FET Phase Grade 11 • Facilitator’s Guide
Mathematical Literacy CAPS IEB
Mathematics Facilitator’s guide
2 0 11 - E - M A M - F G 0 1
Í4+È-E-MAM-FG01/Î
Grade 11
CAPS aligned
DM Oost
TABLE OF CONTENTS 1
Letter of Information
3
Entrance Examination Paper 2 of 2
2
4
©Optimi
Entrance Examination Paper 1 of 2
Memorandum of Study guide
Page 2
10
24
32
1
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
TABLE OF CONTENTS
ȇ
the entrance (readiness) exam to identify possible problem areas, ȇ Year plan for Grade 10 ȇ The memorandums of the activities. y A DVD containing videos referred to in the study guide. Note: This only works on a laptop, not a DVD player 1.2 Entrance (readiness) Examination
Guidelines for facilitators 1
General
2
4
Levels of difficulty
3
2
3
5
Study guide with activities and exercises Portfolio book Year plan
2
3
Included in this facilitator’s guide are two examination papers (Paper 1 and 2) that each learner who did not pass Grade 10 at Impaq must write. If a learner is studying Grade 11 through Impaq for the first time, you must also ensure that he/she takes this Entrance (readiness) Examination. You must grade this paper and make sure that learners catch up on any ‘weak’ points in the work that they may have.
4
The teaching and learning of Mathematics aims to develop: y A critical awareness of how mathematical relationships are used in social, environmental, cultural and economic relations y Confidence and competence to deal with any mathematical situation without being hindered by a fear of Mathematics y An appreciation for the beauty and elegance of Mathematics y A spirit of curiosity and love for Mathematics y Recognition that Mathematics is a creative part of human activity y Deep conceptual understandings in order to make sense of Mathematics y Acquisition of specific knowledge and skills necessary for: ȇ The application of Mathematics to physical, social and mathematical problems ȇ The study of related subject matter (other subjects) ȇ Further study in Mathematics (Specific Aims of Mathematics CAPS 2011) 1.
Success in Grade 11 cannot be achieved if Grade 10 work is not properly understood and handled with ease. Every Impaq Mathematics Study Guide relies on learners’ prior knowledge, insight and ability to identify and complete work accurately. A backlog in Mathematics increases with time and can become impossible to bridge if not dealt with quickly.
2.
General 1.1 Make sure that the following are in your possession: y A study guide covering the work that should be done in Grade 11. y A facilitator’s guide, which includes: ȇ this letter of information, 2
This paper is only meant to assure that learners understand Grade 10 work before they start with Grade 11 work. If learners have a huge backlog, it is recommended the learner first master the Grade 10 work before continuing with Grade 11.
Study guide with activities and exercises The study guide contains references and exercises to develop concepts, comprehension, skills and knowledge of mathematics. ȇ Study the theory and examples with all the written explanations. ȇ Do some of the sums and mark it according to the memorandum. ȇ Use the video material if there is something you do not understand. It is available at the beginning of each exercise, explaining some of the problems.
Recommended calculator for learners: CASIO fx-82ES (Plus), however, any scientific calculator will be sufficient..
4.
Levels of difficulty The number of stars, next to the number of each sum, indicates the level of difficulty. This indication is maintained in the guide, memorandum, tests and examination papers
Level 1 (Basic knowledge)
3.
Level 2 (Routine work)
Portfolio book Refer to the portfolio book for all assessment.
Level 3 (Complex work)
Level 4 (Problem solving) Level 5
*
** ***
****
Knowledge of basic theory and procedures are necessary before any calculations can be done. This section of the question paper has a value of 20% and can be obtained by studying. Applying routine multi-step procedures in various contexts. Continuous practise will enable you to master this section of the question paper. It has a value of 35%.
Applying knowledge and procedures in a variety of contexts. Complex calculations based on a combination of acquired knowledge. Questions are asked indirectly with the emphasis on comprehension. This part of the question paper has a value of 30%. Applying relevant knowledge to unfamiliar, non-routine questions. The most difficult part of the question paper - testing the candidates’ ability to apply all the knowledge they have gained. This part of the question paper has a value of 15%.
***** Enrichment.
For any queries, contact the education specialist at Impaq.
©Optimi
3
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
Year plan: Mathematics Grade 11 (Shortened version)
Year plan Grade 11 Mathematics Term 1
Term
Subject
Unit
T1
Number systems
1
T1
Number patterns
4
T1 T1 T1 T2 T2
Exponents Algebra
Analytic Geometry Functions
Trigonometry
June examinations will include the above topics. It will be 2 papers of 100 marks each Paper 1: Units 1,2,3,4,and 11 Paper 2: Units 12 & 13
13 12
T4
Statistics
9
T3 T4
Financial matters Probability Revision
November Examinations
1
Different types of numbers
3
Rounding off numbers
2
11
5.1 and 5.2
T3
Unit 1: Number Systems
3
Geometry
Subject of exercises These exercises can be seen as lesson units
Time in days
Assuming that learners work in January, February and March on this first 5 Units, a maximum of 54 lessons is required in Term 1.
2
T3
Exercises
5.
4
Unit 2: Exponents
8
10
5
3
Compound bases
5
6
7
4
Mixed Exercises
Basic rules and definitions
4
Note: Unit 7, Proportion and Ratio, is not formally dealt with in Grade 11. However it is applicable up to Grade 12. Unit 6: Transformations is no longer askes as a formal question. The applications of transformation however are required for Functions, Trigonometry and Analytical Geometry.
