Grade 12 REVISION BOOKLET

MATHEMATICS

TABLE OF CONTENTS PAGE

1. Forward 3

2. How to use this booklet 5

3. Key (SUBJECT) concepts 7

4. Revision Questions Set 1 (Master 40%) xx

5. Revision Questions Set 2 (Master an additional 20%) xxx

6. Check your answers Set 1 xxxx

7. Check your answers Set 2 zzzz

8. Study and Examination Tips

9. Message to Grade 12 learners from the Writers

10. Thank you

1. ForwardMessage from the Minister of Basic Education

2. How to use this Revision Booklet Explain use of book Ensure you understand all the relevant concepts, formulae etc. Explain the sections of mastering the 40% first and then the additional 20% Explain how to work with questions and then how to check answers in Section 6 and

7 Explain how to link to Mind the Gap and Textbooks Explain what a learner should do if they get questions wrong (go back to section in

mind the gap/ textbooks/ask questions and re learn the section) then answer the questions again.

Continue with this process until you get every question correct Then find the other past question papers and go through similar questions and check

the memoranda to ensure that you get them right If you don’t go back to the content and go through again Next attempt Section 5 (additional 20%) follow the same process until you have

mastered all concepts You are now ready to answer 60% of the question paper

1. ALGEBRA AND EQUATIONSNotes:

(a) Simplify expressions using the laws of exponents for rational exponents.

(b) Add, subtract, multiply and divide simple surds.

(c) Solve:

(1) Quadratic equations (by factorization and by using the quadratic formula) and quadratic

inequalities in one variable and interpret the solution graphically;

(2) Equations in two unknowns, one of which is linear and one of which is quadratic, algebraically

or graphically.

HINTS TO LEARNERS: QUADRATIC EQUATIONS/INEQUALITIES

If a quadratic equation is already factorized and one side is equal to zero, then there is no

need to expand the equation just write the answers from the given factors

If one side is factorized and the other side is not equal to zero, then first write the question

in standard form and factorize

Simplify until the right hand side is 0

Then factorize the trinomial

You will get at most two solutions

You might want to check your answers

When the question says correct to one OR two decimal places. You are expected to solve

the quadratic equation using the quadratic formula

For the inequality, simplify so that the right hand side is 0

Then use the graphic (draw the sketch of the parabola) or number line Method

Don’t cross multiply against an inequality sign. You will lose roots

HINTS TO LEARNERS: SIMULTANEOUS EQUATIONS

Make x or y the subject of the formula in the linear equation.

Avoid solving for x or y if it has a co-efficient other than 1

Ensure after substitution you have an equation in 1 variable x or y not both.

Then solve the quadratic equation.

Substitute these values in either equation to get the corresponding values.

Example 1

Example 2

Example 3

Exercise 1

1.1 Quadratic Equations

Solve for x:

1.1.1

43

13

OO

1.1.2

1.1.3 Correct to two decimal places

1.1.4

1.1.5

1.1.6

1.1.7

1.2 Quadratic inequalities

Solve for x:

1.2.1

1.2.2

1.2.3

1.2.4

1.2.5

1.3 Simultaneous Equations

Solve the following equations simultaneously:

1.3.1

1.3.2

1.3.3

1.3.4

1.3.5

1.1 Quadratic Equations

1.1.1

1.1.2

1.1.3

1.1.4

1.1.5

1.1.6

1.1.7

1.2 Quadratic Inequalities

1.2.1

1.2.2

1.2.3

1.2.4

1.2.5

2. PATTERNS, SEQUENCE AND SERIES

There are three types of sequences, namely, Arithmetic sequence (AS), Geometric Sequence (GS) and Quadratics Sequence. When working with patterns or sequences, you must know what type of sequence you are working with. Then identify constant difference or constant ratio or second difference.

A list of numbers in order is called a number pattern or number sequence. We need at least three numbers in the list to work out if the numbers form a pattern. If we only have two numbers, we cannot be sure what the pattern is.

A single number in a pattern or sequence is called a term. Term 1 is written as T1, term 2 is written as T2 and so on. The number of the term shows its position in the sequence.

T10 is the 10th term in the sequence. Tn is the nth term in a sequence.

2.1 Arithmetic Sequences and Series

An arithmetic progression is a sequence in which there is a constant difference

between any two consecutive terms. For example:

100; 101; 102; 103; 104 ; ...

3; 7; 11; 15; 19; ...

The distinct feature of these sequences is that each term, after the first, is obtained by

adding a constant, d, to the previous term. In the examples above, ‘d’ is 1 and 4

respectively. In the discussion that follows ‘a’ is used to represent the first term.

In the above sequences if we replace the “;” by “+” the sequence becomes a series.

For example,

100 + 101 + 102 + 103 + 104 +…

3 + 7 + 11 + 15 + 19 +…

There are two important formulae that can be used to solve most arithmetic sequence

and series problems:

The general or n-th term:

The sum of the first n terms: or

Examples1.1 Find the 32nd term of sequence 3; 7; 11; ….1.2 Find the sum of the first 32 terms of the sequence.

Solution

1.1 We may list all 32 terms, but that will take a long time and a lot of space. The simplest or a quickest way is to use the general term, Tn = a + (n – 1)d

Tn = a + (n – 1)dT32 = 3 + (32 – 1)4 = 3 + 31(4) = 3 + 124 = 127

1.2 Similarly, to find the sum we can add all terms together or we can use some formula:

2.1 Consider the sequence: 30 ; 22 ; 14 ; . . . Find the sum of the first 19 terms of the sequence.

Sn=n2

[2a+(n−1 )d ] or

S32=322

[2×3+(32−1 )4 ]

¿16[ 6+(31)4 ]¿16(130 )¿2210

Sn=n2

(a+T n )

S32=322

(3+T 32)

¿16(3+127 )¿16(130 )¿2210

dnaTn )1(

])1(2[2

dnanSn )(2 nn TanS

2.2 The 5th term of an arithmetic sequence is 17 and a common difference of 6, determine the first term of the sequence.

Solution

2.1 To find S19 without first calculating T19 we use the formula:

n=−64

3

2.2 To find T1 given T5, we may subtract d = 6 four times from 17 because T5 = a + 4d, that is,

or

Exercises 1

1.

Given

1.1 Write down the first THREE terms of the series

1.2 Calculate the sum of the series.2 Given the arithmetic series: −7−3+1+. ..+173

2.1 How many terms are there in the series?2.2 Calculate the sum of the series.2.3 Write the series in sigma notation.

3Given the arithmetic sequence:

3.1 Determine the value of 3.2 Write down the common difference of this sequence.

4 The arithmetic sequence 4 ; 10 ; 16 ; ... is the sequence of first differences of aquadratic

sequence with a first term equal to 3. Determine the term of the quadratic

sequence.

])1(2[2

dnanSn

)]18)(119(302[2

1919 S

)]18)(18(60[2

19

)188(2

19

798

7666617 adaT 45

6417 aa 2417

a 7

99

0

)13(t

t

www 23;42;3

w

th50

1.2 Geometric Sequences and Series

A geometric progression is one in which there is a constant ratio between any two

consecutive geometric terms. For example:

1; 10; 100; 1000; 10000;….

3; 9; 27; 81; 243;….

The distinct feature of these sequences is that each term, after the first, is obtained by

multiplying the previous term by a constant, r. In the examples above, ‘r’ is 10 and 3

respectively. Again, in the discussion that follows ‘a’ is used to represent the first

term.

For example, in the second sequence above, the first term, T1, of the sequence is a = 3

and the second term, T2, of the sequence is ar = 3 x 3. The third term, T3, is ar2 = 3 x

32.

In the above sequences if we replace the “;” by “+” the sequence becomes a series.

For example,

1 + 10 + 100 + 1000 + 10000 + ….

3 + 9 + 27 + 81 + 243 + ….

There are two important formulae that can be used to solve most arithmetic sequence

and series problems:

The general or n-th term: Tn = arn – 1

The sum of the first n terms:

1.Examples

Consider the sequence: 3 ; 12 ; 48 ; 192 ; 768; ...1.1 Find the 12th term of the sequence

1.2 Find the sum of the first eight terms

2. Find the sum of the first eleven terms of the following sequence: 40 ; 8 ; 1,6 ; 0,32 ; 0,064;...

3. A geometric sequence has all its terms positive. The first term is 7 and the third term is 28.

3.1 Find the common ratio.3.2 Find the sum of the first 14 terms.

Solutions

1. The sequence can be written as: 3 ; 3 4 = 12 ; 12 4 = 48 ; 48 4 = 192 ; 192

rraSor

rraS

n

n

n

n

1

)1(1

)1(

4 = 768

1.1 It is a geometric sequence with a = 3 and r = 4; Tn = arn – 1

T12 = 3. 412 – 1

= 3. 411

= 3 4 194 304 = 12 582 912

1.2 Sn=

a(r n−1 )r−1

Sn=3(48−1)

4−1 = 48 – 1 = 65 535

2.r= 8

40=1,6

8=1

5=0,2

and a = 40

Sn=a(r n−1 )

r−1

S11=40(1−0,211)

1−0,2 =50 (1−0,211 )

= 50(0,9 999 999 795) = 50,000

3. a = 7 ; T2 = ar2 = 28

3.1 7r2 = 28 r2 = 4

r = 2

3.2 Sn=

a(r n−1 )r−1

S14=

7 (214−1)2−1

¿7 (214−1 ) = 7 16 383 = 114 681

1.

Exercise 2

The tuition fees for the first three years of school are R2 000 ; R2 500 ; R3 125. If these tuition fees form a geometric sequence, find:

1.1 Find the common ratio, r, for this sequence1.2 If fees continue to rise at the same rate, calculate (to the nearest rand) the total

cost of tuition fees for the first six years of school.

2. A geometric sequence has T 3=20 and T 4=40Determine:

2.1 The common ratio

2.2 A formula for T n

3. If 1−5 t , 1−t and t+1 are the first three terms of a convergent geometric series, calculate:

The value of t.The common ratio.The sum to infinity of the series.

4. The first term of a geometric sequence is 3 and the sum of the first 4 terms is 5 timesthe sum of the first 2 terms. The common ratio is greater than 1.Calculate:

4.1 The first three terms of the sequence, and4.2 The value of n for which the sum to n terms will be 765

1.3 Combined Sequences and Series

Examples

Consider the following sequence of numbers:2 ; 5 ; 2 ; 9 ; 2 ; 13 ; 2 ; 17 ; …

1. 1.1 Write down the next TWO terms of the sequence, given that the pattern continues.

1.2 Calculate the sum of the first 100 terms of the sequence.

2.2

Worked Solution2 ; 21

(2+2+. ..+2 )for 50 terms

+(5+9+13+.. . )for 50 terms

¿∑i=1

50

2+∑i=1

50

(4 i+1)

¿2(50 )+[502 (2(5 )+49( 4 ))]

¿100+25(10+196 )¿100+5150¿5250Exercise 3

1.

Consider the sequence:

12

; 4 ; 14

; 7 ; 18

; 10 ; .. .

1.1 If the pattern continues in the same way, write down the next TWO terms in

the sequence.

1.2 Calculate the sum of the first 50 terms of the sequence.

2.

Given: 0 ; −1

2; 0 ; 1

2; 0 ; 3

2; 0 ; 5

2; 0 ; 7

2; 0 ; . . ..

Assume that this number pattern continues consistently.

2.1 Write down the value of the 191st term of this sequence.

2.2 Determine the sum of the first 500 terms of this sequence.

1.4 Quadratic Sequences

A quadratic progression is a sequence in which there is a second constant difference.

That is, between any two consecutive terms is not constant, but the difference

between any two consecutive terms formed by the first difference is constant. For

example:

100; 103; 110; 121; 136 ; ...

