AKK 1
Cambridge IGCSE Extended Mathematics Study Guide
Contents
Topic 1
Number
Topic 2
Algebra and graphs
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Topic 3
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Topic 4
Chapter 26
Chapter 27
Topic 5
Chapter 28
Topic 6
Chapter 29
Chapter 30
Chapter 31
Topic 7
Chapter 32
Chapter 33
Chapter 34
Topic 8
Chapter 35
Chapter 36
Topic 9
Chapter 37
Chapter 38
Chapter 39
Number and language
Accuracy
Calculations and order
Integers, fractions, decimals and percentages
Further percentages
Ratio and proportion
Indices and standard form
Money and finance
Time
Set notation and Venn diagrams
Algebraic representation and manipulation
Algebraic indices
Equations and inequalities
Linear programming
Sequences
Variation
Graphs in practical situations
Graphs of functions
Functions
Geometry
Geometrical vocabulary
Geometrical constructions and scale drawings
Similarity
Symmetry
Angle properties
Loci
Mensuration
Measures
Perimeter, area and volume
Coordinate geometry
Straight-line graphs
Trigonometry
Bearings
Trigonometry
Further trigonometry
Matrices and transformations
Vectors
Matrices
Transformations
Probability
Probability
Further probability
Statistics
Mean, median, mode and range
Collecting and displaying data
Cumulative frequency
AKK 2
Cambridge IGCSE Extended Mathematics Study Guide
1 Number and language
Natural numbers
Integers
Rational numbers
Irrational numbers
Real numbers
•
Venn diagram
Prime numbers
Square numbers
Cube numbers
Triangular numbers
Factors
Prime factors
Highest common factor
ℕ = {1, 2, 3, 4, … }
ℤ = { … , −3, −2, −1, 0, 1, 2, 3, … }
ℚ = ①whole numbers, ②fractions, ③terminating decimals, ④recurring decimals
𝑎𝑎
A rational number can be expressed as where a and b are integers, and b ≠ 0.
1
3
7
𝑏𝑏
153
2
e.g. 0.2 = , 0.3 =
, 7 = , 1.53 =
, 0. 2̇ =
5
10
1
100
9
non-terminating and non-repeating decimals
An irrational number cannot be expressed as a fraction.
3 √5
e.g. π , √2 ,
2
ℝ = the set of rational numbers and irrational numbers together
ℕ⊆ℤ⊆ℚ⊆ℝ
A prime number has factors only 1 and the number itself.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181,
191, 193, 197, 199, …
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, …
1, 8, 27, 64, 125, 216, 343, 512, 729, …
1, 3, 6, 10, 15, 21, 28, 36, …
e.g. The factors of 12 are all the numbers which will divide exactly into 12
i.e. 1, 2, 3, 4, 6, 12
12 = 1 × 12
=2 ×6
=3 ×4
e.g The prime factors of 12 are 2 and 3.
Multiples
Least common multiples
e.g The prime factors of 84 = 22 × 3 × 7
The prime factors of 180 = 22 × 32 × 5
HCF of 84 and 180 = 22 × 3 = 12
e.g. Multiples of 5 are 5, 10, 15, 20, 25, etc.
e.g The prime factors of 84 = 22 × 3 × 7
The prime factors of 180 = 22 × 32 × 5
LCM of 84 and 180 = 22 × 32 × 5 × 7 = 1260
AKK 3
Cambridge IGCSE Extended Mathematics Study Guide
2 Accuracy
Approximation
In many instances exact numbers are not necessary or even desirable. In those circumstances approximations
are given. The common types of approximation are:
• Rounding
to the nearest 10 000
28 571 people ���������������� 30 000
to the nearest 1000
•
•
28 571 people �������������� 29 000
to the nearest 100
28 571 people �������������� 28 600
Decimal places
to 1 d.p.
7.864 ����� 7.9 (The second decimal digit is 5 or more than 5, the first decimal digit is rounded up.)
to 2 d.p.
7.864 ����� 7.86 (The third decimal digit is less than 5, the second decimal digit is not rounded up.)
Significant figures
to 3 s.f.
43.25 ����� 43.3
In the number 43.25, the 4 is the most significant figure as it has a value of 40.)
In contrast, the 5 is the least significant figure as it only has a value of 5 hunderdths.)
1 Round the following numbers to nearest 1000:
(a) 68 786
(b) 74 245
(d) 4020
(e) 99 500
2 Round the following numbers to nearest 100:
(a) 78 540
(b) 6858
(d) 8084
(e) 950
3 Round the following numbers to nearest 10:
(a) 485
(b) 692
(d) 83
(e) 4
4 Give the following to 1 d.p.
(a) 5.58
(d) 157.39
5 Give the following to 2 d.p.
(a) 6.473
(d) 0.088
(c) 89 000
(f) 999 999
(c) 14 099
(f) 2984
(c) 8847
(f) 997
(b) 0.73
(e) 4.04
(c) 11.86
(f) 15.045
(b) 9.587
(e) 0.014
(c) 16.476
(f) 9.3048
6 Write the following to the number of significant figures written in brackets:
(a) 48 599(1 s.f.)
(b) 48 599(3 s.f.)
(c)6841(1 s.f.)
(d) 7538(2 s.f.)
(e) 483.7(1 s.f.)
(f) 2.5728(3 s.f.)
(g) 990(1 s.f.)
(h) 2045(2 s.f.)
(i) 14.952(3 s.f.)
7 Write the following to the number of significant figures written in brackets:
(a) 0.085 62(1 s.f.)
(b) 0.5932(1 s.f.)
(c) 0.942(2 s.f.)
(d) 0.954(1 s.f.)
(e) 0.954(2 s.f.)
(f) 0.003 05(1 s.f.)
(g) 0.003 05(2 s.f.)
(h) 0.009 73(2 s.f.)
(i) 0.00973(1 s.f.)
Appropriate accuracy
In the examination, your answer should be correct to three significant figures.
Answers in degrees should be given to one decimal place.
Example: Calculate 4.64 ÷ 2.3 giving your answer to an appropriate degree of accuracy.
4.64 ÷ 2.3 ≈ 2.0 (correct to 1 d.p.)
AKK 4
Cambridge IGCSE Extended Mathematics Study Guide
Upper and lower bounds
Numbers can be written to different degrees of accuracy. For example 4.5, 4.50 and 4.500 .
4.5 is rounded to one decimal place and therefore any number from 4.45 up to but not including 4.55 would
be rounded to 4.5 . On a number line this would be represented as
4.45 ⩽ x < 4.55
4.45 is known as the lower bound of 4.5 , whilst 4.55 is known as the upper bound.
Example: If the measurements for the length and the width are correct to nearest metre, what is the largest
and smallest possible areas of the rectangle?
Correct to nearest metre ⇒ 1m
1m
= 0.5m
2
The upper bound and the lower bound of the length are
4.5m ⩽ l < 5.5m
The upper and the lower bound of the width are
2.5m ⩽ w < 3.5m
The largest possible area = 5.5m × 3.5m = 19.25m2
The smallest possible area = 4.5m × 2.5m = 11.25m2
•
Ordering symbols
= is equal to
≠ is not equal to
> is greater than
≥ is greater than or equal to
< is less than
≤ is less than or equal to
3 Calculations and order
𝑥𝑥 < 5
𝑥𝑥 ≥ −7
0 ≤ 𝑥𝑥 ≤ 4
The order of operations
The priorities for operations are:
(1) brackets
(2) multiplication/division
(3) addition/subtraction
1 Evaluate the following:
(a) 3 + 4 ÷ 2 × 4
−4 < 𝑥𝑥 ≤ −1
(b) 14 × 2 ÷ (9 − 2)
AKK 5
Cambridge IGCSE Extended Mathematics Study Guide
4 Integers, fractions, decimals and percentages
Fractions
1
2
7
A single unit can be broken into equal parts called fractions, e.g. , ,
2
2
In the fraction :
3
The three is called the denominator.