Rationalise denominators
1
2
Paper 1: Units 1 ; 2 ; 3 ; 4 ; 7; 8 ; 10; 11 Paper 2: Units 5; 6 ; 9 ; 12 ; 13
Notations
Positive Exponents
Factorising exponents
Exponent equations with the unknown in the exponent
Exponent equations with the unknown as base Mixed Exercises
Days/ lessons (1)
(1)
(1)
(1)
(1) [5]
Days/ lessons (1)
(1)
(1)
(1)
(2)
(2)
(1) [9]
Unit 3: Algebra
1
Factorising
(1)
3
Fractions
(2)
2 4 5 6 7 8 9
Simplifying Completing the square Long division
Difficult fractions Equations
Inequalities Invalidity
10 Nature of Roots 11 Modelling
12 Mixed Exercises
Unit 4: Number Patterns 1
Linear patterns
3
Exponential patterns
2 4 5 6 7 8 ©Optimi
Days/lessons
Quadratic patterns
Arithmetic sequences Geometric sequences Other patterns
Recursive formulae Mixed Exercises
Unit 13: Analytical Geometry
(1)
1
Summary of theory
3
Combined applications
2
(2)
4
(2)
5
(2) (4)
Straight lines and angles of inclination Proof of formulae The circle
Days/lessons (1)
(2)
(2)
(2)
(2) [9]
The formal test (assignment 2) must be written after completion of the above Units. The test entails all the previous work. Note that the number of days allocated only serves as a guideline. However, do not waste time as the second term is also heavily loaded. The assignment 1 (investigation) as well as part of assignment 7 (informal assessment) should also be completed
(2) (2) (2) (2)
(1) [23]
Days/lessons (1) (1) (1) (1) (1) (1) (1)
(1) [8] 5
G11 – Mathematics
Term 2
Exercises
G11 – Mathematics – Facilitator’s Guide
Subject of exercises These exercises can be seen as lesson units
Time in days
Unit 12: Trigonometry
2
This work needs more or less 43 lessons or days to complete. If you start at the 12th April then there will be enough time to finish this in time for the June Exams. Unit 11: Functions
1
Revision
3
The straight line
2
4
Basic functions
Functions and transformations
5
The hyperbola
7
The exponential function
6
8
9
The parabola
Trigonometric functions
Applications to and deductions from sketched graphs
10 Functions that has one, two or no intersections 11 Inverse Functions 12 Average Gradient 13 Mixed Exercises
1
3
4
Days/lessons (2)
5
(2)
6
(2)
7
(2)
8
(2)
9
(2)
Definitions and diagrams to determine function values Derivation formulae Special angle sizes
Mixed Exercises with derivation formulae Equations Identities
Sine, Cosine and area rules Trigonometric graphs Mixed Exercises
Days/ lessons (2)
(2)
(2)
(2)
(2)
(2)
(2)
(2)
(1) [17]
At the end of this unit, learners will be able to write the June Examinations. (Assignment 4.1 and 4.2) Paper 1 of 2: 100 marks | Units 1;2;3 ; 4 and 11 Paper 2 of 2: 100 marks | Units 12 and 13 Note that all the work done in the first two terms will be examined Assignment 3 (Research project) should also be completed Assignment 7 (Informal Assessment) should also be completed for the Units done
(2)
(2)
(2)
(2)
(2)
(2)
(2) [26]
6
Exercises
Term 3
Subject of exercises These exercises can be seen as lesson units
Time in days
Prove of logic where 5 every 2 statements 6 result in a third conclusion should be included
In the third term 49 lessons or days are required to complete the required Units From the middle of July until the end of September will be sufficient Unit 6: Transformation Geometry Application of these content in the drawing of graphs in the function section
1
2
Summary of pre-knowledge Mixed Exercises
Unit 5.1: Measuring, space and shape 1
Perimeter, area and volume
3
Pyramids
2
4
5
Cones
Spheres
Mixed Exercises
Unit 5.2: Euclidian geometry Prove of theorems are included Formal reasoning with statements and reasons should always be given.
©Optimi
Days/lessons (2) [3]
Days/lessons (1)
(1)
(1)
Tangents
(5)
4
Similar Triangles
(Decay-formulae)
5 1
3
Days/lessons
3
3
2
(1) [6]
(2)
Circle theorems
Unit 10: Probability
(2)
Theorems already known
Growth formulae
4
4
5
(6)
6
(3)
7
(2)
Recommended abbreviations of (1) theorems grade 8 to 12 [19]
1
2
(1)
1
2
Unit 8: Finance
Mixed Exercises
Depreciation and sinking funds Nominal and effective interest rate
Time lines to calculate interest Basic Probability and Supporting techniques
Complementary events
Dependent and Independent events. Difficult Problems Mixed Exercises
Summary of Theory and formulae
Days/lessons (2)
(2)
(2)
(3)
(2) [11]
Mixed Exercises (2)
(2)
(3)
(3)
(2)
[12]
G11 – Mathematics
Term 4
Exercises
G11 – Mathematics – Facilitator’s Guide
Subject of exercises These exercises can be seen as lesson units
Time in days
Do revision of all the work done through the year by working through all the mixed exercises, tests and exams again. More papers can be found on the internet at (www.thutong.gov.za) or ordered from Examination Aid Tel: (012) 365 1824
12 lessons or days are required before revision can be done Both these Units are revision on word done in grade 10. Try and cover these units quickly to have more time for revision Unit 9: Statistics
1
2
3
4
5
6
7
8
9
Revise measures of central tendencies
(1)
Identify outliers
(2)
Revise measures of dispersion Five-number summary of univariate data
Symmetric or skewed data
Cumulative frequency graphs
Variance and standard deviation Bivariate data
Mixed Exercises
November Examinations Paper 1: Algebra, Functions and Financial matters Unit 1, 2, 3, 4, 7, 8, 10 and 11 (3 hours – 150 marks)
Paper 2: Analytical geometry, Transformation geometry, Trigonometry, Shapes and Statistics Unit 5.1, 6, 9, 12 and 13 (3 hours - 150 marks)
(1)
(2)
The weight of different content in grade 11 is more or less the following. (It is only used as a guideline)
(1)
(1)
Paper 1
Marks
Patterns sequences and series
25
Algebra, equations, factorising, fractions, 45 inequalities and restrictions
(2)
(1)
Finance and growth
(1) [12]
Functions and graphs
Probabilities
All portfolio assignments should be completed at this stage. That includes the following: Assignment 6: Formal Test Assignment 5: Investigation Assignment 7: Informal Tests
Paper 2
Statistics
Analytical Geometry
Trigonometry
Geometry: Euclidean and Form
8
Total
15
45
20
Total 150
Marks
20
30
50
50
150
% in final papers
30%
15%
10%
30%
15%
100%
% in final papers
14%
20%
33%
33%
100%
Before starting a test or exam paper, make sure you have the right calculator. It should be a non-programmable, non-graphic scientific calculator. You should know how to adjust the calculator to scientific -mode and statistic- mode and use it when applicable.