The difference between consecutive terms of the above sequence form another

sequence, obtained as follows: 103 – 100 = 3; 110 – 103 = 7; 121 – 110 = 11; etc.

Thus the sequence is

3; 7; 11; 15; ...

The difference in the new sequence (the first differences of the original sequence) is

constant; this new sequence is arithmetic with d = 4.

Examples

1. Consider the sequence: 5 ; 18 ; 37 ; 62 ; 93 ; …

1.1 If the sequence behaves consistently, determine the next TWO terms of the sequence.

1.2 Calculate a formula for the nth term of the sequence.

1.3 Use your formula to calculate n if the nth term in the sequence is 1 278.

Worked Solution

1.1 130 ; 1731.2 5 18 37 62 93

13 19 25 31 sequence of first difference 6 6 6 second difference is constant

The second difference is constant T n is quadratic an2+bn+c=T n

2 a=6a=3 T n=3n2+bn+c

5=3(1)2+b(1)+c b+c=2 ... (1)

18=3(2)2+b(2)+c 2 b+c=6 ... (2)

(2) – (1):b=4c=−2

T n=3 n2+4 n−2

Alternative method:a (1 )2+b (1)+c=5a+b+c=5 ... (1)

a (2)2+b (2)+c=18 4 a+2b+c=18 ... (2)a (3)2+b(3 )+c=379 a+3 b+c=37 ... (3)

(2) – (1): 3a+b=13 b=13−3 a

Substitute b=13−3 a into (3)9 a+3(13−3 a )+c=379 a+39−9a+c=37 c=−2

Substitute b=13−3 a and c = – 2 into (2)4 a+2(13−3a )+ (−2 )=18 −2 a=−6 a=3 b=4

1.3 3 n2+4n−2=1278

3 n2+4 n−1280=0(3 n+64 )(n−20)=0

n=−643 or n = 20

n=−64

3 is not valid n = 20

1.

Exercise 4

Given the quadratic sequence: – 1 ; – 7 ; – 11 ; p ; …

1.1 Write down the value of p.

1.2 Determine the nth term of the sequence.

1.3 The first difference between two consecutive terms of the sequence is 96.

Calculate the values of these two terms.

2. Given the following quadratic sequence: −2 ; 0 ; 3 ; 7 ; ...

2.1 Write down the value of the next term of this sequence.

2.2 Determine an expression for the nth term of this sequence.

2.3 Which term of the sequence will be equal to 322?

3. Look at the following sequence and answer the questions that follow:10 ; 21 ; 38 ; 61 ; .........3.1 Determine the type of sequence.

3.2 Determine the general term.

3.3 Which term has a value of 1 245?

Answers Exercise 1 1.1 −1+2+5+. .. . or −1 ; 2 ; 5 ; .. .1.2 147502.1 n=42

2.2 3 818

2.3 ∑n=1

46

( 4n−11)

3.1 w=73.2 d=w−1 d=7−1=6

4. 7 255

Exercise 2

1.1 r=1 1

4 1.2 R22 517,58

2.1 r=

T 4

T 3=40

20=2

2.2 T n=5.2n−1

3.1 t=1

3

3.2 1−t=1−1

3=2

3

3.3 1 1

24.1 r=1 (invalid) or r=2 4.2 n=8

Exercise 3

1.1

116 ; 13

1.2 S50=S25+S25=0 , 9997+1000=1000 , 9997

2.1 T 191=0

2.2 S500=31 000

Exercise 4

1.1 p=13

1.2 T n=n2−9 n+7

1.3 n=52 T 52=22432.1 The next term of the sequence is 12

2.2 T n=

12

n2+ 12

n−3 S500=31 0002.3 The 25th term has a value of 322.

3. FUNCTIONS AND GRAPHS3.1 STRAIGHT LINE

General representation or equation

or . is the gradient and is the

Also note the shape of the following linear functions

qaxy xmxy ma or cq or

intercepty

a < 0 a = 0 a > 0 a is undefined

q < 0 y = q q < 0 there is no q-

value

Domain and range is and respectively

3.2 HYPERBOLA

General representation or equation

or or

Dotted lines are asymptotes Dotted lines are asymptotes

is the vertical translation

is the horizontal translation

For , and . The vertical asymptote is and the horizontal

asymptote is . The axis of symmetry are (Positive) and

(Negative)

Domain is and Range is

For , . The vertical asymptote is and the horizontal

x y

xay axy q

xay

qpx

ayqpx

ay

or

0a 0a

q

p

xay 0p 0q 0x

0y xy xy

xx ,0 yy ,0

qxay 0p 0x

asymptote is . The axis of symmetry are (Positive) and

(Negative).

Domain is and Range,

For , the vertical asymptote is and the

horizontal asymptote is . The axis of symmetry is .

Domain is and Range,

For , the vertical asymptote is and

the horizontal asymptote is . The axis of symmetry is .

Domain is and range is

(a)

Example 1

Write down the equations of the asymptotes of

(b) Determine coordinates of B, the x-intercept of f.

(c) Determine the coordinates of D, the y-intercept of

(d) Determine the domain and the range of

(e) Determine the decreasing and increasing functions of the axes of symmetry of

(f) Draw the sketch graph of

Solutions:

a) Vertical asymptote is

Horizontal asymptote is

b) intercept

c) intercept d) Domain is

qy qxy

qxy

xx ,0 yqy ,

apxqyqpx

ay

))((px

qy qpxy

xpx , yqy ,

apxqyqpx

ay

px

qy qpxy

xpx , yqy ,

:Given1

23

)(

x

xf

.f

f

f

f

f

02 x

2x

1y

x 0 y

1132

3)2(1

12

30

xxx

xx

y 0 x xx ;2

Range is

e) Axes of symmetry are:

Increasing (positive

gradient) or

Decreasing (Negative gradient)

f)

3.3 PARABOLA

General representation or Equation

or or or

Important Deductions

for for

For , and , the turning point is and y-intercept is

y = 0

The domain is ℝ and the range is ℝ if or ℝ if

For , , the turning point is and y-intercept is y = q

The domain is ℝ and the range is ℝ if or ℝ if

For , the turning point is and y-intercept is

The domain is ℝ and the range is ℝ if or ℝ if

21

2231

23

y yy ;1

12 xy

1212 xyorx

1xy

3 xy

fy

11

221 x

2axy qaxy 2 qpxay 2

cbxaxy 2

0a 0a

2axy 0p 0q )0;0(

x yy ;0 0a yy ;0

0a

qaxy 2 0p );0( q

x yqy ; 0a yy ;0

0a

qpxay 2 );( qp

qpay 2

x yqy ; 0a yy ;0

0a

For , the turning point is and y-intercept is

y = c

The domain is ℝ and the range is ℝ if or

ℝ if

The roots or x-intercepts are determined by equating y to zero and solve for x.

Example 2:

Sketched below are the graphs of: ; and

h(x) =

A and B are the x- and y - intercepts of h respectively, C (−6 ; 20) and E are the

points of intersection of f and g .

(a) Calculate the coordinates of A, B and E.

(b) Show that the value of

(c) Determine the domain and the range of f

(d) Write down the values of x for which

(e) Determine the equation of the symmetry axis of h if the gradient is negative.

(f) Write down the range of s, if s(x) = f(x) + 2.

(g) Write down the range of t, if t(x) = h(x) + 2

Solutions:

f

D

B

E

gx

O

h

y

A

C(−6 ; 20)

cbxaxy 2

a

bacab

44;

2

2

x

y

abacy ;

44 2

0a

ya

bacy ;4

4 2

0a

82 xxg kxxf 2

12

6

x

16k

0 xfxg

To answer the above questions you need to identify all the functions in order to

apply the deductions indicated above.

A and B are x and y intercepts of g respectively.

at A,

Thus A

at B,

Thus B

E is the x- intercept of the straight line and the parabola. It is easy and straight

forward to use the

equation of the straight line to get the coordinates of E.

At E,

Thus E

b) C(-6; 20) is on f and g ,

substituting the

into

c) Domain is ℝ

Range is ℝ

d) These are values of x for which

the graph of g and f intersect or f

is below g .

It is from C(-6 ; 20) and E(4 ; 0)

That is

e) For negative gradient,

f) + 2 implies the value of p is

increased by 2

The range of s is

g) + 2 implies the value of p is increased by 2

The range of t is ℝ

ℝ

3.4 EXPONENTIAL

General representation or Equation:

or or

The restriction is

0y 012

6

x

44

26

xxx

0;4

12

60

yx

213

yy

2;0

820;0 xy

482

xx

0;4

kkxy 22 620

163620

kk

x

yy ;16

46 x

1)2( xy

312

xyxy

216 y

14y

yy ;21

yy ;3

xaby qaby x qaby px

1;0 bb

Important Deductions

for for for for

For , the asymptote is y = 0 and the y-intercept is

For , the asymptote is y = q and y-intercept is y = a + q

For , the asymptote is y = q and y-intercept is

Example 3

Given: f ( x )=3− x+1−3

(a) Write in the form

(b) Draw the graph of f showing all the intercepts with the axes and the asymptote.

(c) What is the domain and the range of ?

Solutions:

(a)

(b)

The asymptote is , -intercept, y = 0, i.e.

(c)Domain is ℝ and range is ℝ.

3.5 FUNCTIONS AND INVERSES

3

0x

y

f

10 banda 100 banda 10 banda

100 banda

xaby ay

qaby x

qaby px qaby p

)(xf qaby x

f

331333.333.333 1

xxxxy

3y x03

313

x

0311

313

313

0

x

xx

x yy ;3

A function is a relationship between and , where for every -value there is

only y-value. One way to decide whether or not a graph represent a function is to

use the vertical line test. If any line drawn parallel to the y-axis cuts the graph only

once, then the graph represents a function.

Example:

Are the following graphs represent a function or not?

Graph A Graph B Graph C

Answers:

Graph A and Graph B are functions.

Graph C is not a function because the vertical line cuts the graph twice. i.e. for

some x-value on the

graph, there are two y-values.

Inverse of a function is obtained by interchanging x and y of the original function.

i.e. If then its inverse is . Making the subject of the

formula, or

The inverse of a function is the mirror image of the function along the line y = x .

Notation of the inverse is .

Example:

1. Determine the inverse of the following functions in the form

(a) (b)

2. Restrict the domain of such that its inverse will also be a function

Solutions

step 1: Interchange and i.e.

Step 2: Make the subject of the formula:

x y x

baxy bayx y

bxa

y 1

ab

axy

1f

........y

32)( xxf

23)( xxg

23)( xxg

x y 32 yx

y xy 32

(a)

Step 1: interchange x and y, i.e.

Step 2: Make y the subject of the formula:

Since the inverse of a parabola in the syllabus is limited to , The domain

can be restricted to either ℝ or ℝ for all Parabolas in the which

are prescribed by CAPS.

3.6 LOGARITHMIC FUNCTION

is a logarithmic function.

Reads “y is equal to log x base b”

The logarithmic function is only defined if and x > 0

An exponential equation can be written as a logarithmic equation and vice

versa. The base of the exponential equation becomes the base of the

logarithmic equation.

Example

Write each of the following exponential equations as logarithmic equations:

1) 26=64

2) 5³¿125

The inverse of the exponential function is .

Making y the subject of the formula,

23

21

2332

xyorxy

xy

23yx

xy 23

3

32

xy

xy

2axy

xx ;0 xx ;0

xy blog

xy blog

1,0 bb

6

2

1. 2 646 log 64

3

5

2. 5 1253 log 125

xay yax

xa y

Thus logarithmic function is the inverse of exponential function. i.e. If

then

Example

Given: f(x) = 3x

(a)Determine in the form y =……

(b)Sketch the graphs of on the same set of axes.