The two is called the numerator.
3
8
3
A proper fraction has its numerator less than its denominator, e.g. .
4
9
An improper fraction has its numerator more than its denominator, e.g. .
1
• Changing a decimal to a fraction
1 Write the following fractions as decimals:
1
1
(a)
(b)
(c)
(d)
2 Write the following fractions as decimals:
5
27
(a) 4
(b) 3
(c) 9
2
A mixed number is made up of a whole number and a proper fraction, e.g. 4 .
(f)
2
1
(g)
5
3
2
(h)
5
10
100
• Changing a recurring decimal to a fraction
𝑎𝑎
(i) Converting the terminating decimal to the form
0.125 =
125
1000
=
2
3
1
(i)
8
5
100
𝑏𝑏
1
8
𝑎𝑎
(ii) Converting the recurring decimal to the form
𝑏𝑏
Let 𝑥𝑥 = 0. 1̇ 47̇
1000 𝑥𝑥 = 147. 1̇ 47̇
1000 𝑥𝑥 − 𝑥𝑥 = 147. 1̇ 47̇ − 0. 1̇ 47̇
999 𝑥𝑥 = 147
147
𝑥𝑥 =
999
(iii) Rationalising the denominator
2
3√5
=
2
3√5
×
√5
√5
=
2√5
15
(iv) Finding irrational numbers in a given area
For example, find an irrational number between 5 and 6.
√25 = 5
√36 = 6
Therefore, √27 or 3√3 is an irrational number between 5 and 6 .
Otherwise, π + 2 is also a possible number.
Percentages
A fraction whose denominator is 100 can be expressed as a percentage.
29
45
e.g.
= 29% ,
= 45%
100
100
• Calculating a percentage of a quantity
Example: Find 25% of 300 miles.
25% × 300 miles = 0.25 × 300 = 75 miles
2
1
(e)
4
2
(j)
8
(d) 4
6
1000
3
4
1
10
(e) 9
25
100
AKK 6
Cambridge IGCSE Extended Mathematics Study Guide
Fraction
1
2
1
4
1
8
3
8
7
10
4
10
6
10
8
10
10
or
3
10
or
or
7
10
or
9
10
0.75
75%
12.5%
0.625
62.5%
0.875
0.1
1
0.2
5
0.3
2
0.4
5
3
0.6
5
0.7
4
0.8
5
0.9
5 Further percentages
25%
0.125
0.375
8
5
2
Percentage
50%
0.25
4
3
8
1
Decimal
0.5
37.5%
87.5%
10%
20%
30%
40%
60%
70%
80%
90%
• Expressing one quantity as a percentage of another
Example: In an examination a girl obtains 69 marks out of 75. Express this result as a percentage.
69
× 100% = 92%
75
• Percentage increase and decrease
Example 1: A shop assistant has a salary of $16 000 per month. If his salary increases by 8% , calculate:
(i) the amount extra he receives a month,
(ii) his new monthly salary.
(i) Increase = 8% of $16000 = 0.08 × 16000 = $1280
(ii) New salary = old salary + increase = 16000 + 1280 = $17 280 per month
Example 2: A shop is having a sale. It sells a set of tools costing $130 at a 15% discount. Calculate the sale
price of the tools.
The new price = (100% − 15%) of $130 = 85% × 130 = $110.50
10% increase = multiplied by 1.1
50% increase = multiplied by 1.5
10% decrease = multiplied by 0.9
25% decrease = multiplied by 0.75
15% decrease = = multiplied by 0.85
• Reverse percentages
Example: In a test Ahmed answered 92% of the questions correctly. If he answered 23 questions correctly,
how many had he got wrong?
92% of the questions is equivalent to 23 questions.
23
1% the questions is equivalent to questions.
92
23
100% the questions is equivalent to × 100 = 25 questions.
92
Ahmad got 2 questions wrong.
AKK 7
Cambridge IGCSE Extended Mathematics Study Guide
6 Ratio and proportion
Direct proportion is when two quantities increase or decrease in the same ratio.
Example: A machine prints four books in 10 minutes. How many books will it print in 2 hours?
The ratio method:
4 books
𝑥𝑥 books
=
10 minutes
120 minutes
10 𝑥𝑥 = 480
𝑥𝑥 = 48 books
The unitary method:
10 minutes ⟹ 4 books
1 minute
⟹ 0.4 book
120 minutes ⟹ ?
120
⟹0.4×
= 48 books
1
Divide a quantity in a given ratio
Example: Divide 20m in the ratio 3:2 .
3:2 gives 5 parts.
3
× 20m = 12m
5
2
5
× 20m = 8m
Inverse proportion is when one value increases as the other decreases.
As the length of the rectangle doubles, the width has to be halved for the area to stay the same.
Example: If 8 people can pick the apples from the trees in 6 days, how long will it take 12 people?
8 people ⟹ 6 days
1 person ⟹ 6 × 8 days
6×8
12 people ⟹
= 4 days
12
Increase and decrease by a given ratio
Example 1: A photograph is 12 cm wide and 8 cm tall. It is enlarged in the ratio 3:2 . What are the dimensions
of the enlarged photograph?
3
The enlarged width = 12 × = 18 cm
The enlarged height = 8 ×
2
3
2
= 12 cm
Example 2: A photographic transparency 5 cm wide and 3 cm tall is projected onto a screen. If the image is 1.5
m wide:
(i) calculate the ratio of the enlargement,
(ii) calculate the height of the image.
(i) 5 cm width ⟹ 150 cm width
150
1 cm width ⟹
= 30 cm width
5
Therefore, the enlargement ratio is 1:30 .
(ii) Enlarged height = 3 × 30 = 90 cm
AKK 8
Cambridge IGCSE Extended Mathematics Study Guide
7 Indices and standard form
Law of indices
(1) 𝑎𝑎𝑚𝑚 × 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑚𝑚+𝑛𝑛
𝑎𝑎𝑚𝑚
(2) 𝑎𝑎𝑚𝑚 ÷ 𝑎𝑎𝑛𝑛 = 𝑛𝑛 = 𝑎𝑎𝑚𝑚−𝑛𝑛
𝑎𝑎
(3) (𝑎𝑎𝑚𝑚 )𝑛𝑛 = 𝑎𝑎𝑚𝑚𝑚𝑚
The zero index
𝑎𝑎𝑚𝑚
= 1
𝑎𝑎𝑚𝑚
𝑚𝑚−𝑚𝑚
𝑎𝑎
=1
𝑎𝑎0 = 1
Negative indices
𝑎𝑎0
𝑎𝑎−𝑚𝑚 = 𝑎𝑎0−𝑚𝑚 = 𝑚𝑚
𝑎𝑎
1
−𝑚𝑚
= 𝑚𝑚
𝑎𝑎
𝑎𝑎
Exponential equations
Equations that involve indices as unknown are called as exponential equations.
Example: Find the value of x if 2𝑥𝑥 = 32 .
2𝑥𝑥 = 25
𝑥𝑥 = 5
Standard form
Example 1: Write 72 000 in standard form.
72 000 = 7.2 × 10 000 = 7.2 × 104
Example 2: Write 5 × 104 as an ordinary number.
