10 p1 2016 papers/grade 10/maths... · 2020-02-13 · 6 3 3 5 2 4 x y xy solution first equation:6...

PAPER 1 I. EQUATIONS & INEQUALITIES ................................................................................................................. 2

A. One Variable Linear Equations .................................................................................................................................... 2 B. Linear Simultaneous Equations ................................................................................................................................... 5 C. Inequalities on The Number Line ................................................................................................................................. 8 D. Solving Linear Inequalities ......................................................................................................................................... 11 E. Mathematical Modelling ........................................................................................................................................... 13 F. Literal Equations, Changing The Subject of a Formula .............................................................................................. 16

II. EXPONENTS& SURDS ........................................................................................................................... 18 A. Exponents .................................................................................................................................................................. 18 B. Scientific Notation ..................................................................................................................................................... 24 C. Exponential Expressions & Equations ........................................................................................................................ 25 D. Surds .......................................................................................................................................................................... 30 E.Rationalizing Denominators .......................................................................................................................................... 36 E. Trial Tests ................................................................................................................................................................... 38

III. NUMBER PATTERNS ............................................................................................................................. 41 A. Common Number Patterns ........................................................................................................................................ 41 B. Linear/Arithmetic Sequence ...................................................................................................................................... 42 C. Quadratic Sequence .................................................................................................................................................. 45 D. Geometric Sequence ................................................................................................................................................. 48

IV. FACTORIZATION ................................................................................................................................... 50 A. Products of Factors .................................................................................................................................................... 50 B. Factorization .............................................................................................................................................................. 52 C. Factorization of Trinomials ........................................................................................................................................ 56 D. Multiplication and Division of Rational Expressions .................................................................................................. 61 E. Addition and Subtraction of Rational Expressions..................................................................................................... 67

V. QUADRATIC EQUATIONS ..................................................................................................................... 69 A. Solving Quadratic Equations by Factorization ........................................................................................................... 69 B. Solving by Taking Square Root of Both Sides ............................................................................................................. 73 C. Solving by Completing The Square ............................................................................................................................ 75 D. Solving by Using Quadratic Formula .......................................................................................................................... 77 E. Quadratic Equations With Fractions .......................................................................................................................... 79 F. Simultaneous Equations ............................................................................................................................................ 81

VI. FUNCTIONS .......................................................................................................................................... 86 A. Parabola ..................................................................................................................................................................... 86 B. Elements of Parabola and Functions ......................................................................................................................... 93 C. Hyperbola .................................................................................................................................................................. 98 D. Exponential Function ............................................................................................................................................... 104 E. Mixed Functions ...................................................................................................................................................... 110 F. Trial Tests ................................................................................................................................................................. 121

VII. FINANCE ............................................................................................................................................ 123 A. Simple Interest ........................................................................................................................................................ 123 B. Compound Interest .................................................................................................................................................. 124 C. Nominal and Effective Interest Rate ........................................................................................................................ 130

VIII. PROBABILITY ..................................................................................................................................... 135 A. Sets and Venn Diagrams .......................................................................................................................................... 135 B. Probability ................................................................................................................................................................ 139 C. Complement of an Event ......................................................................................................................................... 141 D. Independent Events ................................................................................................................................................. 142 E. Tree Diagrams .......................................................................................................................................................... 145 F. Mutually Exclusive Events ........................................................................................................................................ 149

IX. REVISION AND EXEMPLARS ................................................................................................................ 158 A. Equations and Exponents ........................................................................................................................................ 158 B. Functions ................................................................................................................................................................. 167 C. Exemplar 1 ............................................................................................................................................................... 171 D. Exemplar2 ................................................................................................................................................................ 172 E. Exemplar3 ................................................................................................................................................................ 174

1

I. EQUATIONS & INEQUALITIES A. ONE VARIABLE LINEAR EQUATIONS

The simplest equation to solve is a linear equation. A linear equation is an equation where the power on the variable(letter, e.g. x) is 1(one). This chapter was studied in grade 8; we will give some examples to revise the chapter. QUESTION 1 Solve for x : 6 3 1 2 3 5( ) ( )x x x x

SOLUTION Remove the brackets;

6 3 3 2 3 5x x x x Collect all the terms with variable to one side of the equal sign and the terms with numbers to the other side by changing the signs of the terms moved:

6 3 2 3 5 3x x x x Simplify both sides:

4 8x Divide both sides by 4 to leave x alone:

4 8

4 4

x

2x QUESTION 2 Solve for k :

1 11 1

5 3k k

SOLUTION Multiply each term by 15 ( LCM of 3 and 5 ) to get rid of the denominators.

1 1151 15 1

5 3( ) ( )k k

1 115 15 15 15 1

5 3k k

15 3 5 15k k the unknowns to one side:

3 5 15 15k k

8 30k

8 30

8 8

k

30 15

8 4k

QUESTION 3 Solve for x :

3 8 4 5 87

4 2 6

x x x

SOLUTION Multiply each term by 12 ( LCM of 2, 4 and 6 ) to get rid of the denominators.

Leave the brackets:

3 3 8 12 7 6 4 2 5 8( ) ( ) ( )x x x

Now open and solve: 9 24 84 6 24 10 16

9 6 10 24 16 84 24

13 52

4

x x x

x x x

x

x

2

TASK 1 Solve for the unknowns below

a) 7+5(w‐1)+41=3(3+w) b) 6(3‐x)=5x‐(x‐1)‐3 c) 22‐2(2K‐7)=11K‐(8‐7K) d) 5n + 34 =−2(1 − 7n) e) 2 6 2 5 3 15 9x x x x

f) 1 1

44 2

t

g) 1 1

34 2x x x

h) 3 2

5 52 5d d

i)1 3

23 2

m m

j)2 1

13 4

m m

3

TASK 2 Solve for the unknowns below;

a) 48 28 3 5 9 75( )x x x

b) 4 9 3

7 2

b

b b

c) 3 5 7

22 2

x x

d) 2 3

5 27 2x x

e) 9 20 5 4 6 2b b b

f) 9 7 2

42 7

x xx

g) 5 1 5 5 8 2 4 8x x x x

h) 3 5 3 1 6 3

24 6 2

w w w

4

B. LINEAR SIMULTANEOUS EQUATIONS Thus far, all equations that have been

encountered have one unknown variable, that must be solved for. When two unknown variables need to be solved

for, two equations are required and these equations are known as simultaneous equations. The solutions to the system of simultaneous equations are the values of the unknown variables which satisfy the system of equations simultaneously, that means at the same time. In general, if there are n unknown variables, then

n equations are required to obtain a solution for each of the n variables.

Solution by Substitution A common algebraic technique is the substitution

method: Try to solve easier equation for one of the

variables, preferebly with coefficeient 1. And then substitute the result into the other

equationwhich (hopefully) can be solved. Back substitution then yields the values for the

other variable. QUESTION 1 If a+5=10 and a+2b=25 what is b? SOLUTION Use the first equation ; solve for a:

a=10‐5=5 substitutea into the second equation:

5 2 25

2 20

10

b

b

b

QUESTION 2 Solve the following system of equations: 3x‐5+y=0 and 2x=13‐5y SOLUTION Use the first equation to write y the subject to x.

03 5 x y y =5 ‐3x

Substitute y into the second equation:

2 13 25 15

2 15 13 25

13 12

2 13 5 5 3x x

x x

x x

x

divide both sides by ‐13

13 12

13 1312

13

x

x

Subs. x into 5 3y x

125 3

1329

13

( )y

y

Question 3 Solve the following system of simultaneous equations:

6 3 3

5 2 4

x y

x y

Solution

First equation:6 3 3x y (leave x alone)

6 3 3x y

3+ 3yx =

6

Subs. 3 3

6

yx

into second equation:

5 2 4x y

3 35 2 4

6

yy

Multiple by LCM=6

3 35 6 2 6 4 6

6

5 3 3 12 24

15 15 12 24

3 24 15

3 9

3

yy

y y

y y

y

y

y

Subs. y=3 into3 3

6

yx

3 3 32

6x

5

TASK3 Solve the following system of simultaneous equations:

a) If 1

54x

and ‐x+3z=37 ,

b) x‐z = ‐5 and z+x=11

c) 3

2 14

xx

and 3x+2y=15

d) 4x ‐ 3y = 14, 2x + 3y = ‐2

e) 2x + 6y = 3 4x ‐ 3y = 1

f) 3 2

7 385

rr

and 9r+5k=14

f) 3x + 8y = 24, x + y = 3 h) 11x ‐ 3y = 8, 9x + 4y = 13

6

TASK4 Solve the following system of simultaneous equations:

a) 2 3y 5

3x 2y 1

b) y 13 4x

3x 2y 16

c) 5x ‐ 3y = 1, 4x + 2y = 14

d) 9 15 15

25 12 10

x y

y x

e) 5 17

8 139

yy

and 4x‐9y=10

f) 7 6 18

3 8 13

x y

x y

g) 5 6 27

2 7 20

x y

x y

h) 2x + 3y = ‐8, 3x + 2y = ‐12

7

C. INEQUALITIES ON THE NUMBER LINE

A linear inequality is similar to a linear equation and has the power on the variable is equal to 1. The methods used to solve linear inequalities are identical to those used to solve linear equations. The only difference occurs when there is a multiplication or a division that involves a minus sign. For example, we know that 8 >6. If both sides of the inequality are divided by −2, −4 is not greater than −3. Therefore, the inequality must switch around, making −4 <−3.

The number system: N = { 1;2;3;4;……..} Natural numbers

N 0 = { 0;1;2;3;4;……..} Counting numbers

Z = {…‐3;‐2;‐1 ;0;1;2;3;…} Integers Q = Integer/integer R = Rational numbers+ irrational numbers Inequalities can be shown on a number line:

N; N 0 ; Z : represented by :

R : represented by :

Notation: 1. 1 6{ ; }x x R : means: x is any real

OR 1 6[ ; ) number between 1 and 6;

1 included; 6 excluded 2. [ ] : Included ( ) : Excluded 3. : Included o : Excluded : Infinity are used on the number line.

Examples on the number line: Represent each of the following on the number line. 1. 2 3x ; x N :

2. 2 3x ; x Z :

3. 2 3x ; x R :

4. 3x or 1x ; x R :

Examples: Write down the interval shown on each of the number lines: 1.

3 2x ; OR 3 2( ; ] ; x R

2.

1x or 3x ; x Z OR 2x ; x Z 3.

3x ; x R OR 3( ; ] ; x R

4.

1x ; x R

8

TASK 7Represent each of the following on the number line.

1. 2 5x ; x N :

2. 2x ; x Z :

3. 4x ; x N : 4. 0 4x ; x R : 5. 1x or 3x ; x R :

6. 2x or 4x ; x R : 7. 5 1x ; x R :

8. 1 1

2 34 2

x ; x R :

9. 3x or 2x ; x Z : 10. 3 5( ; ]

9

TASK 8 1‐)Represent each of the following on the number line. a) 2 5x ; x Z :

b) 1 1

3 14 2

x ; x Z

c) 2x or 1x ; x R : d) 5( ; ]

e) 2( ; ]

f) 3 0x or 3x ; x R :

2‐)Write down the interval shown on each of the number lines: a) b) c) d) e)

10

D. SOLVING LINEAR INEQUALITIES The methods used to solve linear inequalities are identical to those used to solve linear equations. The only difference occurs when there is a multiplication or a division that involves a minus sign.

Multiplying or dividing inequality by negative number reverses inequality sign

Example. Solve for x

6 2x Move 6 to the right:

2 6x

4x Multiply both sides by minus and change the direction of the inequality notation:

4x QUESTION 1 Solve for x and illustrate the answer on a number line.

4 3 2 3( )x x

SOLUTION Open the bracket.

4 3 2 6x x 4 2 6 3x x

2 3

3

2

x

x

QUESTION 2 Solve for x and illustrate the answer on a number line.

3 1 2 6 4( )x x

SOLUTION Open the bracket.

3 3 2 6 4x x 3 6 4 3 2x x

3 9x divide both sides by ‐3 and change the direction of the inequality notation;

3x

QUESTION 3 Solve for x and illustrate the answer on a number line.

7 2 3

3 2

x x

SOLUTION Multiply both sides by 6 (LCM of 3 and 2.)

2 14 6 9x x

2 6 9 14x x 4 5x

Divide both sides by ‐3 and change the direction of the inequality notation:

5

4x

QUESTION 4 Solve for x and illustrate the answer on a number line. 5 2 3 9x SOLUTION Send +3 to both sides as ‐3 at the same time(add ‐3 to each place)

5 3 2 9 3x 2 2 6x

Divide each term by ‐2,change the direction!) 1 3x

QUESTION 5 Solve for x and illustrate the answer on a number

line. 4

2 32

x

SOLUTION Multiply each term by 2.

4 4 6x 4 4 6 4x

0 10x

11

TASK 9 Solve for x and illustrate the answer on a number line.

a) 6 3 12x b) 2 1 1 6 5( )x x

c) 1 5

4 3

x x

d) 5 2 0x

e) 4

3 23

x

f) 5 2 3 7x

g) 1 1 1 1

12 3 6 3x x x ( )

h) 1

1 62x

i) 1 1

1 13 3x and

12

E. MATHEMATICAL MODELLING Tom and Jane are friends. Tom picked up Jane’s Physics test paper, but will not tell Jane what her marks are. He knows that Jane hates maths so he decided to tease her. Tom says: “I have 2 marks more than you do and the sum of both our marks is equal to 14. How much did we get?” Let’s help Jane find out what her marks are. We have two unknowns, Tom’s mark (which we shall call t) and Jane’s mark (which we shall call j). Tom has 2 more marks than Jane. Therefore,

t = j + 2 Also, both marks add up to 14. Therefore,

t + j = 14 The two equations make up a set of linear (because the highest power is one) simultaneous equations, which we know how to solve! Substitute for t in the second equation to get:

t + j = 14 j + 2 + j = 14 2j + 2 = 14 j =12 j = 6

Then,

t = j + 2 t = 6 + 2 t = 8

So, we see that Tom scored 8 on his test and Jane scored 6. This problem is an example of a simple mathematical model. We took a problem and we were able to write a set of equations that represented the problem, mathematically. The solution of theequations then gave the solution to the problem.

Problem Solving Strategy The purpose of this section is to teach you the skills that you need to be able to take a problem and formulate it mathematically, in order to solve it. The general steps to follow are: 1. Read ALL of it ! Find out what is requested. 2. Let the requested be a variable e.g. x. 3. Rewrite the information given in terms of x. Means translate the words into algebraic language which is easier than Chinese. 4. Set up an equation (i.e. a mathematical sentence or model) to solve the required variable. 5. Solve the equation algebraically to find the result. Important: Follow the three R’s and solve the problem... Request ‐ Response ‐ Result QUESTION 1 A fruit shake costs R2,00 more than a chocolate milkshake. If three fruit shakes and 5 chocolate milkshakes cost R78,00, determine the individual prices. SOLUTION Summarise the information in a table: Let the price of one chocolate be x .

Price Number Total

Fruit 2x 3 3 2( )x

Chocolate x 5 5 x

Set up an algebraic equation:

3 2 5 78( )x x

Solve the equation:

3 6 5 78x x 3 5 78 6

8 72

9

x x

x

x

Present the final answer: Chocolate milkshake costs R 9,00 and the Fruit shake costs R 11,00.

13

QUESTION 2 Three rulers and two pens cost R 21,00. One ruler and one pen cost R8,00. Find the cost of one ruler and one pen. SOLUTION Translate the problem using variables: Let the cost of one ruler be x rand and the cost of one pen be y rand. Rewrite the information in terms of the variables:

3x+ 2y= 21 x+ y= 8

Solve the equations simultaneously First solve the second equation for y:

y= 8 − x and substitute the result into the first equation:

3x+ 2(8 − x) = 21 3x+ 16 − 2x= 21 x= 5

therefore y= 8 − 5 y= 3

Present the final answers: one Ruler costs R 5,00 and one Pen costs R 3,00 QUESTION 3 The length of a rectangle is 4 cm more than the width. If the perimeter is 32 cm, find the length and the width. SOLUTION Draw a rectangle to summarise the information: Let the width be x.

Set up an algebraic equation: Perimeter : 2(length +with)

2 4 32( )x x

Solve the equation:

2 2 4 32( )x

4 8 32

4 24

6

x

x

x

Present the final answer: The width is 6 cm and the length is 10 cm.

QUESTION 4 The denominator of a fraction is 1 more than the numerator. If 4 is added to the numerator and 6 to the denominator, the new fraction will be equal to the original fraction. Find the original fraction. SOLUTION

Let the original fraction be x

y

Where 1y x .

When 4 is added to the numerator and 6 to the denominator, the new fraction becomes:

4

6

x x

y y

.

Cross‐multiply:

Subst. 1y x into the 4 6y x :

4 1 6( )x x

4 4 6

2 4

2

x x

x

x

Then y: 1

2 1

1

y x

the fraction is : 2

21

.

14

TASK 5 1. Manuel has 5 more CDs than Pedro has. Bob

has twice as many CDs as Manuel has. Altogether the boys have 63 CDs. Find how

many CDs each person has. 2. Three‐eighths of a certain number is 5 more than one‐third of the number. Find the Number. 3. A man runs to a telephone and back in 18

minutes. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone

4. The sum of 27 and an unknown number is 73

more than double of that unknown number. Find the unknown number.

5. In rugby a penalty or drop‐goal counts 3 points; try 5 points and conversion 2 points. A team scored three more tries than conversions and twice as many penalties as conversions and the same number of drop‐goals as conversions. Their total score was 47. How many tries were scored?

6. The length of a rectangle is 2 cm more than the

width of the rectangle. The perimeter of the rectangle is 20 cm. Find the length and the width of the rectangle.

7. I have written down three secret numbers. The

first number is 10 bigger than the second number and the third number is twice the size of the second number. They add up to 310. What is the second number?

8. At the end of a hockey season, Bonginkosi had

scored half his team’s goals and Vukile had scored one third of goals. If Bonginkosi scored 7 goals more than Vukile, how many goals did the team score?

15

F. LITERAL EQUATIONS, CHANGING THE SUBJECT OF A FORMULA

The subject of the formula in;

i) 2 5y x is y

Because we read the expression as y is equal to…

ii) 3k n m is k

Because ...................................................................

iii) 3x y

hx y

is h

Because ...................................................................

iv) 5v xy is v

Because ...................................................................

Changing the subject of a formula means;

re‐writing the same equation leaving another

unknown alone, write in terms of others.

WORKED EXAMPLES i) Change the subject of the formula of

2 5y x .

ii) Solve for m in terms of k and n in

3k n m

iii) If 3k n m , write n in terms of k and m.

iv) Re‐write the following expression making y

as the subject of 3x y

hx y

v) If 5v xyz ,writez in terms of x, y and v.

SOLUTIONS i)

2 5

2 5

5

2

y x

x y

yx

ii)

3

3

3

k n m

k n m

m k n

iii)

3

3

3

k n m

n k m

n k m

iv) 3x y

hx y

Cross multiply:

3

3

1 3

3 3

1 1

( )

( )

hx hy x y

hy y x hx

y h x hx

x hx x hy or

h h

v)

2

2

5

5

5

v xyz

v xyz

vz

xy

16

TASK 6

1. Change the subject of the formula of;

a. 7 1y x

b. 5k n

c. 3 9y x

d. 22 3b a

e. 3 1x y

2. In each of the following expressions write z

interms of other variables;

a. x z y

b. 3

abczh

c. 3a b cz

d. 5

zy

e. 3

12

z mk

f. 3 2z

yx z

g. 3 1

5

z z

a

17

II. EXPONENTS& SURDS A. EXPONENTS

Meaning :52 2 2 2 2 2 32

In 52 x 2: coefficient 5: index x: base The index 5 applies only to x .

Eg. 43 2 3 2 2 2 2 48( )

2 23 3 3 9( )x x x x

Laws of Exponents Multiplication: If the bases are the same, add the powers on the same base.

m n m na a a

Eg. 1. 2 3 5x x x 2. 3 1 4 62 2 2 2 3. 12 2 2 2x y x y ( 12 2 )

4. 3 3 1 5 2 22 2y x y x y x

5. 3 5 5 6 8 112 5 10x y z x y x y z

Multiplication: If the bases are different but the powers are the same,multiply the bases and keep your power the same.

( )m m ma b ab

Eg. 1. 2 2 2( )y x xy

2. 3 3 35 2 10

3. 2 3 6x x x Division: If the bases are the same, subtract the power in denominator from the power in numerator on the same base.

mm n

n

aa

a

Eg. 1. 6

6 2 42

xx x

x

2. 15

213

3 1

24 8

xx

x

3. 4 7 2

6 23

x y zxy z

x y

4. 2 4 5

2 4 5 3 8 123 8

x x xx x

x x

Division: If the bases are different but the powers are the same, divide the base in numerator by the base in denominator and keep your power the same.

( )m

m

m

a a

b b

Eg. 1. 6

6 66

6 63

2 2( )

2. 15

1515

( )a a

b b

Power of 0: Anything to the power of zero is equal to one (1)

0 1a ( 0a )

Eg. 1. 03 1( )y

2. 05 5 1 5x (0 belongs to x not 5)

3. 0 2 25 5a b b

4. 0 1a

5. 0 1( )a

Negative Power: We don’t leave any base with a negative power in the final answer. If the exponential expression moves from the numerator to denominator or vice versa, the sign of the power becomes opposite .

1n

na

a

Eg. 1. 33

1 12

2 8

2. 2

2

1y

y

3. 1 1

1 1

y x x

x y y

4. 2

2

4 4a

b ba

5. 1 13 2 1 1

4 4 3 2 24

Remark: 1 1

1

a b

c

here you cannot cancel out any

term , because there is addition in the numerator.

18

Raise to a power: If there is another power over the original power, we multiply the powers.

( )m n m na a

Eg. 2 3 6

2 4 8

( )

( )

x x

y y

( ) ( )m n n m m na a a

Eg. 22 x can be written as power of 4 2 22 2 4( )x x x

The power of a Product:

( )m n k m k n ka b a b and( )m m k

k

n n k

a a

b b

Eg. 1. 4 3 3 12 122 2 8( )x x x

2. 2 8 4

43 12

( )x y x y

z z

MIXED EXAMPLES

1. 0

0

4 1 4 1 5

4 1 1 1 2( )

a

a

2. 4 3

4 2 8 04

93 4

18

( )( )

xx x y

x

128 8

4

8 8 8

93 4

181 1

7 72 2

xx x

x

x x x

3. 2 3 4 2 2 4 8 12 2 2 4

5 9 7 10 12

2 8 2 8

4 12 4 12

( ) ( )a b ab a b a b

ab a b a b

10 8

10 12

10 10 8 13

4

4

16 64

4 1264

364

364

3

a b

a b

a b

b

b

QUESTION 1

Simplify the following:12

3

512 16

256

.?

SOLUTION The bases 512 , 16 and 256 are the powers of 2. Write each base as power of 2 and use multiplication and division rules.

9 4 12 9 48

8 3 24

9 48 24 33

2 2 2 2

2 2

2 2

.( )

( )

QUESTION 2

Simplify the following:2 3

2 4

128 27 8

32 9

. .?

.

SOLUTION The bases 128, 8 and 32 are the powers of 2, the base 27 and 9 are the powers of 3. ,Write each base as power of 2 and 3. Use multiplication rule.

2 3 7 2 3 3 3 14 9 37

2 4 5 2 2 4 10 8

128 27 8 2 3 2 2 3 22 3

32 9 2 3 2 3

. . ( ) ( )

. ( ) ( )

QUESTION 3

Simplify the following: 1

3

1

18

?

SOLUTION 1 1

3 3

1 33

1 1 1 1

8 2 218

QUESTION 4


3 1 2 3 1

25

5 5 5?

. .

x

x x

SOLUTION

2 4 2 2 4

3 1 2 3 1 3 1 2 3 1

4 8

5 5

4 8 5 5

3

25 5

5 5 5 55

55

5

( )

( )

. .

x x

x x x x

x

x

x x

x

QUESTION 5

Simplify the following:

11 33 62 2

52 4

?xy x xy

x

SOLUTION 1 1 1 3 3

1 1 3 5 1 32 2 2 2 22 2 2 2 2 2

5

2

8 4

2 2

4 2

( )x y x x yx y

x

x y

x y

19

QUESTION 6

Simplify the following: 21 23 2 ?

SOLUTION You can not use the power of the power; the operation in the bracket is addition(two terms). Write the expressions in the bracket in one term.

21 2 2

2

1 13 2

3 47 144

12 49

( )

( )

QUESTION 7


2 1

7 2

3 5?

SOLUTION

1 1

2 1

1 1 2 77 2 7 2 14

1 1 5 93 59 5 45

55 45 22514

14 14 14 19645

QUESTION 8


1

30 27

7125

?

SOLUTION

1133

0 33

13 3

13 3

1

1

27 37 1

125 5

3

5

3

55

3

( )

( )

( )

QUESTION 9


3 13 4

2 3?

SOLUTION

3 1 3 1

3 1

3

3

3 4 3 4

2 3 2 3

2 3

3 4113

108

QUESTION 10


21 1

5

4 9( )

SOLUTION

1 1 1 1

2 2 2 21 1

5 5 5 51 1 9 4 54 94 9 36 36

( ) ( ) ( ) ( )

1 1 1 1

22 2 2 25 5 36

36 6 65 1 5

36

( ) ( ) ( )

QUESTION 11


4 416( )x

SOLUTION

3 34 4 44 4

3 34 44 4

3 3

3

16 2

2

2

8

( ) ( )

( ) ( )

x x

x

x

x

20

TASK 1 1‐) Use exponential laws to simplify the following expressions a) (2u2)3

b) 2 3

2

4p q

pq

c)

05

2ky

d) 2 3 2

4 3

m w m w

m w

e) 32 2

2 7 3

2 2

8

xy x y

xy x y

f)

2 32 4

22 2 4

3

2 3

x y xy

xy x y

2‐) Simplify and express your answer with positive indices

a) 3 13 2 2 4a b a b

b) 2 32 1 42 3x y xy

c) 4 22 2

3 4 7

3 2rs r s

r s s

d) 22 4 6

2 3 2

4

2

x y xy

x y x y

e)

2 22 3 4

2 2

a b ab

b a

21

TASK 2 1‐) Use exponential laws to simplify the following expressions

a)

1 1

3 4

1

6

x x

x

b)

2

327

125

c) 3 42 8 16. . ?

d) 3 4 29 81 27. . ?

e) 5 325 125. ?

f) 1

2 1531024 512 2. . ?

g) 1

2 62243 9 81. . ?