Note that in the new “CAPS” curriculum no formula sheet is given. Pupils should know the different formulae by hart. Also remember that if the question states: “without a calculator” then the different steps should be included so that proof is given that no calculator was used
©Optimi
9
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
MATHEMATICS ENTRANCE EXAMINATION Grade 11 Paper 1 of 2
TIME: 120 MINUTE
Question 2**
The number 4959 is a “cool” number because the following happens: Add the squares of all the digits: 42+ 9 2+ 52+ 92 = 203 Then add the squares of the digits 203: 22+ 0 2+ 32 = 13 Do the same with 13: 12+ 3 2 = 10 2 2 And with 10: 1 + 0 = 1 Because the final answer is one 4959 is a “cool” number
TOTAL Marks: 100
INSTRUCTIONS: y Calculators may be used unless stated otherwise. y If necessary answers should be rounded off to 2 decimal places, unless stated otherwise. y Number the answers correctly according to the numbering system used in this question paper. y Rule off after each answer. y A diagram sheet is attached at the back of this question paper. Write your name and number on the diagram sheet and submit it with your answering sheets. y Use the formula sheet attached to this paper. y Formulae sheets are prohibit. Question 1 1.1*
1.2* 1.3* 1.4*
1.5*
Indicate whether the following statements are true or false. All integers are natural numbers.
All integers are rational numbers. All natural numbers are integers.
All real numbers are rational numbers.
All rational numbers are real numbers.
Study the following
Determine if 141312 is a “cool” number. Question 3 3.1**
3.2***
Simplify the following:
( 2m − 1)2− 2(m + 1)(m − 1)
4𝑦
3𝑦
𝑦−6 𝑦 + 5𝑦 + 6
_ _____ _ 𝑦 + 2 − 𝑦 + 3 − 2
[4]
(3) (5)
[8]
Question 4
(1)
4.1**
(1) (1)
(1)
4.2***
(1)
[5]
10
Solve for 𝑥 in the following equation:
1 _ 2−−𝑥𝑥 = _ 𝑥 − + _ 2 3 𝑥−3
Solve for 𝑥 and y simultaneously
𝑥 + 2𝑦 = 11 and 𝑥 − 2𝑦 = –3
(6) (4)
(10)
Question 5***
Question 6 6.1** 6.2*
Question 7
A man is allowed to shoot 25 shots at a target. For each bulls eye he receives R5, but for each shot that he misses he must pay a fine of R2,50. If the man makes a profit of R5, how many bulls’ eyes did he shoot?
𝑓: 𝑦 = 2𝑥 + 4 and g: 𝑦 = –𝑥2 both for 𝑥 and 𝑦 ∊ R Sketch 𝑓 as well as the straight line that passes through the origin and is perpendicular on 𝑓.
7.1**
7.2*
[5]
Given:
Name this perpendicular line h and determine its equation
Sketch 𝑔 on its own set of axes by making use of the table method. Now translate 𝑔 according to the following rule: (𝑥 ; 𝑦) → (𝑥 – 2; 𝑦 +1), sketch this translation and 𝑔 on the same set of axes. Name this graph j and determine its equation.
8.1*** 8.2***
(8)
(8)
[16]
©Optimi
𝑥-axis
(𝑥 + 4) (𝑥 – 3) = – 10 and 𝑝 > 𝑞,
Determine the value if 𝑝2 + 3𝑞.
(4)
[8]
(6)
Solve for 𝑥: 1–_ 𝑥2 ≥ 6
(3)
Sketch the following on a Cartesian plane: 𝑦 < Sinθ for 0° ≤ θ ≤ 360° and 𝑦 ≥ –2
Question 9
Use the correct formulae and answer the following questions
9.2**
11
If p and q are two solutions of
8.3*
9.1*
𝑦-axis
–3
What is the domain and range of the sketched graph?
(4)
Question 8
a This diagram illustrates the hyperbola 𝑦 = _ 𝑥 + q with 𝑥-intercept (–3 ; 0)
𝑦=2
Determine the values of a and q and write down the equation.
(9)
[18]
What is the simple interest that can be earned, if R1000 is invested at 10% per year over a period of 5 years?
What is the compound interest that can be earned, if R1000 is invested at 10% per year over a period of 5 years? Interest is compounded annually.
(3) (3)
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide 9.3*
Question 10 10.1**
10.2**
What is the difference between the simple and compounded investments calculated in question 9.1 and 9.2? Why is the compounded interest more than the simple interest?
12.1** (1)
3 red
[7]
5 green
Simplify the following exponential expressions:
( 2) 3 _____ ____ 2𝑥 √ 16 𝑥 6
12.2
(3)
m+1 m ________ 4 ( . 16 3m) 2 4 . 2
(4)
[7]
11.2***
Simplify 8 . 2𝑥 + 1– 32 . 2𝑥 Now solve for 𝑥 if 8 . 2
– 32 . 2 = – 64
𝑥 + 1
𝑥
4 blue
3 red
5 green
4 blue
Draw a tree-diagram to determine the probability to draw not a red sweet with the first draw, and not a blue sweet with the second draw.
(3) (4)
[7] Total 100
Question 11 11.1**
Complete the table when two sweets are drawn randomly from the box. The first sweet is placed back and then the second one is drawn.