5 × 104 = 5 × 10 000 = 50 000
Negative indices
A negative index is used when writing a number between 0 and 1 in standard form.
e.g. 100
= 1 × 102
10
= 1 × 101
1
= 1 × 100
0.1
= 1 × 10−1
0.01 = 1 × 10−2
0.001 = 1 × 10−3
0.0001 = 1 × 10−4
Example: Write 1.8 × 10−4 as an ordinary number.
1.8 × 10−4 = 1.8 ÷ 104 = 1.8 ÷ 10000 = 0.00018
Fractional indices
1
𝑛𝑛
𝑎𝑎𝑛𝑛 = √𝑎𝑎
4
𝑚𝑚
𝑛𝑛
𝑛𝑛
𝑎𝑎 𝑛𝑛 = √𝑎𝑎𝑚𝑚 = ( √𝑎𝑎 )𝑚𝑚
Example: Evaluate √4096 without the use of a calculator.
4
4
3
3
4×
√4096 = √163 = 164 = (24 )4 = 2
Evaluate
2
273 ×3−3
1
−
9 2
.
3
4
= 23 = 8
AKK 9
Cambridge IGCSE Extended Mathematics Study Guide
8 Money and finance
• Currency conversions
In 2017, 1 US dollar could be exchanged for 1350 Myanmar Kyats.
Example: (a) How many US dollars can be bought for 100 000 MMK?
1 US$
100 000 MMK = 100 000 MMK ×
= US$ 74.07
1350 MMK
(b) Convert 350 Australian dollars to Myanmar Kyat if 1 AUD = 0.74348 US$ .
0.74348 US$
1350 MMK
350 AUD = 350 AUD ×
×
= 351 294.30 MMK
1 AUD
1 US$
• Earnings
Net pay is what is left after deductions such as tax, insurance and pension contributions are taken from gross
earnings.
Net pay = Gross pay − Deductions
A bonus is an extra payment sometimes added to an employee’s basic pay.
In many companies there is a fixed number of hours that an employee is expected to work. Any work in
excess of this basic week is paid at a higher rate, referred to as overtime.
Overtime may be 1.5 times basic pay, called time and a half, or twice basic pay, called double time.
Piece work is another method of payment. Employees are paid for the number of articles made, not for the
time taken.
1 Mr Ahmet’s gross pay is $188.25 . Deductions amount to $33.43 . What is his net pay?
2 Miss Sally’s basic pay is $128. She earns $36 for overtime and receives a bonus of $18. What is her gross
pay?
3 Mrs Hafar’s gross pay is $203. She pays $54 in tax and $18 towards her pension. What is her net pay?
4 Mr Wong works 35 hours for an hourly rate of $8.30 . What is his basic pay?
5 Maria is paid €5 for every 12 plates that she makes. This is her record for one week.
Mon
240
Tues
360
Wed
288
Thurs 192
Fri
180
How much is she paid?
AKK 10
Cambridge IGCSE Extended Mathematics Study Guide
Profit and loss
Foodstuffs and manufactured goods are produced at a cost, known as the cost price, and sold at the selling
price. If the selling price is greater than the cost price, a profit is made.
Profit = Selling price − Cost price
Loss = Cost price − Selling price
Example 1: A market trader buys oranges in boxes of 12 dozen for $14.40 per box. He buys three boxes and
sells all the oranges for 12c each. What is his profit or loss?
Cost price:
3 × $14.40 = $43.20
Selling price: 3 × 144 × 12c = $51.84
Profit = Selling price − Cost price
= $51.84 − $43.20
= $8.64
Example 2: In a sale a skirt usually costing $35 is sold at a 15% discount. What is the discount?
Discount = 15% of $35 = 0.15 × $35 = $5.25
1 A market trader buys peaches in boxes of 120. He buys 4 boxes at a cost of $13.20 per box. He sells 425
peaches at 12c each— the rest are ruined. How much profit or loss does he make?
2 A car is priced at $7200. The car dealer allows a customer to pay a one-third deposit and 12 payments of
$420 per month. How much extra does it cost the customer?
Percentage profit and loss
profit or loss
× 100%
cost price
Example: A woman buys a car for $7500 and sells it two years later for $4500. Calculate her loss as a
percentage of the cost price.
𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 =
× 100%
Percentage profit or loss =
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
7500 – 4500
=
× 100%
7500
= 40%
When something becomes worth less over a period of time, it is said to depreciate.
Interest
Interest can be defined as money added by a bank to sums deposited by customers.
The money deposited is called the principal.
The percentage interest is the given rate and the money is left for a fixed period of time.
•
Simple interest
r
×t
100
where P = the principal amount
t = time in years
r = interest rate per annum
Example: Find the simple interest rate earned on $ 250 deposited for 6 years at 8% p.a.
8
Simple interest = 250 ×
× 6 = $120
Simple interest = P ×
100
How long will it take for a sum of $250 invested at 8% to earn interest of $80?
AKK 11
Cambridge IGCSE Extended Mathematics Study Guide
•
Compound interest
r t
)
100
r
Compound interest = P(1 +
)t − P
100
Example: Find the compound interest when $3000 is invested for 18 months at an annual payment
rate(APR) of 8.5% . The interest is calculated every six months.
8.5
Interest rate for six months period, r =
= 4.25%
Compounded amount = P(1 +
The number of times that the interest is added, t =
2
18
6
= 3 times
Compounded amount = P(1 +
•
r
100
)t
4.25
= 3000 (1 +
)3 = $3398.99
100
Compound interest = $3399 − $3000 = $399
Debt at compound interest
r t
)
100
Example 1: How long will it take for a debt to double at a compound interest rate of 27% p.a. ?
r t
)
Compounded debt = P(1 +
100
27 t
2P = P(1 +
)
100
t
2 = 1.27
log 2
t=
= 2.89̇
Compounded debt = P(1 +
log 1.27
t = 3 years
Example 2: 8 million tonnes of fish were caught in the North sea in 2005. If the catch is reduced by 20% each
year for 4 years, what weight is caught at the end of this time?
r
Compounded amount = P(1 −
)t
100
20
= 8(1 −
)4
100
= 3.28 million tonnes
1 What rate of simple interest is paid on a deposit of $2000 which earns $400 interest in 5 years?
2 An electrician bought five broken washing machines for $550. He repaired them and sold them for $143
each. What was his percentage profit?
3 A car is bought for $12 500. Its value depreciates by 15% per year.
(a) Calculate its value after:
(i) 1 year
(ii) 2 years
(b) After how many years will the car be worth less than $1000?
AKK 12
Cambridge IGCSE Extended Mathematics Study Guide
9 Time
Times may be given in terms of the 12-hour clock.
e.g. ‘I get up at seven o’clock in the morning, play football at half past two in the afternoon, and go to bed
before eleven o’clock.’
When using the 24-hour clock,
7 a.m. is written as 07 00
2:30 p.m. is written as 14 30
11:00 p.m. is written as 23 00
Example: A train covers 480 km journey from Paris to Lyon at an average speed of 100 km/h. If the train
leaves Paris at 08 35, when does it arrive in Lyon?
distance
time =
=
speed
480 km
100 km/h
= 4.8 hours = 4 hours 48 minutes
Arrival time = 08 35 + 04 48 = 13 23
1 A journey to school takes a boy 22 minutes. What is the latest time he can leave home if he must be at school
at 08 40?
2 A plane leaves London for Boston, a distance of 5200 km, at 09 45. The plane travels at an average speed of
800 km/h. If Boston time is five hours behind British time, what is the time in Boston when the aircraft lands?
10 Set notation and Venn diagrams
A set is a collection of objects, things or symbols which are clearly identified.