(c)Write the domain and range of f (x) and

Solution:

(a) The inverse of is . Changing to logarithmic form, it becomes

(b) Using table method

-2 -1 0 1 2

1 3 9

1 3 9

-2 -1 0 1 2

y = x

xaxy

xayxa

a

y

logloglog

loglogloglog

xaxf )(

xxf alog)(1

1f

xyxfxf and)(),( 1

)(1 xf

xy 3 yx 3 yx 3

13log fxy

)(xf

x

)(xf91

31

)(1 xf

x

91

31

)(1 xf

(c)The domain of is ℝ and the range of is ℝ

The domain of is ℝ and the range of is ℝ.

Notes: As a learner, you should be able to do the following:

Identify different types of graphs in terms of equation and/or sketch

Draw the graph given the equation of the graph

Determine the equation of the function or graph given the sketch and some

points on the sketch

Determine the co-ordinates of the turning point of the parabola

Determine the equation of the asymptote of exponential function

Determine the equation of the asymptotes and lines of symmetry of

hyperbolic functions

Determine the intercepts of each function with axes

Determine the domain and range of all the functions

Determine coordinates of points of intersections of different functions.

EXERCISES

Question 1

The graphs of and are sketched below.

1.1 What is the gradient of g

1.2 Determine the coordinates of A and C.

1.3 Determine the coordinates of D

1.4 What is the y-intercept of f ?

1.5 Determine the axis of symmetry of f.

f x f yy ,0

1f xx ,0 1f y

32)( 2 xxxf 1)( xxg

1.6 Determine the coordinates of the turning point of f.

1.7 What is the domain and the range of f ?

Question 2

The equation of graphs are given for – 3 ≤ x ≤ 3.

2.1 Write down the range of .

2.2 Write down the equation of , the inverse of f.

2.3 Write down the domain and the range of .

2.4Draw f and on the same set of axes, showing intercepts with axes and the

line(s) of symmetry.

2.5Is a function or not. Give reason for your answer.

Question 3

The graphs of the functions qpxaxf 2)( and d

txkxg

are

sketched below.

Both graphs cut the y-axis at 4 . One of the points of intersection of the graphs is

R(1 ; –8), which is also the turning point of f. The horizontal asymptote of g is

2y .

x

y

f

0

g2y

4

)8;1(R .

g

xxf 2)(

f

1f

1f

1f

1f

3.1 Calculate the values of a, p and q.

3.2 Calculate the values of k, t and d.

3.3 Determine the value(s) of x in the interval 1x for which ).x(f)x(g

3.4 Determine the domain and the range of f.

3.5 Write down equations for the axes of symmetry of g, both with negative and

positive gradient.

Question 4

Draw sketch graphs of 2

13)(and22)(

xxhxg x

on the same set of axes.

Show all the intercepts with the axes and the and asymptotes.

Question 5

The graphs of f ( x )=x2+2 x−3and g ( x )= ax+ p

+q are drawn below. A is the

y−¿ intercept of both f and g. The horizontal asymptote of g is also a tangent to

f at B, the turning point of f . The equation of the vertical asymptote of g is

x=−1.

5.1 Write down the coordinates of B

5.2 Determine the equations of the asymptotes of g.

5.3 Write down the coordinates of A.

5.4 Determine the equation of g.

5.5 Determine the equations of axes of symmetry of .

5.6 Write down the range of f (x) and that of −f ( x ) .

Question 6

B

0A

Two functions are defined by f (x)=(x−4)(x+2) and g(x )=¿ 2 x−12.

6.1 Write down the gradient of

6.2 Determine the co-ordinates of the turning point of f

6.3 Determine the range of f.

6.4 Determine the equation of the graph h which is the reflection of f about the y−¿

axis.

6.5 Determine the equation of the graph k which is the reflection of f about the x−¿

axis.

6.6 Determine , the inverse of g, in the form y=¿ ….

Answers to Exercises

Question 1

1.1. mg = 1

1.2. A and C are roots of f Therefore, solve for x: 032 2 xx

0;23Cand0;1A

1or23

0132

xx

xx

1.3. D is the point of intersection of f and g. Solve for x: 132 2 xxx

D(2 ; 3)

1.4. 3y

1.5. 41

2231

x

1.6. Substituting 41x

into 32 2 xxy , 825y

. Turning point is

825;

41

1.7. Domain is x ℝ and range is y;

825

ℝ

Question 2

2.1. f is increasing. range is 8

81isthat22 33 yy

2.2. xy 2log

g

1g

2.3. Domain is 8

81 x

and range is 33 y

2.4.

2.5. It is a function because there is only one y value for each value of x

Question 3

3.1. 8;1 qp and a = 4

3.2. 3and

23;2 ktd

3.3. 10 x

3.4. Domain is x ℝ and range is yy ;8 ℝ

3.5. 21

272

23

xyorxyxy

Question 4

Question 5

5.1. 4;1B

5.2. 4and3 yx

5.3. 3;0A

5.4. 4

11)(

x

xg

5.5. 41 xy that is 53 xyorxy

5.6. Range of f is yy ;4 ℝ

Range of -1f is yy ;4 ℝ

Question 6

6.1. mg = 2

6.2. Turning point is 9;1

6.3. Range is yy ;9 ℝ

6.4. )2)(4()( xxxh

6.5. 82)2)(4()()( 2 xxxxxfxk

6.6. 6

21)(1 xyxg

4. FINANCE, GROWTH AND DECAY

Obtain a clear understanding of Simple and Compound

Interest

Also Know the difference between compound increase and compound decrease.

Differentiate between Straight line(linear depreciation) ; and Reducing

balance depreciation

Convert fluently between nominal/effective interest rates for the following compounding

periods:

- Monthly (divide interest rate by 12 and multiply years by 12)

- Quarterly (divide interest rate by 4 and multiply years by 4)

- Half-Yearly or Semi-annually or bi-annually (divide interest rate by 2 and multiply years by 2

Formulae for converting from nominal to effective interest rate (Memorise):

Study in detail the Future value and present value annuity formulae :

, take note of the signs.

You are expected to calculate any of the following in the above formulae.

A,

P,

i, (but not in the and formulae)

n, (by using logarithms)

x,

4.1 Compound increase and decrease

)1( inPF

niPF )1(

)1( inPF

niPF )1(

nnmm

ni

mi

)()(

11

iixF

n

v]1)1[(

i

ixPn

v])1(1[

vF vP

Simple increase and decrease

Example 1

Thulani wants to sell his caravan. The current value of this caravan is R40

000. Calculate its original value if he bought it 5 years ago and the value

depreciates on the straight line method of 9% per year.

Example 2

A photocopier valued at R24 000 depreciates at a rate of 18% p.a. on the

reducing-balance method. After how many years will its value be R15 000?

4.1.1

Exercise 4.1

A survey conducted in December 2008 determined that 5,7 million South

Africans were living with HIV.

The researchers used a model of exponential growth A=P (1+i )n to predict

that there will be 6 million people living with HIV in December 2015.

Calculate as a percentage, the annual rate of increase that the researchers used

for the 7 years.

4.1.2 A bank offers interest on investment at a rate of 15% p.a. compounded

monthly.

(a) Calculate the effective interest rate equivalent to this.

(b) Determine how much a person must invest now so that they have

R 20 000 in four years’ time.

4.1.3 (a) Daniel has a new trailer worth R8 500. The depreciation rate on this item

is 10% p.a. using the reducing balance method.

Calculate the value of the trailer after 4 years.

(b) Draw a sketch graph showing the reducing value of Daniel’s trailer over

the 4 year period.

4.1.4 A car, costing R210 000 depreciates at a rate of a% per annum, on a reducing

balance basis. Calculate the value of a if it takes 8 years for the value of the

car to decrease to R80 000.

4.1.5 Jenna invests R20 000 into a savings account. The interest paid on money in

the account is 15% p.a. compounded monthly. Calculate how long (in years

and months) it will take for the savings to reach R100 000.

4.1.6 Determine how long, in years, it will take for the value of a motor vehicle to

decrease to 25% of its original value if the rate of depreciation, based on

reducing balance method, is 21% per annum.

4.1.7 Feb/March 2016

Diane invests a lump sum of R5 000 in a savings account for exactly 2 years.

The investment earns interest at 10% p.a., compounded quarterly.

(a) What is the quarterly interest rate for Diane's investment?

(b) Calculate the amount in Diane's savings account at the end of the 2

years.

4.1.8 Nomusa received a valuation of R130 000 for her car that has depreciated at a

rate of 15% p.a. on a reducing balance over the last years.

Determine the value of the car 5 years ago. (Round off your answers to the

nearest thousand rands).

4.2 Nominal and Effective interest rate

Example 1

Given an interest rate of 12% p.a. compounded quarterly; determine the effective

interest rate.

Example 2

Convert an interest rate of 10 p.a., compounded monthly, to an annual interest rate,

compounded semi-annually.

Exercise 4.2

4.2.1 Convert an effective annual interest rate of 14,2% p.a. to a nominal annual

interest rate compounded monthly.

4.2.2 Convert a nominal interest rate of 11% p.a. compounded quarterly to nominal

interest rate per annum compounded monthly.

4.3 Annuities

An annuity is a regular payment made at regular intervals at a particular interest rate.

NB: Compounding period of interest rate should be the same as the interval at which

payments are made e.g. if monthly payments are made and interest is added quarterly

then the quarterly interest rate must first be changed to monthly interest rate.

Future Value

Example

Steven is 20 years old and plans to retire with R8 Million when he turns 60. He starts

paying money into a pension fund one month after his 20th birthday, and his last

payment will be made on his 60th birthday. Calculate his monthly payments if he is to

achieve his goal. The interest rate of money accumulated in the pension fund is 13%

p.a. compounded monthly.

Present Value

Example

Patrick purchases a house for R1 200 000. He pays a deposit of 10% of the value of

the house. The bank grants him a loan for the outstanding amount, at an interest rate

of 8,4% p.a. compounded monthly, payable over a period of 20 years.

(a) Calculate the deposit Patrick pays on the house.

(b) Calculate Patrick’s monthly repayments.

(c) Calculate the balance outstanding at the end of 15 years.

(d) Calculate the total amount Patrick would have paid for the house at the end of

20 years.

Final Payment

Example

A loan of R750 000 is taken out at an interest rate of 12,5% p.a. compounded bi-

annually. It is repaid by equal bi-annual payments of R55 000 and a final payment

less than R55 000.

(a) How many payments are required to settle the loan?

NB: period used here is the number of months remaining240 – 180 = 60

(b) What is the outstanding balance on the loan after the final of R55 000?

(c) What is the value of the final payment?

Sinking Fund

Example

Machinery is purchased at a cost of R550 000 and is expected to rise in cost at 15%

per annum compounded annually and depreciate in value at a rate of 8% p.a.

compounded annually.

A sinking fund is started to make provision for replacing the old machine. The sinking

fund pays 16% p.a. compounded monthly, and you make monthly payments into this

account for 10 years. Determine:

(a) The replacement cost in ten years’ time.

(b) The scrap value of the machine in ten years’ time.

(c) The monthly payment into the sinking fund that will make provision for the

replacement of the new machine.

The balance outstanding after the 31st payment is made will accumulate 1-month interest.

Exercise 4.3

4.3.1 Ms Samuel has finally decided to buy an apartment. She takes out a loan of

R500 000 from the bank and plans to repay the loan in monthly instalments

over a period of 20 years. The interest rate charged is 9% p.a. compounded

monthly.

(a) Calculate her monthly repayments.

(b) What is the total amount she will have paid for the apartment after 20

years.

(c) After 15 years, Ms Samuel wins the lotto and decides to clear the

account. What will the outstanding balance on the loan be?

4.3.2 When Sam started his first job, he decided to save R500 at the end of each

month in an account earning interest at 6% p.a. compounded monthly.