2‐) Simplify and express your answer with positiveexponents

a) 7 7

26 4

8

2

a b

ba

x y

x y

. .?

.

b) 3 2 2 3 ?x y z xy z

c) ?x y y x

a b

b a

d) 3

322

1?a

a

e)

3 1 13 4 4 2

1

2

2 2

4

. .?

a a b

b

...

22

TASK 3 1‐) Use exponential laws to simplify the following expressions

a) 1 13 4 ?

b) 2 1

2

3 5

2

.?

c) 2

3

3 5

2?

d) 1 1

2 1

5 3

2 7?

e) 1a a

f) 1 1

1?

x y

x

g)

2 15 25

4 100?

2‐) Simplify and express your answer with positive indices

a)

43

4

34

3

?

x

x

b) 1 1

2 22 4 ?z z

c)

22 3

2

3

3?

x

x

d) 1 13 2

6

.x x

x

e)

2

30 64

527

f)

11

32

5

6

.?

x x

x

23

B. SCIENTIFIC NOTATION

Scientific Notation is a convenient way to express very large or very small numbers. Scientific Notation requires 1 significant digit in front of the comma. Examples:

1. 30 00318 318 10, ,

2. 91340000000 1 34 10,

2 1

Comma is moved from 1 to2 , that is 9 places.

3. 41 25 10 12500,

4. 65 068 10 0 000005068, ,

5. 4 8 4 32 10 9 1 10 18 2 10 1 82 10, , ,

6.

2 3

2 2

2

2

1 05 10 3 8 10

1 05 10 0 38 10

1 05 0 38 10

1 43 10

, ,

, ,

( , , )

,

TASK 4 1. Write in scientific notation:

a) 3 728 935

b) 173,529

c) 0,000 0605

d) 0,007 832

e) 87 500 000

f) 0,006

g) 345 000 000

h) 158 ,002 3 2. Write these as ordinary numbers:

a) 48 515 10,

b) 56 12 10,

c) 27 632 10,

3. Simplify the following without using a calculator:

a) 3 43 6 10 2 10,

b) 12 72 34 10 5 10,

c) 2 27 23 10 6 9 10, ,

d) 4 54 2 10 51 10,

24

C. EXPONENTIAL EXPRESSIONS & EQUATIONS

1. ADDITION & SUBTRACTION OF EXPONENTS So far we have dealt with multiplication and division of exponential expressions. Now we will study addition and subtraction of exponential expressions.

( )x x xma na m na

( )x x xma na m na

To do addition and subtraction we must have the same base and the same index .Use the common factor when you have the same base and the index.

Eg. 1. 5 5 53 4 7x x x

2. 3 3( )y y yx ax a x ( yx is common)

3. 9 5 2 5 7 5x x x WORKED EXAMPLES QUESTION 1 Simplify the following:

3

3 1

3 2 3

2 3 3

.

.

x x

x x

SOLUTION There are two terms in the numerator and also in the denominator. We need to use the common factor to make one term in each place:

Remark:1 13 3 3x x and

3 33 3 3x x

3 3

3 1 3 1

3 2 3 3 3 2 3

2 3 3 2 3 3 1 327 2 3

24 1 3

251

25

.

.

( )

( )

x x x x

x x x x

x

x

QUESTION 2 Simplify the following:

2 2 2

2

2 3 2

4 5 4

x x

x x

SOLUTION Simplify the expressions in the bracket first.

2 2 2 2 2 2

2 2 4 2

2 2 2

2 4 2

2 2

2 4

2 3 2 2 3 2 2

4 5 4 2 5 22 3 2 2

2 2 5 22 3 2

2 2 5

3 4 1

16 5 11

( )

( )

x x x x

x x x x

x x

x x

x

x


2 3 2 4

2 2

6 2 2

2 2

x x

x

SOLUTION Simplify the expressions in the numerator first; make it one term

2 3 2 4 2 3 2 4

2 2 2 2

2

2

6 2 2 6 2 2 2 2

2 2 2 2 22 6 8 16

2 4 26 8 16

42 4

( )

x x x x

x x

x

x


2

2

3 3

6 3 4 3

x x

x x

SOLUTION

2 2

2 2

2

2

3 3 3 3 3

6 3 4 3 6 3 4 3 33 1 3

3 6 4 3

4

25

( )

( )

x x x x

x x x x

x

x

25

2. EXPONENTIAL EQUATIONS: Organize both sides: try to bring the exponents to the same bases. If a bx x

Then a b

QUESTION 5 Solve for x:

14

64x

SOLUTION

14

64x can be written as power of 2:

26

2 6

12

22 2

2 6

3

( )x

x

x

x


9 314

2( )x x

SOLUTION

9 3

1 9 2 3

9 2 6

9 2 6

14

22 2

2 2

2 2

9 2 6

3 3

1

( )

( ) ( )

x x

x x

x x

x x

x x

x

x


28 4 1x x SOLUTION Remember: 02 1

3 2 2 0

3 2 4 0

3 2 4 0

5 4 0

2 2 2

2 2 2

2 2

2 2

5 4 0

4

5

( ) ( )x x

x x

x x

x

x

x


13 3 4x x SOLUTION Organize left‐ hand side.

1

1

1

3 3 3 4

3 1 3 4

13 1 4

34

3 43

4 33

43 3

1

( )

( )

( )

x x

x

x

x

x

x

x


2

32 8x

SOLUTION Divide both sides by 2.

2

3 4x

to leave x alone, we need to take 3

2

power

of both sides(reciprocal of the power of x) 2 3 3

3 2 2

32 2

3

4

2

12

8

( )

( )

x

x

x


3 2 0 375,x

SOLUTION Divide both sides by 3.

3

3 2 0 375

3 31

28

2 2

3

3

,x

x

x

x

x

26

TASK 5 1) Simplify the following expressions

a) 2 15 5x x

b) 32 2 2

2

.x x

x

c) 2 2 1

2

2 7 7

7

x x

x

d) 2

2

3 4 3

3

.x x

x

e) 2 1

2

6 36

6

x x

x

2) Solve for x.

a) 3 27x

b) 4 8x

c) 15 125 1x x

d) 12 2 6x x

e) 127 81x

f) 15 2 40x

g) 23 3 24x x

27

TASK 6 1) Simplify the following expressions

a) 2 2 22 3 2x x

b) 3

1

3 3 3

3 3

.x x

x x

c) 2 1 22 5 5

25

x x

x

d) 2

2

7 4 7

7

.a a

a

e) 2 1

2

7 49

7

x x

x

2‐) Solve for x.

a) 27 81x

b) 2 4 22 32x x

c) 1 49 27 1x x

d) 12 2 12a a

e) 15 5 125x

f) 25 3 45x

g) 2 2 32b b

28

TASK 7 1‐) Simplify the following expressions

a) 2 1

1

3 3 3

5 3

x x x

x

b) 2 1 2 12 7 7

49

x x

x

c) 2 1 1

2 2

2 5 3 25

25 3 5

x x

x x

d) 1

2 12 2 2( )x x x x

2‐) Solve for x.

a) 1

927

x

b) 5 318

4( )x x

c) 1 425 125 1x x

d) 12 2 6x x

e) 3

43 81x

f) 15 2 1 25,x

g) 23 3 24x x

29

D. SURDS We name the surds according to the degree of the surds:

x is the degree of root in x y :

if x is 2, we name it ‘square root’, if x is 3 we name it ‘cube root’ … If the degree is an even number, the value inside the surd cannot be negative.

i.e 2 not defined, but 3 2 is defined. If the surd is totally cancelled, the number is classified as rational number; otherwise it is an irrational number:

i.e 4 2 is a rational number;

8 2 2 is an irrational number; All the recurring decimals are also irrational numbers:

1 333, ... is a recurring decimal : it is irrational. is a recurring decimal : it is irrational.

Rational numbers: Integers, Fractions, Decimals

Irrational Numbers “Root numbers”, Recurring Decimals,

Non‐Real Numbers “Even Root” of Negative numbers

Laws of Surds Multiplication :If the surds are the same type (having the same degree ;square root, cube root..) multiply the coefficients of the surds independently multiply the numbers inside the surds Keep the same type of surd.

x x xa m b n ab mn

Eg. 1. 2 3 4 7 8 21

2. 4 2 7 3 5 12 70

3. 3 332 4 5 3 10 12

4. 33 2 4 3 = 312 2 3

5. 2a a a a a a

6. 2 7 3 7 6 7 42

Division :If the surds are the same type (having the same degree like square root, cube root..) :

x

xx

a m a m

b nb n

Divide the coefficients of the surds independently Divide the numbers inside the surds Keep the same type of surd.

Eg. 1. 8 15 8 15

4 52 32 3

2. 3

333

10 105

22

Addition and Subtraction :The degrees of the surds and the numbers inside the surds must be the same for addition and subtraction of the surds. Use common factor..

( )x x xa m b m a b m

Eg. 1. 2 3 4 3 2 4 3 6 3( )

2. 6 2 4 2 2 6 4 1 2 3 2( )

3. 3 3 3 32 4 5 4 2 5 4 3 4( )

4. 2 8 4 18 ? Write a surd in the simplest form,

8 4 2 2 2.

18 9 2. .

2 8 4 18 2 4 2 4 9 2

4 2 12 2 16 2 . Surd can be written in exponential form:

mmn na a

m is the power of the number inside , n is the degree of the surd. There must be one term in the surd to use the rule.

1

22 2 (the power of 2 is one here) 2

23 35 5 100 6

100 6 50 32 2a b a b a b

2 2a b a b

2 6 364 8a y ay

100 6100 6 50 32 2a b a b a b

2( )a b a b ( a b )

30

QUESTION 1

Simplify the following: 5 45 2 80( )

SOLUTION Simplify inside the bracket first.

5 9 5 2 16 5 5 3 5 8 5

5 11 5

55

( ) ( )

QUESTION 2

Simplify:98 8

50

SOLUTION Simplify the numerator to one term first.

98 8 49 2 4 2

50 25 2

7 2 2 2

5 2

5 21

5 2

QUESTION 3

Simplify: 2 29 16x x SOLUTION You can’t take square root separately, because rule of exponents and surds don’t work form addition. But you can add like terms:

2 2 29 16 25

5

x x x

x

QUESTION 4

Given M= 1 2

3 5 5x x

a) Show that M is a rational number if x=3 b) Determine values of x for which M is a real number SOLUTION

a) Substitute x with 3;

1 2 1 2

3 3 5 3 5 4 21

2

.

b) Terms inside the square roots must be positive and denominators cannot be zero.

3 5 0x and 5 0x

5

3x and 5x

Someone may show the answer on the number line as well..

31

TASK 81‐) Simplify the following expressions

a) 4 225a b

b) 2( )a b

c) 1616x

d) 4 49 16x x

e) 2 4

121

( )m a b

f) 3 432

8

a b

a

g) 5 3 2

5

20 9

5

a a b

a a

h) 6

9

50

98

a b

b

i) 2x x

2‐) Simplify.

a) 28 63 ?

b) 5 7 3 28 ?

c) 125 20

45

d) 3 3 3 ?

e) 612

4a

f) 4 2 ?a a a

g) 23 2( ) ?

32


a) 6 698 127x x

b) 147 108 ?

c) 27 48

75

d) 5 3 2 5( ) ?

e) 1 5 3 2 4 3( )( ) ?

f) 23 5 2( ) ?

2‐) Simplify.

a) 2 2 1 8( )( ) ?

b) 3 1 3 1( )( ) ?

c) 25 1( ) ?

d) 90 40 8 18 ?

e) 1

2

20 5

4 5 5

f) 11 3 11 3

33

TASK 101‐) Simplify the following expressions

a) 2 50 2 ?

b) 2 29 16x x

c) 3 5 2 125

20

d) 2 3 2 5( ) ?

e) 2 8 2 2( )( ) ?

f) 22 3 3 2( ) ?

2‐) Simplify.

a) 2 2( )( ) ?x x

b) 2 5 3 2 5 3( )( ) ?

c) 72 3 50 5 8

2?

d) 6 8 43 4125 81 36x x x

e) 81

4 24 9 1612

f) 6 1 6 1

34


a) 4

8 2?

b) 6 6128 98x x

c) 50 32 ?

d) 4 419 3 19 3( ). ( )

e) 22 2( ) ?

f) 147 12

27

2‐)Given M= 2 1

2 5 2x x

a) Show that M is a rational number if x=1,5 b) Determine values of x for which M is undefined c) Determine values of x for which M is a real number

35

E.RATIONALIZING DENOMINATORSIt is useful to work with fractions, which have rational denominators instead of surd denominators. We will now see how this can be achieved.

Any expression of the form a (where a is rational) can be changed into a rational number by

multiplying by itself. 2( )a a

Any expression of the form a b can be changed

into a rational number by multiplying by a b

(conjugate of a b ).

Similarly a b can be rationalized by multiplying

by ba (conjugate of ba ).

This is because : 2 2( )( ) ( ) ( )a b a b a b a b

QUESTION 1

Rationalise the denominator: 5 2

5

QUESTION 2

Rationalise the denominator: 1 2 2

3 2

QUESTION 3

Rationalise the denominator: 3

2 2

QUESTION 4

Rationalise the denominator:4

2

y

y

SOLUTION 1 Extend the fraction by multiplying both numerator

and denominator by 5 .

5 2 5 5 2

5 5 5

5 2 5

5

( )


and denominator by 2

1 2 2 2 1 2 2

3 2 3 2 2

2 4

6

2 4

6

( )


and denominator by 22

2 2 3 2 3 6

4 22 2 2 2

2 3 6

2

( )

( ) ( )

SOLUTION 4

Multiply numerator and denominator by 2y

and then simplify common factor

4 2 4 2

2 2 2

( ).( ) ( ).( )

( ) ( )

y y y y

y y y y

2 y 4

4 2

4

2

( ).( )y y

y

y

36

TASK 121. Rationalise the denominators without

calculator

a) 3 1

3

b) 2 3 5

5

c) 3

1 3

d) 2

5 3 3 2

e) 2x

x

f) 1

5

k

k

g) 2

1

n

n

h) 9

3

x

x

2. Prove that 8 5 1 13 2

53 3 6 2 3

3. Write 23 2 2

3

( ) in the form of a b c

where a, b and c are integers, without using

calculator

4. Show x y

x x can be written as

1( )x y

x

.

37

E. TRIAL TESTS TRIAL TEST-1

QUESTION 1 Simplify each of the following:

1.1

11

32

1

4

x x

x

(3)

1.2 8 8162 72x x (3)

[6] QUESTION 2 2.1 Simplify : 9 12 63 427 625 9x x x (4)

2.2 Given M=

3 1

4 7 6x x

2.2.1 Show that M is a rational number if x=2,5

(3)

2.2.2 Determine the values of x for which M is a real number

(3)

2.3 Erin had to find the product of 20072 and

20005 and then calculate the sum of the digits of the answer. Erin arrived at an answer of 11. Is she correct? Show ALL the calculations to motivate your answer.

(5) [15]

QUESTION 3 3.1 Given

3 2

2

xP

3.1.1

Wrıte down the values of x for which P is rational and unequal.

(2)

3.1.2 Write down the values of x for which P is non‐real.

(2)

3.2 Simplify, without using a calculator:

3.2.1 1 1

1

4 2

8

x x

x

(4)

3.2.2 2 212 4 3x x (2)

[10]

TOTAL: 31

TRIAL TEST-2 QUESTION 1 1.1

If M= 1

4 x , determine the values of x

for which 1.1.1 M is undefined. (2)

1.1.2 M is non‐real. (2)


1.2.1 108 2 75 2 27

3?

(4)

1.2.2 3

21

24

( )

(3)

1.2.3 22 4

2 2

x

x

(3) [14]

QUESTION 2 2.1 Simplify, without using a calculator:

2.1.1 Show that

3 2 1 2

1

2 2

2 4

x x

x x

is equal to

4 x

(4)

2.1.2

Hence or otherwise find the value

of : 3 3 2 1 2 3

3 1 3

2 2

2 4

( ) ( )

(2)

2.1.3 2

327

125( )

(2)

2.1.4

Which value is larger ? Show

working details. 6 or 3 16

(3) [11]

TOTAL: 25

38

TRIAL TEST-3 QUESTION 1 1.1 Simplify, without using a calculator:

1.1.1 45 2 20

80?

(4)

1.1.2 1 1

2 1

9 81

27

n n

n

(5)

1.2 Solve for x : 26 3 54x (4)

[13] QUESTION 2 2.1 Simplify, without using a calculator:

2.1.1 2 1 4

0 5

16 81 6

9 ,

. .x x x

x

(5)

2.1.2 23 27 (2)

2.2 Solve for x : 1

23 1 12( )x (3) [10]


3.1

2438

( )a

a

(3)

3.2 3 312 2 12 2( ). ( )

(4) [7]

TOTAL: 30

TRIAL TEST-4 QUESTION 1 1.1

If M= 3

2 1x , determine the values of x

for which 1.1.1 M is undefined. (3)

1.1.2 M is non‐real. (2)


1.2.1 2 1 2 1

2 5

25 5 125

25 5

. .

.

x x x

x x

(5)

1.2.2 2

333

8( )

(3)

1.2.3 10 6 45 332 64 25x x x (4)


2.1.1 21 1

54 216 32( )

(5)

2.2 Solve for x : 12 4 8x x (5)

[10]

TOTAL: 27

39

TRIAL TEST-5 / ADVANCED LEVEL QUESTION 1 1.1 Simplify, without using a calculator: 1.1.1 13 2 13 2 (3)

1.1.2 1 3 2

2

9 27

81

n n

n

(4)

1.1.3 3 45 5

20

(3)

1.2 Solve for x : 21

2 18( )x x

(5) [14]


2.1 20 125 64 16( ) (4)

2.2 2

1

9 10

6 15

x x

x x

(5)

2.3

233

3

3

27

( )[ ) ]x

x

(4)

2.4

1

5 2 2

2

n n

n

(4) [16]

TOTAL: 30

TRIAL TEST-6 / ADVANCED LEVEL QUESTION 1 1.1 Simplify, without using a calculator:

1.1.1 2 8 50

1 2 1 2( )( )

(4)

1.1.2 1 2

15 3

9 5

x x

x x

(4)

1.1.3 1 1

22 21

1( )

( )a b

a b

(4)

1.1.4 2

1

3 4 3

2 3 3

x x

x x

(4)

1.2 Solve for x : 1 1

2 3 3 2 72

x x (4) [20]


2.1 3 2 3 3 2 3

9

( )( )

(4)

2.2 Prove that : 3 62 3 72

(3) 2.3

2 1

3

2 5 6 5

5

x x

x

(3) [10]

TOTAL: 30

40

III. NUMBER PATTERNS A. COMMON NUMBER PATTERNS

Numbers can have interesting patterns. Here we list the most common patterns and how they are made. Examples 1, 4, 7, 10, 13, 16, 19, 22, 25, ... This sequence has a difference of 3 between each consecutive two numbers. The pattern is continued by adding 3 to the last number each time. 3, 8, 13, 18, 23, 28, 33, 38, ... This sequence has a difference of 5 between each consecutive two numbers. The pattern is continued by adding 5 to the last number each time. 2, 4, 8, 16, 32, 64, 128, 256, ... This sequence has a factor of 2 between each consecutive two numbers . The pattern is continued by multiplying the last number by 2 each time. 3, 9, 27, 81, 243, 729, 2187, ... This sequence has a factor of 3 between each consecutive two numbers. The pattern is continued by multiplying the last number by 3 each time.

Square Numbers 1, 4, 9, 16, 25, 36, 49, 64, 81, ... The next number is made by squaring the term number. The second number is 2 squared (2 or 2 × 2) The seventh number is 7 squared (7 or 7 × 7) etc Cube Numbers 1, 8, 27, 64, 125, 216, 343, 512, 729, ...The next term is made by cubing the term number. The second number is 2 cubed (2 or 2 × 2 × 2) The seventh number is 7 cubed (7 or 7 × 7 × 7)... Triangular Numbers 1, 3, 6, 10, 15, 21, 28, 36, 45, ... This sequence is generated from a pattern of dots which form a triangle. By adding another row of dots and counting all the dots we can find the next number of the sequence Fibonacci Numbers 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ... The next number is found by adding the two numbers before it.

General Formula, Tn A sequence does not have to follow a pattern but when it does we can often write down a formula to calculate the nth‐term. In the sequence 1; 4; 9; 16; 25; . . . where the sequence consists of the squares of integers, the formula for the nth‐term is Tn = n

2

You can check this by looking at: T1 = 1

2 = 1 T2 = 22 = 4 T3 = 3

2 = 9 T4= 4

2 = 16 T5 = 52 = 25. . .

Therefore we can generate a pattern, namely squares of integers. QUESTION 1 1.1 Find the sum of the numbers in the 5th row 1.2 Determine a formula for the sum of numbers in the nth row. 1.3 Calculate the sum of the numbers in the (n–1)th row?

SOLUTION 1.1 The relation between the row and the sum of the numbers in the row is the sum is the cube of that row : first row 13=1; second row 23=8 and so on.. The sum of the numbers in the 5th row is 53=125 1.2 Sum of the numbers in the nth row is n3

1.3 The sum in the (n‐1)th row is (n‐1)3

41

B. LINEAR/ARITHMETIC SEQUENCE Definition: A linear sequence is a sequence of numbers in which the differences between each consecutive term is the same, called a common difference.

2 1 3 2T T T T

For example, 5; 8; 11; 14; 17; . . . is a linear sequence. Because;

2 1

3 2

4 3

8 5 3

11 8 3

14 11 3

T T

T T

T T

1 1T and 3d

Let the first term T1=a:

2

3

4

2

3

T a d

T a d d a d

T a d d d a d

As you notice; in the 2nd term there is 1d, In the 3rd term 2d, the 4th term 3d and so on... In the nth term there will be (n‐1)d General term for linear sequence is:

Arithmetic sequence formula:

1( )nT a n d where

Tn = nth term

a = T1 first term d = common difference n = number of the terms

QUESTION 1 Study the following pattern: 1; 6; 11; 16; 21; ... 1.1 What is the next number in the sequence ? 1.2 Use variables to write an algebraic statement to generalise the pattern. 1.3 What will the 100th term of the sequence be? 1.4 Which term of the sequence is 96 ? SOLUTION 1.1 In the sequence the numbers increase by 5

first difference is the same linear Next two numbers are :26 , 31

1.2 Tn= ? a T1 1

d 5

T a n 1 d T 1 n 1 5 T 1 5n 5

5 4nT n

1.3 n=100, T100= ?

T 5.100 4 1.4 T 96, n=?

96 5n 4 100 5n

n=20

QUESTION 2 Given linear sequence:

3 2 1 4 6, ,x x x

Find values of x and first three terms SOLUTION Linear sequence has common difference

2 1 3 2T T T T

Therefore 2 1 3 4 6 2 1

2 2 5

7

( ) ( ) ( ) ( )x x x x

x x

x

And,

1

2

3

7 3 4

2 7 1 13

4 7 6 22

T

T

T

42

TASK 1 QUESTION 1 Consider the following sequences and in each case: ∙ Write down the next two terms. ∙ Try to determine general formula 1.1 3 ; 7 ; 11 ; 15 ; 19 ; ………..

1.2 50 ; 44 ; 38 ; 32 ; 26 ; ……..

1.3 2 ; 3 ; 5 ; 8 ; 13 ; ………..

1.4 1 ; 2 ; 4 ; 8 ;16 ; ………..

1.5 1 ; 4 ; 9 ; 16 ; 25 ; ………..

1.6 – 27 ; – 22 ; – 17 ; – 12 ; – 7 ; ………..

1.7 5 + 7x ; 7 + 9x ; 9 +11x ; 11+13x ;

QUESTION 2 Consider the following sequences and in each case determine: ∙ The next three terms. ∙ A general algebraic formula that you can use to find any term of the sequence.

2.1 – 4, 1, 6 ; 11 ; 16 ; ................ 2.2 3 ; 10 ; 21 ; 36 ; 55 ; …………. 2.3 1 ; 8 ; 27 ; 64 ; 125 ; ……….

2.4 1

3; 2

5; 3

7; 4

9;5

11 ; ……………….

QUESTION 3 3.1 Three consecutive terms of a sequence are 2x‐3; 5x+2; and x‐7. The formula for the general term is linear. Determine the value of x and the three terms. 3.2 Tobias sits and writes: MATHSMATHSMATHS......

If he continues writing this pattern, determine the 2008th letter that he will write down.

3.3 Find the next two terms and the general term of the following sequences: 16 ; 11 ; 6 ......... 3.4 Given the arithmetic sequence : 23; 19; 15;… Find:

a. the 12th term b. which term in the sequence is ‐53.

43

TASK 2 1. Find the 11th term of the arithmetic sequence 4; 7; 10; …

2. Find the 50th term of the arithmetic sequence 13; 10; 7; …

3. Find the 30th term of the arithmetic sequence 4+7x; 5+9x; 6+11x; … 4. Find the 7th term of the arithmetic sequence

if 4; 7. 5. Find the 10th term of the arithmetic sequence

if 2; 5 18T .

6. Which term of the arithmetic sequence 3; 5;

7; … is equal to 27? 7. Given the arithmetic sequence:

1 15 4 3

4 2; ; ; …

a. which term in the sequence is equal to 13?

b. Calculate the 13th term.

8. Which term of the arithmetic sequence 25; 14; 3; … is equal to ‐52?

9. If x+2; 3x‐1and 4x‐3 are the first three terms

of an arithmetic sequence:

a) determine x and then write down the numerical values of the first three terms

b) determine a formula for the thn term of the sequence

c) if the thn term is ‐41, calculate n

d) calculate 20th term. 10. Find the 10th term of the arithmetic

sequence 7; 13; 19; 11. Find the 9th term of the arithmetic sequence

5 ; 8 ; 11 ; …

44

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611

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6

3

6

3

b

b

1

1

c

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quence be ?

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45

TASK 3 QUESTION 1 ‐1; 2; 7; 14; … 1.1 Will the form of the general term be linear or quadratic? Give a reason for your answer. 1.2 Find the general term. 1.3 Extend the sequence by three terms. QUESTION 2

7 ; 13; 23; 37; …

2.1 Determine nT

2.2 Use your formula to write the value of 30T

QUESTION 3 Evaluate:

3 4 5 6 2012 2013 2014

2 3 4 5 2011 2012 2013...........