(2) (2)
[4]
Question 12** A box of sweets contains the following: y 3 red y 5 green y 4 blue sweets. Every sweet has the same possibility to be randomly drawn.
12
Diagram sheet Submit this diagram sheet with your answer sheets Name:
©Optimi
Learner’s nr.
13
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
Memorandum Mathematics Grade 11 Paper 1 of 2 Entrance examination TIME: 120 minutes Total 100
Question 1 Indicate whether the following statements are true or false: 1.1* 1.2* 1.3* 1.4* 1.5*
Question 2**
The number 4959 is a “cool” number because the following happens: Add the squares of all the digits: 42+ 92+ 5 2+ 9 2 = 203 Then add the squares of the digits 203: 22+ 02+ 3 2 = 13 Do the same with 13: 12+ 32 = 10 2 2 And with 10: 1 + 0 = 1 Because the final answer is one 4959 is a “cool” number Determine if 141312 is a “cool” number.
All integers are natural numbers. . False
(1)
All natural numbers are integers. True
(1)
All integers are rational numbers. True
All real numbers are rational numbers. False
All rational numbers are real numbers. True
Study the following:
141312: 32: 13: 10:
(1)
Question 3 ** 3.1**
(1) (1)
[5]
14
1 2 + 4 2 + 1 2 + 3 2 + 1 2 + 2 2 = 32 3 2 + 2 2 = 13 1 2 + 3 2 = 10 141312 is a “cool” number. 2 2 1 + 0 = 1
[4]
Simplify the following
(2m − 1) 2 − 2(m + 1)(m − 1)
(2m − 1) 2 − 2(m + 1)(m − 1) = 4m 2 − 4m + 1 − 2(m 2 − 1) = 4m 2− 4m + 1 − 2m 2 + 2 = 2m 2 − 4m + 3
1 for middle terms 1 for distributive rule 1 for answer
(3)
3.2***
4𝑦 3𝑦 𝑦 − 6 _ 𝑦 + 2 − _ − _____ 2 𝑦 + 3 𝑦 + 5𝑦 +6
Question 4*** 4.1
4𝑦 3𝑦 𝑦−6 _ 𝑦 + 2 − _ − ______ 2 𝑦+3 + 5𝑦 + 6 𝑦
4𝑦 3𝑦 𝑦 − 6 = ___ − ____ − __________ (𝑦 + 2) (𝑦 + 3) ( 𝑦 + 3)(𝑦 + 2) 2
LCM = (𝑥 − 2)(𝑥 − 3) ∴ 𝑥(𝑥 − 3) = (𝑥 − 2) + 2(𝑥 − 3) ∴ 𝑥 2 − 3𝑥 = 𝑥 − 2 + 2𝑥 − 6 ∴ 𝑥 2 – 6𝑥 + 8 = 0 ∴ ( 𝑥 – 4)( 𝑥 – 2) = 0 ∴ 𝑥 = 4 or 𝑥 = 2
y 2 + 5𝑦 + 6
2
= ___________ (𝑦 + 2)(𝑦 + 3) ( 𝑦 + 2)( 𝑦 + 3) = __________ 𝑦 + 2 𝑦 + 3 ( )( )
=1
©Optimi
1 _ 2−−𝑥𝑥 = _ 𝑥 − + _ 2 3 𝑥−3 1 _ 2−−𝑥𝑥 = _ 𝑥 − + _ 2 3 𝑥−3
4𝑦(𝑦 + 3)− 3𝑦(𝑦 + 2)− (𝑦 − 6) = _____________________ (𝑦 + 2)(𝑦 + 3) 4____________________ 𝑦 + 12𝑦 − 3𝑦 − 6𝑦 − 𝑦 + 6 = (𝑦 + 2)(𝑦 + 3)
Solve for 𝑥 in the following equation:
1 for factorising 1 for numerator 1 for denominator 1 for factorising 1 for answer
4.2
(5)
[8]
15
1 for LCD 1 for multiplication of each term 1 for =0 1 for both factors 1 for both answers 1 for final solution
n/a (denominators may not be zero)
∴ Final answer = 4 Solve for 𝑥 and 𝑦 if:
(6)
𝑥 + 2𝑦 = 11 and 𝑥 − 2𝑦 = −3
𝑥 + 2𝑦 = 11 …A 𝑥 – 2𝑦 = –3 …B From A: 𝑥 = 11 – 2𝑦 substitute in B 1 for substituting A in B ∴ ( 11 – 2𝑦) – 2𝑦 = –3 1 for y-answer 1 for substituting in B ∴ 11 – 4𝑦 = –3 1 for both answers ∴ −4𝑦 = −14 −14 ∴ 𝑦 = _ −4 1 _ = 3 2 ∴ 𝑥 + 2(_ 27 ) ∴ = 11 ∴ 𝑥 + 7 = 11 Substitute in A ∴ 𝑥 = 4 (4) 1 _ 𝑥 = 4 and 𝑦 = 3 2 (10)
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide Question 5***
Let bulls eyes = 𝑥 and shots missed = 𝑦
∴ 𝑥 + 𝑦 = 25................A Earns 𝑥 . R5 = 500𝑥 cents Fine = 𝑦 . R2,50 =250𝑦 ∴ 500𝑥 − 250𝑦 = 500..............B From A 𝑥 = 25 − 𝑦 substitute in B ∴ 500 (25 – 𝑦) – 250𝑦 = 500 ∴ 12500 – 500𝑦 – 250𝑦 = 500 ∴ −750𝑦 = −12000 ∴ 𝑦 = 16 Substitute in A: ∴ 𝑥 + 16 = 25 ∴ 𝑥 = 9 He shoots 9 bulls eyes
𝑓: 𝑦 = 2𝑥 + 4 and 𝑔: 𝑦 = − 𝑥 2 both for 𝑥 and 𝑦 ∊ R Sketch 𝑓 as well as the straight line that passes
Question 6 Given:
A man is allowed to shoot 25 shots at a target. For each bulls eye he receives R5, but for each shot that he misses he must pay a fine of R2,50. If the man makes a profit of R5, how many bulls eyes did he shoot?
