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Reg. No.: 2011/011959/07
ISBN: 9781990949821
Mathematics
Facilitator's Guide 1/2 − Grade 12
2512-E-MAM-FG01
CAPS-aligned
Prof. C Vermeulen, Lead author
E du Plessis R Myburgh H Otto M Sherman E van Heerden
Functions (3rd degree polynomials and equations)
PREFACE
Sample
Practise papers are included in
* You will find the latest and most comprehensive information on assessment in the portfolio book and assessment plan.
Optimi's Grade 12 mathematics offering consists of two study guides and two facilitator’s guides, which are based on the concepts of Optimi’s GuidEd Learning™ model to help learners and facilitators achieve success in the study of mathematics. These books cover all work required for Grade 12 mathematics and have been compiled in accordance with the CAPS guidelines as required by the Department of Basic Education.
The study guides are supported by supplementary lesson structures on the Optimi Learning Platform (OLP), which is an online platform. These lesson structures offer continuous guidance to support and enrich the learning process. This guidance is based on the latest insights in education, cognitive psychology and neuroscience. Note that the study guides can also be used independently of the OLP.
In the following section, we explain how the study guides and facilitator’s guides have been compiled and how learners and facilitators can use these to achieve success in mathematics.
The study guides and facilitator’s guides are divided into twelve themes. Study guide 1/2 and facilitator’s guide 1/2 cover themes 1 to 5 (term 1) and theme 6 (term 2) and study guide 2/2 and facilitator’s guide 2/2 cover themes 7 to 12 (term 3) and revision (term 4). The themes correspond with the CAPS guidelines with regard to content and time allocation and represent the year plan.
Time allocation
According to the CAPS requirements, at least 4,5 hours should be spent on teaching mathematics per week. For example, 13,5 hours (three weeks × 4,5 hours per week) will be spent on teaching Theme 1 (patterns, sequences and series). Themes have not been sub-divided into lessons; you and facilitators are at liberty to complete as much content per session and per week as your progress allows.
If learners work at a slower pace, the necessary adaptations should be done so that they will still be able to master all the work in time.
Tip: Use the suggested time allocation along with your learners’ progress to plan your lessons.
Note that the teaching time referred to above does not include the time during which learners should apply and practise the knowledge and concepts they have learned. For this purpose, various exercises are provided throughout each theme.
These exercises involve different ways of applying and practising new knowledge and cover various degrees of difficulty. Learners should try to do all of these exercises. Complete solutions are provided in the facilitator’s guide.
Tip: Ensure that learners do as many of these exercises as possible. Follow up and offer support when learners struggle.
Structure of themes
Learning is a complex process. Millions of brain cells and neural pathways in our brains work together to store new information in the long-term memory so that we will be able to remember it later on.
Long-term memory is not our only type of memory and when we learn, our working memory is just as important. Working memory is different from long-term memory and has a limited capacity. This means that one’s working memory can only handle a small amount of new information at a time.
When one learns mathematics, there is a lot of new information your brain needs to process, which can easily exhaust your working memory. This is related to the cognitive load theory. The study guides have been written and compiled in such a way that it does not overtax the working memory and therefore simplifies the process of learning mathematics. Learners’ cognitive capacity is taken into account at all times.
This means that various strategies are used to ensure that learners have the best possible chance of mastering every section of the work. Ultimately one can say that learning has taken place when learners have stored new information in their longterm memory and have the ability to recall and use this information. The structure of the study guides support this process and helps learners to master mathematics. Tip: Each theme has the same structure in order to make it easier to navigate through them.
Each theme has the following structure:
Introduction
This briefly tells learners what the theme is about without providing details or using ‘difficult’ or unknown concepts. A comprehensive list of the learning outcomes learners need to master in a specific theme is given as a summary at the end of the theme.
Prior knowledge
This section tells learners what existing knowledge they need to master the theme involved.
Revision
This may involve one of the following:
1. revision of the concepts, definitions and procedures required as previous knowledge,
2. an exercise or activity with solutions so that learners can test their prior knowledge by themselves, or
3. a combination of the above.
Do not neglect this revision. It is important to work through this section thoroughly. Mathematical concepts often follow on one another and if basic knowledge is lacking or has not been mastered sufficiently, this will handicap the formation of new knowledge.
Following the introductory part of the theme, new knowledge is dealt with in subthemes.