QUESTION 4 A pattern of double squares is built as shown in the diagram. The dots along the one diagonal of one of the squares are shaded black. Rewrite the table on your answer sheet and complete the table for the number of white dots for each pattern.

Black dots White dots

2 5

3 14

4

5

6

7

10

n

46

QC

1l 1

t QDb

2bn 2t QSDs

QUESTION 1Consider the

1.1 Is theinear, quadr

1.2 Dete

term of the g

QUESTION 2Dots are arrabelow:

2.1 Make a between thenumber.

2.2 Write athe relations

QUESTION 3Study the folDetermine a sequence.

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‐2; 1; 1

e general terratic or neith

ermine a form

given sequen

anged to form

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n algebraic fhip in 2.1.

lowing patteformula for

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TASK 4

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QUESTION 4

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47

D. GEOMETRIC SEQUENCE A geometric sequence is a sequence in which every term in the sequence is equal to the previous term in the sequence, multiplied by a constant number. This means that the ratio between consecutive numbers in the geometric sequence is a constant.

is common ratio To find the common ratio of a geometric sequence, any term is divided by the previous term.

To find any term in the sequence, we use the general formula of :

Geometric sequence formula: 1n

nT a r

Tn = nth term

a = T1 first term r = common ratio n = number of the terms

To find the consecutive term of any geometric sequence, multiply the term by the common ratio. For example:

2; 6; 18; The next term in the sequence will be: 18.3=54 (we multiplied the previous term by the common ratio, 3) QUESTION 1 Continue the sequence to the 7th term: 2; ‐4; 8; SOLUTION

4

2

8

42

We can continue writing the terms by multiplying the terms with ‐2

2; ‐4; 8; ‐16; 32;‐64; 128… QUESTION 2 Find 12th term of the geometric sequence:

5 251

2 4; ; ; …

SOLUTION 1

1

5

2

.

1.5

2

5

2

QUESTION 3 If 1; 1; are the first three terms of a geometric sequence. Find and the first three terms of the sequence. SOLUTION We know that this is a geometric sequence; there is a common ratio: the ratio can be found by dividing the third term by the second term or dividing the second term by the first term

1

1

1

Cross‐multiply to get

2 1 2 1

3 1 1

3

The first three terms are: 4

3 ;

2

3 ; 1

3 ; …

QUESTION 4

The thn term of a geometric sequence is 38 5n

Find the term of the sequence which has a value of 200. SOLUTION The general term is given as a function:

8. 5 n=?

200 8. 5 Divide both sides by 8:

25 5 5 5

3 2 ∴

48

TASK 51. Determine the 4th and 6th terms of the

geometric sequence: 2; 6; 18; …

2. Find the 5th term of the geometric sequence

of which 2and 3.

3. Find 12th term of the geometric sequence:

2 3 42 6 18, , ....x x x …

4. If 4; 2; and 3 1 are the first three

terms of a geometric sequence. Write down

the numerical values of first three terms.

5. Determine the values of x so that the

sequence 3 1

4 3; ;x is a geometric sequence.

6. Find r in the geometric sequence of which the

sixth term is 96 and the first term is 3.

7. Find r in the geometric sequence of which the

fifth term is 7

81 and the first term is 7.

8. Find the 2th term of the geometric sequence of

which 1 1T and 7 729T .

9. Determine the first three terms of the

geometric sequence with fourth term 4

25

and fifth term 4

125

10. Which term of the geometric sequence: 1; 5;

25; … is equal to 625?

49

IV. FACTORIZATION A. PRODUCTS OF FACTORS

The identity is used to express the product of two binomials in the example above. Let us multiply 3 5 2 FOIL method: 3 5 2 6 3 10 5

Combine the like terms.

6 7 5 QUESTION 1 Find the following product

2 3 5 1 SOLUTION 2 3 5 1 10 2 15 3

Combine the like terms 10 13 3

QUESTION 2 Find the following product

3 4 SOLUTION

3 4 3 4 9 12 12 12

Combine the like terms 9 24 12


3 1 2 4 2 SOLUTION

3 1 2 4 2 6 12 6 2 4 2

Combine the like terms 6 12 4 4 2


1

3

1

2

1

2

1

3

SOLUTION

Remember 1

3

1

3.1 3

3 2 2 3

6 9 4 6

6

13

36 6


3 1 2 1 3 SOLUTION Multiply any two factors then the product by the third one.

3 1 2 1 3 3 1 6 2 3 3 1 2 5 3

6 15 9 2 5 3 6 17 4 3

Useful Identities

2 2 2

2 2 2

2 2

( ) 2

( ) 2

( )( )

a b a ab b

a b a ab b

a b a b a b

For example: 1. 5 3 25 30 9

2. 1 1 2

3. 4 8 16

4. 2 2. . 2

5. 3 2 3 2 9 4

6.

50

TASK 1 Find the following products: 1. 2 5 2. 3 5 2 1 3. 5 2 4. 4 2 6 1

5. 4 1 3

6. 1

5

1

4

1

4

1

5

7. 5 2 8. 2 2 1 1

9. 5 5 10. 3 5 3 5 11.

2

12. 1

13. 1

22

14.

15. 22 2 4( )( )( )x x x

51

B. FACTORIZATIONFactorization is a method of writing an expression with more than one term as factors. There will be only one term in the factorized form of that expression.

1. BY COMMON FACTORS: In this method:

the highest common factor of each term HCF is found first.

Then we take out the common factor from each term ;

the other factors of each term are written in a common bracket.

For example: Let us factorize: 6 4 There are two terms namely 6x and ‐4: 2 is the common factor of each term. We can write 6x‐4 as . 3 . 2

Take 2 out of each term and open a bracket; write 2 out of bracket and the other terms of the terms inside the bracket:

∴ 2 3 2 Example 2: Factorize the following:

7 28 35 Let us rewrite the expression as:

. . . . 4. . . . . 5. . . . The HCF of the three terms is 7ab Write 7ab out of the bracket and the other factors left from each term inside the bracket.

∴ 7 4 5 Example 3: Factorize the following:

Let us rewrite the expression as: . The HCF of the terms is Write out of the bracket and the other factors left from each term inside the bracket. Note that : if any term is taken as common factor fully, in the

bracket you have to put 1in that term. If any – sign is taken as a common factor you have to put sign in the place of that term in the bracket after the common factor is taken out.

∴ 1 Example 4: Factorize the following:

4 8 Sometimes the common factor can be binomial(two terms) like (x‐y) in this example.

2. .

The HCF of the terms is 4 Write 4 out of the bracket(don’t open the bracket) and write the other factors left from each term inside the bracket. If there is a factor with bracket inside ,open it .

4 1 2 ∴ 4 1 2 2

Note that:

Example 5: Factorize the following:

2 4

In the second term ,namely– 4 should be written as . The second term becomes: 4 . Rearrange the expression:

. . . 2. . . The HCF of the terms is 2

∴ 2 2

2. BY GROUPING: Group the terms by using common factors After grouping the terms 2 by 2 or 3 by 3 ..,

make sure that each group has the same common factor. Otherwise regroup till you get the same common factors.

Example 1: Factorize the following: There are four terms. Even though some terms have common factor, each term doesn’t have the same factor; common factor method cannot be used. Group them with suitable common factors 2 by 2. (first one with the third one and second term with the fourth one.)

group ‐1 group ‐2

∴


3 4 12 3 3.

group ‐1 group ‐2 4

∴ 3 4

52


4 4 . and 4 4 4 4

. . 1 . 1 group ‐1 group ‐2

4

∴ 1 4 Example 4: Factorize the following:

1 1

group ‐1 group ‐2 1.

∴ 1 1

3. BY DIFFERENCES OF TWO SQUARES

Take the square roots of both terms Multiply the sum of the terms obtained by the

difference of the terms.

2 2b a b b

a b

a a

Example 1: Factorize the following: 4 9 2 3 2 3

2x 3 Example 2: Factorize the following: 9 25 3 5 3 5

3m 5n Example 3: Factorize the following:

1 1 1

x2 1 It is factorized but not fully;

1 can be factorized as 1 1)

∴ 1 1 1 1

Example4: Factorize the following: 25 3 1 4 2 3

5 (3x – 1) 2 (2x + 3) open each bracket separately)

15 5 4 6 15 5 4 6 19 1 11 11

Example‐5:

10 10

5 5

1 1

1 (1 )

x x

x x

Example‐6:

2 22 50 2 25

2 5 ( 5)

x x

x x

Example‐7:

16 6 8 3 8 31 1 1 1 1 1( )( )

4 9 2 3 2 3x y x y x y

4. BY SUM OR DIFFERENCES OF TWO CUBES:

3 3 2 2aa b a b ab b

3 3 2 2aa b a b ab b


23 2

2 32

8 27 2 3 2 2 3 3

2 3 4 6 9x

x x x x

x x x


23 3 2

42 2

64 4 4 4

4 4 16m n

m n m n m m n n

m n m mn n

53

TASK 2Factorize the following expressions: 1. 16 4 2. 48 ‐15 3. 5 20 30

4. 3 12

5. 12 14 16

6. 5 9

7. 14 21

8. 4 4

9. 3 3

10. 2 2

11.

12. 16 25 13. 64 14. 25 49 15. 4 9 16. 3 24 17. 4

9

1

25

18.

3 32 7m n

54

TASK 3 Factorize the following expressions: 1. 3 15 2. 6 4 2 3. 4 4

4. 7 2 3 2 3

5.

6. 2 3

7.

8.

9. 5 5

10. 2 2 2 11. 2 3 6 9

12. 16 4 13. 121 14. 81 100 15. – 16. 5 20 17. 36 24 72 18. 7 7

55

C. FACTORIZATION OF TRINOMIALS

1. FACTORIZING If the leading coefficient of a polynomial is 1(the number in front of x2) it is very easy to factorize that trinomial

To factorize2x bx c :

Identify and , two factors of which are

add up to

. and Write the trinomial in the form of :


5 4 We have to look for two factors of +4 which gives the sum 5. Possible factors of +4 are:

the sum =1+4=5 1 4 4the sum =‐1‐4=‐5 2 2 4the sum =2+2=4

∴ 5 4 4 1 4 1 Example 2: Factorize the following:

5 6 We have to look for two factors of 6 which gives the sum 5. Possible factors of 6 are: 2 3 6the sum =2‐3=‐1 2 3 6the sum =‐2+3=1

the sum =1‐6=‐5 1 6 6the sum =‐1+6=5

∴ 5 6 1 6

1 ‐6 Example 3: Factorize the following:

3 40

∴ 3 40 5 8 5 ‐8


2

∴ 1 2 1 2 1 ‐2 Example 5: Factorize the following:

6 27 Write the trinomial in the standard form:

∴ 6x 27 9 3 9 ‐3 Example 6: Factorize the following:

6 24 30 Take out the common factor of 6, then go on.

∴ 6 4x 5 6 5 1 5 ‐1 Example 7: Factorize the following:

12 Take ‐1 as common factor to make the number in front of .

∴ x 12 3 4 3 ‐4

56

2. FACTORIZING

To factorize2x bx c :

Find two factors of and two factors of .

When “cross‐multiplied”, the results should

add up to


3 10 3

∴ 3 10 3 3 1 3 3 1 3

[3 . 3 . 1 9 10 ]


2 13 21

∴ 2 13 21 2 7 3 2 7 3

[2 . 3 . 7 6 7 13 ]


4 7 2

∴ 4 7 2 4 1 2 4 1 ‐2

[4 . 2 . 1 8 7 ]


10 7 3 Write the trinomial in the standard form and take

out – 1 common factor. 3 10 7 3 10 7

∴ 3 10 7 3 7 1

3 ‐7 ‐ 1

[ 3 7 10 ]


4 5 take out as common factor.

4 5 1

∴ 4 5 1 4 1 1 4 ‐1 ‐1

[4 . 1 . 1 4 5 ]


2 5 3 This is still a trinomial, but instead of we have

2 5 3 2 ‐ 3 ( ‐ 1

2 2 3 1 Example 7: Factorize the following:

12 32 ∴ 12 32 8 4

8 4

[4 8 12


5 24

∴ 5 24 8 3 ‐ 8 3

[3 8 5

57

TASK 4

Factorize the following expressions: 1. 7 12

2. 20

3. 15 14

4. 8 12

5. 14 48

6. 7 10

7. 11 12 8. 2 18 40

9. 22 13 1

10. 5 28 15

11. 3 20 32

12. 4 4 1

13. 2 1

14. 6 5 6

15. 45 3 18

58

TASK 5

Factorize the following expressions: 1. 5 24

2. 6 8

3. 11 12

4. 6

5. 2 24 70

6. 2 15

7. 14 4

8. 7 16 15

9. 5 13 28

10. 11 70 24

11. 5 14

12. 9 42 45

13. 4 11 6

14. 5 4

15. 7 19 6

59

TASK 6 Factorize the following expressions: 1. 10 21

2. 6 9

3. 3 21 24

4. 27 6

5. 16 15

6. 4 2 5 3

7. 2 26 24

8. 2 35

9. 8 10 3

10. 2 10 12

11. 7 2 4

12. 9 42 45

13. 21 19 22

14. 4 3

15. 12 7

60

D. MULTIPLICATION AND DIVISION OF RATIONAL EXPRESSIONS A rational expression is in its simplest form when the numerator and the denominator have no common factor other than 1.

In order to write a rational expression in the simplest form we use the following procedures:

Factorize both the numerator and the

denominator completely.

Cancel out any common factors

If there is division, convert the division to

the multiplication and swap the numerator

and the denominator. Factorize and cancel

out the same factors.

QUESTION 1 Simplify the following expression.

2 2a x ax

a x

SOLUTION Step1: Factorize both the numerator and the denominator with suitable method.

2 2 ( )a x ax ax a x

a x a x

Step2: Cancel out the same factors.

( )ax a xax

a x


2

2

2

2

a ab

b ab


2

2

2 2

2 2

2

2

( )

( )

( )

( )

a ab a a b

b ab b b a

a b a

b b a


2

2

( )

( )

a b a a

b b a b


2 2

2

2x xy y

x xy


2 2

2

2 ( )( )

( )

x xy y x y x y

x xy x x y


( )( )

( )

x y x y x y

x x y x


2 2

2

3 4

4

x x x

x x x


2 2

2

3 4 4 1

4 1 4

( )( )

( )

x x x x x x x

x x x x x x


4 1

1 4

( )( )

( )

x x x xx

x x x


2 2

3 3

3 6

2 8

ax a x

ax a x


2 2

3 3 2 2

3 6 3 2

2 8 2 4

3 2

2 2 2

( )

( )

( )

( )( )

ax a x ax x a

ax a x ax x a

ax x a

ax x a x a


3 2 3

2 2 2 2 2

( )

( )( ) ( )

ax x a

ax x a x a x a

61


2 2

2

3 2 2 6

2 3 3 6

x x x x

x x x


2 2

2

3 2 2 6 1 2 2 3

2 3 3 6 3 1 3 2

( )( ) ( )

( )( ) ( )

x x x x x x x x

x x x x x x


1 2 2 3 2

3 1 3 2 3

( )( ) ( )

( )( ) ( )

x x x x x

x x x


21 4

3

( )x

x


21 4 1 2 1 2

3 31 3

3

( ) ( )( )

( )( )

x x x

x x

x x

x


1 31

3

( )( )x xx

x


4 2

2

1

1 4 4

x x x

x x x SOLUTION Step1: Change the division to the multiplication by inverting numerator with the denominator. Factorize both the numerator and the denominator with suitable method.

4 3

2 2 2

2

2

4 4 1 4 1

1 1 1 1 1

1 1 4 1

1 1 1

( ) ( )

( )( )

( )( ) ( )

( )( )

x x x x x x

x x x x x x x

x x x x x

x x x x


4x


2 2 2

2 2 2

2 1 4 1 5 6

9 2 1 2

x x x x x

x x x x x

SOLUTION Step1: Change the division to the multiplication by inverting numerator with the denominator. Factorize both the numerator and the denominator with suitable method.

2 2 2

2 2 2

2 1 2 1 5 6

9 4 1 2

x x x x x x

x x x x

1 2 1 1 2 1 3 2

3 3 2 1 2 1 2 1

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

x x x x x x

x x x x x x


1 2 1 1 2 1 3 2

3 3 2 1 2 1 2 1

( )( ) ( )( ) ( )( )

( )( ) ( )( ) ( )( )

x x x x x x

x x x x x x

1 2

3 2

( )( )

( )( )

x x

x x


2 3 2

3 3

x x x x

x x

SOLUTION Step1: Change the division to the multiplication by inverting numerator with the denominator. Factorize both the numerator and the denominator with suitable method.

2

3 2 2

3 1 3

3 3 1

( )

( )

x x x x x x

x x x x x x


1 3 1

3 1

( )

( )

x x x

x x x x x

62

TASK 7 Simplify the following expressions: 1. 5 5

7

5 25

14

2. 3 10

2 10

5 10

4

3. 45 5

5 1 3.3

1 3

4.

5. 2 3 6

.3 3

6.

63

TASK 8 Simplify the following expressions: 1. 2

2 1 4 1 .2 1

2. 9

3

3. 2

4. 2 2

1

5. 3

5 6.

4

2

6. 2 2

2 2

7. 3 2

16 4

8. 2 18

4 12

64

TASK 9 Simplify the following expressions: 1. 1 2 2

4

2. 9

4 12

5 15

8

3. 3

9 6

4. 3 2

6

5. 2 7 4

10 5.8 2

16

6. 16 4

8 4

2

2 2

7. 10 25

25.

5

5

8. 1000

10 100

100

10 100

65

TASK 10 Simplify the following expressions: 1.

2

2. 2

4.

2

3.

1.

2 1

4.

5. 2 8

7 10

5 4

6 5

6. 9

3.

4

2 28

66

E. ADDITION AND SUBTRACTION OF RATIONAL EXPRESSIONS In order to add/subtract and write a rational expressions in the simplest form we use the following procedures:

To add/subtract rational expressions:

Factorize the denominators if possible

Find the LCM of the denominators.

After making the denominators the same

by using LCM, combine the terms under

the same denominator.

QUESTION 1

Simplify the following expression.1 1

2 2 4

x x

x x

SOLUTION Step1.Factorize the denominators and find the LCM.

1 1

2 2 2( )

x x

x x

LCM : 2 2( )x

Step2.Multiply the first term by 2:

1 1 2 2 1 3 1

2 2 2 2 2 2 2( ) ( ) ( )

x x x x x

x x x x

(2) QUESTION 2 Simplify the following expression.

2 3 1

2 2 3x x x

SOLUTION Remember 2 2( )x x

2 3 1 1 1

2 2 3 2 3x x x x x

LCM = (x‐2)(x+3)

1 1

2 3x x

(x+3) (x‐2)

3 2 5

2 3 2 3( )( ) ( )( )

x x

x x x x


2 23 2 4

x x

x x x

SOLUTION

2 22

2 1 2 2 1 2 2( )( ) ( )( ) ( )( )( )

x x x x x x

x x x x x x x

(x‐

2) (x+1)

LCM = (x+1)(x+2)(x‐2)

22 2

1 2 2 1 2 2 1 2

( )

( )( )( ) ( )( )( ) ( )( )

x x x x x

x x x x x x x x


2 4 5

2 3 4x x x

SOLUTION

2

2

4

3

5

4

(x‐3)(x+4) (x+2)(x+4) (x‐3)(x+2)

LCM = (x+2)(x‐3)(x+4) 2 3 4 4 2 4 5 3 2

2 3 4

2 12 4 6 8 5 6

2 3 4

2 2 24 4 24 32 5 5 30

2 3 4

67

TASK 11 Simplify the following expressions: 1. 1

3

2

6

3

2.

3

3

5

3.

11

1

2

4. 1 1 1

1

5.

9 3

3

6. 3

5

1

7 10

1

2

68

V. QUADRATIC EQUATIONS A. SOLVING QUADRATIC EQUATIONS BY FACTORIZATION

To solve a quadratic equation by using factorization:

Write the equation in the form of 2 0ax bx c standard form

Use one of the methods to factorize theequation.

Write the answers.

QUESTION 1 Solve for x: 3 4 5 0( )( )x x

SOLUTION The equation is already factorised. Either 3 4x is zero or 5x .

3 4 0

3 4

4

3

x

x

x

Or5 0

5

x

x


2 4 5x x SOLUTION

Write equation in the form of2 0ax bx c

2 4 5 0x x

Factorise the trinomial.

5 0x or 1 0x

5x or 1x QUESTION 3 Solve for x:

25 6 4x x SOLUTION

Write the equation in the form of 2 0ax bx c

26 5 4 0x x Factorise the trinomial:

2 1 3 4 0( )( )x x

2 1 0x or 3 4 0x 1

2x or

4

3x

QUESTION 4 Solve for x: 7 12( )x x

SOLUTION Open the bracket:

2 7 12x x

Write the equation in the form of 2 0ax bx c

2 7 12 0x x 3 4 0( )( )x x

3 0x or 4 0x


21 10

2 3x x

SOLUTION Multiply each term by minus(‐) sign to make the sign of the coefficient of the leading term ( 2x ) positive. Then multiply each term by 6 (LCM of 2 and 3) to get rid of the denominators:

21 16 6 6 0

2 3( ) ( ) ( )x x

We get 23 2 0x x

As seen from the equation, the constant term is zero(c=0).In this case we factorise the binomial by using common factors:x is the common factor

3 2 0( )x x

0x or 3 2 0x

0x or2

3x

69

QUESTION 6

Solve for x: 23 12x

SOLUTION Divide each term by 3:

2 4x 2 4 0x (standard form)

Factorize the 2 4x by using difference of two squares:

Note: 2 2a b = ( )( )a b a b

a b (differences of two squares)

2 2 0( )( )x x

2 0x or 2 0x 2x or 2x

QUESTION 7

Solve for x: 2 9 0x

SOLUTION

2 9x Square of a real number cannot be negative, no solution! QUESTION 8 Solve for x:

2 90

4x

SOLUTION Multiply each term by minus(‐) sign to make the sign of the coefficient of the leading term ( 2x ) positive. Then multiply each term by 4 (LCM) to get rid of the denominator:

24 9 0x use differences of two squares to factorise it: 2 3 2 3 0( )( )x x

2 3 0x or 2 3 0x

3

2x or

3

2x

Or the equation can be solved by multiplying each term by minus sign:

2 90

4x

3 3 30

2 2 2( )( )x x x or

3

2x


23 2 16( )x

SOLUTION Open the perfect square first:

2 2 22( )a b a ab b

2 23 2 9 12 4 16( )x x x

Write the equation in the form of 2 0ax bx c 29 12 12 0x x

Divide each term by 3:

23 4 4 0x x Factorise the trinomial.

3 2 0x or 2 0x

2

3x or 2x

70

TASK 1 Solve for x

1. 3 7 0( )( )x x

2. 3 5 4 5 0( )( )x x

3. 2 4 0x x

4. 23 15 0x x

5. 2 5 14 0x x

6. 2 8 12x x

7. 216 0x

8. 22 5 12 0x x

9. 215 4 3 0x x 10. 3 4 20( )( )x x

11. 3 1 1( )( )x x x

12. 22 5 9( )x

13. 3 212 4 5 0x x x

14. 22 61 101x

71

TASK 2 Solve for x

1. 3 4 0( )( )x x

2. 2 3 3 0( )( )x x

3. 24 0x x

4. 2 6 7 0x x

5. 22 5 3 0x x

6. 2 100 0x

7. 2 2 3 12x x

8. 3 2 12( )( )x x

9. 213 42x x

10. 22 50 0x

11. 2 25 4 0x kx k

12. 21 3 24( )x

13. 22 4 68x

72

B. SOLVING BY TAKING SQUARE ROOT OF BOTH SIDES

To solve an quadratic equation by taking Square root of both sides: Make sure that the unknown which you will

find should be in a perfect square.

e.g 24 2 9 0( )x

Leave the term with perfect square alone by sending the other terms to the other side.

e.g 24 2 9( )x

Leave the perfect square alone and use the rule: take square root of both sides.

2x a

x a


2 16 0x SOLUTION As seen from the equation x is a perfect square. When 2x is left alone we get:

2 16

16

x

x

.

4x or 4x (easy one!) QUESTION 2 Solve for x:

24 12 0x SOLUTION Divide each term by 4:

2 3 0x 2 3

3

x

x

3 1 73,x or 3 1 73,x


226 0

3x

SOLUTION Multiply each term by 3 :

22 18 0x Leave the term with perfect square ( 22 x )alone:

22 18x then leave the perfect square( 2x ) alone:divide both sides by 2;

2 9

9

x

x

.


212 8 0

2( )x

SOLUTION

Leave the term with perfect square 212

2( )x

alone:

212 8

2( )x

then leave the perfect square22( )x alone

:multiply both sides by ‐2; 22 16( )x

take square root of both sides:

2 16x . 2 4x or 2 4x


26 12 0x SOLUTION

26 12x divide both sides by 6:

2 2x There is no real solution because the result of a perfect square cannot be a negative number.

73

TASK 3 Solve for x

1. 23 75x

2. 24 9 0x ifx is real.

3. 23 16 0( )x

4. 212 0

3x

5. 2 14 0

3x

6. 21 1 0( )x

7. 24 2 9 0( )x

8. 214 0

2x

9. 24 3 0( )x

10. 224 10 0

5( )x

74

C. SOLVING BY COMPLETING THE SQUARE

To solve a quadratic equation by using completing the square: Write the equation in the form of

2 .....ax bx

takethe constant term to the right side 2

2

2 8 10 0

2 8 10

x x

x x

Divide each term by the coefficient of 2x

2 4 5x x .

Halve the coefficient of x and square it. Add the result toboth sides.

2 242

2( ) ( ) should be added;

2 2 23 2 5 2( ) ( )x x

Write the left‐hand side as a perfect square: then follow the method of taking square root

of both sides. use calculator 22 9

2 9

( )x

x

2 9

5

x

x

OR 2 9

1

x

x

QUESTION 1 Solve for x by completing the square.

2 2 1 0x x SOLUTION Take the constant term to other side:

2 2 1x x Halve the coefficient of x and square it. Add the result in both sides.

2 221

2( ) ( ) should be added in both sides;

2 2 22 1 1 1( ) ( )x x

Write the left‐hand side as a perfect square: then follow the method of taking square root of both sides.

