Advanced Maths Book : Summary of Grade 12 Work

Page 1

M. Malan

St udy G u id e St udy G u id e

Via Afrika Mathematics Grade 12

Exponents and Surds Contents Introduction ......................................................................................................................... 3

Chapter 1 Number patterns, sequences and series ........................................ 4 OVERVIEW ............................................................................................................................... 4 Unit 1 Arithmetic sequences and series Unit 2 Geometric sequences and series Unit 3 The sum to n terms(Sn): Sigma notation Unit 4 Convergence and sum to infinity Mixed exercises .................................................................................................................... 8

Chapter 2 Functions ..................................................................................... 10 OVERVIEW ............................................................................................................................... 10 Unit 1 The definitions of a function Unit 2 The inverse of a function Unit 3 The inverse of y = ax + q Unit 4 The inverse of the quadratic function y = ax2 Mixed exercises .................................................................................................................... 21

Chapter 3 Logarithms ................................................................................... 24 OVERVIEW ............................................................................................................................... 24 Unit 1 The definition of a logarithm Unit 2 Solving exponential equations using logarithms Unit 3 The graph of y = logbx where b > 1 and 0 < b < 1 Mixed exercises .................................................................................................................... 28

Chapter 4 Finance, growth and decay ............................................................ 29 OVERVIEW ............................................................................................................................... 29 Unit 1 Future value annuities Unit 2 Present value annuities Unit 3 Calculating the period Unit 4 Analysing investments and loans Mixed exercises .................................................................................................................... 35

Chapter 5 Compound angles ......................................................................... 37 OVERVIEW ............................................................................................................................... 37 Unit 1 Deriving a formula for cos(đ?›ź − đ?›˝) Unit 2 Formula for cos(đ?›ź + đ?›˝) and đ?‘ đ?‘–đ?‘–(đ?›ź Âą đ?›˝)

Unit 3 Unit 4 Unit 5 Unit 6

Double angles Identities Equations Trigonometric graphs and compound angles

Mixed exercises .................................................................................................................... 46

Chapter 6 Solving problems in three dimensions ........................................... 48 OVERVIEW ............................................................................................................................... 48 Unit 1 Problems in three dimensions Unit 2 Compound angle formulae in three dimensions Mixed exercises .................................................................................................................... 51

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Exponents and Surds Chapter 7 Polynomials .................................................................................. 53 OVERVIEW ............................................................................................................................... 53 Unit 1 The Remainder Theorem Unit 2 The Factor Theorem Mixed exercises .................................................................................................................... 57

Chapter 8 Differential calculus ...................................................................... 58 OVERVIEW ............................................................................................................................... 58 Unit 1 Limits Unit 2 The gradient of a graph at a point Unit 3 The derivative of a function Unit 4 The equation of a tangent to a graph Unit 5 The graph of a cubic function Unit 6 The second derivative (concavity) Unit 7 Applications of differential calculus Mixed exercises .................................................................................................................... 70

Chapter 9 Analytical geometry ...................................................................... 72 OVERVIEW ............................................................................................................................... 72 Unit 1 Equation of a circle with centre at the origin Unit 2 Equation of a circle centred off the origin Unit 3 The equation of the tangent to the circle Mixed exercises .................................................................................................................... 77

Chapter 10 Euclidean geometry ..................................................................... 81 OVERVIEW ............................................................................................................................... 81 Unit 1 Proportionality in triangles Unit 2 Similarity in triangles Unit 2 Theorem of Pythagoras Mixed exercises .................................................................................................................... 94

Chapter 11 Statistics: regression and correlation ........................................... 97 OVERVIEW ............................................................................................................................... 97 Unit 1 Symmetrical and skewed data Unit 2 Scatter plots and correlation Mixed exercises .................................................................................................................... 106

Chapter 12 Probability .................................................................................. 108 OVERVIEW ............................................................................................................................... 108 Unit 1 Solving probability problems Unit 2 The counting principle Unit 3 The counting principle and probability Mixed exercises .................................................................................................................... 112 ANSWERS TO MIXED EXERCISES ...................................................................................................... 113 EXEMPLAR PAPER 1 ..................................................................................................................... 151 EXEMPLAR PAPER 2 .................................................................................................................... 168

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Exponents and Surds Introduction to Via Afrika Mathematics Grade 12 Study Guide Woohoo! You made it! If you’re reading this it means that you made it through Grade 11, and are now in Grade 12. But I guess you are already well aware of that… It also means that your teacher was brilliant enough to get the Via Afrika Mathematics Grade 12 Learner’s Book. This study guide contains summaries of each chapter, and should be used side-by-side with the Learner’s Book. It also contains lots of extra questions to help you master the subject matter. Mathematics – not for spectators You won’t learn anything if you don’t involve yourself in the subject-matter actively. Do the maths, feel the maths, and then understand and use the maths. Understanding the principles •

Listen during class. This study guide is brilliant but it is not enough. Listen to your teacher in class as you may learn a unique or easy way of doing something.

•

Study the notation, properly. Incorrect use of notation will be penalised in tests and exams. Pay attention to notation in our worked examples.

•

Practise, Practise, Practise, and then Practise some more. You have to practise as much as possible. The more you practise, the more prepared and confident you will feel for exams. This guide contains lots of extra practice opportunities.

•

Persevere. We can’t all be Einsteins, and even old Albert had difficulties learning some of the very advanced Mathematics necessary to formulate his theories. If you don’t understand immediately, work at it and practise with as many problems from this study guide as possible. You will find that topics that seem baffling at first, suddenly make sense.

•

Have the proper attitude. You can do it!

The AMA of Mathematics ABILITY is what you’re capable of doing. MOTIVATION determines what you do. ATTITUDE determines how well you do it. “Pure Mathematics is, in its way, the poetry of logical ideas.” Albert Einstein

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1 Number patterns, sequences and series Chapter

Overview

Unit 1 Page 10 •

Arithmetic sequences and series

Formula for an arithmetic sequence

Unit 2 Page 14 Chapter 1 Page 8 Number patterns, sequences and series

Geometric sequences and series

•

Formula for the nth term of a sequence

•

The sum to đ?‘– terms in an arithmetic sequence The sum to đ?‘– terms in a geometric sequence

Unit 3 Page 18 The sum to � terms (�� ): Sigma notation

•

Unit 4 Page 28 Convergence and sum to infinity

•

Convergence

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Bok.

2 Work through the notes in this chapter of this study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in this study guide.

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1 Number patterns, sequences and series Chapter

TABLE 1: SUMMARY OF SEQUENCES AND SERIES TYPE Arithmetic Sequence (AS) (also named the linear sequence) Constant st 1 difference

GENERAL TERM: đ?‘‡đ?‘› đ?‘‡đ?‘› = đ?‘Ž + (đ?‘– − 1)đ?‘‘ đ?‘Ž = đ?‘“đ?‘–đ?‘&#x;đ?‘ đ?‘Ą đ?‘Ąđ?‘?đ?‘&#x;đ?‘š đ?‘‡1

đ?‘‘ = đ?‘?đ?‘œđ?‘–đ?‘ đ?‘Ąđ?‘Žđ?‘–đ?‘Ą đ?‘‘đ?‘–đ?‘“đ?‘“. đ?‘‘ = đ?‘‡2 − đ?‘‡1

đ?‘œđ?‘&#x; đ?‘‡3 − đ?‘‡2 etc.

SUM OF TERMS: �� �� =

or

đ?‘– [2đ?‘Ž + (đ?‘– − 1)đ?‘‘] 2

�� =

where

đ?‘– [đ?‘Ž + đ?‘™] 2

đ?‘™ = the last term of the sequence

EXAMPLES A)

2 ; 5 ; 8 ; 11 ; ... +3 +3 +3

đ?‘‘=

đ?‘‡đ?‘› = 2 + (đ?‘– − 1)(3) = 2 + 3đ?‘– − 3 = 3đ?‘– − 1 B) 1 ; -4 ; -9 ; ... đ?‘‘ = -5

Geometric Sequence (GS) (also named exponential sequence)

Constant ratio

Quadratic Sequence (QS)

Constant

nd

2 difference

đ?‘‡đ?‘› = đ?‘Žđ?‘&#x; đ?‘›âˆ’1

đ?‘Ž = đ?‘“đ?‘–đ?‘&#x;đ?‘ đ?‘Ą đ?‘Ąđ?‘?đ?‘&#x;đ?‘š đ?‘‡1 đ?‘&#x; = đ?‘?đ?‘œđ?‘–đ?‘ đ?‘Ąđ?‘Žđ?‘–đ?‘Ą đ?‘&#x;đ?‘Žđ?‘Ąđ?‘–đ?‘œ

đ?‘‡2 đ?‘‡3 đ?‘&#x;= đ?‘œđ?‘&#x; đ?‘‡1 đ?‘‡2

đ?‘‡đ?‘› = đ?‘Žđ?‘–2 + đ?‘?đ?‘– + đ?‘?

�� = Or �� =

đ?‘Ž(đ?‘&#x; đ?‘› − 1) đ?‘&#x;−1 đ?‘›

đ?‘Ž(1 − đ?‘&#x; ) 1−đ?‘&#x;

Or đ?‘†âˆž =

đ?‘Ž 1−đ?‘&#x;

Where −1 < đ?‘&#x; < 1 (Converging series)

-5

đ?‘‡đ?‘› = 1 + (đ?‘– − 1)(−5) = 1 − 5đ?‘– + 5 = −5đ?‘– + 6 A)

2 ; -4 ; 8 ; -16 ; ...

đ?‘&#x; = x-2 x-2 x-2

đ?‘‡đ?‘› = 2(−2)đ?‘›âˆ’1

NOT CONVERGING as đ?‘&#x; < −1 B) 3 ;

3 2

;

3 4

1

đ?‘&#x;= x

2

3

;

; ...

8

1

x

2

1

x

2

1 đ?‘›âˆ’1 đ?‘‡đ?‘› = 3 ďż˝ ďż˝ 2 CONVERGING as −1 < đ?‘&#x; < 1 3 ; 8 ; 16 ; 27 ; ...

đ?‘“= 1st difference đ?‘ = 2nd difference

đ?‘“:

using simultaneous equations (see example)

Setup three equations using the first three terms: �1 = 3:

Determine đ?‘Ž, đ?‘? and đ?‘?

Alternatively: đ?‘Ž =đ?‘ á2 đ?‘? = đ?‘“1 − 3đ?‘Ž đ?‘? = đ?‘‡1 − đ?‘Ž − đ?‘?

where đ?‘“1 = first term of first differences

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đ?‘ :

5

8 3

11 3

3=đ?‘Ž+đ?‘?+đ?‘? đ?‘‡2 = 8:

‌(1)

8 = 4đ?‘Ž + 2đ?‘? + đ?‘? ‌(2) đ?‘‡3 = 16:

16 = 9đ?‘Ž + 3đ?‘? + đ?‘? ‌(3) Solving simultaneously leads to: đ?‘‡đ?‘› = 32đ?‘–2 + 12đ?‘– + 1

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1 Number patterns, sequences and series Chapter

TYPES OF QUESTIONS YOU CAN EXPECT

STRATEGY TO ANSWER THIS TYPE OF QUESTION

EXAMPLE(S) OF THIS TYPE OF QUESTION

Identify any of the following three types of sequences: Arithmetic (AS), Geometric (GS) and Quadratic (QS)

Determine whether sequence has a • constant 1st difference (AS) • constant ratio (GS) • constant 2nd difference (QS)

See Table 1 above

Determine the formula for the general term, �� , of AS, GS

You need to find: • � and � for an AS

See Table 1 above

Determine any specific term

Substitute the value of � into ��

See Text Book : Example 1, nr. 1 d and 2 d, p.8 (AS) Example 1, nr. 1 b, 3 b, p.11 (AS) Example 1, nr. 1, p. 15 (GS)

Substitute all known variables into the general term to get an equation with � as the only unknown. Solve for �. OR Substitute all known variables into the �� -formula to get an equation with � as the only unknown. Solve

See Text Book: Example 1, nr.1 c, p.8 Example 1, nr.1 c, p.11 Example 1, nr. 3, p.15

and QS (from Grade 11)

for a sequence e.g. �30

Determine the number of terms in a sequence, đ?‘–, for an AS, GS and QS or the position, đ?‘–, of a specific given term or when the sum of the series is given

When given two sets of information, make use of simultaneous equations to solve: đ?’‚ and đ?’… (for an AS) đ?’‚ and đ?’“ (for a GS) Determine the value of a variable (đ?‘Ľ) when given a sequence in terms of đ?‘Ľ.

• •

đ?‘Ž and đ?‘&#x; for a GS đ?‘Ž, đ?‘? and đ?‘? for a QS

for đ?‘–.

Example 2, nr.3, p.20 Example 3, nr. 2, p.24

For each set of information given,

See Text Book: Example 1, nr. 3, p.11 (AS) Example 1, nr.2, p.15 (AS) Example 3, nr.3, p.24 (GS)

Remember: đ?‘– must be a natural number (not negative, not a fraction)

substitute the values of đ?‘– and đ?‘‡đ?‘› or đ?‘– and đ?‘†đ?‘› . You then have 2 equations which you can solve simultaneously (by substitution) For AS use constant difference: đ?‘‡3 − đ?‘‡2 = đ?‘‡2 − đ?‘‡1 For GS use constant ratio: đ?‘‡2 đ?‘‡3 = đ?‘‡1 đ?‘‡2

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The first three terms of an AS are given by 2đ?‘Ľ − 4; đ?‘Ľ − 3; 8 − 2đ?‘Ľ Determine đ?‘Ľ: 8 − 2đ?‘Ľ − (đ?‘Ľ − 3) = đ?‘Ľ − 3 − (2đ?‘Ľ − 4) ∴đ?‘Ľ=5

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1 Number patterns, sequences and series Chapter

For a series given in sigma notation: • Determine the number of terms

Remember: The “counter� indicates the number of terms in the series

∑đ?‘›đ?‘˜=1 đ?‘‡đ?‘˜ has đ?‘– terms

(counter đ?‘˜ runs from 1 to đ?‘–) ∑đ?‘›đ?‘˜=0 đ?‘‡đ?‘˜ has (đ?‘– + 1) terms

(counter runs from 0 to đ?‘–; so one term extra) ∑đ?‘›đ?‘˜=5 đ?‘‡đ?‘˜ has (đ?‘– − 4) terms ( four terms not counted )

•

Determine the value of the series, in other words, �� .

Write a given series in sigma notation. Determine the sum, đ?‘şđ?’” , of an AS and a GS (when the number of terms are given or not given) Determine whether a GS is converging or not Determine đ?‘†âˆž for a converging GS

Determine the value of a variable (�) for which a series will converge, e.g. (2� + 1) + (2� + 1)2 +‌ Apply your knowledge of sequences and series on an applied example (often involving diagram/s)

Remember the expression next to the ∑-sign is the general term, đ?‘‡đ?‘› .

See Text Book: Example 1, p.19

Determine the general term, đ?‘‡đ?‘˜ and number of terms, đ?‘– and substitute into ∑đ?‘› đ?‘˜=1 đ?‘‡đ?‘˜

Example 1, p.19

This will help you to determine đ?‘Ž and đ?‘‘ or đ?‘&#x;.

In some cases you have to first determine the number of terms, đ?‘– using đ?‘‡đ?‘› . Substitute the values of đ?‘Ž, đ?‘– and đ?‘‘/đ?‘&#x; into the formula for đ?‘†đ?‘›

See Text Book:

Substitute vales of đ?‘Ž and đ?‘&#x; Into formula for đ?‘†âˆž

See Text Book: Example 1, nr. 1, p.29

Determine đ?‘&#x; in terms of đ?‘Ľ and use −1 < đ?‘&#x; < 1

See Text Book: Example 1, nr. 3, p.29

Generate a sequence of terms from the information given. Identify the type of sequence.

See Text Book: Exercise 5, nr. 6, p.30

Converging if −1 < đ?‘&#x; < 1

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Example 2, nr.1 & 2, p.20 Example 3, nr. 1, p.24

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1 Number patterns, sequences and series Chapter

Mixed Exercise on sequences and series 1

Consider the following sequence:

5; 9; 13; 17; 21; ‌

a

Determine the general term.

b

Which term is equal to 217?

2

a b

T5 of a geometric sequence is 9 and T9 is 729. Determine the constant ratio. Determine T10.

3

The following is an arithmetic sequence: 2 x − 4 ; 5 x ; 7 x − 4

4

5

a

Determine the value of x .

b

Determine the first 3 terms.

Consider the following sequence: a

Determine the general term.

b

Which term is equal to 260?

2 ; 7 ; 15 ; 26 ; 40 ; ‌

How many terms are there in the following sequence? 17 ; 14 ; 11 ; 8 ; ‌ ; -2785

6

Tom links balls with rods in arrangements as shown below:

Arrangement 1 Arrangement 2

1 ball, 4 rods 4 balls, 12 rods

7

Arrangement 3

Arrangement 4

9 balls, 24 rods

16 balls 40 rods

a

Determine the number of balls in the nth arrangement.

b

Determine the number of rods in the nth arrangement.

Determine the following: 30

a

∑ k =1

10

(8 − 5k)

b

∑ k =2

1+5+9+‌+21

1 4

(2)đ?‘˜âˆ’1

8

Write the following in sigma notation:

9

The 5th term of an arithmetic sequence is zero and the 13th term is equal to 12. Determine: a

the constant difference and the first term.

b

the sum of the first 21 terms.

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1 Number patterns, sequences and series Chapter

10

The first two terms of a geometric sequence are: (đ?‘Ľ + 3) and (đ?‘Ľ 2 − 9) a

b

For which value of đ?‘Ľ is this a converging sequence?

Calculate the value of đ?‘Ľ if the sum of the series to infinity is 13.

11

Calculate the value of:

12

đ?‘†đ?‘› = 3đ?‘–2 − 2đ?‘–. Determine đ?‘‡9 .

13

99+97+95+â‹Ż+1 299+297+295+â‹Ż+201

The first four terms of a geometric sequence are 7; đ?‘Ľ ; đ?‘Ś ; 189.

a

b

Determine the values of đ?‘Ľ and đ?‘Ś.

If the constant ratio is 3, make use of a suitable formula to determine the number of terms in the sequence that will give a sum of 206 668.

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2 Functions

Chapter

Overview Unit 1 Page 40 • • •

The definition of a function

Relations and functions Type of relations Which relations are functions? Definition of a function? Function notation

• • Unit 2 Page 44 Chapter 2 Page 36 Functions

The inverse of a function

•

The concept of inverses by studying sets of ordered number pairs

•

Graphs of đ?‘“and đ?‘“ −1 on the same set of axes

Unit 3 Page 46 The inverse of đ?‘Ś = đ?‘Žđ?‘Ľ + đ?‘ž Unit 4 Page 48 The inverse of the quadratic function đ?‘Ś = đ?‘Žđ?‘Ľ 2

•

Restricting the domain of the parabola

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Book.

2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in the study guide.

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2 Functions

Chapter

TYPES OF RELATIONS BETWEEN TWO VARIABLES TYPE

DESCRIPTION

NON-FUNCTIONS One-to-many

FUNCTIONS

One-to-one

PROPERTIES

TYPICAL EXAMPLES

• One �-value in domain has MORE THAN ONE �-value • Does NOT pass vertical line test

• Inverse of a parabola (See Unit 4)

• Each �-value has a unique �value

• Straight line graph and its inverse • Hyperbola and its inverse • Exponential graph and its inverse, the logarithmic function

• No �-value appears more than once in domain • More than one �-value maps onto the same �-value • Passes VERTICAL line test

• Parabola • Graph of the cubic function • Trigonometric graphs

• No �- or �-value appear more than once in domain or range • Passes VERTICAL line test

Many-to- one

REVISION OF THE STRAIGHT LINE GRAPH

• •

đ?’Ž

Standard form: đ?’š = đ?’Žđ?’™ + đ?’„

Gradient of line Indicates “steepness� and direction of line:

•

đ?‘š > 0 (+)

đ?‘š < 0(−)

�=0 •

•

đ?’„

đ?‘Ś-intercept

Where đ?‘Ľ = 0

đ?‘Ś −đ?‘Ś

đ?‘š = đ?‘Ľ2 −đ?‘Ľ1 2

1

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2 Functions

Chapter

PARALLEL AND PERPENDICULAR LINES Let đ?‘Ś = đ?‘š1 đ?‘Ľ + đ?‘?1 and

đ?‘Ś = đ?‘š2 đ?‘Ľ + đ?‘?2 be two lines.

If the lines are PARALLEL, then: đ?‘š1 = đ?‘š2

If the lines are PERPENDICULAR, then:

đ?‘š1 Ă— đ?‘š2 = −1

TO DETERMINE THE EQUATION OF A STRAIGHT LINE GIVEN: 1. Gradient and a point

2. đ?’š-intercept and a point

3. Two points on the line

EXAMPLES

1 2

A line has a gradient of and goes through the point (4;1): 1 đ?‘š= 2 1 Substitute point (4;1) into đ?‘Ś = đ?‘Ľ + đ?‘? 2 1 1 = (4) + đ?‘? 2 đ?‘? = −1 1 đ?‘Ś = đ?‘Ľâˆ’1 2 A line has a đ?‘Ś-intercept 3 and goes through the point (-2;1): đ?‘?=3 Substitute point (-2;1) into đ?‘Ś = đ?‘šđ?‘Ľ + 3 1 = đ?‘š(−2) + 3 đ?‘š=1 đ?‘Ś =đ?‘Ľ+3 A line goes through the points (4;-3) and (2;1). đ?‘Ś −đ?‘Ś 1−(−3) đ?‘š = đ?‘Ľ2 −đ?‘Ľ1 = 4−(2) = 2 2

4. A point or đ?’š-intercept plus information regarding relationship to another line

1

Substitute any one of the two points into đ?‘Ś = 2đ?‘Ľ + đ?‘? 1=2(2)+c đ?‘? = −3 đ?‘Ś = 2đ?‘Ľ − 3

a) A line is parallel to the line đ?‘Ś = −đ?‘Ľ + 3 and goes through the point (5;-2). Parallel lines have same gradients; so đ?‘š = −1 Sub (5;-2) into đ?‘Ś = −đ?‘Ľ + đ?‘? −2 = −(5) + đ?‘? đ?‘?=3

b) A line is perpendicular to the line đ?‘Ś = 2đ?‘Ľ − 1 and has a đ?‘Ś-intercept of 4. Š Via Afrika ›› Mathematics Grade 12

12

2 Functions

Chapter

Perpendicular lines have gradients with a product of −đ?&#x;?. 1 đ?‘š Ă— 2 = −1 ∴đ?‘š=− 2 −1 đ?‘Ś= đ?‘Ľ+4 2

REVISION OF THE PARABOLA

EQUATION IN STANDARD FORM

đ?’š = đ?’‚đ?’™đ?&#x;? + đ?’ƒđ?’™ + đ?’„

(đ?’‚ ≠đ?&#x;Ž)

đ?’‚

đ?’„

•

Indicates shape of parabola

đ?‘Ś-intercept

•

đ?’‚ > 0 (+)

Where đ?‘Ľ = 0

Concave up

đ?’ƒ

ď Š Remember: Positive(+) people smile!

•

Affects the axis of symmetry and turning point (TP)

đ?’‚ < 0(−)

•

Concave down

•

ď Œ

Equation of axis of symmetry: đ?‘Ľ = − đ?‘?

Coordinates of TP ďż˝âˆ’ 2đ?‘Ž ;

Remember: Negative (−) people are sad!

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4đ?‘Žđ?‘Žâˆ’đ?‘?2 ďż˝ 4đ?‘Ž

đ?‘? 2đ?‘Ž

đ?’™-intercepts

•

Also called roots/zeroes

•

Substitute đ?‘Ś = 0

13

2 Functions

Chapter

EQUATION IN TURNING POINT FORM

đ?’š = đ?’‚(đ?’™ − đ?’‘)đ?&#x;? + đ?’’ (đ?’‚ ≠đ?&#x;Ž)

đ?’‚

đ?’‘ and đ?’’

Indicates shape of parabola đ?’‚ > 0 (+)

Concave up

ď Š Remember: Positive(+) people smile! đ?’‚ < 0(−)

• •

Equation of axis of symmetry đ?’™

=đ?’‘

Coordinates of turning point (đ?’‘; đ?’’)

Intercepts

Concave down

ď Œ Remember:

• •

đ?‘Ľ-intercepts (make đ?‘Ś = 0) đ?‘Ś-intercept (make đ?‘Ľ = 0)

Negative (−) people are sad! DOMAIN: đ?‘Ľ ∈ đ?‘… RANGE:

đ?‘Ś ∈ (−∞; đ?‘ž)

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đ?‘Ś ∈ (đ?‘ž; ∞)

14

2 Functions

Chapter

DETERMINE THE EQUATION OF A PARABOLA GIVEN: 2 ROOTS (𝒙-INTERCEPTS) PLUS 1 POINT

GIVEN: TURNING POINT PLUS 1 POINT

FORM OF EQUATION:

FORM OF EQUATION:

𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝒙𝟏 and 𝒙𝟐 are the roots

𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 (𝑝; 𝑞) is die turning point of the parabola

EXAMPLE:

EXAMPLE: y

y

5 (2;6)

(−1;2)

x

−1

3 x

𝒙𝟏 = −𝟏

𝒚 = 𝒂(𝒙 − 𝒙𝟏 )(𝒙 − 𝒙𝟐 ) 𝒚 = 𝒂(𝒙 − (−𝟏))(𝒙 − 𝟏) 𝒚 = 𝒂(𝒙 + 𝟏)(𝒙 − 𝟏)

𝒙𝟐 = 𝟏

Now substitute the other point (𝟐; 𝟔): 𝟔 = 𝒂(𝟐 + 𝟏)(𝟐 − 𝟏) 𝟔 = 𝒂(𝟏)(−𝟏) 𝟔 = −𝟏𝒂 −𝟐 = 𝒂

𝒚 = −𝟐(𝒙 + 𝟏)(𝒙 − 𝟏) 𝒚 = −𝟐�𝒙𝟐 − 𝟐𝒙 − 𝟏�

𝒚 = −𝟐𝒙𝟐 + 𝟒𝒙 + 𝟔 (standard form)

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(𝑝; 𝑞) = (−1; 2)

𝑦 = 𝑎(𝑥 − 𝑝)2 + 𝑞 2 𝑦 = 𝑎�𝑥 − (−1)� + 2 𝑦 = 𝑎(𝑥 + 1)2 + 2

Now substitute the point (0;5): 5 = 𝑎(0 + 1)2 + 2 5=𝑎+2 3=𝑎

𝑦 = 3(𝑥 + 1)2 + 2 𝑦 = 3(𝑥 2 + 2𝑥 + 1) + 2 𝑦 = 3𝑥 2 + 6𝑥 + 3 + 2 𝑦 = 3𝑥 2 + 6𝑥 + 5 (standard form)

15

2 Functions

Chapter

REVISION OF THE HYPERBOLA

đ?’‚

đ?’‚

đ?’š = đ?’™âˆ’đ?’‘ + đ?’’

đ?’’

Horizontal asymptote

đ?’š=đ?’’

Indicates shape of hyperbola (with respect to asymptotes) đ?’‚ > 0 (+)

đ?’‚ < 0(−)

đ?’‘

Intercepts

Vertical asymptote

đ?’™=đ?’‘

•

đ?‘Ľ-intercept (make đ?‘Ś = 0)

•

đ?‘Ś-intercept (make đ?‘Ľ = 0)

Domain: đ?‘Ľ ∈ đ?‘…; đ?‘Ľ ≠đ?‘? Range: đ?‘Ś ∈ đ?‘…; đ?‘Ś ≠đ?‘?

2

EXAMPLE: đ?‘Ś = đ?‘Ľâˆ’1 − 2 Axes of symmetry:

Axes of symmetry (AS) •

Two axes of symmetry

•

AS go through intersect of

• •

asymptotes (đ?‘?; đ?‘ž)

Equations: đ?‘Ś = đ?‘Ľ + đ?‘˜1 and đ?‘Ś = −đ?‘Ľ + đ?‘˜2

Substitute the point (đ?‘?; đ?‘ž) to calculate đ?‘˜1 and đ?‘˜2

Š Via Afrika ›› Mathematics Grade 12

đ?‘Ś-intercept: 2 đ?‘Ś= − 2 = −4 −1 đ?‘Ľ- intercept: 2 0= − 2 ;đ?‘Ľ = 2

Substitute(1; −2) into đ?‘Ś = đ?‘Ľ + đ?‘˜1 đ?‘Žđ?‘–đ?‘‘ đ?‘Ś = −đ?‘Ľ + đ?‘˜2

đ?‘Ľâˆ’1

Asymptotes: đ?‘Ľ = 1 đ?‘Žđ?‘–đ?‘‘ đ?‘Ś = −2

−2 = 1 + đ?‘˜1 đ?‘Žđ?‘–đ?‘‘ − 2 = −1 + đ?‘˜2 đ?‘˜1 = −3 đ?‘Žđ?‘–đ?‘‘ đ?‘˜2 = −1 đ?‘Ś = đ?‘Ľ − 3 đ?‘Žđ?‘–đ?‘‘ đ?‘Ś = −đ?‘Ľ − 1

y

x

1

2

−2

−4

16

2 Functions

Chapter

REVISION OF THE EXPONENTIAL GRAPH

đ?’š = đ?’‚đ?’™âˆ’đ?’‘ + đ?’’

đ?’‚

Indicates shape of hyperbola đ?’‚>đ?&#x;?

•

đ?&#x;Ž<đ?’‚<1

•

đ?’’

Horizontal asymptote: đ?‘Ś = đ?‘?

Indicates that the graph đ?‘Ś = đ?‘Ž đ?‘Ľ was

translated (shifted) vertically up/down

đ?’’ > 0: shifted upwards

đ?’’ < 0: shifted downwards

đ?’‘

đ?‘Ľ

Indicates that the graph đ?‘Ś = đ?‘Ž was

translated (shifted) horizontally left/right đ?’‘ > 0: shifted left

đ?’‘ < 0: shifted right

EXAMPLE:

đ?‘Ś = 2đ?‘Ľ+1 − 1

Asymptote: đ?‘Ś = −1 đ?‘Ľ-intercept (đ?‘Ś = 0): 2đ?‘Ľ+1 − 1 = 0 ∴ đ?‘Ľ = −1 0+1 đ?‘Ś-intercept: (đ?‘Ľ = 0): đ?‘Ś = 2 −1=1 y

Intercepts 1

• •

đ?‘Ľ-intercept (make đ?‘Ś = 0) đ?‘Ś-intercept (make đ?‘Ľ = 0)

x

−1 −1

Domain: đ?‘Ľ ∈ đ?‘…

Range: đ?‘Ś ∈ (đ?‘ž; ∞) Š Via Afrika ›› Mathematics Grade 12

17

2 Functions

Chapter

EXAMPLES OF SYMMETRICAL EXPONENTIAL GRAPHS

SYMMETRICAL IN THE 𝑦 −axis y

𝑥

𝑦 = 3𝑥

1 𝑦 = � � = 3−𝑥 3

x

SYMMETRICAL IN THE 𝑥 −axis y

𝑦 = 3𝑥 x

𝑦 = −3𝑥

© Via Afrika ›› Mathematics Grade 12

18

2 Functions

Chapter

INTERSECTS OF TWO GRAPHS To determine the coordinates of the point where two graphs INTERSECT: Use SIMULTANEOUS EQUATIONS

EXAMPLE Determine the coordinates of the points of intersection of đ?‘“(đ?‘Ľ) = 3đ?‘Ľ + 6 and đ?‘”(đ?‘Ľ) = −2đ?‘Ľ 2 + 3đ?‘Ľ + 14

Equate the two equations and solve for đ?’™: 3đ?‘Ľ + 6 = −2đ?‘Ľ 2 + 3đ?‘Ľ + 14 2đ?‘Ľ 2 − 8 = 0 đ?‘Ľ2 − 4 = 0 (đ?‘Ľ − 2)(đ?‘Ľ + 2) = 0 đ?‘Ľ = 2 or đ?‘Ľ = −2

Substitute đ?’™-values back into one of equations

(choose the easier one):

If đ?‘Ľ = 2 then đ?‘Ś = 3(2) + 6 = 12

So one point of intersection is (2; 12). If đ?‘Ľ = −2 then đ?‘Ś = 3(−2) + 6 = 0

The other point of intersection is (−2; 0) which

is also the đ?‘Ľ-intercept of both graphs.

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19

2 Functions

Chapter

THE INVERSE OF A FUNCTION • The inverse of a function, đ?‘“, is denoted by đ?‘“ −1 . • đ?‘“ −1 is a reflection of đ?‘“in the line đ?‘Ś = đ?‘Ľ

• To determine the equation of đ?‘“ −1 , swop đ?‘Ľ and đ?‘Ś in the equation of đ?‘“

• The đ?‘Ľ-intercept of đ?‘“ is the đ?‘Ś-intercept of đ?‘“ −1 FUNCTION đ?’‡

Straight line

INVERSE OF FUNCTION, đ?’‡âˆ’đ?&#x;?

Straight line

đ?’‡: đ?’š = đ?’Žđ?’™ + đ?’„

Exponential graph đ?’™

đ?’‡: đ?’š = đ?’‚

Parabola đ?’‡: đ?’š = đ?’‚đ?’™đ?&#x;?

EXAMPLES

DIAGRAM

đ?‘“: đ?‘Ś = 2đ?‘Ľ + 3

y

f

Inverse: 2đ?‘Ś + 3 = đ?‘Ľ đ?‘“

Logarithmic function đ?‘“

−1

: đ?‘Ś = log đ?‘Ž đ?‘Ľ

The inverse of a parabola is NOT A FUNCTION NB: The DOMAIN of the parabola has to be RESTRICTED to đ?‘Ľ ≼ 0 or đ?‘Ľ ≤ 0 so that đ?‘“ −1 is also a function

Š Via Afrika ›› Mathematics Grade 12

−1

:đ?‘Ś =

x

1 3 đ?‘Ľâˆ’ 2 2

đ?‘“: đ?‘Ś = 3đ?‘Ľ

y

f

Inverse:

đ?‘“ −1 : đ?‘Ś = log 3 đ?‘Ľ

đ?‘“: đ?‘Ś = 2đ?‘Ľ 2

x

y

Inverse:

đ?‘Ľ = 2đ?‘Ś 2 1 đ?‘Ś2 = đ?‘Ľ 2

x

1 đ?‘“ −1 : Âąďż˝ đ?‘Ľ 2

20

2 Functions

Chapter

Mixed Exercise on Functions 1

2

Determine the coordinates of the intercept of the following two lines: 2đ?‘Ľ − 3đ?‘Ś = 17 3đ?‘Ľ − đ?‘Ś = 15 a

b c

Determine the equation of line đ?‘“.

f

3

Determine the equation of line đ?‘”.

P

Determine the co-ordinates of point, P,

1

where the two lines intersect.

d

y

4

x −4

Are these two lines perpendicular?

−3

4

(2;-1)

Give a reason for your answer. e

Write down the equation of the line

g

which is parallel to line đ?‘” with a đ?‘Ś-intercept

−4

of -2.

3

The diagram shows the graphs of y = x 2 − 2 x − 3

y

and y = mx + c . a

Determine the lengths of OA, OB and OC.

b

Determine the co-ordinates of the turning point D.

c

Determine m and c of the straight line.

d

Use the graph to determine for which values

A

B

x

C

of k for which the equation x 2 − 2 x + k = 0 would

D

have only one real root.

C

4

The diagram shows the graph of f ( x) = −2( x + 1) + 8 . 2

E

D

C is the turning point. E is the mirror image of the y-intercept of đ?‘“. Determine: a

the length of AB.

b

the co-ordinates of C.

c

the length of DE.

Š Via Afrika ›› Mathematics Grade 12

A

B

21

x

2 Functions

Chapter

5

a

b c

6

1 đ?‘Ľ 2

Consider the function đ?‘”(đ?‘Ľ) = ďż˝ ďż˝ − 2.

Make a neat drawing of đ?‘”. Clearly show the asymptote and intercepts with the axes. Determine the domain of đ?‘”.

For which values of � would �(�) ≼ 0.

The graph of đ?‘“(đ?‘Ľ) =

đ?‘Ž đ?‘Ľ

y

; � ≠0 is shown.

đ??´(−2; 2) is a point on the graph where it cuts the line đ?‘Ś = −đ?‘Ľ. a

b c d 7

Determine the value of đ?‘Ž .

A(−2;2) x

Write down the coordinates of B.

f

B

Graph đ?‘“ is translated 2 units up and 1 unit right. Write down the equation of the new graph.

The graphs of the following are shown : 1 đ?‘“(đ?‘Ľ) = −đ?‘Ľ 2 − 2đ?‘Ľ + 8 and đ?‘”(đ?‘Ľ) = đ?‘Ľ − 1 2

Determine: a

the coordinates for A

b

the coordinates for B and C

c

the length of CD

d

the length of DE which is parallel to the đ?‘Ś-axis

e f g

E

R f

the length of GH which is parallel to the đ?‘Ś-axis

i

the đ?‘Ľ-values for which đ?‘“(đ?‘Ľ) − đ?‘”(đ?‘Ľ) > 0.

x

H

the đ?‘Ľ-value for which RS would have a maximum length.

the maximum length of RS.