Each sub-theme has the following structure:
SUBTHEME
Introduction
New concepts and procedures are explained. Relevant previous knowledge is also dealt with here if necessary.
Worked examples
Worked examples show learners how the new concepts and procedures are applied and help them understand and apply the newly taught concepts and procedures.
Exercises
The exercises give learners the opportunity to practise the concepts and procedures taught. It is important for learners to try and complete all exercises. Complete solutions are provided in the facilitator’s guides. Questions usually progress from easy (in order to master and practise basic concepts and procedures) to difficult (more complex operations).
Mixed exercises are also provided, where learners get the opportunity to practise different concepts and procedures and integrate these with previous themes.
Summary of theme
End of theme exercise
Sample
Here learners will find a summary of what they should have mastered in the theme. This is expressed in more formal mathematical language in order to be in keeping with the CAPS (the curriculum statement).
This is a mixed exercise involving all concepts and procedures dealt with in the theme, where this work can also be integrated with previous work. The degree of difficulty of this exercise varies. It is important that learners try and complete all the exercises. Complete solutions can be found in the facilitator’s guides.
Mixed exercises such as these in this textbook form a very important component of mastering mathematics. There is a big difference between the ability to recognise one’s work and the ability to recall it. When learners are able to recognise their work, they will often say ‘Oh, of course!’ but they may struggle to remember this when writing an examination.
When learners are able to recall their work, this means that they have captured that knowledge in their long-term memory and are able to remember and use it. Mixed exercises enable learners to not only recognise the work, but also recall it from their long-term memory.
When learners practise the same type of sum or problem over and over, they often get lazy and do not reflect upon the exercise anymore. Learners are convinced that they know exactly what type of sum or problem they need to solve. But in a test or exam, all these problems are mixed up and then it might be difficult to know what to do.
When mixed exercises form part of the learning process, learners learn to identify and complete a sum or problem correctly. This means that they are truly prepared for tests or exams, because they can recall their work instead of merely recognising it.
Self-evaluation
In each theme, and usually following each sub-theme, there is an activity where you need to reflect critically about the extent to which you have mastered certain concepts and procedures. This activity has the following format:
Use the following scale to determine how comfortable you are with each topic in the table below:
1. Alarm! I don’t feel comfortable, but I just need more time to work through the topic again.
2. Help! I don’t feel comfortable with the topic at all. I need help.
3. OK! I feel moderately comfortable with the topic, but I still struggle sometimes.
4. Sharp! I feel comfortable with the topic.
5. Party time! I feel totally comfortable with the topic and can even answer more complicated questions about it.
Complete the table:
Facilitators should use this evaluation to ascertain whether learners need more help in the theme or sub-theme involved. If so, it is recommended to do revision or more exercises immediately in order to ensure that learners master the essential concepts and procedures. The self-evaluation can also be used to plan enrichment. If learners have mastered the work in that theme or sub-theme, enrichment exercises can be done.
It is important not to move on to the next theme or sub-theme before the topic involved has been completely taught and mastered, even if this means that learners spend more time on a specific theme than recommended by the CAPS. Be flexible in adapting the time allocation according to learners’ needs. However, it is also important to complete the themes involved before a test or exam is written.
Tip: Use learners’ self-evaluation to decide whether they need assistance with the section involved, what the nature of such assistance should be, and whether they could move on to the next section.
Assessment criteria
Visit Impaq’s online platform for the assessment plan and comprehensive information about the compilation and mark allocation of tests, assignments and examinations. The number of assignments, mark allocation and relative weighting are subject to change.
Tip: Focus on the CAPS requirements and plan the year’s assessment accordingly. Learners must complete seven formal assessment tasks for school-based assessment.
Note:
• Only one project/investigation should be done per year.
• No graphing or programmable calculators are allowed (for example to factorise or find the roots of equations). Calculators should only be used to do standard numeric calculations and to verify calculations done by hand.
• Formula sheets are not provided during tests and final examinations in Grade 12.
Tip: This table only indicates the formal assessment (i.e. the assessment used for promotion). Informal continuous assessment should also take place to monitor learners’ progress so that gaps in their knowledge are seen and rectified timeously.
Note:
The themes covered in the examination papers are subject to change. Always refer to the portfolio book and assessment plan for updated information about the composition of the examination papers.