21 2

1 2

( )x

x

2 1x or 2 1x

QUESTION 2 Solve for x by completing the square.

25 3 4 0x x SOLUTION Take the constant term to other side and rearrange the terms:

23 5 4x x

Divide each term by the coefficient of 2x . i.e

divide by ‐3.

2 5 4

3 3x x

Halve the coefficient of x (half of 5

3 is

5

6 )and

square it. Add 25

6( ) to both sides;

2 2 25 5 4 5

3 6 3 6( ) ( )x x =

Write the left‐hand side as a perfect square: then follow the method of taking square root of both sides.

25 73

6 36

5 73

6 36

( )x

x

51 42 2 25

6, ,x or

51 42 0 59

6, ,x

QUESTION 3

If 2 2( )f x x x , show by completing the square

that 21 1( ) ( )f x x

SOLUTION

Let us rearrange ( )f x by using the completing

square: 2 2( )f x x x

Halve the coefficient of x and square it. Add the result in both sides.

21( ) should be added in both sides; 2 2 21 2 1( ) ( ) ( )f x x x

2

2

1 1

1 1

( ) ( )

( ) ( )

f x x

f x x

75

TASK 4 Solve for x by completing the square.

1. 2 10 2 0x x

2. 2 4 3 0x x

3. 22 12 4 0x x

4. 2 5 9 0x x

5. 23 6 2 0x x

6. 22 11 0x x

7. 25 4 0x x

8. If 2 4( )g x x x , show by completing the

square that 22 4( ) ( )g x x .

9. Given : 2 4 2x x

9.1 Write the above expression in the

form 2( )a x p q .

9.2 Hence, determine the value of x which will maximise 2 4 2x x

76

D. SOLVING BY USING QUADRATIC FORMULA

To solve an quadratic equation by using quadratic formula: Write the equation in the form of

2 0ax bx c standard form

Use the following quadratic formula after identifying the values of a ,b and c .

2 4

2

b b acx

a

The expression in the square root is called

discriminant where 2 4b ac .

If the result of is negative , there are no real

roots e.g 9 no real roots. When you have difficulty to factorise the trinomials , apply quadratic formula. In this method, we use decimal places because calculator is needed in most of the cases. QUESTION 1

Solve for x;22 3 7 0x x

SOLUTION When it is tried to factorise the trinomial above you see that it does not work; here we use the quadratic formula

2a 3b 7c 2 4

2

b b acx

a

23 3 4 2 7

2 2

3 65

4

( )( )

( )x

3 65

4x

or

3 65

4x

İf the question indicates two decimal places, use calculator:

1 27,x or 2 77,x

QUESTION 2

Solve for xif 21 54 0

2 2x x

SOLUTION Multiply each term by ‐2:

2 5 8 0x x Use the quadratic formula because factorisation method does not work for this equation.

1a 5b 8c

2 4

2

b b acx

a

25 5 4 1 8

2 1

5 25 32

2

5 7

4

( )( )

( )x

No real solutions. 7 is undefined for real numbers. QUESTION 3 Use quadratic formula to solve

27 2 5 0x x SOLUTION

7a 2b 5

2 4

2

b b acx

a

22 2 4 7 5

2 7

2 144

14

( ) ( )x

2 12

14x

or

2 12

14x

1x or5

7x

77

TASK 5 Solve for x by using the quadratic formula (two decimal places if needed)

1. 23 4 0x x

2. 22 3 2 0x x

3. 2 2 1 0x x

4. 24 5 0x x

5. 25 3 14x x

6. 4 2 0( )x x

7. 26 19 10 0x x

8. 212 2

2x x

9. 21 25 0

3 3x x

10. 24 9 1 1( )( )x x x

78

E. QUADRATIC EQUATIONS WITH FRACTIONS

To solve an quadratic equation with fractions: Factorise each denominator and by

finding LCM, cancel out the denominators.

Do not forget to indicate the restrictions: the values which make the denominators

zero undefined

After cancelling the denominators, use one of the methods to solve the equation.


31 2

1x

x

SOLUTION Restriction: 1x Move to the LHS

33

1x

x

Cross multiply

2

3 1 3

4 3 3

x x

x x

The standard form:

2 4 0x x Common factor

4 0( )x x

0x or 4x

QUESTION 2 Solve for x :

1 21

1 1

x

x x

SOLUTION Restrictions: 1x and 1x

Add fractions on the left: 1( )x ( 1x ) =LCM

1 2 11

1 1

( ) ( )

( )( )

x x x

x x

2

2

1 2 21

1

x x x

x

Cross multiply: 2 22 3 1 1x x x

The standard form :

2 3 0x x

3 0( )x x


1 2 2

1 2 1

x x

x x x

SOLUTION Restrictions: 2x , 1x

Remember 1 1( )x x

1 2 2

1 2 1

x x

x x x

Move fractions with same denominator to LHS:

1 2 2

1 1 22 3 2

1 2

x x

x x xx

x x

Cross multiply: 22 7 6 2 2x x x

The standard form :

22 9 4 0x x

2 1 4 0( )( )x x

1

2x or 4x

79

TASK 6 Solve for x(two decimal places when needed)

1. 8

2xx

2. 1 1 2

2 15x x

3. 4

11

xx

4. 3

2 12 1

xx

5. 1 2

3 3

xx

x x

6. 2 2 1

1 7

x x

x x

7. 1 2

11 1

x

x x

8. 1 2

2 3 2

x

x x x

80

F. SIMULTANEOUS EQUATIONS In this section, you will learn how to solve sets of simultaneous equations where one is linear and one is a quadratic:

To solve simultaneous equation: Starting from the linear equations , make the

suitable unknown subject to the other unknown.

Substitute it into the quadratic equation to make one variable equation.And solve the quadratic equation.

QUESTION 1 Solve for x and y:

2 4 0y x and2 4x y

SOLUTION

Make y subject to x in the linear: 2 4y x

Subs. into we get: 2

2

2 4 4

2 8 0

x x

x x

Solve for x: 2 4 0( )( )x x

2x and 4x

Use the equation 2 4y x to find the y values for

each x.

If 2x , 2 2 4 0( )y

If 4x , 2 4 4 12( )y

QUESTION 2 Solve for xandy:

2 2 0x y and28 8 0x y

SOLUTION

Make y subject to x in the linear: 2 2x y

Subs. 2 2y x into we get: 28 2 2 8 0( )x x

28 4 8 4 8 0x x x 24 4 0x

Divide both sides by 4 : 2 1 0

1 1 0( )( )

x

x x

1x and 1x

Use the equation 2 2x y to find the y values

for each x.

If 1x , 2 2 1 4( )y

If 1x , 2 21 0( )y

QUESTION 3 Solve for x and y

2

2 3

2 2 0

x y

x xy

SOLUTION Make x subject to y in the linear: 3 2x y

Subs. it into 2 2 2 0x xy we get:

23 2 2 3 2 2 0( ) ( )y y y

Write 2 and y together

2 2

2

3 2 3 2 2 3 2 2 0

9 12 4 6 4 2 0

8 18 7 0

4 7 2 1 0

( ) ( ) ( )

( )( )

y y y y

y y y y

y y

y y

7

4y and

1

2y

Use the equation 3 2x y to find the value of x:

1

2x and 2x

QUESTION 4 The sum and product of two numbers are equal to ‐10 and ‐600 respectively. Determine the numbers. Show all calculations. SOLUTION Let the numbers be x and y. Sum is ‐10:

10

10

x y

x y

Product is ‐600

600xy

Substitute 10x y into second equation

10 600( )y y 210 600y y

Make standard form and factorise:

2 10 600 0y y 30y or 20y

Use first equation to find x values

20x 30x

42 xy 42 yx

088 2 yx

81

TASK 7 1. Solve for x and y

2

3 2

2

x y

x xy x

2. Solve for x and y:

2 2

2

52 0

x y

x y


2 2

3

2 2 5

x y

x y xy

4. Solve for x and y

2 2

2 1 0

9 2

x y

x y xy


2

2 6 3

2 2 18

y x

x x y


2 2

2 1

2 5 2 3

y x

x xy x y y

82

TASK 8


2

3 6 0

4 0

x y

x y y


2

6 4 0

12 2 0

x y

x y


2

2 10 0

5 0

x y

x y y

4. A group of students went out for lunch after their examination. They were to share the expenses equally. The total bill was R120. Ten of the students did not have any money, so each of the others had to pay an extra R2. Determine the number of students in the group.

5. A piece of wire of length 500 mm is bent to

form a rectangle with area 1225 mm2. One side of the rectangle has length x mm. Calculate the value(s) of x.

6. A man bought a certain amount of golf

balls for R 400. The next week the shop had a sale and each ball cost R 4 less. For the new price he would have received 5 more balls for the same money. How many balls could he buy with the sale

83

TRIAL TEST 1 QUESTION 1 1.1 Solve for x, rounded off to TWO decimal

places where necessary:

1.1.1 1 5

61x x

(6)

1.1.2 2 3 28x x (5)

1.2 Solve for x and y simultaneously: 2 3x y

2 25 15x xy y

(7) [18]

QUESTION 2 2.1 Solve for x Solve for x, rounded off to

TWO decimal places where necessary: 2.1.1 4 3 12( )( )x x (4)

2.1.2 23 2 6 10x x (5)

2.1.3 23 2 3( )x x (6)

2.2 Solve for x and y simultaneously: 4 3 50y x and

2 2 100x y (7) [22]

TOTAL: 40


places where necessary: 1.1.1 2 2 24x x (3)

1.1.2 6

2xx

(5)

1.1.3 1 2 6( )( )x x (4) 1.2 Solve for x and y simultaneously: 3 5x y and

2 3xy y (7) [19]

QUESTION 2 2.1 Solve for 2.1.1 4 3 8( )( )x x (4)

2.1.2 1 2

11 1

x

x x

(5)

2.1.3 23 2 4( )x (5)

2.2 Solve for x and y simultaneously:

2 2

7 2

3 15

y x

x xy y

(7) [21]

TOTAL: 40

x

84

TRIAL TEST 3

QUESTION 1 1.1 Solve for x, rounded off to TWO decimal

places where necessary: 1.1.1 2 3 10 0x x (3)

1.1.2 3

1 21

xx

(6)

1.1.3 2 7 5x x (4)

1.2 The sum of the first number and half of the

second number is 57. The difference between half the first number and the second is 6. Let numbers be x and y where x; y are natural numbers.

1.2.1 Write down two equations in terms of x and y.

(2)

1.2.2 Hence or otherwise, calculate the values of the two numbers.

(3) [18]


places where necessary: 2.1.1 9 14 0( )x x (3)

2.1.2 2

6 3 2

1 1 1

x x

x x x

(6)

2.2 If

5

3

xB

x

, determine the values of x

for which: 2.2.1 B is undefined (2)

2.2.2 B is non‐real (3)

[14]

TOTAL: 32


places where necessary: 1.1.1 22 4x x (2)

1.1.2 5

3xx

(4)

1.1.3 (5)

1.2 Solve for x and y simultaneously: 3x y and

2 22 2 5x y xy (7) [18]


places where necessary: 2.1.1 3 1 1( )( )x x x (4)

2.1.2 2 3 1x x (4)

2.2 If

2 2( )f x x x , show by completing the

square that 21 1( ) ( )f x x .

(4)

[15]

TOTAL: 30

2 3x x

85

VI. FUNCTIONS A. PARABOLA

Parabola is a graph of a quadratic function with

the standard form of 2y ax bx c .To sketch

any parabola in the standard form, we should follow the steps below:

To sketch any parabola 2y ax bx c

Find the x‐intercepts(roots) by allowingy=0.You will solve a quadratic equation. Indicate the x‐intercepts : 0(?; )

Find the y‐intercept by allowing 0x .

Indicate the y‐intercept : 0( ;?)

Find the turning point of parabola by using the

formula: 2

bx

a

.

After find the y‐coordinate of the turning point by substituting x‐value into the function. Turning point: TP(x;y)

Decide the direction of parabola by checking

the coefficient of 2x which is a . If 0a , the arms of parabola are upwards(happy face);parabola has minimum value at turning point. If 0a , the arms of parabola are downwards (sad face);parabola has maximum value.

Make a sketch graph of the function by

showing the intercepts as well as the turning point.

QUESTION 1

Make a sketch graph of the function: 2 6 7y x x

SOLUTION

x‐intercepts: when 0y , x=? 2 6 7 0

7 1 0( )( )

x x

x x

7x or 1x

7 0( ; ) and 1 0( ; )

y‐intercept: when 0x , y =?

0 7( ; )

Turning point:2

bx

a

6

32 1

( )

( )TPx

Subs. 3x into the function: 23 6 3 7 16( ) ( )TPy

TP 3 16( ; )

Because 1 0a , arms are upwards; happy face

QUESTION 2

Make a sketch graph of the function: 22 8 6( )f x x x

SOLUTION

x‐intercepts: when 0y , x=? 22 8 6 0x x

Divide each term by 2: 2 4 3 0

1 3 0( )( )

x x

x x

1x or 3x

1 0( ; ) and 3 0( ; )

y‐intercept: when 0x , 0( )f =?

0 6( ; )

Turning point:2

bx

a

8

22 2

( )

( )TPx

Subs. 2x into the function: 22 2 8 2 6 2( ) ( )TPy

TP 2 2( ; )

86

QUESTION 3 Make a sketch graph of the function:

213 4

2y x x

SOLUTION

x‐intercepts: when 0y , x=?

213 4 0

2x x

multiply both sides by ‐2. 2 6 8 0

2 4 0( )( )

x x

x x

2x or 4x

2 0( ; ) and 4 0( ; )

y‐intercept: when 0x , y =? 0 4( ; )

for each step we should use the original function.

Turning point:2

bx

a

33

12

2

( )

( )TPx

Subs. 3x into the original function:

213 3 3 4 0 5

2( ) ( ) ,TPy

TP 3 0 5( ; , )

Because 1

02

a

, the arms are downwards;

sad face

QUESTION 4

Sketch the graph of function2 4( )g x x

SOLUTION

x‐intercepts: when 0y , x=? 2 4 0x

Multiply both sides by ‐1. 2 4 0

2 2 0( )( )

x

x x

2x or 2x

2 0( ; ) and 2 0( ; )

y‐intercept: when 0x ,y=?

0 4( ; )

Turning point:2

bx

a

0

02 1

( )

( )TPx

Subs. 0x into the function: 4TPy

TP 0 4( ; ) the same as y‐intercept(when b=0)

QUESTION 5

Make a sketch graph of 2 1y x x

SOLUTION

x‐intercepts: when 0y , x=? 2 1 0x x

Use the quadratic formula: 21 1 4 1 1

2

1 3

2

( )( )x

Because 3 is not real , there is no solution No x‐intercept; the graph does not cut x‐axes. y‐intercept: 0 1( ; )

Turning point:

1 1

2 1 2( )TPx

21 1 31

2 2 4( )TPy

TP1 3

2 4( ; )

87

1. TP FORM To sketch a parabola in form of

2( )y a x p q Finding the turning point is very easy!

p is the x‐coordinate of TP

q is the y‐coordinate of TP

Find the x‐intercepts(roots) by allowing 0y .Leave the perfect square

2( )x p alone by

sending to the other side and dividing both

sides by a . Then take square root of both sides to find the x‐values.

Find the y‐intercept by allowing 0x . Decide the direction of parabola by checking

the coefficient of 2x which is a . If 0a , the arms of parabola are upwards(happy face); parabola has minimum value at turning point. If 0a , the arms of parabola are downwards (sad face); parabola has maximum value.

Make a sketch graph of the function by

showing the intercepts as well as the turning point.

QUESTION 6

Make a sketch graph of the function: 22 3 2( )y x

SOLUTION You can open the bracket and you write the function in standard form and go on normally. But it is easy to work with turning point method.

TP 3 2( ; ) ( ; )p q

x‐intercepts: 22 3 2 0( )x

leave23( )x alone:

2

2

2 3 2

3 1

( )

( )

x

x

,

take square root of both sides: 23 1

3 1

3 1

( )x

x

x

4x or 2x

4 0( ; ) and 2 0( ; )

y‐intercept: when 0x , y =? 22 0 3 2 16( )y

0 16( ; )

Because 2 0a , the arms are upwards;

QUESTION 7

Make a sketch graph of the function:

212 1

2( )y x

SOLUTION

TP 2 1( ; ) ( ; )p q

x‐intercepts:

212 1 0

2( )x .

leave22( )x alone:

212 1

2( )x

multiply both sides by ‐2 22 2( )x

take square root of both sides: 22 2

2 2

2 2

( )x

x

x

0 59,x or 3 41,x

0 59 0( , ; ) and 3 41 0( , ; )


210 2 1 1

2( )y

0 1( ; )

q

88

QUESTION 8

Make a sketch graph of the function: 23( )y x

SOLUTION The function is with TP form but q value is missing:

23 0( )y x

TP 3 0( ; ) ( ; )p q

x‐intercepts:

23 0

3 3 0

( )

( )( )

x

x x

.

bothx‐values are ‐3 (double root) also TP.

3 0( ; ) and 3 0( ; )


20 3 9( )y

0 9( ; )

QUESTION 9

Sketch the graph of function 212

2y x x

SOLUTION x‐intercepts:

212 0

2x x

2 4 0

4 0( )

x x

x x

0x or 4x y‐intercept: 0 0( ; )

Turning point:

2

22 0 5

( )

( , )TPx

212 2 2 2 4 2

2( ) ( )TPy

2. ROOT FORM Another useful form is root (x‐intercept form

The function is in the form 1 2( )( )y a x x x x is

called the root form: where 1x and 2x are the

roots(x‐intercepts). When function in this form, it is very easy to find the x‐intercepts.

QUESTION 10

Draw the graph of function1

3 42( )( )y x x

SOLUTION x‐intercepts:

10 3 4

2( )( )x x

3 0( ; ) and 4 0( ; )

y‐intercept:

10 3 0 4 6

2( )( )y

0 6( ; )

Turning point:

TPx is exactly between roots, so we can find it by

adding x intercepts divided by 2:

4 30 5

2,TPx

Substitutex in

1 1 13 4 6 1

2 2 2( )( ) ,TPy

TP 0 5 6 1( , ; , )

89

TASK 1 Make a sketch graph of the functions below by showing the turning point as well as the intercepts.

1. 2 8 9y x x

2. 214 6

2y x x

3. 2 9( )g x x

4. 22 8 10y x x

90


1. 23 1 3( )y x

2. 212 3

3( )y x

3. 214

3y x x

4. 23 1( )y x

91


1. 1

1 33( )( )y x x

2. 23 12 16y x x

3. 21 4

5 5y x x

4. 24 9( )y x

92

B. ELEMENTS OF PARABOLA AND FUNCTIONSThe axis of symmetry of a parabola is a vertical line that divides the parabola into two congruent halves.

Axis of Symmetry: The line equals tox‐coordinate of the turning point:

x p Depending on shape of its shape parabola has Maximum or Minimum:

Maximum or Minimum ( )f x is equals to the y‐

coordinate of the turning point: y q

Any parabola has a max or min point, which is its turning point If a parabola has positive coefficient of 2x (happy face) then it has min value, else (sad face) it has max value.

Domainis the possible x‐values of a function.

In a parabola, domain is element of all real numbers. It is indicated as x R

Range: Range is the possible y‐values of a function.

In a parabola, range is related to the turning point of parabola. If the turning point has a maximum value,

range is y q .

If the turning point has a minimum value, range is y q .

Point on the graph: Substitute the x and y values into equation

This method is used to find unknown(s)

EXAMPLE

Given 2y ax c

Find the values of c and a SOLUTION Take the point (0;‐12)

212 0

12

a c

c

Or you can directly say c is ‐12 because it’s y‐intercept Now take the point (‐2;0)

20 2 12

12 4

3

( )a

a

a

Shifted Parabola, Reflected Parabola:

Let ( )f x be the original function.

( )f x a shifting ( )f x a units to the right.

( )f x a shifting ( )f x a units to the left.

( )f x a shifting ( )f x a units downwards.

( )f x a shifting ( )f x a units upwards.

( )f x a b shifting ( )f x a units to the

right and b units upwards.

( )f x reflecting ( )f x in the y‐axis.

symmetrical to y‐axis . Change x with –x

→reflecting in the

.(symmetrical to x‐axis.).

The average gradient between two points:

2 1

2 1

y ym

x x

= 2 1

2 1

( ) ( )f x f x

x x

The average gradient between any two points on a curve is the gradient of a line which passes through these two points.

93

QUESTION 1

Given :2 2 1( )f x x x

1.1. Find the axes of symmetry of ( )f x .

1.2. Write the equation of ( )g x obtained by

shifting ( )f x by 1 unit to the right.

1.3. Write the equation of ( )h x obtained by

shifting ( )f x by 2 units down.

1.4. Write the equation of ( )k x obtained by

reflecting in the y‐axis.

1.5. Find the domain and range of ( )f x .

1.6. Find the minimum value of ( )f x .

1.7. If ( )f x is shifted by 2 units to the left,

find the coordinates of new turning point. SOLUTION

1.1. Find x‐coordinate of turning point, use

21

2 2 1

( )

( )

bx

a

Axes of symmetry is 1x

1.2. Shifting ( )f x by 1 unit right 1( )f x

2

2

1 2 1 1

4 2

( ) ( ) ( )g x x x

x x

1.3. Shifting ( )f x by 2 units down: 2( )f x

2

2

2 1

2 3

2( ) ‐h x x x

x x

1.4. Reflecting in the y‐axis: ( )f x

2

2

2 1

2 1

( ) ( ) ( )k x x x

x x

1.5. Domain: x R Range: 2y

For range find the y‐coordinate of TP. 21 1 2 1 1 2( ) ( )f

TP is the minimum; arms of parabola up(a>0)

1.6. Minimum value is at TP: 2y

1.7. Only x‐coordinate of ( )f x will change: It will

decrease by 2 units. New TP : 1 2( ; ) .

QUESTION 2 2y x graph is given.

Hence sketch the following:

2.1 2 3y x

The original function shifted 3 units upwards

2.2 22( )y x

The original function shifted 2 units to the right

2.3 2 1y x

The original function shifted 1 unit downwards

2.4 23 2( )y x

The original function shifted 3 units to the right and 2 units upwards

2.5 2 2y x

The function 2y x

shifted 2 units upwards

2.6 23( )y x

The function 2y x

shifted 3 units to the left

2.7 21 4( )y x

The function 2y x

shifted 1 unit to the right and 4 units upwards

94

TASK 4 1. Given :

22 8 3( )f x x x


1.2. Find the average gradient of ( )f x between

1x and 3x .

1.3. The function ( )g x is obtained by shifting

( )f x by 3 units to the left. Write the

equation of ( )g x .

1.4. The function ( )h x is obtained by shifting

( )f x by 4 units upwards. Write the

equation of ( )h x .

1.5. The function ( )k x is obtained by reflecting

( )f x in the x‐axis. Write the equation of

( )k x .


1.7. Find the maximum value of ( )f x .

1.8. If ( )f x is shifted by 3 units down and 2units

to the left, find the coordinates of new

turning point.

2. Given : 21 3( ) ( )f x x


2.2. Find the average gradient of ( )f x between

2x and 0x .

2.3. The function ( )g x is obtained by shifting

( )f x by 4 units to the right. Write the

equation of ( )g x .

2.4. The function ( )h x is obtained by shifting

( )f x by 2 units downwards. Write the

equation of ( )h x .

2.5. The function ( )k x is obtained by reflecting

( )f x in the x‐axis. Write the equation of

( )k x .


2.7. Find the minimum value of ( )f x .

2.8. If ( )f x is shifted by 1 unit to the right and 3

units upwards, find the coordinates of the

new turning point.

95

TASK 5 1. The figure shows the graph of C is the turning

point of 24 8 5( )g x x x , and D is the y‐

intercept

Determine:

1.1. AB

1.2. The co‐ordinates ofC,D and E (ED//AB).

2. Given 2( )f x ax c and the graph of it below.

Find values of a and c.

3. The figure represents the graph of y=(x+2)2+5.

3.1. Calculate the lengths of AB if A is the

turning point.

3.2. Calculate the coordinates of C.

3.3. G is the point (x ;30), calculate the value of

x.

3.4. Determine the average gradient between A

and C.

3.5. What is minimum value of expression

(x+2)2+5

96

TASK 6 QUESTION 1

2y x graph is given.

Hence sketch the following

functions.

1.12 3y x

1.2 22( )y x

1.3 2 2y x

1.4 21 1( )y x

1.5 2 2y x

1.6 21( )y x

1.722 1( )y x

QUESTION 2

The sketch, not drawn according to scale, show the parabola y=‐x2+2x+k that intercepts with the x‐axis at A and C. A straight line through A cuts the parabola in B (3; 5). 2.1 Show by calculations that k = 8 2.2 Determine the co‐ordinates of A. 2.3 Determine the gradient of AB. 2.4 A straight line ED with E and D (2; p) on the

parabola is parallel to AB. Determine: 2.4.1 the value of p. 2.4.2 the equation of ED. 2.4.3 the co‐ordinates of E.(intersection point of two graphs). (Solve simultaneously)

x

y 2y x

97

C. HYPERBOLA Hyperbola is a function with the standard form of:

ay q

x p

where a is constant.

Hyperbola y does not have any x‐intercept

nor y‐intercept. The value of x can not be zero neither y.The graph doesn’t touch either of axes.

Domain of y : x ∈ R but x 0

Range of y : y ∈ R but y 0

x 0 (y‐axis) is vertical asymptote. y 0 (x‐axis) is horizontal asymptote. If a 0, the function is in the first and

third quadrants.

If a 0, the function is in the second and

fourth quadrants.

Function of the Form :y q

The vertical asymptote: x p (the x‐value which makes the denominator zero) The horizontal asymptote: y q

To sketch hyperbola, a

y qx p

we need to determine four characteristics: Asymptotes: x p , y q

y‐intercept: make 0x

x‐intercept: make 0y

The shape: if 0a in 1st and 3rd quadrant*

if 0a in 2nd and 4th quadrant*

Domain x p (all x values but p)

Range y q (all y values but q)

*Here, the quadrant of asymptotes are implied.

1. QUESTION 1

Given: 1

24

( )h xx

1.1. Determine the equations of asymptotes of h

1.2. Determine the coordinates of the intercepts

of h with the x‐and y‐axes

1.3. Sketch the graph of h showing clearly the

asymptotes and ALL intercepts with the

axes.