A

F

the length of AF which is parallel to the đ?‘Ľ-axis

h

y

g

B

G

D

C

S

y

8

The diagram alongside shows the graphs of the functions of

a b c d e f

đ?‘Ž +đ?‘ž đ?‘Ľ+đ?‘? Write down the equation of the asymptote of đ?‘“. đ?‘“(đ?‘Ľ) = đ?‘? đ?‘Ľ + đ?‘? đ?‘Žđ?‘–đ?‘‘ đ?‘”(đ?‘Ľ) =

Determine the equation of đ?‘“. Write down the equations of the asymptotes of đ?‘”. Determine the equation of đ?‘”. Determine the equations of the axes of symmetry of đ?‘” . For which values of đ?‘Ľ is đ?‘“(đ?‘Ľ) > đ?‘”(đ?‘Ľ)?

Š Via Afrika ›› Mathematics Grade 12

(2 ;5 )

f

B(−6 ;0 )

−2

x

−1

g A(0 ;−3 ) −4

22

2 Functions

Chapter

9

The graph of đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 2 is given. a

b

10

y

Determine the equation of đ?‘“ −1 in the form đ?‘“ −1 : đ?‘Ś = â‹Ż How can one restrict the domain of đ?‘“ so that đ?‘“ −1 will be a function?

f(x)=2x²

x

The graph of đ?‘“(đ?‘Ľ) = đ?‘Ž đ?‘Ľ is given. The point A (-1; 3) lies on the graph.

3

y

2

1

a b c d

11

Determine the equation of đ?‘“. Determine the equation of đ?‘“ −1 in the form đ?‘“ −1 : đ?‘Ś = â‹Ż Make a neat drawing of the graph of đ?‘“ −1 . Determine the domain of đ?‘“ −1 .

x −3

−2

−1

1

2

3

−1

−2

−3

A straight line graph has an đ?‘Ľ-intercept of -2 and a đ?‘Ś-intercept of 3. Write down the coordinates of the đ?‘Ľ- and đ?‘Ś-intercepts of đ?‘“ −1 .

Š Via Afrika ›› Mathematics Grade 12

23

3 Logarithms Chapter

Overview

Unit 1 Page 60 The definition of a logarithm

• •

Changing exponents to the logarithmic form Proofs of the logarithmic laws

Unit 2 Page 64 Chapter 3 Page 58 Logarithms

Solve exponential equations using logarithms

•

Using logarithms

•

Inverse of đ?‘Ś = đ?‘“(đ?‘Ľ) = 2đ?‘Ľ Inverse of the function 1 đ?‘Ľ đ?‘Ś = đ?‘“(đ?‘Ľ) = ďż˝ ďż˝ 2

Unit 3 Page 66 The graph of đ?‘Ś = log đ?‘? đ?‘Ľ where đ?‘? > 1 and 0 < đ?‘? < 1

•

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Bok.

2 Work through the notes in this chapter of this study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in this study guide.

Š Via Afrika ›› Mathematics Grade 12

24

3 Logarithms Chapter

Definition of logarithm If log đ?’ƒ đ?’™ = đ?’š, then đ?’ƒđ?’š = đ?’™.

EXAMPLES: Converting from one form to another Logarithmic form

Exponential form

đ?&#x;?đ?&#x;“ = đ?&#x;?đ?&#x;’đ?&#x;? đ?&#x;Ž, đ?&#x;“đ?&#x;? = đ?&#x;Ž, đ?&#x;?đ?&#x;?đ?&#x;“ đ?&#x;?đ?&#x;Žđ?&#x;? = đ?&#x;?đ?&#x;Žđ?&#x;Žđ?&#x;Ž

log đ?&#x;? đ?&#x;?đ?&#x;’đ?&#x;? = đ?&#x;“ log đ?&#x;Ž,đ?&#x;“ đ?&#x;Ž, đ?&#x;?đ?&#x;?đ?&#x;“ = đ?&#x;? log đ?&#x;?đ?&#x;Ž đ?&#x;?đ?&#x;Žđ?&#x;Žđ?&#x;Ž = đ?&#x;? đ?&#x;? log đ?&#x;? √đ?&#x;? = đ?&#x;?

LOGARITHMIC LAW

Law 1: đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘¨. đ?‘Š = đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘¨ + đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘Š đ?‘¨

Law 2: đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘Š = đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘¨ − đ??Ľđ??Ľđ??Ľ đ?’Ž đ?‘Š

đ?&#x;?

đ?&#x;?đ?&#x;? = √đ?&#x;?

• • • •

• Law 3: đ??Ľđ??Ľđ??Ľ đ?’™ đ?‘ˇđ?’š = đ?’š đ??Ľđ??Ľđ??Ľ đ?’™ đ?‘ˇ đ??Ľđ??Ľđ??Ľ đ?’‚

Law 4: đ??Ľđ??Ľđ??Ľ đ?’ƒ đ?’‚ = đ??Ľđ??Ľđ??Ľ đ?’ƒ

• • •

EXAMPLES log đ?‘˜ đ?‘Žđ?‘?đ?‘? = log đ?‘˜ đ?‘Ž + log đ?‘˜ đ?‘? + log đ?‘˜ đ?‘?

log 5 25.5 = log 5 25 + log 5 5 = 2 + 1 = 3 đ?‘Ś

log đ?‘š đ?‘§ = log đ?‘š đ?‘Ś − log đ?‘š đ?‘§ log 5

0,2 25

= log 5 0,2 − log 5 25 = log 5 5−1 − log 5 25 = −1 − 2 = −3 log đ?‘Ś đ?‘Ž3 = 3 log đ?‘Ś đ?‘Ž

log 5 0,04 = log 5 5−2 = −2 log 5 5 = −2 log đ?‘Ž

log đ?‘? đ?‘Ž = log đ?‘? log 5

log 2 5 = log 2 = 2,32

Note that: • log đ?‘Ž đ?‘Ž = 1 (đ?‘Ž ≠0) • log đ?‘Ž 1 = 0 • log đ?‘Ž = log đ?&#x;?đ?&#x;Ž đ?‘Ž Š Via Afrika ›› Mathematics Grade 12

25

3 Logarithms Chapter

USING LOGARITHMS TO SOLVE EQUATIONS We know that equations involving exponents can be solved using exponential laws: đ?&#x;?đ?’™ = đ?&#x;?đ?&#x;?đ?&#x;? đ?&#x;?đ?’™ = đ?&#x;?đ?&#x;• (prime factorise) ∴đ?’™=đ?&#x;•

But, what if we cannot use prime factors? đ?&#x;?đ?’™ = đ?&#x;?đ?&#x;? đ??Ľđ??Ľđ??Ľ đ?&#x;?đ?’™ = đ??Ľđ??Ľđ??Ľ đ?&#x;?đ?&#x;? đ?’™đ??Ľđ??Ľđ??Ľ đ?&#x;? = đ??Ľđ??Ľđ??Ľ đ?&#x;?đ?&#x;? đ??Ľđ??Ľđ??Ľ đ?&#x;?đ?&#x;? đ?’™= = đ?&#x;‘, đ?&#x;• đ??Ľđ??Ľđ??Ľ đ?&#x;?

Š Via Afrika ›› Mathematics Grade 12

26

3 Logarithms Chapter

THE INVERSE OF THE 𝑓: 𝑦 = 𝑎 𝑥

EXPONENTIAL GRAPH

is the logarithmic function 𝑓 −1 : 𝑦 = log 𝑎 𝑥 ; 𝑥 > 0 EXAMPLES

𝑓(RED GRAPH)

𝒚=𝟒

𝒙

𝑓

−1

(BLUE GRAPH)

DIAGRAM y

1

𝒚 = 𝐥𝐥𝐥 𝟒 𝒙

x

1

y

𝟏𝒙 𝒚= 𝟒

𝒚 = 𝐥𝐥𝐥 𝟏 𝒙 𝟒

1 x

1

y

𝒚 = −𝟒𝒙

𝒚 = 𝐥𝐥𝐥 𝟒 (−𝒙)

−1

x

−1

y

𝟏𝒙 𝒚=− 𝟒

𝒚 = 𝐥𝐥𝐥 𝟏 (−𝒙)

© Via Afrika ›› Mathematics Grade 12

𝟒

−1

x

−1

27

3 Logarithms Chapter

Mixed Exercise on Logarithms

1

Make use of the definition of the logarithm to solve for đ?‘Ľ: a

b c d e f g 2

log 1 đ?‘Ľ = 2 3

− log 4 đ?‘Ľ = 2

log 5 đ?‘Ľ = −2

log đ?‘Ľ 3 = 6

log 3 81 = đ?‘Ľ 1 9

log 3 = đ?‘Ľ

b c d

Determine the value of đ?‘Ž.

Determine the equation of đ?‘“ −1.

Determine the equation of đ?‘” if đ?‘“ and đ?‘” are symmetrical in the đ?‘Ś-axis.

Determine the equation of â„Ž, the reflection of đ?‘“ −1 in đ?‘Ľ-axis.

The function đ?‘“ is given by the graph đ?‘“(đ?‘Ľ) = log 2 đ?‘Ľ. a

Determine the equations of the following graphs: i

ii iii iv v vi b c 4

9 4

The graph of �(�) = � � goes through the point �2; �. a

3

log 3 đ?‘Ľ = 2

đ?‘”, the reflection of đ?‘“ in the đ?‘Ľ-axis

đ?‘?, the reflection of đ?‘“ in the đ?‘Ś-axis

đ?‘ž, the reflection of đ?‘” in the đ?‘Ś-axis

đ?‘“ −1 , the inverse of đ?‘“

đ?‘”−1 , the inverse of đ?‘”

â„Ž, the translation of đ?‘“ two units left

Sketch the graphs of đ?‘“, đ?‘“ −1, đ?‘” and đ?‘”−1 on the same system of axes. Determine the domain and range of đ?‘“ −1 and đ?‘”−1 .

y

B

The graph of đ?‘Ś = log đ?‘? đ?‘Ľ is shown in the diagram alongside.

a

b c d

Determine the coordinates of point A. How do we know that đ?‘? > 1.

3 2

Determine đ?‘? if B is the point ďż˝8; ďż˝.

Determine the equation of đ?‘”, the inverse of this graph.

e Determine the value of đ?‘Ž if C is the point (đ?‘Ž; −2). Š Via Afrika ›› Mathematics Grade 12

x

A

C

28

4 Finance, growth and decay Chapter

Overview

Unit 1 Page 78 •

Future value annuities

Deriving the future value formula

Unit 2 Page 82 Present value annuities Chapter 4 Page 76 Finance, growth and decay

•

Deriving the present value formula

•

Finding the value of đ?‘–

Unit 3 Page 86 Calculating the period Unit 4 Page 88 Analysing investments and loans

• • •

Outstanding balances on a loan Sinking fund Pyramid schemes

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Book.

2 Work through the notes in this chapter of this study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter.

Š Via Afrika ›› Mathematics Grade 12

29

4 Finance, growth and decay Chapter

HIRE PURCHASE AGREEMENTS

Example:

đ??´ = đ?‘ƒ(1 + đ?‘–đ?‘–)

Kelvin buys computer equipment on hire purchase for R20 000.

NB: Simple Interest

He has to put down 10% deposit and repays the amount monthly over 3 years. The interest rate is 15% p.a. Deposit = 10% of R20 000 = R2 000. He has to repay đ??´ = 18000(1 + 0,15 Ă— đ?&#x;?) = đ?‘…26 100 in total. 36 monthly payments of R26 100á 36 =R725 each.

INFLATION / INCREASE IN PRICE OR VALUE

NB: Compound Interest

đ??´ = đ?‘ƒ(1 + đ?‘–)đ?‘›

đ?‘–= number of years

DEPRECIATION Choose the correct formula!

Straight line method

Reducing-balance method

đ??´ = đ?‘ƒ(1 − đ?‘–đ?‘–)

đ??´ = đ?‘ƒ(1 − đ?‘–)đ?‘› đ?‘–= number of years

Š Via Afrika ›› Mathematics Grade 12

30

4 Finance, growth and decay Chapter

NOMINAL AND EFFECTIVE INTEREST RATES NB: đ?’Ž = the number of times per year interest is added

Daily:

đ?‘–đ?‘›đ?‘œđ?‘š đ?’Ž ďż˝ ďż˝1 + đ?‘–đ?‘’đ?‘“đ?‘“ ďż˝ = ďż˝1 + đ?’Ž

Monthly:

đ?‘š = 365 đ?‘š = 12

Quarterly:

đ?‘š=4

Half-yearly (semi-annually):

đ?‘š=2

EXAMPLE: What is the effective rate if the nominal rate is 18% p.a. compounded quarterly? In other words: Which rate compounded annually will give me the same return as 18% compounded quarterly? 0,18 4 ďż˝ −1 4 =0,1925186...

���� = �1 +

Effective rate = 19,25%

FUTURE VALUE ANNUITIES đ??ˇ=

đ?‘Ľ[(1+đ?‘–)đ?‘› −1] đ?‘–

Choosing the value of đ?’” is very important!

KEY WORDS: • Regular investments (monthly/quarterly etc.) • Sinking funds

Example 1

• Annuity/pension • Savings plan

First payment in one month’s time. Last payment in one year’s time. Now � = 12

First payment Š Via Afrika ›› Mathematics Grade 12

Last payment 31

4 Finance, growth and decay Chapter

Example 2 First payment immediately. Last payment in one year’s time. Now � = 13

First payment

Last payment

Example 3 Assume investment pays out in one year’s time, but the first payment was made 2 months from now and the last payment in one year’s time. Now � = 11

First payment

Last payment

Example 4 (Watch out!) First payment immediately, but last payment in 9 months’ time. Now � = 10

First payment

đ?‘­=

đ?’™[(đ?&#x;?+đ?’”)đ?&#x;?đ?&#x;Ž −đ?&#x;?] đ?’”

Last payment

(đ?&#x;? + đ?’”)đ?&#x;?

Š Via Afrika ›› Mathematics Grade 12

BUT, the investment still earns interest for another 3 months before paying out

32

4 Finance, growth and decay Chapter

FUTURE VALUE ANNUITIES đ?‘ƒ=

đ?‘Ľ[1−(1+đ?‘–)−đ?‘› ] đ?‘–

KEY WORDS:

NB: There must always be ONE GAP between the P-value and the first payment!

•

Regular payments (monthly/quarterly etc.)

•

Loan (NOT HIREPURCHASE)

•

Bond/home loan

•

Repayment of debt

•

How long will money be enough to provide regular income?

Example 1

Payment starts one month after the granting of the loan. Last payment in one year’s time. GAP Now � = 12

First payment

Last payment

đ?‘ƒ=

Granting of loan Example 2

đ?‘Ľ[1−(1+đ?‘–)−12 ] đ?‘–

Payment starts in 3 months’ time. Last payment in one year’s time. GAP Now đ?‘– = 12 − 2 = 10

First payment

Last payment

Granting of loan NB: Loan amount accumulates interest for 2 months: Š Via Afrika ›› Mathematics Grade 12

đ?‘ƒ(đ?&#x;? + đ?’”)đ?&#x;? =

đ?‘Ľďż˝1−(1+đ?‘–)−10 ďż˝ đ?‘–

33

4 Finance, growth and decay Chapter

OUTSTANDING BALANCE OF LOAN

Option 2

Option 1

Use A- and F-formula

Use P-formula

đ?’” = number of payments

đ?’” = number of payments left

already made

Example A loan of is being repaid over 20 years in monthly payments of R6 000. The interest rate is 15% p.a. compounded monthly. What is the outstanding balance after 12½ years? Option 1 Outstanding period = 7½ years = 90 months

đ??´đ?‘Žđ?‘™đ?‘Žđ?‘–đ?‘?đ?‘? =

6000ďż˝1âˆ’ďż˝1+ 0,15 12

0,15 −đ?&#x;—đ?&#x;Ž ďż˝ ďż˝ 12

Option 2 Payments already made = 12½X12=150 payments already paid Outstanding balance = đ??´ − đ??ˇ

Balance = đ?‘ƒ ďż˝1 +

0,15 đ?&#x;?đ?&#x;“đ?&#x;Ž 12

ďż˝

−

Š Via Afrika ›› Mathematics Grade 12

0,15 đ?&#x;?đ?&#x;“đ?&#x;Ž ďż˝ −1ďż˝ 12 0,15 12

6000��1+

where P is the initial loan amount.

34

4 Finance, growth and decay Chapter

Mixed Exercise on Finance, growth and decay 1

Determine through calculation which of the following investments is the best, if R15 000 is invested for 5 years at: a 10,6% p.a. simple interest b 9,6% p.a., interest compounded quarterly.

2

An amount of money is now invested at 8,5% p.a compounded monthly to grow to R95 000 in 5 years. a Is 8,5% called the effective or nominal interest rate? b Calculate the amount that must be invested now. c Calculate the interest earned on this investment.

3

Shirley wants to buy a flat screen TV. The TV that she wants currently costs R8 000. a The TV will increase in cost according to the rate of inflation, which is 6% per annum. How much will the TV cost in two years’ time? b For two years Shirley puts R2 000 into her savings account at the beginning of every six month period (starting immediately). Interest on her savings is paid at 7% per annum, compounded six-monthly. Will she have enough to pay for the TV in two years’ time? Show all your calculations.

4

Calculate: a the effective interest rate to 2 dec. places if the nominal interest rate is 7,85% p.a., compounded monthly. b the nominal interest rate if interest on an investment is compounded quarterly, using an effective interest rate of 9,25% p.a.

5

Equipment with a value(new) of R350 000 depreciated to R179 200 after 3 years, based on the reducing balance method. Determine the annual rate of depreciation.

6

R20 000 is deposited into a new savings account at 9,75% p.a., compounded quarterly. After18 months, R10 000 more is deposited. After a further 3 months, the interest rate changes to 9,95% p.a., compounded monthly. Determine the balance in the account 3 years after the account was opened.

7

A company recently bought new equipment to the value of R900 000 which has to be replaced in 5 years’ time. The value of the equipment depreciates at 15% per year according to the reduced-balance method. After 5 years the equipment can be sold second hand at the reduced value. The inflation rate on the equipment is 18% per year.

© Via Afrika ›› Mathematics Grade 12

35

4 Finance, growth and decay Chapter

a

The company wants to establish a sinking fund to replace the equipment in 5 years’ time. Calculate what the value of the sinking fund should be to replace the equipment.

b

Calculate the quarterly amount that the company has to pay into the sinking fund to be able to replace the equipment in 5 years’ time. The company makes the first payment immediately and the last payment at the end of the 5 year period. The interest rate for the sinking fund is 8% per year compounded quarterly.

8

Goods to the value of R1 500 is bought on hire purchase and repaid in 24 monthly payments of R85. Calculate the annual interest rate that applied for the hire purchase agreement.

9

Peter makes a loan to buy a house. He pays back the loan over a period of 20 years in monthly payments of R6 500. Peter qualifies for an interest rate of 12% per years compounded monthly. He makes his first payment one month after the loan was granted. a Calculate the amount Peter borrowed. b Calculate the amount that Peter still owes on his house after he has been paying back the loan for 8 years.

10

Megan’s father wants to make provision for her studies. He starts paying R1000 on a monthly base into an investment on her 12th birthday. He makes the last payment on her 18th birthday. She needs the money 5 months after her 18th birthday. The interest rate on the investment is 10% per annum compounded monthly. Calculate the amount Megan has available for her studies.

11

Stephan starts investing R300 into an investment monthly, starting one month from now. He earns interest of 9% per annum compounded monthly. For how long must he make these monthly investments so that the total value of his investment is R48 000? Give your answer as follows: …. years and …. Months

12

Carl purchases sound equipment to the value of R15 000 on hire purchase. The dealer expects him to put down a 10% deposit. The interest rate is 12% per annum and he has to repay the money monthly over 4 years. It is compulsory for him to insure the equipment through the dealer at a premium of R30 per month. Calculate the total amount Carl has to pay the dealer monthly.

13

Tony borrows money to the value of R400 000. He has to pay back the money in 16 quarterly payments, but only has to make his first payment one year from now. The interest rate is 8% per annum compounded quarterly. Calculate the quarterly payment Tony has to pay.

© Via Afrika ›› Mathematics Grade 12

36

5 Compound angles Chapter

Overview Unit 1 Page 108 •

Deriving a formula for đ?‘?đ?‘œđ?‘ (đ?›ź − đ?›˝)

How to deriving a formula for đ?‘?đ?‘œđ?‘ (đ?›ź − đ?›˝)

Unit 2 Page 112

Formulae for cos(� + �) and sin(� ¹ �)

• • •

Double angles

• •

Unit 3 Page 116

Chapter 5 Page 102 Compound angles

Unit 4 Page 120 Identities

Formula for cos(đ?›ź + đ?›˝) Formula for sin(đ?›ź + đ?›˝) Formula for sin(đ?›ź − đ?›˝) Formula for sin2đ?›ź Formula for đ?‘?đ?‘œđ?‘ 2đ?›ź

• •

Proving identities Finding the value(s) for which the identity is not defined

•

Equations with compound and double angles

•

Drawing and working with graphs of compound angles

Unit 5 Page 124 Equations Unit 6 Page 128 Trigonometric graphs and compound angles

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1 Work through all examples in this chapter of your Learner’s Book. 2 Work through the notes in this chapter of this study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in this study guide.

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5 Compound angles Chapter

REVISION OF TRIGONOMETRY

BASIC TRIGONOMETRIC RATIOS Ratio

𝑜 𝑠𝑠𝑠𝑠 = ℎ 𝑎 𝑐𝑐𝑐𝑐 = ℎ 𝑜 𝑡𝑡𝑡𝑡 = 𝑎

Inverse

ℎ 𝑜 ℎ 𝑠𝑠𝑠𝑠 = 𝑎 𝑎 𝑐𝑐𝑐𝑐 = 𝑜

Ratio

𝑜 𝑠𝑠𝑠𝑠 = ℎ 𝑎 𝑐𝑐𝑐𝑐 = ℎ 𝑜 𝑡𝑡𝑡𝑡 = 𝑎

𝒉

𝑐𝑐𝑐𝑐𝑐𝑐 =

𝜽 𝒂

Γ̅

𝒄

Inverse

ℎ 𝑜 ℎ 𝑠𝑠𝑠𝑠 = 𝑎 𝑎 𝑐𝑐𝑐𝑐 = 𝑜

𝑐𝑐𝑐𝑐𝑐𝑐 =

𝜽

𝒓

𝒙

Γ̅

𝒚

YOU HAVE TO KNOW IN WHICH QUADRANT AN ANGLES LIES AND WHICH RATIO (AND ITS INVERSE) IS POSITIVE THERE: 2nd quadrant (𝟏𝟏𝟎° − 𝜽)

(−𝟏𝟏𝟎° − 𝜽)

(𝟏𝟏𝟎° + 𝜽)

(𝜽 − 𝟏𝟏𝟎°)

3rd quadrant

𝒚

1st quadrant

𝜽

(𝟏𝟔𝟎° + 𝜽)

(𝟏𝟔𝟎° − 𝜽)

𝑥

(−𝜽)

4th quadrant

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5 Compound angles Chapter

REDUCTION FORMULAE

𝑠𝑖𝑖(180° − 𝑠)

𝑐𝑜𝑠(180° − 𝑠)

𝑠𝑖𝑖𝑠

−𝑐𝑜𝑠𝑠

𝑡𝑎𝑖(180° − 𝑠)

−𝑡𝑎𝑖𝑠

𝑠𝑖𝑖(180° + 𝑠)

−𝑠𝑖𝑖𝑠

𝑐𝑜𝑠(180° + 𝑠)

𝑡𝑎𝑖(180° + 𝑠)

𝑦

−𝑐𝑜𝑠𝑠

𝑠𝑖𝑖(360° − 𝑠)

−𝑠𝑖𝑖𝑠

𝑡𝑎𝑖(360° − 𝑠)

−𝑡𝑎𝑖𝑠

𝑐𝑜𝑠(360° − 𝑠)

𝑡𝑎𝑖𝑠

𝑥

𝑐𝑜𝑠𝑠

CO-RATIOS/CO-FUNCTIONS Ratio

Co-ratio

𝑠𝑠𝑠(90° − 𝜃)

𝑐𝑐𝑐𝑐

𝑡𝑡𝑡(90° − 𝜃)

𝑐𝑐𝑐𝑐

Ratio

Co-ratio

𝑠𝑠𝑠(90° + 𝜃)

𝑐𝑐𝑐𝑐

𝑐𝑐𝑐(90° − 𝜃)

𝑐𝑐𝑐(90° + 𝜃)

𝑡𝑡𝑡(90° + 𝜃)

𝑠𝑠𝑠𝑠

−𝑠𝑠𝑠𝑠

(90° − 𝑠) Is in 1st quadrant

(90° + 𝑠) Is in 2nd quadrant

−𝑐𝑐𝑐𝑐

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5 Compound angles Chapter

KNOW YOUR SPECIAL TRIANGLES!

√𝟐

𝟒𝟓°

𝟔𝟎°

Γ̅

𝟏𝟎°

𝟐 𝟏

Γ̅

𝟏

√𝟏

IDENTITIES 𝒕𝒕𝒕𝒕 =

𝒔𝒔𝒔𝒔

𝒄𝒄𝒄𝒄

and 𝒄𝒄𝒄𝒄 =

𝟏

𝒕𝒕𝒕𝒕

=

𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔

SQUARE IDENTITIES: 𝒔𝒔𝒔𝟐 𝜽 + 𝒄𝒄𝒄𝟐 𝜽 = 𝟏

From this follows that: ∴ 𝒄𝒄𝒄𝟐 𝜽 = 𝟏 − 𝒔𝒔𝒔𝟐 𝜽 ∴ 𝒔𝒔𝒔𝟐 𝜽 = 𝟏 − 𝒄𝒄𝒄𝟐 𝜽

Note that the two identities above can both be FACTORISED as differences of two squares: 𝒄𝒄𝒄𝟐 𝜽 = 𝟏 − 𝒔𝒔𝒔𝟐 𝜽 = (𝟏 − 𝒔𝒔𝒔𝒔)(𝟏 + 𝒔𝒔𝒔𝒔) 𝒔𝒔𝒔𝟐 𝜽 = 𝟏 − 𝒄𝒄𝒄𝟐 𝜽 = (𝟏 − 𝒄𝒄𝒄𝒄)(𝟏 + 𝒄𝒄𝒄𝒄) _____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 40

5 Compound angles Chapter

COMPOUND ANGLE-IDENTITIES 𝒄𝒄𝒄(𝜶 − 𝜷) = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 + 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄(𝜶 + 𝜷) = 𝒄𝒄𝒄𝒄 𝒄𝒄𝒄𝒄 − 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔(𝜶 + 𝜷) = 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄𝒄 + 𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔 𝒔𝒔𝒔(𝜶 − 𝜷) = 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄𝒄 − 𝒄𝒄𝒄𝒄 𝒔𝒔𝒔𝒔 DOUBLE ANGLE-IDENTITIES 𝒔𝒔𝒔 𝟐𝟐 = 𝟐 𝒔𝒔𝒔𝒔 𝒄𝒄𝒄 𝜶 𝒄𝒄𝒄 𝟐𝟐 = 𝒄𝒄𝒄𝟐 𝜶 − 𝒔𝒔𝒔𝟐 𝜶 = 𝟏 − 𝟐 𝒔𝒔𝒔𝟐 𝜶 = 𝟐𝒄𝒄𝒄𝟐 𝜶 − 𝟏 𝒕𝒕𝒕 𝟐𝟐 =

𝟐 𝒕𝒕𝒕 𝜶 𝟏 − 𝒕𝒕𝒕𝟐 𝜶

TIPS FOR PROVING IDENTITIES •

Work with LHS and RHS separately

•

Write DOUBLE angles as SINGLE angles

•

Watch out for SQUARE IDENTITIES

•

Write everything in terms of 𝒔𝒔𝒔 and 𝒄𝒄𝒄

• •    •

When working with fractions, put EVERYTHING over the LCD Be on the look out for opportunities to FACTORISE, e.g. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 − 𝒔𝒔𝒔𝒔 = 𝒔𝒔𝒔𝒔(𝟐𝟐𝟐𝟐𝟐 − 𝟏) 𝒄𝒄𝒄𝟐 𝜶 − 𝒔𝒔𝒔𝟐 𝜶 = (𝒄𝒄𝒄𝒄 + 𝒔𝒔𝒔𝒔)(𝒄𝒄𝒄𝒄 − 𝒔𝒔𝒔𝒔) 𝟐𝒔𝒔𝒔𝟐 𝜶 + 𝒔𝒔𝒔𝒔 − 𝟏 = (𝟐𝟐𝟐𝟐𝟐 − 𝟏)(𝒔𝒔𝒔𝒔 + 𝟏)

It is sometimes necessary to replace 𝟏 with 𝒔𝒔𝒔𝟐 𝜶 + 𝒄𝒄𝒄𝟐 𝜶

E.g. 𝒔𝒔𝒔𝟐𝜶 + 𝟏 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 + 𝒔𝒔𝒔𝟐 𝜶 + 𝒄𝒄𝒄𝟐 𝜶 = (𝒔𝒔𝒔𝒔 + 𝒄𝒄𝒄𝒄)𝟐

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 41

5 Compound angles Chapter

FINDING THE GENERAL SOLUTION OF A TRIGONOMETRIC EQUATION

STEP

EXAMPLES OF HOW TO APPLY STEP

Get trig ratio (sin/cos/tan) alone on LHS

A

One value alone on RHS

B C

Now use RHS consisting of a:

A

SIGN (+ or -) and a VALUE Indicates Quadrant

Get reference angle using: đ?’”đ?’”đ?’”−đ?&#x;? (+đ?’—đ?’‚đ?’?đ?’–đ?’†) đ?’„đ?’„đ?’”−đ?&#x;? (+đ?’—đ?’‚đ?’?đ?’–đ?’†) Or đ?’•đ?’‚đ?’”−đ?&#x;? (+đ?’—đ?’‚đ?’?đ?’–đ?’†)

The angle in the trig equations will be equated to the following in the respective quadrants: 1st = Ref ∠2nd = đ?&#x;?đ?&#x;?đ?&#x;Ž° −Ref ∠3rd = đ?&#x;?đ?&#x;?đ?&#x;Ž° +Ref ∠4th = đ?&#x;?đ?&#x;”đ?&#x;Ž° −Ref ∠Then solve for đ?’™

+đ?‘˜360°; đ?‘˜ ∈ đ?‘? for sin/cos or +đ?‘˜180°; đ?‘˜ ∈ đ?‘? for tan

B

C

A

B

C

2 sin 3đ?‘Ľ = 0,4 đ?‘ đ?‘–đ?‘–3đ?‘Ľ = 0,2 1 đ?‘?đ?‘œđ?‘ đ?‘Ľ = −0,2 3 đ?‘?đ?‘œđ?‘ đ?‘Ľ = −0,6 2 tan(đ?‘Ľ − 10°) + 3 = 0 3 tan(đ?‘Ľ − 10°) = − 2

đ?‘ đ?‘–đ?‘–3đ?‘Ľ = +0,2 The + indicates the 1st and 2nd quadrant, where sin is positive. Reference ∠= đ?‘ đ?‘–đ?‘–−1 (0,2) = 11,54° đ?‘?đ?‘œđ?‘ đ?‘Ľ = −0,6 The – indicates the 2nd and 3rd quadrant, where cos is negative. Reference ∠= đ?‘?đ?‘œđ?‘ −1 (0,6) = 53,13° 3 tan(đ?‘Ľ − 10°) = − 2 The − indicates the 2nd and 4th quadrant, where tan is negative. 3

Reference ∠= đ?‘Ąđ?‘Žđ?‘–−1 ďż˝2ďż˝ = 56,31°

2 sin 3đ?‘Ľ = 0,4 đ?‘ đ?‘–đ?‘–3đ?‘Ľ = 0,2 1st : 3đ?‘Ľ = 11,54° + đ?‘˜360° ; đ?‘˜ ∈ đ?‘? đ?‘Ľ = 3,85° + đ?‘˜120° OR 2nd : 3đ?‘Ľ = 180° − 11,54° + đ?‘˜360° đ?‘Ľ = 56,15° + đ?‘˜120° 1 đ?‘?đ?‘œđ?‘ đ?‘Ľ 3

= −0,2 đ?‘?đ?‘œđ?‘ đ?‘Ľ = −0,6 2nd : đ?‘Ľ = 180° − 53,13° + đ?‘˜360°; đ?‘˜ ∈ đ?‘? đ?‘Ľ = 126,87° + đ?‘˜360° OR 3rd : đ?‘Ľ = 180° + 53,13° + đ?‘˜360° đ?‘Ľ = 233,13° + đ?‘˜360° 2 tan(đ?‘Ľ − 10°) + 3 = 0 3 tan(đ?‘Ľ − 10°) = − 2 2nd : đ?‘Ľ − 10° = 180° − 56,31° + đ?‘˜180°; đ?‘˜âˆˆđ?‘? đ?‘Ľ = 133,69° + đ?‘˜180°

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5 Compound angles Chapter

EQUATIONS INVOLVING TWO TRIGONOMETRIC FUNCTIONS

1

EXAMPLES

𝒔𝒔𝒔𝒙 = 𝒄𝒄𝒔𝒙

𝒔𝒔𝒔𝒙 𝒄𝒄𝒔𝒙 = 𝒄𝒄𝒔𝒙 𝒄𝒄𝒔𝒙

2

or

3

or

𝒕𝒂𝒔𝒙 = 𝟏 𝒙 = 𝟒𝟓° + 𝒌. 𝟏𝟏𝟎° ; 𝒌𝝐𝒁 𝒔𝒔𝒔𝒙 = 𝒄𝒄𝒔𝟏𝒙 𝒄𝒄𝒔(𝟗𝟎° − 𝒙) = 𝒄𝒄𝒔𝟏𝒙

𝟗𝟎° − 𝒙 = 𝟏𝒙 + 𝒌. 𝟏𝟔𝟎° ; 𝒌𝝐𝒁 −𝟒𝒙 = −𝟗𝟎° + 𝒌. 𝟏𝟔𝟎° 𝒙 = 𝟐𝟐, 𝟓° − 𝒌. 𝟗𝟎°

𝟗𝟎° − 𝒙 = −𝟏𝒙 + 𝒌. 𝟏𝟔𝟎° ; 𝒌𝝐𝒁 𝟐𝒙 = −𝟗𝟎° + 𝒌. 𝟏𝟔𝟎° 𝒙 = −𝟒𝟓° + 𝒌. 𝟏𝟏𝟎°

𝐬𝐥𝐧(𝒙 + 𝟐𝟎°) = 𝐜𝐥𝐬(𝟐𝒙 − 𝟏𝟎°) 𝒄𝒄𝒔[𝟗𝟎° − (𝒙 + 𝟐𝟎°)] = 𝒄𝒄𝒔(𝟐𝒙 − 𝟏𝟎°) 𝒄𝒄𝒔(𝟕𝟎° − 𝒙) = 𝒄𝒄𝒔(𝟐𝒙 − 𝟏𝟎°) 𝟕𝟎° − 𝒙 = 𝟐𝒙 − 𝟏𝟎° + 𝒌. 𝟏𝟔𝟎° −𝟏𝒙 = −𝟏𝟎𝟎° + 𝒌. 𝟏𝟔𝟎° 𝒙 = 𝟏𝟏, 𝟏𝟏° − 𝒌. 𝟏𝟐𝟎° ; 𝒌 ∈ 𝒁

𝟕𝟎° − 𝒙 = −(𝟐𝒙 − 𝟏𝟎°) + 𝒌. 𝟏𝟔𝟎° 𝒙 = −𝟒𝟎° + 𝒌. 𝟏𝟔𝟎°

COMMENTS

÷ by 𝑐𝑜𝑠𝑥 on both sides

May NOT divide by 𝑐𝑜𝑠𝑥 both sides Trig function on both sides should be the same Angles on LHS and RHS should either be the same or be in two different quadrants where 𝑐𝑜𝑠 have the same sign (1st and 4th quadrant) Alternative: sin on both sides

𝑠𝑖𝑖(𝑥 + 20°) = 𝑐𝑜𝑠(2𝑥 − 30°) 𝑠𝑖𝑖(𝑥 + 20°) = sin[90° − (2𝑥 − 30°)] 𝑠𝑖𝑖(𝑥 + 20°) = 𝑠𝑖𝑖(120° − 2𝑥) or

𝑥 + 20° = 120° − 2𝑥 + 𝑘. 360° 3𝑥 = 100° + 𝑘. 360° 𝑥 = 33,33° − 𝑘. 120° ; 𝑘 ∈ 𝑍

𝑥 + 20° = 180° − (120° − 2𝑥) + 𝑘. 360° −𝑥 = 40° + 𝑘. 360° 𝑥 = −40° − 𝑘. 360°

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5 Compound angles Chapter

EXAMPLES OF EQUATIONS INVOLVING DOUBLE ANGLES

1st quadrant:

4th quadrant:

𝒄𝒄𝒔𝜽. 𝒄𝒄𝒔𝟏𝟒° + 𝒔𝒔𝒔𝜽. 𝒔𝒔𝒔𝟏𝟒° = 𝟎, 𝟕𝟏𝟓 𝒄𝒄𝒔(𝜽 − 𝟏𝟒°) = 𝟎, 𝟕𝟏𝟓 Ref ∠= 𝟒𝟒, 𝟏𝟔° 𝜽 − 𝟏𝟒° = 𝟒𝟒, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° 𝜽 = 𝟓𝟏, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁

𝜽 − 𝟏𝟒° = −𝟒𝟒, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° 𝜽 = −𝟏𝟎, 𝟏𝟔° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁

𝒔𝒔𝒔𝟐𝜽 + 𝟐𝒔𝒔𝒔𝜽 = 𝟎 𝟐𝒔𝒔𝒔𝜽𝒄𝒄𝒔𝜽 + 𝟐𝒔𝒔𝒔𝜽 = 𝟎 𝟐𝒔𝒔𝒔𝜽(𝒄𝒄𝒔𝜽 + 𝟏) = 𝟎 𝒔𝒔𝒔𝜽 = 𝟎 or 𝒄𝒄𝒔𝜽 = −𝟏 𝜽 = 𝒌. 𝟏𝟏𝟎° ; 𝒌 ∈ 𝒁 or 𝜽 = 𝟏𝟏𝟎° + 𝒌. 𝟏𝟔𝟎° 𝟐𝒔𝒔𝒔𝟐 𝜽 + 𝒔𝒔𝒔𝜽 = 𝟏 𝟐𝒔𝒔𝒔𝟐 𝜽 + 𝒔𝒔𝒔𝜽 − 𝟏 = 𝟎 ∴ (𝟐𝒔𝒔𝒔𝜽 + 𝟏)(𝒔𝒔𝒔𝜽 − 𝟏) = 𝟎 𝟐𝒔𝒔𝒔𝜽 + 𝟏 = 𝟎 or 𝒔𝒔𝒔𝜽 + 𝟏 = 𝟎 𝟏 𝒔𝒔𝒔𝜽 = − or 𝒔𝒔𝒔𝜽 = −𝟏

No solution

𝟐

𝜽 = 𝟐𝟕𝟎° + 𝒌. 𝟏𝟔𝟎° ; 𝒌 ∈ 𝒁

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 44

5 Compound angles Chapter

PROBLEMS WITH COMPOUND-ANGLES TO BE DONE WITHOUT A CALCULATOR •

Write given information in form where trig function is ALONE on LHS

•

Select QUADRANT and draw TRIANGLE in correct quadrant (2 sides of triangle will be known)

•

Use the Theorem of PYTHAGORAS to determine 3rd side

•

Now work with the expression of which you need to find the value: write all compound or double angles in terms of SINGLE ANGLES

•

Now SUBSTITUTE VALUES from diagram(s) and SIMPLIFY

Example: đ?&#x;“

If đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;? + đ?&#x;?đ?&#x;? = đ?&#x;Ž and đ?œś ∈ [đ?&#x;—đ?&#x;—°; đ?&#x;?đ?&#x;?đ?&#x;?°] and đ?œˇ = đ?&#x;?đ?&#x;? ; đ?œˇ > 90°, determine without the use of a calculator the value of:

a c

b

đ?’”đ?’”đ?’”(đ?œś − đ?œˇ)

d

�����

Solution:

đ?&#x;?đ?&#x;?