The individual objects in the set are called the elements or members of the set.
Elements may be specified in two ways:
• by description
• by listing the elements.
For example,
Description
The set of vowels in the alphabet
The set of positive even numbers
Listing
{a, e, i, o, u}
{2, 4, 6, 8, 10, 12, …}
Number of elements in set A is written as n(A).
If A = {1, 3, 5, 7} , then n(A) = 4.
Membership of the set is written as:
∈ (is a member of) (belongs to)
∉ (is not a member of) (does not belong to)
Let A = {1, 3, 5, 7}
1∈A
6∉A
Finite sets are the sets in which it is possible to list all the elements.
For example, the set of days in a week is a finite set.
AKK 13
Cambridge IGCSE Extended Mathematics Study Guide
Infinite sets are the sets in which it is impossible to list all the elements.
For example, the set of prime numbers is an infinite set.
The universal set(E ) is the set which contains all the possible elements for a particular case.
For example, E is the set of all letters in the alphabet.
V is the set of vowels in the alphabet.
C is the set of consonants in the alphabet.
E
Subsets
If all the elements of one set X are also elements of another set Y, then X is said to be a subset of Y.
Y = {1, 2, 3, 4, 5}
X = {1, 2, 3}
Z = {8, 9}
W = {1, 2, 3, 4, 5}
Then, W ⊆ Y (W is a subset of Y)
X ⊂ Y (X is a proper subset of Y)
Z ⊈ Y (Z is not a subset of Y)
The empty set { } or ∅ is one which contains no elements.
An empty set is a subset all sets.
∅⊆X
∅⊆Y
Number of all subsets of a set containing n elements is given by the formula 𝟐𝟐𝒏𝒏 .
For example,
• A = {a}
n(A) = 1
The all subsets of A are {a} and ∅ .
The number of all subsets of A = 21 = 2
• B = {a, b}
n(B) = 2
The all subsets of B are {a}, {b}, {a, b} and ∅ .
The number of all subsets of B = 22 = 4
• C = {a, b, c}
n(C) = 3
The all subsets of C are {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c} and ∅ .
The number of all subsets of C = 23 = 8
The complement of a set(Aˈ) is the set of elements which are in E but not in A .
E
The intersection of sets is used to denote common elements.
The union of sets is everything which belongs to either or both sets.
For example, A = {1, 2, 3, 4, 5} and B = {1, 4, 9}
A ∩ B = {1, 4}
A ∪ B = {1, 2, 3, 4, 5, 9}
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Cambridge IGCSE Extended Mathematics Study Guide
A∩B
A∪B
A∩B∩C
A∪B∪C
Bˈ ∩ A
(A ∩ B)ˈ
(A ∪ B)ˈ
Aˈ ∩ B
Aˈ ∪ B
(A ∪ B)ˈ ∪ (A ∩ B)
(A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C)
De Morgan’s Laws
(A ∩ B)ˈ = Aˈ ∪ Bˈ
(A ∪ B)ˈ = Aˈ ∩ Bˈ
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
(A ∪ C)ˈ ∩ B
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
1 Describe the following sets in words:
(a) {2, 4, 6, 8}
(b) {2, 4, 6, 8, … }
(c) {1, 4, 9, 16, 25, … }
2 Calculate the value of n(A) for each of the sets shown below:
(a) A = {days in a leap year}
(b) A = {prime numbers between 50 and 60}
(c) A = {x: x is an integer and 5 ⩽ x ⩽ 10}
3 In a class of 31 students, some study Physics and some study Chemistry. If 22 study Physics, 20 study
Chemistry and 5 study neither, calculate the number of students who take both subjects.
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11 Algebraic representation and manipulation
(a) Formulae
( 𝑎𝑎 + 𝑏𝑏 )2 = 𝑎𝑎2 + 2𝑎𝑎𝑎𝑎 + 𝑏𝑏 2
( 𝑎𝑎 − 𝑏𝑏 )2 = 𝑎𝑎2 − 2𝑎𝑎𝑎𝑎 + 𝑏𝑏 2
( 𝑎𝑎 + 𝑏𝑏 )3 = 𝑎𝑎3 + 3𝑎𝑎2 𝑏𝑏 + 3𝑎𝑎𝑏𝑏 2 + 𝑏𝑏 3
( 𝑎𝑎 − 𝑏𝑏 )3 = 𝑎𝑎3 − 3𝑎𝑎2 𝑏𝑏 + 3𝑎𝑎𝑏𝑏 2 − 𝑏𝑏 3
𝑎𝑎2 − 𝑏𝑏 2 = ( 𝑎𝑎 + 𝑏𝑏 )( 𝑎𝑎 − 𝑏𝑏 )
𝑎𝑎3 + 𝑏𝑏 3 = ( 𝑎𝑎 + 𝑏𝑏 )(𝑎𝑎2 − 𝑎𝑎𝑎𝑎 + 𝑏𝑏 2 )
𝑎𝑎3 − 𝑏𝑏 3 = ( 𝑎𝑎 − 𝑏𝑏 )(𝑎𝑎2 + 𝑎𝑎𝑎𝑎 + 𝑏𝑏 2 )
(b) Simplifying: Multiply out brackets and gather up like terms
For example, 3𝑥𝑥( 𝑥𝑥 + 2𝑦𝑦 ) − 2𝑦𝑦( 3𝑥𝑥 − 𝑦𝑦 ) = 3𝑥𝑥 2 + 6𝑥𝑥𝑥𝑥 − 6𝑥𝑥𝑥𝑥 + 2𝑦𝑦 2 = 3𝑥𝑥 2 + 2𝑦𝑦 2
(c) Factoring
(i) extracting the highest common factor
For example, 6𝑥𝑥 2 − 3𝑥𝑥𝑥𝑥 = 3𝑥𝑥(2𝑥𝑥 − 𝑦𝑦)
(ii) quadratics
(a) no number term:
(b) no 𝑥𝑥 term:
(c) full quadratic:
(d) grouping:
1 Factorise the following fully:
(a) 𝑚𝑚𝑚𝑚 − 5𝑚𝑚 − 5𝑛𝑛𝑛𝑛 + 25𝑛𝑛
𝑥𝑥 2 − 2𝑥𝑥 = 𝑥𝑥(𝑥𝑥 − 2)
𝑥𝑥 2 − 16 = (𝑥𝑥 + 4)(𝑥𝑥 − 4)
2𝑥𝑥 2 + 9𝑥𝑥 + 4 = (2𝑥𝑥 + 1)(𝑥𝑥 + 4)
𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑎𝑎 − 𝑏𝑏 2 − 𝑏𝑏𝑏𝑏 = 𝑎𝑎(𝑏𝑏 + 𝑐𝑐) − 𝑏𝑏(𝑏𝑏 + 𝑐𝑐) = (𝑏𝑏 + 𝑐𝑐)(𝑎𝑎 − 𝑏𝑏)
