ( constant ratio = −2 ) T1 −5 T2 10 (b) Tn = ar n −1 with a = −5, r = −2 and n = 20 ∴ T20 = (−5)(−2)20−1 = (−5)(−2)19 = 2 621 440 Revision of basic exponential equations and other types Exponential equations and equations involving odd or even exponents form an integral part of the examples and exercises which follow in this chapter. It is therefore advisable to revise these basic equations before proceeding. Consider the following examples of exponential equations: Solve for n: (a) (c) 64 = 2n (b) ∴ 26 = 2 n ∴ (−2)6 = (−2) n ∴6 = n ∴n = 6 ∴6 = n ∴n = 6 1 1 = 81 3 4 n 1 1 ∴ = 3 3 ∴4 = n (d) n 1 1 = − 81 3 4 ∴n = 4 1 1 = 2 − 128 2 n n 1 1 1 1 ∴ × = 2 − × 128 2 2 2 1 1 ∴ = − 256 2 8 n 1 1 ∴ − = 3 3 ∴4 = n ∴n = 4 (d) 64 = (−2) n n 1 1 ∴ − = − 2 2 ∴n = 8 n 7 n Consider the following equations involving odd or even exponents: (a) r2 = 9 This equation can be solved by taking the square root on on both sides ∴r = ± 9 (Don't forget to include ± since there are two solutions) ∴ r = ±3 (b) r 3 = 27 In this equation, there is no need to include ± since there is only one solution. Simplify take the cube root on both sides: ∴ r = 3 27 ∴r = 3 (c) r 3 = −27 ∴ r = 3 −27 ∴ r = −3 (d) r 4 = 16 This equation can be solved by taking the fourth root on on both sides ∴ r = ± 4 16 (Don't forget to include ± since there are two solutions) ∴ r = ±2 Summary Consider the equation r n = a If n is odd, then there is one solution: r=na r = ± n a provided a > 0 If n is even, then there are two solutions: EXAMPLE 7 Determine which term in the sequence 4 ; 12 ; 36 ; 108 ; ... is equal to 8748. Solution T3 36 T2 12 = = 3 and = =3 T1 4 T2 12 ∴ r = 3 (the sequence is geometric) a = 4, r = 3 and Tn = 8748 (we know the actual term's value but not the position or n value) Tn = ar n −1 (state the formula) n−1 ∴ 8748 = 4(3) 8748 ∴ = (3) n−1 4 (substitute a = 4, r = 3 and Tn = 8748 ) ∴ 2187 = (3) n−1 ∴ 37 = 3n−1 ∴7 = n −1 ∴n = 8 ∴ T8 = 8748 8 EXAMPLE 8 k + 1 ; k − 1 ; 2 k − 5 are the first three terms of a geometric sequence. Calculate the value of k and hence determine the possible sequences. Solution T2 T3 = T1 T2 k − 1 2k − 5 = k + 1 k −1 ∴ (k − 1) 2 = (2k − 5)(k + 1) ∴ ∴ k 2 − 2k + 1 = 2k 2 − 3k − 5 ∴0 = k 2 − k − 6 ∴ 0 = (k − 3)(k + 2) ∴ k = 3 or k = −2 For k = 3 : T1 = k + 1 = 3 + 1 = 4 For k = −2 : T2 = k − 1 = 3 − 1 = 2 T3 = 2k − 5 = 2(3) − 5 = 1 The sequence is therefore: 4 ; 2 ; 1 ; ......... T1 = k + 1 = −2 + 1 = −1 T2 = k − 1 = −2 − 1 = −3 T3 = 2k − 5 = 2(−2) − 5 = −9 The sequence is therefore: −1 ; − 3 ; − 9 ; ......... EXERCISE 2 1. 2. Determine the general term for each of the following geometric sequences: 1 2 (a) (b) 2; − 1; ;..... 2;8;32;..... (c) − ; 2; − 6;..... 2 3 (d) 1 ; 0, 2 ; 0, 04;..... Determine the 9th term for each of the following geometric sequences: (a) 128; 64; 32;..... (b) 0, 25; 0, 5; 1;..... 6−k 4. 5. 4 1 ; 1 ; 4 ;..... 9 3 2 1 (d) (e) (f) Tk = 3 Tk = 2400 3 2 (a) Which term of the geometric sequence 2; 6;18;.... is equal to 4374? 3 3 3 (b) Determine the number of terms in the sequence − ; ; − ;....;192 8 4 2 Consider the following sequence: 0, 625 ; 1, 25 ; 2,5 ; 5 ; ..... (a) Determine the value of the 10th term. (b) Determine which term will equal 80. Consider the following sequences: sequence 1: sequence 2: 2 ; 2 ; 2 2 ; ........ . 2 ; 2 2 ; 3 2 ; ....... (a) Show that sequence 1 is arithmetic. (b) Show that sequence 2 is geometric. (c) Which term of sequence 1 will be equal to 200 ? (d) Which term of sequence 2 will be equal to 256 ? 2 3 ; − 1; ;..... 3 2 3. (c) 9 k −1 6. 7. x − 4 ; x + 2 ; 3 x + 1 ; ... are the first three terms of a geometric sequence. Determine the sequence if x is positive. t + 1 ; 1 − t ; 1 − 5t ; ... are the first three terms of a geometric sequence. (a) Determine the numerical value of t where t ≠ 0. (b) Determine the sequence. (c) Determine the 10th term. 2 (d) Which term equals 10 ? 3 DETERMINING THE SUM OF THE TERMS OF A SEQUENCE If T1 ; T2 ; T3 ; T4 ; ... denotes a sequence then the sum T1 + T2 + T3 + T4 + ... is called a series. A series is formed by adding the terms of a sequence. EXAMPLE 9 (a) Sequence: 1 ; 3 ; 5 ; 7 ; 9 ; …. Corresponding series: 1 + 3 + 5 + 7 + 9 + ...... (b) Sequence: 5 ; 2 ; −1; − 4 ; …. Corresponding series: 5 + 2 + (−1) + ( −4 ) + ..... or : 5 + 2 − 1 − 4 + ..... We will use the symbol Sn to denote the sum of the first n terms of a series. ∴ S1 = T1 ∴ S2 = T1 + T2 The sum of the first 2 terms ∴ S3 = T1 + T2 + T3 The sum of the first 3 terms ∴ S4 = T1 + T2 + T3 + T4 The sum of the first 4 terms ∴ Sn = T1 + T2 + T3 + ... + Tn The sum of the first n terms EXAMPLE 10 Consider the series: 1 + 4 + 7 + 10 + .... Determine: (a) S3 (b) S7 Solutions (a) S3 = 1 + 4 + 7 = 12 (b) S7 = 1 + 4 + 7 + 10 + 13 + 16 + 19 = 70 EXAMPLE 11 Determine the sum of the first 7 terms of the series with Tk = 5k − 3. Solution S7 = T1 + T2 + T3 + T4 + T5 + T6 + T7 ∴ S7 = [5(1) − 3] + [5(2) − 3] + [5(3) − 3] + [5(4) − 3] + [5(5) − 3] + [5(6) − 3] + [5(7) − 3] ∴ S7 = 2 + 7 + 12 + 17 + 22 + 27 + 32 = 119 10 SERIES AND SIGMA NOTATION The mathematical symbol ∑ is the capital letter S in the Greek alphabet. It is used as the symbol for summing a series. Consider the following series: n ∑T k = T1 + T2 + T3 + ... + Tn k =1 This is read as follows: The sum of all the terms Tk (general term) from k = 1 to k = n where n ∈ N . EXAMPLE 12 7 Calculate: ∑ (5k − 3) k =1 Solution Translating this expression is as follows: Start with k = 1 and substitute all integers from 1 up to and including 7 into the expression 5k − 3 . Simplify each substitution and then determine the sum of the numbers. Last value to be substituted into the general term General term 7 ∑ (5k − 3) = [5(1) − 3] + [5(2) − 3] + [5(3) − 3] + [5(4) − 3] + [5(5) − 3] + [5(6) − 3] + [5(7) − 3] k =1 = 2 + 7 + 12 + 17 + 22 + 27 + 32 = 119 First value to be substituted into the general term EXAMPLE 13 7 Calculate: ∑ 3.2r r=0 Solution 7 ∑ 3.2r = 3.20 + 3.21 + 3.22 + 3.23 + 3.24 + 3.25 + 3.26 + 3.27 r=0 = 3 + 6 + 12 + 24 + 48 + 96 + 192 + 384 = 765 Take note that the number of terms in the above series is actually 8 and not 7. Consider some further examples: 11 8 (a) ∑ 2k = 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + 2(6) + 2(7) + 2(8) k =1 = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 Number of terms: 8 ( 8 − 1 + 1 = 8) = 72 8 (b) ∑ 2k = 2(0) + 2(1) + 2(2) + 2(3) + 2(4) + 2(5) + 2(6) + 2(7...
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