The two papers at the end of the year are compiled as follows: Paper 1: 3 hours Paper 2: 3 hours
Algebra, equations and inequalities (Themes 3 and 7)
Patterns and series (Theme 1)
Finance, growth and decay (Theme 4)
Functions and graphs (Themes 2 and 3)
Differential calculus (Theme 8)
Probability (Theme 12)
Statistics (Theme 11)
Analytical geometry (Theme 9)
Trigonometry (Themes 5 and 6)
geometry and measurement (Theme 10)
Final mark 150 Final mark
Supplementary books
Any other books can be used along with this textbook for extra exercises and explanations, including:
• Maths 4 A��rica, available at www.maths4africa.co.za
Sample
Tip: You need to know which themes are covered in which papers, as well as the relative weighting of each. Make sure that papers meet the requirements of this distribution.
• The Si��avula textbook, available online for free at www.siyavula.com
• P��thagoras, available at www.fisichem.co.za.
Tip: Help learners obtain and use supplementary resources efficiently.
Calculator
We recommend the CASIO fx-82ES (Plus) or CASIO fx-82ZA. However, any scientific, non-programmable and non-graphing calculator is suitable.
Tip: Ensure that learners have a suitable calculator.
THEME 1
PATTERNS, SEQUENCES AND SERIES
CAPS learning requirements
Learners should know and be able to apply the following:
1. Number patterns, including arithmetic and geometric sequences and series.
2. Sigma notation.
3. Derivation and application of the formulae for the sum of arithmetic and geometric series:
3.1 S
Paper 1
Weight 25 ± 3 marks
Facilitator tips
• Note that the derivation of the various formulae is examinable:
◦ General term of arithmetic sequence
◦ General term of geometric sequence
◦ Sum of arithmetic series
◦ Sum of geometric series
Sample
• Learners often confuse the term number with the term itself. They should practise identifying both n and T n using many examples. For example, in the sequence 1 4; 1; 4; 16; …, T4 = 16 because the fourth term is 16. Compare this to T3 = 4, where the third term is 4.
• Remind learners that n represents a term number, so it can only be a natural number. When solving for n, reject all non-natural solutions.
Introduction
In this theme learners will learn more about: Sequences
• Linear number patterns. These increase or decrease by a constant amount.
• Quadratic number patterns. These change by a steadily increasing or decreasing amount.
• Geometric number patterns. These increase or decrease by a constant ratio.
Series
• Sigma notation, which is a shorthand way of writing the sum of a series.
• The sum of arithmetic series
• The sum of geometric series
• The sum to infinity of certain geometric series
Prior
knowledge
To master this theme, learners should already know the following:
• Number patterns
◦ Notation for the nth term: T n
◦ Example: In the pattern 1 2; 1; 2; 4; 8; …
The third term or the term in the third position, is 2
We write T3 = 2
• Linear patterns
The difference between successive terms is constant. For example: 1; 3; 5; … (the common difference is 2).
The general term is given by T n = dn + c where d = common difference and c is a constant.
Linear patterns will be dealt with again in Grade 12 under the name of arithmetic sequences.
• Quadratic patterns
The first difference changes by a regular amount, and the second difference is constant. For example: 1; 3; 7; 13; … (the second difference is 2).
The general term is given by T n = a n 2 + bn + c
• Exponential equations
How to solve exponential equations of the form
b n = b p
Revision
Linear number patterns
In a linear number pattern, the general term is T n = dn + c where d = first difference and c is a constant.
Revision example 1:
Determine the general term of a linear number pattern
Find the general term of the number pattern 8 ; 3; 2; …
Solution
Find the first difference between successive terms:
3 ( 8) = 5
5
Theme 1: Patterns, sequence and
The first difference is constant, so the number pattern is linear.
Substitute d = 5 in T n = dn+ c:
T n = 5n + c
To find c, use one of the given terms, e.g. for n = 2 : T2 = 3
T2 = 5(2) + c 3 = 10 + c c = 13
Substitute in the original equation:
T n = 5n − 13
Revision example 2:
Sample
Use the general term of a linear number pattern
The general term of a number pattern is T n = 1 2 n + 3.
a) Find the first three terms of the number pattern.
b) Which term of the number pattern is 40?
Solution
a) T n = 1 2 n + 3
Substitute n = 1, n = 2 and n = 3 in the given formula: T1 = 1 2(1) + 3 = 3 1 2 T2 = 1 2(2) + 3 = 4 T3 = 1 2(3) + 3 = 4 1 2
The first three terms are: 3 1 2; 4; 4 1 2 b) T n = 1 2 n + 3
Substitute T n = 40 and solve for n: 40 = 1 2 n + 3 1 2 n = 37
n = 2(37)
n = 74
The 74th term is 40.