(a) Calculate the amount of money he could expect to be in his account at

the end of 15 years.

(b) At the end of the 15 years, Sam added money to his account so that the

total was R150 000. He no longer made monthly payments but the

money remained in the account earning interest at 8,5% p.a.

compounded quarterly.

Calculate the amount in the account after a further 5 years.

4.3.3 Thulani buys a house and takes out a loan

for R450 000. The interest is 9,5% p.a.

compounded monthly.

(a) Calculate his monthly repayments if the loan is repaid over 20 years.

(b) How much money does he pay in total over 20 years to repay the loan?

(c) Calculate the balance on the loan after 8 years.

(d) Calculate his monthly repayments if he chose to repay the loan over

15 years.

(e) How much money does he pay over 15 years to repay the loan and how

much money does he save by reducing the repayment period by 5 years?

4.3.4 Jill negotiates a loan of R300 000 with a bank which has to be paid by means

of R5 000 and a final payment which is less than R5 000. The repayments start

one month after the granting of the loan. Interest is fixed at 18% per annum

compounded monthly.

(a) Determine the number of payments required to settle the loan.

(b) Calculate the balance outstanding after Jill has paid the last R5 000.

(c) Calculate the value of the final payment made by Jill to settle the loan.

4.3.5 Peter purchases a car for R200 000. He expects to replace the car in 5 years’

time. The replacement cost of the car increases at a rate of 11% p.a.

compounded annually and the rate of depreciation of his current car is 18%

p.a. on a reducing balance. Peter sets up a sinking fund to pay for a new car in

5 years. Calculate:

(a) The trade-in value of Peter’s car in 5 years’ time.

(b) The cost of a new car in 5 years’ time.

(c) The value of the sinking fund in 5 years’ time if Peter trades in his old

car.

(d) The monthly payments into the sinking fund if it earns an interest of 7%

p.a. compounded monthly.

ANSWERS

4.1.1

4.1.2 (a)

(b)

4.1.3 (a)

(b)

4.1.4

4.1.5

4.1.6

4.1.7 (a)

(b)

4.1.8

4.2.1

4.2.2

4.3.1 (a)

(b)

(c)

4.3.2 (a)

(b)

4.3.3 (a)

(b)

(c)

8 500

5 500

4years

Rand

s

(d)

(e)

4.3.4 (a)

(b)

(c)

4.3.5 (a)

(b)

(c)

(d)

5. Differential CalculusHINTS TO LEARNERS:

1. There are two ways of finding the derivative- using the power rule and using first principles. First principles involves limits.

2. Differentiate by using the power rule (If f ( x )=axn, thenf ¿( x )=anxn−1

)

3. Don’t forget, derivative notation,f¿( x ) .

4. Simplify the expression first, removing any surds, any quotients etc.

5. The following simplification rule will prove helpful :

a+bc

=ac+ b

c and that xy + 4y = 2x2 +5x -12 is the same as y(x+4) = (2x -3)(x+4) hence y = 2x-3, x≠4

6. It useful to know, 3√ x5=x

53

5.1.1

QUESTION 5. 1

Differentiate, using the first Principles, the functions:f ( x )=−2 x2

Solution

f ( x )=−2 x2

f ( x+h)=−2( x+h)2

¿−2( x2+2 xh+h2)

¿−2 x2−4 xh−2 h2

f ' ( x )=limh→0

f ( x+h)−f ( x )h

¿ limh→0

−2 x2−4 xh−2h−(−2 x2 )h

¿ limh→0

−2 x2−4 xh−2h2+2 x2

h

¿ limh→0

−4 xh−2 h2

h

¿ limh→0

h(−4 x−2 h)h

¿ limh→0

−4 x−2h

¿−4 x

5.1.2 f ( x )=−x2+2 x

Solution

f ( x )=−x2+2 x

f ( x+h)=−( x+h)2+2( x+h)¿−( x2+2 xh+h )2+2( x+h )¿−x2−2 xh−h+2 x+2 h

f ' ( x )=limh→0

−x2−2 xh−h2+2 x+2 h−(−x2+2 x )h

¿ limh→0

−x2−2 xh−h2+2 x+2 h+x2−2 xh

¿ limh→0

−2 xh−h2+2 hh

¿ limh→0

h(−2 x−h+2 )h

¿ limh→0

−2 x−h+2

¿−2 x+2

5.1.3 f ( x )=12

x2+x−2

Solution

f ( x )= 12

x2+x−2

f ( x )=12

( x+h )2+x+h−2

¿12

( x2+2 xh+h2 )+x+h−2

¿12

x2+xh+12

h2+ x+h−2

f ' ( x )=limh→0

12

x2+xh+12

h2+x+h−2−(12

x2+x−2 )

h

¿ limh→0

12

x2+xh+12

h2+ x+h−2−12

x2−x+2

h

¿ limh→0

xh+12

h2+h

h

¿ limh→0

h( x+12

h+1)

h

¿ limh→0

x+12

h+1

¿ x+1

5.1.4 f ( x )=−2x

Solution

f ( x )=− 2x

f ( x+h)=− 2x+h

f ' ( x )= limh→0

−2x+h

−(−2x

)

h

¿ limh→0

−2x+h

+2x

h

¿ limh→0

−2 xx ( x+h)

+2( x+h )x ( x+h)

h

¿ limh→0

−2 x+2( x+h )x ( x+h)

÷h

¿ limh→0

−2 x+2 x+2hx ( x+h)

÷h

¿ limh→0

2 hx ( x+h)

×1h

¿ limh→0

2x ( x+h)

¿2x2

¿2 x−2

5.2.1 QUESTION 5. 2

Determine

dydx if y=( x+1)( x−2)

Solutiony=x2+ x−2x−2¿ x2−x−2dydx

=2 x−1

5. 2.2 Differentiate:

f ( x )=( x+2)3

√x

Solution

f ( x )=(x+2 )3

√ x

¿( x+2 )( x+2 )(x+2 )

x12

¿( x2+4 x+4 )( x+2)

x12

¿x ( x2+4 x+4 )+2( x2+4 x+4 )

x12

¿ x3+4 x2+4 x+2 x2+8 x+8

x12

¿ x3+6 x2+12 x+8

x12

¿ x3

x12

+6 x2

x12

+12 x

x12

+8

x12

¿ x3−1

2 +6 x2−1

2 +12 x1−1

2 +8 x−1

2

¿ x52+6x

32 +12 x

12 +8 x

−12

f ' ( x )=52

x52

−1+6 .3

2x

32

−1+12.1

2. x

12

−1+8 .(−1

2) .x

−12

−1

¿52

x32 +9 x

12 +6 x

−12 −4 x

−32

5.2.3

Determine Dt (

t2−12 t +2

)

Solution

Dt (t2−12 t+2

)

¿ Dt (( t +1)( t−1)2( t+1 )

¿ Dtt−12

¿ Dt (t2

−12

)

¿ Dt (t2

)−Dt (12

)

¿12

5.2.4

Determine f' ( x ) if

f ( x )=2 xx2

− 12√x

x

Solution

f ( x )=2 xx2

−12√x

¿2 x−1−12

x12

f ' ( x )=2(−1) x−2−12

(12

) x12−1

¿−2x−2−14

x−

12

5.2.5 Determine

D x[3√x2−1

2x ]

¿ Dx [ x23 −1

2x ]

¿23

x23

−1−1

2

¿23

x−1

3 −12

5.3.1

QUESTION 5.3

A rectangular box has a length of 5x units, breadth of (9−2 x )units and its height of x units.

5.3.1.1 Show that the volume V of the box is given by V=10 x2−4 x3.

SolutionV=x .2 x (5−2 x )=2 x2(5−2x )¿10 x2−4 x3

5.3.1.2 Determine the value of x for which the box will have a maximum volume.Solution

2x(5 – 2x)

dVdx

=20 x−12 x2

For maximum volume

dVdx

=0

So

20 x−12 x2=04 x(5−3 x )=0

x=53

, x≠0

5.3.2 A 330 ml can of container with height h and radius r is shown below.

5.3.2.1 Determine the height of the can in terms of the radius r.

Solution

Volume = base area. Height

330=πr2×h

h=330πr 2

5.3.2.2 Show that the surface area of the container is A=2πr2+660

r

Solution

A=πr 2+πr2+2 πr . 330πr2

¿2πr2+660r

5.3.2.3 Determine the radius of the container in cm, if the surface area of the container has to be as small as possible.Solution

dAdr

=4 πr−660r2

For area to be small

dAdr

=0

4 πr−660r2

=0

4 πr3=660

r=3√165π

The cubic function is important. You can either be asked to sketch the graph of a cubic function, or determine its equation if given the graph. Remember that at a turning point the derivative is zero.QUESTION 5.4.

4.1 5.4.1Given f ( x )=x3−4 x2−11 x+30

5.4.1.1 Use the fact that f (2)=0 to write a factor of f ( x ) .Solution( x−2 ) is a factor of f .

5.4.1.2 Calculate the coordinates of the intercepts of f .Solutionf ( x )=x3−4 x2−11x+30¿( x−2)( x2−2x−15)¿( x−2)( x+3 )(s−5 )f ( x )=0( x+3)( x−2 )( x−5)=0x=−3 or x=2 or x=5x-intercepts (-3, 0); (2, 0); (5,0)

5.4.1.3 Calculate the coordinates of the stationary points of f .f ( x )=x3−4 x2−11 x+30f ' ( x )=3 x2−8 x−11At turning points f

' ( x )=0( x+1)(3 x−11 )=0

x=−1 or x=−11

3

y=36 y=−400

27

TP’s are (-1;36) and (11

3;−14 , 81)

5.4.1.4 Sketch the curve of f . Show all intercepts with the axes.Solution

5.4.1.5 For which values of x will f' ( x )<0 ?

f ( x )<0 if −1<x<3 ,67

QUESTION 9(Nov 2012)5.4.2 The graph of the function f ( x )=−x3−x2+16 x+16 is sketched below.

x

y

0

f

9.1 Calculate the x-coordinates of the turning points of f.9.2 Calculate the x-coordinate of the point at which f

'( x ) is a maximum.

C Consider the graph of g( x )=−2x2−9 x+5 .9.1 Determine the equation of the tangent to the graph of g at x = –1.9.2 For which values of q will the line y = –5x + q not intersect the parabola?

Solutions5.4.2.1

f ( x )=−x3−x2+16 x+16f ' ( x )=−3 x2−2 x+160=−3 x2−2 x+163 x2+2 x−16=0(3 x+18 )( x−2 )=0

x=− 83 or x=2

5.4.2.2

f // ( x )=0−6 x−2=0

x=−13

5.4.3.1

g( x )=−2 x2−9 x+5g(−1)=−2(−1 )2−9(−1)+15¿12g¿ (x )=−4 x−9mtan=−4(−1)−9¿−5y=−5 x+c12=−5 (−1 )+cc=7y=−5 x+75.4.3.2

y=−5x+q and y=−2x2−9 x+3−5 x+q=−2x2−9 x+3q=−2( x+1)2+7q>7

PAST PAPERS

QUESTION 5.5

5.5.1 Determine f' ( x ) from first principles if f ( x )=−x2+4 .

5.5.2 Determine the derivative of:

5.5.2.1 y=3 x2+10 x

5.5.2.2 f ( x )=(x−3x )

2

5.5.3 Given: f ( x )=2 x3−23 x2+80 x−84

5.5.3.1 Prove that ( x−2 ) is a factor of f.

5.5..3.2 Hence, or otherwise, factorise f ( x ) fully.

5.5..3.3 Determine the x-coordinates of the turning points of f.