6.1**
1 for equation A 1 for equation B 1 for substitution A in B 2 for answers
through the origin and is perpendicular on f. Name this perpendicular line h and determine its equation. For 𝒇:
𝑥-intercept let 𝑦 = 0 ∴ 0 = 2𝑥 + 4 ∴ −4 = 2𝑥 ∴ 𝑥 = −2 (−2 ; 0) 𝑦-intercept let 𝑥 = 0 ∴ 𝑦 = 2(0) + 4 = 4 (0 ; 4) For 𝒉: ℎ and 𝑓 are perpendicular ∴ 𝑚 ℎ.𝑚 𝑓 = −1 ∴ 𝑚 ℎ(2) = −1 ∴ 𝑚 ℎ = _ − 1 2
1 for 𝑥-intercept 1 for y-intercept 2 for gradient 2 for each sketch Note: Subtract 1 if general information is not all given
Through (0 ; 0) count 1 downwards and 2 to the right 𝑦-axis
[5]
h
–2
16
f –1
4
2
𝑥-axis (8)
6.2*
𝑎 Question 7 This diagram illustrates the hyperbola 𝑦 = _ 𝑥 + 𝑞 with
Sketch g on its own set of axes by making use of the table method. Now translate g according to the following rule:
𝑥-intercept (−3 ; 0)
(𝑥 ; 𝑦) → (𝑥 – 2 ; 𝑦 +1), ), sketch this translation and g on the same set of axes. Name this graph j and determine its equation.
y=2
𝑦-axis
j: y = –(𝑥 + 2) 2 + 1
1
2
g: 𝑦 = –𝑥 2
𝑥-axis
–3
–1
𝑦-axis
𝑥-axis
7.1**
(8)
[16]
7.2*
Determine the values of a and q and write down the equation
1 for asymptote q 2 for substituting in formula 1 for a
𝑞 = 2 the asymptote is 𝑦 = 2 (−3 ; 0) is on the graph 𝑎 substitute (𝑥 ; 𝑦) = (−3 ; 0) in 𝑦 = _ 𝑥 + 𝑞 𝑎 _ ∴ 0 = − 3 + 2 6 = 𝑎 6 _ Equation 𝑦 = 𝑥 + 2
(4)
What is the domain and the range of the graph? Domain: D = {𝑥 / 𝑥 є R ; 𝑥 ≠ 0}
Range: W = {𝑦 / 𝑦 є R ; 𝑦 ≠ 2}
2 for each Notations should be correct
©Optimi
17
(4)
[8]
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide Question 8 8.1***
If p and q are the two solutions of
(𝑥 + 4) (𝑥 – 3) = – 10 and 𝑝 > 𝑞 ,
Determine 𝑝 + 3𝑞.
Given 𝑝 > 𝑞
∴ 𝑝 = 1 and 𝑞 = –2 ∴ 𝑝 2 + 3𝑞 = (1) 2 + 3(–2) =1–6 = –5
Solve for 𝑥:
𝑦 < Sinθ for 0° ≤ θ ≤ 360° and 𝑦 ≥ –2 𝑦y== Sin Sinθ θ
11
–2
360°
180°
–1 1 for = 0 1 for factors 1 for both answers 1 for p and q 1 for substituting 1 for answer
θ-axis
𝑦 = –2
4 Marks for x-intercepts, min/max and form 4 Marks for 4 boundaries 1 Mark for general information
(9)
[18]
Question 9 Use the correct formula and determine the following.
(6)
9.1*
1–_ 𝑥2 ≥ 6
1–_ 𝑥2 ≥ 6 ∴ LCM = 2 ∴ 2 – 𝑥 ≥ 12 ∴ – 𝑥 ≥ 10 ∴ 𝑥 ≤ 10
Sketch the following on a Cartesian plane: (900 ; 1)
2
(𝑥 + 4)(𝑥 – 3) = –10 𝑥 2 + 𝑥 –12 = –10 𝑥 2 + 𝑥 – 2 = 0 (𝑥 + 2)(𝑥 – 1) = 0 𝑥 = –2 of 𝑥 = 1
8.2***
8.3*
What is the simple interest that can be earned, if R1000 is invested at 10% per year over a period of 5 years?
A = P(1 + ni) 10 = R 1 000 ( 1 + 5 . _ 100 ) = R 1 500 Interest = R 1 500 – 1 000 = R 500
1 for LCD multiplication 1 for simplifying 1 for answer
1 for substituting in correct formula 1 for answer 1 for interest
(3) 18
(3)
9.2**
What is the compound interest that can be earned, if R1000 is invested at 10% per year over a period of 5 years? Interest is compounded annually..
A = P(1 + i) n
9.3**
Question 10** 10.1**
10 = 1 000(1 + _ 100 ) = R 1 610,51 Interest = R 1 610,51 – R 1 000 = R 610,51
1 for substituting in correct formula 1 for answer 1 for interest
The difference is R 110, 51 because interest on interest is earned with compound interest.
m+1
m
1 for prime base numbers 11 for numerator 1 for denominator 1 for answer
(4)
[7]
Question 11 11.1** (1)
[7]
( 2) 3 _____ ____ 2𝑥 √ 16 𝑥 6 1 for numerator 1 for denominator 1 for answer
= ________ 4 ( . 16 3m) 2
=1
(3)
Simplify the following exponential expressions:
( 2) 3 _____ ____ 2𝑥 √ 16 𝑥 6 3 6 ____ 2 𝑥 3 4𝑥
m+1 m ________ 4 ( . 16 3m) 2 4 . 2
4 . 2 2) m + 1 ( 2) m (__________ 2 = 2 2 . 6m 2 2 2m + 2 . 2 4m = ________ 2 2 + 6m 2 6m + 2 = _____ 26m 2 + 2
5
What is the difference between the simple and compounded investments calculated in question 9.1 and 9.2? Why is the compounded interest more than the simple interest?