Revision example 3:
Determine the number of terms in a linear number pattern
Given the linear number pattern: 13; 4; 5; … ; 113
Determine the number of terms in the pattern.
Solution
First, find the general term.
The first difference is T2 T1 = 4 13
d = 9
T n = dn+ c
T n = 9n+ c
Substitute one of the terms. We will use the first term, T1 = 13. 13 = 9(1)+ c
c = 13 + 9
c = 22
The general term is:
T n = 9n+ 22
Substitute T n = 113 and solve for n :
113 = 9n+ 22
9n = 22 + 113
9n = 135
n = 135 9 = 15
The 15th term is 113.
Quadratic number patterns
Sample
In a quadratic number pattern, the general term is: T n = a n 2 + bn + c
Second difference = 2a
First of first differences = 3a + b
First term = a + b + c
Revision example 4:
Determine the general term of a quadratic number pattern
Find the general term of the quadratic number pattern 15 ; 8; 2; 15; …
Solution
Find the first and second differences between the terms of the quadratic number pattern: –15 7 3 First difference: Second difference: 10 3 13 – 8 2 15
Second difference: 2a = 3 ∴ a = 3 2
First of first differences: 7 = 3a + b 7 = 3(3 2) + b b = 5 2
First term:
Substitute the values of a, b and c in the formula for the general term:
= a n 2 + bn + c
The general term of this pattern is T n = 3 2 n 2 + 5 2 n − 19
Revision example 5:
Find the value of a term, given the general term of a quadratic number pattern
Determine the value of the 40th term of the quadratic number pattern with general term
T n = 3n 2 + 3n 12.
Solution
Substitute n = 40 and determine the value of T40: T40 = 3 (40) 2 + 3(40) 12 = 4 908
Revision example 6:
Find the term number, given the general term of a quadratic number pattern
Which term of the quadratic pattern with general term T n = 5 n 2 79n 16 will be equal to 404?
Solution
Substitute T n = 404 and solve for n :
5 n 2 79n 16 = 404
Sample
Thus the 20th term is 404.
Note that n can only be a natural number, so we reject all other solutions.
Revision example 7: Find the next term of a quadratic
Given the following quadratic number pattern, write down the next term: 16; 13; 8; 1; …
Solution
Find the first and second differences between the terms of the pattern and extend the pattern:
The second difference is 2. This provides a clue to the missing term, as the pattern of first differences requires that 7 +
9, so − 1 +
= 8. The next term in the pattern is 8.
Revision example 8:
Find an unknown term of a quadratic number pattern
3; 12; k; 48… is a quadratic number pattern. Determine the value of k.
Solution
Make a diagram of differences:
Second differences are equal, so k 21 = 60 2k
Solve for k:
3k = 81 k = 27
Revision exercise
1. Find the general term of the following number patterns: 1.1 33 ; 55; 77; … 1.2 30; 50; 130; … 1.3 7 1 4 ; − 1 1
2. The general term of a number pattern is T n = 1 5 n + 2.
2.1 Find the first three terms of the number pattern.
2.2 Which term of the number pattern is − 2?
3. Given the linear number pattern: 17; 36; 55; … ; 473
Determine the number of terms in the pattern.
4. Study the number patterns and calculate the following:
◦ Determine if the sequence is linear or quadratic.
◦ Find the formula for the general term.
◦ Calculate the following three terms in the sequence by using the formula for the general term.
◦ Calculate the 100th term in each case.
4.1 − 5; − 1; 3; … 4.2 1; 4; 9; 16; 25; …
2; 5; 16; 31; …
5. Given the quadratic number pattern: 2; 3; 5; 8; …
5.1 Write down the next term.
5.2 Determine the general term.
6. Given the following quadratic number pattern: 17; 12; 11; 14; …
6.1 Determine the general term.
6.2 Find the 30th term.
6.3 Which term of the pattern is equal to 182?
7. Give the first three terms of the quadratic number pattern with general term T n = 1 4 n 2 5n + 13
8. The general term of a quadratic number pattern is T n = 13 n 2 5n + 6. Determine the second difference of the pattern.
9. 1; 4; x; 22; … is a quadratic number pattern. Determine the value of x.