5.5.3.4 Sketch the graph of f , clearly labelling ALL turning points and intercepts with the axes.

5.5.3.5 Determine the coordinates of the y-intercept of the tangent to f that has a slope of 40 and touches f at a point where the x-coordinate is an integer.

Answers5.5.1 −2 x

5.5.2.1 6 x+10

5.5.2.2 2 x2−18 x−3

5.5.3.1 0

5. 5.3.2 ( x−2 )(2 x−7 )( x−6 )

5.5.3.3

83 or 5

5.5.3.45.5.3.5 (0, -65)

QUESTION 6.6

A rain gauge is in the shape of a cone. Water flows into the gauge. The height of the water is h cm when the radius is r cm. The angle between the cone edge and the radius is 60°, as shown in the diagram below.

5.6.1 Determine r in terms of h. Leave your answer in surd form.

5.6.2 Determine the derivative of the volume of water with respect to h when h is equal to 9 cm.

Answers

5.6.1

h√3

5.6.2 27 π

QUESTION5.7

The number of molecules of a certain drug in the bloodstream t hours after it has been taken

is represented by the equation M (t )=−t3 + 3 t 2+72 t , 0<t <10 .

5.7.1 Determine the number of molecules of the drug in the bloodstream 3 hours after the drug was taken.

5.7.2 Determine the rate at which the number of molecules of the drug in the bloodstream is

Formulae for volume:

V=π r2 h

V=13

π r2 h

V =lbhV= 4

3π r3

r

60°

h

changing at exactly 2 hours after the drug was taken.

5,7.3 How many hours after taking the drug will the rate at which the number of molecules of the drug in the bloodstream is changing, be a maximum? (3)

Answers5.7.1 216 molecules5.7.2 72 molecules per hour5.7.3 maximum rate of change of the molecules in the bloodstream is after one hour

7. STATISTICS(a) Collect, organise and interpret univariate numerical data in order to determine:

measures of central tendency (mean, median, mode) of grouped and ungrouped

data,

five number summary

box and whisker diagrams

measures of dispersion: range, percentiles, quartiles, inter-quartile and semi-inter-

quartile range.

(b) Represent measures of central tendency and dispersion in univariate numerical data

by:

using ogives; and

calculating the variance and standard deviation of sets of data manually (for small

sets of data) and using calculators (for larger sets of data) and representing results

graphically.

(c) Represent skewed data in box and whisker diagrams, and frequency polygons.

Identify outliers.

(d) Represent bivariate numerical data as a scatter plot and suggest intuitively and by

simple investigation whether a linear, quadratic or exponential function would best fit

the data.

(e) Use a calculator to calculate the linear regression line which best fits a given set of

bivariate numerical data.

(f) Use a calculator to calculate the correlation coefficient of a set of bivariate numerical

data and make relevant deductions.

HINTS TO LEARNERS:

1. Clarity with respect to

1.1 the concepts MEAN and MEDIAN

1.2 an understanding of the concept STANDARD DEVIATION

1.3 the meaning of the standard deviation away from the MEDIAN

1.4 calculation (use of correct calculator keys)

2. Familiarise yourself on how to use the calculator for statistical functions.

Understand clearly the number of data within one, two or three standard deviation of the

mean.

3. Learners must attempt every question in this section, and it is possible to easily score

marks.

Example 1

A street vendor has kept a record of sales for November and December 2007.

The daily sales in rands is shown in the histogram below.

(a) Complete the cumulative frequency table for the sales over November and December.

(b) Draw an ogive for the sales over November and December.

(c) Use your ogive to determine the median value for the daily sales. Explain how you obtain

your answer.

(d) Estimate the interval of the upper 25% of the daily sales.

(a)

Daily sales (in Rand) Frequency Cumulative Frequency

5 5

11 16

22 38

13 51

7 58

3 61

Solutions:

Daily sales (in rands)

Frequency(in days)

3

0

6

9

12

15

18

21

24

60 70 80 90 100 110 120

(b)

40 50 60 70 80 90 100 110 120 1300

10

20

30

40

50

60

70

Sales in Nov and Dec 2007

Daily sales (rands)

Cum

ulati

ve F

requ

ency

Hints:

x-coordinate - use upper limit of each interval

y-coordinate – cumulative frequency

if the frequency of the first interval is not 0, then include an interval before the given

one and use 0 as its frequency

(c) Median = R87. There are 61 data points, so the median in on the 31st position. On the y-axis

put a ruler at 31; move horizontally until you touch the graph, then move vertically down to

read the x-coordinate.

(d) The upper 25% lies above 75%. 75% of 61 = 45,75. Read 45,75 from the y-axis across to the

graph and down to the x-axis. Therefore the upper 75% of sales lies in the :

Example 2: Feb/March 2016

The box and whisker diagram below shows the marks (out of 80) obtained in a History test

by a class of nine learners.

(a) Comment on the skewness of the data.

(b) Write down the range of the marks obtained.

(c) If the learners had to obtain 32 marks to pass the test, estimate the percentage of the class that

failed the test.

(d) In ascending order, the second mark is 28, the third mark 36 and the sixth mark 69. The

seventh and eighth marks are the same. The average mark for this test is 54.

28 36 69

Fill in the marks of the remaining learners in ascending order.

(a)

Solutions:

Skewed to the left or negatively skewed.

(b) Range = 80-20

= 60

(c) 32 is the lower quartile hence 25% of the class failed the test

(d) Median is 62 hence T5=62

Upper quartile is 75which is a mark between seventh and eighth marks and the seventh and

eighth marks are the same

20000020

30 400

500

60 70 80756232

20 28 36 41 62 69 75 75 80

Example 3:

The data below shows the energy levels, in kilocalories per 100 g, of 10 different snack foods.

440 520 480 560 615 550 620 680 545 490

(a) Calculate the mean energy level of these snack foods.

(b) Calculate the standard deviation.

(c) The energy levels, in kilocalories per 100 g, of 10 different breakfast cereals had a

mean of 545,7 kilocalories and a standard deviation of 28 kilocalories. Which of the

two types of food show greater variation in energy levels? What do you conclude?

Solution

(a ) Mean =550010

=550

(b) kilocalories

(c) Snack foods have a greater variation. The standard deviation for snack foods

is 69,03 kilocalories whilst the standard deviation for breakfast cereals is 28

kilocalories. i.e. energy levels of breakfast cereals is spread closer to the mean than in

those of the snack food.

Example 4: Nov 2016

A survey was conducted at a local supermarket relating the distance that shoppers lived from

the store to the average number of times they shopped at the store in a week. The results are

shown in the table below.

Distance from the store

in km1 2 3 4 5 7 8 10

Average number of

times shopped

per week

12 10 7 7 6 2 3 2

σ=69 , 03

0 1 2 3 4 5 6 7 8 9 100

2

4

6

8

10

12

SCATTER PLOT

Distance from the store in kmAve

rage

num

ber

of ti

mes

sh

oppe

d pe

r w

eek

(a) Use the scatter plot to comment on the strength of the relationship between the distance a

shopper lived from the store and the average number of times she/he shopped at the store in a

week.

(b) Calculate the correlation coefficient of the data.

(c) Calculate the equation of the least squares regression line of the data.

(d) Use your answer at QUESTION (c) to estimate the average number of times that a shopper

living 6 km from the supermarket will visit the store in a week.

(e) Sketch the least squares regression line on the scatter plot.

(a) Strong

(b) r = – 0,95 (–0,9462..)…

(c) a = 11,71 (11,7132…)

b = –1,12 (–1,1176…)

y=−1, 12 x+11 , 71

(d

)

y=−1, 12(6 )+11 ,71 = 5 times

(e)

Using CASIO fx 82 ZA plus calculator- Mode - 2: STAT- 2: A + B x- Enter the x values first:

1 =; 2 =; 3 =; 4 =; 5 =; 7=; 8=; 10= - Use arrows to move right to y column

and up to start next to 1. - Enter y values:

12 =; 10 =; 7 =; 7=; 6=; 2=; 3=; 2=- Press (orange) AC button- Press SHIFT STAT (at 1)- Press 5: Reg

Press 3: r = and get r = – 0,95 (–0,9462..)…

To get equation of regression line:- Press (orange) AC button- Press SHIFT STAT (at 1)- Press 5: Reg- Press 1: A = and get 11,7132… This is the y-intercept of the regression line- Press orange AC button- Press SHIFT STAT - Press 5: Reg- Now press 2: B = and get -1,1176…This is the gradient of the regression lineAnswer: The least squares regression line: y=−1, 12 x+11 , 71 (correct to 2 decimal places)

Using SHARP EL-W53HT- Mode - 1: STAT- 1: LINE- Enter the values in coordinate form:- 1 (x,y) 12change; 2 (x,y) 10 change;- 3 (x,y) 7 change; 4 (x,y) 7 change;- 5 (x,y) 6 change; 7 (x,y) 2 change;- 8 (x,y) 3 change; 10 (x,y) 2 changePress On: It goes back to Stat 1 (LINE)Press ALPHA (÷): r _ appears on the screen Press =: the value of r appears on the screen.

To get equation of regression line:Press ALPHA(a): a appears on the screenPerss =: the value of a appears 11,7132… This is the y-intercept of the regression line

To get equation of regression line:Press ALPHA(b): b appears on the screenPerss =: the value of b appears -1,1176… This is the gradient of the regression lineAnswer: The least squares regression line: y=−1, 12 x+11 , 71 (correct to 2 decimal places)

0 1 2 3 4 5 6 7 8 9 100123456789

101112

SCATTER PLOT/SPREIDIAGRAM

Distance from the store in kmAve

rage

num

ber

of ti

mes

sh

oppe

d pe

r w

eek

Practice Exercises:

1. Feb/ March 2016

A company recorded the number of messages sent by e-mail over a period of 60 working

days. The data is shown in the table below.

NUMBER OF MESSAGES NUMBER OF DAYS

10 < x ≤ 20 2

20 < x ≤ 30 8

30 < x ≤ 40 5

40 < x ≤ 50 10

50 < x ≤ 60 12

60 < x ≤ 70 18

70 < x ≤ 80 3

80 < x ≤ 90 2

1.1 Estimate the mean number of messages sent per day, rounded off to TWO

decimal places.

1.2 Draw a cumulative frequency graph (ogive) of the data.

1.3 Hence, estimate the number of days on which 65 or more messages were sent.

2 As part of an environmental awareness initiative, learners of Greenside High School were

requested to collect newspapers for recycling. The cumulative frequency graph (ogive) below

shows the total weight of the newspapers (in kilograms) collected over a period of 6 months

by 30 learners.

2.1 Determine the modal class of the weight of the newspapers collected.

2.2 Determine the median weight of the newspapers collected by this group of learners.

2.3 How many learners collected more than 60 kilograms of newspaper?

3 A group of 30 learners each randomly rolled two dice once and the sum of the values on the

uppermost faces of the dice was recorded. The data is shown in the frequency table below.

Sum of the values Frequency

on uppermost faces

2 0

3 3

4 2

5 4

6 4

7 8

8 3

9 2

10 2

11 1

12 1

3.1 Calculate the mean of the data.

3.2 Determine the median of the data.

3.3 Determine the standard deviation of the data.

3.4 Determine the number of times that the sum of the recorded values of the dice is within ONE

standard deviation from the mean. Show your calculations.

4 At a certain school only 12 candidates take Mathematics and Accounting. The marks, as a

percentage, scored by these candidates in the preparatory examinations for Mathematics and

Accounting are shown in the table and scatter plot below.

Mathematics 52 82 93 95 71 65 77 42 89 48 45 57

Accounting 60 62 88 90 72 67 75 48 83 57 52 62

40 50 60 70 80 90 10040

50

60

70

80

90

100

Scatter Plot

Percentage achieved in Mathematics

Perc

enta

ge a

chie

ved

in A

ccou

nting

4.1 Calculate the mean percentage of the Mathematics data

4.2 Calculate the standard deviation of the Mathematics data

4.3 Determine the number of candidates whose percentages in Mathematics lie within ONE

standard deviation of the mean.