SOLUTION

1.1. Horizontal asymptote : y 2

Vertical asymptote :x 4

1.2. x‐intercept:

10 2

41

24

2 8 1

3 5,

x

xx

x

y‐intercept:

12

0 41 75,

y

1.3. Draw the asymptotes first (use dotted line):

Use the first and third quadrants (a 1

0)

98

2. QUESTION 2

Given: 1

12

( )f xx

2.1. Give the domain of f.

2.2. For what value of x is f(x)=0?

2.3. Determine the y intercept of f.

2.4. Write the equations of the asymptotes of f.

2.5. Draw a neat graph of , indicating the

asymptotes and intercepts with the axes.

SOLUTION 2.1. Domain: x ∈ R but x 2

Range: y ∈ R but y 1

2.2. x‐intercept: f(x)=y=0

10 1

21

12

2 1

3

x

xx

x

2.3. y‐intercept: (0;?)

11

0 21 5,

y

2.4. Horizontal asymptote : y 1

Vertical asymptote :x 2

2.5. Draw the asymptotes first (use dotted line) :

Use the first and third quadrants

( 1 0)

3.

QUESTION 3

Given: 2

3.1. Write down equations of the asymptotes

off.

3.2. Determine the coordinates of the intercepts

of h with the x‐and y‐axes

3.3. Draw a neat graph of f, indicating the


SOLUTION 3.1. Horizontal asymptote: y 2

Vertical asymptote : x 0

3.2. x‐intercept: f(x)=y=0 (?;0)

40 2

42

2

x

xx

y‐intercept: (0;?)

f 0 24

0

Since the denominator cannot be zero, there is no x‐intercept(x=0 is an asymptote)

3.3. Draw the asymptotes’ first (use dotted line)

: Use the second and fourth quadrants

( 2 0)

99

4. QUESTION 4

4.1. Find the equations of asymptotes

4.2. Hence, determine the equation of hyperbola

SOLUTION 4.1. From the graph equations of asymptotes:

4

3

x

y

4.2. The equation of hyperbola is

34

ay

x

To find a use the point (‐3;0)

0 33 4

313

a

a

a

5. QUESTION 5

A hyperbola, ( )f x , is described with the following

properties:

The equation of the vertical asymptote is x ‐2

The range ( )f x of is 1 1 ( ; ) ( ; )

The x intercept of is ‐3;0

5.1. Write the domain of f.

5.2. Determine the horizontal asymptote of f.

5.3. Find the equation of f.

5.4. Calculate the y intercept of f.

5.5. Draw a neat graph of f.

SOLUTION 5.1. 2 2 ( ; ) ( ; ) or 2 ,x R x

5.2. 1y

5.3. The equation of hyperbola is

12

a

yx

To find a use the point (‐3;0)

0 13 2

11

1

a

a

a

11

2

( )f x

x

5.4. 1 3

0 10 2 2

( )f

30

2( ; )

5.5.

100

TASK 7 QUESTION 1

Given: 1

1.1 Determine equations of asymptotes of g. 1.2 Determine the coordinates of the intercepts of g with the x‐and y‐axes 1.3 Sketch the graph of g showing clearly the asymptotes and ALL intercepts with the axes.

QUESTION 2

Given:

2.1 Write down equations of the asymptotes of f.

2.2 Determine the coordinates of the intercepts of h with the x‐and y‐axes 2.3 Draw a neat graph of f, indicating the


101

QUESTION 3

Write the equation of asymptotes and hence find the equation of the graph f

QUESTION 4 Find the equation of the graph

x

y

1

3

QUESTION 5 Find an equation of f

QUESTION 6 Find an equation of g

102

TASK 8 QUESTION 1

Given: 2

1.1 Give the domain of f.

1.2 Determine the y intercept of f.

1.3 For what value of x is f(x)=0?


1.5 Draw a neat graph of f, indicating the


QUESTION 2

Given: 1


2.2 Determine the coordinates of the intercepts of h with the x‐and y‐axes 2.3 Draw a neat graph of f, indicating the


103

D. EXPONENTIAL FUNCTION Basic exponential function is a function with the standard form of :

y a ory a where a>0. To draw the graph of an exponential function it is very useful to use point to point plotting (table method) Let’s look at four different types of exponential functions

Example 1: 3xy (a>1 and b>0)Type 1

Domain: ∈ Range: 0 Horizontal asymptote : 0 (x‐axes) Vertical asymptote: does not exist x‐intercept: none y‐intercept: (0;1) This type is an increasing function (it goes up when looked from left to right), above the x‐axes. We notice that when x is increasing, the increase in the y value is very rapid. Also the result cannot be zero or a negative number. The bigger the a value, the steeper the graph is. The smaller the a value, the wider the graph is.

Example 2: 1

3

x

y

(0<a<1 and b>0)Type 2

can be expressed as 3 as well.

Domain: ∈ Range: 0 Horizontal asymptote : 0 (x‐axes) Vertical asymptote: does not exist x‐intercept: none y‐intercept: (0;1) This type is an decreasing function ( it goes down when looked from left to right), above the x‐axes.

Example 3: 3xy (a>1 and b<0)Type 3

Domain: ∈ Range: 0 Horizontal asymptote : 0 (x‐axes) Vertical asymptote: does not exist x‐intercept: none y‐intercept: (0;1) We notice that when x is increasing, the decrease in the y value is very rapid. Also the result cannot be zero or a positive number.

104

Example 2: 1

3

x

y

(0<a<1 and b<0)Type 4

Domain: ∈ Range: 0 Horizontal asymptote : 0 (x‐axes) Vertical asymptote: does not exist x‐intercept: none y‐intercept: (0;1)

To sketch graphs of functionx py b a q

we need to determine four characteristics: Asymptote:, only y q y‐intercept:

make 0x

x‐intercept: make 0y

The shape: if 1a increasing, when b 0

decreasing when b 0

if 0 1a decreasing ,when b 0

increasing when b 0

Domain x R Range: y q if b is positive

y q ifb is negative

STEPS to DRAW in GENERAL 1. Draw the horizontal asymptote

2. Find and plot y intercept

3. Find and plot x intercept if possible

4. Find and plot any other point on the

graph to decide the type shape

QUESTION 1 Draw the following functions;

5( ) xf x , 2 5( ) xg x , 15( ) xh x , 5 1( ) xm x ,

5 5( ) xk x SOLUTION

5( ) xf x is an exponential

function with y‐intercept is (0;1) and horizontal asymptote: y 0.

2 5( ) xg x has the same

asymptote as f(x) y‐intercept:

00 2 5 2 1 2( )g

15( ) xh x has

the same asymptote as f(x) y‐intercept:

0 10 5 5( )h

5 1( ) xm x has different

asymptote than f(x) Asymptote: y 1. Draw the asymptote y‐intercept:

00 5 1 1 1 2( )m

(0;2)

5 5( ) xk x

Asymptote: 5 Draw the asymptote y‐intercept:

00 5 5 4( )k

there is also x‐intercept in this function; x‐intercept:

0 5 5

5 5

1

x

x

x

105

QUESTION 2

Given 1

2( )

x

f x

+1 ,13( ) xh x

2.1 Draw all the graphs (show the intercepts if possible)

2.2 If ( )n x is reflection of ( )f x in the y‐axis and

( )b x is reflection of ( )h x in the x‐axis

Write the equations of ( )n x and ( )b x

SOLUTION 2.1 f(x) is a type‐2 (a=0,5<1) exp. function Horizontal asymptote :y 1

y‐intercept: 00 0 5 1 2( ) ,f

h(x) seems a type‐1 exp. function but because the negative sign in front of x, it is a type‐2 function. Horizontal asymptote :y 0

y‐intercept: 1 00 3 3( )h

2.2 ( )n x is reflection of ( )f x in the y‐axis

Change x with –x: 0,5 +1

( )b x is reflection of ( )h x in the y‐axis

Change y with –y: 3

QUESTION 3

The graph of 1 2( ) . xf x a (a is a constant)

passes through the origin as shown below:

fx

y

O

3.1 Show that a=‐1

3.2 Determine the value of f(‐15) correct to FIVE

decimal places

3.3 Determine the value of x, if P(x; 0,5) lies on

the graph of f.

3.4 If the graph of f is shifted 2 units to the right

to form the function h, write equation of h

SOLUTION 3.1 f(x) goes through the origin (0;0) which satisfies the equation:

00 1 2

1

a

a

3.2 15 1 2 32767

3.3 1 2

0 5 1 2

2 0 5

1

.

.

x

x

x

3.4 Shifting 2 units to the right : f(x‐2)

2

1 2

106

QUESTION 4

The following graph is for ( ) xf x a b ( 0a )

f

x

y

O

(2;144)P

4.1 If 3

4b , find a.

4.2 Hence write down equation of f.

4.3 Determine, correct to TWO decimal

places, the value of f(13)

4.4 Describe the transformation of the curve

of f to h if h(x)=f(‐x)

SOLUTION 4.1 The point P is on the graph ; satisfies the equation:

144 .3

4

1449

16

∴ 256

4.2 256.

4.3 256.

10775,65

4.4 ( )f x reflection of ( )f x in the y‐axis.

(symmetrical to y‐axis) .

QUESTION 5 2 . t intersect the x‐axis at (‐1;0).

5.1 Write down the value of q.

5.2 Calculate the value of p.

5.3 Write the equation of 3.

SOLUTION 5.1 q is the horizontal asymptote :

q = ‐2 5.2 The point (‐1;0) is on the graph which satisfies the equation:

2 2x py

1

1 1

0 2 2

2 2

1 1

2

p

p

p

p

5.3 2 2

3.

2 2 3

∴ 2 1

107

TASK 9 QUESTION 1

x

y2xy

1

2xy graph is given. Hence sketch the following

functions.

1.1 2 1xy

1.2 2 4xy

1.3 12xy

1.4 12 2xy

1.5 22xy

1.6 2xy

1.7 2 3xy

1.8 12 3xy

1.9 2 1

1.10 2 2

108

TASK 10 QUESTION 1

Given 5( ) xf x +3, 5( ) xg x , 3 3( ) xh x

1.1 Draw all the graphs(show the intercepts if possible)

1.2 ( )n x is reflection of ( )f x in the y‐axis ;

( )b x is reflection of ( )h x in the x‐axis

Write the equations of ( )n x and ( )b x

QUESTION 2 The graph of 1 . 3 (a is a constant)

passes through the origin as shown below:

2.1 Show that t =‐1

2.2 Determine the value of x, if A(x; ‐2) lies on

the graph of h.

2.3 Write the equation of k, if the graph of h is

shifted 5 units to the right to give function k.

QUESTION 3

3 .

3.1 Write down the value of n.

3.2 Calculate the value of k.

109

E. MIXED FUNCTIONS When we are working with more than one function, you mainly use the following concepts: Vertical Distance : All the points on a vertical

line have the same x ‐coordinates. When you are finding the distance between any two points on a vertical line, you should use the difference between two y‐coordinates of the points, the top point minus the bottom point:

Vertical Distance top bottomy y

Horizontal Distance : All the points on a

horizontal line have the same y‐coordinates. When you are finding the distance between any two points on a horizontal line, you should use the difference between two x‐coordinates of the points, the right point minus the left point:

Horizontal Distance right leftx x

Intersection points of two graphs: To find the

intersection points of two graphs, you equate the graphs:

1 2y y or ( ) ( )f x g x

Firstly you find the x‐coordinates of the graphs by solving the new equation generated, then substitute the x values into one of the graphs to find the y values of the points.

Inequalities: To find the values of x, for which

0( )f x or 0( )f x look for the part of graph

where function above or below x‐axis. Example:

Answer:

2x or 3x 2 3x

Inequalities: To find the values of x, for which

( ) ( )f x g x or ( ) ( )f x g x look for the part of

graph where function f above or below g .

Example:

2 8 20( )f x x x and 4 8( )g x x

Answer: A and B are point of intersection, x of A is ‐1 and x of B is 4.

1 4x Shifted and Reflected Functions;

Let ( )f x be the original function.

( )f x a shifting ( )f x a units to the right.

( )f x a shifting ( )f x a units to the left.

( )f x a shifting ( )f x a units downwards.

( )f x a shifting ( )f x a units upwards.

( )f x a b shifting ( )f x a units to the right and b units upwards.

( )f x reflecting ( )f x in the y‐axis. (symmetrical

to y‐axis) .Change x with ‐x

( )f x reflecting ( )f x in the x‐axis. (symmetrical

to x‐axis) .Change y with ‐y

110

QUESTION 1

The figure shows graphs of

ky

x on (0; ) and y=mx+c. Accordingly find:

1.1 the values of k, m and c. 1.2 the co‐ordinates of A 1.3 the distance BC if OD=6 units and BDC is perpendicular to the x‐axis. 1.4 the equation of the new function that is

formed when k

yx

is shifted 1 unit to the

left and 2 units upwards 1.5 the equation of the new function that is

formed when k

yx

is shifted vertically so

that it passes through the point (2;2) SOLUTION

1.1 The graph ofk

yx

is the hyperbola with

the point (1 ; ‐6) on it. For k ;

61

∴ 6

To find the equation of straight line weneed two points on the line : (1 ; ‐6) and (3 ; 0).

0 6

3 13

m

3y x c

For point (3 ; 0): 0 3 3

9

c

c

1.2 Point A is the intersection of two graphs ; should be equated.

3 96

3 9 6 Divide each term by 3;

3 2 3 2 0

1 2 0 1 2

Point A(2 ; ?). To find the y‐coordinate of A, use any of the equation of the graphs :

6

6

23

∴ A(2 ; ‐3) 1.3The point D(6 ; 0) (OD is 6 units) . The length BC is a vertical distance:

(line‐hyperbola) The point B(6 ; y2) on the line y=3x‐9;

2 3 6 9 9.y

The point C(6 ; y1) on the hyperbola

6

61

9 1 10

BC = 10 units 1.4We have already found the value of k.

6

1 →shifted 1 unit to the left (a=‐1) 2 →shifted 2 units upwards (b=2)

The new equation is :

6

12

1.5When is shifted vertically , new

equation is in the form of

To find b in the equation use the point (2;2)given.

26

2

2 3 5

The new equation: ∴ 5

111

QUESTION 2

The graphs of 3( ) ( )f x x x and

12

2( )g x x

, are represented below

x

yL

MP

f

g

2.1 Determine the values of x for which

f(x)=0

2.2 Calculate the coordinates of P, the

turning point of f.

2.3 Determine the average gradient of the curvef

between x= ‐5 and x= ‐3.

2.4 Determine the values of x for which 0( )f x

2.5 Give the coordinates of turning point of f(x‐2)

2.6 L is point on the straight line and M is a point

on the parabola such that LM is perpendicular

to the x‐axis. Show that expression for LM can

be written as: 212 3

2LM x x x

SOLUTION

2.1 x‐intercepts : when 0( )y f x , x=?

0 3( )x x

0x or 3x 2.2 Turning point: First open the bracket.

2 3( )f x x x 3 3

2 2 1 2

( )

( )

bx

a

Subs. 3

2x into the function:

23 3 3 273

2 2 2 4( ) ( ) ( )f

TP3 27

2 4( ; )

2.3 Gradient formula

2 1

2 1

5 3

5 3

10 05

2

( ) ( ) ( ) ( )

( )

f x f x f fm

x x

2.4 0( )f x means:

solve for x when all the y‐values are positive (above the x‐axis)

3x or 0x

2.5 f(x‐2) : shifting ( )f x by 2 units right.

TP3 27 1 27

22 4 2 4

( ; ) ( ; )

2.6 LM is a vertical line where the point L is on the line and the point M is on the parabola. Vertical

Distance 2 1y y (line ‐parabola)

2

2

2

12 3

21

2 32

72

2

( )LM x x x

LM x x x

LM x x

QESTION 3 The diagram below shows the graphs of

and . The point M(1;‐2) is the point of

intersection of f and g. Determine the value of .

SOLUTION The graph o is the parabola with the point (1 ; ‐2) on it. For :

2 1 ∴ 2

112

QUESTION 4 Below is a sketch graph of parabola

22 1 8( ) ( )f x x and straight line g. P is the

turning point of f. g cuts the y‐axis at (0;‐1). f and g

intersect at B and D.

4.1 Find coordinates of P.

4.2 Find the length of OB.

4.3 Show that the equation of g is1

13

( )g x x .

4.4 Calculate the coordinates of D.

4.5 If ( ) ( )h x f x , explain how the graph of h

can be obtained from the graph f.

4.6 Write down the equation of h.

4.7 For which x, 0( )f x

4.8 For which x, ( ) ( )f x g x

SOLUTION

4.1 Our equation is 22 1 8( )y x is TP form

1 8( ; )P to find a, use the point 0 6( ; )

You may open the bracket to put the equation in standard form:

2 22 1 8 2 2 1 8( ) ( )y x x x 22 4 6y x x

4.2 The points A and B are x ‐intercepts : when

0( )y f x , x=? 20 2 4 6x x (divide both sides by ‐2)

20 2 3

0 3 1( )( )

x x

x x

3 0( ; )B and 1 0( ; )A

OB=3 units

4.3 Use y mx c where c is the y‐intercept to

find the equation of the straight line:

0 1 1

3 0 3m

gradient between (0;‐1) and (3;0)

1c (y intercept is ‐1)

11

3( )g x x

4.4 The point D is the intersecting point of two

graphs: ( ) ( )f x g x

2 12 4 6 1

3x x x

(multiply both sides ‐3) 2

2

6 12 18 3

6 11 21 0

3 6 7 0( )( )

x x x

x x

x x

3x or7

6x

The x‐coordinate of the point D is 7

6x

Subs. x into any equation (straight line is easier)

7 1 7 251

6 3 6 18( ) ( )g

7 25

6 18( ; )C

4.5 ( ) ( )h x f x is the reflection of ( )f x in the y‐

axis.

4.6 ( )k x = ( )f x . 2

2

2 4 6

2 4 6

( ) ( ) ( )k x x x

x x

4.7 0( )f x means above the a‐axis:

1 3x

4.8 ( ) ( )f x g x means f above g, from D to B

Therefore:

73

6x

113

QUESTION 5 The diagram represents the graphs of the following: f=y=‐2x2+8x+10 and g=y=‐2x‐2 . The point E is the turning point of f.

5.1 Determine the co‐ordinates of:

4.1.1 A 4.1.2 B and C 4.1.3 R

5.2 If OD =1 unit, calculate the length of PQ. 5.3 What will the equation of the new parabola be

if y=‐2x2+8x+10 is translated 3 units up?

5.4 for which x, ( ) ( )f x g x ?

SOLUTION 5.1.1 Turning point: y=‐2x2+8x+10

82

2 2 2

( )

( )

bx

a

Subs. 2x into the function: 22 2 8 2 10

18

( ) ( )y

E 2 8( ; ) . The y‐coordinate of the point E is the same as

the y‐coordinate of the point A. 0 18( ; )A

5.1.2 The points B and C are the x‐intercepts:

20 2 8 10x x (divide both sides by ‐2) 20 4 5

0 5 1( )( )

x x

x x

5x or 1x 5 0( ; )B and 1 0( ; )C

5.1.3 The point R is the intersecting point of two

graphs: ( ) ( )f x g x 2

2

2

2 8 10 2 2

2 10 12 0

5 6 0

6 1 0( )( )

x x x

x x

x x

x x

6x or 1x The x‐coordinate of the point R is 6x Subs. x into any equation (str. line is easier)

6 2 6 2 12 2

14

( ) ( )g

6 14( ; )R

5.2 The PQ is a vertical line( top bottomy y )

The points P, D and Q are on the same vertical line: they have the same x‐coordinates: 1x

The point P is on the parabola:2

2

2 8 10 2 2

2 1 8 1 10 2 1 2

16 4

20

[ ] [ ]

[ ( ) ( ) ] [ ( ) ]

( )

PQ x x x

PQ

PQ

PQ

20PQ units

5.3 When 22 8 10( )f x x x is translated 3

units up, the new equation becomes 2

2

2 8 10

2 8 13

3y x x

y x x

5.4 ( ) ( )f x g x means f below g, which is x less

point C or x is greater than R.

1x or 6x

114

QUESTION 6 The diagram below shows the graphs of

4 and .

The point P(1;4) is the intersection of f and g.

6.1 Determine the equation of h(x), the resultant

function when f(x) is reflected about the x=0

6.2 Determine the value of in g(x).

6.3 Determine the equation of m(x), the resultant

function g(x) is shifted horizontally 2 units to

the right and vertically 1 unit down.

6.4 For which x, ( ) ( )g x f x ?

6.5 Calculate the coordinates of the intercepts of

m(x) with the axes.

SOLUTION 6.1 Reflection about y‐axis: f(‐x)

4

6.2 The point P(1;4) is on g(x).

41

∴ 4 .

6.3 Horizontally 2 units right : x→ x‐2

Vertically 1 unit down: y → y‐1

4

4

21

6.4 ( )g x is above ( )f x , when x from 0 ( 0x ) to

point P:

0 1x

6.5 x‐intercept:

04

21

14

2

2 4

6

∴ 6 ; 0 y‐intercept :

04

0 21

3

∴ 0 ; 3 QUESTION 7

Given: 2( ) xf x 2( )g x

x 2( )h x x

Sketch the above graphs on the same set of axes, showing all intercepts with the axes, and other significant points. SOLUTION

is a increasing exponential function with y‐intercept (0;1) and horizontal asymptote y=0. g(x) is the basic hyperbola with a=2 > 0. h(x) is a straight line goes through origin with positive gradient

2 2 (intersection point : A(1,2)

115

TASK 11 QUESTION 1

The sketch shows the graphs of f(x)=x2‐2x‐3 and g(x)=mx +c 1.1 Determine the lengths of OA, OB and OC. 1.2 Determine the coordinates of D the turning

point of the parabola. 1.3 Determine the values of m and c.

1.4 For which values of x, ( ) ( )f x g x ?

QUESTION 2

Given: ; 2

2.1 If A(‐2;2) is a point on f, determine the value

of a.

2.2 If E(‐1;k) then determine the value of k. 2.3 Write down the equations of asymptotes of g.

2.4 Determine the equation of g.

116

TASK 12 QUESTION 1

The sketch shows the graphs of 2 2 3( ) f x x x

and g( x) mx c .

A and B are the intercepts on the x‐axis. C and D are

the intercepts on the y–axis. T is the

turning point of the graph of f.

1.1 Determine the lengths of OC and AB.

1.2 Determine the equation of the axis of

symmetry of the graph of f.

1.3 Show that the length of ST = 4 units.

1.4 The graph of g is parallel to AC.

Calculate:

(a) The gradient of AC

(b) The values of m and c

QUESTION 2

Sketched are graphs of f= y=2x+1 and g=3

yx

.

Find:

x

y

A

B

C

D

E

F

O

2.1 The coordinates of A and B

2.2 The coordinates of C and then distance CD *2.3 The coordinates of E and F if EF=4.

117

TASK 13QUESTION 1

Sketched are 3 and

1

25

1.1 Determine the co‐ordinates of A and B. 1.2 Determine the co‐ordinates of C and D. 1.3 Is ga increasing or decreasing function? Motivate your answer.

1.4 If 0x , for which values of x, ( ) ( )f x g x ?

QUESTION 2

Sketched are graphs of 2 3 4( )f x x x and

33( )g x

x

.

x

y

A B

CD

EF

D is the turning point of the parabola.

2.1 If the x‐coordinates of E and F are 1 and 3 respectively, then write the y‐coordinates of E and F.

2.2 Calculate the coordinates of A.

2.3 Write down the equation of

2.3.1 the axis of symmetry of the parabola.

2.3.2 the reflection of 3

3yx

in the y‐axis.

2.4 If 0x , for which values of x, ( ) ( )f x g x

118

TASK 14QUESTION 1 1.1 Draw neat sketch graphs of the functions

33

2( )f x

x

and 3( )g x x showing all

intercepts with the axes and any asymptotes.

1.2 Write down the domain of ( )f x .

1.3 Determine algebraically for which values of

x , ( )f x and ( )g x intersect

QUESTION 2

Sketched are 4

12

( )i xx

and ( )j x mx k .

Calculate:

2.1 the asymptotes of the hyperbola.

2.2 the co‐ordinates of A and B

2.3 For which x, 0( )i x

119

TASK 15QUESTION 1 Given the graphs illustrated of a parabola

5 1( ) ( )( )f x x x with turning point A and y–

intercept Band 5xg x k , also passing

through the point A.

1.1 Show that A is (2;9) and B is (0;5)

1.2 Find the value of k

1.3 Find the co‐ordinates of C , D and E.

1.4 State the range of f and g.

QUESTION 2

Functions 2. 3 4 and

2.1 Find the asymptote of

2.2 Determine the y‐intercept of f.

2.3 Find one other point on the graph of f.

2.4 Sketch the graph of f.

2.5 What is the range of f ?

2.6 Find the asymptotes of g .

2.7 Determine the y‐intercept of g .

2.8 Sketch the graph of g (on the same diagram)

120

F. TRIAL TESTS TRIAL TEST 1

QUESTION 1

Given: 22 3 8( ) ( )f x x

1.1 Write down the co‐ordinates of the

turning point of f.

(2)

1.2 Draw a sketch graph of f. Clearly indicate

the co‐ordinates of the turning point as

well as the intercepts with the axes.

(6)

1.3 Without any further calculations, sketch

the graph of 2( )y f x on a the same

set of axes. Indicate only the co‐ordinates

of the turning point.

(2) [10]

(2)

QUESTION 2

The sketch shows the graphs of2 8 20( )f x x x and 4 8( )g x x

2.1 Determine the length of OA and OB.

(4)

2.2 Find the co‐ordinates of the point C and

M(Turning point)

(6)

2.3 Calculate the length of EF if is G(6;0) (5) [15]

TOTAL: 25

TRIAL TEST 2 QUESTION 1 1.1 Find values of p, q and a for

( )a

f x qx p

(6)

1.2 The straight line defined by g(x)= mx +c

which intersects the f at (0; 0) and (‐4;t).

Calculate the values of t, m and c

(3)

1.3 Draw g on the same system of the axes (3)

QUESTION 2

A sketch of22 9: ( )f x x and

:g x mx k are given.

x

y

A

E D C

BF

2.1 Find the co‐ordinates of A,B,C,D(turning

point) and E if EC//AB

(6)

2.2 Find the values of m and k (3)

2.3 Find the distance DF, where F lies on AC

and DF is parallel to the y‐axis.