(b) 3𝑥𝑥 2 + 3𝑥𝑥 − 18
(c) 4𝑥𝑥 2 − 81𝑦𝑦 2
(d) 𝑥𝑥 4 − 𝑦𝑦 4
2 Make the letter in bold the subject of the formula:
(a) 𝑚𝑚𝒇𝒇2 = 𝑝𝑝
(c)
1
𝒑𝒑 − 3𝑞𝑞
=
5
𝑝𝑝 + 𝑞𝑞
3 Simplify the following algebraic fractions:
(a)
(c)
7𝑏𝑏3
𝑐𝑐
÷
4𝑏𝑏2
3𝑐𝑐 3
2(𝑦𝑦 + 4)
3
− (𝑦𝑦 − 2)
(b) 𝐴𝐴 = 𝜋𝜋 𝑟𝑟�𝒑𝒑 + 𝑞𝑞
(d) 𝑟𝑟(𝑠𝑠 − 𝒕𝒕) = 2𝑡𝑡 + 𝑟𝑟
(b)
(d)
−(𝑥𝑥 − 2)
6
𝑎𝑎2 − 𝑏𝑏2
(𝑎𝑎 + 𝑏𝑏)2
−
3𝑥𝑥 + 2
2
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Cambridge IGCSE Extended Mathematics Study Guide
12 Algebraic indices
• Positive indices
Simplify the following using indices:
(a) 𝑑𝑑 × 𝑑𝑑 × 𝑒𝑒 × 𝑒𝑒 × 𝑒𝑒 × 𝑒𝑒 × 𝑒𝑒
•
Since
Negative indices
𝑎𝑎−𝑚𝑚 = 𝑎𝑎0−𝑚𝑚
Therefore
Simplify:
(a)
(b) 𝑝𝑝3 × 𝑝𝑝2 × 𝑞𝑞4 × 𝑞𝑞5
(𝑒𝑒 4 )5
𝒂𝒂
−𝒎𝒎
=
=
=
𝑎𝑎0
𝑎𝑎𝑚𝑚
1
𝑎𝑎𝑚𝑚
𝟏𝟏
𝒂𝒂𝒎𝒎
(b)
𝑒𝑒 14
•
Since
(ℎ−2 ×ℎ−5 )−1
ℎ0
Zero index
𝑎𝑎𝑚𝑚 ÷ 𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑚𝑚−𝑛𝑛
𝑎𝑎𝑚𝑚 ÷ 𝑎𝑎𝑚𝑚 = 𝑎𝑎𝑚𝑚−𝑚𝑚
𝑎𝑎𝑚𝑚
Therefore
•
= 𝑎𝑎0
𝒂𝒂 = 1
Fractional indices
Since
162 = √16
In general:
𝑎𝑎𝑚𝑚
𝟎𝟎
1
𝟏𝟏
𝒏𝒏
𝑚𝑚
𝑛𝑛
𝒏𝒏
and
𝒂𝒂 = √𝒂𝒂
𝑛𝑛
𝑛𝑛
𝑎𝑎 = √𝑎𝑎𝑚𝑚 = ( √𝑎𝑎 )𝑚𝑚
1
3
273 = √27
𝑛𝑛
Simplify the following algebraic expressions, giving your answer in the form ( √𝑡𝑡)𝑚𝑚 :
(i)
2
√𝑡𝑡 × 𝑡𝑡 3
1
−
𝑡𝑡 3
• Simple linear equations
Solve the following linear equations:
(i) 2(𝑥𝑥 + 1) = 3(𝑥𝑥 − 5)
(ii)
3
√𝑡𝑡
2
−
𝑡𝑡 2 ×𝑡𝑡 5
13 Equations and inequalities
• Constructing equations
1 Find the value of x in the following isosceles triangle.
(ii)
2𝑥𝑥 + 3
4
=
4𝑥𝑥 − 2
6
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Cambridge IGCSE Extended Mathematics Study Guide
2 Using angle properties, calculate the value of x in the following diagram:
• Simultaneous equations
1 Solve the following simultaneous equations by elimination method.
2𝑥𝑥 − 5𝑦𝑦 = −8
−3𝑥𝑥 − 2𝑦𝑦 = −26
2 Find the value of x and y by substitution method.
3 Calculate the value of x.
• Solving quadratic equations by factorizing
1 Solve the following quadratic equations:
(i) 𝑥𝑥 2 − 2𝑥𝑥 = 63
(ii) 16𝑥𝑥 2 −
1
25
=0
2 A triangle has a base length of (𝑥𝑥 − 8) cm and a height of 2𝑥𝑥 cm. If its area is 20 cm2 calculate its height and
base length.
• The quadratic formula
A quadratic equation in the form 𝑎𝑎𝑎𝑎 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 can be solved by the use of the quadratic formula:
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
𝑥𝑥 =
2𝑎𝑎
Solve the quadratic equation 𝑥𝑥 2 + 7𝑥𝑥 + 3 = 0.
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Cambridge IGCSE Extended Mathematics Study Guide
• Completing the square
Example: Solve the quadratic equation by completing the square.
𝑥𝑥 2 − 𝟒𝟒𝑥𝑥 − 3 = 0
𝐴𝐴(𝑥𝑥 − 𝐵𝐵)2 − 𝐵𝐵2 + 𝐶𝐶 = 0
4
4 ⟹ 𝐵𝐵 = = 𝟐𝟐
2
(𝑥𝑥 − 𝟐𝟐)2 − 𝟐𝟐2 − 3 = 0
(𝑥𝑥 − 𝟐𝟐)2 − 7 = 0
(𝑥𝑥 − 2)2 = 7
𝑥𝑥 − 2 = ±√7
𝑥𝑥 = 2 ± √7
𝑥𝑥 = 4.656 or 𝑥𝑥 = −0.646
Solve the quadratic equation by completing the square.
(i) 𝑥𝑥 2 + 5𝑥𝑥 − 7 = 0
(ii) −7𝑥𝑥 2 − 𝑥𝑥 + 15 = 0
• Linear inequalities
Example 1: Solve the following inequalities and represent the solution on a number line.
(i) 15 + 3𝑥𝑥 < 6
(ii)
17 ⩽ 7𝑥𝑥 + 3
3𝑥𝑥 < 6 − 15
17 − 3 ⩽ 7𝑥𝑥
3𝑥𝑥 < −9
14 ⩽ 7𝑥𝑥
−9
𝑥𝑥 <
2 ⩽ 𝑥𝑥
3
𝑥𝑥 < −3
𝑥𝑥 ⩾ 2
Solve the following inequalities and represent the solution on a number line.
(a) 5 < −𝑥𝑥 − 7
(b) 1 ⩾ −3𝑥𝑥 + 7
Example 2: Find the range of values for which 7 < 3𝑥𝑥 + 1 ⩽ 13 and illustrate the solutions on a number line.
7 < 3𝑥𝑥 + 1
and
3𝑥𝑥 + 1 ⩽ 13
6 < 3𝑥𝑥
3𝑥𝑥 ⩽ 12
2 < 𝑥𝑥
𝑥𝑥 ⩽ 4
𝑥𝑥 > 2
and
𝑥𝑥 ⩽ 4
2 < 𝑥𝑥 ⩽ 4
Solve the inequality 5 < 3𝑥𝑥 + 2 ⩽ 17. Illustrate your answer on a number line.
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14 Linear programming
• Graphing an inequality
Example 1: On a pair of axes, shade the region which satisfies the inequality ⩾ 3 .
The line 𝑥𝑥 = 3 is drawn first.
Example 1: On a pair of axes, shade the region which satisfies the inequality ⩽ 𝑥𝑥 + 2 .
The line 𝑦𝑦 = 𝑥𝑥 + 2 is drawn first.
x 0 5
y 2 7
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Cambridge IGCSE Extended Mathematics Study Guide
• Graphing more than one inequality
Example: On the same pair of axes leave unshaded the regions which satisfy the following inequalities
simultaneously.
𝑥𝑥 ⩽ 2
𝑦𝑦 > −1
𝑦𝑦 ⩽ 3
𝑦𝑦 ⩽ 𝑥𝑥 + 2
On the same pair of axes leave unshaded the regions which satisfy the following inequalities simultaneously.