đ?’”đ?’”đ?’”đ?’” = − đ?&#x;?đ?&#x;?

đ?&#x;“

đ?’„đ?’„đ?’„ positive in 1st and 4th quad

But đ?œś ∈ [đ?&#x;—đ?&#x;—°;đ?›ź đ?&#x;?đ?&#x;?đ?&#x;?°], so 3rd quad 13

đ?’„đ?’„đ?’„đ?’„đ?’„

đ?’„đ?’„đ?’„đ?’„ = đ?&#x;?đ?&#x;?

��� negative in 3rd and 4th quad

−12

đ?’„đ?’„đ?’„(đ?œś + đ?œˇ)

th But đ?œˇ > đ?›˝ 90°,5so 4 quadrant

13

đ?’™ = −đ?&#x;“ −đ?&#x;?đ?&#x;?

đ?&#x;“

đ?’š = −đ?&#x;?đ?&#x;? −đ?&#x;“

−đ?&#x;?đ?&#x;?

−đ?&#x;?đ?&#x;?

−đ?&#x;?đ?&#x;?

a đ?’”đ?’”đ?’”(đ?œś − đ?œˇ) = đ?’”đ?’”đ?’”đ?’” đ?’”đ?’”đ?’”đ?’” − đ?’„đ?’„đ?’„đ?’„ đ?’„đ?’„đ?’„đ?’„ = ďż˝ đ?&#x;?đ?&#x;‘ ďż˝ ďż˝đ?&#x;?đ?&#x;?ďż˝ − ďż˝ đ?&#x;?đ?&#x;? ďż˝ ďż˝ đ?&#x;?đ?&#x;? ďż˝ = −đ?&#x;“

đ?&#x;“

−120 169

b đ?’„đ?’„đ?’„(đ?œś + đ?œˇ) = đ?’„đ?’„đ?’„đ?’„ đ?’„đ?’„đ?’„đ?’„ − đ?’”đ?’”đ?’”đ?’” đ?’”đ?’”đ?’”đ?’” = ďż˝ đ?&#x;?đ?&#x;? ďż˝ ďż˝đ?&#x;?đ?&#x;?ďż˝ − ďż˝ đ?&#x;?đ?&#x;? ďż˝ ďż˝ đ?&#x;?đ?&#x;? ďż˝ = −1 −đ?&#x;?đ?&#x;?

−đ?&#x;“

120

c đ?’”đ?’”đ?’”đ?’”đ?’” = đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;?đ?&#x;? đ?’„đ?’„đ?’„đ?’„ = đ?&#x;? ďż˝ đ?&#x;?đ?&#x;? ďż˝ ďż˝ đ?&#x;?đ?&#x;? ďż˝ = 169

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 45

5 Compound angles Chapter

Mixed Exercise on Compound angles 1

Solve the following equations for . Give the general solution unless otherwise stated. Answers should be given correct to 2 decimal places where exact answers are not possible. a b c d e f

2

2đ?‘?đ?‘œđ?‘ 2đ?‘Ľ + 1 = 0

đ?‘ đ?‘–đ?‘– x = 3 cos x for đ?‘Ľ ∈ [90°; 360°] đ?‘ đ?‘–đ?‘–đ?‘Ľ = cos 3đ?‘Ľ

6 − 10đ?‘?đ?‘œđ?‘ đ?‘Ľ = 3đ?‘ đ?‘–đ?‘–2 đ?‘Ľ ; đ?‘Ľ ∈ [−360°; 360°]

2 − đ?‘ đ?‘–đ?‘–đ?‘Ľ đ?‘?đ?‘œđ?‘ đ?‘Ľ − 3đ?‘?đ?‘œđ?‘ 2 đ?‘Ľ = 0

3đ?‘ đ?‘–đ?‘–2 đ?‘Ľ − 8đ?‘ đ?‘–đ?‘–đ?‘Ľ + 16đ?‘ đ?‘–đ?‘–đ?‘Ľ đ?‘?đ?‘œđ?‘ đ?‘Ľ − 6đ?‘?đ?‘œđ?‘ đ?‘Ľ + 3đ?‘?đ?‘œđ?‘ 2 đ?‘Ľ = 0

Prove the following identities, stating any values of đ?‘Ľ or đ?‘ for which the identity is not valid:

1 cos x

a

cos x + tan x sin x =

b

sin θ cosθ 1 − = 1 − cosθ sin θ sin θ

c

1 − cos 2 x = tan x sin x cos x

d

sin 3 x + sin x cos 2 x = tan x cos x

e

1 + tan x 1 + 2 sin x cos x = 1 − tan x cos 2 x − sin 2 x

f g h

1

sin(45° + đ?‘Ľ) . sin(45° − đ?‘Ľ) = 2 đ?‘?đ?‘œđ?‘ 2đ?‘Ľ sin 2đ?œƒâˆ’cos đ?œƒ sin đ?œƒâˆ’cos 2đ?œƒ

=

cos đ?‘Ľâˆ’cos 2đ?‘Ľ+2 3 sin đ?‘Ľâˆ’sin 2đ?‘Ľ

cos đ?œƒ

1+sin đ?œƒ

=

1+cos đ?‘Ľ sin đ?‘Ľ

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 46

5 Compound angles Chapter

3

4

Simplify:

b

(

(

(

) ( )

) (

) )

c d

cos 73°

đ?‘?đ?‘œđ?‘ (−163°) đ?‘Ąđ?‘Žđ?‘–197° đ?‘?đ?‘œđ?‘ 326°

Given that 5đ?‘?đ?‘œđ?‘ đ?‘Ľ + 4 = 0 , calculate, without the use of a calculator, the value(s) of : a

b

5đ?‘ đ?‘–đ?‘–đ?‘Ľ + 3đ?‘Ąđ?‘Žđ?‘–đ?‘Ľ tan 2đ?‘Ľ

If 3đ?‘ đ?‘–đ?‘–đ?‘Ľ = −1 ; đ?‘Ľ ∈ [90°; 270°] and đ?‘Ąđ?‘Žđ?‘–đ?‘Ś =

calculator the value of: a b 7

(without using a calculator)

Given that đ?‘ đ?‘–đ?‘–17° = đ?‘˜, express in terms of đ?‘˜:

b

6

)

sin 180 0 + x tan x − 360 0 tan 360 0 − x cos 240 0 tan 225 0

a

5

(

a

sin 180 0 − x tan (− x ) tan 180 0 + x cos x − 90 0

3 4

; đ?‘Ś ∈ [90°; 360°]. Determine without the use of a

cos(đ?‘Ľ − đ?‘Ś)

đ?‘?đ?‘œđ?‘ 2đ?‘Ľ − đ?‘?đ?‘œđ?‘ 2đ?‘Ś

Simplify without the use of calculator: a b c

đ?‘?đ?‘œđ?‘ 2 22,5° − đ?‘ đ?‘–đ?‘–2 22,5°

đ?‘ đ?‘–đ?‘–22,5° đ?‘?đ?‘œđ?‘ 22,5° 2đ?‘ đ?‘–đ?‘–15°đ?‘?đ?‘œđ?‘ 15°

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 47

6 Solving problems in three dimensions Chapter

Overview

Unit 1 Page 146 Problems in three dimensions Chapter 6 Page 142 Solving problems in three dimensions

•

Trigonometry in real life

Unit 2 Page 150 Compound angle formulae in three dimensions

•

Using compound angle formulae in three dimensions

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Book.

2 Work through the notes in this chapter of this study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 48

6 Solving problems in three dimensions Chapter

REVISION ON THE USE OF THE đ?’”đ?’”đ?’”đ?’–đ?’” −, đ?’„đ?’„đ?’”đ?’”đ?’”đ?’–đ?’” − and the đ?’‚đ?’“đ?’†đ?’‚ −FORMULAE

INFORMATION GIVEN

UNKNOWN

FORMULA TO USE

2 angles and 1 side (∠∠s)

s

đ?‘ đ?‘–đ?‘–-rule

2 sides and a not- included ∠(ss∠)

âˆ

2 sides and an included ∠(s∠s)

s

3 sides (sss)

âˆ

2 sides and an included âˆ

Area

Area, side and âˆ

s

đ?‘ đ?‘–đ?‘–-rule Watch out for ambiguous case! ∠can be acute or obtuse

FORM OF FORMULA đ?’‚

đ?‘ đ?‘–đ?‘›đ??´

đ?‘?

= sin đ??´

đ?’‚ is unknown đ?‘ đ?‘–đ?‘–đ?‘¨ đ?‘ đ?‘–đ?‘–đ??´ = đ?‘Ž đ?‘?

� is unknown

đ?‘?đ?‘œđ?‘ -rule

đ?’‚2 = đ?‘? 2 + đ?‘? 2 − 2đ?‘?đ?‘?đ?‘?đ?‘œđ?‘ đ??´ đ?’‚ is unknown

đ?‘?đ?‘œđ?‘ -rule

đ?‘? 2 + đ?‘? 2 − đ?‘Ž2 đ?‘?đ?‘œđ?‘ đ??´ = 2đ?‘?đ?‘? đ?‘¨ is unknown

Area-rule

Area of ∆= 2 đ?‘Žđ?‘?đ?‘ đ?‘–đ?‘–đ??ś

1

Area is unknown

Area-rule

2 Ă— đ??´đ?‘&#x;đ?‘?đ?‘Ž đ?‘Žđ?‘ đ?‘–đ?‘–đ??ś đ?’ƒ is unknown

đ?’ƒ=

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 49

6 Solving problems in three dimensions Chapter

TIPS FOR SOLVING PROBLEMS IN THREE DIMENSIONS • •

Where there are 3 triangles, start with the ∆ with the most information and

work via the 2nd ∆ to the 3rd ∆ which contains the unknown to be calculated.

Indicate all RIGHT angles – remember they don’t always look like đ?&#x;—đ?&#x;—° angles

•

Shade the horizontal plane in the diagram (e.g. floor, ground)

•

Be on the lookout for reductions like đ?’„đ?’„đ?’„(đ?&#x;—đ?&#x;—° − đ?œś) = đ?’”đ?’”đ?’”đ?’” and

• •

đ?’”đ?’”đ?’”(đ?&#x;?đ?&#x;?đ?&#x;?° − đ?œś) = đ?’”đ?’”đ?’”đ?’” to simplify expressions

Use compound and double angle formulae to convert to single angles When writing out the solution – always indicate in which ∆ you are working

EXAMPLE P, Q and R are in the same horizontal plane. TP is a vertical tower 5,9 m high. The angle of elevation of T from Q is 65°. đ?‘ƒđ?‘„ďż˝ đ?‘… = đ?‘ƒđ?‘…ďż˝ đ?‘„. a b c

Calculate the length of PQ to the nearest meter. Hence show that đ?‘…đ?‘„ = 5,5 đ?‘?đ?‘œđ?‘ đ?‘Ľ. If it is further given that đ?‘Ľ = 42°, calculate the area of ∆đ?‘ƒđ?‘„đ?‘….

Solution: a b

5,9 đ?‘ƒđ?‘„

= đ?‘Ąđ?‘Žđ?‘–65°

∴ đ?‘ƒđ?‘„ =

5,9 đ?‘Ąđ?‘Žđ?‘›65°

= 2,75 đ?‘š

đ?‘„đ?‘ƒďż˝đ?‘… = 180° − 2đ?‘Ľ

đ?‘…đ?‘„ đ?‘ đ?‘–đ?‘›đ?‘ƒ

=

đ?‘ƒđ?‘„ đ?‘ đ?‘–đ?‘›đ?‘…

đ?‘…đ?‘„ 2,75 = sin(180°âˆ’2đ?‘Ľ) đ?‘ đ?‘–đ?‘›đ?‘Ľ đ?‘…đ?‘„ 2,75 = sin 2đ?‘Ľ đ?‘ đ?‘–đ?‘›đ?‘Ľ đ?‘…đ?‘„ 2,75 = 2sin đ?‘Ľđ?‘Žđ?‘œđ?‘ đ?‘Ľ đ?‘ đ?‘–đ?‘›đ?‘Ľ

c

∴ đ?‘…đ?‘„ = 2 Ă— 2,75đ?‘?đ?‘œđ?‘ đ?‘Ľ đ?‘…đ?‘„ = 5,5 đ?‘?đ?‘œđ?‘ đ?‘Ľ

1 2 1 Ă— 2,75 Ă— 2

Area of ∆đ?‘ƒđ?‘„đ?‘… = Ă— đ?‘ƒđ?‘„ Ă— đ?‘„đ?‘… Ă— đ?‘ đ?‘–đ?‘–đ?‘„ =

(5,5đ?‘?đ?‘œđ?‘ 42°) Ă— đ?‘ đ?‘–đ?‘–42°

= 3,76 square units

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 50

6 Solving problems in three dimensions Chapter

Mixed Exercise on Problems in Three Dimensions 1

a

In the diagram alongside B, D and E are in the same horizontal plane. đ??´đ??ˇďż˝ đ??ˇ = 120° AB and CD are two vertical towers. AB = 2CD = 2â„Ž meter The angle of elevation from E to A is đ?›ź. The angle of elevation from E to C is (90° − đ?›ź). Determine the length BE in terms of â„Ž and đ?›ź.

b

Show that the distance between the two towers can be given as:

c

đ??´đ??ˇ =

2

a

ℎ√đ?‘Ąđ?‘Žđ?‘›4 đ?›ź+2đ?‘Ąđ?‘Žđ?‘›2 đ?›ź+4 đ?‘Ąđ?‘Žđ?‘›đ?›ź

Hence determine the height of the tower CD, rounded to the nearest meter, if đ?›ź = 42° and BD=400 m. B, C and D are three points in the same horizontal plane and AB is a vertical pole of length đ?‘? metres. The angle of elevation of A from C is đ?‘ and đ??´đ??śĚ‚ đ??ˇ = đ?‘ . Also, đ??śđ??´ďż˝đ??ˇ = 30° and đ??´đ??ˇ = 8 đ?‘š. ďż˝ đ??´ in terms of đ?‘ . Express đ??śđ??ˇ

8sin(30°+đ?œƒ) . đ?‘Žđ?‘œđ?‘ đ?œƒ

b

Hence show that đ?‘? =

3

In the diagram alongside, AB is a vertical flagpole 5 metres high. AC an AD are two stays. B, C and D are in the same horizontal plane. ďż˝ đ??ś = đ?›˝. đ??´đ??ˇ = 12 đ?‘š, đ??´đ??śĚ‚ đ??ˇ = đ?›ź and đ??´đ??ˇ Show that đ??śđ??ˇ =

13sin(đ?›ź+đ?›˝) đ?‘ đ?‘–đ?‘›đ?›ź

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 51

6 Solving problems in three dimensions Chapter

4 a b c

ďż˝ đ??ś=đ?‘ . In ∆đ??´đ??´đ??ś AD = ; DB = đ?‘– ; CD = đ?‘? and đ??´đ??ˇ

Complete in terms of đ?‘š , đ?‘? and đ?‘ : Area ∆đ??´đ??ˇđ??ś = â‹Ż 1

Show that the area of ∆đ??´đ??´đ??ś = 2 đ?‘?(đ?‘š + đ?‘–) sin đ?‘ . If the area of ∆đ??´đ??´đ??ś = 12,6 đ?‘?đ?‘š2 ; đ??´đ??´ = 5,9 đ?‘?đ?‘š and đ??ˇđ??ś = 8,1 đ?‘?đ?‘š, calculate the value(s) of đ?‘ .

5

ďż˝ = 90° , đ??´đ??śĚ‚ đ??ˇ = đ?‘ In the diagram, đ??ˇ

a

đ??´đ??śĚ‚ đ??´ = đ?›ź ; AB = BC and đ??´đ??ˇ = đ?‘? units.

b

Express BC in terms of đ?‘? and đ?‘ .

Determine, without stating reasons,

c

the size of đ??´ďż˝1 in terms of đ?›ź.

Hence , prove that AC = sin đ?œƒ.sin đ?›ź

6

In the diagram PQ is a vertical building.

đ?‘?.sin 2đ?›ź

Q, R and S are points in the same horizontal plane. The angle of elevation of P, the top of the building, measured from R, is �.

đ?‘…đ?‘„ďż˝đ?‘† = 30°

đ?‘„đ?‘†đ?‘… = 150° − đ?›ź

a

�� = 12 �

b

Hence show that the height PQ of the building is given by

c

Show that đ?‘„đ?‘… =

6(đ?‘Žđ?‘œđ?‘ đ?›ź+√3đ?‘ đ?‘–đ?‘›đ?›ź) đ?‘ đ?‘–đ?‘›đ?›ź

đ?‘ƒđ?‘„ = 6 + 6√3đ?‘Ąđ?‘Žđ?‘–đ?›ź

Hence calculate the value of � if PQ = 23 m.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 52

7 Polynomials Chapter

Overview

Unit 1 Page 158 Chapter 7 Page 156 Polynomials

The Remainder Theorem

•

The Remainder theorem

Unit 2 Page 160 The Factor Theorem

•

The Factor Theorem

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1

Work through all examples in this chapter of your Learner’s Book.

2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 53

7 Polynomials Chapter

THE REMAINDER THEOREM đ?‘“(đ?‘Ľ) = (đ?‘Žđ?‘Ž + đ?‘?). đ?‘ž(đ?‘Ľ) + đ?‘&#x;(đ?‘Ľ) divisor

quotient

remainder

The remainder theorem can be used to calculate the remainder when a polynomial đ?‘“(đ?‘Ľ)is divided by (đ?‘Žđ?‘Ž + đ?‘?) đ?‘? ∴ đ?‘“ ďż˝âˆ’ ďż˝ = đ?‘&#x;(đ?‘Ľ) đ?‘Ž

Choosing the correct value to substitute is very important: If you divide đ?’‡ by đ?’ˆ(đ?’™) = (đ?’™ − đ?&#x;?)

Value to substitute into đ?’‡(đ?’™) đ?‘“(2) =? 1 đ?‘“ ďż˝ ďż˝ =? 2 đ?‘“(−3) =? 2 đ?‘“ ďż˝âˆ’ ďż˝ =? 3

(đ?&#x;?đ?&#x;? − đ?&#x;?) (đ?’™ + đ?&#x;‘)

(đ?&#x;‘đ?&#x;‘ + đ?&#x;?)

THE FACTOR THEOREM đ?‘?

• •

If đ?‘“ ďż˝âˆ’ đ?‘Žďż˝ = 0 then:

(đ?‘Žđ?‘Ž + đ?‘?) is a FACTOR of đ?‘“(đ?‘Ľ) and đ?‘“(đ?‘Ľ) is DIVISIBLE by (đ?‘Žđ?‘Ž + đ?‘?)

When trying out đ?‘Ľ −values that give 0, try them in the following order: 1; −1; 2; −2; 3; −3 etc.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 54

7 Polynomials Chapter

DIFFERENT METHODS TO FACTORISE A CUBIC POLYNOMIAL (3RD DEGREE) METHOD AND DESCRIPTION OF STEPS SUM AND DIFFERENCE OF CUBES

FACTORISE BY GROUPING • Group terms in two pairs • Take out common factor from each pair • Two sets of brackets now become common factor • Factorise bracket further if possible FACTORISE BY INSPECTION • Find one linear factor using factor theorem • Find other factor (quadratic expression) by inspection

EXAMPLES 3

A) đ?‘“(đ?‘Ľ) = đ?‘Ľ + 27 = (đ?‘Ľ + 3)(đ?‘Ľ 2 − 3đ?‘Ľ + 9)

Cannot factorise further B)đ?‘“(đ?‘Ľ) = 8đ?‘Ľ 3 − 1 = (2đ?‘Ľ − 1)(4đ?‘Ľ 2 + 2đ?‘Ľ + 1) Cannot factorise further đ?&#x;? đ?‘“(đ?‘Ľ) = đ?’™ + đ?&#x;?đ?’™đ?&#x;? − đ?&#x;’đ?’™ − đ?&#x;?đ?&#x;? = đ?‘Ľ 2 (đ?’™ + đ?&#x;?) − 4(đ?’™ + đ?&#x;?) = (đ?’™ + đ?&#x;?)(đ?’™đ?&#x;? − đ?&#x;’) = (đ?‘Ľ + 3)(đ?’™ + đ?&#x;?)(đ?’™ − đ?&#x;?) đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 3 − đ?&#x;?đ?’™đ?&#x;? − 10đ?‘Ľ − 6

đ?‘“(−1) = 2(−1)3 − 2(−1)2 − 10(−1) −6=0

∴ (đ?‘Ľ + 1) is a factor đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(đ?’‚đ?‘Ľ 2 + đ?’ƒđ?‘Ľ + đ?’„)

Now find these coefficients Start with đ?’‚ and đ?’„: 1Ă—đ?‘Ž = 2 ∴đ?‘Ž =2 1 Ă— đ?‘? = −6 ∴ đ?‘? = 6 You now need to find đ?’ƒ: Multiply the two brackets; the two đ?‘Ľ 2 -terms need to give you −đ?&#x;?đ?’™đ?&#x;? : đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(đ?&#x;?đ?‘Ľ 2 + đ?’ƒđ?‘Ľ + đ?&#x;”)

SYNTHETIC OR LONG DIVISION • Find one linear factor using factor theorem • Find other factor (quadratic expression) by long division or synthetic division (SEE NEXT PAGE)

đ?‘?đ?‘Ľ 2 + 2đ?‘Ľ 2 = −đ?&#x;?đ?’™đ?&#x;? ∴ đ?‘? = −4 2 ∴ đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(2đ?‘Ľ − 4đ?‘Ľ + 6) = (đ?‘Ľ + 1)(2đ?‘Ľ + 2)(đ?‘Ľ − 3) đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 3 − đ?&#x;?đ?’™đ?&#x;? − 10đ?‘Ľ − 6

đ?‘“(−1) = 2(−1)3 − 2(−1)2 − 10(−1) −6=0

∴ (đ?‘Ľ + 1) is a factor đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(đ?’‚đ?‘Ľ 2 + đ?’ƒđ?‘Ľ + đ?’„) Find đ?‘Ž, đ?‘?, đ?‘? using synthetic division

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 55

7 Polynomials Chapter

SYNTHETIC DIVISION

đ?‘“(−1) = 0 , so (đ?‘Ľ + 1)is a factor

đ?‘“(đ?‘Ľ ) = 2đ?‘Ľ 3 − 2đ?‘Ľ 2 − 10đ?‘Ľ − 6

đ?‘“(−1) = 0 −đ?&#x;?

đ?&#x;? đ?&#x;?

−đ?&#x;? −đ?&#x;? −đ?&#x;’

−đ?&#x;?đ?&#x;Ž đ?&#x;’ −đ?&#x;”

This method is called SYNTHETIC division, because we don’t really divide.

−đ?&#x;” đ?&#x;” đ?&#x;Ž

We actually multiply and add. Note the following: • • • •

The đ?‘Ľ-value of −1 that gave us the factor (đ?‘Ľ + 1) is written on the LHS The coefficients of the cubic polynomial are written in the top row The first coefficient, 2, is carried down to the last row Now starting from the left:

MULTIPLY along the dotted arrow

repeat

and write the ANSWER in the block one row up and one column right

steps

•

Now ADD DOWN in the column (the two values underneath each other)

•

You MUST get 0 in the last block

•

The 3 values in the bottom row are the coefficients of the quadratic factor.

So, đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(2đ?‘Ľ 2 − 4đ?‘Ľ − 6)

You can now complete the factorising: đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(2đ?‘Ľ + 2)(đ?‘Ľ − 3) _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 56

7 Polynomials Chapter

Mixed Exercise on Polynomials 1

Factorise the following expressions completely: a c e g i k

2

b c d e f g

4

b

4đ?‘Ľ 3 − 2đ?‘Ľ 2 + 10đ?‘Ľ − 5

f

3đ?‘Ľ 3 − 7đ?‘Ľ 2 + 4

j

đ?‘Ľ 3 + 3đ?‘Ľ 2 + 2đ?‘Ľ + 6

d

đ?‘Ľ 3 − đ?‘Ľ 2 − 22đ?‘Ľ + 40

h

đ?‘Ľ3 − đ?‘Ľ2 − đ?‘Ľ − 2

5đ?‘Ľ 3 + 40

4đ?‘Ľ 3 − đ?‘Ľ 2 − 16đ?‘Ľ + 4 đ?‘Ľ 3 + 2đ?‘Ľ 2 + 2đ?‘Ľ + 1 đ?‘Ľ 3 + 2đ?‘Ľ 2 − 5đ?‘Ľ − 6 đ?‘Ľ 3 − 19đ?‘Ľ + 30

Solve for đ?‘Ľ:

a

3

27đ?‘Ľ 3 − 8

đ?‘Ľ 3 + 2đ?‘Ľ 2 − 4đ?‘Ľ = 0

đ?‘Ľ 3 − 3đ?‘Ľ 2 − đ?‘Ľ + 6 = 0

2đ?‘Ľ 3 − 12đ?‘Ľ 2 − đ?‘Ľ + 6 = 0 2đ?‘Ľ 3 − đ?‘Ľ 2 − 8đ?‘Ľ + 4 = 0

đ?‘Ľ3 + đ?‘Ľ2 − 2 = 0 đ?‘Ľ 3 = 16 + 12đ?‘Ľ

đ?‘Ľ 3 + 3đ?‘Ľ 2 = 20đ?‘Ľ + 60

Show that đ?‘Ľ − 3 is a factor of đ?‘“(đ?‘Ľ) = đ?‘Ľ 3 − đ?‘Ľ 2 − 5đ?‘Ľ − 3 and hence solve đ?‘“(đ?‘Ľ) = 0.

Show that 2đ?‘Ľ − 1 is a factor of đ?‘”(đ?‘Ľ) = 4đ?‘Ľ 3 − 8đ?‘Ľ 2 − đ?‘Ľ + 2 and hence solve đ?‘”(đ?‘Ľ) = 0.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 57

8 Differential calculus Chapter

Overview Unit 1 Page 170 •

Limits

Investigating limits

Unit 2 Page 172 •

The gradient of a graph at a point

•

Function notation and the average gradient The gradient of a graph at a point

Unit 3 Page 176 • • •

The derivative of a function Chapter 8 Page 166 Differential calculus

First principles Rules for differentiation The derivative at a point

Unit 4 Page 182 The equation of a tangent to a graph

•

Calculating the equation of a tangent to a graph

•

Plotting a cubic function

•

Change in concavity

•

Modelling real-life problems

Unit 5 Page 184 The graph of the cubic function Unit 6 Page 188 The second derivative (concavity) Unit 7 Page 192 Application of differential calculus

REMEMBER YOUR STUDY APPROACH SHOULD BE:

1 Work through all examples in this chapter of your Learner’s Book. 2

Work through the notes in this chapter of the study guide.

3

Do the exercises at the end of the chapter in the Learner’s Book.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 58

8 Differential calculus Chapter

THE CONCEPT OF A LIMIT Notation: đ??Ľđ??Ľđ??Ľđ?’™â†’đ?&#x;’ đ?’‡(đ?’™)

We say: “The limit of đ?’‡ as đ?’™ approaches đ?&#x;’â€?

What does it mean?

The limit is the đ?’š-value (remember đ?’š = đ?’‡(đ?’™)) which the function

approaches as the đ?’™-value approaches (gets closer to) a certain value from the left or the right . Examples: a

Let đ?’‡(đ?’™) = đ?&#x;?đ?’™đ?&#x;? + đ?&#x;’

đ??Ľđ??Ľđ??Ľđ?’™â†’đ?&#x;? đ?’‡(đ?’™) = đ??Ľđ??Ľđ??Ľđ?’™â†’đ?&#x;’ đ?&#x;?đ?’™đ?&#x;? + đ?&#x;’ = đ?&#x;?(đ?&#x;?)đ?&#x;? + đ?&#x;’ = đ?&#x;”

Before calculating the limit, it is sometimes necessary to FACTORISE and SIMPLIFY first: An examples of this is: lim�→3 =

đ?‘Ľ 2 −9

đ?‘Ľâˆ’3 (đ?‘Ľâˆ’3)(đ?‘Ľ+3) limđ?‘Ľâ†’3 đ?‘Ľâˆ’3

= lim (� + 3) �→3

= (3 + 3) =6

Substituting đ?‘Ľ = 3 now will cause division by 0

First factorise the numerator and cancel out

Note that “lim� falls away in the step where you substitute

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 59

8 Differential calculus Chapter

AVERAGE GRADIENT BETWEEN 2 POINTS From previous grades you know you can calculate the gradient between two đ?‘Ś −đ?‘Ś

points (đ?‘Ľ1 ; đ?‘Ś1 ) and (đ?‘Ľ2 ; đ?‘Ś2 ) using the formula: đ?‘š = đ?‘Ľ2−đ?‘Ľ1 2

1

In the diagram below the points A(đ?‘Ľ; đ?‘“(đ?‘Ľ)) and B(đ?‘Ľ + â„Ž; đ?‘“(đ?‘Ľ + â„Ž)) are indicated. The AVERAGE GRADIENT between A and B is given by: đ?‘šđ??´đ??´ =

đ?‘“(đ?‘Ľ+â„Ž)−đ?‘“(đ?‘Ľ) â„Ž

THE GRADIENT OF A GRAPH AT A POINT By letting ℎ approach 0, the distance between point A and B will become smaller and smaller. A and B will almost “become one point�.

The average gradient then becomes the GRADIENT OF THE GRAPH AT A POINT = limℎ→0 NOTATION: This is denoted by �′ (�)

đ?‘“(đ?‘Ľ+â„Ž)−đ?‘“(đ?‘Ľ) â„Ž

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 60

8 Differential calculus Chapter

The formula � ′ (�) = limℎ→0

đ?‘“(đ?‘Ľ+â„Ž)−đ?‘“(đ?‘Ľ) â„Ž

can be used to find any of the

following from FIRST PRINCIPLES:

Look out for the words: FIRST PRINCIPLES

• The derivative of � at any point

• The gradient of the tangent to graph � at any point • The gradient of the function � at any point

• The rate of change of � at any point

�′ (�) can also be determined using DIFFERENTIATION RULES Function �

đ?’‡(đ?’™) = đ?’Œ where đ?’Œ is a constant

Derivative � ′ (�) �′(�) = 0

đ?’‡(đ?’™) = đ?’™đ?’” ; đ?’™đ???đ?‘š

đ?‘“ ′ (đ?‘Ľ) = đ?‘–đ?‘Ľ đ?‘›âˆ’1

đ?’‡(đ?’™) = đ?’Œđ?’™đ?’Ž ; đ?’Ž ∈ đ?‘š where đ?’Œ is a constant

đ?‘“ ′ (đ?‘Ľ) = đ?‘˜ Ă— đ?‘šđ?‘Ľ đ?‘šâˆ’1

When functions are added/subtracted, apply the rule to each function separately: đ?‘Ťđ?’™ [đ?’‡(đ?’™) Âą đ?’ˆ(đ?’™)] = đ?‘Ťđ?’™ [đ?’‡(đ?’™)] Âą đ?‘Ťđ?’™ [đ?’ˆ(đ?’™)]

• • • •

Examples đ?‘“(đ?‘Ľ) = −5 đ?‘“ ′ (đ?‘Ľ) = 0 đ?‘Ś=4 đ?‘‘đ?‘Ś =0 đ?‘‘đ?‘Ľ đ??ˇđ?‘Ľ [đ?‘Ľ 6 ] = 6đ?‘Ľ 5 1 đ?‘“(đ?‘Ľ) = 3 = đ?‘Ľ −3

đ?‘“

′ (�)

đ?‘Ľ

= −3đ?‘Ľ −4 = 4

• đ?‘“(đ?‘Ľ) = 2đ?‘Ľ đ?‘“ ′ (đ?‘Ľ) = 2 Ă— 4đ?‘Ľ 4−1 = 8đ?‘Ľ 3 1

5 2

1

5

• đ??ˇđ?‘Ľ ďż˝ đ?‘Ľ ďż˝ = Ă— đ?‘Ľ 2 2 2 5 3 = đ?‘Ľ2 4 2 • đ??ˇđ?‘Ľ [5đ?‘Ľ − 4đ?‘Ľ + 6] = 5 Ă— 2đ?‘Ľ − 4 + 0 = 10đ?‘Ľ − 4

−3 đ?‘Ľ4

5 −1 2

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 61

8 Differential calculus Chapter

BEFORE YOU APPLY THE DIFFERENTIATION RULES, MAKE SURE THERE ARE: •

No brackets : a

•

No đ?‘Ľ under a fraction line: b

c •

đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(2đ?‘Ľ − 1) = 2đ?‘Ľ 2 + đ?‘Ľ − 1 đ?‘“ ′ (đ?‘Ľ) = 4đ?‘Ľ + 1 đ?‘“(đ?‘Ľ) =

3đ?‘Ľ 2 −2 đ?‘Ľ

=

3đ?‘Ľ 2 đ?‘Ľ

2

đ?‘“ ′ (đ?‘Ľ) = 3 − 2(−1)đ?‘Ľ −2 = 3 + đ?‘Ľ 2 đ?‘“(đ?‘Ľ) =

đ?‘Ľ 2 −đ?‘Ľâˆ’6 đ?‘Ľ+2

� ′ (�) = 1

=

No đ?‘Ľ under a root sign: d

2

− đ?‘Ľ = 3đ?‘Ľ − 2đ?‘Ľ −1

(đ?‘Ľâˆ’3)(đ?‘Ľ+2) đ?‘Ľ+2

1

=đ?‘Ľâˆ’3

đ?‘“(đ?‘Ľ) = 3√đ?‘Ľ − 4đ?‘Ľ = 3đ?‘Ľ 2 − 4đ?‘Ľ 1

1

3

1

đ?‘“ ′ (đ?‘Ľ) = 3 Ă— đ?‘Ľ 2−1 − 4 = đ?‘Ľ −2 − 4

NB: NOTE THE DIFFERENCE BETWEEN THE FOLLOWING: đ?‘“(4) is the đ?’š-VALUE of the function at đ?‘Ľ = 4

� ′ (4) is the GRADIENT of the function at � = 4

As well as the gradient of the TANGENT at đ?‘Ľ = 4

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 62

8 Differential calculus Chapter

THE CUBIC GRAPH đ?’š = đ?’‚đ?’™đ?&#x;? + đ?’ƒđ?’™đ?&#x;? + đ?’„đ?’™ + đ?’… đ?’‚ indicates THE SHAPE đ?’‚ > 0 (+) or

đ?‘Ž

đ?’‚ < 0(−) or

STATIONARY POINTS TURNING POINTS

POINTS OF

INFLECTION Where � ′ (�) = 0 Local maximum

Where � ′′ (�) = 0

Local minimum How do I determine whether it is: A LOCAL MAXIMUM or A LOCAL MINIMUM? � ′′ ( )

�′

�′′

0

What does �′ and �′′ tell me? =0 Negative (< 0) � decreases

Local maximum Concave down

đ?‘“ turns

� ′′ ( )

0

Positive (> 0) đ?‘“ increases

Point of inflection Local minimum Concave up

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 63

8 Differential calculus Chapter

• •

đ?’™ −INTERCEPTS/ROOTS AND SHAPE

For đ?‘Ľ −intercepts: solve đ?‘“(đ?‘Ľ) = 0

A cubic graph can have either one(only) or two or three đ?‘Ľ −intercepts

EXAMPLES: a

đ?‘“(đ?‘Ľ) = đ?‘Ľ 3 + 4đ?‘Ľ 2 − 11đ?‘Ľ − 30

đ?‘“(3) = (3)3 + 4(3) − 11(3) − 30 = 0

-5 -2

3

∴ (đ?‘Ľ − 3) is a factor

đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 3)(đ?‘Ľ 2 + 7đ?‘Ľ + 10) = (đ?‘Ľ − 3)(đ?‘Ľ + 2)(đ?‘Ľ + 5)

b

The roots are −5; −2 and 3. đ?‘“(đ?‘Ľ) = đ?‘Ľ 3 − 3đ?‘Ľ + 2

đ?‘“(1) = (1)3 − 3(1) + 2 = 0 ∴ (đ?‘Ľ − 1) is a factor

đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 1)(đ?‘Ľ 2 + đ?‘Ľ − 2)

đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 1)(đ?‘Ľ − 1)(đ?‘Ľ + 2) = (đ?‘Ľ − 1)2 (đ?‘Ľ + 2)

If TWO FACTORS are the SAME, then the đ?‘Ľ −INTERCEPT is also a TURNING POINT. The graph BOUNCES at đ?‘Ľ = 1

-2

1

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 64

8 Differential calculus Chapter

EQUATION OF TANGENT TO GRAPH AT A SPECIFIC POINT • Substitute đ?‘Ľ −value into đ?’‡(đ?’™) to find coordinates of POINT of TANGENCY • Determine đ?‘“ ′ (đ?‘Ľ) using differentiation rules

• Substitute đ?‘Ľ −value into đ?’‡â€˛ (đ?’™) to find GRADIENT of TANGENT, đ?’Ž

• Substitute gradient into đ?‘Ś = đ?’Žđ?‘Ľ + đ?‘?