(3) [12]

TOTAL: 24

121

TRIAL TEST 3 QUESTION 1

Given:3

25

( )f xx

1.1 Determine equation of asymptotes

(2)

1.2 Determine the x‐ and y‐intercepts of the

function.

(4)

1.3 For which values of x is 0( )f x (2)

1.3 Write down equation of f reflected in

y‐axis.

(3) [11]

(2)

QUESTION 2

The graph of 2 20y x x is shown. Find:

2.1 The length of OA, OB and OC (5)

2.2 The value of k, if D is (k;10) (3)

2.3 The length of FG when OF=4 (3)

2.4 The maximum length of FG if G is variable

point on the parabola between A and C

with FG perpendicular to the x axis.

(3) [14]

TOTAL: 30

TRIAL TEST 4 QUESTION 1 1.1 Sketch the graph of 2 4( ) xf x

Indicate the co‐ordinates of the

intercepts and the turning point on

the graph. Show ALL the calculations.

(8)

1.2 For which values of x is 0( )f x ? (3)

[11]

QUESTION 3 Given the functions

211 2

2( ) ( )y f x x and 2 6( )y g x x

3.1 Write down the co‐ordinates of the

turning point of f.

(2)

3.2 Calculate the roots of the equation

0( )f x

(4)

3.3 Write down the equation of the axis of

symmetry of f

(1)

3.4 Sketch the graphs of ( )y f x and

( )y g x on the same system of axes.

Show ALL intercepts with the axes.

(7) [11]

TOTAL: 25

122

VII. FINANCE A. SIMPLE INTEREST

Simple Interest is where you earn or pay the same amount of interest every year (interest on the initial amount that you invested, but not interest on interest). As an easy example of simple interest, consider how much you will get by investing R1 000 for 1 year with a bank that pays you 7% simple interest. At the end of the year, you will get an interest of7% =7/100=0,07 The amount of interest is R1 000 × 0,05= R70 So, with an principal of R1 000 at the start of the year, your accumulated amount at the end of the year will therefore be:

= R1 000 + R70= R1 070

Simple Interest formula: 1( )A P i n

A is “After”, a final Amount P is “Previous”, a starting amount i is an interest rate nis a number of years

Keywords: Simple Interest Straight line method Hire purchase

Depreciation means i

QUESTION 1 If I deposit R1 000 into a special bank account which pays a Simple Interest of 7% for 3 years, how much will I get back at the end? SOLUTION Step 1 : • P = R1 000 • interest rate, i = 7% =0,07 • period of time, n = 3 years We need to find the Accumulated amount (A). Step 2 : Determine how to approach the problem We know from that: A = P(1 + in) Step 3 : Solve the problem A = P(1 + in)= 1 000(1 + 0,07.3)= R1 210 QUESTION 2 If I deposit R30 000 into a special bank account which pays a Simple Interest of 7.5% ,for how

many years must I invest this amount to generate R45 000? SOLUTION Step 1 : Determine what is given and what is required • P = R30 000 • interest rate, i = 7,5% = 0,075 • A = R45 000 We are required to find the number of years.(n=?) Step 2 : Determine how to approach the problem A = P(1 + in) Step 3 : Solve the problem A = P(1 + in)

45 000 = 30 000(1 + 0,075n)

45 000

30 0001 0,075

0,075n = 1,5– 1 n = 6,6666667

n has to be a whole number, therefore n = 7. QUESTION 3 Troy is keen to buy an additional hard drive for his laptop advertised for R 2 500 on the internet. There is an option of paying a 10% deposit then making 24 monthly payments using a hire‐purchase agreement where interest is calculated at 7,5% p.a. simple interest. Calculate what Troy’s monthly payments will be. SOLUTION Step 1 : A new opening balance is required, as the 10% deposit is paid in cash. • 10% of R 2 500 = R250 • new opening balance, P = R2 500 − R250= R2 250 • interest rate, i = 7,5% = 0,075pa • period of time, n = 2 years We are required to find final amount (A) and then the monthly payments. Step 2 : Determine how to approach the problem Closing Balance A = P(1 + in) Step 3 : Solve the problem A = P(1 + in) = R2 250(1 + 2 × 7,5%)= R2 587,50 Monthly payment = 2587,50÷ 24= R107,81

123

B. COMPOUND INTEREST With Compound Interest, you work out the interest for the first period, add it to the total, and then calculate the interest for the next period, and so on ... Example: If I deposit R1 000 into a special bank account which pays a Simple Interest of 7%. What if I empty the bank account after a year, and then take the principal and the interest and invest it back into the same account again. Then I take it all out at the end of the second year, and then put it all back in again? And then I take it all out at the end of 3 years? SOLUTION Determine what is given and what is required • opening balance, P = R1 000 • interest rate, i = 7% • period of time, 1 year at a time, for 3 years We are required to find the closing balance at the end of three years. Determine the accumulated amount at the end of the first year A = P(1 + i )= R1 000(1 + 0,07)= R1 070 Determine the accumulated amount at the end of the second year, when P = 1070 A = P(1 + i ) = R1 070(1 + 0,07) = R1 144 ,90 Table below shows other steps

Year Amount in Rands

0 1 000

1 1 070

2 1 144,90

3 1 225,04

In this situation, when I took the money out and then re‐invested it, I was actually earning interest in the second year on my interest (R70) from the first year. (And interest on the interest on my interest in the third year!) Compound interest is the interest payable on the principal and any previous interest.

Compound Interest formula:

1( )nA P i

A is “After”, a final Amount P is “Previous”, a starting amount i is an interest rate nis a number of years

Keywords: Compound Interest Interest of previous value or year Inflation Population Reducing balance ( i )

Depreciation means i

QUESTION 1 An amount of R3 500 is invested into account which pays compound interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years. SOLUTION Compound interest :A P 1 i

A 3 500 1 0,075 A = R4044,69

QUESTION 2 South Africa’s population is increasing by 2,5% per year. If the current population is 43 million, what was the population two years ago ? SOLUTION • final population A = 43 000 000 • starting population P is unknown • period of time, n = 2 year • interest rate, i = 2,5% per year Population is compound formula: A P 1 i

243000000 1 2 5( , %)P

Divide by number in front of P

2

43000000

1 2 5

40 928 019

( , %)P

P

QUESTION 3 3000A , 2000P , 5n . Find i, for compound

interest. SOLUTION Substitute values into formula

5

5

3000 2000 1

30001

2000

( )

( )

i

i

Take fifth root of both sides:

555 1 5 1

1 084472 1

0 084472

8 45

, ( )

,

,

, %

i

i

i

i

124

TASK 1 QUESTION 1 Calculate the simple interest for the following problems. 1.1 A loan of R300 at a rate of 8% for l year. 1.2 An investment of R225 at a rate of 12,5% for 6 years. QUESTION 2 I made a deposit of R5 000 in the bank for my 5 year old son’s 21st birthday. I have given him the amount of R 18 000 on his birthday. At what rate was the money invested, if simple interest was calculated?

QUESTION 3 Bongani buys a dining room table costing R 8 500 on Hire Purchase. He is charged simple interest at 17,5% per annum over 3 years. 3.1 How much will Bongani pay in total ? 3.2 How much interest does he pay ?

QUESTION 4 If the average rate of inflation for the past few years was 7,3% and your water and electricity account is R 1 425 on average, what would you expect to pay in 6 years time ? QUESTION 5 Shrek wants to invest some money at 11% per annum compound interest. How much money (to the nearest rand) should he invest if he wants to reach a sum of R 100 000 in five year’s time ? QUESTION 6 Bianca has R1 450 to invest for 3 years. Bank A offers a savings account which pays simple interest at a rate of 11% per annum, whereas Bank B offers a savings account paying compound interest at a rate of 10,5% per annum. Which account would leave Bianca with the highest accumulated balance at the end of the 3 year period? QUESTION 7 A business buys a truck for R560 000. Over a period of 10 years the value of the truck depreciates to R0 (using the straight‐line method). What is the value of the truck after 8 years ?

125

TASK 2 QUESTION 1 On January 1, 2017 the value of my Kia Sorento is R320 000. Each year after that, the cars value will decrease 20% of the previous year’s value. What is the value of the car on January 1, 2021. QUESTION 2 The population of Bonduel decreases at a rate of 9,5% per annum as people migrate to the cities. Calculate the decrease in population over a period of 5 years if the initial population was 2 178 000. QUESTION 3 A 20 kg watermelon consists of 98% water. If it is left outside in the sun itloses 3% of its water each day. How much does in weigh after a month of 31 days ? QUESTION 4 A computer depreciates at x% per annum using the reducing‐balance method. Four years ago the value of the computer was R10 000 and is now worth R4 520.Calculate the value of x correct to two decimal places.

QUESTION 5 Spiderman buys a Mercedes worth R385 000 in 2015. What will the value of the Mercedes be at the end of 2020 if 5.1the car depreciates at 6% p.a. straight‐line depreciation 5.2 the car depreciates at 12% p.a. reducing‐ balance depreciation. QUESTION 6 Fiona buys a DsTV satellite dish for R3 000. Due to weathering, its value depreciates simply at 15% per annum. After how long will the satellite dish be worth nothing ?

QUESTION 7 Shrek wants to buy his grandpa’s donkey for R800. His grandpa is quite pleased with the offer, seeing that it only depreciated at a rate of 3% per year using the straight‐line method. Grandpa bought the donkey 5 years ago. What did grandpa pay for the donkey then?

126

TASK 3 QUESTION 1 Mpumelelo invested R6 000 at 12% simple interest per annum for 4 years. Kathleho invested R6 000 at 10% compound interest per annum for 4 years. Which person gained more interest on their investment? Substantiate your answer. Show all calculations. QUESTION 2 Bula wants to buy a video camera for R9 500. He has deposit of R1 500. He wishes to pay the balance using a hire purchase agreement over 2.5 years. The interest charged on the loan is 16% per annum. Included in the agreement, is an insurance cost of 2% per annum on the purchase price of the video camera. Calculate his monthly instalment. QUESTION 3

3.1 A photocopier is sold after 4 years for R12 250. What was the purchase price if

the depreciation is 25% p.a. on a reducing

balance?

3.2 How long will it take a sum of money to treble itself at 16% per annum simple

interest.

QUESTION 4 4.1 In 10 years the urban population f the

Western Cape grew from 984 372 to 5 million. Use the formula for compound interest to calculate the growth rate.

4.2 Monique bought a stove for R3 750. After

3 years she paid for it and the R956.25 interest that was charged. Determine the simple rate of interest that was charged.

QUESTION 5 How much compound interest is payable on a loan of R2 000 for a year, if the interest rate is 10%? QUESTION 6 Calculate the compound interest for the following problems. 6.1 A R2 000 loan for 2 years at 5%. 6.2 A R1 500 investment for 3 years at 6%.

127

TASK 4 QUESTION 1

The diagram displays two methods of depreciation for farming equipment. 1.1 What is the initial value of the equipment? 1.2 What type of decrease does Method A

illustrate? 1.3 What type of decrease does Method B

illustrate? 1.4 Determine the value of depreciation rate

for method B. QUESTION 2 After 5 years of reducing balance depreciation, an

office equipment has a of its original value. The

original value of the equipment was R80000. Calculate the depreciation interest rate, as a percentage.

QUESTION 3 Anna bought a car, the graph shows how the value of the car decreases yearly.

3.1 What is the initial value of the

vehicle? 3.2 What is the value of the car after 2

years? 3.3 What type of depreciation is this? 3.4 Determine the rate of decrease as

a percentage. 3.5 Write down a formula for working

out the depreciated value after n years.

QUESTION 4 Office equipment is presently valued at R36 000. Calculate the value of the equipment at the end of 5 years if depreciation is calculated at 8% p.a. on a straight line basis.

128

TASK 5 1. Mpumelelo invested R6 000 at 12% simple

interest per annum for 4 years. Kathleho

invested R6 000 at 10% compound interest per

annum for 4 years. Which person gained more

interest on their investment? Substantiate your

answer. Show all calculations.

4. Bula wants to buy a video camera for R9 500. He

has deposit of R1 500. He wishes to pay the

balance using a hire purchase agreement over

2.5 years. The interest charged on the loan is

16% per annum. Included in the agreement, is

an insurance cost of 2% per annum on the

purchase price of the video camera. Calculate his

monthly instalment.

5. Examination Aid bought furniture valued at

R15000. The depreciation is calculated at a rate

12% p.a. on a straight line basis. Calculate the

value of the furniture at the end of 5 years.

6. A photocopier is sold after 4 years for R12 250.

What was the purchase price if the depreciation

is 25% p.a. on a reducing balance?

7. How long will it take a sum of money to treble

itself at 16% per annum simple interest.

8. In 10 years the urban population f the Western

Cape grew from 984 372 to 5 million. Use the

formula for compound interest to calculate the

growth rate.

9. Monique bought a stove for R3 750. After 3

years she paid for it and the R956.25 interest

that was charged. Determine the simple rate of

interest that was charged.

10. Angelina invested R13 000 in shares. It was not a

very good investment. The value of the

investment depreciated at 6.5% p.a. calculated

on a straight line basis. After how many years

will her investment be less than R10 000?

11. The value of a car which initially cost R189 000 is

decreased on a reducing balance at r% p.a. and

after 5 years the car has a value of R100 833.34.

Calculate the rate, r, by which the car decreased

annually.

12. A computer was purchase for R5 500. After 3

years its book value was R1787,50.

12.1. Determine the decrease rate is the

depreciation is simple decrease.

12.2. What is the annual amount of

depreciation?

129

C. NOMINAL AND EFFECTIVE INTEREST RATE Sometimes interest is charged yearly compounded many times within the year. This type of interest rate called a nominal interest. Let's see what happens if it is compounded twice per year. Example:R1000 invested with "10% per annum, compounded semi‐annually". Semi‐annual means twice a year. So the 10% is split into two:

5% halfway through the year,

and another 5% at the end of the year,

but each time it is compounded (meaning the interest is added to the total):

1 10001 5 1050( %)A R

And then second time compounded,

2 10501 5 1102 50( %) ,A R

Yes, there are two annual interest rates: In this example we have two rates:

10% The Nominal Rate (the rate they mention)

10.25% The Effective Annual Rate (the rate after compounding)

The Effective Annual Rate is what actually gets paid! If interest is compounded within the year, the Effective Annual Rate will be higher than the Nominal rate.

Compound Interest formula:

1( )nmi

A Pm

A is “After”, a final Amount P is “Previous”, a starting amount i is an interest rate nis a number of years m: how many times it is compounded

If semi‐annually m 2 , 212

( ) niA P

If quarterly m 4 , 414

( ) niA P

If monthly m 12 , 12112

( ) niA P

If daily m 365 , 3651365

( ) niA P

Converting Effective Nominal Interest Rates:

1( )nP i 1( )ni

Pm

m

1 1( ) ( )nom meff

ii

m

QUESTION 1 On their saving accounts, Bank offers an interest rate of 18% p.a., compounded monthly. If you save R100 in such an account now, how much would the amount have accumulated to in 3 years’ time? SOLUTION It is a nominal interest rate question.

We use the formula: A P 1 .

m= 12 (compounded monthly, 12 times a year) i=0,18 per annum.

A 100 10,18

12.

A = R 170,91 QUESTION 2 Cebela received R120 000 on her investment of R80 000 in five years. Calculate the interest rate per annum compounded quarterly. SOLUTION

Using the formula: 1( )nmi

A Pm

5 4

20

20

120000 80000 14

1 5 14

1 5 14

1 1 0204804

0 0204804

0 081921 8 19

( )

, ( )

,

,

,

, , %

i

i

i

i

i

i

QUESTION 3 Calculate the effective rate equivalent to a nominal interest rate of 8,75% p.a. compounded monthly. SOLUTION The nominal rate is given . We need to find the effective rate by using a conversion formula :

Conversion formula : 1 1( ) ( )nom meff

ii

m

1 i 10,0875

1212

i= 0,091=9,1%

130

TASK 61. Calculate the final amount to be paid on a loan

of R15 000 at 12% per annum for 4 years, for

each of the following conditions:

1.1. simple interest

1.2. compound interest calculated annually

1.3. compound interest calculated quarterly

1.4. compound interest calculated monthly

1.5. compound interest calculated daily.

2. The nominal yearly rate of interest is 7.79%,

calculate the effective yearly rate if interest is

compounded daily

3. How much money must you invest at 12.5%

per annum, compounded monthly, if you know

you will need R15 000 in 3 years time?

4. Paul invested money in a bank for 4 years. The

nominal interest rate on the account was 6,1%

per annum compounded monthly. This is

equivalent to an effective interest rate of

6,29% per annum.

Seth invested money in a different bank for 4

years. The nominal interest rate on his

investment was 6% per annum compounded

daily.

Seth thinks he has a better deal than Paul. Do

you agree? Justify your answer by comparing

their effective interest rates.

5. Paris opens accounts at a number of clothing

stores and spends freely. She gets herself into

terrible debt and she cannot pay off her

accounts. She owes Hilton Fashion world

R5 000 and the shop agrees to let Paris pay the

bill at a nominal interest rate of 24%

compounded monthly.

5.1. How much money will she owe Hilton

Fashion World after two years ?

5.2. What is the effective rate of interest that

Hilton Fashion World is charging her?

131

TASK 7 1. R55 000 is deposited into a savings account.

Calculate the value of the savings after 8 years

in each of the following cases if the interest

rates are:

1.1. 2% p.a. compounded annually

1.2. 12% p.a. compounded semi‐annully

1.3. 12% p.a. compounded monthly

2. A company takes out a loan of R2,8 million to

expand the business. The loan is repaid in one

amount at the end of 5 years. The interest on

the loan is calculated at 8,5% p.a.

compounded quarterly. Calculate how much

money the company owes at the end of 5

years.

3. R5 600 is deposited in a savings account.

Calculate how much money is in the savings

account at the end of 18 months if the interest

is 9.25% p.a. compounded half yearly.

4. R2500 is deposited into a savings account at

15% interest per annum compounded

monthly.

4.1. What is the monthly interest rate?

4.2. Determine the yearly effective interest

rate, correct to 1 decimal digit.

4.3. Calculate the amount of money in the

savings account at the end of three years.

5. The inflation rate in Zimbabwe was 10% per

month for 2006. What was the effective

annual inflation rate for 2006?

6. After 5 years of reducing balance depreciation,

an office equipment has a 1/5 of its original

value. The original value of the equipment was

R80 000. Calculate the depreciation interest

rate, as a percentage.

132

TASK 81. Nicholas wants to buy a fridge costs R18 000.

He has to pay a deposit of 20% of the cost and

the balance by means of a hire‐purchase

agreement. The rate of interest on the loan is

20,25% pa simple interest. The repayment

period of the loan is 42 months. In addition to

the hire‐purchase agreement, an annual

insurance premium of 1,5% of the total cost of

the fridge should be added. The annual

insurance premium should be paid in monthly

payments.

1.1. Calculate the value of the loan that

Nicholas will take.

1.2. Calculate the total amount that must be

repaid on the hire purchase agreement.

1.3. Calculate the monthly repayment, which

includes the monthly insurance premium.

2. R8 000 is deposited into a savings account for

one year at an interest rate of 13,5% p.a.

compounded quarterly.

2.1. What is the quarterly interest rate?

2.2. Calculate the effective annual interest

rate.

3. On her 5th birthday, Bianca’s father invested a

certain sum of money in order to pay for a

university education. Interest of 12% was paid,

compounded semi‐annually. On her 17th

birthday, the investment has grown to R13

214.60. How much did he initially invest?

4. R1950 at a compound interest rate 9% p.a.

Determine the amount of interest earned at

the end of a year if it is compounded

4.1. Quarterly.

4.2. Yearly.

4.3. Which investment is the best?

4.4. What is the effective yearly rate if it is

compounded monthly?

5. Calculate the amount of money that could be

borrowed at 5,5% per annum compound

interest, compounded quarterly, over 4 years,

so that the total repayments do not exceed

R50 000 .

133

TASK 9

RATE OF EXCHANGE & CURRENCY 1. Use the table below to answer the questions

that follows:

1.1 You have R5000 to spend in Switzerland.

How much Francs can you buy? 1.2 What will it cost you in Rands to purchase

4500 dollars? 1.3 If you exchange 600 pounds how much

Rands will you get? 2. Below is a table with the buying and selling

prices of different currencies:

2.1 You have R5000 to spend in Switzerland.

How much Francs can you buy? 2.2 What will it cost you in Rands to purchase

5000 Yen? 2.3 If you exchange 1000 NZD how much

Rands will you get? 2.4 You want to import a personal computer

from Japan at a total cost of 96180 Yen. The equivalent computer cost R8 500 in South Africa. Will you import or buy locally?

3. I want to buy an IPOD that costs £100,

with the exchange rate currently at

£1 = R14. I believe the exchange rate will

reach R12 in a month.

3.1 How much will the MP3 player cost in

Rands, if I buy it now?

3.2 How much will I save if the exchange rate

drops to R12?

3.3 How much will I lose if the exchange rate

moves to R15?

4. Study the following exchange rates:

Country Currency Exchange Rate

United Kingdom (UK) Pounds(£) R14,13

United States (USA) Dollars ($) R7,04

4.1 In South Africa the cost of a new Honda

Civic is R173 400. In England the same

vehicle costs £12 200 and in the USA $ 21

900. In which country is the car the

cheapest if you compare it to the South

African Rand ?

4.2 Sollie and Arinda are waiters in a South

African Restaurant attracting many

tourists from abroad. Sollie gets a £6 tip

from a tourist and Arinda gets $ 12. How

many South African Rand did each one

get?

Country Currency Value of Unit (in Rand)

USA Dollar 7,081 Switzerland Franc 5,892 United Kingdom Pound 13.982

Country Currency Symbol Exchange Rate (units per R1)

Switzerland Franc Swiss Franc

0.1697

New Zealand

Dollar NZD 0.21

Japan Yen Y 16.03

134

VIII. PROBABILITY A. SETS AND VENN DIAGRAMS

A set is a collection of things.

For example, the items you wear is a set: these would include shoes, socks, hat, shirt, pants, and so on. You write sets inside curly brackets like this:

{socks, shoes, pants, watches, shirts, ...} You can also have sets of numbers: Set of whole numbers: {0, 1, 2, 3, ...} Set of prime numbers: {2, 3, 5, 7, 11, 13, 17, ...}

The Empty Set has no elements: {}

Sample Space (S) or Universal Set is the set that contains everything. Well, not exactly everything. Everything that we are interested in now.

1. VENN DIAGRAM A Venn diagram is a graphical way of representing the relationship between sets.

In each Venn diagram a set is represented by a closed curve. The region inside the curve represents the elements that belong to the set, while the region outside the curve represents the elements that are excluded from the set. Example 1. You could have a set made up of your ten best friends: {alex, blair, casey, drew, erin, francis, glen, hunter, ira, jade} Now let's say that alex, casey, drew and hunter play Soccer: Soccer = {alex, casey, drew, hunter} And casey, drew and jade play Tennis: Tennis = {casey, drew, jade} You could put their names in two separate circles:

2. UNION The union of two sets A and B is the set of all elements of A or B, also denoted by A∪ B

You can now list your friends that play Soccer OR Tennis. Soccer OR Tennis = {alex, casey, drew, hunter, jade} Not everyone is in that set ... only your friends that play Soccer or Tennis. We can also put it in a "Venn Diagram":

A Venn Diagram is clever because it shows lots of information:

Do you see that Alex, Casey, drew and

hunter are in the "Soccer" set?

And that Casey, drew and jade are in the

"Tennis" set?

And here is the clever thing: Casey and

drew are in BOTH sets!

3. INTERSECTION The intersection of two sets A and B is the set of all elements of A and B together, also denoted by A ∩ B.

"Intersection" is when you are in BOTH sets. In our case that means they play both Soccer AND Tennis ... which is casey and drew. And this is how we write it down: Soccer AND Tennis = {casey, drew}

135

4. COMPLEMENT The complement of a set A is the set of all the elements of the sample space that lies outside of set A and denoted by A′ (or AC).

When the Event is {Monday, Wednesday} the complement is {Tuesday, Thursday, Friday, Saturday, Sunday} Example 2. Consider the events A= {1, 2, 3, 4} and B= {4, 5, 6} in the experiment of rolling a die. Write the events A∪ B, A ∩ B and A′. The sample space is {1, 2, 3, 4, 5, 6}. Therefore, A ∪ B= {1, 2, 3, 4, 5, 6} (all outcomes of A or B) A ∩ B= {4} (all common outcomes of A and B) A′ = {5, 6} (all outcomes of the sample space those do not lie in the event A) Example 3. In a group of 40 learners, 25 play chess and 22 play rugby, while 7 play neither of the two. How many students play both chess and rugby?

Let’s say number of student who play both x. Then who play chess only: 25‐x Then who play rugby only: 22‐x On Venn Diagram:

And finally, total:

25 22 7 40

54 40

14

x x x

x

x

5. THREE SETS The following strategies are useful in solving survey problems:

Draw a Venn diagram in a universal set. Label the sets. Write the given information in the regions: Start with the intersection of all sets. Write the sets only A and B, only A and C,

only B and C. Write the sets only A, only B and only C.

Example 4. From a group of 80 grade 11 learners, 40 of them can speak Afrikaans(A), 30 can speak English (E) and 30 can speak Zulu (Z), but:

10 cannot speak any of the three

languages.

15 speak both Afrikaans and English.

10 speak both Afrikaans and Zulu.

10 speak both English and Zulu.

Find number of speaking all three languages.

Draw Venn Diagram and start from x as intersection of ALL sets

Add all values to get 80 total:

75 80

5

x

x

136

TASK 11. Let S denote the set of whole numbers from 1

to 16, X denote the set of even numbers from

1 to 16 and Y denote the set of prime numbers

from 1 to 16. Draw a Venn diagram depicting

S, X and Y .

2. There are 79 Grade 10 learners at school. All of

these take some combination of Maths,

Geography and History. The number who take

Geography is 41, those who take History is 36,

and 30 take Maths. The number who take

Maths and History is 16; the number who take

Geography and History is 6, and there are 8

who take Maths only and 16 who take History

only.

2.1. Draw a Venn diagram to illustrate all this

information.

2.2. How many learners take Maths and

Geography but not History?

2.3. How many learners take Geography only?

2.4. How many learners take all three

subjects?