4.4 Calculate the equation of the least squares regression line for the data.

4.5 If a candidate from this group scored 60% in the Mathematics examination but was absent for

the Accounting examination, predict the percentage that this candidate would have scored in

the Accounting examination, using your equation in QUESTION 4.4. (Round off your answer

to the NEAREST INTEGER.)

4.6 Use the scatter plot and identify any outlier(s) in the data.

5 A restaurant wants to know the relationship between the number of customers and the number

of chicken pies that are ordered.

number of customers (x) 5 10 15 20 25 30 35 40

number of chicken pies (y) 3 5 10 10 15 20 20 24

5.1 Determine the equation of the regression line correct to two decimal places.

5.2 Determine the value of r, the correlation coefficient. Describe the type and strength of the

correlation between the number of people and the number of chicken pies ordered.

5.3 Determine how many chicken pies 100 people would order.

5.4 If they only have 12 pies left, how many people can they serve?

ANSWERS

1.1

Mean = 2(15)+8 (25 )+. .. 2(85 )60

=308060 = 51,33 messages per day

1.2

0 10 20 30 40 50 60 70 80 90 1000

5

10

15

20

25

30

35

40

45

50

55

60

65

70

OGIVE/OGIEF

Number of messages/Getal boodskappe

Cum

ulati

ve F

requ

ency

/Kum

ulati

ewe

Frek

wen

sie

1.3 Number of days = 60 – 46 (see on graph above)

= 14 days

2.1

2.2 Approximately 53

2.3

3.1 mean = 6,73

3.2

median =

T15+T 16

2=7+7

2 = 7

3.3 SD = 2,264... 2,26

3.4 (6,73 – 2,26 ; 6,73 + 2,26)

= (4,47 ; 8,99)

4 + 4 + 8 + 3 = 19 times

4.1

4.2

4.3

4.4

4.5

4.6

5.1

5.2

5.3

5.4

8. ANALYTICAL GEOMERY

LEARNING HINTS

N.B. Mathematical language and terminology must be learnt in more detail

NB “Only Analytical methods must be used:”

1. Learners must learn which formula is to be used to prove the most basic aspects of

Analytical Geometry.

Eg. Bisect is 2 mid-points

Perpendicular is the product of 2 gradients = - 1

2. Learners should then follow the method laid out below:

Select the correct formula from the data sheet

Label the ordered pairs using the correct two points, eg A and C.

Substitute correctly and accurately into your chosen formula

Perform the arithmetic, preferably without a calculator

3. Often Analytical Geometry questions follow on, (scaffolding). Look out for that, as you

might have already proven an aspect above, that you will require for the next sub-

question

4. Use the diagram more effectively.

e.g. Highlight the sides you are going to use for proving perpendicular, so you can see

clearly which points you are going to use for the substitution.

5. You must answer the question, and remember to conclude, exactly what you were asked

to show / prove / conclude. Use wording to do this.

6. Learners need to know the 2 forms of the equations of a circle, by alluding to several

different types of examples and exercises. Furthermore this formula needs to be shown to

the candidates by training them to use the Data Sheets. Completing the square in terms of

x and y need to be emphasized.

Where questions required learners to prove or show a certain expected response, you must

convince the marker of their proof.

7. Such questions often require more calculations than those expecting mere calculations, to

be given at the end.

8. Learners need to remember that the product of gradients equals -1, is accepted to prove ¿

If asked to prove ¿ , every effort must be made to show that this product = -1. Repetition

of exam type questions such as this must be practised with learners.

9. With regards to determining the equation of a line, educators must cover all aspects of the

equation of a line including those passing through the origin, as well as the equations of

horizontal and vertical lines in their teaching.

10. Practice exercises are often required to teach the above points.

11. Grade 11 work must NOT be ignored. Circle centre (0 ; 0) was no problem in Grade 11,

but can be completely forgotten in Grade 12.

QUESTION 8.1

In the diagram below, P(1 ; 1), Q(0 ; –2) and R are the vertices of a triangle and P R Q = θ.

The x-intercepts of PQ and PR are M and N respectively. The equations of the sides PR

and QR are y = –x + 2 and x + 3y + 6 = 0 respectively. T is a point on the x-axis, as shown.

8.1.1 Determine the gradient of QP.

Solution:

mPQ=1−(−2)

1−0

O Nx

y

x + 3y + 6 = 0

y = –x + 2

θ

R

Q(0 ; –2)

P(1 ; 1)

TM

= 3

8.1.2 Prove that P Q R = 90°.

Solution:

QR: y=−1

3x−2

mQR=−13

mPQ×mQR=3×−13

=−1

PQ QR ∴P Q R=90 °

8.1.

3

Determine the coordinates of R.

Solution:

−13

x−2=−x+2

23

x=4

x = 6

y = –4

R(6 ; –4)

8.1.4 Calculate the length of PR. Leave your answer in surd form.

Solution

PR=√(1−6)2+(1−(−4 ))2

¿√50=5√2

8.1.5 Determine the equation of a circle passing through P, Q and R in the form

( x−a )2+( y−b )2=r2.

Solution

PR is a diameter [chord subtends 90]

Centre of circlel: ( 1+6

2; 1−4

2 )

=(3 1

2; −1 1

2 )r=√50

2 OR 5√2

2 OR 3,54

∴(x− 72 )

2+( y+ 3

2 )2=50

4 OR 25

2 OR 12,5

8.1.6 Determine the equation of a tangent to the circle passing through P, Q and R at

point P in the form y = mx + c.

Solution

m of radius = –1

m of tangent = 1

Equation of tangent:

y− y1=( x−x1 )y−1=x−1∴ y=x

8.1.7 Calculate the size of θ.

Soluion:

tan P N T=mPR=−1

∴P N T=135°

tan P M T=mPQ=3

∴P M T=71 , 57 °

P=63,43° [ext of ]

∴θ=26 , 57 ° [sum of s in ∆]

QUESTION 8.2

In the diagram below, the equation of the circle with centre O is x2+ y2=20 . The tangent

PRS to the circle at R has the equation y=1

2x+k

. PRS cuts the y-axis at T and the x-

axis at S.

y=12

x+k

x2+ y2=20

S

T

V

O

P

R

x

y

8.2.1 Determine, giving reasons, the equation of OR in the form y = mx + c.

Solution:

OR TR [radius tangent/raakl]

mTR×mOR=−1

mOR = –2

y = –2x

8.2.2 Determine the coordinates of R.

Solution:

x2+( − 2 x )2=20x2+4 x2=205 x2−20=0x2−4=0( x+2)( x−2 )=0

x = 2

y = –2(2) = –4

R(2 ; –4)

8.2.3 Determine the area of OTS, given that R(2 ; –4).

Solution:

Subst R(2 ; –4) into the equation of PRS:

−4=12

(2)+k

k=−5

OT = 5

0=12

x−5

x = 10

OS = 10

Area = 12 OS . OT

= 12 (10)(5)

= 25 sq units

8.2.4 0=

xV +22 and

0=yV −4

2

V(–2 ; 4)

T(0 ; –5) .... from

VT=√(−2−0 )2+(4−(−5 ))2

¿√4+81¿√85

QUESTION 8.3

In the diagram, A(–7 ; 2), B, C(6 ; 3) and D are the vertices of rectangle ABCD.

The equation of AD is y = 2x + 16. Line AB cuts the y-axis at G. The x-intercept of line 

BC is F(p ; 0) and the angle of inclination of BC with the positive x-axis is α . The

diagonals of the rectangle intersect at M.

y

xO

A( –7 ; 2)

B

C(6 ; 3)

D

M

G

F(p ; 0)

y = 2x + 16

8.3.1 Calculate the coordinates of M.

8.3.2 Write down the gradient of BC in terms of p.

8.3.3 Hence, calculate the value of p.

8.3.4 Calculate the length of DB.

8.3.5 Calculate the size of .

8.3.6 Calculate the size of O G B .

8.3.7 Determine the equation of the circle passing through points D, B and C in the form

( x−a )2+( y−b )2=r2 .

8.3.8 If AD is shifted so that ABCD becomes a square, will

BC be a tangent to the circle passing through points A,

M and B, where M is now the intersection of the

diagonals of the square ABCD? Motivate your answer.

ANSWERS

8.3.1

(−12

; 52

)

8.3.2− 3

P−6

8.3.3 p=4 1

2

8.3.4 √170

8.3.5 α=63 , 430

8.3.6 116 ,570

8.3.7 42.5

8.3.8 BC is a tangent

QUESTION 8.4

In the diagram, M is the centre of the circle passing through T(3 ; 7), R and S(5 ; 2). RT is

a diameter of the circle. K(a ; b) is a point in the 4th quadrant such that KTL is a tangent to

the circle at T.

8.4.1 Give a reason why T SR = 90° .

8.4.2 Calculate the gradient of TS.

8.4.3 Determine the equation of the line SR in the form y = mx + c.

8.4.4The equation of the circle above is

( x−9 )2+( y−6 12 )

2=36 1

4.

K(a ; b)

R

S(5 ; 2)

T(3 ; 7)M

x

y

O

L

8.4.4.1

8.4.4.2

8.4.4.3

8.4.4.4

8.4.4.5

Calculate the length of TR in surd form.

Calculate the coordinates of R.

Calculate sin R .

Show that b = 12a – 29.

If TK = TR, calculate the coordinates of K.

QUESTION 8.5

Determine:

8.5.1 The gradient of PQ

8.5.2 The equation of MN in the form y = mx + c and give reasons

8.5.3 The length of MN

8.5.4 The coordinates of S such that PQRS, in this order, is a parallelogram

O 45°

N(7 ; 1)

S

RM

T

Q(–2 ; –3)

P(a ; b)

x

y

8.5.5 The coordinates of P

QUESTION 8.6

In the diagram below, Q(5 ; 2) is the centre of a circle that intersects the y-axis at P(0 ; 6) and S. The tangent APB at P intersects the x-axis at B and makes the angle α with the

positive x-axis. R is a point on the circle and P R S=θ .

8.6.1 Determine the equation of the circle in the form ( x−a )2+( y−b )2=r2 .8.6.2 Calculate the coordinates of S.

8.6.3 Determine the equation of the tangent APB in the form y = mx + c.

8.6.4 Calculate the size of α.

α

R

S

P(0 ; 6)

Q(5 ; 2)

A

Bx

y

O

8.6.5 Calculate, with reasons, the size of .

8.6.6 Calculate the area of PQS.

ANSWERS

8.4.1 in semi circle/ at centre = 2 on circle

8.4.2 −5

2

8.4.3 y=2

5x

8.4.4.1 √145

8.4.4.2 R(15 ;6 )

8.4.4.3 √55

8.4.4.4 b=12 a−29

8.4.4.5 k (2 ;−5)

8.5.1 18.5.2 y = x – 6

8.5.3

7√22

8.5.4 S(16 ; 5) 8.5.5 P(5 ; 4)

8.6.1 ( x−5 )2+( y−2)2= 418.6.2 (0 ; –2) or y = –2S

8.5.3 y =

54

x + 6

8.6.4 α = 51,34°8.6.5 20 sq units8.6.6

9. Trigonometry Please make sure that you know the names of the sides of a right-angled triangle.

Trigonometry is the study of the relationship between the sides and angles of triangles.

The word trigonometry means ‘measurement of triangles’.

The trigonometric ratios

Using as the reference angle in ABC

The side opposite the 90° is the hypotenuse side, therefore side AC is the hypotenuse

side.

The side opposite is the opposite side, therefore AB is the opposite side.