• Substitute point of tangency in đ?‘Ś = đ?’Žđ?‘Ľ + đ?‘? to find the value of đ?‘?. EXAMPLE Determine equation of tangent to đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 3 − 5đ?‘Ľ 2 − 4đ?‘Ľ + 3 at đ?‘Ľ = 1 Substitute đ?‘Ľ = 1: đ?‘“(1) = 2(1)3 − 5(1)2 − 4(1) + 3 = −4

∴ Point of tangency is (1; −4)

đ?‘“ ′ (đ?‘Ľ) = 6đ?‘Ľ 2 − 10đ?‘Ľ − 4 đ?‘“ ′ (1) = 6(1)2 − 10(1) − 4 = −8

∴ Gradient of tangent at đ?‘Ľ = 1 is −8 ; so đ?‘Ś = −8đ?‘Ľ + đ?‘?

Substitute (1; −4) into đ?‘Ś = −8đ?‘Ľ + đ?‘?: Equation of tangent is đ?‘Ś = −8đ?‘Ľ + 4

−4 = −8(1) + đ?‘? ∴đ?‘?=4

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 65

8 Differential calculus Chapter

SKETCHING THE CUBIC GRAPH EXAMPLE: đ?’‡(đ?’™) = đ?’™đ?&#x;? − đ?’™đ?&#x;? − đ?&#x;?đ?’™ + đ?&#x;?đ?&#x;?

DESCRIPTION OF STEP

Determine shape (using đ?’‚) Determine đ?’š −intercept Make đ?’™ = đ?&#x;Ž

Determine đ?’™ −intercepts Solve đ?’‡(đ?’™) = đ?&#x;Ž

Determine turning points and their đ?’š −values Solve đ?’‡â€˛(đ?’™) = đ?&#x;Ž Substitute đ?’™ −values into đ?’‡(đ?’™)

Make a neat drawing

STEP APPLIED TO THIS EXAMPLE

đ?‘Ž = 1 (positive)

đ?‘Ś = (0)3 − (0)2 − 8(0) + 12 = 12

đ?‘“(2) = 0 ∴ (đ?‘Ľ − 2) is a factor đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 2)(đ?‘Ľ 2 + đ?‘Ľ − 6) đ?‘“(đ?‘Ľ) = (đ?‘Ľ − 2)(đ?‘Ľ − 2)(đ?‘Ľ + 3) Roots are −3 and 2. đ?‘Ľ = 2 is also a turning point where graph bounces đ?‘“ ′ (đ?‘Ľ) = 3đ?‘Ľ 2 − 2đ?‘Ľ − 8 = 0 (3đ?‘Ľ + 4)(đ?‘Ľ − 2) = 0 4 đ?‘Ľ = − 3 or đ?‘Ľ = 2

We already know from the previous step that (2; 0) is one turning point (local minimum). Let us now find the other TP’s đ?‘Ś −coordinates −4 2 −4 −4 3 đ?‘“(đ?‘Ľ) = ďż˝ ďż˝ − ďż˝ ďż˝ − 8 ďż˝ ďż˝ + 12 = 18,52 3 3 3 Local maximum at (−1,33 ; 18,52) y

f

(−1,33 ; 18,52)

12

x

−3

2

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 66

8 Differential calculus Chapter

FINDING THE EQUATION OF A CUBIC GRAPH IN THE FORM đ?’š = đ?’‚đ?’™đ?&#x;? + đ?’ƒđ?’™đ?&#x;? + đ?’„đ?’™ + đ?’…

INFORMATION GIVEN (CAN BE SHOW ON A GRAPH OR NOT)

STEPS

From the đ?‘Ś −intercept we already know that đ?‘‘ = −8.

đ?’š −intercept: đ?’š = −đ?&#x;? and đ?’™ −intercepts: đ?’™ = −đ?&#x;?; −đ?&#x;? and đ?&#x;’

But we are going to use the three roots: đ?‘Ś = đ?‘Ž(đ?‘Ľ + 2)(đ?‘Ľ + 1)(đ?‘Ľ − 4) Substitute the point (0; −8): −8 = đ?‘Ž(0 + 2)(0 + 1)(0 − 4) −8 = −8đ?‘Ž đ?‘Ž=1 ∴ đ?‘Ś = 1(đ?‘Ľ + 2)(đ?‘Ľ + 1)(đ?‘Ľ − 4) Removing the brackets gives: đ?‘Ś = đ?‘Ľ 3 − đ?‘Ľ 2 − 10đ?‘Ľ − 8

We were given two roots (of which one is also a turning point) and the other turning point. y

x

1

NB: The graph BOUNCES at đ?‘Ľ = 1. This factor will therefore have to be squared.

4

đ?‘Ś = đ?‘Ž(đ?‘Ľ − 1)2 (đ?‘Ľ − 4)

( 3;−4)

Substitute the other turning point (3; −4): −4 = đ?‘Ž(3 − 1)2 (3 − 4) −4 = −4đ?‘Ž đ?‘Ž=1 ∴ đ?‘Ś = đ?‘Ž(đ?‘Ľ − 1)2 (đ?‘Ľ − 4) Removing the brackets gives: đ?‘Ś = đ?‘Ľ 3 − 6đ?‘Ľ 2 + 9đ?‘Ľ − 4

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 67

8 Differential calculus Chapter

SPECIAL APPLICATIONS OF DERIVATIVES RATES OF CHANGE • Distance/Displacement • Speed/Velocity • Acceleration

đ?‘ (đ?‘Ą)

EXAMPLE

đ?‘ ′ (đ?‘Ą)

đ?‘ ′′ (đ?‘Ą)

The displacement of a moving object is described by the equation đ?‘ (đ?‘Ą) = 10đ?‘Ą − đ?‘Ą 2 where đ?‘ , represents displacement in metres and đ?‘Ą, time in

seconds.

a Determine the displacement after 2 seconds. b What time will it take for the object to reach a maximum displacement? c

Determine the velocity of the object after 3 seconds.

d Determine the acceleration of the object. Is it going faster or slower? SOLUTIONS a đ?‘ (2) = 10(2) − (2)2 = 16 đ?‘š

b đ?‘ ′ (đ?‘Ą) = 10 − 2đ?‘Ą = 0 ∴ 10 − 2đ?‘Ą = 0 ∴đ?‘Ą =5đ?‘ c

đ?‘ ′ (3) = 10 − 2(3) = 4 đ?‘š. đ?‘ −1

d đ?‘ ′′ (đ?‘Ą) = −2 đ?‘š. đ?‘ −2

The object is going slower because the acceleration is negative.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 68

8 Differential calculus Chapter

USING FIRST DERIVATIVE TO DETERMINE MINIMUM OR MAXIMUM • For area, đ??´(đ?‘Ľ), to be a min/max, solve đ??´â€˛ (đ?‘Ľ) = 0

• For volume, �(�), to be a min/max, solve � ′ 4(�) = 0

• For cost, đ??ś(đ?‘Ľ), to be a minimum, solve đ??ś ′ (đ?‘Ľ) = 0

• For profit, đ?‘ƒ(đ?‘Ľ), to be a maximum, solve đ?‘ƒâ€˛(đ?‘Ľ) = 0

EXAMPLE

The volume of water in a water reservoir is given by: đ?‘‰(đ?‘Ą) = 60 + 8đ?‘Ą − 3đ?‘Ą 2

where �(�) is the volume in thousands of litres and � is the number of days water is pumped into the reservoir. a

Determine the rate of change of the volume after 3 days.

b

When will the volume of water in the reservoir be a maximum?

c

What will the maximum level of water in the reservoir be?

SOLUTIONS a

đ?‘‰ ′ (đ?‘Ą) = 8 − 6đ?‘Ą

b

∴ đ?‘‰ ′ (3) = 8 − 6(3) = −10 thousand liters/day

c

đ?‘Ą = 3 = 1,3 days

đ?‘‰ ′ (đ?‘Ą) = 8 − 6đ?‘Ą = 0 ∴ 8 − 6đ?‘Ą = 0 4

4

4

4 2

đ?‘‰ ďż˝3ďż˝ = 60 + 8 ďż˝3ďż˝ − 3 ďż˝3ďż˝ = 58,67 thousand litres

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 69

8 Differential calculus Chapter

Mixed Exercise on Differential Calculus 1

Determine �′ from first principles if a

b 2

3

đ?‘“(đ?‘Ľ) = 1 − đ?‘Ľ 2

đ?‘“(đ?‘Ľ) = −3đ?‘Ľ 2

Determine: a

đ?‘‘đ?‘Ś

b

đ??ˇđ?‘Ľ ďż˝

đ?‘‘đ?‘Ľ

1

if đ?‘Ś = √đ?‘Ľ − 2đ?‘Ľ 2 2đ?‘Ľ 2 −đ?‘Ľâˆ’15 đ?‘Ľâˆ’3

ďż˝

Determine the equation of the tangent to the curve đ?‘“(đ?‘Ľ) = −2đ?‘Ľ 3 + 3đ?‘Ľ 2 + 32đ?‘Ľ + 15 at the point đ?‘Ľ = −2.

4

Sketch the graph with the following properties showing all the key points on the graph: � ′ (�) < 0 when 1 < � < 5

� ′ (�) > 0 when � < 1 and � > 5 � ′ (5) = 0 and � ′ (1) = 0

5

đ?‘“(0) = −6 and đ?‘“(3) = 0 đ?‘“ ′′ (3) = 0

(2; 9) is a turning point of the graph đ?‘“(đ?‘Ľ) = đ?‘Žđ?‘Ľ 3 + 5đ?‘Ľ 2 + 4đ?‘Ľ + đ?‘?. Determine the

values of đ?‘Ž and đ?‘? in the equation of đ?‘“. 6

The diagram below represents the graph of � = � ′ (�), the derivative of �. 4

a b c

Write down the đ?‘Ľ-values of the turning point of đ?‘“.

Write down the đ?‘Ľ-value of the point of inflection of đ?‘“. For which values of đ?‘Ľ will đ?‘“(đ?‘Ľ) decrease?

−2

y

2

x 4

6

8

−4

−8

−12

y=f '(x)

−16

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 70

8 Differential calculus Chapter

7

The graph of đ?‘“(đ?‘Ľ) = đ?‘Ľ 3 + đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + đ?‘? is drawn. The curve has turning points at B and

(1; 0). The points (−1; 0) and (1; 0) are đ?‘Ľ-intercepts.

a

Show that đ?‘Ž = −1; đ?‘? = −1 and đ?‘? = 1.

b

Determine the coordinates of B.

y

B

x

(−1;0)

8

The distance covered in metres by an object is given as đ?‘ (đ?‘Ą) = đ?‘Ą 3 − 2đ?‘Ą 2 + 3đ?‘Ą + 5. Determine: a

9

(1;0)

an expression for the speed of the object at any time đ?‘Ą.

b

the time at which the speed of the object is at a minimum.

c

the time at which the acceleration of the object will be 8 đ?‘š. đ?‘ −2.

The sketch below shows a rectangular box with base ABCD. AB = 2đ?‘Ľ metres and BC = đ?‘Ľ metres. The volume of the box is 24 cubic metres. Material to cover the top (PQRS) of the box costs R25 per square metre. Material to cover the base ABCD and the four sides costs R20 per square metre.

a b c

Show that the height(â„Ž) of the box is given by â„Ž = 12đ?‘Ľ −2 .

Show that the total cost (C) in rand is given by: đ??ś(đ?‘Ľ) = 90đ?‘Ľ 2 + 1440đ?‘Ľ −1 .

Determine the value of đ?‘Ľ for which the cost will be a minimum.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 71

9 Analytical geometry Chapter

Overview Unit 1 Page 204 Equation of a circle with centre at the origin

• •

Chapter 9 Page 202 Analytical geometry

Finding the equation of a circle Symmetrical points on a circle

Unit 2 Page 208 Equation of a circle centred off the origin

•

•

Finding the equation of a circle with any given centre General form

• •

Lines on circles Equation of a tangent

Unit 3 Page 214 The equation of the tangent to the circle

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1 Work through all examples in this chapter of your Learner’s Book. 2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the text book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 72

9 Analytical geometry Chapter

REVISION OF CONCEPTS FROM PREVIOUS GRADES CONCEPT Distance between two points đ?‘¨(đ?’™đ?&#x;? ; đ?’šđ?&#x;? ) and đ?‘Š(đ?’™đ?&#x;? ; đ?’šđ?&#x;? ) Coordinates of midpoint

Average gradient between two points đ?‘¨(đ?’™đ?&#x;? ; đ?’šđ?&#x;? ) and đ?‘Š(đ?’™đ?&#x;? ; đ?’šđ?&#x;? )

Gradient of straight line through đ?‘¨ and đ?‘Š Equation of a straight line Angle of inclination, đ?œ˝

NB: Angle between line and POSITIVE đ?’™ −axis

To prove that points �, � and � are collinear (i.e. arranged in a straight line)

Parallel lines

Perpendicular lines

FORMULA / METHOD đ?‘‘ = ďż˝(đ?‘Ľ2 − đ?‘Ľ1 )2 + (đ?‘Ś2 − đ?‘Ś1 )2 đ?‘Ľ1 + đ?‘Ľ2 đ?‘Ś1 + đ?‘Ś2 ďż˝ ; ďż˝ 2 2 đ?‘š=

đ?‘Ś2 − đ?‘Ś1 đ?‘Ľ2 − đ?‘Ľ1

Or when given the angle of inclination, đ?‘ , use đ?‘š = đ?‘Ąđ?‘Žđ?‘–đ?‘ đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘? Or đ?‘Ś − đ?‘Ś1 = đ?‘š(đ?‘Ľ − đ?‘Ľ1 ) đ?‘š = đ?‘Ąđ?‘Žđ?‘–đ?‘ If đ?‘š > 0(+) , then đ?‘ is an acute angle (smaller than 90°)

If đ?‘š < 0(−) , then đ?‘ is an obtuse angle (bigger than 90° but )

Prove that đ?‘šđ??´đ??´ = đ?‘šđ??´đ??¸ Or đ?‘šđ??´đ??´ = đ?‘šđ??´đ??¸ Or đ?‘šđ??´đ??¸ = đ?‘šđ??´đ??¸

Two lines đ?‘Ś = đ?‘š1 đ?‘Ľ + đ?‘?1 and đ?‘Ś = đ?‘š2 đ?‘Ľ + đ?‘?2 are parallel if đ?‘š1 = đ?‘š2 Two lines đ?‘Ś = đ?‘š1 đ?‘Ľ + đ?‘?1 and đ?‘Ś = đ?‘š2 đ?‘Ľ + đ?‘?2 are perpendicular if đ?‘š1 Ă— đ?‘š2 = −1

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9 Analytical geometry Chapter

OTHER DEFINITIONS/CONCEPTS YOU HAVE TO KNOW DEFINITION

EXAMPLES A (1;4)

C(3;-2) K Altitude of a triangle = line from one vertex perpendicular to opposite side

B(-3;4) To determine the equation of altitude AK: • Determine gradient of BC • Determine gradient of AK • Determine equation of AK (substitute A) −2 − 4 đ?‘šđ??´đ??¸ = = −1 3 − (−3) But BCâ?ŠAK, so đ?‘šđ??´đ??ž = 1 Substitute point A(1;4): đ?‘Ś − 4 = −1(đ?‘Ľ − 1) Equation of AK: đ?‘Ś = −đ?‘Ľ + 5 K(1;6)

M(4;2) A Median = line joining vertex of triangle to midpoint of opposite side

L(-4;-4) To determine the equation of median KA: • Determine coordinates of midpoint A • Determine gradient of KA • Determine equation of KA −4 + 4 −4 + 2 đ??´ďż˝ ; ďż˝ 2 2 đ??´(0; −1) 6 − (−1) đ?‘šđ??žđ??´ = =7 1−0 đ?‘Ś − 6 = 7(đ?‘Ľ − 1) đ?‘Ś = 7đ?‘Ľ − 1

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 74

9 Analytical geometry Chapter

C(4;6)

B(-2;4)

Perpendicular bisector = the line through the midpoint of a line and perpendicular to that line

To determine equation of perpendicular bisector of BC: • Determine gradient of BC • Determine gradient of bisector • Determine equation of bisector 6−4 1 đ?‘šđ??´đ??¸ = = 4 − (−2) 3 Product of gradients must be −1: đ?‘šđ?‘?đ?‘’đ?‘&#x;đ?‘? đ?‘?đ?‘–đ?‘ đ?‘’đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; = −3 đ?‘Ś − 6 = −3(đ?‘Ľ − 4) đ?‘Ś = −3đ?‘Ľ + 18

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 75

9 Analytical geometry Chapter

THE CIRCLE : (đ?’™ − đ?’‚)đ?&#x;? + (đ?’š − đ?’ƒ)đ?&#x;? = đ?’“đ?&#x;? (centre-radius form)

SUMMARY ON CIRCLES Equation of circle with radius đ?’“ and centre at the origin

Equation of circle with radius đ?’“ and centre (đ?’‚; đ?’ƒ)

To determine radius and centre of circle when given equation

đ?‘Ľ2 + đ?‘Ś2 = đ?‘&#x;2 (đ?‘Ľ − đ?‘Ž)2 + (đ?‘Ś − đ?‘?)2 = đ?‘&#x; 2 Example A: Determine the radius and centre of the circle with equation (đ?‘Ľ + 1)2 + (đ?‘Ś − 3)2 = 16 (đ?‘Ľ + 1)2 + (đ?‘Ś − 3)2 = 16 can be written as (đ?‘Ľ − (−1))2 + (đ?‘Ś − (3))2 = 16 Centre: (−1; 3)

Radius = √16 = 4

Example B: Determine the radius and centre of the circle with equation đ?‘Ľ 2 + đ?‘Ś 2 + 4đ?‘Ľ + 6đ?‘Ś − 10 = 0 We are going to use COMPLETION OF THE SQUARE • • •

Constant term to RHS; group đ?‘Ľ- and đ?‘Ś-terms đ?‘Ľ 2 + 4đ?‘Ľ + â‹Ż + đ?‘Ś 2 + 6đ?‘Ś + â‹Ż = 10 1

2

Complete square for đ?‘Ľ and đ?‘Ś - add ďż˝2 Ă— đ?‘?đ?‘œđ?‘?đ?‘“đ?‘“đ?‘–đ?‘?đ?‘–đ?‘?đ?‘–đ?‘Ąďż˝ đ?‘Ľ 2 + 4đ?‘Ľ + đ?&#x;’ + đ?‘Ś 2 + 6đ?‘Ś + đ?&#x;— = 10 + đ?&#x;’ + đ?&#x;— Write in centre-radius form (đ?‘Ľ + 2)2 + (đ?‘Ś + 3)2 = 23

Centre: (−2; −3)

Radius = √23

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 76

9 Analytical geometry Chapter

NB: Radius is PERPENDICULAR to tangent

Equation of tangent to circle at given point Determine the equation of the tangent to the circle (đ?’™ + đ?&#x;?)đ?&#x;? + (đ?’š − đ?&#x;?)đ?&#x;? = đ?&#x;?đ?&#x;” through the point (đ?&#x;?; đ?&#x;“). The steps are: • Determine centre of circle • Determine gradient of RADIUS • Determine gradient of TANGENT: Remember: đ?’Žđ?’“đ?’‚đ?’… Ă— đ?’Žđ?’•đ?’‚đ?’” = −đ?&#x;? • Determine equation of tangent by substituting point of tangency The centre of circle (đ?’™ + đ?&#x;?)đ?&#x;? + (đ?’š − đ?&#x;?)đ?&#x;? = đ?&#x;?đ?&#x;” is (−đ?&#x;?; đ?&#x;?). Radius joins centre (−đ?&#x;?; đ?&#x;?) with point of tangency (đ?&#x;?; đ?&#x;“). đ?&#x;“−đ?&#x;? đ?&#x;? đ?&#x;? đ?’Žđ?’“đ?’‚đ?’… = = = đ?&#x;? − (−đ?&#x;?) đ?&#x;’ đ?&#x;? ∴ đ?’Žđ?’•đ?’‚đ?’”đ?’ˆđ?’†đ?’”đ?’• = −đ?&#x;? Equation of tangent: đ?’š − đ?&#x;“ = −đ?&#x;?(đ?’™ − đ?&#x;?) đ?’š = −đ?&#x;?đ?’™ + đ?&#x;—

Mixed Exercise on Analytical Geometry 1

A (−2; 1), B(đ?‘?; −4), C(5; 0) and D (3; 2) are the vertices of trapezium ABCD in a Cartesian plane with đ??´đ??´Ç đ??śđ??ˇ. a

b c

Show that đ?‘? = 3.

Calculate AB:CD in simplest form. If N (đ?‘Ľ; đ?‘Ś) is on AB and NBCD is a parallelogram, determine the coordinates of N.

d

Determine the equation of the line passing through B and D.

e

What is the angle of inclination of line BD?

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9 Analytical geometry Chapter

2

f

Calculate the area of parallelogram NBCD.

g

R (−1; đ?‘ž), A and C are collinear. Calculate the value of đ?‘ž.

� 2 + 4� + � 2 + 2� – 8 = 0 is the equation of a circle with centre M in a Cartesian plane.

a b c d e

3

Prove that the circle passes through the point N(1; −3) Determine the equation of PN, the tangent to the circle at N. Calculate đ?‘ , the angle of inclination of the tangent, rounded off to one decimal place. Determine the coordinates of the point where the tangent in 2 b intersects the đ?‘Ľ-axis. Calculate the coordinates of the point(s) where the circle with centre M cuts the đ?‘Ś-axis.

In the diagram, P, R and S are vertices of ∆PRS . P is a point on the y-axis. The coordinates of R is (-6; -12). The equation of PR is 3đ?‘Ľ – đ?‘Ś + 6 = 0. The median SM and the altitude RN intersect at the origin O.

a

Calculate the gradient of RO.

b

Calculate the gradient of PS.

c

Determine the equation of PS.

d

Calculate the inclination of PS rounded of to one decimal digit.

4

(

)

e

If the coordinates of N are 2n; 3 5 + n , determine

f

the value of đ?‘–.

3

Calculate the coordinates of S.

The equation of a circle is � 2 + � 2 + 4� – 2� – 4 = 0. a

Determine the coordinates of M, the centre of the circle, as well as the length of the radius.

b c

Calculate the value of p if N(đ?‘?; 1) with đ?‘? > 0, is a point on the circle.

Write down the equation of the tangent to the circle at N.

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9 Analytical geometry Chapter

5

A (-3; 3), B(2; 3), C(6; - 1) and D(đ?‘Ľ; đ?‘Ś) are the vertices of quadrilateral ABCD in a Cartesian plane. a

Determine the equation AD.

b

Prove that the coordinates of D are

( ) if D is equidistant from B and C. 3 3 ;− 2 2

c

Determine the equation of BD.

d

Determine the size of θ , the angle between BD and BC, rounded off to one decimal digit.

e 6

Calculate the area of ∆BDC rounded off to the nearest square unit.

In the diagram, points A(2; 3), B(đ?‘?; 0) and C(5; - 3) are the vertices of ∆ABC in a

Cartesian plane. AC cuts the đ?‘Ľ-axis at D. a

Calculate the coordinates of D.

b

Calculate the value of p if BC = AC and đ?‘? < 0.

c

Determine the angle of inclination of straight line AC, rounded off to one decimal place.

d

If đ?‘? = −1, calculate the size of AĚ‚ , rounded off to one decimal digit.

7

In the Cartesian plane the equation of a circle with centre M is given by: � 2 + � 2 + 6� – 7 = 0

Determine, by calculation, whether the straight line đ?‘Ś = đ?‘Ľ + 1 is a tangent to the circle, or not.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 79

9 Analytical geometry Chapter

8

In the diagram, centre C of the circle lies on the straight line 3đ?‘Ľ + 4đ?‘Ś + 7 = 0. The straight line cuts the circle at D and E(−1; −1). The circle touches the y-axis at P(0; 2).

a

Determine the equation of the circle

b

in the form: (đ?‘Ľ – đ?‘–)2 + (đ?‘Ś – đ?‘ž)2 = đ?‘&#x; 2

Determine the length of diameter DE.

c

Determine the equation of the perpendicular bisector of PE.

d

Show that the perpendicular bisector of PE and straight line DE intersect at C.

9

a b

In the diagram, P, R(4; −4), S and T (0; 4) are the vertices of a rectangle. P and S lie on the đ?‘Ľ – axis. The diagonals intersect at W.

Show that the coordinates of S are (2 + 2√5; 0). Determine the gradient of TS rounded off to two decimals.

c

Calculate ���� rounded off to two decimals.

10 a b

Show that the equation of the tangent to the circle � 2 + � 2 – 4� + 6� + 3 = 0

at the point (5; -2) is đ?‘Ś = −3đ?‘Ľ + 13

If T(đ?‘Ľ; đ?‘Ś) is a point on the tangent in 10.a, such that its distance from the centre of the circle is 20 units, determine the values of đ?‘Ľ and đ?‘Ś.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 80

10 Euclidean geometry Chapter

Overview

Unit 1 Page 244 Proportionality in triangles

Chapter10 Page 236 Euclidean geometry

• •

Ratio Theorem 1

Unit 2 Page 250 Similarity in triangles

• •

Theorem 2 Theorem 3

•

Prove of Theorem of Pythagoras

Unit 3 Page 256 Theorem of Pythagoras

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1 Work through all examples in this chapter of your Learner’s Book. 2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 81

10 Euclidean geometry Chapter

REVISION of geometry From previous years

CONGRUENCY SSS

AAS

SAS (included angle)

RHS

SIMILARITY AAA

SSS

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 82

10 Euclidean geometry Chapter

PROPERTIES OF SPECIAL QUADRILATERALS PARALLELOGRAM • Both pairs of opposite sides are parallel • Both pairs of opposite side are equal • Both pairs of opposite angles are equal • Diagonals bisect each other RECTANGLE All properties of parallelogram Plus: • Both diagonals are equal in length • All interior angles are equal to 90°

RHOMBUS All properties of parallelogram Plus: • All sides are equal • Diagonals bisect each other perpendicularly • Diagonals bisect interior angles SQUARE All properties of a rhombus Plus: • All interior angles are 90° • Diagonals are equal in length KITE • • • • •

Two pairs of adjacent sides are equal Diagonal between equal sides bisects other diagonal One pair of opposite angles are equal (unequal sides) Diagonal between equal sides bisects interior angles (is axis of symmetry) Diagonals intersect perpendicularly

TRAPEZIUM • One pair of opposite sides are parallel

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 83

10 Euclidean geometry Chapter

HOW TO PROVE THAT A QUADRILATERAL IS A PARALLELOGRAM Prove any ONE of the following (most often by congruency): • Prove that both pairs of opposite sides are parallel • Prove that both pairs of opposite sides are equal • Prove that both pairs of opposite angles are equal • Prove that the diagonals bisect each other

HOW TO PROVE THAT A PARALLLELOGRAM IS A RHOMBUS Prove ONE of the following: • Prove that the diagonals bisect each other perpendicularly • Prove that any two adjacent sides are equal in length

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 84

10 Euclidean geometry Chapter

MIDPOINT THEOREM The line segment joining the midpoints of two sides of a triangle, is parallel to the 3rd side of the triangle and half the length of that side.

1

If AD = DB and AE = EC, then DE ǁ BC and DE = 2BC

CONVERSE OF MIDPOINT THEOREM

If a line is drawn from the midpoint of one side of a triangle parallel to another side, that line will bisect the 3rd side and will be half the length of the side it is parallel to.

1

If AD = DB and DE ǁ BC, then AE = EC and DE = 2BC.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 85

10 Euclidean geometry Chapter

REVISION OF CIRCLE GEOMETRY (FROM GRADE 11) Theorem 1 If AC = CB in circle O, then OC â?Š AB.

Converse of Theorem 1 If OC â?Š chord AB, then AC = BC.

Theorem 2 The angle at the centre of a circle subtended by an arc/a chord is double the angle at the ďż˝ B = 2 Ă— ACďż˝ B circumference subtended by the same arc/chord. AO

Theorem 3 The angle on the circumference subtended by the diameter, is a right angle. (The angle in a semi-circle is 90°).

Converse of Theorem 3 If đ??śĚ‚ = 90°, then AB is the diameter of the circle.

Theorem 4 The angles on the circumference of a circle, subtended by the same arc or chord, are equal.

Converse of Theorem 4 If a line segment subtends equal angles at two other points, then these four points lie on the circumference of a circle.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 86

10 Euclidean geometry Chapter

Corollaries of Theorem 4 Equal chords subtend equal angles at the circumference of the circle.

Equal chords subtend equal angles at the centre of the circle.

Equal chords of equal circles subtend equal angles at the circumference.

Theorem 5 The opposite angles of a cyclic quadrilateral are supplementary. đ??´Ě‚ + đ??śĚ‚ = 180° ďż˝ = 180° đ??´ďż˝ + đ??ˇ

Converse of Theorem 5 If the opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.

Theorem 6 The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.

Converse of Theorem 6 If the exterior angle of a quadrilateral is equal to the opposite interior angle, then it is a cyclic quadrilateral.

Theorem 7 The tangent to a circle is perpendicular to the radius at the point of tangency.

Converse of Theorem 7 If a line is drawn perpendicularly to the radius through the point where the radius meets the circle, then this line is a tangent to the circle.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 87

10 Euclidean geometry Chapter

Theorem 8 If two tangents are drawn from the same point outside a circle, then they are equal in length.

Theorem 9 (Tan chord theorem) The angle between the tangent to a circle and a chord drawn from the point of tangency, is equal to the angle in the opposite circle segment. Acute angle Obtuse angle

Converse of Theorem 9 If a line is drawn through the endpoint of a chord to form an angle which is equal to the angle in the opposite segment, then this line is a tangent.

THREE WAYS TO PROVE THAT A QUADRILATERAL IS A CYCLIC QUADRILATERAL Prove that: • one pair of opposite angles are supplementary • the exterior angle is equal to the opposite interior angle • two angles subtended by a line segment at two other vertices of the quadrilateral, are equal.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 88

10 Euclidean geometry Chapter

Example 1 In the diagram alongside< O is the centre of circle DABMC. BC and DM are diameters. AC and DM intersect at T. OT =3DT ABÇ DM a

Prove that T is the midpoint of AC.

b

Determine the length of MC in terms of DT.

c

ďż˝ in terms of đ?‘‚ďż˝2. Express đ??ˇ

Solution: a

đ??´Ě‚1 = 90° đ?‘‡ďż˝1 = 90°

∠in semi ⊙ int. ∠s suppl

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 89

10 Euclidean geometry Chapter

REVISING THE CONCEPT OF PROPORTIONALITY

A

6 cm

B

D

4 cm

9 cm

C E

6 cm

F

AB : BC = 6 : 4 = 3 : 2 DE : EF = 9 : 6 = 3 : 2 Although, AB : BC = DE : EF it does NOT mean that AB = DE, AC = DF or BC = EF.

GRADE 12 GEOMETRY Theorem 1

Converse of Theorem 1

A line drawn parallel to one side of a triangle that intersects the other two sides, will divide the other two sides proportionally.

If a line divides two sides of a triangle proportionally, then the line is parallel to the third side of the triangle.

If DE Ç BC then

đ??´đ??´ đ??ˇđ??ˇ

=

or AD : DB = AE : EC

đ??´đ??´ đ??¸đ??¸

If

đ??´đ??´ đ??ˇđ??ˇ

=

đ??´đ??´ đ??¸đ??¸

then DE Ç BC.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 90

10 Euclidean geometry Chapter

Theorem 2 (Midpoint Theorem) (Special case of Theorem 1) The line segment joining the midpoints of two sides of a triangle, is parallel to the 3rd side of the triangle and half the length of that side.

1 2

Converse of Theorem 2 If a line is drawn from the midpoint of one side of a triangle parallel to another side, that line will bisect the 3rd side and will be half the length of the side it is parallel to.

If AD = DB and AE = EC, then DE Ç BC and DE = BC. If AD = DB and DE Ç BC, then AE = EC 1 2

and DE = BC.

Theorem 3

Converse of Theorem 3

The corresponding sides of two equiangular proportional, triangles are proportional.

If the sides of two triangles are then the triangles are equiangular.

If ∆đ??´đ??´đ??´ ||| ∆đ??ˇđ??ˇđ??ˇ then

đ??´đ??´ đ??ˇđ??ˇ

=

đ??ľđ??ľ đ??¸đ??¸

=

đ??´đ??´ đ??ˇđ??ˇ

If

đ??´đ??´ đ??ˇđ??ˇ

=

đ??ľđ??ľ đ??¸đ??¸

=

đ??´đ??´ đ??ˇđ??ˇ

then ∆đ??´đ??´đ??´ |||∆đ??ˇđ??ˇđ??ˇ

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 91

10 Euclidean geometry Chapter

Theorem 4 The perpendicular drawn from the vertex of the right angle of a right-angled triangle, divides the triangle in two triangles which are similar to each other and similar to the original triangle.