2𝑦𝑦 ⩾ 𝑥𝑥 + 4
𝑦𝑦 ⩽ 2𝑥𝑥 + 2
𝑦𝑦 < 4
𝑥𝑥 ⩽ 3
AKK 21
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15 Sequences
A sequence is a collection of terms arranged in a specific order according to a rule.
2, 4, 6, 8, 10
1, 4, 9, 16, 25
1, 2, 4, 8, 16
5 5 5
1, 1, 2, 3, 5, 8
1, 8, 27, 64, 125
10, 5, , ,
2 4 8
• Arithmetic sequences
In an arithmetic sequence there is a common difference(d) between successive terms.
Formulae for the terms of an arithmetic sequence
① The term-to-term rule
② The formula for the 𝑛𝑛𝑡𝑡ℎ term
𝑈𝑈𝑛𝑛+1 = 𝑈𝑈𝑛𝑛 + 5
1 Find the rule for the 𝑛𝑛𝑡𝑡ℎ term of the sequence 12, 7, 2, −3, −8, …
2 (i) Find the rule for the 𝑛𝑛𝑡𝑡ℎ term of the sequence 0, 4, 8, 12, 16 .
(ii) Calculate the 10𝑡𝑡ℎ term of the sequence.
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Cambridge IGCSE Extended Mathematics Study Guide
• Sequences with quadratic and cubic rules
Example 1: Deduce the rule for the 𝑛𝑛𝑡𝑡ℎ term for the sequence 0, 7, 18, 33, 52, …
Let 𝑈𝑈𝑛𝑛 = 𝑎𝑎𝑛𝑛2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
2𝑎𝑎 = 4
3𝑎𝑎 + 𝑏𝑏 = 7
𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 = 0
therefore
therefore
therefore
𝑎𝑎 = 2
𝑏𝑏 = 1
𝑐𝑐 = −3
𝑈𝑈𝑛𝑛 = 2𝑛𝑛2 + 𝑛𝑛 − 3
Example 2: Deduce the rule for the sequence −6, −8, −6, 6, 34, …
Let 𝑈𝑈𝑛𝑛 = 𝑎𝑎𝑛𝑛3 + 𝑏𝑏𝑛𝑛2 + 𝑐𝑐𝑐𝑐 + 𝑑𝑑
6𝑎𝑎 = 6
12𝑎𝑎 + 2𝑏𝑏 = 4
7𝑎𝑎 + 3𝑏𝑏 + 𝑐𝑐 = −2
𝑎𝑎 + 𝑏𝑏 + 𝑐𝑐 + 𝑑𝑑 = −6
𝑎𝑎 = 1
therefore
therefore
𝑏𝑏 = −4
therefore
𝑐𝑐 = 3
𝑑𝑑 = −6
therefore
𝑈𝑈𝑛𝑛 = 𝑛𝑛3 − 4𝑛𝑛2 + 3𝑛𝑛 − 6
Use a table to find the formula for the 𝑛𝑛𝑡𝑡ℎ term of the following sequences:
(i) 2, 5, 10, 17, 26
(ii) – 4, 3, 22, 59, 120
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Cambridge IGCSE Extended Mathematics Study Guide
• Geometric sequences
In a geometric sequence there is a common ratio(r) between the successive terms.
① The term-to-term rule
𝑈𝑈𝑛𝑛+1 = 2𝑈𝑈𝑛𝑛
② The formula for the 𝑛𝑛𝑡𝑡ℎ term
3 = 3 × 20
6 = 3 × 21
12 = 3 × 22
24 = 3 × 23
In general 𝑈𝑈𝑛𝑛 = 𝑎𝑎 × 𝑟𝑟 𝑛𝑛−1
…
𝑈𝑈𝑛𝑛 = 3 × 2𝑛𝑛−1
• Applications of geometric sequences
Compound interest
Example: Alex deposits $100 in his savings account. The interest rate offered by the savings account is 10%
each year. Calculate the amount in the account after 5 years.
𝑈𝑈𝑛𝑛 = 𝑎𝑎 × 𝑟𝑟 𝑛𝑛−1
𝑈𝑈6 = 100 × 1.16−1 = $161.05
Adrienne deposits $1500 in her savings account. The interest rate offered by the savings account is 6% each
year. Calculate how much interest she has gained after 10 years.
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Cambridge IGCSE Extended Mathematics Study Guide
•
Direct variation
16 Variation
In each case y is directly proportional to x. This is written 𝑦𝑦 ∝ 𝑥𝑥.
The graphs are straight lines passing through the origin.
𝑦𝑦 = 𝑘𝑘𝑘𝑘 where k is equal to the gradient of the graph and is called the constant of proportionality.
In the above cases, y is directly proportional to 𝑥𝑥 𝑛𝑛 , where 𝑛𝑛 > 0 .
𝑦𝑦 ∝ 𝑥𝑥 𝑛𝑛
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The graphs with (x, y) plotted are not linear. However if the graph of 𝑦𝑦 = 2𝑥𝑥 2 is plotted as (𝑥𝑥 2 , 𝑦𝑦), then the
graph is linear and passes through the origin.
• Inverse variation
1
𝑘𝑘
If y is inversely proportional to x, then 𝑦𝑦 ∝ and = .
𝑥𝑥
1
𝑥𝑥
If a graph of y against is plotted, the graph is a straight line passing through the origin.
𝑥𝑥
Example 1: ∝ 𝑥𝑥 . If 𝑦𝑦 = 7 when = 2 , find y when 𝑥𝑥 = 5 .
𝑦𝑦 = 𝑘𝑘𝑘𝑘
7 = 𝑘𝑘 × 2
so
𝑘𝑘 = 3.5
When = 5 ,
𝑦𝑦 = 3.5 × 5 = 17.5
1
Example 2: ∝ . If 𝑦𝑦 = 5 when = 3 , find y when 𝑥𝑥 = 30 .
𝑦𝑦 =
𝑘𝑘
𝑥𝑥
𝑘𝑘
𝑥𝑥
5=
3
When = 30 ,
15
𝑦𝑦 = = 0.5
so
𝑘𝑘 = 15
30
y is inversely proportional to 𝑥𝑥 3 . If 𝑦𝑦 = 3 when = 2 , find:
(a) the constant of propotionality
(b) the value of y when 𝑥𝑥 = 4
(c) the value of y when 𝑥𝑥 = 6
(d) the value of x when when 𝑦𝑦 = 24 .
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Cambridge IGCSE Extended Mathematics Study Guide
17 Graphs in practical situations
• Conversion graphs
Example: The graph below converts South African rand into euros based on an exchange rate of
€1 = 8.80 rand.
(a) Using the graph, what would be the cost in euros of a drink costing 25 rand?
25 rand = €2.80
(b) If a meal cost 200 rand, what would be the cost in euros?
25 × 8 rand = €2.80 × 8
200 rand = €22.4
•
Speed, distance and time
𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝
𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭
Find the average speed of a car travelling 110 km in 2 h 12 min.
𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
• Travel graphs
The gradient of a distance-time graph represents the speed.
speed = gradient of distance-time graph
rise
speed =
run
30 m
=
2.5 s
= 12 m/s
AKK 27
Cambridge IGCSE Extended Mathematics Study Guide
The gradient of a velocity-time graph represents acceleration.
acceleration = gradient of velocity-time graph
rise
acceleration =
run
250 m/s
=
50 s
= 50 m/s2
The area under a velocity-time graph represents the distance travelled.
distance travelled = area under velocity-time graph
1
distance travelled = × base ×height
2
1
= × 50 s × 250 m/s
2
= 6250 m
18 Graphs of functions
• Quadratic functions
The general expression of a quadratic function is = 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 .