= 2𝑥 3 ©Optimi
10.2**
Simplify 8 . 2𝑥 + 1– 32 . 2𝑥
= 8 . 2 – 32 . 2 = 8(2 𝑥 + 1 − 4 . 2 𝑥) = 8(2 𝑥 . 2 1 – 4 . 2 𝑥) = 8 . 2 𝑥 . 2(1 – 2) = –2 3 2 𝑥 . 2 1 = – 2 4 + 𝑥 𝑥 + 1
𝑥
Alternative:
8 . 2𝑥 + 1– 32 . 2𝑥 = 232𝑥 + 1– 252𝑥 = ( 2321– 25)2𝑥 = – 16 . 2𝑥 = – 24 + 𝑥
(3)
(3) 19
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide 11.2***
Now solve for 𝑥 if 8 . 2 𝑥 + 1 – 32 . 2 𝑥 = – 64
8 . 2 𝑥 + 1 – 32 . 2 𝑥 = – 64 From above mentioned
∴ –2 4 + 𝑥 = – 2 6 ∴ 2 4 + 𝑥 = 2 6 ∴ 4 + 𝑥 = 6 ∴ 𝑥 = 2
1 for prime factor bases 1 for answer
[5]
A box of sweets contains 3 red; 5 green; 4 blue sweets. Every sweet has the same possibility to be randomly drawn.
12.1**
Complete the table when two sweets are drawn randomly from the box. The first sweet is placed back and then the second one is drawn. 3 red
5 green
12.2
4 blue
3 _ 3 _ 9 _ 12 . 12 = 144 5 _ 3 _ 15 _ 12 . 12 = 144 3 _ 4 _ 12 _ 12 . 12 = 144
5 green
5 _ 3 _ 15 _ 12 . 12 = 144 3 _ 4 _ 12 _ 12 . 12 = 144 5 _ 20 4 _ _ 12 . 12 = 144
4 blue
3 _ 4 _ 12 _ 12 . 12 = 144 5 _ 20 4 _ _ 12 . 12 = 144 16 4 _ 4 _ _ 12 . 12 = 144
Draw a tree-diagram to determine the probability to draw not a red sweet with the first draw, and not a blue sweet with the second draw.
Red
4 _ 12
9 12 _
(2)
Question 12**
3 red
3 _ 12
Not Red
Blue
8
12 _
Not Blue
P(not red and not blue) 9 8 72 =_ 12 × _ 12 = _ 144 = _ 12 [7] Total 100
(3) (4)
20
MATHEMATICS ENTRANCE EXAMINATION Grade 11 PAPER 2 of 2 TIME: 120 MINUTES
TOTAL MARKS: 100
INSTRUCTIONS: y Calculators may be used unless stated otherwise. y If necessary answers should be rounded off to 2 decimal places, unless stated otherwise. y Number the answers correctly according to the numbering system used in this question paper. y Formulae sheets are prohibited. Question 1
1.1**
1.2**
Question 2**
2.1***
2.2***
©Optimi
If it is given that 10° < 2𝑥 + 10 < 190° , determine the variable (unknown) in each of the following: Give your answer rounded off to 2 decimal places.
sin _ 𝑥2 = 0,588
2 tan 𝑥 = 2,456
In which quadrant does  lie?
Draw a Cartesian plane and indicate A as well as the values of 𝑥, 𝑦 and 𝑟 on the terminal arm.
Use the sketch and determine the value of:
2.4**
Sin A Use the sketch and prove that: Tan A =_ Cos A
Question 3
Study the right angled triangle with AC = b, AB = c and BC = a.
2 __ 2 − 1
3.1*
(2)
[4]
3.2**
(6)
3.3****
(4)
21
(5)
[18]
α
c
(2)
(3)
cos A
A
Do not use_a calculator to answer this question.
√ 3 Given: ______ + 2 = 0 and Cos A < 0 sin A
2.3**
B
b
a
θ C
Complete the following in terms of the sides of the triangle: Cos α = ... Cos θ = ...
Cos β = ...
Now write down AB, BC and AC in terms of Cos α or Cos θ and the sides a or b or c.
Now prove that the area of
∆ ABC = _ 12 𝑏 2Cos θ Cos α
(3)
(3)
(4)
[10] G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide Question 4
4.1***
4.2**
4.3***
4.4**** Question 5
5.6**
Use analytical methods to answer this question. Round off all calculations to 2 decimal places. Sketch the following points on a Cartesian plane: A (−3 ; −1), B(1 ; 0) , C(2 ; 5) and D (−2 ; 4)
5.7** 5.8**
Prove that the diagonals of ABCD bisect each other.
(7)
Prove that DA || CB.
(5)
What is the distance from the above mentioned intersect point to point A?
Determine the gradient of the perpendicular line on AB.
5.9***
Question 6
(4)
6.1***
(4)
[20]
Study the sketch and complete the given questions
If D̂ 1= B ̂1+ B̂2 then ...
(1)
If AD = DB and DE || BC then… If F̂ + Ĉ + Ĉ = 180° then... 2
1
(1)
(1)
2
If Area ∆DBC = Area ∆EBC then…
(1)
[9]
Study the sketch and answer the questions formally. Use “statements” and “reasons”
ABEF is a quadrilateral. Determine the value of k and m if SA = SF Given: B̂1= 2k, F̂1= k, Ê1= 7k, and Ŝ1 = m AF||BE and AB||SF 2
1
1
1
2
1
2
1
5.1*
If  1= Ĉ 1 then...
(1)
If AD = AE then…
(1)
5.2**
If AD = DB and AE = EC then...
5.5**
If DB = FC and BD || FC then
5.3*
5.4**
If AE = EC and DE = EF then...