Corollaries of Theorem 4 ∆đ??´đ??´đ??´|||∆đ??ˇđ??ˇđ??ˇ ∴

đ??´đ??´ đ??ˇđ??ˇ

=

đ??ľđ??ľ đ??ľđ??ľ

=

đ??´đ??´ đ??ˇđ??ˇ

∴ đ??´đ??´2 = đ??ľđ??ľ. đ??ľđ??ľ

∆đ??´đ??´đ??´|||∆đ??ˇđ??ˇđ??ˇ

∴

đ??´đ??´ đ??ˇđ??ˇ

=

đ??ľđ??ľ đ??´đ??´

=

đ??´đ??´ đ??ˇđ??ˇ

∴ đ??´đ??´ 2 = đ??śđ??ś. đ??śđ??ś

∆đ??ˇđ??ˇđ??ˇ|||∆đ??ˇđ??ˇđ??ˇ

∴

đ??ˇđ??ˇ đ??ˇđ??ˇ

=

đ??ľđ??ľ đ??´đ??´

=

đ??ˇđ??ˇ đ??ˇđ??ˇ

∴ đ??´đ??´ 2 = đ??ľđ??ľ. đ??ˇđ??ˇ

Theorem 5 (The Theorem of Pythagoras) Using the corollaries of Theorem 4, it can be proven that: đ??ľđ??ľ 2 = đ??´đ??´2 + đ??´đ??´ 2

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 92

10 Euclidean geometry Chapter

Example 4

Given: đ??´đ??ˇ: đ??ˇđ??´ = 2: 3 and đ??´đ??ˇ = 3 đ??ˇđ??ś.

Instruction: Determine the ratio of đ??śđ?‘ƒ: đ?‘ƒđ??ˇ. Solution:

đ??´đ??´

In ∆đ??´đ??´đ??ˇ

đ??žđ??´

5

=2

4

But it was given that đ??´đ??ˇ = 3 đ??ˇđ??ś 4

5

∴ 3 đ??ˇđ??ś = 2 đ??žđ??ˇ đ??´đ??¸

đ??žđ??´

5

4

=2á3 =

In ∆đ??śđ??ˇđ??ž

15

8 đ??¸đ?‘ƒ

đ?‘ƒđ??´

∴ đ??śđ?‘ƒ: đ?‘ƒđ??ˇ = 15: 8

đ??¸đ??´

= đ??´đ??ž =

5

∴ đ??´đ??ˇ = 2 đ??žđ??ˇ

15 8

TIPS TO SOLVE A GEOMETRY PROBLEM • READ-READ-READ the information next to the diagram thoroughly • TRANSFER all given information on the DIAGRAM • Look for KEYWORDS, e.g. TANGENT: What do the theorems say about tangents? CYCLIC QUADRILATERAL: What are the properties of a cyclic quad? • Set yourself “SECONDARYâ€? GOALS, e.g. - To prove that two sides of triangle are equal (primary goal), first prove that there are two equal angles (secondary goal) - To prove that a line is a tangent, the secondary goal can be to prove that the line is perpendicular to a radius • For questions like: Prove that đ??´Ě‚1 = đ??śĚ‚2 . Start with ONE PART. Move to the OTHER PART step-by-step stating reasons.

ďż˝đ?&#x;? ; ∴ đ?‘¨ ďż˝đ?&#x;? ďż˝ đ?&#x;? = đ??´Ě‚2 ; đ??´Ě‚2 = đ??śĚ‚1 ; đ??śĚ‚1 = đ?‘Ş ďż˝đ?&#x;? = đ?‘Ş E.g. đ?‘¨

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 93

10 Euclidean geometry Chapter

Mixed Exercise on Euclidian Geometry

1

In the diagram, TBD is a tangent to circles BAPC and BNKM at B. AKC is a chord of the larger circle and is also a tangent to the smaller circle at K. Chords MN and BK intersect at F. PA is produced to D. BMC, BNA and BFKP are straight lines. Prove that:

a b c d

MN Ç CA ∆đ??žđ?‘€đ?‘ is isosceles

đ??´đ??ž đ??žđ?‘ƒ

đ??´đ?‘€

= đ?‘€đ??¸

DA is a tangent to the circle passing through points A, B and K.

2

In the diagram below, chord BA and tangent TC of circle ABC are produced to meet at R. BC is produced to P with RC=RP. AP is not a tangent.

Prove that: a

ACPR is a cyclic quadrilateral.

b

∆đ??śđ??´đ??´|||∆đ?‘…đ?‘ƒđ??´

c d e

đ?‘…đ??ś =

đ??¸đ??´.đ?‘…đ??´ đ??´đ??¸

đ?‘…đ??´. đ??´đ??ś = đ?‘…đ??ś. đ??śđ??´

Hence prove that đ?‘…đ??ś 2 = đ?‘…đ??´. đ?‘…đ??´

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 94

10 Euclidean geometry Chapter

3

In the diagram alongside, circles ACBN and AMBD Intersect at A and B. CB is a tangent to the larger circle at B. M is the centre of the smaller circle. CAD and BND are straight lines. Let đ??´Ě‚3 = đ?‘Ľ

a b

4

ďż˝ in terms of đ?‘Ľ. Determine the size of đ??ˇ

Prove that: i

CB Ç AN

ii

AB is a tangent to circle ADN.

In the diagram below, O is the centre of circle ABCD. DC is extended to meet circle BODE at point E. OE cuts BC at F. Letđ??ˇďż˝1 = đ?‘Ľ.

a b

Determine đ??´Ě‚ in terms of đ?‘Ľ. Prove that: i

BE=EC

ii

BE is NOT a tangent to circle ABCD.

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10 Euclidean geometry Chapter

5

In the diagram alongside, medians AM and CN of ∆đ??´đ??´đ??ś intersect at O.

BO is produced to meet AC at P. MP and CN intersect in D. ORÇ MP with R on AC. a b 6

đ?‘ đ??´

Calculate, giving reasons, the numerical value of đ?‘ đ??¸ .

Use đ??´đ?‘‚: đ??´đ?‘€ = 2: 3, to calculate the numerical value of In the diagram, AD is the diameter of circle ABCD.

đ?‘…đ?‘ƒ đ?‘ƒđ??¸

.

AD is extended to meet tangent NCP in P. Straight line NB is extended to Q and intersect AC in M with Q on straight line ADP. ACâ?ŠNQ at M.

a

Prove that NQÇ CD.

b

Prove that ANCQ is a cyclic quadrilateral.

c

i ii

d e

Prove that ∆đ?‘ƒđ??śđ??ˇ|||∆đ?‘ƒđ??´đ??ś.

Hence, complete: đ?‘ƒđ??ś 2 = â‹Ż

Prove that đ??´đ??ś 2 = đ??śđ??ˇ. đ?‘ đ??´

If it is further given that PC=MC, prove that 1−

đ??´đ?‘€2 đ??´đ??¸ 2

đ??´đ?‘ƒ.đ??´đ?‘ƒ

= đ??¸đ??´.đ?‘ đ??´

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 96

11 Statistics: regression and correlation Chapter

Overview

Unit 1 Page 266 Symmetrical and skewed data

• •

Symmetrical data Skewed data

Unit 2 Page 270 Scatter plots and correlation Chapter 11 Page 262 Statistics: regression and correlation

• • • • • • •

Bivariate data Correlation Examples of scatter plots and correlation Drawing scatter plots The least squares method The correlation coefficient Using a calculator to find the regression line

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1 Work through all examples in this chapter of your text book. 2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the text book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 97

11 Statistics: regression and correlation Chapter

Measures of dispersion (indicates spread of data)

Measures of central tendency

UNGROUPED DATA Mode = most frequent number Mean =

GROUPED DATA Modal class = interval with highest frequency Estimated mean =

đ?‘†đ?‘˘đ?‘š đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘’đ?‘&#x; đ?‘œđ?‘“ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’đ?‘

NB: Data has to be arranged in ascending order đ?‘„2 , Median = Middle value (for an odd number of values)

where đ?‘Ľđ?‘– = midpoint of class đ?‘– and đ?‘“đ?‘– = frequency of class đ?‘– Median class interval = class/interval in which middle value lies

đ?‘†đ?‘˘đ?‘š đ?‘œđ?‘“ đ?‘Ąđ?‘¤đ?‘œ đ?‘šđ?‘–đ?‘‘đ?‘‘đ?‘™đ?‘’ đ?‘Łđ?‘Žđ?‘™đ?‘˘đ?‘’đ?‘ 2

Or (for an even number of values) Percentiles (divide data into 100 equal parts) 30 E.g. the position of đ?‘ƒ30 = 100 (đ?‘– + 1)

đ?‘„1 , Lower quartile = Middle value of all the values below the median (excluding median) đ?‘„3 , Upper quartile = Middle value of all the values above the median (excluding median) Range = Maximum – Minimum Inter quartile range (IQR) = đ?‘„3 − đ?‘„1 Semi Inter quartile range =

∑ đ?‘Ľđ?‘– Ă—đ?‘“đ?‘– ∑ đ?‘“đ?‘–

Position of đ?‘„2 =

(đ?‘›+1)

Position of đ?‘„1 =

(đ?‘›+1)

Position of đ?‘„3 =

2

4

3(đ?‘›+1) 4

đ?‘„3 −đ?‘„1 2

Five point summary (used to draw box-and-whisker diagram): Min, đ?‘„1 , Median, đ?‘„3 , Max

SYMMETRICAL DISTRIBUTION NORMAL DISTRIBUTION ���� = ������ = ����

DISTRIBUTION OF DATA ASYMMETRICAL DISTRIBUTIONS NEGATIVELY SKEWED POSITIVELY SKEWED đ?‘šđ?‘?đ?‘Žđ?‘– − đ?‘šđ?‘?đ?‘‘đ?‘–đ?‘Žđ?‘– < 0 đ?‘šđ?‘?đ?‘Žđ?‘– − đ?‘šđ?‘?đ?‘‘đ?‘–đ?‘Žđ?‘– > 0

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 98

11 Statistics: regression and correlation Chapter

OGIVE Ogive = cumulative frequency graph

80

Ages

70

NB: When drawing the ogive:

60

50

•

plot the (upper class boundary ; cumulative frequency)

•

the graph has to be grounded

30

•

the shape of the graph has to be smooth rather

20

than consist of “connected dots�

10

40

Q3

Q1 5

10

15

20

Freq 25

THE OGIVE CAN BE USED TO DETERMINE THE MEDIAN AND QUARTILES.

MEASURES OF DISPERSION AROUND THE MEAN

VARIANCE đ?œŽ 2

Variance, đ?œŽ 2 , is an indication of how far each value in the data set is from the mean, đ?‘ĽĚ… . đ?œŽ2 =

∑(đ?‘Ľđ?‘– −đ?‘ĽĚ… )2 đ?‘›

(for population)

STANDARD DEVIATION đ?œŽ

Standard deviation (SD), đ?œŽ: SD = √đ?‘Łđ?‘Žđ?‘&#x;đ?‘–đ?‘Žđ?‘–đ?‘?đ?‘?

The larger the standard deviation, the larger the deviation from the mean would be. A normal distribution is shown below:

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 99

30

11 Statistics: regression and correlation Chapter

USING A TABLE TO CALCULATE VARIANCE AND STANDARD DEVIATION

UNGROUPED DATA First calculate the mean,đ?‘ĽĚ… then the following columns.

Variance =

(đ?‘Ľ − đ?‘ĽĚ… )2

(đ?‘Ľ − đ?‘ĽĚ… )

DATA VALUES, đ?‘Ľ

Calculate the total of this column,∑(đ?‘Ľ − đ?‘ĽĚ… )2

∑(đ?‘Ľâˆ’đ?‘ĽĚ… )2 đ?‘›

Standard variance = √đ?‘Łđ?‘Žđ?‘&#x;đ?‘–đ?‘Žđ?‘–đ?‘?đ?‘? GROUPED DATA

First calculate the estimated mean, đ?‘ĽĚ… = Class Frequency đ?‘“ Interval

Midpoint đ?‘š

∑ đ?‘“Ă—đ?‘š ∑đ?‘“

đ?‘“Ă—đ?‘š

đ?‘š − đ?‘ĽĚ…

(đ?‘š − đ?‘ĽĚ… )2

đ?‘“ Ă— (đ?‘š − đ?‘ĽĚ… )2

Calculate the total of this column Variance =

∑ đ?‘“(đ?‘Ľâˆ’đ?‘ĽĚ… )2 đ?‘›

Standard variance = √đ?‘Łđ?‘Žđ?‘&#x;đ?‘–đ?‘Žđ?‘–đ?‘?đ?‘? _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 100

11 Statistics: regression and correlation Chapter

USING CASIO ��-82ZA PLUS CALCULATOR TO CALCULATE STANDARD DEVIATION

MODE 2 : STAT 1 : 1 – VAR Enter the data points: Push = after each data point AC SHIFT STAT (above the 1 button)

To switch on the frequency column when calculating the SD for a frequency table, first do the following: Shift Setup; Down arrow (on big REPLAY button); 3: STAT; 2: ON

4 : VAR 3: đ?œŽđ?‘Ľđ?‘– To clear screen: MODE 1: COMP

DETERMINING OUTLIERS

Inter quartile range, IQR = đ?‘„3 − đ?‘„1 An outlier is identified if it is: • •

Less than đ?‘„1 − đ??źđ?‘„đ?‘… Ă— 1,5 or Larger than đ?‘„3 + đ??źđ?‘„đ?‘… Ă— 1,5

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 101

11 Statistics: regression and correlation Chapter

SCATTER DIAGRAMS (SCATTER PLOTS) FOR BIVARIATE DATA

Scatter diagrams are used to graphically determine whether there is an association between two variables. By investigation one can determine which of the following curves (regression functions) would best fit the diagram:

Linear (straight line)

Quadratic (parabola)

Exponential function

USING A CALCULATOR TO DETERMINE THE EQUATION OF THE REGRESSION LINE (LEAST SQUARES REGRESSION LINE)

The standard form of a straight line equation is: where đ?‘š is the gradient and đ?‘? is the đ?‘Ś-intercept.

đ?‘Ś = đ?‘šđ?‘Ľ + đ?‘?

NB: On the calculator the regression line is determined in the form: đ?’š = đ?‘¨ + đ?‘Šđ?’™ (In this form đ??´ = the gradient of the line and đ??´ = the đ?‘Ś-intercept) _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 102

11 Statistics: regression and correlation Chapter

On the calculator press: MODE 2 2: A+Bx Enter the data points (column X and Y): Push = after each data point Press AC. SHIFT STAT 5: REG 1: A = (to determine the đ?‘Ś- intercept of the line) SHIFT STAT 5: REG 2: B = (to determine the gradient of the line) SHIFT STAT 5: REG 3: đ?‘&#x; = (to determine correlation coefficient) EXAMPLE đ?‘Ľ

đ?‘Ś

5

10

15

20

25

30

35

40

20

223

25

35

30

40

50

55

Using the calculator, the equation for the line of best fit (or regression line) can be determined giving:

100

đ?‘Ś = 1đ?‘Ľ + 12,25

y

95 90 85 80 75 70

NB: The line of best fit ALWAYS goes through the point (đ?‘ĽĚ… ; đ?‘Śďż˝).

65 60 55 50 45

In this case it goes through the point ( 23 ; 35 )

40 35 30 25 20 15 10 5

x 5

10

15

20

25

30

35

40

45

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 103

11 Statistics: regression and correlation Chapter

CORRELATION

The strength of the relationship between the two variables represented in a scatter diagram, depends on how close the points lie to the line of best fit. The closer the points lie to this line, the stronger the relationship or correlation. Correlation (tendency of the graph) can be described in terms of the general distribution of data points, as follows:

Strong positive Correlation

Strong negative

Fairly strong positive

Fairly strong negative

Perfect positive

No correlation

Perfect negative

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 104

11 Statistics: regression and correlation Chapter

CORRELATION COEFFICIENT

The correlation between two variables can also be described in terms of a number, called the correlation coefficient. The correlation coefficient, đ?‘&#x;, indicates the strength and the direction of the correlation between two variables. This number can be anything between −1 and 1.

đ?’“

đ?&#x;?

Perfect positive relationship

đ?&#x;Ž, đ?&#x;—

Strong positive relationship

đ?&#x;Ž, đ?&#x;?

Weak positive relationship

đ?&#x;Ž, đ?&#x;“ đ?&#x;Ž

Fairly strong positive relationship No relationship

−đ?&#x;Ž, đ?&#x;?

Weak negative relationship

−đ?&#x;Ž, đ?&#x;—

Strong negative relationship

−đ?&#x;Ž, đ?&#x;“

Example

Interpretation

−đ?&#x;?

Fairly weak negative relationship Perfect negative relationship

Refer to the previous example again. For the given data set đ?‘&#x; = 0,958 which means that there is a strong positive relationship between the two variables.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 105

11 Statistics: regression and correlation Chapter

Mixed Exercise on Statistics 1

A national soccer team has participated against teams of other countries in a competition for the past 14 years. Their results were as follows:

a

b

YEAR

MATCHES PLAYED

WINS

DRAWS LOSSES GOALS GOALS FOR AGAINST

1999

5

3

2

0

11

3

2000

3

1

1

1

2

22

2001

5

3

1

1

10

4

2002

4

2

0

2

8

6

2003

7

2

3

2

5

4

2004

7

6

1

0

14

5

2005

5

2

0

3

8

7

2006

7

5

1

1

15

4

2007

6

1

2

3

9

11

2008

4

2

1

1

4

2

2009

3

1

1

1

2

3

2010

3

1

0

2

5

10

2011

1

0

0

1

2

3

2012

5

4

0

1

18

9

Determine the quartiles for: i

the matches played

ii

the wins

iii

the goals scored against the soccer team.

Draw a box and whisker plot for the goals against the soccer team and comment on the distribution of the data.

c

Calculate the mean of the number of matches played.

d

Calculated the standard deviation of the number of matches played.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 106

11 Statistics: regression and correlation Chapter

2

Fifty people were asked what percentage of their December holiday expenses were related to transport costs. The responses were as follows: PERCENTAGE 10 < � ≤ 20

FREQUENCY (f) 6

20 < � ≤ 30

14

50 < � ≤ 60

3

30 < � ≤ 40

16

40 < � ≤ 50

11

a

Draw an ogive to represent the data above.

b

Use your ogive to determine the median percentage of the holiday expenses spent on travel expenses.

c

Calculate the estimated mean.

d

Calculate the standard deviation of the data.

3

An athlete’s ability to take and use oxygen is called his VO2 max. The following table shows the VO2 max and the distance eleven atheletes can run in an hour. VO2 max

a

20

Distance(km) 8

55

30

25

40 30

50

40 35

30

50

18

13

10

11

16

14

9

15

12

13

Represent the data on a scatter graph. b

Determine the equation of the line of best fit.

c

Draw the line of best fit on the scatter graph.

d

Use your line of best fit to predict the VO2 max of an athlete that runs 19 km.

e

Determine the correlation coefficient of the data and comment on the correlation.

4

Five number 4; 8; 10; đ?‘Ľ and đ?‘Ś have a mean of 10 and a standard deviation of 4. Find đ?‘Ľ

5

and đ?‘Ś.

The standard deviation of five numbers is 7,5. Each number is increased by 2. What will the standard deviation of the new set of numbers be? Explain your answer.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 107

12 Probability Chapter

Overview

Unit 1 Page 282 Solving probability problems

Chapter 12 Page 280 Probability

• • •

Venn diagrams Tree diagrams Two-way contingency tables

Unit 2 Page 288 The counting principle

•

The fundamental counting principle

•

Using the counting principle to calculate probability

Unit 3 Page 292 The counting principle and probability

REMEMBER YOUR STUDY APPROACH SHOULD BE: 1 Work through all examples in this chapter of your Learner’s Book. 2 Work through the notes in this chapter of the study guide. 3 Do the exercises at the end of the chapter in the Learner’s Book. 4 Do the mixed exercises at the end of this chapter in the study guide.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 108

12 Probability Chapter

SUMMARY OF THEORY ON PROBABILITY CONCEPT/DEFINITION

MATHEMATICAL NOTATION/RULE

Probability = the chance that an event will occur

đ?‘ƒ

Sample Space = the set of all possible outcomes

�

The number of elements in the sample space General rule for A and B inside the sample space S Intersection

Union

�(�)

đ?‘ƒ(đ??´đ?‘œđ?‘&#x;đ??´) = đ?‘ƒ(đ??´) + đ?‘ƒ(đ??´) − đ?‘ƒ(đ??´đ?‘Žđ?‘–đ?‘‘đ??´)

Values of probability can range from 0 to1 • For an event, K, that is certain NOT to happen đ?‘ƒ(đ??ž) = 0 • For an event, K that is CERTAIN to happen đ?‘ƒ(đ??ž) = 1 If đ?‘† = {2; 4; 6} then đ?‘–(đ?‘†) = 3

đ??´ đ?‘Žđ?‘–đ?‘‘ đ??´ or đ??´ ∊ đ??´

đ??´ đ?‘œđ?‘&#x; đ??´ or đ??´ âˆŞ đ??´

Inclusive events have elements in common

đ?‘ƒ(đ??´ ∊ đ??´) ≠0

Mutually exclusive/disjoint events DON’T INTERSECT, i.e. have NO elements in common

đ?‘ƒ(đ??´ ∊ đ??´) = 0

Exhaustive events = together they contain ALL elements of S

EXAMPLE

A

B

∴ đ?‘ƒ(đ??´đ?‘œđ?‘&#x;đ??´) = đ?‘ƒ(đ??´) + đ?‘ƒ(đ??´) ∴ đ?‘ƒ(đ??´ ∊ đ??´) = 1

Complement of A = all elements which are NOT in A

Complement of A = A’

Complementary events = mutually exclusive and exhaustive (everything NOT in A, is in B)

đ?‘ƒ(đ?‘–đ?‘œđ?‘Ą đ??´) = 1 − đ?‘ƒ(đ??´) đ?‘ƒ(đ??´â€˛ ) = 1 − đ?‘ƒ(đ??´) Or đ?‘ƒ(đ??´â€˛ ) + đ?‘ƒ(đ??´) = 1

Independent events = outcome of 1st event DOES NOT influence the outcome of 2nd event

đ?‘ƒ(đ??´ ∊ đ??´) = đ?‘ƒ(đ??´) Ă— đ?‘ƒ(đ??´)

Tossing a coin and throwing a die

Dependent events = outcome of 1st event DOES influence the outcome of 2nd event

đ?‘ƒ(đ??´ ∊ đ??´) ≠đ?‘ƒ(đ??´) Ă— đ?‘ƒ(đ??´)

Choosing a ball from a bag, not replacing it, then choosing a 2nd ball

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 109

12 Probability Chapter

FACTORIAL NOTATION The product 5 Ă— 4 Ă— 3 Ă— 2 Ă— 1 can be written as 5! ∴ đ?‘–! = đ?‘– Ă— (đ?‘– − 1) Ă— (đ?‘– − 2) Ă— ‌ Ă— 3 Ă— 2 Ă— 1 The Fundamental Counting Principle RULE

EXAMPLE

RULE 1 Where there are đ?’Ž ways to do one thing and đ?’” ways to do another, then there are đ?’Ž Ă— đ?’” ways to do both

a) You have 3 pants and 4 shirts. That means you have 3 Ă— 4 = 12 different outfits.

RULE 3 Where đ?’” different things have to be placed in đ?’“ đ?’”! positions, the number of arrangements is (đ?’”−đ?’“)!

c) 8 students participated in a 100 m race. The first three positions can be occupied in 8! 8! = 5! = 8 Ă— 7 Ă— 6 = 336 ways. (8−3)!

RULE 2 Where đ?’” different things have to be placed in đ?’” positions, the number of arrangements is đ?’”!

b) 5 children have to be seated on 5 chairs in the front row of a class. The number of ways they can be seated is 5!=120

RULE 4 When seating đ?’ƒ boys and đ?’ˆ girls in a row, the number of arrangements are: • Boys and girls in any order: (đ?’ƒ + đ?’ˆ)! ways • Boys together and girls together: đ?&#x;? Ă— đ?’ƒ! Ă— đ?’ˆ! ways • Only girls together: (đ?’ƒ + đ?&#x;?)! Ă— đ?’ˆ! arrangements • Only boys together: (đ?’ˆ + đ?&#x;?)! Ă— đ?’ƒ! arrangements

d) 3 girls and 4 boys have to be seated on 7 chairs, with girls together and boys together. Number of ways = 2 Ă— 3! Ă— 4! = 288 e)5 Maths books and 2 Science books have to be place on a shelf, but the Maths books have to be placed together. Number of ways = (2 + 1)! Ă— 5! = 360

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 110

12 Probability Chapter

LETTER ARRANGEMENTS

When making new words from the letters in a given word , one has to distinguish between:

Treating repeated letters as

Treating repeated letters as IDENTICAL.

DIFFERENT letters.

The following rule applies:

The normal counting

For � letters of which �1 are identical, �2 are identical, ‌ and �� are identical, the number of arrangements is given by:

principle (Rule 2) applies here.

�! �1 ! × �2 ! × ‌ × �� !

Examples: 1

How many different arrangements can be made with the letters of the word MATHEMATICS, if repeated letters are treated as different letters. The letters are regarded as 11 different letters. Number of arrangements 11!

2

How many different arrangements can be made with the letters of the word MATHEMATICS, if repeated letters are treated as identical. The letters are regarded as 11 different letters. 11!

Number of arrangements = 2!Ă—2!Ă—2! = 6 652 800 (The M, A and T repeat)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 111

12 Probability Chapter

Mixed Exercise on Probability

1

How many different 074- cell phone numbers are possible if the digits may not repeat?

2

How many different 082- cell phone numbers are possible if the digits may only be integers?

3

What is the probability that you will draw a queen of diamonds from a pack cards?

4

How many different arrangements can be made with the letters of the word TSITSIKAMMA, if:

5

a

repeating letters are regarded as different letters

b

repeating letters are regarded as identical.

Four different English books, three different German books and two different Afrikaans books are randomly arranged on a shelf. Calculate the number of arrangements if:

6

a

the English books have to be kept together

b

all books of the same language have to be kept together

c

the order of the books does not matter.

In how many different ways can a chairman and a vice-chairman be chosen from a committee of 12 people?

7

The letters of the word MATHEMATICS have to be rearranged. Calculate the probability that the “word” formed will not start and end with the same letter.

8

In how many different ways can the letters of the word MATHEMATICS rearranged so that a

the H and the E stay together.

b

the E keep its position.

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 112

Answers to Mixed Exercises

Chapter 1: Number patterns, sequences and series 1

a

b

2

a

đ?‘‡đ?‘› = đ?‘Ž + (đ?‘– − 1)đ?‘‘ đ?‘Ž=5; đ?‘‘ =4 đ?‘‡đ?‘› = 5 + (đ?‘– − 1)4 = 4đ?‘– + 1 217 = 4đ?‘– + 1 4đ?‘– = 216 ∴ đ?‘– = 54 9 = đ?‘Žđ?‘&#x; 4 729 = đ?‘Žđ?‘&#x; 8

729 9 4

b

3

a

b 4

a

đ?‘Žđ?‘&#x; 8

= đ?‘Žđ?‘&#x; 4

đ?‘&#x; = 81 đ?‘&#x; = Âą3 đ?‘‡10 = đ?‘&#x; Ă— đ?‘‡9 đ?‘‡10 = Âą2187

đ?‘‡2 − đ?‘‡1 = đ?‘‡3 − đ?‘‡2 ďż˝5đ?‘Ľ − (2đ?‘Ľ − 4)ďż˝ = ďż˝(7đ?‘Ľ − 4) − 5đ?‘Ľďż˝ 5đ?‘Ľ − 2đ?‘Ľ − 7đ?‘Ľ + 5đ?‘Ľ = −4 − 4 đ?‘Ľ = −8 −20 ; −40 ; −60

đ?‘‡đ?‘› = đ?‘Žđ?‘–2 + đ?‘?đ?‘– + đ?‘? 3

đ?‘Ž=2

3

1

đ?‘? = 5 − 3 ďż˝2ďż˝ = 2 3

1

đ?‘? =2−2−2=0 b

3

1

�� = �2� �2 + �2� � 3

Note: alternative methods can be used

1

260 = �2� �2 + �2� �

3đ?‘–2 + đ?‘– − 520 = 0 (3đ?‘– + 40)(đ?‘– − 13) = 0 đ?‘– = 13 13đ?‘Ąâ„Ž đ?‘Ąđ?‘?đ?‘&#x;đ?‘š đ?‘–đ?‘ đ?‘?đ?‘žđ?‘˘đ?‘Žđ?‘™ đ?‘Ąđ?‘œ 260.

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 113

Answers to Mixed Exercises

5

6

7

𝑇𝑛 = 𝑎 + (𝑖 − 1)𝑑 𝑎 = 17 ; 𝑑 = −3 −2785 = 17 + (𝑖 − 1)(−3) −2802 = (𝑖 − 1)(−3) 934 = (𝑖 − 1) 𝑖 = 935 The sequence has 935 terms. a b

a

𝑇𝑛 = 𝑖2 𝑇𝑛 = 𝑎𝑖2 + 𝑏𝑖 + 𝑐 𝑎 =4÷2=2 𝑏 = 8 − 3(2) = 2 𝑐 =4−2−2 =0 ∴ 𝑇𝑛 = 2𝑖2 + 2𝑖

𝑇1 = 3 ; 𝑇2 = −2 ; 𝑇3 = −7 𝑛 𝑆𝑛 = [2𝑎 + (𝑖 − 1)𝑑] 2

𝑎 = 3 ; 𝑑 = −5 b

𝑆30 =

9

2

[2(3) + (30 − 1)(−5)]

𝑆30 = −2085 1

𝑇1 = 2 ; 𝑇2 = 1 ; 𝑇3 = 2 𝑆9 =

8

30

1 9 �2 −1� 2

2−1

𝑆9 = 255,5

𝑖=6 𝑇𝑛 = 1 + (𝑖 − 1)4 = 4𝑖 − 3 1 + 5 + 9 + ⋯ + 21 = ∑6𝑘=1 4𝑘 − 3

a

𝑇5 = 0 ; 𝑇13 = 12 0 = 𝑎 + 4𝑑 …(1) 12 = 𝑎 + 12𝑑 …(2) (2)-(1): 12 = 8𝑑 3

𝑑=2

b

𝑆21 =

21 2

3

𝑎 = −4 �2� = −6 3

�2(−6) + (21 − 1) �2��

𝑆21 = 189

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 114

Answers to Mixed Exercises

10

a

For it to be a converging sequence −1 < đ?‘&#x; < 1. đ?‘‡

đ?‘&#x; = đ?‘‡2 = đ?‘&#x;= b

ďż˝đ?‘Ľ 2 −9ďż˝

đ?‘Ľ+3 1 (đ?‘Ľ+3)(đ?‘Ľâˆ’3) đ?‘Ľ+3

đ?‘&#x; =đ?‘Ľâˆ’3 ∴ −1 < đ?‘Ľ − 3 < 1 2<đ?‘Ľ<4 đ?‘Ž đ?‘†âˆž = 1−đ?‘&#x; 13 =

(đ?‘Ľ+3)

1−(đ?‘Ľâˆ’3) (đ?‘Ľ+3)

13 = (−đ?‘Ľ+4)

13(−đ?‘Ľ + 4) = (đ?‘Ľ + 3) −13đ?‘Ľ + 52 = đ?‘Ľ + 3 −13đ?‘Ľ − đ?‘Ľ = 3 − 52 −14đ?‘Ľ = −49 7

11

đ?‘Ľ=2

For series in numerator: 99 = 1 + (đ?‘– − 1)2 đ?‘– = 50 terms đ?‘†50 =

50

[2(1) + (50 − 1)2] = 2 500

�50 =

50

[2(201) + (50 − 1)2] = 12 500

2

For series in denominator: 299 = 201 + (đ?‘– − 1)2 đ?‘– = 50 terms

12

13

2

Value =

2 500

12 500

1

=5

đ?‘‡9 = đ?‘†9 − đ?‘†8 đ?‘†9 = 3(9)2 − 2(9) = 225 đ?‘†8 = 3(8)2 − 2(8) = 176 ∴ đ?‘‡9 = 225 − 176 = 49 a

Let đ?‘&#x; =constant ratio 7đ?‘&#x; 3 = 189 đ?‘&#x; 3 = 27 đ?‘&#x;=3 đ?‘Ľ = 7 Ă— 3 = 21 đ?‘Ś = 21 Ă— 3 = 63

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 115

Answers to Mixed Exercises

b

206 668 = 206 668 =

7(3đ?‘› −1) 3−1 7(3đ?‘› −1) 2 đ?‘›

413 336 = 7(3 − 1) 59 048 = 3đ?‘› − 1 3đ?‘› = 59 049 3đ?‘› = 310 ∴ đ?‘– = 10

Chapter 2: Functions 1

2

2đ?‘Ľ − 3đ?‘Ś = 17 ‌ (1) 3đ?‘Ľ − đ?‘Ś = 15 ‌ (2) (2) Ă— 3: 9đ?‘Ľ − 3đ?‘Ś = 45 ‌ (3) (1) − (3) : − 7đ?‘Ľ = −28 đ?‘Ľ=4 Substitute into (1): 2(4) − 3đ?‘Ś = 17 đ?‘Ś = −3 Intercept is (4; −3) a

b

c

d

đ?‘Ś = đ?‘šđ?‘Ľ + 3 Substitute (−3; 0) 0 = đ?‘š(−3) + 3 đ?‘š=1 ∴ đ?‘Ś =đ?‘Ľ+3 đ?‘Ś = đ?‘šđ?‘Ľ + 1 Substitute (2; −1): −1 = đ?‘š(2) + 1 đ?‘š = −1 đ?‘”: đ?‘Ś = −đ?‘Ľ + 1 đ?‘Ľ + 3 = −đ?‘Ľ + 1 2đ?‘Ľ = −2 đ?‘Ľ = −1 Substitute đ?‘Ľ = −1: đ?‘Ś = −1 + 3 = 2 ∴ đ?‘ƒ(−1; 2) Yes, because the products of their gradients is −1. (−1 Ă— 1 = −1)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 116

Answers to Mixed Exercises

e 3

a

b

c

d

4

a

b c

đ?‘Ś = −đ?‘Ľ − 2

Let đ?‘Ś = 0: 0 = đ?‘Ľ 2 − 2đ?‘Ľ − 3 0 = (đ?‘Ľ − 3)(đ?‘Ľ + 1) ∴ đ?‘Ľ = 3 đ?‘œđ?‘&#x; đ?‘Ľ = −1 đ??´(−1; 0) and đ??´(3; 0) Let đ?‘Ľ = 0: đ?‘Ś = (0)2 − 2(0) − 3 đ?‘Ś = −3 ∴ đ??ś(0; −3) đ?‘‚đ??´ = 1 đ?‘˘đ?‘–đ?‘–đ?‘Ą đ?‘‚đ??´ = 3 đ?‘˘đ?‘–đ?‘–đ?‘Ąđ?‘ đ?‘‚đ??ś = 3 đ?‘˘đ?‘–đ?‘–đ?‘Ąđ?‘ đ?‘Ľ=

−đ?‘? 2đ?‘Ž

2

= 2(1) = 1

Substitute đ?‘Ľ = 1: đ?‘Ś = (1)2 − 2(1) − 3 = −4 đ??ˇ(1; −4) đ?‘? = −3 đ?‘š=

−3−0 0−3

đ?‘š=1 For the graph to have only one real root it has to move 4 units up. đ?‘Ś = đ?‘Ľ 2 − 2đ?‘Ľ − 3 + 4 = đ?‘Ľ 2 − 2đ?‘Ľ + 1 ∴đ?‘˜=1 Let đ?‘Ś = 0: 0 = −2(đ?‘Ľ + 1)2 + 8 0 = −2đ?‘Ľ 2 − 4đ?‘Ľ + 6 0 = (−2đ?‘Ľ + 2)(đ?‘Ľ + 3) đ?‘Ľ = 1 đ?‘œđ?‘&#x; đ?‘Ľ = −3 đ??´(−3; 0) and đ??´(1; 0) đ??´đ??´ = 4 units đ??ś(−1; 8) đ?‘Ľ = 0, đ?‘Ś = 6 đ??ˇ(0 ; 6) đ??ˇ(−2; 6) ∴ đ??ˇđ??ˇ = 2 units

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 117

Answers to Mixed Exercises 5

a

y

x

−1 −1

−2

b c 6

a

𝑥∈𝑅 𝑥 ≤ −1

𝑎

Substitute the point A into the equation 𝑦 = 𝑥 𝑎

b c 7

a b

c d

e

f

2 = −2

𝑎 = −4 𝐴(2; −2) −4

𝑦 = 𝑥−1 + 2

𝑦 = −(0)2 − 2(0) + 8 = 8 𝐴(0; 8) 0 = −𝑥 2 − 2𝑥 + 8 0 = (−𝑥 + 2)(𝑥 + 4) 𝑥 = 2 𝑜𝑟 𝑥 = −4 𝐴(−4; 0) and 𝐶(2; 0) 𝐷(−1; 0) 𝐶𝐷 = 3 𝑢𝑖𝑖𝑡𝑠 𝑥 = −1 𝑦 = −(−1)2 − 2(−1) + 8 𝑦 = −1 + 2 + 8 = 9 𝐷(−1; 9) 𝐷𝐷 = 9 𝑢𝑖𝑖𝑡𝑠 𝐴(0; 8) 𝐷(−2; 8) 𝐴𝐷 = 2 𝑢𝑖𝑖𝑡𝑠 1