The side adjacent to is called the adjacent side, therefore BC is the adjacent side.

Trigonometry involves the ratios of the sides of right triangles.

sin θ = oppositehypotenuse

cos θ = adjacenthypotenuse

tan θ = oppositeadjacent

PRACTICAL PROBLEMS

PRE – KNOWLEDGE NEEDED:

θAdjacent

Opposite Hypotenuse

CB

A

Right-angled triangle

Adjacent; Opposite and Hypotenuse sides in a right-angled triangle

Pythagoras theorem

We work with the ratios of the sides of the triangle:

The ratio opposite

hypotenuse is called sine (abbreviated to sin θ)

The ratio adjacent

hypotenuse is called cosine (abbreviated to cos θ)

The ratio oppositeadjacent is called tangent (abbreviated to tan θ)

9.1 THE CARTESIAN PLANE AND IDENTITIES

The Cartesian plane is divided into 4 quadrants by two coordinate axes. These 4 quadrants are

labelled 1, 2, 3 and 4 respectively.

When working with ratios on the Cartesian plane we will make use of the symbols x , y and r (radius).

y

x

cosine is positiveCT

Ssine is positive

All functions are positiveA

43

2 1

tangent is positive

y

θ

P ( x ; y )

xx

r

y

sin θ = yr

cos θ = xr

tan θ = yx

We can find any trigonometric functions if the co-ordinates of the terminal side is given.

Example 1.

If sin θ = −4

5 and 900 ≤ θ ≤ 2700. Determine

1.1 cos 2 θ 1.2 5 sin θ − 3 cos θ

Solution

You must find first the length of the other side,

Using Pythagoras:

x2 + y2 = r2

x2 + (−4 )2 = (5 )2

x2 = 25−16x2 = 9x =± 3

4 −

2700

−90 °

O

1800

−180 °

00≤θ≤900900≤θ≤1800

2700≤θ≤3600

900

00

360 °

1

1800≤θ≤2700

32 Angles are measured in this

+

y=−4

3600

2700

1800

y900

00;

x

P

T

r=5

x=?θ

∴ x = −3( x is negative in the 3rd quadrant)

1.1cos 2 θ = (−3

5 )2= 9

25

1.2

Example 2

If tan 1270 = −P , express each of the following in terms of P.

2.1 sin 1270

2.2 cos 1270

2.3

sin 2 1270

cos 2 1270

Solutions

tan 1270 = P−1

= yx ; Since x

2 + y2 = r2;

2.1sin 1270 = P

√1 + P2

2.2cos 1270 =− 1

√1 + P2

2.3

sin2 1270

cos2 1270 =tan2 1270

= P2

8.2 USING SPECIAL ANGLES TO GET EXACT VALUES

300 450 600

x √3 1 1

y 1 1 √3

5 sin θ −3 cos θ = 5(−45 ) − 3(−3

5 )=−4+9

5

=−115

=−2 15

i . e (−1 )2 + P2=r2

√1+P2=√r2

√ 1 +P2=r

r 2 √2 2

sin 00 = 0 cos 00 = 1 tan 00 = 0

sin 900 = 1 cos 900 = 0 tan 900 = undefined

sin 1800 = 0 cos 1800 = −1 tan 1800 = 0

sin 2700 = −1 cos 2700 = 0 tan 2700 = undefined

sin 3600 = 0 cos 3600 = 1 tan 3600 = 0REDUCTION FORMULAE

00 ≤ θ ≤ 3600use:

sin (1800 + θ ) = −sin θ sin (1800 − θ ) = sin θ

cos ( 1800 ± θ ) = −cos θ

sin (3600 − θ ) = −sin θ cos ( 3600 − θ ) = cos θ

1

1

√2450

450

21

√3300

600

00

(1 ;0 )

(0 ; 1 )

(−1 ; 0 )

(0 ; −1 )

y900

1800

2700

3600x

0 °360 ° − θ1800 − θ

180 ° + θ 360 ° − θ

y

x 0 °

tan (1800 − θ ) = − tan θ tan (1800 + θ ) = tan θ

tan (3600 − θ ) = − tan θ

Co-functions:

sin ( 900 ± x ) = cos x cos ( 900 + x ) = −sin x cos (90° − x ) = sin x

θ > 3600: reduce angles to less than 3600

by subtracting multiples of 3600.

e.g tan (θ + k . 3600) = tan θ

θ < 00: sin (− x ) = − sin x ; cos (− x ) = cos x tan (− x ) = − tan x

Examples:

Simplify without using a calculator:

1 . cos (90° − x ) . sin (360°+ x )sin (−90 ° )

+cos 2 (360 °+ x ) . tan 150 °tan (−120 ° )

2.

sin 220 ° . cos 130 ° − cos 320 ° . cos 220° . tan 40 °tan 220 °

Solution

1.

cos ( 90° − x ) . sin ( 360°+ x )sin (−90° )

+ cos 2 (360 °+ x ) . tan 150°tan (−120° )

=(sin x)(sin x )−1

+ (cos x )2 tan(180°−30 ° )tan (−90 °−30 ° )

¿−sin2 x+cos2 x (− tan30 ° )tan 30°

¿−sin2 x−cos2 x¿−(sin2 x+cos2 x )=−1

2.

sin 220 ° . cos 130 ° − cos 320 ° . cos 220° . tan 40 °tan 220 °

=sin(180°+ 40 ° ) .cos (90 °+40 ° )−cos (360°−40 ° ) . cos(180 °+40 ° ) . tan 40 °tan(180°+ 40 ° )

=(−sin 40 ° )(−sin 40 °)−(cos 40 ° )(−cos40 ° )( tan 40 ° )tan 40 °

=sin240 °+cos2 40 °¿1

Exercise 1

1. Given that sin 23 °=√k , determine, in its simplest form, the value of each of the

following in

terms of k, WITHOUT using a calculator:

1.1 sin 203°

1.2 cos 23°

1.3 tan(–23°)

2. Given: sin 16°= p

Determine the following in terms of p, without using a calculator.

2.1 sin196 °

2.2 cos16 °

3. Given: cos 2B=3

5 and 0° ≤ B ≤ 90°

Determine, without using a calculator, the value of EACH of the following in its simplest

form:

3.1 cos B

3.2 sin B

3.3 cos (B + 45°)

Simplify the following expression to a single trigonometric function:

4.1

4 cos(−x ) .cos (90 °+ x )sin(30 °−x ). cos x+cos(30 °−x ). sin x

4.2 tan 480 ° . sin 300° . cos14 ° . sin(−135 ° )sin 104 ° .cos 220°

REMEMBER TRIGONOMETRIC IDENTITIES (work with the LHS and RHS

separately)

Choose most difficult side and use identities to simplify it.

Look for square identities

Change functions to sin x and cos x

If there are fractions: Get LCM

Factorize or simplify if necessary

If the expression contains cos x − 1and cannot be solved, then multiply above and

below by cos x + 1

See which value(s) ofx the expression is not defined.

NB: “hence” means that the previous answer must be used.

Identities:

tan x = sin xcos x ; sin2 x + cos2 x = 1

Compound angles:

o cos (α + β ) = cos α . cos β − sin α . sin β

o cos (α − β ) = cos α . cos β + sin α . sin β

o sin (α + β ) = sin α . cos β + sin β . cos α

o sin (α − β ) = sin α . cos β − sin β . cos α

Doubles angles

o sin 2 A = 2 sin A . cos A

o cos 2 A = 2 cos 2 A − 1

o cos 2 A = 1 − 2 sin 2 A

o cos 2 A = cos 2 A − sin 2 A

Examples

Prove the identity:

1 + sin x1 −sin x

− 1 − sin x1 + sin x

= 4 tan xcos x

Hints:

Simplify the LHS

Get an LCD

Then simplify

GENERAL SOLUTION OF TRIGONOMETRIC EQUATIONS

Write the equation on its own on one side of the equation

Start by simplifying an equation as far as possible. Use identities, double and

compound angle formulae, and factorisation where possible. You want to have one

trig ratio and one angle equal to a constant, for example cos x = 1

2

Find the reference angle

Identify the possible quadrants in which the terminal rays of the angles could be,based

on the sign of the function

Because trig functions are periodic, there will be a number of possible solutions to an

equation. You will need to write down the general solution of the equation.

Once the solution has been solved, write down the general solution by adding the

following:

+ k 3600 for cosine and sine, because they repeat every 3600

+ k 1800 for tangent, because it repeats every 1800

If an equation contain double angles, for example, sin 3 θ , cos 2 x and tan 5 y , find

the general solutions for 3 θ , 2 x and 5 y first. Only then divide by 3, 2 or 5 to find

the final solutions. If you divide first, you will lose valid solutions.

Apply restrictions

Example: Find the general solutions of the following:

2 cos x sin x −cos x = 015 sin2 x−7 cos x+15 cos2 x+2

Hint: Factorize; equate factors to zero; look for quadrants and find solution.

9.3 TRIGONOMETRIC GRAPHS

Know to sketch and interpret the graphs of sine, cosine and tangent

Example

1. In the diagram below, the graphs of f ( x )=cos x+q and g( x )=sin ( x+ p ) are drawn

on the same system of axes for –240° x 240°. The graphs intersect at (0 ° ; 1

2 ), (–

120° ; –1) and (240° ; –1).

1.1 Determine the values of p and q.

1.2 Determine the values of x in the interval –240° x 240° for which f (x) > g(x).

1.3 Describe a transformation that the graph of g has to undergo to form the graph of h,

where h( x )=−cos x .

240°

y

x

f

g

–240° –180° –120° –60° 0° 60° 120° 180°

1

–1

–1

Solutions

1.1 f (x) = cos x –

12 and/en g(x) = sin(x + 30°)

p = 30° and/en q = –

12

1.2 x (–120° ; 0°) OR/OF –120° < x < 0°

1.3 The graph of g has to shift 60° to the left and then be reflected about the x-axis.

9.4 SOLVING TRIANGLES THAT ARE NOT RIGHT – ANGLED

2D AND 3D PROBLEMS

1. PROBLEMS IN TWO DIMENSIONS Problems in two dimensions usually require that you solve a combination of

triangles in one plane ( a flat surface ) It is easier to work with right angled triangles, so identify them first. The triangles usually share a common side, which you can then use in the second

triangle.

Sine Rule:

asin A

= bsin B

= csin C OR

sin Aa

= sin Bb

= sin Cc

The first arrangement rule is recommended to calculate a side, and,

The second arrangement is recommended to calculate an angle.

Use the sine rule for triangles for which you are given: Two angles and a side Two sides and the non – included angle.

B

A

C

b

a

c

Cosine rule There are different arrangements of the cosine rule depending on what you are trying

to find:

For calculating a side, use a2 = b2 + c2 − 2 b c cos A OR

For calculating the included angle, use cos A = b2 + c2 − a2

2 b c

Use the cosine rule when you have been given: Three sides Two sides and the included angle.

Area rule

Area Δ ABC = 1

2a b sin C = 1

2b c sin A= 1

2a c sin B

2. PROBLEMS IN THREE DIMENSIONS Three-dimensional problems use the same principles as two – dimensional ones.

However, triangles are oriented in two planes (horizontal and vertical) and the difficulty can be visualising the situation.

Draw a sketch of the situation if you have not been given one, and shade the horizontal plane to make it easier to see.

Fill in all the given information and fill in every possible angle Look out for right –angled triangles as this is often the place you will start. Use trig

ratios and Pythagoras to solve these triangles. If there are no right-angled triangles, start with the triangle with the most given

information. Look for common sides so that you use one triangle to help you to solve another side.

B

CA

ac

b

A

B

C

b

ca

Use the correct rule for the given situation and the correct arrangement of that rule.

Exercise 2.