3. Pieces of paper labelled with the numbers 1 to

12 are placed in a box and the box is shaken.

One piece of paper is taken out and then

replaced.

3.1. What is the sample space, S?

3.2. Write the set A, representing the event of

taking a piece of paper labelled with a

factor of 12.

3.3. Write the set B, representing the event of

taking a piece of paper labelled with a

prime number.

3.4. Represent A, B and S by means of a Venn

diagram.

3.5. Find

i. n (S) ii. n (A) iii. n (B)

4. In a group of 50 learners, 35 take Mathematics

and 30 take History, while 12 take neither of

the two. How many students take both

Mathematics and History?

137

TASK 21. A survey of 80 students at a local library

indicated the reading preferences below:

44 read the National Geographic magazine 33 read the Getaway magazine 39 read the Leadership magazine 23 read both National Geographic and Leadership magazines 19 read both Getaway and Leadership magazines 9 read all three magazines 69 read at least one magazine

1.1. How many students did not read any

magazine? (1)

1.2. Let the number of students who read

National Geographic and Getaway, but

not Leadership, be represented by x.

Draw a Venn diagram to represent

reading preferences. (5)

1.3. Hence show that x = 5. (3)

1.4. How many read at least two of the three

magazines? (3)

2. A school organised a camp for their 103 Grade

12 learners. The learners were asked to

indicate their food preferences for the camp.

They had to choose from chicken, vegetables

and fish.

The following information was collected: • 2 learners do not eat chicken, fish or vegetables • 5 learners eat only vegetables • 2 learners only eat chicken • 21 learners do not eat fish • 3 learners eat only fish • 66 learners eat chicken and fish • 75 learners eat vegetables and fish

Let the number of learners who eat chicken, vegetables and fish be x.

2.1. Draw an appropriate Venn diagram to

represent the information. (7)

2.2. Calculate x. (2)

2.3. Calculate number of learners which:

1.1.1. Eats only chicken and fish, and no

vegetables. (2)

1.1.2. Eats any TWO of the given food

choices. (2)

138

B. PROBABILITY Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

Probability is how likely something is to happen.

When a coin is tossed, there are two possible outcomes:

heads (H) or tails (T)

The probability of the coin landing H is ½. And the probability of the coin landing T is ½. When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6. The probability of any one of them is 1/6. We use "P" to mean "Probability Of",

Experiment is an action where the result is uncertain.

Tossing a coin, throwing dice, seeing what pizza people choose are all examples of experiments. A probability is a real number between 0 and 1 that describes how likely it is that an event will occur. It can be described as fraction (0,75 can also be written as 3/4 ). You can show probability on a Probability Line:

Sample Space is all the possible outcomes of an experiment), denoted with n(S).

Example: choosing a card from a deck There are 52 cards in a deck (not including Jokers) So the Sample Space is all 52 possible cards: {Ace of Hearts, 2 of Hearts, etc... }

Event is a single result of an experiment

Example Events:

Getting a Tail when tossing a coin is an event

Rolling a "5" is an event.

An event can include one or more possible outcomes:

Choosing a "King" from a deck of cards

any of the 4 Kings is an event

Rolling an "even number" 2, 4 or 6 is

also an event

In general

Number of ways it can happenProbability of event =

Total number of outcomesOR

( )( )

( )

n EP E

n S

Example 1.A die is thrown once. What is the probability that the score is a factor of 6? The factors of six are 1, 2, 3 and 6, so the Number of ways it can happen = 4 There are six possible scores when a die is thrown, so the Total number of outcomes = 6 So the probability that the score is a factor of six = 4/6 = 2/3 Example 2. A fair coin is tossed three times. What is the probability of obtaining one Head and two Tails? Represent 'Heads up' by H and 'Tails up' by T. There are 8 possible ways the coins can land: (H, H, H), (H, H, T), (H, T, H), (H, T, T), (T, H, H), (T, H, T), (T, T, H) (T, T, T) Of these, 3 have one Head and two Tails: (H, T, T), (T, H, T) and (T, T, H) So: The Number of ways it can happen = 3 The Total number of outcomes = 8

3

8(Head and Two Tails)P

139

TASK 3 1. A bag contains 6 red, 3 blue, 2 green and 1

white balls. A ball is picked at random.

Determine the probability that it is:

1.1. Red

1.2. blue or white

1.3. not green

1.4. not green or red

2. A playing card is selected randomly from a

pack of 52 cards. Determine the probability

that it is:

2.1. the 2 of hearts

2.2. a red card

2.3. a picture card

2.4. an ace

2.5. a number less than 4

3. Each of the letters of the word MISSISSIPPI are

written on separate pieces of paper that are

then folded, put in a hat, and mixed

thoroughly. One piece of paper is chosen

(without looking) from the hat. What is the

probability it is an I?

4. Even numbers from 2 to 100 are written on

cards. What is the probability of selecting a

multiple of 5, if a card is drawn at random?

5. A card is chosen at random from a deck of 52

playing cards. There are 4 Queens and 4 Kings

in a deck of playing cards. What is the

probability it is a Queen or a King?

6. A fair coin is tossed three times. What is the

probability of obtaining one Head and two

Tails?

7. A coin is tossed three times. Find the

probability that head and tail show alternately.

8. A committee of three is chosen from five

councilors ‐ Adams, Burke, Cobb, Dilby and

Evans. What is the probability Burke is on the

committee?

140

C. COMPLEMENT OF AN EVENT

The complement of an event A is the set of all outcomes of the sample space that lies outside of event A and denoted by A′ (not(A) or AC).

When the event is Heads, the

complement is Tails

When the event is Monday, Wednesday

the complement is Tuesday, Thursday,

Friday, Saturday, Sunday

When the event is Hearts the

complement is Spades, Clubs,

Diamonds, Jokers

So the Complement of an event is all the other outcomes (not the ones you want).

And together the Event and its Complement make all possible outcomes.

P(A) + P(A') = 1

Example 1. Rolling a "5" or "6" Event A:A={5, 6} Number of ways it can happen: 2 Total number of outcomes: 6

2 1

6 3( )P A

The Complement of Event A is A’ = {1, 2, 3, 4} Number of ways it can happen: 4 Total number of outcomes: 6

4 2

6 3( ')P A

Let us add them:

1 21

3 3( ) ( ')P A P A

It makes sense, right? Event A plus all outcomes that are not Event A make up all possible outcomes.

Why is the Complement Useful? It is sometimes easier to work out the complement first. Example 2. Throw two dice. What is the probability that the two scores are different? Different scores are like getting a 2 and 3, or a 6 and 1. It is quite a long list: A = { (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), ... etc ! } But the complement (which is when the two scores are the same) is only 6 outcomes: A' = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) } And the probability is easy to work out:

P(A') = 6/36 = 1/6 Knowing that P(A) and P(A') together make 1, we can calculate:

P(A) = 1 ‐ P(A') = 1 ‐ 1/6 = 5/6 So in this case it's easier to work out P(A') first, then find P(A)

141

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142

TASK 4

1. If 3

4( )P A what is probability of ( ')P A

2. Two fair coins are tossed. What is the

probability at least one coin lands heads up?

3. A spinner is made from a piece of card in the

shape of a regular pentagon with a toothpick

pushed through the center. When the spinner

is spun and it lands on an edge, each of the

numbers from 1 to 5 is equally likely. If the

spinner is spun twice, what is the probability

the two scores are different?

4. Two fair dice are thrown. What is the

probability that two scores do not add to 7?

5. A code consists of a digit chosen from 0 to 9

followed by a letter of the alphabet. What is

the probability the code is 9Z?

6. Two cards are drawn from the top of a well‐

shuffled deck. What is the probability that they

are both black aces?

7. A survey was conducted about the

broadcasting of a certain television

programme. 150 males and 100 females were

interviewed. The table below shows some of

the results:

Male Female Total

Liked it 60 a 130

Did not like it b 30 c

TOTAL 150 100 d

7.1. Calculate the values of the letters (a, b, c

and d) in the table.

7.2. Calculate the probability of choosing at

random in the survey, a female who

didn’tlike a programme.

7.3. Is a person's preference for the

programme independent of the person's

gender? Support your answer with

appropriate calculations.

8. Two events, A and B are independent events

and P(A) = 1

3 and P(A and B) =

1

12 ,

determine P(B)

9. A bag contains 5 red marbles, 4 green marbles

and 1 blue marble. A marble is chosen at

random from the bag and not replaced; then a

second marble is chosen. What is the

probability both marbles are green?

143

TASK 51. A box contains 24 purple, 32 orange and 41

white blocks. A block is selected randomly.

What is the probability that the block will be:

1.1. Purple

1.2. purple or white

1.3. purple and orange

1.4. not orange?

2. A small school has a class with children of

various ages. The table gives the number of

pupils of each age in the class:

3 years old 4 years old 5 years old

Male 2 7 6

Female 6 5 4

If a pupil is selected at random what is the probability that the pupil will be:

2.1. a female

2.2. a 4 year old male

2.3. aged 3 or 4

2.4. aged 3 and 4

2.5. not 5

2.6. either 3 or female?

3. A bag contains 5 red marbles, 4 green marbles

and 1 blue marble.

A marble is chosen at random from the bag

and not replaced; then a second marble is

chosen. What is the probability that neither

marble is blue?

4. Fiona has 85 labelled discs, which are

numbered from 1 to 85. If a disc is selected

at random what is the probability that the disc

number:

4.1. ends with 5

4.2. is a multiple of 3

4.3. is a multiple of 6

4.4. is number 65

4.5. is not a multiple of 5

4.6. is a multiple of 4 or 3

4.7. is a multiple of 2 and 6

4.8. is number 1?

5. Two cards are drawn from the top of a well‐

shuffled deck. What is the probability that they

are both Diamonds?

6. Imagine there are four groups with 5 members

each:

A member of each group gets randomly

chosen for the winners circle,

then one of those gets randomly chosen

to get the big money prize:

6.1. What is the probability of someone to go

to winner circle?

6.2. What is the probability of someone from

1st group to win?

6.3. What is the probability of random person

to win?

144

E. TREE DIAGRAMS Calculating probabilities can be hard, sometimes you add them, sometimes you multiply them, and often it is hard to figure out what to do ... tree diagrams to the rescue! Here is a tree diagram for the toss of a coin:

There are two "branches" (Heads and Tails)

The probability of each branch is written

on the branch

The outcome is written at the end of the

branch

We can extend the tree diagram to two tosses of a coin

How do you calculate the overall probabilities?

You multiply probabilities along the

branches

You add probabilities down columns

Now we can see such things as:

The probability of "Head, Head" is

0.5 0.5 0.25

All probabilities add to 1.0 which is

always a good check

The probability of getting at least one

Head from two tosses is

0.25 0.25 0.25 0.75

... and more

Example. You are off to soccer, and love being the Goalkeeper, but that depends who is the Coach today:

with Coach Sam the probability of being

Goalkeeper is 0.5

with Coach Alex the probability of being

Goalkeeper is 0.3

Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6). So, what is the probability you will be a Goalkeeper today? Let's build the tree diagram.

Now we add the column:

0.3 + 0.12 = 0.42 probability of being a Goalkeeper today Check

complete calculations and make sure they add to 1

0.3 + 0.3 + 0.12 + 0.28 = 1 Conclusion: So there you go, when in doubt draw a tree diagram, multiply along the branches and add the columns. Make sure all probabilities add to 1

145

TASK 6 1. One die and a coin are thrown together. Use a

tree diagram to find

1.1. The probability that number is even and

the Heads.

1.2. The probability that numbers is less than

five and Tails.

2. Jack wakes up late on average 3 days in every

5 days.

If Jack wakes up late, the probability he’s late

for school is 9

10

If Jack does not wake up late, the probability

he’s late for school is 3

10

On what percent of days does Jack get to

school on time?

3. Tina's favorite meal is pasta, followed by ice

cream for dessert.

Tina's Mom cooks pasta once a week.

If she cooks pasta, then the probability Tina

gets ice cream for desert is 2

3

If she doesn’t cook pasta, then the

probability Tina gets ice cream is 1

4

What is the probability that Tina gets ice

cream for dessert?

4. Teddy has a two pairs of black shoes and three

pairs of brown shoes. He also has three pairs

of red socks, four pairs of brown socks and six

pairs of black socks.

If Teddy chooses a pair of shoes at random and

a pair of socks at random:

4.1. what is the probability that he chooses

shoes and socks of the same color?

4.2. what is the probability that the colors he

chooses are black and brown?

146

5. The probability of a fine day is 3

7and the

probability of a wet day is 4

7.

If it’s a fine day:

The probability Sipho walks to school is 7

10

The probability Sipho mother drives him to

school is 2

10

And the probability Sipho takes a taxi to school

is 1

10

If it’s a wet day:

The probability Sipho walks to school is 1

9

The probability Sipho mother drives him to

school is 5

9

And the probability Sipho takes a taxi to school

is 3

9

5.1. For a day selected at random, what is

probability Sipho takes a taxi to school?

5.2. If Sipho goes to school 215 days in a year,

estimate how many days is his mother

drives him to school?

6. A bag contains 3 white balls, 4 green balls and

5 red balls. Two balls are drawn from the bag

without replacement. Use a tree diagram to

find

6.1. the probability that the balls are different

colors.

6.2. find the probability that the balls are the

same color

7. Figures obtained from a city's police

department seem to indicate that ofall the

motor vehicles reported stolen, 80% were

stolen by syndicates to be sold off and 20%

were stolen by individual persons for their

own use.

Of those vehicles presumed stolen by

syndicates:

• 24% were recovered within 48 hours

• 16% were recovered after 48 hours

• 60% were never recovered

Of those vehicles presumed stolen by

individual persons:

• 38% were recovered within 48 hours

• 58% were recovered after 48 hours

• 4% were never recovered

7.1. Draw a tree diagram for the above

information. (5)

147

7.2. Calculate the probability that if a vehicle

werestolen in this city, it would be stolen

by a syndicate and recovered within 48

hours. (2)

7.3. Calculate the probability that a vehicle

stolen will not be recovered. (3)

8. There are 20 boys and 15 girls in a class. The

teacher chooses individual learners at random

to deliver a speech.

8.1. Calculate the probabilitythat the first

learner chosen is a boy. (1)

8.2. Draw a tree diagram to represent the

situation if the teacher chooses three

learners, one after the other. Indicate on

your diagram ALL possible outcomes. (4)

8.3. Calculate the probabilitythat a boy, then a

girl and then another boy is chosen in

that order. (3)

8.4. Calculate the probability that all three

learners chosen are girls. (2)

8.5. Calculate the probability that at least one

of the learners chosen is a boy. (3)

9. The probability that it will rain on a given day

is 63%. A child has a 12% chance of falling in

dry weather and is three times aslikely to fall

in wet weather.

9.1. Draw a tree diagram to represent all

outcomes of the above information. (6)

9.2. What is the probability that a child will

not fall on any given day? (3)

9.3. What is the probability that a child will fall

in dry weather? (2)

10. The swimming team of South Africa consists of

EIGHT swimmers, namely three men and five

women. They have to swim in two qualifying

rounds before they can go through to the

finals. As part of their rotation policy they

must select a different captain for each

qualifying round.

Draw a tree diagram to determine all the

probabilities of men or women being selected

as captain – with a column for the first round,

then the second round, then the possible

outcomes and then probabilities as fractions.

Label each branch with its probability. (7)

148

F. MUTUALLY EXCLUSIVE EVENTS Mutually Exclusive events: can't happen at the same time. Examples:

Turning left and turning right are

Mutually Exclusive you can't do both at

the same time

Tossing a coin: Heads and Tails are

Mutually Exclusive

Cards: Kings and Aces are Mutually

Exclusive

What is not Mutually Exclusive:

Turning left and scratching your head

can happen at the same time

Kings and Hearts, because you can have

a King of Hearts!

Aces and Kings are Mutually Exclusive(can't be both):

Hearts and Kings are not Mutually Exclusive (can be both):

1. MUTUALLY EXCLUSIVE

For mutually exclusive events it is impossible for them to happen together:

P(A and B) = 0 The probability of A and B together is impossible.

But the probability of A or B is the sum of the individual probabilities:

P(A or B) = P(A) + P(B)

Example 1. In a Deck of 52 Cards:

the probability of a King is 1/13, so

P King 1/13

the probability of an Ace is also 1/13, so

P Ace 1/13

When we combine those two Events:

The probability of a card being a King

and an Ace is 0 Impossible

The probability of a card being a King or

an Ace is 1/13 1/13 2/13

Which is written like this: P(King and Ace) = 0

P(King or Ace) = (1/13) + (1/13) = 2/13

2. NOT MUTUALLY EXCLUSIVE

When events are not Mutually Exclusivewe use the formula:

P(A or B) = P(A) + P(B) ‐ P(A and B)

Pay attention the formula above may be used for mutually exclusive events as well. Compare those formulas to see why. Example 2. How many Hearts or Kings are there in 52 card deck?

Hearts or Kings:

all the Hearts 13 of them

all the Kings 4 of them

But that counts the King of Hearts twice!

So we correct our answer, by subtracting the extra "and" part:

16 Cards = 13 H + 4 K ‐ the 1 extra King of Hearts

Count them to make sure this works!

As a probability this is:

P(H or K) = P(H) + P(K) ‐ P(H and K)

13 4 1

52 52 5216

52

P H or K –

149

1. QUESTION 1

P(A) = 0,5, P(B) = 0,3 and P(A or B) = 0,65 are given. 1.1. Are the events A and B mutually exclusive?

Justify your answers. 1.2. Calculate 1.3. Are the events A and B independent? Justify

your answers

SOLUTION 1.1. If two event are mutually exclusive then

P A or B P A P B

Let’s use this rule and substitute:

0 65 0 5 0 3, , ,

This is wrong! Therefore, No

1.2. Use the rule:

P A or B P A P B P A and B Substitute:

0 65 0 5 0 3

0 15

0 15

, , ,

,

,

P A and B

P A and B

P A and B

1.3. If two event independent then:

P A and B P A P B

Substitute:

0 15 0 5 0 3, , ,

This is right! Therefore, Yes

2. QUESTION 2

What is the probability of throwing at least one six in four rolls of a regular six sided die? SOLUTION This question is difficult straight forward, but if you think opposite : What is probability not having any six in four rolls

is 5 5 5 5 625

6 6 6 6 1296 .

Therefore,

6551

1295128

259

(at least one six)P

3. QUESTION 3

In a factory, three machines, A, B and C, are used to manufacture plastic bottles. They produce20%, 30% and 50% respectively of the total production. 1%, 2% and 6% respectively of the plastic bottles produced by machines A, B and C are defective. 3.1. Represent the information by means of a

tree diagram (4)

3.2. What is the probability that it was produced

by machine B and it is not defective?

(3)

3.3. What is the probability that the bottle is

defective? (3)

SOLUTION 3.1. First choose machine, then it may be

defective or not defective

3.2. P (Not defective and from B) =0,294 3.3. P(Defective) = 0,001 + 0,006 + 0,03 = 0,038

4. QUESTION 4 Given result of a research in the table:

Don’t play sport Play sport Total

Male 51 69 120

Female 49 67 116

Total 100 136 236

4.1. What is the probability that person is male?

Female who plays sport?

4.2. Are the events 'male' and 'do not play sport'

independent?

SOLUTION

4.1. 120

236(Male)P ,

67

236(F and Play S)P

4.2. (Male and not Play) (Male) (not Play)P P P

51 120 100

236 236 236

This is wrong! Therefore, No. (if you round off answer to 2 decimals then Yes)

150

TASK 7 1. P(A) = 0,45, P(B) = 0,3 and P(A or B) = 0,615.

1.1. Are the events A and B mutually

exclusive?

1.2. Are the events A and B independent?

2. A card is chosen at random from a pack of 52

playing cards.What is the probability of a King

or a Queen?

3. A card is chosen at random from a pack of 52

playing cards.What is the probability of a King

or a Heart?

4. A number is chosen at random from the set of

two‐digit numbers from 10 to 99 inclusive.

4.1. List all such numbers starting with 2.

4.2. Lisr all such numbers ending with 2

4.3. What is the probability the number

contains at least one digit 2?

5. Two fair dice are thrown.What is the

probability that the score on the first die is 6

or the score on the second die is 5?

There are 30 children in a class and they all have at least one cat or dog.14 children have a cat, 19 children have a dog.

5.1. Draw a venn diagram and calculate the

number of children having dog and cat

5.2. What is the probability that a child chosen

at random from the class has both a cat

and a dog?

6. In a group of 25 boys, 20 play ice hockey and

17 play baseball. They all play at least one of

the games.

6.1. Draw a venn diagram and calculate the

number of boys playing both.

6.2. What is the probability that a boy chosen

at random from the class plays ice hockey

but not baseball?

7. Let A and B be two events in a sample space.

Suppose that P(A) = 0,4; P(A or B) = 0,7

andP(B) = k.

7.1. For what value of k, A and B are mutually

exclusive? (2)

7.2. For what value of k, A and B are

independent? (4)

151

TASK 8 1. A group of 45 children were asked if they eat

Frosties and/or Strawberry Pops. 31 eat both

and 6 eat only Frosties. What is the probability

that a child chosen at random will eat only

Strawberry Pops?

2. In a group of 42 pupils, all but 3 had a packet

of chips or a Fanta or both. If 23 had a packet

of chips and 7 of these also had a Fanta, what

is the probability that one pupil chosen at

random has:

2.1. both chips and Fanta

2.2. only Fanta

3. Use a Venn diagram to work out the following

probabilities for a die being rolled:

3.1. a multiple of 5 and an odd number

3.2. a number that is neither a multiple of 5

nor an odd number

3.3. a number which is not a multiple of 5, but

is odd

4. A packet has yellow and pink sweets. The

probability of taking out a pink sweet is 7/12.

What is the probability of taking out a yellow

sweet?

5. In a car park with 300 cars, there are 190

Opels. What is the probability that the first car

to leave the car park isnot an Opel?

6. Tamara has 18 loose socks in a drawer. Eight

of these are orange and two are pink.

Calculate the probability that the first sock

taken out at random is:

6.1. orange

6.2. not orange

6.3. pink

6.4. not pink

6.5. orange or pink

6.6. neither orange nor pink

7. A plate contains 9 shortbread cookies, 4 ginger

biscuits, 11 chocolate chip cookies and 18

Jambos.

If a biscuit is selected at random, what is the

probability that:

7.1. it is either a ginger biscuit of a Jambo

7.2. it is not a shortbread cookie

8. 280 tickets were sold at a raffle. Ingrid bought

15 tickets. What is the probability that Ingrid:

8.1. wins the prize

8.2. does not win the prize

152

TASK 9 1. The children in a nursery school were classified

by hair and eye colour. 44 had red hair and not

brown eyes, 14 had brown eyes and red hair, 5

had brown eyes but not red hair and 40 did

not have brown eyes or red hair.

1.1. How many children were in the school?

1.2. What is the probability that a child chosen

at random has:

i. brown eyes ii. red hair

1.3. A child with brown eyes is chosen

randomly. What is the probability that

this child will have red hair?

2. A jar has purple, blue and black sweets in it.

The probability that a sweet chosen at random

will be purple is 1/7 and the probability that it

will be black is 3/5 .

2.1. If I choose a sweet at random what is the

probability that it will be:

i. purple or blue ii. black iii. purple

2.2. If there are 70 sweets in the jar how many

purple ones are there?

2.3. 2/5 of the purple sweets in (b) have

streaks on them and the rest do not. How

many purple sweets have streaks?

3. For each of the following, draw a Venn

diagram to represent the situation and find an

example to illustrate the situation.

3.1. a sample space in which there are two

events that are not mutually exclusive

3.2. a sample space in which there are two

events that are complementary

4. Use a Venn diagram to prove that the

probability of either event A or B occurring (A

and B are not mutually exclusive) is given by:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

5. All the clubs are taken out of a pack of cards.

The remaining cards are then shuffled and one

card chosen. After being chosen, the card is

replaced before the next card is chosen.

5.1. What is the sample space?

5.2. Find a set to represent the event, P, of

drawing a picture card.

5.3. Find a set for the event, N, of drawing a

numbered card.

5.4. Represent the above events in a Venn

diagram.

153

TASK 10 1. Hospital records indicated that maternity

patients stayed in the hospital for the number

of days shown in the distribution.

Number of days stayed Frequency

3 15 4 32 5 56 6 19 7 5

127

Find these probabilities:

1.1. A patient stayed exactly 5 days

1.2. A patient stayed less than 6 days.

1.3. A patient stayed at most 4 days.

1.4. A patient stayed at least 5 days.

2. In a group of 61 students, number of girls with

brown eyes is 14 and number of boys with

blue eyes is 29. The number of boys with

brown eyes is 2 times the number girls with

blue eyes.

2.1. Complete table below

Blue eyes Brown eyes Total

Boys

Girls

Total

2.2. If teacher choose one learner from that

group, what is the probability that learner

is girl and has blue eyes.

3. P(A) = 0,3 and P(B) = 0,5. Calculate P(A or B) if:

3.1. A and B are mutually exclusive events

3.2. A and B are independent events

4. Match the following statements of chance

with one of the given probabilities:

Statement of chance Probability

This event is impossible, it can’t happen

0.35

This event is certain, it will always occur

0.1

This event is very unlikely, it may occur, but hardly ever

0

This event will occur almost every time

0.55

Approximatly 50/50 chance 0.9

More unlikely than likely 1

5. A cup has 3 red and 3 blue discs placed in it.

The cup is shaken and a disc is drawn. The

colour is noted and the disc is replaced. This is

repeated until 5 discs have been drawn.

5.1. What is the probability of drawing a red

disc on draw 1?

5.2. What is the probability of drawing a red

disc on draw 2? And draw 3, 4 and 5?

5.3. Suppose that a red disc was drawn in

each of the 5 draws. What is the

probability of drawing a red disc in the

next draw?

5.4. If we draw 10 discs as described above,

how many red discs would you expect to

draw?

6. Suppose that you read the “Racing News”

where it is stated that a particular race horse

has odds of 4 to 1 against winning in a

particular race. What is the probability that

the horse wins the race?

154

TASK 111. Suppose that we have a spinner which can

land on any number 1, 2 or 3. We spin the

spinner once and note where it stops. (Note

that each section below, given a number, is

supposed to be of equal size.)

Suppose that the spinner is spun twice and the

number is noted where the spinner stops each

time.