−𝑥 2 − 2𝑥 + 8 = 2 𝑥 − 1

−2𝑥 2 − 4𝑥 + 16 = 𝑥 − 2 −2𝑥 2 − 5𝑥 + 18 = 0 (2𝑥 + 9)(−𝑥 + 2) = 0 𝑥= 𝑥=

−9 2 −9 2

𝑜𝑟 𝑥 = 2

at H

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 118

Answers to Mixed Exercises Substitute đ?‘Ľ = 1 −9

đ?‘Ś = 2ďż˝ 2 ďż˝âˆ’ 1 đ?‘Ś=−

13

−9 2

1

into the equation đ?‘Ś = 2 đ?‘Ľ − 1

4 −9 −13

g

∴ đ??şďż˝2 ;

4

ďż˝

đ??şđ??ť = 3,25 đ?‘˘đ?‘–đ?‘–đ?‘Ąđ?‘

1

đ?‘“(đ?‘Ľ) − đ?‘”(đ?‘Ľ) = −đ?‘Ľ 2 − 2đ?‘Ľ + 8 − ďż˝2 đ?‘Ľ − 1ďż˝ 5

= −đ?‘Ľ 2 − 2 đ?‘Ľ + 9

Minimum at turning point:

h i

đ?‘Ľ=

5 2

5

= −4

−2

đ?‘“(đ?‘Ľ) − đ?‘”(đ?‘Ľ) > 0 9

8

a b

c d

5 2

5 −5

đ?‘…đ?‘†đ?‘šđ?‘Žđ?‘Ľ = − ďż˝âˆ’ 4ďż˝ − 2 ďż˝ 4 ďż˝ + 9 = −2 < đ?‘Ľ < 2

169 16

∴ đ?‘“(đ?‘Ľ) > đ?‘”(đ?‘Ľ)

đ?‘Ś = −4 đ?‘Ś = đ?‘?đ?‘Ľ + đ?‘? đ?‘? = −4 đ?‘Ś = đ?‘?đ?‘Ľ − 4 Substitute the point (2; 5) into the equation: 5 = đ?‘?2 − 4 đ?‘?2 = 9 đ?‘?=3 đ?‘Ś = 3đ?‘Ľ − 4 đ?‘Ś = −1 ; đ?‘Ľ = −2 đ?‘Ž

đ?‘Ś = đ?‘Ľ+2 − 1

Substitute the point đ??´(0; −3): đ?‘Ž

−3 = 0+2 − 1 đ?‘Ž

−3 = 2 − 1 đ?‘Ž 2

e

f

= −2

đ?‘Ž = −4 −4

đ?‘Ś = đ?‘Ľ+2 − 1

Substitute (−2; −1) into đ?‘Ś = đ?‘Ľ + đ?‘˜1 and đ?‘Ś = −đ?‘Ľ + đ?‘˜2 −1 = −2 + đ?‘˜1 −1 = 2 + đ?‘˜2 đ?‘˜1 = 1 đ?‘˜2 = −3 đ?‘Ś =đ?‘Ľ+1 đ?‘Ś = −đ?‘Ľ − 3 đ?‘Ľ > −2; đ?‘Ľ ≠0

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 119

Answers to Mixed Exercises

9

a

b

10

a

đ?‘Ś = 2đ?‘Ľ 2 đ?‘Ľ = 2đ?‘Ś 2 đ?‘Ľ đ?‘Ś2 = 2

đ?‘Ľ

� = ¹�2

đ?‘Ľ ≤ 0 đ?‘œđ?‘&#x; đ?‘Ľ ≼ 0 đ?‘Ś = đ?‘Žđ?‘Ľ Substitute point đ??´: 3 = đ?‘Žâˆ’1 1

đ?‘Ž=3

c

f

y

1 đ?‘Ľ

x

� = �3� b 11

y=x

đ?‘“ −1 : đ?‘Ś = log ďż˝1ďż˝ đ?‘Ľ

d

3

đ?‘Ľ −intercept: (3; 0) đ?‘Ś −intercepđ?‘Ą: (0; −2)

đ?‘Ľ>0

Chapter 3: Logarithms 1

a b c d e

f

đ?‘Ľ = 32 = 9 1 2

1

đ?‘Ľ = ďż˝3ďż˝ = 9 log 4 đ?‘Ľ = −2

1

1

đ?‘Ľ = (4)−2 = (4)2 = 16 1

1

đ?‘Ľ = (5)−2 = (5)2 = 25 đ?‘Ľ 3 = 106 đ?‘Ľ = 102 đ?‘Ľ = 100 81 = 3đ?‘Ľ 3đ?‘Ľ = 34 đ?‘Ľ=4

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 120

Answers to Mixed Exercises

g

2

a b c d

3

a

c

4

a

b c

1 9

= 3đ?‘Ľ

3đ?‘Ľ = 3−2 đ?‘Ľ = −2 3

đ?‘Ž=2

9

4

= đ?‘Ž2

đ?‘“ −1 : đ?‘Ś = log ďż˝3ďż˝ đ?‘Ľ 2

3 −đ?‘Ľ

�(�) = �2�

â„Ž(đ?‘Ľ) = − log ďż˝3ďż˝ đ?‘Ľ 2

i) đ?‘”(đ?‘Ľ) = − log 2 đ?‘Ľ ii) đ?‘?(đ?‘Ľ) = log 2 (−đ?‘Ľ) iii) đ?‘ž(đ?‘Ľ) = − log 2 (−đ?‘Ľ) iv) đ?‘“ −1 : đ?‘Ś = 2đ?‘Ľ v) đ?‘”−1 : đ?‘Ś = 2−đ?‘Ľ vi) â„Ž(đ?‘Ľ) = log 2 (đ?‘Ľ + 2) For đ?‘“ −1 đ?‘Žđ?‘–đ?‘‘ đ?‘”−1 : Domain đ?‘Ľ ∈ đ?‘… ; Range đ?‘Ś > 0

b

đ?‘Ś −coordinate = 0 0 = log đ?‘? đ?‘Ľ đ?‘Ľ = đ?‘?0 = 1 đ??´(1; 0) Because graph is increasing as đ?‘Ľ increases. Substitute đ??´: 3

8 = đ?‘?2 2 3

d e

9

Substitute �2; 4�:

2 3 3 2

(8) = ďż˝đ?‘? ďż˝ 2

3 2

= log đ?‘? 8

2

đ?‘? = (8)3 = (23 )3 = 22 = 4 đ?‘”(đ?‘Ľ) = 4đ?‘Ľ Substitute đ?‘Ś = −2: −2 = log 4 đ?‘Ľ 1

đ?‘Ľ = 4−2 = 16

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 121

Answers to Mixed Exercises

Chapter 4: Finance, growth and decay 1

a

b

đ??´ = đ?‘ƒ(1 + đ?‘–. đ?‘–)

đ??´ = 15 000ďż˝1 + (0,106)(5)ďż˝ đ??´ = đ?‘…22 950 đ??´ = đ?‘ƒ(1 + đ?‘–)đ?‘› 20

2

a b

đ??´ = 15 000ďż˝1 + (0,024)ďż˝ đ??´ = đ?‘…24 104,07 It is better to invest it at 9.6% p.a , interest compounded quarterly. Nominal interest rate đ??´ = đ?‘ƒ(1 + đ?‘–)đ?‘› 95 000 = đ?‘ƒ ďż˝1 + đ?‘ƒ=

c 3

a

b

95 000

�1+

0,085 60 12

0,085 60 ďż˝ 12

ďż˝

đ?‘ƒ = đ?‘…62 202,48 đ?‘…95 000 − đ?‘…62 202,48 = đ?‘…32 797,52 đ??´ = đ?‘ƒ(1 + đ?‘–)đ?‘›

đ??´ = 8 000(1 + 0.06)2 đ??´ = đ?‘…8 988,80

đ??ˇ= đ??ˇ=

đ?‘Ľ[(1+đ?‘–)đ?‘› −1] đ?‘–

[1 + đ?‘–]

0,07 4 2 000�� ďż˝ −1ďż˝ 2

0.035

�1 +

0,07 2

ďż˝

đ??ˇ = đ?‘…8 724,93 She will NOT have enough money to buy the TV in two years. 4

a

1 + ���� = �1 + 1 + ���� = �1 +

đ?‘–đ?‘›đ?‘œđ?‘š đ?‘š đ?‘š

ďż˝

0.0785 12 12

đ?‘–đ?‘’đ?‘“đ?‘“ = 0.08138 ‌ đ??ˇđ?‘“đ?‘“. đ?‘&#x;đ?‘Žđ?‘Ąđ?‘? = 8.14%

ďż˝

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 122

Answers to Mixed Exercises b

1 + 𝑖𝑒𝑓 = �1 +

𝑖𝑛𝑜𝑚 𝑚 𝑚

1 + 0,0925 = �1 + 4

�1,0925 = �1 +

1.022 − 1 = 5

𝑖𝑛𝑜𝑚

�

𝑖𝑛𝑜𝑚 4 4

𝑖𝑛𝑜𝑚

4

4

�

�

𝑖𝑛𝑜𝑚 = 0.0894 … Nom. rate= 8,95% p.a. compounded quarterly

𝐴 = 𝑃(1 + 𝑖)𝑛

179 200 = 350 000(1 − 𝑖)3 0,512 = (1 − 𝑖)3 3

6

1 − 𝑖 = �0,512 𝑖 = 0.2 Dep. rate= 20%

𝐴 = �20 000 �1 +

0.0975 7 4

� + 10 000 �1 +

0.0975

𝐴 = �23 672.43 + 10 243.75�(1.13185 … ) 𝐴 = 𝑅38 388,36 OR 𝐴 = �20 000 �1 +

7

4

� + 10 000� �1 +

0.0975 4

� �1 +

𝐴 = �23 109.142 + 10 000�(1.024375)(1.13 … ) 𝐴 = 𝑅38 388,36 a

b

0.0995 15 12

�

0.0995 15 12

�

𝐴 = 900 000(1 − 0.15)5 𝐴 = 𝑅399 334,78 𝐴 = 900 000(1 + 0.18)5 𝐴 = 𝑅2 058 981,98 𝑅2 058 981,98 − 𝑅399 334,78 = 𝑅1 659 647,20 1 659 647,20 = 𝑥=

8

0.0975 6

4

�� �1 +

𝑥�(1+0.02)61 −1�

0,02×1659647,20 [(1+0,02)61 −1]

0,02

𝑥 = 𝑅14 144,81

𝐴 = 𝑃(1 + 𝑖. 𝑖) 2 yrs = 24 months (24 × 85) = 1 500(1 + 𝑖. 2) 𝑖 = 0.18 rate= 18%

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 123

Answers to Mixed Exercises

9

a

đ?‘ƒ= đ?‘ƒ=

b

đ?‘Ľ[1−(1+đ?‘–)−đ?‘› ]

đ?‘– ďż˝6 500��1−(1+0,01)−240 ďż˝ 0,01

đ?‘ƒ = đ?‘…590 326,21 đ?‘ƒ=

ďż˝6 500��1−(1+0,01)−96 ďż˝ 0,01

đ?‘ƒ = đ?‘…399 930,07

10

đ??ˇ=

0,01 73 ďż˝ −1ďż˝ 12 0,01 12

1 000��1+

đ??ˇ = đ?‘…99 915,81 đ??´ = đ?‘ƒ(1 + đ?‘–)đ?‘›

đ??´ = 99 915,81 ďż˝1 + 11

đ??´ = đ?‘…104 147,21 đ??ˇ=

đ?‘Ľ[(1+đ?‘–)đ?‘› −1] đ?‘–

48 000 =

0,01 5 12

ďż˝

0,09 đ?‘› ďż˝ −1ďż˝ 12 0,09 12

300��1+

2,2 = (1,0075)đ?‘› đ?‘– = log1,0075 2,2 đ?‘– = 106 8 years and 10 months 12

đ??´ = đ?‘ƒ(1 + đ?‘–. đ?‘–)

đ??´ = 13 500ďż˝1 + (0.12)(4)ďż˝

13

đ??´ = đ?‘…19 980 Repayment = đ?‘…19 980 á 48 = đ?‘…416,25 Including insurance= đ?‘…416,25 + đ?‘…30 = đ?‘…446,25

đ??´ = 400 000(1 + 0,02)4 đ??´ = đ?‘…432 972,86 (amount owing after 1 year) đ?‘ƒ=

đ?‘Ľ[1−(1+đ?‘–)−đ?‘› ] đ?‘–

432 972,86 =

đ?‘Ľďż˝1−(1+0,02)−16 ďż˝

đ?‘Ľ = đ?‘…31 888,51

0,02

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 124

Answers to Mixed Exercises

Chapter 5: Compound angles 1

a

2 cos 2 x = −1 ∴ 𝑐𝑜𝑠2𝑥 = −

b

2𝑥 = ±120° + 𝑘. 360°; 𝑘 ∈ 𝑍 ∴ 𝑥 = ±60° + 𝑘. 180°; 𝑘 ∈ 𝑍 𝑠𝑖𝑖𝑥 = 3𝑐𝑜𝑠𝑥

𝑠𝑖𝑛𝑥 𝑎𝑜𝑠𝑥

c

d

e

f

1 2

=3

𝑡𝑎𝑖𝑥 = 3 ∴ 𝑥 = 71,47° + 𝑘. 180°; 𝑘 ∈ 𝑍 𝑠𝑖𝑖𝑥 = 𝑐𝑜𝑠3𝑥 ∴ cos(90° − 𝑥) = 𝑐𝑜𝑠3𝑥 90° − 𝑥 = ±3𝑥 + 𝑘. 360° 4𝑥 = 90° + 𝑘. 360° or 2𝑥 = −90° + 𝑘. 360° 𝑥 = 22,5° + 𝑘. 90° 𝑥 = −45° + 𝑘. 180°; 𝑘 ∈ 𝑍 3 6 − 10𝑐𝑜𝑠𝑥 − 3(1 − 𝑐𝑜𝑠 𝑥) = 0 ∴ 3𝑐𝑜𝑠 3 𝑥 − 10𝑐𝑜𝑠𝑥 + 3 = 0 ∴ (3𝑐𝑜𝑠𝑥 − 1)(𝑐𝑜𝑠𝑥 − 3) = 0 ∴ 𝑐𝑜𝑠𝑥 =

1 3

or 𝑐𝑜𝑠𝑥 = 3 (no solution)

𝑡𝑎𝑖𝑥 = −

1 2

or 𝑡𝑎𝑖𝑥 = 1

∴ 𝑥 = ±70,53° + 𝑘. 360°; 𝑘𝜖𝑍 For 𝑥 ∈ [−360°; 360°] 𝑥 ∈ {−289,47°; −70,53°; 289,47°} 2(𝑠𝑖𝑖2 𝑥 + 𝑐𝑜𝑠 2 𝑥) − 𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 3𝑐𝑜𝑠 2 𝑥 = 0 2𝑠𝑖𝑖2 𝑥 − 𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠 2 𝑥 = 0 (2𝑠𝑖𝑖𝑥 + 𝑐𝑜𝑠𝑥)(𝑠𝑖𝑖𝑥 − 𝑐𝑜𝑠𝑥) = 0 𝑥 = −26,57° + 𝑘. 180° or 𝑥 = 45° + 𝑘. 180°; 𝑘 ∈ 𝑍 3(𝑠𝑖𝑖2 𝑥 + 𝑐𝑜𝑠2 𝑥) − 8𝑠𝑖𝑖𝑥 + 16𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 − 6𝑐𝑜𝑠𝑥 = 0 ∴ 3 − 6𝑐𝑜𝑠𝑥 − 8𝑠𝑖𝑖𝑥 + 16𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 = 0 3(1 − 2𝑐𝑜𝑠𝑥) − 8𝑠𝑖𝑖𝑥(1 − 2𝑥𝑜𝑠𝑥) = 0 (1 − 2𝑐𝑜𝑠𝑥)(3 − 8𝑠𝑖𝑖𝑥) = 0 𝑐𝑜𝑠𝑥 =

1 2

or 𝑠𝑖𝑖𝑥 =

3 8

∴ 𝑥 = ±60° + 𝑘. 360° or 𝑥 = 22,02° + 𝑘. 360° or 𝑥 = 157,98° + 𝑘. 360°; 𝑘 ∈ 𝑍

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 125

Answers to Mixed Exercises 2

a

LHS= cos x +

sin x × sin x cos x

cos 2 x + sin 2 x cos x 1 = cos x =

=RHS Not valid for x = 90 0 + k .180 0 ; k ∈ Z b

LHS =

sin 2 θ − cos θ (1 − cos θ ) sin 2 θ + cos 2 θ − cos θ 1 − cos θ 1 = = = (1 − cos θ )sin θ sin θ (1 − cos θ )sin θ (1 − cos θ )sin θ

=RHS Not valid for θ = k .180 0 ; k ∈ Z c

LHS =

1 − cos 2 x sin 2 x sin x . sin x = tan x. sin x = RHS = = cos x cos x cos x

Not valid for x = 90 0 + k .180 0 ; k ∈ Z d

LHS =

(

)

sin x sin 2 x + cos 2 x sin x = = tan x =RHS cos x cos x

Not valid for x = 90 0 + k .180 0 ; k ∈ Z

sin x cos x cos x + sin x cos x = cos x + sin x × × LHS = sin x cos x cos x − sin x cos x + sin x 1− cos x 2 cos x + 2 sin x cos x + sin 2 x 1 + 2 sin x cos x = = = RHS cos 2 x − sin 2 x cos 2 x − sin 2 x 1+

e

Not valid for x = ±45 0 + k .180 0 ; k ∈ Z

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 126

Answers to Mixed Exercises

đ??żđ??ťđ?‘† = sin(45° + đ?‘Ľ). sin(45° − đ?‘Ľ)

f

= (sin 45°. cos đ?‘Ľ + sin đ?‘Ľ. đ?‘?đ?‘œđ?‘ 45°) Ă— (sin 45°. cos đ?‘Ľ − sin đ?‘Ľ . cos 45°) √2

= ďż˝ 2 cos đ?‘Ľ + √2

2

√2 √2 sin đ?‘Ľďż˝ ďż˝ 2 cos đ?‘Ľ 2 √2

2

= ďż˝ 2 cos đ?‘Ľďż˝ − ďż˝ 2 sin đ?‘Ľďż˝ 1 2

−

√2 sin đ?‘Ľďż˝ 2

1 2

= đ?‘?đ?‘œđ?‘ 2 đ?‘Ľ − đ?‘ đ?‘–đ?‘–2 đ?‘Ľ 1 2

= (đ?‘?đ?‘œđ?‘ 2 đ?‘Ľ − đ?‘ đ?‘–đ?‘–2 đ?‘Ľ) 1 2

= cos 2đ?‘Ľ g

= đ?‘…đ??ťđ?‘† =

đ??żđ??ťđ?‘† =

sin 2đ?œƒâˆ’cos đ?œƒ sin đ?œƒâˆ’cos 2đ?œƒ

2 sin đ?œƒ.cos đ?œƒâˆ’cos đ?œƒ sin đ?œƒâˆ’(1−2đ?‘ đ?‘–đ?‘›2 đ?œƒ) cos đ?œƒ(2 sin đ?œƒâˆ’1)

= 2đ?‘ đ?‘–đ?‘›2 đ?œƒ+sin đ?œƒâˆ’1

cos đ?œƒ(2 sin đ?œƒâˆ’1) sin đ?œƒâˆ’1)(sin đ?œƒ+1)

= (2

= h

cos đ?œƒ sin đ?œƒ+1

= đ?‘…đ??ťđ?‘† = = =

=

đ??żđ??ťđ?‘† =

cos đ?‘Ľâˆ’cos 2đ?‘Ľ+2 3 sin đ?‘Ľâˆ’sin 2đ?‘Ľ

cos đ?‘Ľâˆ’ďż˝2đ?‘Žđ?‘œđ?‘ 2 đ?‘Ľâˆ’1ďż˝+2 3 sin đ?‘Ľâˆ’2 sin đ?‘Ľ.cos đ?‘Ľ

−2đ?‘Žđ?‘œđ?‘ 2 đ?‘Ľ+cos đ?‘Ľ+3 3 sin đ?‘Ľâˆ’2 sin đ?‘Ľ.cos đ?‘Ľ

(−2 cos đ?‘Ľ+3)(cos đ?‘Ľ+1) sin đ?‘Ľ(3−2 cos đ?‘Ľ)

cos đ?‘Ľ+1 sin đ?‘Ľ

= đ?‘…đ??ťđ?‘†

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 127

Answers to Mixed Exercises 3

4

)

sin 180 0 − x tan (− x ) sin x(− tan x ) = = −1 0 0 tan x(sin x ) tan 180 + x cos x − 90

b

sin 180 0 + x tan x − 360 0 sin x. tan x = = 2 sin x 0 0 0 − tan x(− 0,5)(1) tan 360 − x − cos 60 tan 45

a

cos 730 = cos 90 0 − 17 0 = sin 17 0 = k

b

cos(−163°) = đ?‘?đ?‘œđ?‘ 163° = −đ?‘?đ?‘œđ?‘ 17° = −√1 − đ?‘˜ 2

c d

5

(

a

a

(

(

) (

(

)(

) (

(

)

)(

)

)

)

đ?‘Ąđ?‘Žđ?‘–197° = đ?‘Ąđ?‘Žđ?‘–17° =

đ?‘˜

√1−đ?‘˜ 2

đ?‘?đ?‘œđ?‘ 326° = đ?‘?đ?‘œđ?‘ 34° = đ?‘?đ?‘œđ?‘ 2(17°) = 1 − 2đ?‘ đ?‘–đ?‘–2 17° = 1 − 2đ?‘˜ 2 4

đ?‘?đ?‘œđ?‘ đ?‘Ľ = − 5

3  3  5 sin x + 3 tan x = 5  + 3  5  − 4

= 3− b

6

9 3 = 4 4

or

 − 3  − 3 = 5  + 3   5  −4

or

= −5 +

9 3 =− 4 4

2 tan x 1 − tan 2 x  3  2  24 3 16 −4 3 16 24  =− Ă— =− ∴ tan 2 x = = or tan 2 x = Ă— 2 2 6 7 7 2 7  3  1−   −4 tan 2 x =

a

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 128

Answers to Mixed Exercises

cos(đ?‘Ľ − đ?‘Ś) = cos đ?‘Ľ. cos đ?‘Ś + sin đ?‘Ľ. sin đ?‘Ś =

= b

=

−2√2 −4 −1 −3 Ă— + Ă— 3 5 3 5

8√2 1 + 15 5 8√2+3 15

cos 2đ?‘Ľ − cos 2đ?‘Ś

= 1 − 2đ?‘ đ?‘–đ?‘–2 đ?‘Ľ − (1 − 2đ?‘ đ?‘–đ?‘–2 đ?‘Ś) = 2đ?‘ đ?‘–đ?‘–2 đ?‘Ś − 2đ?‘ đ?‘–đ?‘–2 đ?‘Ľ −3 2 5

−1 2 3

= 2ďż˝ ďż˝ − 2ďż˝ ďż˝ =

7

a b

c

18 2 − 25 9

=

112 225

cos 2(22,5°) = cos 45° =

1 2

1

√2

1 2

Ă— 2 sin 22,5°. cos 22,5° = Ă— sin 2(22,5°) 1 2

= sin 45° =

√2 4

sin 2(15°) = sin 30° =

1 2

Chapter 6: Solving problems in three dimensions 1

a b

2â„Ž

đ??źđ?‘– ∆đ??´đ??´đ??ˇ âˆś tan đ?›ź = đ??´đ??´

2â„Ž

∴ đ??´đ??ˇ = tan đ?›ź

â„Ž

đ??źđ?‘– ∆đ??śđ??ˇđ??ˇ: tan(90° − đ?›ź) = đ??´đ??´ đ??źđ?‘– ∆đ??´đ??ˇđ??ˇ:

∴ đ??ˇđ??ˇ = â„Ž tan đ?›ź

đ??´đ??ˇ2 = đ??´đ??ˇ 2 + đ??ˇđ??ˇ2 − 2(đ??´đ??ˇ)(đ??ˇđ??ˇ). cos đ??ˇ

= (2â„Ž. cot đ?›ź)2 + (â„Ž. tan đ?›ź)2 − 2(2â„Ž. cot đ?›ź)(â„Ž. tan đ?›ź) cos 120° 1

= 4â„Ž2 . đ?‘?đ?‘œđ?‘Ą 2 đ?›ź + â„Ž2 đ?‘Ąđ?‘Žđ?‘–2 đ?›ź − 4â„Ž2 (cot đ?›ź . tan đ?›ź) ďż˝âˆ’ 2ďż˝

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 129

Answers to Mixed Exercises = ℎ2 (4𝑐𝑜𝑡 2 𝛼 + 𝑡𝑎𝑖2 𝛼 + 2) 4

= ℎ2 �𝑡𝑎𝑛2 𝛼 + 𝑡𝑎𝑖2 𝛼 + 2�

=

ℎ2 �𝑡𝑎𝑛4 𝛼+2𝑡𝑎𝑛2 𝛼+4� 𝑡𝑎𝑛2 𝛼

ℎ√𝑡𝑎𝑛4 𝛼+2𝑡𝑎𝑛2 𝛼+4

𝐴𝐷 =

c

tan 𝛼

𝐴𝐴 tan 𝛼

ℎ = √𝑡𝑎𝑛4

𝛼+2𝑡𝑎𝑛2 𝛼+4

509.tan 42°

= √𝑡𝑎𝑛4

42°+2𝑡𝑎𝑛2 42°+4

𝐶𝐷 = 182,90 𝑚 2

� 𝐴 = 180° − 𝑠 − 30° = 150° − 𝑠 𝐶𝐷

a

𝑝

b

tan 𝑠 = 𝐸𝐴 𝐸𝐴

sin(150°−𝜃)

𝐶𝐴 = = =

8.sin(150°−𝜃) sin 𝜃

8.sin�180°−(150°−𝜃)� sin 𝜃

8.sin(30°+𝜃) sin 𝜃

8.sin(30°+𝜃)

𝑝=� 3

8

= sin 𝜃

∴ 𝑝 = 𝐶𝐴. tan 𝑠

sin 𝜃

� tan 𝑠 =

8.sin(30°+𝜃) cos 𝜃

𝐴𝐷 = 13(𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑎𝑠) 𝐴̂ = 180° − (𝛼 + 𝛽) 𝐸𝐴

13

∴ sin[180°−(𝛼+𝛽)] = sin 𝛼 𝐸𝐴

13

∴ sin(𝛼+𝛽) = sin 𝛼 ∴ 𝐶𝐷 =

13 sin(𝛼+𝛽) sin 𝛼

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 130

Answers to Mixed Exercises

4

a b

1

𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 = 2 𝑚. 𝑝 sin(180° − 𝑠) 1

𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 = 2 𝑖. 𝑝 sin 𝑠

𝐴𝑟𝑐𝑎 ∆𝐴𝐴𝐶 = 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 + 𝐴𝑟𝑐𝑎 ∆𝐴𝐷𝐶 1

1

= 2 𝑚𝑝 sin(180° − 𝑠) + 2 𝑖𝑝 sin 𝑠 1

1

= 2 𝑚𝑝. sin 𝑠 + 2 𝑖𝑝 . sin 𝑠 1

c

= 2 𝑝(𝑚 + 𝑖) sin 𝑠

1

12,6 = 2 (8,1)(5,9) sin 𝑠 sin 𝑠 = 0,527306968 …

𝑠 = 31,82° 𝑂𝑅 𝑠 = 180° − 31,82° = 148,18°

5

a

𝑝

sin 𝑠 = 𝐴𝐸

𝑝

b c

∴ 𝐴𝐶 = sin 𝜃

𝐴�1 = 180° − 2𝛼 𝐴𝐸

sin 𝐴�1

𝐴𝐸

= sin 𝐴

𝐴𝐸

= sin(180°−2𝛼)

𝐴𝐶 = 6

a

𝑝 sin 𝜃

sin 𝛼

𝑝 sin(180°−2𝛼) sin 𝜃.sin 𝛼

𝑝.sin 2𝛼

= sin 𝜃.sin 𝛼

𝑅� = 180° − 30° − (150° − 𝛼) = 𝛼 12

sin 𝑅� 12

sin 𝛼

𝑄𝑅

= sin(150°−𝛼) 𝑄𝑅

= sin(30°+𝛼)

𝑄𝑅 =

12 sin(30°+𝛼) sin 𝛼

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 131

Answers to Mixed Exercises = = b

=

12(sin 30°.cos đ?›ź+cos 30°.sin đ?›ź) 1 2

12� .cos �+

sin �

√3 sin đ?›źďż˝ 2

sin �

6ďż˝cos đ?›ź+√3 sin đ?›źďż˝ sin đ?›ź

đ?‘ƒďż˝ = 180° − 90° − đ?›ź = 90° − đ?›ź đ?‘„đ?‘…

sin đ?‘ƒďż˝

đ?‘ƒđ?‘„

= sin �

6ďż˝cos đ?›ź+√3 sin đ?›źďż˝ sin đ?›ź

sin(90°âˆ’đ?›ź)

đ?‘ƒđ?‘„

= sin �

đ?‘ƒđ?‘„ sin(90° − đ?›ź) = đ?‘ƒđ?‘„ = c

6 cos � cos �

+

sin đ?›ź.6ďż˝cos đ?›ź+√3 sin đ?›źďż˝

6√3 sin đ?›ź

sin �

cos �

đ?‘ƒđ?‘„ = 6 + 6√3 tan đ?›ź 23 = 6 + 6√3 tan đ?›ź 17 = 6√3 tan đ?›ź đ?‘Ąđ?‘Žđ?‘–đ?›ź = 1,64 đ?›ź = 58,56°

Chapter 7: Polynomials 1

a b c

27đ?‘Ľ 3 − 8 = (3đ?‘Ľ − 2)(9đ?‘Ľ 2 + 6đ?‘Ľ + 4)

5đ?‘Ľ 3 + 40 = 5(đ?‘Ľ 3 + 8) = 5(đ?‘Ľ + 2)(đ?‘Ľ 2 − 2đ?‘Ľ + 4) đ?‘Ľ 3 + 3đ?‘Ľ 2 + 2đ?‘Ľ + 6

= đ?‘Ľ 2 (đ?‘Ľ + 3) + 2(đ?‘Ľ + 3) = (đ?‘Ľ + 3)(đ?‘Ľ 2 + 2)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 132

Answers to Mixed Exercises d

4ЁЭСе 3 тИТ ЁЭСе 2 тИТ 16ЁЭСе + 4

= ЁЭСе 2 (4ЁЭСе тИТ 1) тИТ 4(4ЁЭСе тИТ 1) = (4ЁЭСе тИТ 1)(ЁЭСе 2 тИТ 4)

e

= (4ЁЭСе тИТ 1)(ЁЭСе тИТ 2)(ЁЭСе + 2) 4ЁЭСе 3 тИТ 2ЁЭСе 2 + 10ЁЭСе тИТ 5

= 2ЁЭСе 2 (2ЁЭСе тИТ 1) + 5(2ЁЭСе тИТ 1) f

= (2ЁЭСе тИТ 1)(2ЁЭСе 2 + 5)

ЁЭСе 3 + 2ЁЭСе 2 + 2ЁЭСе + 1

= (ЁЭСе 3 + 1) + (2ЁЭСе 2 + 2ЁЭСе)

= (ЁЭСе + 1)(ЁЭСе 2 тИТ ЁЭСе + 1) + 2ЁЭСе(ЁЭСе + 1)

g

= (ЁЭСе + 1)(ЁЭСе 2 + ЁЭСе + 1)

ЁЭСе 3 тИТ ЁЭСе 2 тИТ 22ЁЭСе + 40

= (ЁЭСе тИТ 2)(ЁЭСе 2 + ЁЭСе тИТ 20)

h

= (ЁЭСе тИТ 2)(ЁЭСе + 5)(ЁЭСе тИТ 4) ЁЭСе 3 + 2ЁЭСе 2 тИТ 5ЁЭСе тИТ 6

= (ЁЭСе тИТ 2)(ЁЭСе 2 + 4ЁЭСе + 3) i

= (ЁЭСе тИТ 2)(ЁЭСе + 3)(ЁЭСе + 1) 3ЁЭСе 3 тИТ 7ЁЭСе 2 + 4

= (ЁЭСе тИТ 1)(3ЁЭСе 2 тИТ 4ЁЭСе тИТ 4) j

= (ЁЭСе тИТ 1)(3ЁЭСе + 2)(ЁЭСе тИТ 2)

ЁЭСе 3 тИТ 19ЁЭСе + 30

= (ЁЭСе тИТ 2)(ЁЭСе 2 + 2ЁЭСе тИТ 15)

k

= (ЁЭСе тИТ 2)(ЁЭСе + 5)(ЁЭСе тИТ 3) ЁЭСе3 тИТ ЁЭСе2 тИТ ЁЭСе тИТ 2

= (ЁЭСе тИТ 2)(ЁЭСе 2 + ЁЭСе + 1)

_____________________________________________________________________________________ ┬йVia Afrika >> Mathematics Grade 12 133

Answers to Mixed Exercises

2

a

b

c

ЁЭСе(ЁЭСе 2 + 2ЁЭСе тИТ 4) = 0

ЁЭСе = 0 ЁЭСЬЁЭСЯ ЁЭСе = тИТ1 ┬▒ тИЪ5

(ЁЭСе тИТ 2)(ЁЭСе 2 тИТ ЁЭСе тИТ 3) = 0 ЁЭСе = 2 ЁЭСЬЁЭСЯ ЁЭСе =

1┬▒тИЪ3 2

(2ЁЭСе 3 тИТ 12ЁЭСе 2 ) тИТ (ЁЭСе тИТ 6) = 0 2ЁЭСе 2 (ЁЭСе тИТ 6) тИТ (ЁЭСе тИТ 6) = 0

(ЁЭСе тИТ 6)(2ЁЭСе 2 тИТ 1) = 0 1

d

ЁЭСе = 6 ЁЭСЬЁЭСЯ ЁЭСе = ┬▒я┐╜2

(2ЁЭСе 3 тИТ ЁЭСе 2 ) тИТ (8ЁЭСе тИТ 4) = 0

ЁЭСе 2 (2ЁЭСе тИТ 1) тИТ 4(2ЁЭСе тИТ 1) = 0 (ЁЭСе 2 тИТ 4)(2ЁЭСе тИТ 1) = 0

1

e

ЁЭСе = 2 ЁЭСЬЁЭСЯ ЁЭСе = тИТ2 ЁЭСЬЁЭСЯ ЁЭСе = 2

f

ЁЭСе=1

(ЁЭСе тИТ 1)(ЁЭСе 2 + 2ЁЭСе + 2) = 0 (ЁЭСе + 2)(ЁЭСе 2 тИТ 2ЁЭСе тИТ 8) = 0

(ЁЭСе + 2)(ЁЭСе тИТ 4)(ЁЭСе + 2) = 0 g

ЁЭСе = тИТ2 ЁЭСЬЁЭСЯ ЁЭСе = 4

(ЁЭСе 3 тИТ 20ЁЭСе) + (3ЁЭСе 2 тИТ 60) = 0

ЁЭСе(ЁЭСе 2 тИТ 20) + 3(ЁЭСе 2 тИТ 20) = 0 (ЁЭСе 2 тИТ 20)(ЁЭСе + 3) = 0

ЁЭСе = ┬▒2тИЪ5 ЁЭСЬЁЭСЯ ЁЭСе = тИТ3

_____________________________________________________________________________________ ┬йVia Afrika >> Mathematics Grade 12 134

Answers to Mixed Exercises 3

𝑓(3) = 33 − 32 − 5(3) − 3

= 27 − 9 − 15 − 3 = 0

(𝑥 − 3) is a factor

(𝑥 − 3)(𝑥 2 + 2𝑥 + 1) = 0 (𝑥 − 3)(𝑥 + 1)2 = 0

𝑥 = 3 𝑜𝑟 𝑥 = −1 4

1 3

1

1 2

1

𝑔 �2� = 4 �2� − 8 �2� − 2 + 2

=

1 1 −2− +2=0 2 2

(2𝑥 − 1)(2𝑥 2 − 3𝑥 − 2) = 0

(2𝑥 − 1)(2𝑥 + 1)(𝑥 − 2) = 0 1

1

𝑥 = 2 𝑜𝑟 𝑥 = − 2 𝑜𝑟 𝑥 = 2

Chapter 8: Differential calculus 1

a

𝑓 ′ (𝑥) = limℎ→0 = limℎ→0 = limℎ→0 = limℎ→0 = limℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ

1−(𝑥+ℎ)2 −�1−𝑥2 � ℎ

1−�𝑥 2 +2𝑥ℎ+ℎ2 �−�1−𝑥2 � −2𝑥ℎ−ℎ2

ℎ

ℎ

ℎ(−2𝑥−ℎ) ℎ

= lim (−2𝑥 − ℎ) ℎ→0

= −2𝑥

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 135

Answers to Mixed Exercises b

𝑓 ′ (𝑥) = limℎ→0 = limℎ→0 = limℎ→0 = limℎ→0 = limℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥) ℎ