10. A quadratic number pattern has general term T n = 3 (n 14) 2 + 8. What is the value of the smallest term of the pattern, and which term has this value?
Solutions
1. 1.1
55 – 33 = 22; 77 – 55 = 22
Linear number pattern with d = 22
T n = dn + c
T n = 22n+ c
T1 = 33 = 22(1) + c
c = 33 22 = 11
T n = 22n + 11
1.2 50 − ( 30) = 80; 130 − 50 = 80
Linear number pattern with d = 80
T n = dn+ c
T n = 80n+ c
T2 = 50 = 80(2)+ c
c = 50 160 = 110
T n = 80n − 110 1.3 − 1 1 4 − (− 7 1 4) = 6
4 3 4 ( 1 1 4) = 6
Linear number pattern with d = 6
T n = dn+ c
T n = 6n+ c
T1 = 7 1 4 = 6(1)+ c
c = 13 1 4
2. T n = 1 5 n + 2 2.1 T1 = − 1 5(1) + 2 = 1 4 5
T2 = 1 5(2) + 2 = 1 3 5 T3 = 1 5(3) + 2 = 1 2 5 2.2 T n = − 1 5 n + 2 = − 2 1 5 n = 4
n = 20
The 20th term is 2.
Theme 1: Patterns, sequence and series
Sample
3. 36 17 = 19; 55 36 = 19
Linear number pattern with d = 19
T n = dn+ c
T1 = 17 = 19(1)+ c
c = 2
T n = 19n 2
473 = 19n − 2
19n = 475
n = 475 19 = 25
There are 25 terms in the pattern 4.
T n = 6n 13 1 4 4.1 T2 − T1 = − 1 − ( 5) = 4
T3 T2 = 3 ( 1) = 4
The first differences are equal, so the number pattern is linear and d = 4. Formula for the general term:
T n = dn+ c
T n = 4n+ c
Substitute n = 1 for the 1st term:
T1 = 4(1) + c = − 5
Solve for c :
c = 5 4
c = − 9
T n = 4n 9
For the 4th, 5th and 6th terms, substitute n = 4; 5; 6:
T4 = 4(4) − 9 = 7
T5 = 4(5) 9 = 11
T6 = 4(6) 9 = 15
To find the 100th term, substitute n = 100:
T n = 4n 9 T100 = 4(100) 9 = 391
Second differences are constant. Sequence is quadratic.
T n = a n 2 + bn + c
From 2nd difference:
2a = 2
∴ a = 1
From 1st of the 1st difference: 3a + b = 3
3(1) + b = 3 ∴ b = 0
From given terms: T1 = 1 a + b + c = 1
1 + 0 + c = 1
c = 0
Next three terms are:
Second differences are constant. Sequence is quadratic. T n = a n 2 + bn + c
5. 5.1 Find the first and second differences between the terms of the pattern and extend the pattern: 2 1
5.2 Refer to the diagram of differences:
The second difference is 1. Follow the pattern:
3 + 1 = 4, so 8 + 4 = 12. The next term in the pattern is 12.
Second difference: 2a = 1 ∴ a = 1 2
First of first differences: 1 = 3a + b
Substitute the values of a, b and c in the formula for the general term: T n = a n 2 + bn + c The general term of this pattern is T n = 1 2 n 2 − 1 2 n + 2 6.
6.1 Find the first and second differences between the terms of the pattern:
Second difference: 2a = 4 ∴ a = 2
First of first differences: 5 = 3a + b
− 5 = 3a + b
5 = 3(2) + b
b = 11
First term: 17 = a + b + c 17 = 2 11 + c
∴ c = 26
Substitute the values of a, b and c in the formula for the general term:
T n = a n 2 + bn + c
The general term of this pattern is T n = 2 n 2 11n + 26
6.2 Substitute n = 30 and determine the value of T30:
T30 = 2 (30) 2 11(30) + 26 = 1 496
6.3 Substitute T n = 182 and solve for n :
2 n 2 − 11n + 26 = 182
2 n 2 11n + 26 182 = 0
2 n 2 11n 156 = 0 n = ( 11) ± √ ( 11) 2 4(2)( 156) 2(2) n = 11 ± √ 1 369 4 n = 12 OR 6,5 (N / A)
∴ n = 12
Thus the 12th term is 182.
7. Substitute n = 1, n = 2 and n = 3: T1 = 1 4 (1) 2 5(1) + 13 = 8 1 4