If a > 0 , the graph is concave upwards. (smile face)
If a < 0 , the graph is concave upwards. (sad face)
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• Graphical solution of a quadratic equation
Example: Draw the graph of 𝑦𝑦 = 𝑥𝑥 2 − 4𝑥𝑥 + 3 , and hence, solve the equation 𝑥𝑥 2 − 4𝑥𝑥 + 3 = 0 .
x −2 −1 0 1 2
3 4 5
y 15 8
3 0 −1 0 3 8
The graph crosses the x-axis at two points.
The solutions of the equation are 𝑥𝑥 = 1 and = 3 .
• The reciprocal function
2
Example: Draw the graph of 𝑦𝑦 = for −4 ≤ 𝑥𝑥 ≤ 4 .
x
y
−4
−0.5
−3
−0.7
−2
−1
−1
−2
0
𝑥𝑥
1
2
2
1
3
0.7
4
0.5
This is a reciprocal function giving a hyperbola.
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• Types of graph
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 0
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 1
This is a linear function
giving a straight line.
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3
𝑓𝑓(𝑥𝑥) =
This is a linear function
giving a straight line.
1
𝑥𝑥
This is a reciprocal function
giving a hyperbola.
This is a cubic function
giving a cubic curve.
• Exponential functions
Example: Plot the graph of the function 𝑦𝑦 = 2𝑥𝑥 for −3 ≤ 𝑥𝑥 ≤ 3 .
x
y
−3
0.125
−2
0.25
−1
0.5
0
1
1
2
2
4
3
8
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2
This is a quadratic function
giving a parabola.
𝑓𝑓(𝑥𝑥) =
1
𝑥𝑥 2
This is a reciprocal function,
shown on the graph above.
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• Gradients of curves
Example: For the function = 2𝑥𝑥 2 , calculate the gradient of the curve at the point where 𝑥𝑥 = 1 .
Gradient =
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
10−(−2)
=
3−0
=4
• Solving equations by graphical methods
Example: Plot a graph of 𝑦𝑦 = 3𝑥𝑥 2 − 𝑥𝑥 − 2 , and use the graph to solve the equation 3𝑥𝑥 2 − 7 = 0 .
x
y
−3
28
−2
12
−1
2
0
−2
1
0
2
8
3
22
3𝑥𝑥 2 − 7 = 0
3𝑥𝑥 2 = 7
3𝑥𝑥 2 − 𝑥𝑥 − 2 = −𝑥𝑥 − 2 + 7
3𝑥𝑥 2 − 𝑥𝑥 − 2 = −𝑥𝑥 + 5
2
The solutions are where the curve 𝑦𝑦 = 3𝑥𝑥 − 𝑥𝑥 − 2 and the line 𝑦𝑦 = −𝑥𝑥 + 5 intersect.
The solutions are 𝑥𝑥 ≈ −1.5 and ≈ 1.5 .
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19 Functions
Functions are rules which require an input, x, and give a single output, f(x) which is also called y.
Domain is the set of input values. This may be given in a question or the natural domain which is the set of
all possible input values.
Example: The natural domain of f(𝑥𝑥) = √𝑥𝑥 − 3 is 𝑥𝑥 ≥ 3 , since any values of x below 3 do not give a real
output.
Domain = {𝑥𝑥: 𝑥𝑥 ≥ 3}
Range is the set of output values.
Example: For f(𝑥𝑥) = √𝑥𝑥 − 3 , the output f(𝑥𝑥) = 𝑦𝑦 values are all the numbers between 0 and infinity.
Range = {𝑦𝑦: 𝑦𝑦 ≥ 0}
Inverse functions
The inverse of a function is its reverse, i.e. it ‘undoes’ the function’s effects.
To find the inverse of a function:
• rewrite the function replacing f(𝑥𝑥) with y
• interchange x and y
• rearrange the equation to make y the subject.
Example: Find the inverse of the function f(𝑥𝑥) = 2𝑥𝑥 − 3 .
f(𝑥𝑥) = 2𝑥𝑥 − 3
𝑦𝑦 = 2𝑥𝑥 − 3
𝑥𝑥 = 2𝑦𝑦 − 3
𝑥𝑥+3
𝑦𝑦 =
So 𝑓𝑓 −1 (𝑥𝑥) =
2
𝑥𝑥+3
2
.
Composite functions
If the output from one function f is used as the input for another function g, giving the composite function
g(f(x)) .
Example: If f(x) = 𝑥𝑥 2 and g(x)= 𝑥𝑥 + 3 , evaluate gf(2).
f(𝑥𝑥) =𝑥𝑥 2
g(𝑥𝑥)=𝑥𝑥+3
2�⎯⎯⎯⎯� 4 �⎯⎯⎯⎯⎯� 7
The composite function gf(x)= g(f(𝑥𝑥))
= g(𝑥𝑥 2 )
= 𝑥𝑥 2 + 3
gf(2) = 22 + 3 = 7
20 Geometrical vocabulary
Angles and lines
Acute angles are those between 0˚ and 90˚ .
Right angles are exactly 90˚ .
Obtuse angles lie between 90˚ and 180˚ .
Reflex angles are between 180˚ and 360˚ .
To find the shortest distance between two points, you measure the length of straight line which joins them.
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In the diagram below, CD is perpendicular to AB.
In the triangle below, CD is called the height or altitude of ∆ ABC .
Parallel lines are straight lines which never meet. They are marked with arrows as shown:
Triangles
Types of triangle
Acute-angled triangle
Figures
Properties
Every angle is less than 90˚ .
Right-angled triangle
It has an angle of 90˚.
Obtuse-angled triangle
It has one angle greater than 90˚.
Isosceles triangle
It has two sides of equal length,
and the angles opposite the equal sides are equal.
Equilateral triangle
It has three sides of equal length and equal
angles.
Scalene triangle
It has three sides of different lengths,
and all three angles are different.
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Congruent triangles
Congruent triangles have the same shape and size; they are identical.
They can be proved as follows:
SSS
Three corresponding sides are equal.
SAS
Two corresponding sides and the included angle are equal.
AAS
Two angles and the corresponding side are equal.
ASA
Two angles and the corresponding side are equal.
RHS Each triangle has a right angle, and the hypotenuse and a corresponding side are equal in length.
Similar triangles
Similar triangles have the same shape but may have different sizes; related by enlargement.
They can be proved as follows:
AA
Two pairs of angles are equal.
AAA Three pairs of angles are equal.
Quadrilaterals
A quadrilateral is a plane shape consisting of four angles and four sides.
Square
• Opposite sides are parallel.
• All sides are equal in length.
• All angles are equal in size.
• Diagonals bisect each other at right angles.
Rectangle
Rhombus
Parallelogram
Trapezium
Kite
• Opposite sides are parallel.
• Opposite sides are equal in length.
• All angles are equal in size.
• Diagonals bisect each other.
• Opposite sides are parallel.
• All sides are equal in length.
• Opposite angles are equal in size.
• Diagonals intersect at right angles.
• Two pair of parallel sides.
• Opposite sides are equal in length.
• Opposite angles are equal in size.
• One pair of parallel sides.
• An isosceles trapezium has one pair of parallel sides and
the other pair of sides are equal in length.
• Two pairs of equal sides.
• One pair of equal angles.
• Diagonals intersect at right angles.
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Polygons
Any closed figure made up of straight lines is called a polygon.
If the sides are the same length and the interior angles are equal, the figure is called a regular polygon.