−3(𝑥+ℎ)2 −�−3𝑥2 � ℎ

−3�𝑥 2 +2𝑥ℎ+ℎ2 �−�−3𝑥2 � −6𝑥ℎ−3ℎ2

ℎ

ℎ

ℎ(−6𝑥−3ℎ) ℎ

= lim (−6𝑥 − 3ℎ) ℎ→0

= −6𝑥

2

a

1

1

1

𝑦 = √𝑥 − 2𝑥 2 = 𝑥 2 − 2 𝑥 −2 𝑑𝑦 𝑑𝑥

1

1

1

= 2 𝑥 −2 − 2 × −2𝑥 −3 1

1

= 2 𝑥 −2 + 𝑥 −3 b

=2

1

√𝑥

𝐷𝑥 �

1

+ 𝑥3

2𝑥 2 −𝑥−15

= 𝐷𝑥 �

𝑥−3

�

(2𝑥+5)(𝑥−3) 𝑥−3

�

= 𝐷𝑥 [2𝑥 + 5] = 2 3

𝑓(𝑥) = −2𝑥 3 + 3𝑥 2 + 32𝑥 + 15

𝑓(−2) = −2(−2)3 + 3(−2)2 + 32(−2) + 15 = −21

𝑓 ′ (𝑥) = −6𝑥 2 + 6𝑥 + 32

𝑓 ′ (−2) = −6(−2)2 + 6(−2) + 32 = −4 Sub (−2; −21) into 𝑦 = −4𝑥 + 𝑐 −21 = −4(−2) + 𝑐

𝑐 = −29

𝑦 = −4𝑥 − 29

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 136

Answers to Mixed Exercises

4

5

Point of inflection

(2; 9) is a point on the graph and a turning point ∴ đ?‘“(2) = 9 and đ?‘“ ′ (2) = 0 đ?‘“ ′ (đ?‘Ľ) = 3đ?‘Žđ?‘Ľ 2 + 10đ?‘Ľ + 4 0 = 3đ?‘Ž(2)2 + 10(2) + 4 ∴ đ?‘Ž = −2 9 = (−2)(2)3 + 5(2)2 + 4(2) + đ?‘? ∴ đ?‘? = −3

6

a b c

7

a

b

Turning point where đ?‘“ ′ (đ?‘Ľ) = 0 ∴ đ?‘Ľ = −2 and đ?‘Ľ = 5 ′′ (đ?‘Ľ) Point of inflections is where đ?‘“ = 0 , therefor where graph of đ?‘“′ turns đ?‘“ will decrease where its gradient đ?‘“ ′ is negative (đ?‘“ ′ < 0) −2 < đ?‘Ľ < 5 The graph bounces at đ?‘Ľ = 1 and has an đ?‘Ľ-intercept at đ?‘Ľ = −1 ∴ đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(đ?‘Ľ − 1)2 đ?‘“(đ?‘Ľ) = (đ?‘Ľ + 1)(đ?‘Ľ 2 − 2đ?‘Ľ + 1) = đ?‘Ľ 3 − đ?‘Ľ 2 − đ?‘Ľ + 1 ∴ đ?‘Ž = −1; đ?‘? = −1; đ?‘? = 1 B is a turning point where đ?‘“ ′ (đ?‘Ľ) = 0 3đ?‘Ľ 2 − 2đ?‘Ľ − 1 = 0 (3đ?‘Ľ + 1)(đ?‘Ľ − 1) = 0) 1

At B đ?‘Ľ = − 3 1

1 3

1 2

1

32

đ?‘“ ďż˝âˆ’ 3ďż˝ = ďż˝âˆ’ 3ďż˝ − ďż˝âˆ’ 3ďż˝ − ďż˝âˆ’ 3ďż˝ + 1 = 27 1 32

đ??´ ďż˝âˆ’ 3 ; 27ďż˝

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 137

Answers to Mixed Exercises

8

a b c

9

a b

c

If 𝑠(𝑡) is distance, then 𝑠 ′ (𝑡) is speed. 𝑠 ′ (𝑡) = 3𝑡 2 − 4𝑡 + 3 Speed is a minimum where 𝑠 ′′ (𝑡) = 6𝑡 − 4 = 0 2

𝑡=3

6𝑡 − 4 = 8 𝑡 = 2𝑠

Volume = 2𝑥 2 ℎ = 24 12

ℎ = 𝑥 2 = 12𝑥 −2

𝐶(𝑥) = 2𝑥 2 × 25 + 2𝑥 2 × 20 + 2 × 𝑥ℎ × 20 + 2 × 2𝑥ℎ × 20 = 90𝑥 2 + 120𝑥ℎ = 90𝑥 2 + 120𝑥(12𝑥 −2 ) = 90𝑥 2 + 1440𝑥 −1 𝐶 ′ (𝑥) = 180𝑥 − 1440𝑥 −2 = 0 1440

180𝑥 −

𝑥2

=0

180𝑥 3 − 1440 = 0 𝑥3 = 8 𝑥=2

Chapter 9: AnalyticalgGeometry 1

a

𝑚𝐴𝐴 = 𝑚𝐸𝐴 1−(−4) −2−𝑝 5

−2−𝑝

b

c

0−2

= 5−3

=

−2 2

2+𝑝=5 𝑝=3

= −1

𝐴𝐴 = �(3 − (−2))2 + (−4 − 1)2 = 5√2 𝐶𝐷 = �(5 − 3)2 + (0 − 2)2 = 2√2 𝐴𝐴: 𝐶𝐷 = 5√2: 2√2 = 5: 2 𝑚𝑁𝐴 = 𝑚𝐸𝐴 𝑦+4 𝑥−3

= −1

𝑚𝑁𝐴 = 𝑚𝐴𝐸 𝑦−2 𝑥−3

=2

(1)-(2):

𝑁(1; −2)

∴ 𝑦 = −𝑥 − 1 … (1) ∴ 𝑦 = 2𝑥 − 4 … (2)

0 = 3𝑥 − 3 𝑥 = 1 𝑦 = −2

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 138

Answers to Mixed Exercises

d e f

g

B and D have the same đ?‘Ľ-coordinate, so it is a vertical line with equation đ?‘Ľ = 3. Angle of inclination of a vertical line is 90°. Area of parallelogram = base x perpendicular height đ??´đ?‘&#x;đ?‘?đ?‘Ž đ?‘œđ?‘“ đ?‘ đ??´đ??śđ??ˇ = đ??śđ??ˇ Ă— đ?‘?đ?‘?đ?‘&#x;đ?‘? â„Žđ?‘?đ?‘–đ?‘”â„Žđ?‘Ą = 2√2 Ă— 6 = 12√2 đ?‘šđ??´đ?‘… = đ?‘šđ??´đ??¸ đ?‘žâˆ’1

1−0

= −2−5 −2+1

đ?‘žâˆ’1 −1

2

a

b

1

= −7 8

∴đ?‘ž=7

Substitute đ?‘Ľ = 1 and đ?‘Ś = −3 in LHS. If LHS=0, then the point đ?‘ (1; −3) lies on the circle. đ??żđ??ťđ?‘† = đ?‘Ľ 2 + 4đ?‘Ľ + đ?‘Ś 2 + 2đ?‘Ś – 8 = (1)2 + 4(1) + (−3)2 + 2(−3)– 8 = 0 ∴ đ?‘ lies on the circle First determine the centre of the circle: đ?‘Ľ 2 + 4đ?‘Ľ + 4 + đ?‘Ś 2 + 2đ?‘Ś + 1 = 8 + 4 + 1 (đ?‘Ľ + 2)2 + (đ?‘Ś + 1)2 = 13 Centre of circle is đ?‘€(−2; −1) −1+3

2

đ?‘šđ?‘€đ?‘ = −2−1 = − 3

MNâ?ŠPN (radiusâ?Štangent) 3

∴ đ?‘šđ?‘ƒđ?‘ = 2

3

Substitute đ?‘ (1; −3): đ?‘Ś = 2 đ?‘Ľ + đ?‘? 3

−3 = 2 (1) + đ?‘?

c d e

3

9

đ?‘Ś = 2đ?‘Ľ − 2 3

9

∴ đ?‘? = −2

đ?‘ = đ?‘Ąđ?‘Žđ?‘–−1 ďż˝2ďż˝ = 56,3°

đ?‘Ľ −intercept where đ?‘Ś = 0: 3

9

0 = 2đ?‘Ľ − 2

∴đ?‘Ľ=3

đ?‘Ś −intercepts are where đ?‘Ľ = 0: (0)2 + 4(0) + đ?‘Ś 2 + 2đ?‘Ś – 8 = 0

∴ đ?‘Ś 2 + 2đ?‘Ś – 8 = 0 (đ?‘Ś + 4)(đ?‘Ś − 2) = 0 The points are (0; −4) and (0; 2). _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 139

Answers to Mixed Exercises

3

a b

c d e

đ?‘šđ?‘…đ?‘‚ =

−12 −6

=2

PSâ?ŠRN (RN is altitude of ∆) đ?‘šđ?‘ƒđ?‘† Ă— đ?‘šđ?‘…đ?‘ = −1 1

∴ đ?‘šđ?‘ƒđ?‘† = − 2

đ?‘ƒ(0; 6) (đ?‘Ś −intercept of PR) 1

∴ đ?‘Ś = −2đ?‘Ľ + 6 1

đ?‘Ąđ?‘Žđ?‘–−1 ďż˝2ďż˝ = 26,57°

Inclination of PS = 180° − 26,57° = 153,43° 3

Substitute đ?‘ (2đ?‘–; 3 5 + đ?‘–) into equation of PS 3

1

3 5 + đ?‘– = − 2 (2đ?‘–) + 6 3

3 5 + đ?‘– = −đ?‘– + 6 2

2đ?‘– = 2 5 = f

6

đ?‘–=5

12 5

Find equation of SM. SM is the median, so M is the midpoint of PR. −6+0 −12+6

��

;

2

2

ďż˝ = (−3; −3)

��� = 1 so equation of SM: � = � Solve equations of SM and PS simultaneously to calculate coordinates of S 1

đ?‘Ľ = −2đ?‘Ľ + 6 4

a

b

c

�(4; 4)

∴ đ?‘Ľ = 4; đ?‘Ś = 4

� 2 + 4� + � 2 – 2� = 4

đ?‘Ľ 2 + 4đ?‘Ľ + 4 + đ?‘Ś 2 – 2đ?‘Ś + 1 = 4 + 4 + 1 (đ?‘Ľ + 2)2 + (đ?‘Ś − 1)2 = 9 Centre đ?‘€(−2; 1) radius= 3 Substitute đ?‘ (đ?‘?; 1) into equation of circle. đ?‘? 2 + 12 + 4(đ?‘?) – 2(1) – 4 = 0 đ?‘?2 + 4đ?‘? − 5 = 0 (đ?‘? + 5)(đ?‘? − 1) = 0 ∴ đ?‘? = 1 as đ?‘? > 0 Radius through N is horizontal. Therefore the tangent will be vertical. Equation of tangent: đ?‘Ľ = 1

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 140

Answers to Mixed Exercises 5

a b

3−0

𝑚𝐴𝐴 = −3−0 = −1

AD goes through origin: So, equation is 𝑦 = −𝑥 𝐴𝐷2 = 𝐷𝐶 2 (𝑥 − 2)2 + (𝑦 − 3)2 = (𝑥 − 6)2 + (𝑦 + 1)2 Substitute 𝑦 = −𝑥 (𝑥 − 2)2 + (−𝑥 − 3)2 = (𝑥 − 6)2 + (−𝑥 + 1)2 𝑥 2 − 4𝑥 + 4 + 𝑥 2 + 6𝑥 + 9 = 𝑥 2 − 12𝑥 + 36 + 𝑥 2 − 2𝑥 + 1 16𝑥 = 24 3

c

d

𝑥=2

𝑚𝐴𝐴 =

=9

Substitute 𝐴(2; 3) into 𝑦 = 9𝑥 + 𝑐 3 = 9(2) + 𝑐 ∴ 𝑐 = −15 𝑦 = 9𝑥 − 15 Inclination of BD= 𝑡𝑎𝑖−1 (9) = 83,7° 𝑚𝐴𝐸 =

e

3 2 3 2− 2

3−(− )

3

∴ 𝑦 = −2

3−(−1) 2−6

= −1

Inclination of BC= 135° ∴ 𝑠 = 135° − 83,7° = 51,3° 3 2

3 2

𝐴𝐷 = ��2 − 2� + �3 + 2� =

√82 2

𝐴𝐶 = �(3 + 1)2 + (2 − 6)2 = 4√2 1

𝐴𝑟𝑐𝑎 𝑜𝑓 ∆𝐴𝐷𝐶 = 2 𝐴𝐷 × 𝐴𝐶 × 𝑠𝑖𝑖𝑠 1

=2× 6

a

√82 2

= 10 𝑠𝑞 𝑢𝑖𝑖𝑡𝑠

First determine equation of AC 𝑚𝐴𝐸 =

b

c

× 4√2 × 𝑠𝑖𝑖51,3°

3−(−3) 2−5

= −2

Substitute (2; 3): 3 = −2(2) + 𝑐 7

𝑥 − 𝑖𝑖𝑡𝑐𝑟𝑐𝑐𝑝𝑡(𝑦 = 0): 𝑥 = 2

∴ 𝑦 = −2𝑥 + 7

𝐴𝐶 2 = 𝐴𝐶 2 (𝑝 − 5)2 + (0 + 3)2 = (5 − 2)2 + (−3 − 3)2 𝑝2 − 10𝑝 + 25 = 9 + 36 𝑝2 − 10𝑝 − 20 = 0 𝑝=

10±√180 2

7

𝐷 �2 ; 0�

= 5 ± 3√5

𝑝 = 5 − 3√5 𝑚𝐴𝐸 = −2 𝐼𝑖𝑐𝑙𝑖𝑖𝑎𝑡𝑖𝑜𝑖 𝑜𝑓 𝐴𝐶 = 180° − 𝑡𝑎𝑖−1 (2) = 116,6°

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 141

Answers to Mixed Exercises

d

đ??´(−1; 0)

3−0

đ?‘šđ??´đ??´ = 2+1 = 1

Inclination of đ??´đ??´ = 45° đ??´Ě‚ = đ?‘–đ?‘–đ?‘?đ?‘™đ?‘–đ?‘–đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘– đ?‘œđ?‘“ đ??´đ??ś − đ?‘–đ?‘–đ?‘?đ?‘™đ?‘–đ?‘–đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘– đ?‘œđ?‘“ đ??´đ??´ = 116,6° − 45° = 71,6°

7

The line will be a tangent if it intersects the circle in only one point. Substitute đ?‘Ś = đ?‘Ľ + 1 into equation of circle and solve for đ?‘Ľ. There should be only one solution. đ?‘Ľ 2 + (đ?‘Ľ + 1)2 + 6(đ?‘Ľ + 1) − 7 = 0 đ?‘Ľ 2 + đ?‘Ľ 2 + 2đ?‘Ľ + 1 + 6đ?‘Ľ + 6 − 7 = 0 2đ?‘Ľ 2 + 8đ?‘Ľ = 0 đ?‘Ľ = 0 or đ?‘Ľ = −4 The line is NOT a tangent.

8

a

b c

đ?‘Ś = 2 at C. Substitute into 3đ?‘Ľ + 4đ?‘Ś + 7 = 0 3đ?‘Ľ + 4(2) + 7 = 0 3đ?‘Ľ = −15 đ?‘Ľ = −15 ∴ đ??ś(−5; 2) and the radius is 5. (đ?‘Ľ + 5)2 + (đ?‘Ś − 2)2 = 25 length of đ??ˇđ??ˇ = 10 2+1

đ?‘šđ?‘ƒđ??´ = 0+1 = 3

1

đ?‘šđ?‘?đ?‘’đ?‘&#x;đ?‘? đ?‘?đ?‘–đ?‘ đ?‘’đ?‘Žđ?‘Ąđ?‘œđ?‘&#x; = − 3 Midpoint of PE= ďż˝

0−1 2−1 2

;

2

1 1

ďż˝ = ďż˝âˆ’ 2 ; 2ďż˝ 1

Substitute midpoint into đ?‘Ś = − 3 đ?‘Ľ + đ?‘? 1 2

1

đ?‘?=3 d

1

1

= − 3 ďż˝âˆ’ 2ďż˝ + đ?‘? 1

1

đ?‘Ś = −3đ?‘Ľ + 3 1

1

3đ?‘Ľ + 4 ďż˝âˆ’ 3 đ?‘Ľ + 3ďż˝ + 7 = 0 4

4

3đ?‘Ľ − 3 đ?‘Ľ + 3 + 7 = 0 5

đ?‘Ľ=− 3 đ?‘Ľ = −5

1

25 3

1

đ?‘Ś = − 3 (−5) + 3 = 2

The lines intersect at (−5; 2)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 142

Answers to Mixed Exercises

9

a

Let the coordinates of S be (𝑥; 0) ST⏊SR 4

4

𝑚𝑆𝑇 × 𝑚𝑆𝑅 = −𝑥 × 𝑥−4 = −1 𝑥(𝑥 − 4) = 16 𝑥 2 − 4𝑥 − 16 = 0 b c

𝑥=

4±�(−4)2 −4(−16) 2

= 2 ± 2√5

But S is on positive 𝑥 −axis, so 𝑆�2 + 2√5; 0� 4−0

𝑚𝑆𝑇 = 0−�2+2√5� = −0,62

Inclination of TS= 180° − 𝑡𝑎𝑖−1 (0,62) = 148,20° 4+4

𝑚 𝑇𝑅 = 0−4 = −2

10

a

Inclination of TR= 180° − 𝑡𝑎𝑖−1 (2) = 116,57° 𝑅𝑇�𝑆 = 148,20° − 116,57° = 31,63° 𝑥 2 + 𝑦 2 – 4𝑥 + 6𝑦 + 3 = 0 𝑥 2 − 4𝑥 + 𝑦 2 + 6𝑦 = −3 𝑥 2 − 4𝑥 + 4 + 𝑦 2 + 6𝑦 + 9 = −3 + 4 + 9 (𝑥 − 2)2 + (𝑦 + 3)2 = 10 Centre is (2; −3) 𝑚𝑟𝑎𝑑𝑖𝑢𝑠 =

b

−2+3 5−2

1

=3

𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = −3 Substitute (5; −2) into 𝑦 = −3𝑥 + 𝑐 −2 = −3(5) + 𝑐 𝑐 = 13 ∴ 𝑦 = −3𝑥 + 13

�(𝑥 − 2)2 + (𝑦 + 3)2 = √20 (𝑥 − 2)2 + (𝑦 + 3)2 = 20 Substitute 𝑦 = −3𝑥 + 13 into equation above: (𝑥 − 2)2 + (−3𝑥 + 13 + 3)2 = 20 (𝑥 − 2)2 + (−3𝑥 + 16)2 = 20 𝑥 2 − 4𝑥 + 4 + 9𝑥 2 − 96𝑥 + 256 = 20 10𝑥 2 − 100𝑥 + 240 = 0 𝑥 2 − 10𝑥 + 24 = 0 (𝑥 − 6)(𝑥 − 4) = 0 𝑥 = 6 or 𝑥 = 4 𝑦 = −3(6) + 13 = −5 or 𝑦 = −3(4) + 13 = 1 𝑇(6; −5) or 𝑇(4; 1)

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 143

Answers to Mixed Exercises

Chapter 10: Euclidian geometry 1

a

b

c

ďż˝1 đ??´ďż˝1 = đ?‘€ đ??´ďż˝1 = đ??śĚ‚ ďż˝1 = đ??śĚ‚ ∴đ?‘€

corr ∠s alt ∠s tan chord ∴ ∆đ??žđ?‘€đ?‘ is isosceles ďż˝4 = đ?‘ ďż˝2 alt ∠s đ??ž ďż˝2 = đ??´ďż˝3 đ?‘ ∠s in same segment đ??´Ě‚3 = đ??´ďż˝3 ∠s in same segment Ě‚ ďż˝ ∴ đ??ž4 = đ??´3 ∴ đ?‘ đ??ž||đ??´đ?‘ƒ alt ∠s= ∴ đ?‘€đ?‘ ||đ??śđ??´ ďż˝1 = đ?‘€ ďż˝2 đ??ž ďż˝1 = đ?‘ ďż˝2 đ??ž

đ??´đ?‘

đ??´đ??ž

∴ đ?‘ đ??´ = đ??žđ?‘ƒ But

d

2

a

b

tan chord

đ??´đ??ž

đ??´đ?‘

đ?‘ đ??´

∴ đ??žđ?‘ƒ =

đ??´đ?‘€

line || to one side of ∆

= đ?‘€đ??¸ line || to one side of ∆ đ??´đ?‘€ đ?‘€đ??¸

đ??´Ě‚3 = đ??´ďż˝3 ∠s in same segment đ??´ďż˝3 = đ??´ďż˝2 equal chords subt equal ∠s Ě‚ ďż˝ ∴ đ??´3 = đ??´2 ∴ đ??ˇđ??´ is a tangent to the circle through A, B and K đ??śĚ‚3 = đ??śđ?‘ƒďż˝ đ?‘… ∠s opp equal sides Ě‚ Ě‚ Ě‚ đ??ś3 + đ??ś2 = đ??´1 + đ??´ďż˝ ext ∠of ∆ đ??śĚ‚2 = đ??´ďż˝ tan chord Ě‚ Ě‚ ∴ đ??ś3 = đ??´1 ∴ đ??´Ě‚1 = đ??śđ?‘ƒďż˝đ?‘… both = đ??śĚ‚3

ACPR is a cyclic quadrilateral (ext ∠of quad) In ∆đ??śđ??´đ??´ and ∆đ?‘…đ?‘ƒđ??´: đ?‘ƒďż˝2 = đ??śĚ‚2 ∠s in same segment = đ??´ďż˝ proven in 2 a ∴ đ??´ďż˝ = đ?‘ƒďż˝2 đ??śĚ‚1 = đ??´đ?‘…ďż˝ đ?‘ƒ ext ∠of cyclic quad đ??´Ě‚1 = đ??´Ě‚3 3rd ∠of ∆ ∴ ∆đ??śđ??´đ??´|||∆đ?‘…đ?‘ƒđ??´ ∠∠âˆ

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 144

Answers to Mixed Exercises

c

𝑅𝑃 𝐸𝐴

𝑅𝐴

𝑅𝑃 = d

from 2 b

= 𝐸𝐴

𝐸𝐴.𝑅𝐴

but 𝑅𝑃 = 𝑅𝐶

𝐸𝐴 𝐸𝐴.𝑅𝐴

∴ 𝑅𝐶 =

𝐸𝐴

In ∆𝑅𝐴𝐶 and ∆𝑅𝐶𝐴: 𝐶̂2 = 𝐴� tan chord 𝑅�1 is common 𝑅𝐶̂ 𝐴 = 𝑅𝐴̂𝐶 3rd angle ∴ ∆𝑅𝐴𝐶|||∆𝑅𝐶𝐴 ∠∠∠

e

𝐴𝐸

𝑅𝐸

𝐸𝐴

= 𝑅𝐴

= 𝑅𝐴 𝐸𝐴

∆𝑠 |||

𝑅𝑃 𝐸𝐴

RC=RP

𝑅𝐴. 𝐴𝐶 = 𝑅𝐶. 𝐶𝐴 𝑅𝐸

𝐸𝐴

𝐸𝐴

= 𝑅𝐴

𝐴𝐶 =

from 2 b

𝐸𝐴.𝑅𝐴 𝑅𝐸

…(i)

From 2 d 𝐴𝐶 = ∴ 3

a

bi

b ii

𝐸𝐴.𝑅𝐴 𝑅𝐸 2

=

𝑅𝐸.𝐸𝐴

𝑅𝐸.𝐸𝐴

𝑅𝐴

∴ 𝑅𝐶 = 𝑅𝐴. 𝑅𝐴 𝐴�2 = 𝐴̂3 = 𝑥 �1 = 180° − 2𝑥 𝑀 � = 2𝑥 ∴𝐷 �

𝑀 𝐶̂ = 21

𝑅𝐴

∠s opp equal sides sum ∠s of ∆

∠ at centre =2x∠circ

= 90° − 𝑥 𝐶𝐴� 𝐷 = 180° − (90° − 𝑥 + 2𝑥) sum ∠s of ∆ = 90° − 𝑥 �1 = 𝐶̂ = 90° − 𝑥 ext ∠ of cyclic quad 𝑁 �1 ∴ 𝐶𝐴� 𝐷 = 𝑁 ∴ 𝐶𝐴||𝐴𝑁 corr ∠s � � 𝐶𝐴 𝐴 = 𝐷 = 2𝑥 tan chord 𝐶𝐴� 𝐴 = 𝐴̂2 alt ∠s � 𝐴̂2 = 𝐷 ∴AB is a tangent (∠betw line&chord= ∠sub chord)

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 145

Answers to Mixed Exercises

4

a

bi

ii

5

a

đ??´ďż˝3 = đ??ˇďż˝1 = đ?‘Ľ ∠s in same segment ďż˝2 = đ?‘Ľ đ??´ďż˝3 = đ??ˇ ∠s opp = sides ďż˝ đ??´đ?‘‚đ??ˇ = 180° − 2đ?‘Ľ sum ∠s of ∆ đ??´Ě‚ = 90° − đ?‘Ľ ∠at centre =2x∠circ đ??śĚ‚1 = 90° − đ?‘Ľ ext ∠of cyclic quad đ??ˇďż˝2 = 180° − (đ?‘Ľ + 90° − đ?‘Ľ) sum ∠s of ∆ = 90° In ∆đ??´đ??ˇđ??ˇ and ∆đ??śđ??ˇđ??ˇ: đ??ˇďż˝1 = đ??ˇďż˝2 = 90° adj ∠s str line BF = FC FE is common ∆đ??´đ??ˇđ??ˇ ≥ ∆đ??śđ??ˇđ??ˇ s∠s BE = EC (≥) sum ∠s of ∆ đ??´ďż˝1 = 90° − đ?‘Ľ ∴ đ??´ďż˝1 = đ??´Ě‚ ∴ BE is not a tangent ďż˝đ??´ďż˝1 + đ??´ďż˝2 ≠đ??´Ě‚ďż˝

P is midpoint of AC AB||PM In ∆đ??´đ?‘ đ??ś: đ?‘ đ??´ đ?‘ đ??¸

b

=

đ??´đ?‘€

đ??´đ?‘€ đ??´đ??¸

medians concur midpt theorem

đ??´đ?‘ƒ

= đ??´đ??¸ line || 1 side of ∆ 1

= 2đ??´đ?‘€ = 2 In ∆đ??´đ?‘€đ?‘ƒ: đ??´đ?‘‚

= đ?‘‚đ?‘€ đ?‘…đ?‘ƒ đ?‘ƒđ??¸

2đ?‘‚đ?‘€

đ?‘‚đ?‘€ đ?‘…đ?‘ƒ

= đ??´đ?‘ƒ

đ?‘‚đ?‘€

= đ??´đ?‘€

đ?‘‚đ?‘€

= 3đ?‘‚đ?‘€

BP is a median line || 1 side of ∆

1

6

a

b

=3

đ??śĚ‚2 = 90° ďż˝2 = 90° đ?‘€

∴ đ?‘ đ?‘„||đ??śđ??ˇ ďż˝ đ??śĚ‚1 = đ?‘ đ??´Ě‚2 = đ??śĚ‚1 ďż˝ =đ?‘

∠in semi ⊙ AM�NM corr ∠s= || lines, corr ∠s tan chord

∴ ANCQ is a cyclic quad

∠s subt by same line segm

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 146

Answers to Mixed Exercises

ci

ii d

In ∆đ?‘ƒđ??śđ??ˇ and ∆đ?‘ƒđ??´đ??ś: đ??śĚ‚1 = đ??´Ě‚2 tan chord đ?‘ƒďż˝ is common ďż˝1 = đ??´đ??śĚ‚ đ?‘ƒ đ??ˇ 3rd âˆ

∴ ∆đ?‘ƒđ??śđ??ˇ ||| ∆đ?‘ƒđ??´đ??ś ∠∠∠đ?‘ƒđ??ś 2 = đ??´đ?‘ƒ. đ??ˇđ?‘ƒ In ∆đ?‘ đ??´đ??ś and ∆đ??´đ??śđ??ˇ: ďż˝ = đ??´Ě‚2 đ?‘ ∠s in same segm = đ??´ďż˝2 ∠s in same segm đ??śĚ‚4 = đ??´Ě‚1 tan chord ďż˝2 =đ??ˇ ∠s in same segm đ??´ďż˝1 = đ??´đ??śĚ‚ đ??ˇ 3rd ∠∴ ∆đ?‘ đ??´đ??ś ≥ ∆đ??´đ??śđ??ˇ ∠∠∠đ??´đ??¸

e

đ??¸đ??´

∴ đ?‘ đ??´ = đ?‘ đ??´

đ??´đ??ś 2 = đ??śđ??ˇ. đ?‘ đ??´ 1−

đ??´đ?‘€2 đ??´đ??¸ 2

=

=

đ??´đ??¸ 2 −đ??´đ?‘€2

đ?‘€đ??¸ 2 đ??´đ??¸ 2 đ?‘ƒđ??¸ 2

đ??´đ??¸ 2

= đ??´đ??¸ 2

Pyth.

đ??´đ?‘ƒ.đ??´đ?‘ƒ

= đ??¸đ??´.đ?‘ đ??´

Chapter 11: Statistics: regression and correlation 1

a Matches played Wins Goals scored against

Lower Q 3 1 3

Median 5 7 4,5

Upper Q 6 3 9

b

Positively skewed (skewed to the right) c d

65 14

= 4,64

Standard deviation = 1,72

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 147

Answers to Mixed Exercises

2

a

% transport costs 60 50 40 30 20 10 0

10

20

30

40

50

60

% transport costs

b c

d

Median = Âą32%

Class midpoint Frequency 15 6 25 14 35 16 45 11 55 3 TOTAL 50 1660 Estimated mean đ?‘ĽĚ… = 50 = 33,2% Class midpt đ?‘Ľđ?‘– 15 25 35 45 55 TOTAL Standard deviation= ďż˝

Freq đ?‘“ 6 14 16 11 3 50 5938 50

FreqxMidpoint 90 350 560 495 165 1660

đ?‘ĽĚ… − đ?‘Ľđ?‘– -18,2 -8,2 1,8 11,8 21,8

(đ?‘ĽĚ… − đ?‘Ľđ?‘– )2 331,24 67,24 3,24 139,24 475,24

đ?‘“(đ?‘ĽĚ… − đ?‘Ľđ?‘– )2 1987,44 941,36 51,84 1531,64 1425,72 5938

= 10,90

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 148

Answers to Mixed Exercises

3

a&c

20

Di sta nce

15

10

5

VO2 10

b d e

20

30

40

50

60

đ?‘Ś = 0,2432đ?‘Ľ + 3,6834 Substitute đ?‘Ś = 19 then đ?‘Ľ = 62,98 (VO2) đ?‘&#x; = 0,8985 ‌ Strong positive correlation

4 (đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘?đ?‘&#x; − đ?‘šđ?‘?đ?‘Žđ?‘–)2 36 4 0 (đ?‘Ľ − 10)2 (đ?‘Ś − 10)2

đ?‘ đ?‘˘đ?‘šđ?‘?đ?‘?đ?‘&#x; 4 8 10 đ?‘Ľ đ?‘Ś

Mean= 10 ∴

4+8+10+đ?‘Ľ+đ?‘Ś 5

= 10

Which simplifies to: � + � = 28 ‌ . (1) Standard deviation = 4

36+4+0+(đ?‘Ľâˆ’10)2 +(đ?‘Śâˆ’10)2

âˆ´ďż˝

5

=4

Which simplifies to: (đ?‘Ľ − 10)2 + (đ?‘Ś − 10)2 = 40 ‌ (2) Substitute đ?‘Ś = 28 − đ?‘Ľ from (1) into (2): đ?‘Ľ 2 − 28đ?‘Ľ + 192 = 0 (đ?‘Ľ − 12)(đ?‘Ľ − 16) = 0 đ?‘Ľ = 12 or đ?‘Ľ = 16 đ?‘Ś = 16 or đ?‘Ś = 12 _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 149

Answers to Mixed Exercises

5

The standard deviation will remain 7,5. If all the numbers are 2 bigger, then the mean will also be 2 bigger. The difference between each number and the mean will therefore remain the same leaving the standard deviation unchanged.

Chapter 12: Probability 1

2

3 4

7 spaces that have to be filled using 7 digits without repetition(as 0,7 and 4 may not be used again) ∴ 7! = 5040 7 spaces have to be filled – 10 digits are available for each space

∴ 107 = 10 000 000

a

11!

11!

b 5

2!2!2!2!2!

a

Total number of arrangements= 4! × 6! = 17 280 4! × 3! × 2! × 3! = 1728

c

7

= 1 247 400 (5 letters repeat)

Regard the 4 English books as a unit. The number of arrangements for the English books is 4!=24

b

6

1

P(Queen of diamonds)= 52

9! = 362 880

12 × 11 = 132

11!

First calculate the total number of words: 2!2!2! = 4 989 600

Now calculate how many of these WILL start and end on the same letter. It can start and end with M, A or T 9!

∴ 2!2! = 90720 8

90 720

54

P(not start and end on same letter)= 1 − 4 989 600 = 55 a

10!×2 2!2!2!

= 907 200

b

10!

2!2!2!

= 453 600

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 150

Exemplar Paper 1

Exemplar Paper 1 (3 hours; 150 marks) 1

a

Solve for đ?‘Ľ: i

ii iii iv b c

2

đ?‘Ľ + 2 = đ?‘Ľ+1

(4)

đ?‘Ľ − √đ?‘Ľ = 6

(4)

5đ?‘Ľâˆ’2 + 5đ?‘Ľ+1 = 126

(5)

(đ?‘Ľ 2 +4)(2−đ?‘Ľ) đ?‘Ľ+2

≼0

(6)

Consider the equation: đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 3 + đ?‘?đ?‘Ľ 2 + đ?‘?đ?‘Ľ − 9đ?‘?

If (2đ?‘Ľ + đ?‘?) is a factor of đ?‘“(đ?‘Ľ) and đ?‘? ≠0, determine the value(s) of đ?‘?. (5) 2 is a root of 2đ?‘Ľ 2 − 3đ?‘Ľ − đ?‘? = 0. Determine the value of đ?‘? and hence the

other root.

(4) [28]

2

a

The sum of the first 20 terms of an arithmetic progression is 410, while the sum of the next 30 terms is 2865. Determine the first three terms of the

b

progression.

(7)

3; đ?‘Ľ; 15; đ?‘Ś; 35 is a quadratic sequence.

(4)

ii

(4)

i c d

Determine the values if đ?‘Ľ and đ?‘Ś.

Determine formula for �� .

Find đ?‘– such that ∑đ?‘›đ?‘˜=7(2đ?‘˜ − 3) is equal to the sum of the first 6 terms of the sequence −24; 48; −96; ‌

For which value(s) of đ?‘Ľ will the following series be convergent? (đ?‘Ľ + 2) + (đ?‘Ľ + 2)2 + (đ?‘Ľ + 2)3 + â‹Ż

3

a

(7)

(2) [24]

Melissa decides to save R1 200 per month for a certain period. The bank offers her an interest rate of 12% p.a. compounded monthly for this period. Determine how long Melissa has to make this monthly payment if she wants to have a lump sum of R200 000.

(5)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 151

Exemplar Paper 1 b

Richard plans to buy a house on a 20 year mortgage and can only afford to pay R5 000 per month. If the interest rate is currently 12% per annum compounded monthly, determine the size of the mortgage he can take, if he starts paying one month after the mortgage was approved.

c

(3)

An amount of R300 000 is to be used to provide quarterly withdrawals for the next 10 years. The withdrawal amount will remain fixed and the first withdrawal will be in 3 months’ time. An interest rate of 15% p.a. compounded quarterly applies. Determine the value of each quarterly withdrawal. (4) [12]

4

1

1

In the diagram đ?‘“ is the graph of đ?‘Ś = − 2 đ?‘Ľ 2 + 2 đ?‘Ľ + đ?‘˜ cuts the đ?‘Ľ −axis at B and C 3

and the đ?‘Ś −axis at D. đ?‘” is the graph of đ?‘Ś = đ?‘Žđ?‘Ľ − 2 and cuts the đ?‘Ľ −axis at B. â„Ž is

the graph of đ?‘Ś = đ?‘š đ?‘Ľ and cuts the đ?‘Ś −axis at D. QR and ST are parallel to the đ?‘Ś −axis. 1

đ??´ ďż˝đ?‘Ľ; 4ďż˝ is a point on â„Ž and vertically above C. y

h

g

Q

D

S A

B

P

F

x

C

T R

a

f

(6)

b

Determine the values of đ?‘˜ and đ?‘š.

c

Calculate the length of QR if OP = 2 units.

(4)

d

Determine the length op OF, if ST = 4 units.

(4)

e

Determine the equation of ℎ−1.

(2)

f

Determine the value of đ?‘Ž.

Write down the domain of ℎ−1 .