(7)
(1) (1) (1)
22
6.2**
Asked: Prove AECF a quadrilateral
Question 7
6
7
7
38
38
38
21
36
7.1*
22
36
24
37
(6)
The following are the test marks of 39 learners. The marks are out of 50. 9
11
15
16
20
20
39
39
43
45
45
50
24
37
24
37
24
37
24
37
32
37
Determine the mode of the data.
33
37
Determine the median of the data.
7.4**
Determine the upper quartile of the data.
7.5*
©Optimi
7.7***
[13]
7.2*
7.3**
7.6**
Given: ABCD is a quadrilateral with BE = DF
Determine the lower quartile of the data. Determine the inter quartile range.
36
What do you think is the outcomes of this test? Discuss the following
7.8.2**
How many learners achieved distinctions?
7.8.3****
38 (1)
Display the data on a histogram.
7.8***
7.8.1**
20
Draw the frequency table by classifying the possible outcomes in 5 groups. Use the following headings: ȇ Class ȇ Score ȇ Frequency ȇ Median ȇ Class limits ȇ Cumulative frequency
How many learners scored less than 40%?
Was the test difficult or easy? Motivate your answer.
(10) (6) (1) (1) (1)
[26] Total 100
(1) (2) (2)
(1)
23
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
MEMORANDUM MATHEMATICS GRADE 11 PAPER 2 OF 2 ENTRANCE EXAMINATION TIME: 120 MINUTES TOTAL 100
Question 2** 2.1***
Do not use_a calculator to answer this question.
√ 3 Given: ______ + 2 = 0 and Cos A < 0, sin A In which quadrant does A ̂ lie? _
√ 3 ____ sinA + 2 = 0
_
√ 3 ____ sinA = − 2
Cos A < 0
in 2 nd of 3 rd ∴ −2sin A = √ 3
∴ Question 1
1.1**
1.2**
∴
As it is given that 10° < 2𝑥 + 10 < 190° , determine the variable (unknown) in each of the following. Give your answer rounded off to 2 decimal places.
2.2***
∴ tan 𝑥 = 1,228° ∴ 𝑥 ≈ 50,84°
in 3 rd or 4 th A lies in the 3rd quadrant
Draw a Cartesian plane and indicate A as well as the values of x, y and r on the terminal arm. 𝑦-axis A
(2)
r= 2 _
P(𝑥 ; 𝑦) = (−1 ; −√ 3 )
2 tan 𝑥 = 2,456
2 tan 𝑥 = 2,456
_
√3 sin A = ____ −2
(6)
sin _ 𝑥2 = 0,588
sin _ 𝑥2 = 0,588 ∴_ 𝑥2 = 36,0152 ∴ 𝑥 ≈ 72,03° to 2 decimal places
_
(2)
[4]
24
𝑥-axis
Pythagoras: 𝑟2 = 𝑥2 + 𝑦2 _ ∴ 22 = 𝑥2 + (− √ 3 )2 ∴ 𝑥 2 = 4 − 3 ∴𝑥=±1 But in the 3rd quadrant ∴𝑥=−1
(4)
2.3**
Use the sketch and determine the value of: 2 __ − 1 Cos2A
__ 2 2 − 1 Cos A
2.4**
(_ −1 2) 2
=8−1 =7
_
−1
_
=√ 3
(3)
_
⤡ LHS = RHS ⤢
−1 = _____ – √2 3 ÷ _ 2 _
= √ 3
3.2** (5)
[18]
©Optimi
3.3****
(3)
Cos 𝛼 of Cos θ and the sides a, b or c.
Cos θ = _ BC ∴ BC = bCosθ b
Now prove that the area of
(3)
∆ ABC = _ 12 b 2Cos θ Cos α
Area of ∆ABC = _ 1 BC . AB 2
b
a
Now write AB , BC and AC in terms of
a a Cos θ = _ AC ∴ AC = _ Cos θ
A
B
Cos α = _ bc Cos θ = _ ba Cos β = Cos 90° = 0
Cos α = _ AB ∴ AB = b Cos α b
Study the right angled triangle with AC = b, AB = c and BC = a.
c
Cos α = ...
Cos β = ...
sin A RHS: = _ Cos A
LHS: tan A
Complete the following in terms of the sides of the triangle: Cos θ = ...
Sin A Use the sketch and prove that: Tan A =_ Cos A −√ 3 = _____
Question 3
2 = _____ − 1
3.1*
12 b Cos θ . b Cos α =_
12 b 2Cos θ Cos α =_
C 25
(4)
[10]
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide Question 4
4.1***
4.2**
Use analytical methods to answer this question. Sketch the following points on a Cartesian plane: A (−3 ; −1), B(1 ; 0), C(2 ; 5) and D (−2 ; 4)
√
2 _ 5 − 1 = (_ −32+ ; 2 )
− 1 ; _4 = (_ 2 2)
____________
−2 _12 ) + (−3) 2 = ( 2
___
= √ 6 _14 + 9
D(–2 ; 4)
Midpoint of AC
It is distance AP _____________________
= √ ( − 3 + _12 )2 + ( − 1 − 2) 2
Prove that the diagonals of ABCD bisect each other. 𝑦-axis C(2;5)
A(–3 ; –1)
What is the distance from the above mentioned intersect point to point to A?
____
B(1 ; 0)
𝑥-axis
4.3***
Prove that DA || CB.
4+0 = (_ −22+ 1 ; _ 2 ) − 1 ; _ 4 = ( _ 2 2)
4.4**** (7)
(4)
m DA = _ ––31+– 24
m CB = _ 10 –– 25
=5 ∴ m DA = m CB ∴ DA || CB
=5
=_ –– 15
Midpoint of DB
−1 ; 2 − 1 ; 2 = ( _ = ( _ ) ) 2 2 ∴ Bisecting points are equal ∴ AC and DB bisect each other.
= √ 15 _14 = 3,91 units
=_ –5 –1
Determine the gradient of the perpendicular line on AB.