1.1. Draw a tree diagram to show all possible

outcomes of the experiment.

1.2. Write out the sample space of this

experiment.

1.3. Define the following event

A: The spinner stopped on an odd

number both times

B: The spinner stopped on an

even number both times

C: The spinner stopped on the

numbers that add up to 4

D: The spinner stopped on the

numbers that add up to an odd

number

Find the probabilities of the events A, B, C

and D.

2. Suppose that a die is rolled once. List the

outcomes in the sample space that would

define the following events:

A: A number greater than 2 but less

than 5 is obtained

B: An odd number is obtained

2.1. Give the Venn diagram that depicts this

situation.

2.2. Find the following probabilities:

(i) P(A)

(ii) P(B)

(iii) P(A or B)

(iv) P(A and B)

(v) P(not A)

3. Estimate the probability for each event given

below. State if the probability is below

average, a 50‐50 chance, above average,

certain, impossible or if you cannot tell with

the given information:

3.1. Event A: The next baby to be born will be

a boy.

3.2. Event B: A triangle will have three sides.

3.3. Event C: Bafana‐Bafana will win the 2018

World Cup Soccer.

3.4. Event D: A frog will become the next

captain of the football team

155

TASK 12 1. Suppose that there are 24 children in a class.

Each child is given the following instructions:

Tick one of the following:

I have a dog but no cat

I have a cat but no dog

I have a dog and a cat

I don’t have a dog and don’t have a cat

From this information received, the following summary was made:

Children

Dog, no Cat 6

Cat, no dog 3

Dog, Cat 5

No Dog, no Cat 8

1.1. Depict the corresponding Venn diagram

using D for Dog and C for Cat.

1.2. What is the probability that a randomly

selected child owns a dog?

1.3. What is the probability that a randomly

chosen child from the class has no cat?

2. A boy has 0.19 probability of winning an art

competition and a 0.13 probability of winning

a talent competition where he will perform a

song. Suppose he has a 0.11 chance of winning

both the art competition and the talent

competition.If the events are defined as

A: The boy wins the art competition

T: The boy wins the talent competition

2.1. Then find P(A), P(T), P(A or T).

2.2. Are events A and T mutually exclusive?

Why?

2.3. What is the probability that the boy does

not win the art competition?

2.4. What is the probability that he wins the

art competition or the talent

competition?

3. The following (Stats SA release PO 305), gives

the registered births by province as recorded

for 2002. Figures presented below are in

thousands.

Eastern Cape Free State

Gauteng KZN Limpopo

288 66 204 395 215

Mpumalanga North Cape

North West

Western Cape

Other

114 21 105 102 8

If a randomly chosen person in South Africa is

identified, what is the probability that this

person comes from

3.1. Mpumalanga?

3.2. Mpumalanga or Northern Cape?

3.3. Gauteng and the Free State?

156

TASK 13 1. There are 130 Grade 10 learners in a school.

Sixty‐eight learners take Mathematics and

50learners take Physical Science. Thirty‐two

learners take Mathematics and Physical

Science.

1.1. Draw a Venn diagram to illustrate the

given data.

1.2. Hence calculate the probability that a

learner in Grade 10, chosen at random:

1.1.3. Takes Physical Science

1.1.4. Takes Mathematics but not Physical

Science

1.1.5. Takes Mathematics or Physical

Science

2. Mrs James has a CD player to give to one of

the learners in her class who participated in a

competition. Here are the names of the

learners and their ages:

NAME AGE

Mishin 15

Damien 16

Jane 16

Kuveshan 17

Muhammed 16

Benedict 17

Angelina 15

Mike 16

Events are defined as follows:

E1 : A boy is selected E2 : A 16‐year‐old must win E3 : The winner's name starts with an M E4 : A 17‐year‐old must win

2.1. Give n(S).

2.2. Determine P(E1).




2.6. Determine P(E1 or E4).

2.7. Determine P(E2 and E3).

3. A survey was conducted at Mutende Primary

School to establish how many of the 650

learners buy vetkoek and how many buy

sweets during break. The following was found:

• 50 learners bought nothing • 400 learners bought vetkoek • 300 learners bought sweets 3.1. Represent this information with a Venn

diagram

3.2. If a learner is chosen randomly, calculate

the probability that this learner buys:

i. sweets only ii. vetkoek only iii. neither vetkoek nor sweets iv. vetkoek and sweets v. vetkoek or sweets

4. In a survey at Lwandani’s Secondary School 80

people were questioned to find out how many

read the Sowetan and how many read the

Daily Sun newspaper or both. The survey

revealed that 45 read the Daily Sun, 30 read

the Sowetan and 10 read neither. Use a Venn

diagram to find the percentage of people that

read:

4.1. Only the Daily Sun

4.2. Only the Sowetan

4.3. Both the Daily Sun and the Sowetan

157

IX. REVISION AND EXEMPLARS A. EQUATIONS AND EXPONENTS

REVISION 1 QUESTION 1 Simplify each of the following:

1.1

1 1

3 4

1

2

y y

y

(3)

1.2 16 1650 18x x (3)

QUESTION 2 2.1 Simplify : 6 8 43 464 16 49x x x (4)

2.2 Given K=

2 1

3 4 2x x

2.2.1

Show that K is a rational number if x=1,25

(3)

2.2.2

Determine the values of x for which K is a real number

(3)

2.3

Jared had to find the product of 20092 and

20055 and then calculate the sum of the digits of the answer. Jared arrived at an answer of 7. Is she correct? Show ALL the calculations to motivate your answer.

(5)


3.1.1 1 1

1

9 27

81

x x

x

(4)

3.1.2

3

481

16( )

(2)

QUESTION 4 4.1

If 2 , what is x ?

(4)

4.2 If 2 8 and 4 11, what

is ?

(5)

4.3 Solve for x and illustrate the answer on a

number line. 1

3 52

x

(4)

158

REVISION 2 QUESTION 1 1.1 Simplify, without using a calculator: 1.1.1 58 3 58 3 (3)

1.1.2 433

3

2

8

( )[ ) ]x

x

(3)

1.1.3 2

1

4 15

12 5

x x

x x

(4)

1.2 Solve for x : 21

3 127( )x x

(5) [14]

QUESTION 2 2.1

If A= 5

3 x , determine the values of x

for which 2.1.1 A is undefined. (3)

2.1.2 A is non‐real. (2)

QUESTION 3 3.1 If5 3 , what is T ?

(4)

3.2 If 3 2 2 and

3 3 , what is ? (5)


number line 1 1

2 25 3x and 0x

(4)

159

REVISION 3QUESTION 1 1.1 Simplify, without using a calculator:

1.1.1 28 3 63

700?

(4)

1.1.2 1 1

1

4 8

32

n n

n

(4)

1.2 Solve for x : 12 3 2 112x x (4)


2.1.1 2 1 5

0 5

32 125 10

25 ,

. .x x x

x

(3)

2.1.2 24 81 (2)

2.2 Solve for x : 1

34 1 20( )x

(3)

QUESTION 3 3.1 If ,what is x ?

(4)

3.2 If 3 4 13and 4 3 1, what

is ?

(5)


number line 2 2

5 7

x x

(4)

160

REVISION 4 QUESTION 1 1.1

Show that 3 2 1 2

1

3 3

3 9

x x

x x

is equal to 9x (4)

1.2 Hence or otherwise find the value of :

3 2 1 2

1

3 3

3 9

x x

x x

(2)

1.3 Which values is larger ? Show working

details. 2 or 5 6 (3)


2.1.1 75 2 3 3 27

12?

(4)

2.1.2 2

333

8( )

(3)

2.1.3 23 9

3 3

x

x

(3) [14]

2.2 Rationalise the denominator:

2 1

4 2

(3)

QUESTION 3 3.1 6 3 1 2 3 5( ) ( )x x x x (4) Solve for x. 3.2 If 2 4 and

4 5 1, find and ? (5)


number line

12 2 1

3x

(4)

161

REVISION 5QUESTION 1 1.1 Simplify, without using a calculator:

1.1.1 28

63 28?

(4)

1.1.2

2 1 5

42

32 27 6

9

. .x x x

x

(5)

1.2 Determine the values of x for which 3 4x

x

is unreal

(4) [13]


2.1.1 2 33 20 125 0 25( , ) ( . )

(4)

2.1.2 4

0319 11 3 427( ) ( ) (4)

2.2 Solve for and illustrate the answer on a

number line 1 3 4 11x (4) [12]

162

REVISION 6 QUESTION 1 1.1 Calculate the following products: 1.1.1 2 23 5( )x y (4)

1.1.2 2 3( )( )b a (4)

1.1.3 22 2 4( )( )p p p (4)

1.1.4 2 3 2 3( ) (2)

QUESTION 2 2.1 Factorize fully: 2.1.1 4 16x (4)

2.1.2 22 5 3m m (4)

QUESTION 3 The lengths of the perpendicular sides of aright

angled triangle are √2 1 and √2 1 units .

Calculate the length of the hypotenuse (in the surd form)

(4)

QUESTION 4 2 5 4

2 10 3 15

a a a

a a

(4)

163

REVISION 7QUESTION 1 1.1 Simplify

122

4

9( )x

y

(4)

1.2 It is given that

3 35 5000M N ,where M and N are whole numbers. Determine the value of M N

(3)

1.3 Simplify

1 1

1 1

x y

x y y x

(4)

QUESTION 2 2.1 Simplify : 10 6 45 332 27 9x x x (4)

2.2 Given T=

4 1

1 8 4x x

2.2.1

Show that T is a rational number if 0,25

(3)

2.2.2

Determine the values of for which T is a real number

(3)

QUESTION 3 3.1 Solve for if 5 5 5x (3)

3.2 Solve for if 3 62 8x (3)

164

REVISION 8 QUESTION 1 1.1 Calculate the following products: 1.1.1 21 1( )( )x x x (3)

1.1.2 21

2( )x h (3)

1.1.3 5 1 5 15 5( )( )x x x x (3)

1.1.4 21( )x

x (3)

QUESTION 2 2.1 Factorize fully: 2.1.1 28 6 1x x (3)

2.1.2 5 3 10 6xy y x (4)

QUESTION 3 Calculate the values of

2 25

5( )

xf x

x

if x = 12345654321 (4)

QUESTION 4 Calculate the value of

2009 2008 (4)

165

REVISION 9 QUESTION 1 1.1 Solve for x, rounded off to TWO decimal

places where necessary: 1.1.1 4 3 0( )( )x y if y = 3 (2)

1.1.2 2 2 0( )x x (2) 1.1.3 22 5 0x x (4)

1.2

if 1

5ax

, then determine the value(s)

of x for which

1.2.1 a = 0 (1)

1.2.2 a is undefined (3) 1.3 If 3x and 1y satisfy the equation

and 2 23 5x xy by determine the

values of b

(4) [16]

QUESTION 2 2.1

Given: 2 3 3 1 2( ) ( )( )( )f x x x x

Solve 0( )f x if :

2.1.1 x is an integer. (1)

2.1.2 x is a rational number (1)

2.1.3 x is a real number. (2)

2.2 Solve for x, rounded off to TWO decimal

places where necessary:

2.2.1 3 3

22 2x x

(5)

2.2.2 15

2x

xx

(4)

2.3 Solve for x correct to two decimal places

without using a formula: 23 7 1 0x x

(use completing square method)

(4) [18]

TOTAL: 34

166

B. FUNCTIONS REVISION 10

QUESTION 1

Given: 2

1.1 Give the domain of f. (1) 1.2 For what value of x is f(x)=0? (2)

1.3 Determine the y intercept of f.

(2)

1.4 Write down the equations of the

asymptotes of f.

(2)

1.5 Draw a neat graph of f, indicating the asymptotes

and intercepts with the axes.

(3) [10]

QUESTION 2 The graph below is a sketch of the function:

2 4

2.1 Write down the equation of the

asymptote of this function.

(1) 2.2 Give the co‐ordinates of points A and B.

(3) 2.3 Give the equations of the following

graphs: (give your answer in the form y = ….. )

2.3.1 a reflection in the x‐axis (2)

2.3.2 a reflection in the y‐axis (2)

2.3.3 a shifted function 3 units to the right.

(2) [10]

TOTAL : 20

167

REVISION 11QUESTION 1 Given

3 1( ) xf x , 3 1( ) xg x , 13( ) xh x

1.1 Draw all the graphs (show the intercepts

if possible)

(6)

1.2 Write the relation between ( )f x and

( )g x

(1)

1.3 ( )t x is reflection of ( )f x in the x=0; ( )b x

is reflection of ( )h x in the y=0. Write the

equations of ( )t x and ( )b x

(3) [10]

QUESTION 2 2.1 Draw a neat sketch graph of

. Clearly indicate all

intercepts with the axes.

(2)

2.2 Draw the following on the same system

of axes as in f(x) 1 ;

.

(4)

2.3 Describe the transformation from f to g . (2) 2.3 Describe the transformation from f to h . (2)

[10]

TOTAL : 20

168

REVISION 12 QUESTION 1

Consider the function: 5( ) xh x

1.1 Draw a rough sketch of h(x) (2) 1.2 What will be the equation of the graph

which is a reflection of ( )h x in the y ‐

axis?

(2)

1.3 What will be the equation of the graph

which is a reflection of ( )h x in the x ‐

axis?

(2)

1.4 Describe the transformations of ( )h x

which will result in the graph of 15 3( ) xp x

(3) [9]

QUESTION 2 Consider the functions:

2 2 3( )f x x x and

3 2 1( ) xg x

2.1 Draw both functions on the same set of

axes (6)

2.2 Write down the axis of symmetry of f . (1)

2.3 What is the range of f . (1) 2.4 What is the range of g . (1) 2.5 Use the graphs to determine the value of

0 0 .

(2) [11]

TOTAL : 20

169

REVISION 13QUESTION 1 Given:

2 and

1.1 Calculate the values of p and q . (4) 1.2 Write down the domain of g(x).

(1)

1.3 Write down the range of f.

(1)

1.4 Write down the equation of the asymptote of f.

(1)

1.5 Write down the equations of the axes of symmetry of g.

(2)

1.6 EDF is a line perpendicular to the x‐axis. Calculate the distance EF if the distance OD =2 units.

(4)

1.7 For which x, f g (3) [16]

QUESTION 2 In the sketch f is the graph of the function

2.1 Calculate the value of aand qif the

point P(3 ; 9) lies on the graph.

(4)

2.2 Write down the equation of the

asymptote of f

(1) 2.3 Write down the range of f. (1) 2.3 For which x, f < 0. (1) 2.2 Write down the equation of h(x) if it is the

reflection of f in the y‐axis

(2) [9]

TOTAL : 25

170

C. EXEMPLAR 1 2 HOURS QUESTION 1 EQUATIONS AND INEQUALITIES 1.1 If3 2 2 4 5 1( ) ( )x x x x ,

what is x ?

(4)

1.2 Solve for and

2 3 2 and 3 8 (5)

1.3 Solve for and illustrate the answer on a

number line. 2 4

2 63

x

(4) [13]

QUESTION 2 EXPONENTS Simplify each of the following:

2.1

11612

1

4

a a

a

(3)

2.2

1 1

2

25 125

625

x x

x

(3)

2.3

1

21 1

1

4 9( )

(4)

2.4

1

7 3 3

3

n n

n

(4) [14]

QUESTION 3 SURDS 3.1 Simplify : 8 827 75x x (3)

3.2 Simplify : 5 45 3 20 2 5( ) (4)

3.3

Simplify : 3

502 (3)

3.4 Given Y=

8 2

3 2x x

3.4.1

Show that Y is a rational number if x=2,5

(3)

3.4.2

Determine the values of x for which Y is a real number

(3) [16]

QUESTION 4 PRODUCTS Find the following products: 4.1 1 6 2 4 (2) 4.2 5 6

(3)

4.3 3

(3) [8]

QUESTION 5 FACTORIZATION

Factorize the following expressions fully: 5.1 5 20 (2) 5.2 12 3 2 6 2 3 (2) 5.3 11 11 (2) 5.4 5 5 (3) 5.5 4 21 (2) 5.6 14 13

(3)

5.7 6 13 5 (3)

[17] QUESTION 6 SIMPLIFICATION

Simplify the following expressions fully: 6.1 2 3

10 10.3 2

4 9

(6)

6.2

21

2

1

(5) [11]

QUESTION 7 QUADRATIC EQUATIONS

Solve for , rounded off to TWO decimal places where necessary: 7.1 2 2 24x x (3)

7.2 2 2 0( )x x (2)

7.3 6x x (4)

7.4 22 5 0x x (4)

7.5 24

3 12 03( )x (4)

7.5 Solve for x by completing square method)

23 7 1 0x x

(4) [21]

TOTAL MARKS: 100

171

D. EXEMPLAR2 2 HOURS QUESTION 1 1.1 Solve for the unknown, rounded off to

TWO decimal places where necessary: 1.1.1 3 2 1 0( )( )y y (2)

1.1.2 2 24 0( )x x (4)

1.1.3 23 4 2x x (5)

1.1.4 12

1 05x

(3)

1.2 Solve for x and y simultaneously if

2 1x y and 2 22 3 0x y

(8)

1.3 What is nature of root for

22 5 7( )f x x x

(3)

1.4 Determine the values of m for which

1 2m

m

is unreal

(3)

1.5

Calculate the value of 2 9

3( )

xf x

x

, if

31264599999x

(4) [32]

QUESTION 2 2.1 Simplify without the use of a calculator.

2.1.1

2

30 125( , )

(3)

2.1.2 3

041 15 2 416

( ) ( ) (4)

2.1.3

1 1

12 2

4 9 2

6 4 3

x x x

xx

(6)

2.1.4 2

1

3 3

3 3

x x

x x

(4)

2.2 Solve for :

3 4127

9( )x x

(4)

2.3

Simplify and write (show all working) 24

√96 √24 n the form of √

(5) [26]

QUESTION 3 Find the following products: 3.1 3 2 5 1 (2) 3.2 22

23( )x

(3)

3.3 2√3 2√3 (3)

[8] QUESTION 4 Simplify the following expressions fully: 3 2

5 5

9 4

10

(6) [6]

QUESTION 5 5.1 Consider the sequence 5 ; 13 ; 21 ;…

5.1.1

Write down the next two terms of the sequence.

(2)

5.1.2 Determine the expression for the nth term, Tn

(4)

5.1.3 Which term is 93 (2)

5.2 Temba opens a savings account and

decides the follow a pattern of savings his money so that he can boost his saving power each week. R28 ; R38; R44; R58; ...

5.2.1

Determine the next two amounts that he saves.

(2)

5.2.2 Determine the expression for the nth term, Tn

(7)

5.2.3

Calculate the amount that he would save in the tenth week. i.e.

(2)

5.2.4 Calculate the week that he would save R314.

(4) [23]

172

QUESTION 6 The denominator of a fraction is 2 less

than the numerator. If 5 is added to the numerator and 3 to the denominator, the new fraction will be equal to the original fraction. Find the original fraction.

(6) [6]

QUESTION 7

2( ) xf x k intersects 2( )g x ax bx c at A and

at B. B lies on the y‐axis. A (‐2; 8) is also turning point of g.

7.1) Determine the value of k (2)

7.2) The co‐ordinates of B. (1)

7.3)If 2c and 1 5,a find equation of g. (4)

7.4) For which values of x, ( ) ( )g x f x (2)

[10]

QUESTION 8

Given: 2 2 3( )f x x x and 2 1( ) xg x

8.1 Draw both functions on the same set of axes with showing all x and y intercept points and Turning point and asymptote (7)

8.2 Write down the axis of symmetry of f . (1) 8.3 What is the range of f (1) 8.4 What is the range of g (1)

[10]

TOTAL MARKS : 120

173

E. EXEMPLAR3 1. QUESTION 1

1.1. Solve for unknown:

1.2. 3 2 3 0( )( )x x (2)

1.3. 3 1

7 52

x (4)

1.4. 9 4

12 1x x

(5)

1.5. Determine the values of x and y:2 3x y

and 2 25 15x xy y (7)

[22]

2. QUESTION 2

2.1. Consider the sequence 2 ; ‐4 ; ‐4 ; 2 ; 14 ...

2.1.1. Write down next term of the

sequence. (1)

2.1.2. Find the general term of sequence.

(5)

2.1.3. Calculate the 30th term of the

sequence (2)

Consider the following; (Figure 1 has a perimeter of 4 units and an area of 1 unit square)

Figure 1 Figure 2 Figure 3 2.2. Write down the area and the perimeter of

Figure 4. (2)

2.2.1. Determine the general rule of Pn ,

the perimeter of the nth figure. (2)

2.2.2. Determine the general rule of An ,

the area of the nth figure. (2)

2.2.3. Which is the first figure having a

perimeter of more than 400 units. (2)

2.2.4. Find the perimeter of the figure that

has an area of 161 units square (3)

[19]

3. QUESTION 3

Simplify, without using calculator, the following completely;

3.1. 2 2 2

2

9 5 6 9

5 2 2 3 3

x x x x x

x x y x y xy

(7)

3.2. 2 1

2 1 1

6 50

15 8

.

.

x x

x x

(5)

3.3. 12 2

1 2

2

2

x y

y

(5)

[17]

4. QUESTION 4

4.1. If x<0, write the following expression in its

simplest form, without use of a calculator:

34

2 2

10000 8

25 9x x

(5)

4.2. Solve for x if 3 14 32x (5)

4.3. Given; 9 1

2 6 4K

x x

4.3.1. Check whether K is an integer or a

rational number if x=3,5 (3)

4.3.2. Determine the values of x for which

K is real number (3)

[16]

174

5. QUESTION 5

The graphs of and

are sketched below.

5.1. Write down the range of . (1)

5.2. Determine the values of a, p and q then

write the equation of f. (4)

5.3. Write equation of the axes of symmetry of

the function f, which has no intersection

with f. (3)

5.4. Determine the values of m and c then write

the equation of . (3)

5.5. Calculate the length of CD, distance

between the points C and D. (3)

5.6. Find the average gradient of in

between C and B. (2)

5.7. Determine the coordinates of the point A,

the intersection of f and g. (4)

5.8. Write the function ( )k x if f is translated 2

units down and 3 units right. (2)

5.9. Describe the transformation of g to h if

(2)

[24]

6. QUESTION 6

Given: and

6.1. Write down the coordinates of the turning

point of f. (2)

6.2. Draw a sketch graph of f on the set of axes

given in the DIAGRAM. (3)

Clearly indicate the coordinates of the

turning point as well as the intercepts with

the axes

6.3. Determine the range of the function f (1)

6.4. Write the equation of the axes of symmetry

of the function f. (1)

6.5. Find the x‐coordinates of intersections of f

and h, where . (3)

6.6. Draw a sketch graph of h on the same set of

axes in the DIAGRAM SHEET.

Indicate the turning point as well as the

intercepts with the axes and intersections

with f. (3)

6.7. Write the maximum value of the function h.

(1)

6.8. Determine the x values for which both f

and h increase as x increases (2)

6.9. Write a function k(x) for the distance

between f(x) and h(x) in between their

intersections, and then calculate the

maximum distance. (4)

[20]

( )a

f x qx p

( )g x mx c

( )f x

( )g x

( )f x

3 2( )h x x

2 2 8( )f x x x 22 2( ) ( )h x x

( ) ( )f x h x

175

7. QUESTION 7

7.1. A grade‐10 learner in Star College, Grayson,

wants to invest his R5000 for 3 years so that

he can pay the varsity fees after his

graduation from Star College. Bank A has

the following three options:

i) Tahir suggests a compound interest

rate of 15% p.a.

ii) Siviwe suggests a simple interest

rate of 16% p.a.

iii) Malusi suggests an interest rate of

14,5% p.a. compounded monthly

Which friend of Grayson recommends for

the best investment? Explain your answer

briefly. (5)

7.2. In 2010, Luvuyo buys a new Mercedes‐E200

for R600 000 which has the depreciation on

a reducing balance rate of 8% p.a. and an

inflation rate of 5% p.a.

7.3. Calculate the amount of money that will be

needed to re‐new the Mercedes in 2015 (4)

7.4. Determine the interest rate p.a.

compounded quarterly, if Luvuyo invests

R100 000 so that he will be able to re‐new

his car in 2015. (3)

7.5. If Akshay wants to buy the Mercedes‐E200

from Luvuyo for a R250 000, for how long

must he wait and in which year he is going

to buy that car? (3)

7.6. Ryan, Yasar and Nikhil have R100 000 all

together. They want to invest this money to

start up a business in 2020. Vahdet offers

them a compounded daily interest rate of

10,5% p.a. for the first 7 years and a

compounded semi‐annually interest rate of

13% p.a. for the rest of the period.

Accordingly;

7.6.1. How much does Vahdet promise to

give these 3 friends in 2020? (3)

7.6.2. Calculate the effective interest rate

p.a. for the first year of their

investment. (3)

[21]

8. QUESTION 8

8.1. Merdan is at a café. The waiter offers him

either a hot or a cold drink. The hot drinks

are tea, coffee and cappuccino and cold

drinks are iced tea and ayran. Each drink has

three sizes: Regular or large Draw tree

diagram and find number of choices of drink

(5)

8.2. How many different three‐digit number can

be formed using the digits in the set

{0;4;5;6;7;8;9} if :

8.2.1. no restriction? (2)

8.2.2. without repetition? (2)

[9]

9. QUESTION 9

9.1. The following (Stats SA release PO 305),

gives the registered births by province as

recorded for 2002. (in thousands.)

E Cape Limpopo Free State KZN N Cape

288 215 66 395 21

N West Gauteng Mpumalanga W Cape Other

105 204 114 102 8

If a randomly chosen person in SA, what is

the probability that this person comes

from

9.1.1. Mpumalanga? (1)

9.1.2. Mpumalanga or Northern Cape? (2)

9.1.3. Gauteng and the Free State? (2)

9.2. There are 130 Grade 10 learners in a school.

Sixty‐eight learners take Mathematics and

50 learners take Physical Science. Thirty‐two

learners take Mathematics and Physical

Science.

9.2.1. Draw a Venn diagram to illustrate

the given data. (3)

9.2.2. Hence calculate the probability that

a learner in Gr10, chosen at random:

a) Takes Physical Science (1)

b)Takes Mathematics but not Physical Science (1)

c)Takes Mathematics or Physical Science (1)

[11]

TOTAL 150

176

10 p1 2016 papers/grade 10/maths... · 2020-02-13 · 6 3 3 5 2 4 x y xy solution first equation:6...

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