(2)

(2) [20]

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 152

Exemplar Paper 1 5

đ?‘Ž

The functions đ?‘“(đ?‘Ľ) = đ?‘Ľ+đ?‘? + đ?‘? and đ?‘”(đ?‘Ľ) = 2đ?‘Ľ − 13 intersect each other. The asymptotes of đ?‘“(đ?‘Ľ) intersect in(6; −8). đ?‘“(đ?‘Ľ) goes through (7; −4). y

y = 2x - 13 x

¡

(7;-4)

(6;-8)

a b

Determine the values of a, đ?‘? and đ?‘?.

(4)

For which values of � would �(�) ≼ �(�)?

(3)

hyperbola.

(3)

c

Determine the co-ordinates of the intersects of đ?‘“ and đ?‘”.

(5)

d

Determine the equation of the dotted line which is the axis of symmetry of the [15]

6

Determine: a b c

lim�→1

đ?‘Ľ 2 −1

(3)

1−đ?‘Ľ

đ?‘“ ′ (đ?‘Ľ) from first principles if đ?‘“(đ?‘Ľ) = −2đ?‘Ľ 2 . 1

đ?‘”′ (đ?‘Ą) if đ?‘”(đ?‘Ą) = 2√đ?‘Ą + 2đ?‘Ą 2 ; đ?‘Ą ≠0

(4) (4) [11]

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 153

Exemplar Paper 1

7

The figure shows the graph of đ?‘“(đ?‘Ľ) = 2đ?‘Ľ 3 + đ?‘Žđ?‘Ľ 2 + đ?‘?đ?‘Ľ + 3. The curve has a local minimum turning point F at (2; −9).

y

D

E A

x

C

B

F (2;−9)

a b c

8

Show that đ?‘Ž = −5 and đ?‘? = −4.

(6)

Determine the equation of the tangent to the graph at đ?‘Ľ = 3.

(4)

If it is given that đ??´(−1; 0), calculate the coordinates of B and C.

(5) [15]

A container firm is designing an open-top rectangular box that will hold 108 đ?‘?đ?‘š3 . The box has a square base with sides đ?‘Ľ and height â„Ž.

a

Show that the total outside surface area of the

b

box will be � = � 2 +

432 đ?‘Ľ

â„Ž

.

For which value of đ?‘Ľ and â„Ž will the outer surface area be a minimum.

đ?‘Ľ

đ?‘Ľ

(4) (5) [9]

9

a

A six-member working group is to be selected from five teachers and nine students. If the working group is randomly selected, what is the probability that it will include at least two teachers?

(4)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 154

Exemplar Paper 1 b

đ?‘ƒ(đ??´ đ?‘œđ?‘&#x; đ??´) = 0,6 and đ?‘ƒ(đ??´) = 0,2

(2)

ii

Find đ?‘ƒ(đ??´) given that events A and B are mutually exclusive.

(4)

i

In how many ways can the letters of the word PROBABILITY

i

c

Find đ?‘ƒ(đ??´) given that events A and B are independent.

be arranged to form different “words� – the word “probability� itself is included? ii

(3)

In how many ways can the letters of the word PROBABILITY be arranged to form different “words� if the R and O have to be kept together?

(3) [16]

MEMORANDUM: Exemplar Paper 1 1

a

i

ii

2

đ?‘Ľ + 2 = đ?‘Ľ+1

(đ?‘Ľ + 2)(đ?‘Ľ + 1) = 2 đ?‘Ľ 2 + 3đ?‘Ľ = 0 đ?‘Ľ(đ?‘Ľ + 3) = 0 ∴ đ?‘Ľ = 0 đ?‘œđ?‘&#x; đ?‘Ľ = −3 đ?‘Ľ − √đ?‘Ľ = 6

Let = √đ?‘Ľ , then đ?‘˜ 2 = đ?‘Ľ đ?‘˜2 − đ?‘˜ − 6 = 0 (đ?‘˜ − 3)(đ?‘˜ + 2) = 0 đ?‘˜ = √đ?‘Ľ = 3 iii

đ?‘Ľ=9

(đ?‘Ľ 2 +4)(2−đ?‘Ľ) đ?‘Ľ+2

or đ?‘˜ = √đ?‘Ľ = −2 Not valid

≼0

(đ?‘Ľ 2 + 4) > 0 for all values of đ?‘Ľ ∈ đ?‘… (2−đ?‘Ľ)

(đ?‘Ľ+2) (đ?‘Ľâˆ’2) (đ?‘Ľ+2)

≼0 ≤0

∴ −2 < đ?‘Ľ ≤ 2

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 155

Exemplar Paper 1 5𝑥−2 + 5𝑥+1 = 126 5𝑥 . 5−2 + 5𝑥 . 51 = 126

iv

1

5𝑥 �25 + 5� = 126 126

5𝑥 � 25 � = 126 b

5𝑥 = 52 𝑥=2

− c

𝑝

−𝑝 3

𝑝 2

𝑝

If (2𝑥 + 𝑝) is a factor, then 𝑓 �− 2� = 2 � 2 � + 𝑝 �− 2� + 𝑏 �− 2� − 9𝑝 = 0 𝑝3 4

+

𝑝3 4

−

𝑏𝑝 2

− 9𝑝 = 0

× 2) 𝑏𝑝 = −18𝑝 ÷ 𝑝) 𝑏 = −18

Substitute 𝑥 = 2: 2(2)2 − 3(2) − 𝑝 = 0 ∴𝑝=2 2𝑥 2 − 3𝑥 − 2 = 0 (2𝑥 + 1)(𝑥 − 2) = 0 1

∴ 𝑥 = − 2 is the other root 2

a

𝑆20 = 410 𝑆50 = 𝑆20 + 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑐𝑥𝑡 30 𝑡𝑐𝑟𝑚𝑠 = 410 + 2865 = 3275 410 =

20 2

[2𝑎 + 19𝑑]

41 = 2𝑎 + 19𝑑 … . (1) 3275 =

50 2

[2𝑎 + 49𝑑]

131 = 2𝑎 + 49𝑑 … . (2) (2)-(1): 30𝑑 = 90

b

i

∴ 𝑑 = 3 en 𝑎 = −8

ii

T:

𝑥 = 8 ; 𝑦 = 24 𝑓:

3 ; 8 ; 15 ; 24 ; 35 5 ; 7 ; 9 ; 11

𝑠: 2 ; 2 ; 2 𝑎 =2÷2=1 𝑏 = 5 − 3(1) = 2 2 𝑇𝑛 = 𝑖 + 2𝑖

𝑐 =3−1−2 =0

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 156

Exemplar Paper 1 c

−24; 48; −96; ‌ is a geometric series with đ?‘Ž = −24 and đ?‘&#x; = −2 đ?‘†6 =

−24ďż˝(−2)6 −1ďż˝

= 504

−2−1

∑đ?‘›đ?‘˜=7(2đ?‘˜ − 3) = 11 + 13 + 15 + â‹Ż + (2đ?‘– − 3) đ?‘› 504 = [2(11) + (đ?‘– − 1)(2)] 2

2

d

đ?‘– + 10đ?‘– − 504 = 0 (đ?‘– + 28)(đ?‘– − 18) = 0 ∴ đ?‘– = 18 đ?‘&#x; =đ?‘Ľ+2

For convergent series −1 < đ?‘&#x; < 1 −1 < đ?‘Ľ + 2 < 1 −3 < đ?‘Ľ < −1 3

a

200 000 =

0,12 đ?‘› ďż˝ −1] 12 0,12 12

1200[�1+

5

(1,01)đ?‘› = 3 đ?‘™đ?‘œđ?‘”

5

3 đ?‘– = đ?‘™đ?‘œđ?‘”1,01 = 51,33755

Melissa must make at least 52 payments b c

đ?‘ƒ=

300000 = đ?‘Ľ=

4

a

0,12 −240 ďż˝ ďż˝ 12 0,12 12 0,15 −40

5000ďż˝1âˆ’ďż˝1+

đ?‘Ľďż˝1âˆ’ďż˝1+

4 0,15 4

0,15 4 0,15 −40 ďż˝1âˆ’ďż˝1+ ďż˝ ďż˝ 4

300000Ă—

ďż˝

= đ?‘…454 097,08 ďż˝

= đ?‘…14957,84

D is the đ?‘Ś −intercept of đ?‘“ and â„Ž . Substitute đ?‘Ľ = 0 into đ?‘Ś = đ?‘š đ?‘Ľ ∴đ?‘Ś=1 ∴đ?‘˜=1

To find đ?‘š we need the coordinates of A.

First find the roots of đ?‘“ os we can get the đ?‘Ľ-value of A. 1

1

− 2 đ?‘Ľ2 + 2 đ?‘Ľ + 1 = 0 đ?‘Ľ2 − đ?‘Ľ − 2 = 0 (đ?‘Ľ − 2)(đ?‘Ľ + 1) = 0

đ?‘Ľ = 2 at C and A

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 157

Exemplar Paper 1 1

Sub 𝐴 �2; 4� into 𝑦 = 𝑚 𝑥 1 4

b c

= 𝑚2

1

∴𝑚=2

3

3

Sub 𝐴(−1; 0): 0 = 𝑎(−1) − 2 𝑥 = −2 at Q and R 𝑄𝑅 = 𝑦𝑄 − 𝑦𝑅 3

3

∴ 𝑎 = −2

1

1

= − 2 (−2) − 2 − [ − 2 (−2)2 + 2 (−2) + 1]

d

7

=2 1

1

3

1

3

− 2 𝑥 2 + 2𝑥 − 2 = 0

e f 5

a

3

− 2 𝑥2 + 2 𝑥 + 1 + 2 𝑥 + 2 = 4 𝑥 2 − 4𝑥 + 3 = 0 (𝑥 − 3)(𝑥 − 1) = 0 ∴ 𝑂𝐷 = 1 ℎ−1 = log 1 𝑥 2

𝑥 > 0; 𝑥 ∈ 𝑅

Asymptotes go through (6; −8) ∴ 𝑏 = −6 and 𝑐 = −8

𝑎

Substitute (7; −4) into 𝑓(𝑥) = 𝑥−6 − 8 𝑎

b

−4 = 7−6 − 8 ∴𝑎=4 4

𝑥−6 4 𝑥−6

− 8 = 2𝑥 − 13 = 2𝑥 − 5

4 = (2𝑥 − 5)(𝑥 − 6) 2𝑥 2 − 17𝑥 + 26 = 0 (2𝑥 − 13)(𝑥 − 2) = 0 𝑥=

13 2

or 𝑥 = 2

𝑦 = 0 or 𝑦 = −9 c d

13

Intersects are � 2 ; 0� and (2; −9) 13

𝑥 ∈ [2; 6) or 𝑥 ∈ [ 2 ; ∞)

Substitute (6; −8) into 𝑦 = 𝑥 + 𝑐 −8 = 6 + 𝑐 ∴ 𝑐 = −14

𝑦 = 𝑥 − 14

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 158

Exemplar Paper 1

6

a b

lim𝑥→1

𝑥 2 −1 1−𝑥

= lim𝑥→1

−(𝑥−1)

= lim𝑥→1 −(𝑥 + 1) = −2 𝑓 ′ (𝑥) = limℎ→0 = lim

ℎ→0

= lim

ℎ→0

𝑓(𝑥+ℎ)−𝑓(𝑥)

ℎ −2(𝑥+ℎ)2 −�−2𝑥 2 � ℎ −2𝑥ℎ−2ℎ2 ℎ

= lim − 2𝑥 − ℎ ℎ→0

c

(𝑥−1)(𝑥+1)

= −2𝑥

1

1

1

𝑔(𝑡) = 2√𝑡 + 2𝑡 2 = 2𝑡 2 + 2 𝑡 −2 1

1

1

𝑔′ (𝑡) = 2 × 2 𝑡 −2 + 2 × −2𝑡 −3 = 7

a

1

√𝑡

1

− 𝑡3

𝑓(2) = −9 and 𝑓 ′ (2) = 0 𝑓 ′ (𝑥) = 6𝑥 2 + 2𝑎𝑥 + 𝑏 0 = 6(2)2 + 2𝑎(2) + 𝑏 4𝑎 + 𝑏 = −24 … . (1) −9 = 2(2)3 + 𝑎(2)2 + 𝑏(2) + 3 2𝑎 + 𝑏 = −14…(2)

b

(1)-(2): 2𝑎 = −10 ∴ 𝑎 = −5 𝑏 = −2𝑎 − 14 = −2(−5) − 14 = −4

From (a) it follows that 𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3

If 𝑥 = −1 is a root, then (𝑥 + 1) is a factor of 𝑓 𝑓(𝑥) = 2𝑥 3 − 5𝑥 2 − 4𝑥 + 3 = (𝑥 + 1)(2𝑥 2 − 7𝑥 + 3 = (𝑥 + 1)(2𝑥 − 1)(𝑥 − 3) 1

𝑥 = −1, or 𝑥 = 2 or 𝑥 = 3 c

1

𝐴 �2 ; 0� and 𝐶(3; 0)

𝑓 ′ (𝑥) = 6𝑥 2 − 10𝑥 − 4 𝑓 ′ (3) = 6(3)2 − 10(3) − 4 = 20 Sub (3; 0) into 𝑦 = 20𝑥 + 𝑐

Eq of tangent: 𝑦 = 20𝑥 − 60

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 159

Exemplar Paper 1

8

a

Volume = 108 đ?‘Ľ 2 â„Ž = 108 ∴ℎ=

108

2

đ?‘Ľ2

� = � + 4�ℎ

108

= đ?‘Ľ 2 + 4đ?‘Ľ ďż˝ đ?‘Ľ 2 ďż˝

b

= đ?‘Ľ2 +

432 đ?‘Ľ

đ?‘† will be a minimum where đ?‘† ′ (đ?‘Ľ) = 0 đ?‘† = đ?‘Ľ 2 + 432đ?‘Ľ −1 đ?‘† ′ (đ?‘Ľ) = 2đ?‘Ľ − đ?‘Ľ 3 = 216

432 đ?‘Ľ2

=0 108

đ?‘Ľ = 6 đ?‘š and â„Ž = (6)2 = 3 đ?‘š 9

a

Total number of different six-member groups= 9!

14! 7!

Number of groups with no teacher= 3! = 60 480

= 17 297 280

Number of groups with one teacher only= 5 Ă— 9 Ă— 8 Ă— 7 Ă— 6 Ă— 5 = 75 600 Total number of groups with less than two teachers= 136 080

Total number of groups with two or more teachers = 17 297 280 − 136 080 = 17 161 200 P(two or more teachers)=

bi

ii

ci ii

17 161 200 17297280

= 0,99

đ?‘ƒ(đ??´đ?‘œđ?‘&#x;đ??´) = đ?‘ƒ(đ??´) + đ?‘ƒ(đ??´) for mutually exclusive 0,6 = 0,2 + đ?‘ƒ(đ??´) đ?‘ƒ(đ??´) = 0,4

đ?‘ƒ(đ??´ đ?‘Žđ?‘–đ?‘‘ đ??´) = đ?‘ƒ(đ??´) Ă— đ?‘ƒ(đ??´) for independentevents đ?‘ƒ(đ??´ đ?‘œđ?‘&#x; đ??´) = đ?‘ƒ(đ??´) + đ?‘ƒ(đ??´) − đ?‘ƒ(đ??´ đ?‘Žđ?‘–đ?‘‘ đ??´) đ?‘ƒ(đ??´ đ?‘œđ?‘&#x; đ??´) = đ?‘ƒ(đ??´) + đ?‘ƒ(đ??´) − đ?‘ƒ(đ??´) Ă— đ?‘ƒ(đ??´) 0,6 = 0,2 + đ?‘ƒ(đ??´) − 0,2đ?‘ƒ(đ??´) 0,4 = 0,8đ?‘ƒ(đ??´) đ?‘ƒ(đ??´) = 0,5 11!

2!2! 10! 2!2!

= 9 979 200 = 907 200

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 160

Exemplar Paper 2

Exemplar Paper 2 (3 hours; 150 marks)

1

Given the following box-and-whisker plot:

a

Which quarter has the smallest spread of data? What is the spread?

(2)

b

Determine the inter quartile range.

(2)

c

Are there more data in the interval 5-10 or in the interval 10-13? How do you know this?

d

(2)

Which interval has the fewest data in it? Is it 0-2, 2-4, 10-12 or 12-13? How do you know it?

(2) [8]

2

A factory produces and stockpiles metal sheets to be shipped to a motor vehicle manufacturing plant. The factory only ships when there is a minimum of 3254 sheets in stock at the beginning of that day. The table shows the day and the number of sheets in stock at the beginning of that day.

a b

Day

1

2

3

4

5

6

Sheets

854

985

1054

1195

1204

1384

Determine the equation of the least squares regression line for this set of data rounding coefficients to three decimal places.

(3)

Use this equation to determine the day the sheets will be shipped.

(3) [6]

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 161

Exemplar Paper 2 3

The ogive below represents the results of a survey amongst first year students on the average time per day they spend exercising. Answer the questions that follow.

a

How many students participated in the survey?

b

Approximately how many students spend more between 10 and 20 minutes per day exercising?

c

(1) (1)

Use the ogive to determine the median time spent on daily exercise. (2) [4]

4

In the diagram, KC is a diameter of the circle and đ??ž(1; 4); đ??ś(7; 2) and đ??´(đ?‘Ľ; đ?‘Ś) are points

on the circle. Determine: a b

the equation of the circle

(5) 1

point B if the gradient of KB= 2

(9) [14]

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 162

Exemplar Paper 2 5

In the diagram, ABCD is a quadrilateral with đ??´(4; 12), đ??´(1; 3), đ??ś(4; 2) and D(8;4).

a

Determine the gradients of BC and CD.

(4)

b

Show that AB â?Š BC.

(3)

quadrilateral.

(4)

Determine the equation of the circle ABCD.

(7)

c d

Prove that ABCD is a cyclic

[18]

6

Given the vertices đ??´(2; 3), đ??´(5; 4), đ??ś(4; 2) and đ??ˇ(1; 1) of parallelogram ABCD. Determine: a b

7

a

the coordinates of M, the point of intersection of diagonals AC and BD

(2)

the equation of the median PM of ∆DMC

(5) [7]

5

If đ?‘ đ?‘?đ?‘?đ??´ = 4 and 180° < đ??´ < 360°, determine the following without the use of a calculator: i ii

b

8

(1)

đ?‘?đ?‘œđ?‘ đ??´

đ?‘ đ?‘–đ?‘–2đ??´

(4)

If đ?‘ đ?‘–đ?‘–17° = đ?‘˜, express

đ?‘Žđ?‘œđ?‘ đ?‘’đ?‘Ž73° đ?‘Žđ?‘œđ?‘ 343°

in terms of đ?‘˜.

(3) [8]

a

Determine the value of the following without using a calculator:

b

đ?‘?đ?‘œđ?‘ 69°. đ?‘?đ?‘œđ?‘ 9° + đ?‘?đ?‘œđ?‘ 81°. đ?‘?đ?‘œđ?‘ 21°

(4)

Consider the following identity: i i

1+đ?‘Žđ?‘œđ?‘ đ?‘Ľ+đ?‘Žđ?‘œđ?‘ 2đ?‘Ľ

1

= ����

(4)

Prove the identity.

(4)

đ?‘ đ?‘–đ?‘›đ?‘Ľ+đ?‘ đ?‘–đ?‘›2đ?‘Ľ

For which values of đ?‘Ľ will the identity be undefined?

[12] _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 163

Exemplar Paper 2

9

a

Solve the following equations for the interval [−90°; 90°]:

(2)

ii

(2)

i b

2đ?‘Ąđ?‘Žđ?‘–đ?‘Ľ = −0,6842

đ?‘ đ?‘–đ?‘–2đ?‘Ľ. đ?‘?đ?‘œđ?‘ đ?‘Ľ − đ?‘ đ?‘–đ?‘–đ?‘Ľ. đ?‘?đ?‘œđ?‘ 2đ?‘Ľ = 0,5

Determine the general solution of: 1

(5)

cos ďż˝2 đ?‘Ľ + 15°� = sin(2đ?‘Ľ − 15°) 10

[9]

The graphs of đ?‘Ś = đ?‘Žđ?‘ đ?‘–đ?‘–đ?‘Ľ and đ?‘Ś = đ?‘?đ?‘œđ?‘ đ?‘?đ?‘Ľ are drawn over the interval y 2

1

x −180

−150

−120

−90

−60

−30

30

60

90

120

150

180

−1

−2

a b

Write down the values of đ?‘Ž and đ?‘?.

(2)

Use your graph to determine approximate values of đ?‘Ľ; đ?‘Ľ ∈ [−180°; 180] for 1

which đ?‘?đ?‘œđ?‘ 2 đ?‘Ľ − đ?‘ đ?‘–đ?‘–đ?‘Ľ = 2

(5)

[7]

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 164

Exemplar Paper 2

11

In the diagram below, Q is the base of a vertical tower PQ, while R and S are points in the same horizontal plane as Q. The angle of elevation of P, the top of the tower, as measured from R, is đ?‘Ľ. Furthermore, đ?‘…đ?‘„ďż˝ đ?‘† = đ?‘Ś, đ?‘„đ?‘† = đ?‘Ž đ?‘šđ?‘?đ?‘Ąđ?‘&#x;đ?‘?đ?‘ and the area of ∆đ?‘„đ?‘…đ?‘† = đ??´ đ?‘š2.

a b

Show that đ?‘ƒđ?‘„ =

2đ??´đ?‘Ąđ?‘Žđ?‘›đ?‘Ľ đ?‘Žđ?‘ đ?‘–đ?‘›đ?‘Ś

Calculate the value of đ?‘Ś if đ?‘ƒđ?‘„ = 76,8đ?‘š; đ?‘Ž = 87,36; đ??´ = 480,9đ?‘š2 and

đ?‘Ľ = 46,5°. 12

a

(5)

(3) [8]

Write down the converse of the following theorem: The angle between a tangent to a circle and a chord drawn through the point of contact, is equal to an angle in the alternate segment.

(2)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 165

Exemplar Paper 2 b

The diagonal AC of quadrilateral ABCD bisects đ??´đ??śĚ‚ đ??ˇ while AD is

a tangent to the circle ABC at point A. Prove that AB is a tangent to circle ACD.

(5)

c

Two circles intersect at A and B. AB is produced to P. PQ is a tangent to the smaller circle at Q. QB produced meets the larger circle at R. PR cuts the larger circle at X. AX and AQ are drawn. Prove that: i

Points A, X, P and Q are on the circumference of the same circle. (5)

ii

PQ is a tangent to the circumscribed circle of ∆đ?‘„đ?‘…đ?‘‹.

(3)

[15] _____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 166

Exemplar Paper 2

13

a b

ďż˝ and đ??´ďż˝ = đ??ˇďż˝ . In ∆đ??´đ??´đ??ś and ∆đ??ˇđ??ˇđ??ˇ, đ??´Ě‚ = đ??ˇ đ??´đ??´

đ??´đ??¸

đ??´đ??¸

Prove the theorem that đ??´đ??´ = đ??´đ??¸ = đ??´đ??¸ .

(8)

In the diagram A, B, C and E are points on a circle. AE bisects đ??´đ??´Ě‚đ??ś and BC. AE intersect in D.

Prove that:

14

i

∆đ??´đ??´đ??ˇ///∆đ??´đ??ˇđ??ś

(4)

ii

đ??´đ??´. đ??´đ??ś = đ??´đ??ˇ2 + đ??´đ??ˇ. đ??ˇđ??ś

(7) [19] đ??´đ?‘…

3

In ∆đ??´đ??´đ??ś , P is the midpoint of AC, RS//BP and đ??´đ??´ = 5. CR and BP intersect at T.

Determine, giving reasons, the following ratios: a b c d

đ??´đ?‘†

(4)

đ?‘†đ??¸ đ?‘…đ?‘‡

(3)

đ?‘†đ?‘ƒ đ??´đ?‘†

đ?‘‡đ??¸ đ??´đ?‘&#x;đ?‘’đ?‘Ž ∆đ?‘‡đ?‘ƒđ??¸ đ??´đ?‘&#x;đ?‘’đ?‘Ž ∆đ?‘…đ?‘†đ??¸

(3)

(6) [15] TOTAL: 150

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 167

Exemplar Paper 2

MEMORANDUM: Exemplar Paper 2 1

a b

Fourth quarter. Spread = 13 − 12 = 1

c

More data in 10-13

IQR= 12 − 2 = 10

Median = 10 and Max=13. Therefore 50% of the data lies in interval 10-13 25% of data lies between 2-10. Therefore less than 50% in 5-10 d

2-4 has fewest data 0-2, 2-10 ,10-12 and 12-13 all represent 25% of the data 2-4 will only be a part of 25% (less than 25%)

2

a b

đ?‘Ś = 767,867 + 98,514đ?‘Ľ

3254 = 767,867 + 98,514đ?‘Ľ 98,514đ?‘Ľ = 2486,133 đ?‘Ľ = 25,236

Shipping will be done on the 26th day.

3

4

a

100

b

80-20=60

c

14 minutes

a

Midpoint= ďż˝

b

1+7 4+2 2

;

2

ďż˝ = (4; 3)

Radius = ďż˝(4 − 1)2 + (3 − 4)2 = √10 (đ?‘Ľ − 4)2 + (đ?‘Ś − 3)2 = 10 KB â?Š BC (đ??´ďż˝ = 90°; angle in semi circle) đ?‘šđ??´đ??¸ = −2

đ?‘Śâˆ’4

1

∴ đ?‘šđ??žđ??´ = đ?‘Ľâˆ’1 = 2

đ?‘Śâˆ’2

∴ đ?‘šđ??´đ??¸ = đ?‘Ľâˆ’7 = −2

2(đ?‘Ś − 4) = đ?‘Ľ − 1

‌..(1)

(đ?‘Ś − 2) = −2(đ?‘Ľ − 7)

Solving equations (1) and (2) simultaneously yields: đ?‘Ľ = 5 ;đ?‘Ś = 6

‌ ‌ (2)

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 168

Exemplar Paper 2 5

a b c

3−2

1

4−2

đ?‘šđ??´đ??¸ = 1−4 = − 3 12−3

đ?‘šđ??´đ??´ =

4−1

∴ AB � BC

12−4

đ?‘šđ??´đ??´ =

4−8

1

đ?‘šđ??¸đ??´ = 8−4 = 2

=3

1

∴ đ?‘šđ??´đ??´ Ă— đ?‘šđ??´đ??¸ = − 3 Ă— 3 = −1 1

= −2

∴ đ?‘šđ??¸đ??´ Ă— đ?‘šđ??´đ??´ = 2 Ă— −2 = −1

∴ AD â?Š CD đ??´ďż˝ = 90° from 5 b ďż˝ = 180° đ??´ďż˝ + đ??ˇ

ďż˝ = 90° đ??ˇ

ABCD is a cyclic quad (opp angles supp) d

AC is diameter of circle( angles in semi circle =90°) Midpoint of AC= �

6

a b

2+4 3+2

Mďż˝

2

;

2

i

2

;

2

5

(diagonals bisect each other)

5 3 1

4

đ?‘?đ?‘œđ?‘ đ??´ = đ?‘ đ?‘’đ?‘Žđ??´ = 5

đ?‘ đ?‘–đ?‘–2đ??´ = 2đ?‘ đ?‘–đ?‘–đ??´đ?‘?đ?‘œđ?‘ đ??´ = 17°

a

ďż˝ = (4; 7)

� = �2 ; 2�

=2Ă—

8

2

Median PM join M with point P on DC, where P is the midpoint of DC

ii

b

;

� = �3; 2�

1+4 1+2

a

2

Radius = 12 − 7 = 5 (đ?‘Ľ − 4)2 + (đ?‘Ś − 7)2 = 25

Pďż˝ 7

4+4 12+2

−24 25

−3 5

4

Ă—5

đ?‘Žđ?‘œđ?‘ đ?‘’đ?‘Ž73° đ?‘Žđ?‘œđ?‘ 343°

=

đ?‘Žđ?‘œđ?‘ đ?‘’đ?‘Ž17° cos 17°

=

1 đ?‘˜ ďż˝đ?‘˜2 −1 1

=

1

đ?‘˜âˆšđ?‘˜ 2 −1

đ?‘?đ?‘œđ?‘ 69°. đ?‘?đ?‘œđ?‘ 9° + đ?‘?đ?‘œđ?‘ 81°. đ?‘?đ?‘œđ?‘ 21° = đ?‘ đ?‘–đ?‘–21°. đ?‘?đ?‘œđ?‘ 9° + đ?‘ đ?‘–đ?‘–9°. đ?‘?đ?‘œđ?‘ 21° = sin(21° + 9°) = sin 30 ° =

1 2

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 169

Exemplar Paper 2 b

i

Undefined at asymptotes of tan 𝑥: 𝑥 = 90° + 𝑘. 180°; 𝑘 ∈ 𝑍 Also undefined where 𝑠𝑖𝑖𝑥 + 𝑠𝑖𝑖2𝑥 = 0 𝑠𝑖𝑖𝑥 + 2𝑠𝑖𝑖𝑥𝑐𝑜𝑠𝑥 = 0 𝑠𝑖𝑖𝑥(1 + 2𝑐𝑜𝑠𝑥) = 0 1

∴ 𝑠𝑖𝑖𝑥 = 0 or 𝑐𝑜𝑠𝑥 = − 2

𝑥 = 𝑘. 360° or 𝑥 = ±120° + 𝑘. 360°; 𝑘 ∈ 𝑍 1+𝑎𝑜𝑠𝑥+𝑎𝑜𝑠2𝑥

ii

𝑠𝑖𝑛𝑥+𝑠𝑖𝑛2𝑥

=

= =

1+𝑎𝑜𝑠𝑥+2𝑎𝑜𝑠2 𝑥−1

𝑠𝑖𝑛𝑥+2𝑠𝑖𝑛𝑥𝑎𝑜𝑠 𝑎𝑜𝑠𝑥(1+2𝑎𝑜𝑠𝑥) 𝑠𝑖𝑛𝑥(1+2𝑎𝑜𝑠𝑥) 𝑎𝑜𝑠𝑥 𝑠𝑖𝑛𝑥 1

= 𝑡𝑎𝑛𝑥 9

a

i

2𝑡𝑎𝑖𝑥 = −0,6842 𝑡𝑎𝑖𝑥 = −0,3421 𝑥 = −18,89°

ii

b

1

𝑠𝑖𝑖2𝑥. 𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑖𝑥. 𝑐𝑜𝑠2𝑥 = 0,5 sin(2𝑥 − 𝑥) = 0,5 𝑠𝑖𝑖𝑥 = 0,5 𝑥 = 60°

cos �2 𝑥 + 15°� = sin(2𝑥 − 15°) 1

cos �2 𝑥 + 15°� = cos[90° −(2𝑥 − 15°)] 1

cos �2 𝑥 + 15°� = cos[90° −(2𝑥 − 15°)] 1

cos �2 𝑥 + 15°� = cos(105° − 2𝑥) 1

�2 𝑥 + 15°� = (105° − 2𝑥) + 𝑘. 360° 1

or �2 𝑥 + 15°� = −(105° − 2𝑥) + 𝑘. 360° 10

a b

𝑥 = 36° + 𝑘. 144°; 𝑘 ∈ 𝑍

𝑎 = 2; 𝑏 = 2

𝑥 = 80° − 𝑘. 240°; 𝑘 ∈ 𝑍

1

𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑖𝑥 = 2

2𝑐𝑜𝑠 2 𝑥 − 2𝑠𝑖𝑖𝑥 = 1 2𝑐𝑜𝑠 2 𝑥 − 1 = 2𝑠𝑖𝑖𝑥

∴ It is where the two graphs meet. 𝑥 = 20° 𝑜𝑟 160°

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 170

Exemplar Paper 2

11

a

đ?‘ƒđ?‘„

đ?‘Ąđ?‘Žđ?‘–đ?‘Ľ = đ?‘„đ?‘…

∴ đ?‘ƒđ?‘„ = đ?‘„đ?‘…đ?‘Ąđ?‘Žđ?‘–đ?‘Ľ

1

Area of ∆đ?‘„đ?‘…đ?‘† = 2 đ?‘„đ?‘†. đ?‘„đ?‘…đ?‘ đ?‘–đ?‘–đ?‘„đ?‘…ďż˝ đ?‘† 1

∴ đ??´ = 2 đ?‘Ž Ă— đ?‘„đ?‘…đ?‘ đ?‘–đ?‘–đ?‘Ś 2đ??´

đ?‘„đ?‘… = đ?‘Žđ?‘ đ?‘–đ?‘›đ?‘Ś b

76,8 =

đ?‘ đ?‘–đ?‘–đ?‘Ś = 12

a

2đ??´

đ?‘ƒđ?‘„ = đ?‘Žđ?‘ đ?‘–đ?‘›đ?‘Ś đ?‘Ąđ?‘Žđ?‘–đ?‘Ľ =

2đ??´đ?‘Ąđ?‘Žđ?‘›đ?‘Ľ

đ?‘Žđ?‘ đ?‘–đ?‘›đ?‘Ś 2(480,9)đ?‘Ąđ?‘Žđ?‘›46,5° 87,36đ?‘ đ?‘–đ?‘›đ?‘Ś 2(480,9)đ?‘Ąđ?‘Žđ?‘›46,5° 87,36(76,8)

đ?‘Ś = 8,69° đ?‘œđ?‘&#x; 171,31°

= 0,151064

If a line is drawn through the endpoint of a chord to form an angle which is

equal to the angle in the opposite segment, then this line is a tangent. b

đ??´Ě‚2 = đ??´ďż˝ = đ?‘Ľ tan chord đ??śĚ‚1 = đ??śĚ‚2 = đ?‘Ś given

đ??´Ě‚1 = 180° − (đ?‘Ľ + đ?‘Ś) sum of angles of ∆đ??´đ??´đ??ś

c

ďż˝ = 180° − (đ?‘Ľ + đ?‘Ś) sum of angles of ∆đ??´đ??ˇđ??ś đ??ˇ ďż˝ ∴ đ??´Ě‚1 = đ??ˇ

∴ AB is a tangent to the circle i

đ?‘‹ďż˝1 = đ??´ďż˝2 angles in same segm

But

ii

= đ??´Ě‚2 + đ?‘„ďż˝3 ext angle of triangle

đ??´Ě‚2 = đ?‘„ďż˝1 + đ?‘„ďż˝2 tan chord ∴ đ?‘‹ďż˝1 = đ?‘„ďż˝1 + đ?‘„ďż˝2 + đ?‘„ďż˝3

∴ A,X,P,Q concyclic (ext angle = opp int angle) đ?‘„ďż˝1 = đ??´Ě‚1

= đ?‘…ďż˝

AXPQ cyclic quad

angles in same segm

∴ PQ is a tangent

_____________________________________________________________________________________ ŠVia Afrika >> Mathematics Grade 12 171

Exemplar Paper 2 13

a

Book work

b

i

In ∆ABD and ∆AEC: 𝐴̂1 = 𝐴̂2 𝐴� = 𝐷�

given angles in same segm

∴∆ABD⦀∆AEC (AAA)

ii

In ∆ABD and ∆CED: 𝐴� = 𝐷�

proven

�1 = 𝐷 �2 𝐷

vert opp ⊾s

∴∆ABD⦀∆CED (AAA) 𝐴𝐴

∴𝐴𝐴 =

𝐴𝐴 𝐴𝐸

∴𝐴𝐴. 𝐴𝐶 = 𝐴𝐷. 𝐴𝐷 = (𝐴𝐷 + 𝐷𝐷)𝐴𝐷 = 𝐴𝐷2 + 𝐴𝐷. 𝐷𝐷 𝐴𝐴

But AD.DE=BD.DC

14

𝐴𝑅

b

Let 𝐴𝑆 = 3𝑚 and 𝐴𝑃 = 5𝑚 but AP=PC (given) ∴ 𝐴𝑃 = 𝑃𝐶 = 5𝑚

c d

3

∴𝐴𝐴. 𝐴𝐶 = 𝐴𝐷2 + 𝐴𝐷. 𝐷𝐶

a

𝐴𝐴

= 5 given

𝐴𝐴

(𝐴𝐸 = 𝐴𝐴 )

Let 𝐴𝑅 = 3𝑘 and 𝐴𝐴 = 5𝑘 𝐴𝑆

3

𝐴𝑆

3𝑚

∴ 𝑆𝑃 = 2

RS//BP

3

∴ 𝑆𝐸 = 7𝑚 = 7 𝑅𝑇 𝑇𝐸

2𝑚

= 5𝑚 2

=5

𝐴𝑟𝑒𝑎∆𝑇𝑃𝐸 𝐴𝑟𝑒𝑎∆𝑅𝑆𝐸

1

RS//TP

𝑇𝐸.𝑃𝐸.𝑠𝑖𝑛𝐴𝐸̂ 𝑅

= 21 2

𝑅𝐸.𝑆𝐸.𝑠𝑖𝑛𝐴𝐸̂ 𝑅

𝑇𝐸 𝑃𝐸

5 5

25

= 𝑅𝐸 . 𝑆𝐸 = 7 . 7 = 49

_____________________________________________________________________________________ ©Via Afrika >> Mathematics Grade 12 172

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