(5)
m perpendicular line . m AB = –1
∴ m perpendicular line . _ 0 + 1 = – 1 1+3 ∴ m perpendicular line . _ 14 = – 1
26
∴ m perpendicular line = – 4
(4)
[20] Question 5
Study the following sketch. Complete the appropriate Theorem in each question. Give reasons for your answers.
5.1*
If  1= Ĉ 1 then...
5.2**
If AD = DB and AE = EC then...
5.3* 5.4**
©Optimi
5.5**
AB || CF alternate angles are equal
(1)
1 DE || BC and DE = _ BC Line from midpoint of 2 one side of triangle to midpoint of second side is parallel to the third and half the length.
(1)
ADCF a quadrilateral. Two pairs of diagonals biceps
(1)
D̂ 1= Ê 1 angles opposite equal sides
(1)
If AE = EC and DE = EF then... If AD = AE then…
27
If DB = FC and BD || FC then
5.6**
BDFC a quadrilateral. One pair of opposite sides equal and parallel If D̂ = B̂ + B̂ then ...
5.7**
If AD = DB and DE || BC then…
1
1
2
(1)
DF||BC corresponding angles equal
(1)
5.8**
AE = EC Line from midpoint of one side of triangle parallel to second side biceps the third. If F̂ + Ĉ + Ĉ = 180° then...
(1)
5.9***
If Area ∆DBC = Area ∆EBC then…
2
DF||BC
1
2
co-interior angles 180°
The basis and area of the two triangles are equal. This means that the height must be equal too and therefore the triangles lie between two parallel lines. DE||BC
(1)
(1)
[9]
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide Question 6 6.1***
For k:
Study the sketch and answer the questions formally. Use “statement” and “reason
Statement
B̂1= F̂2 but B̂ = 2k
ABEF is a quadrilateral. Determine the value of k and m of SA = SF Given: B̂1= 2k, F̂1= k, Ê1= 7k, and Ŝ1 = m AF||BE and AB||SF 2
1
1
1
1
∴ F ̂2 = 2k............................. F̂1= B ̂2 but F̂ = k 1
∴B ̂2 = k ..............................
2
alternate angles equal AB || SF Given ..1 alternate angles equal AF||BE Given ..2
In ∆FBE: Interior angles of B̂2+ F̂2+ Ê1= 180° But Ê 1= 7k and from 1 and 2 triangle k + 2k + 7k = 180° ∴ 10k = 180° ∴ k = 18°
1
2
Reason
1
For m:
(7)
Statement
SA = SF ∴A ̂ 1= F̂1+ F ̂2 But F̂1= k and F ̂2 = 2k ̂ ∴A 1 = 3k
With k = 18° ∴ Â 1 = 3(18) = 54° In ∆ASF: Â 1+ F̂1+ F̂2+ Ŝ1= 180° but Ŝ 1= m 0
28
∴ 3k + k + 2k + m = 180⁰ ∴ 6(18) + m = 180⁰ ∴ m = 72⁰
Reason
Given Equal sides subtend equal angles already proven
Interior angles of triangle
6.2**
Question 7
Given: ABCD is a quadrilateral with BE = DF
6
7
7
38
38
38
21
36
7.1*
Asked: Prove AECF a quadrilateral.
Connect A and C so that AC and BD intersect at M
7.2*
M
Statement
BM = MD and
7.3**
Reason
Diagonals of quad ABCD biceps AM = MC .......................... ...1 But BM = BE + EM Given and MD = MF + FD ∴ BE + EM = MF+FD And BE = DF ∴ EM = MF ......................... ...2 From 1 and 2: AECF is a quadrilateral
©Optimi
2 diagonals biceps at M
7.4**
(6)
[13] 29
22
36
24
37
The following are the test marks of 39 learners. The marks are out of 50. 9
11
15
16
20
20
39
39
43
45
45
50
24
37
24
37
24
37
24
37
32
37
Determine the mode of the data.
33
37
20
36
38
The mode is the most frequently occurring score and in this case it is 37.
(1)
The median is the middlemost score and in this case it is the 20th score and that is 36.
(1)
Determine the median of the data.
Determine the lower quartile.
The lower quartile Q1 = 10th term and that is the number 20 20 36 38 10th Q1
Determine the upper quartile
20th Q2
30th Q3
The upper quartile Q3 = 30ste term and that is the number 38.
(2) (2)
G11 – Mathematics
G11 – Mathematics – Facilitator’s Guide
Determine the interquartile range.
7.6**
Draw the frequency table by classifying the possible outcomes in 5 groups. Use the following headings: ȇ Class ȇ Score ȇ Frequency ȇ Middle value ȇ Class boundaries ȇ Cumulative frequency
Class
Q3– Q1= 38 – 20 = 18
Score
|||| 11–20 |||| | 0–10
21–30
|||| ||
31– 40 |||| |||| |||| ||| 41–50
||||
Total
Frequency
4 6
7
18 4
39
Middle values 5
15,5
25,5
35,5 45,5
7.7***
(1)
10,5 – 20,5
20,5 – 30,5
30,5 – 40,5 40,5 – 50
Histogram of Marks
.
1 Mark for each bar 1 Mark for general information
18
(10)
16 14 12 10 8 6 4
ClassCumulative boundaries frequency 0 – 10,5
Display the data on a histogram.
Frequency
7.5*
2
4
10
17
7.8
35 39
7.8.1**
2 Marks for each correct row.
7.8.2**
30
10,5 20,5 30,5 40,5 Class boundaries: Marks out of 50
What do you think is the outcomes of this test? Discuss the following: How many learners scored less than 40%?
50,5 (6)
10 learners scored less than 20 out of 50 = 40%
(1)
On the frequency table it is not clear enough who scored 40 out of 50. The raw data however, is arranged and it is thus possible to know that there are only 4 learners that scored 80%.
(1)
How many learners achieved distinctions?