1. In the diagram, BC is the diameter of circle BCD. D = 900

BC D = θA C B = αAB = BCBD = p units

1.1 Express BC in term of p and θ .

1.2 Determine, without stating reasons, the size of B1 in terms of α .

1.3Hence, prove that AC = p . sin 2 α

sin θ . sin α

2. AB is a vertical lighthouse 100m high with B

at sea level. At a point P the captain of a fishing boat measures the angle of the angle

of elevation of A from P to be 160.

Travelling in a straight line from P to Q , the captain now measures the angle of

elevation of A from Q to be 140.

He also determines that P B Q = 1100.

2.1 Calculate to the nearest metre, how far from B the boat is at P and Q .

2.2 If the boat took 20 minutes to travel from P

to Q

, calculate, to the nearest whole

number, the average speed of the boat in km per hour.

Calculate the area of triangle PBQ

3. In the diagram below, the graphs of f ( x )=cos x+q and g( x )=sin ( x+ p ) are drawn on the same system of axes for –240° x 240°. The graphs intersect at

(0 ° ; 12 )

, (–120° ; –1) and (240° ; –1).

p

θα

2

1B

D

C

A

1100 160

140

100

P

Q

B

A

3.1 Draw the graph of g(x) = sin(x + 60°) for –120° ≤ x ≤ 240°. 3.2 Determine the values of x in the interval –120° ≤ x ≤ 240° for which sin(x + 60°) + 2cos x > 0.

Answers for exercise 1

1.1 −√k

1.2 √1−k

1.3 −√ k

1−k

2.1 sin 196 °=−sin 16 °=−p

2.2 cos16 °= √1−sin2 16 °=√1− p2

3.1 cos B=√ 4

5or 2

√5 or [ 0° ≤ B ≤ 90°]

3.2 sin B= 1

√5or √5

5

3.3

√22√5

or √1010

4.1 −4 sin 2x

4.2

32=1 1

2

10. EUCLIDEAN GEOMETRY10. NOTES

1

Euclidean Geometry builds from lower grades. For learners to be able to succeed in Euclidean

geometry, they must be able to know definitions, corollaries and statements of theorems from

grade 9. Some of the important deductions are the following which learners should be able to:

Parallel lines cut by a transversal line

Identify Corresponding, Alternating and co-interior angles

Know that corresponding angles are equal

Know that Alternate angles are equal

Know that the sum of co-interior angles is equal to 1800

Triangles

Know different types of triangles

Know that sum of interior angles of a triangle is equal to 1800

Know properties of different types of triangles

Know that all sides of each equilateral triangle are equal.

Know that each of the angles of an equilateral triangle is equal to 600

Know the three congruency axioms

The exterior angle of a triangle is equal to the sum of interior opposite angles

The line joining the midpoints of a triangle is parallel and half the length of the third

side (midpoint theorem)

Quadrilateral

Know different types of quadrilaterals

Know properties of all quadrilaterals

Know that all sides of the Square are equal

All angles of the Square and the Rectangle are equal and each equal to 900

Opposite angles of the Parallelogram and the kite are equal

Opposite sides of the Parallelogram are equal and parallel

Square, Rhombus and Rectangle are Parallelogram

Diagonals of the Square and the Rectangle are equal

Diagonals of the Square, the Rhombus and the Kite bisect each other at right angles

Theorems (learners should be able to prove the following theorems)

The line drawn from the centre of a circle perpendicular to a chord bisects the chord;

The angle subtended by an arc at the centre of a circle is double the size of the angle

subtended by the same arc at the circle (on the same side of the chord as the centre);

The opposite angles of a cyclic quadrilateral are supplementary;

The angle between the tangent to a circle and the chord drawn from the point of contact

is equal to the angle in the alternate segment;

A line drawn parallel to one side of a triangle divides the other two sides proportionally;

Equiangular triangles are similar.

NOTE: Proofs of these theorems are provided in “Mind the Gap” guide and all CAPS

textbooks

10.

2

Some important points about proofs in Geometry

Read the problem carefully for understanding. You may need to underline important points and

make sure you understand each term in the given information.

Draw the sketch if it is not already drawn. The sketch need not be accurately drawn but must be

as close as possible to what is given i.e. lines and angles which are equal must look equal, lines

which are parallel must appear parallel etc. Also indicate further observations based on

previous theorems.

Indicate on the figure or diagram all the equal lines and angles, lines which are parallel,

measures of angles given if not already indicated in the question. It might be more helpful to

have a variety of colour pens or highlighters for this purpose

Usually you can see the conclusion before you actually start your formal proof of a rider.

Don’t forget to write the reason for each important statement you make, quoting in brief the

theorem or another result as you proceed.

Sometimes you may need to work backwards, asking yourself what do I need to show or to

prove (required to be proved) and then see if you can prove that as you reverse.

Examples 1

In the diagram below, tangent KT to the circle at K is parallel to the chord NM. NT cuts the circle

at L. Δ KML is drawn. M 2=400 and M K T=840

Determine, giving reasons, the size of:

1.1 K2

1.2 N1

1.3 T1.4 L2

1.5 L1

Solutions:

From the given information, key words are tangent parallel lines and chords. Therefore statements and

reasons will be based on theorems which have these words. Use those theorems to determine the sizes

of angles which their sizes are not given. When you put the size of an angle, write a short hand reason.

i.e. M1 = 840 Alternate angles, NM // KT , M1 = L2 subtended by the same arc KN, K2 = 400 tan chord

theorem, K1 = N1 = 440 subtended by ML, L1 = 1800 – (124 + 400) = 160 sum of angles of a triangle

etc.

1.1 K2=M 2=400 tan chord theorem

1.2 N1=K1=840−400=440 subtended by chord ML

1.3 T=N1=440Alternate angles NM // KT

1.4 M 1=840 Alternate angles, NM // KT

L2= M 1=840 subtended by the same arc KN

1.5 L1=1800−( M + N1 ) = 1800 – (124 + 400) = 160 sum of angles of a triangle

Example 2

In the diagram below, AB and DC are chords of a circle. E is a point on AB such that BCDE is a

parallelogram. D E B=108 ° and D A E=2 x+40 ° .

Calculate, giving reasons, the value of x.

Solution:

Properties of parallel gram and cyclic quadrilateral.

C=1080 Opposite angles of a parallelogram are equal

C+ D A E=1800 ; 1080+2 x+400=1800 Opposite angles of cyclic quadrilateral ABCD

∴2 x=1800−1480⇒ 2x=320

x=160

Example 3

In the diagram below, ABC is drawn in the circle. TA and TB are tangents to the circle. The

straight line THK is parallel to AC with H on BA and K on BC. AK is drawn. Let A3=x .

3.1Prove that K3=x .

3.2 Prove that AKBT is a cyclic quadrilateral.

3.3 Prove that TK bisects A K B .

Solutions:

Look for Tan chord, Corresponding angles, alternating as we have tangents and parallel

lines

3.1 K3=C [corresp s/ooreenk e ; CA| |KT]

= A3 [tan-chord th/raakl-koordst]

= x

3.2 K3=x= A3 [proved/bewys in 9.1]

AKBT is cyc quad [line (BT) subtends equal s/

lyn (BT) onderspan gelyke e]

OR/OF

[converse s in same segment/

omgek e in dies segment]

3.3 K3=C [proven in 9.1]

= B2 [tan-chord th/raakl-koordst]

=K 2 [s in the same segm/e in dies segm]

TK bisects/halveer A K B

OR/OF

K2=B2 [s in the same seg/e in dies segm]

= A3 [tans from same pt; s opp equal sides/

rkle v dies pt; e to gelyke sye]

= K3 [proven in 9.1]

TK bisects/halveer A K B

Example 4

4.1 Give reasons for the following statements

4.1.1 B1=x

4.1.2 BC D=B1

4.2 Prove that BCDE is a cyclic quadrilateral.

4.3 Which TWO other angles are each equal to x?

4.4 Prove that B2=C1 .

Solutions

4.1 4.1.1 tangent chord theorem/raaklyn-koordstelling

4.1 4.1.2 corresponding/ooreenkomstige s/e; FB || DC

4.2 E1=BC D

∴BCDE = cyclic quad [converse ext cyc quad/omgek: buitekdvh]

4.3 D2=E2 [∠ s in the same segment/∠ e in dies segment ]D2=F B D [alt ∠s , BF ||CD/ verwiss ∠e,BF||CD ]

4.4 B3= y OR B3=C2 [s in the same segment/e in dies segment]

B2=x− y OR B3+ B2=x [from 9.3 and 9.4]

C1=x− y [from 9.2 and 9.3]

∴ B2=C1

OR/OF

In Δ BFE and Δ BEC

E1=E2 [=x ]F=B3+ B4 [ tan-chord theorem ]∴ ΔBFE/// ΔCBE [∠ ,∠ ,∠ ]∴ B2=C1

Exercises

Question 1

In the diagram, the vertices of Δ PNR lie on the circle with centre O. Diameter SR and chord NP

intersect at T. Point W lies on NR. OT¿ NP. R2=30 ° .

1.1 S1.2 R1

1.3 N1

1.4 If it is further given that NW = WR, prove that TNWO is a cyclic quadrilateral.

Question 2

VN and VY are tangents to the circle at N and Y. A is a point on the circle, and AN, AY and

NY are chords so that A = 65°. S is a point on AY so that AN ¿∨¿ SV. S and N are joined

2.1 Write down, with reasons, THREE other angles each equal to 65°

2.2 Prove that VYSN is a cyclic quadrilateral.

2.3 Prove that ∆ASN is isosceles.

Solutions

Question 1

1.1 P1+ P2=90 ° ∠ in semi circle

S+ P1+ P2+ R2=180 ° ∠ ' s in Δ

S+90°+30 °=180 °

∴ S=60°

1.2 S=N 1 + N2 PR subt = ∠ ' s

N1+ N 2=60 °

N1+ N 2+T 1+ R 1=180 ° ∠ ' s in Δ

60 °+90 °+ { R 1 =180° ¿

∴ R 1=30 °

OR

T 4+ P1+ S=180° ∠ ' s in Δ

90 °+ { P1+60 °=180 °¿

P1=30 °

P1=R1 NR subtends = ∠ ' s

∴ R1=30 °

1.3 O 1=2 R1 ∠ at centre = 2.∠ on

circumference

∴ O 1=2(30 °)=60 °

N1+T 1+O 1=180 ° ∠ ' s in Δ

N1+90°+ 60°=180 °

N1=30 °

OR

O 1=2(30 ° )=60 ° ∠at center = 2 . ∠at circumference

N O1 R=120 ° adj. on str. line

N2=R1 ∠ ' s opp = sides

N O1 R+2 N 2 =180° ∠ ' s in Δ

120 °+2 N 2=180°

2 N 2 =180°−120 °

N 2=30 °

But N 1+ N2=60°

∴ N 1=30 °

1.4 NW = WR given

∴ W 1=90 ° line from centre, midpoint chord

T 1=90°

∴ W 1+ T1=180 °

∴ TNWO is a cyclic quad opp ∠ ' ssupplQuestion 2

2.1 S1= 65° (corr ∠’s; AN || SV )/(oor.∠’e; AN || SV)Y 3 = 65° (tan-chord th)(∠ tussen rkl en koord)N 1 = 65° (tan-chord th)¿ tussen rkl en koord)

2.2 S1 = N 1

VYSN is a cyclic quad/is ’n koordevierhoek (YV subtends equal angles)/ YV onderspan gelyke ∠’e)

2.3 S2 = 65° (∠’s in same segment)/ ( ∠’e in dieselfde sirkelsegment)

N 3 = 65° (alt. ∠’s; AN || SV)/(verw∠’s AN || SV)∴ A = N 3

AS = SN (sides opp equal angles)/ (sye teenoor gelyke hoeke)