3 sides
triangle
4 sides
quadrilateral
5 sides
pentagon
6 sides
hexagon
7 sides
heptagon
8 sides
octagon
9 sides
nonagon
10 sides
decagon
11 sides
undecagon
12 sides
dodecagon
Nets
The net of a three-dimensional shape like a cube shows the faces of the tube opened out into a twodimensional plan.
Opposite
sides equal
in length
All sides
equal in
length
All angles
right angles
Both pairs of
opposite
sides parallel
Diagonals
intersect at
right angles
Diagonals
intersect at
right angles
All angles
equal
Rectangle
Yes
Square
Yes
Parallelogram
Kite
Rhombus
Equilateral
triangle
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21 Geometrical constructions and scale drawings
Constructing triangles
Example: Construct the triangle ABC given that:
AB = 8 cm, BC = 6 cm and AC = 7 cm.
Constructing simple geometric figures
Example: Construct a parallelogram ABCD in which AB = 6 cm, AD = 3 cm and ∠DAB = 40˚ .
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Bisecting lines and angles
The word bisect means ‘to divide in half ’.
(a) Perpendicular bisector of a segment
(b) Bisector of an angle
(c) 60⁰ angle construction
Scale drawings
Construct a drawing of a triangular enclosure shown below using a scale of 1 cm for each metre.
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Similar shapes
Two polygons are said to be similar if:
• they are equi-angular and
• corresponding sides are in proportion.
Volume of similar shapes
22 Similarity
𝑎𝑎 𝑏𝑏 𝑐𝑐 ℎ
= = = = 𝑘𝑘(scale factor of enlargement)
𝑝𝑝 𝑞𝑞 𝑟𝑟 𝐻𝐻
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝟏𝟏
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝟏𝟏 𝟐𝟐
𝒉𝒉 𝟐𝟐
=�
� =� �
𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝑨𝟐𝟐
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐
𝑯𝑯
𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝟏𝟏
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝟏𝟏 𝟑𝟑
=�
�
𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝑽𝟐𝟐
𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝟐𝟐
23 Symmetry
Reflective symmetry
A line of symmetry divides a two-dimensional(flat) shape into two congruent shapes.
1 line of symmetry
2 lines of symmetry
no lines of symmetry
A plane of symmetry divides a three-dimensional(solid) shape into two identical solid shapes.
Rotational symmetry
A two-dimensional shape has rotational symmetry if, when rotated about a central point, it fits its outline.
rotational symmetry of order 2
rotational symmetry of order 4
A three-dimensional shape has rotational symmetry if, when rotated about a central line, it looks the same at
certain intervals.
rotational symmetry of order 2 about the axis shown
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Circle terms
Equal chords and perpendicular bisectors
If AB = XY ,
OA = OY (equal radii)
OB = OX ,
then ∆ OAB ≅ ∆ OYX (SSS congruency)
OM = ON , and
OM and ON are perpendicular bisectors of AB and XY respectively.
Radius-tangent theorem
The tangent to a circle is perpendicular to the radius at the point of contact.
OT⟂PT
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Tangents from an external point
Tangents drawn from an external point are equal in length.
∠OAC = ∠OBC (90˚each; radius ⟂ tangent)
OC = OC (common side)
OA = OB (equal radii)
∆ OAC ≅ ∆ OBC (RHS congruency)
Therefore, CA = CB
24 Angle properties
Angles at a point
One complete revolution is equivalent to a rotation of 360˚.
Example: Calculate the size of x in the diagram below:
𝑥𝑥 + 120 + 170 = 360 (The sum of all angles around a point is 360˚.)
𝑥𝑥 = 70
Angles on a line
Half a complete revolution is equivalent to a rotation of 180˚ about a point.
Example: Calculate the size of angle a in the diagram below:
40 + 88 + 𝑎𝑎 + 25 = 180 (The sum of all the angles on a straight line is 180˚.)
𝑎𝑎 = 27
𝑎𝑎 + 𝑏𝑏 = 180
𝑐𝑐 + 𝑏𝑏 = 180
Therefore, 𝑎𝑎 = 𝑐𝑐 (Vertically opposite angles are equal in size.)
Angles formed within parallel lines
𝑎𝑎 = 𝑏𝑏 (Corresponding angles are equal in size.)
𝑎𝑎 = 𝑐𝑐 (Alternate angles are equal in size.)
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Angles in a triangle
Example: Calculate the size of angle x in the triangle below:
𝑥𝑥 + 37 + 64 = 180 (The sum of the angles in a triangle is 180˚.)
𝑥𝑥 = 79
Angles in a quadrilateral
Example: Calculate the size of angle p in the quadrilateral below:
𝑝𝑝 + 90 + 80 + 60 = 360 (The sum of four angles in a quadrilateral is 360˚.)
𝑝𝑝 = 130
The sum of all interior angles
The number of triangles is two less than the number of sides of the polygon.
Therefore, the sum of all interior angles = (𝒏𝒏 − 𝟐𝟐) × 𝟏𝟏𝟏𝟏𝟏𝟏˚
The sum of all exterior angles
For any convex polygon, the sum of all exterior angles = 360˚
Example: Calculate the size of angle x in the regular hexagon below:
6𝑥𝑥 = 360
𝑥𝑥 = 60
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The angles in a semi-circle
The angle in a semi-circle is 90˚.
The angle between a tangent and a radius of a circle
The angle between a tangent and the radius at the same point is 90˚.
Angle properties of irregular polygons
Example: Calculate the value of x in the pentagon below:
Sum of all interior angles = ( 𝑛𝑛 − 2 ) × 180˚
𝑥𝑥 + 2𝑥𝑥 + 3𝑥𝑥 + 𝑥𝑥 + 3𝑥𝑥 = ( 5 − 2 ) × 180
10𝑥𝑥 = 360
𝑥𝑥 = 36
Angle at the centre theorem
The angle at the centre of a circle is twice the angle on the circle subtended by the same arc.
∠ AOB = 2 × ∠ ACB
Angles subtended by the same arc
Angles subtended by an arc on the circle are equal in size.
∠ ADB = ∠ ACB
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The following diagrams show other cases of the angle at the centre theorem.
Opposite angles of a cyclic quadrilateral theorem
The opposite angles of a cyclic quadrilateral are supplementary or add to 180˚.
𝛂𝛂 + 𝛃𝛃 = 𝟏𝟏𝟏𝟏𝟏𝟏˚
𝛉𝛉 + 𝛟𝛟 = 𝟏𝟏𝟏𝟏𝟏𝟏˚
Exterior angle of a cyclic quadrilateral theorem
𝛉𝛉 = 𝛅𝛅
25 Loci
The locus of a point is the path of a point moving along a particular description.
•
The locus of the points which are at a given distance from a given point is the circumference of
a circle centre A.
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•
The locus of the points which are at a given distance from a given straight line
•
The locus of the points which are at a given distance from a given segment
•
The locus of the points which are equidistant from two given points
•
The locus of the points which are equidistant from two given intersecting straight lines
Locus of points equidistant from:
1 fixed point
a circle
2 fixed points
perpendicular bisector
3 fixed points, A, B and C
circumcentre of ABC
2 fixed lines
the angle bisector
3 fixed lines
the incentre of the triangle
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26 Measures
Metric units
• The common units of length are: kilometer, metre, centimetre and millimeter.
1 km = 1000 m
1 m = 100 cm
1 cm = 10 mm
• The common units of mass are: tonne, kilogram, gram and milligram.
1 tonne = 1000 kg
1 kg = 1000 g
1 g = 1000 mg
• The common units of capacity are: litre and millilitre.
1 litre = 1000 ml
Area and volume conversions
24 Angle properties
• Circles
(a) Arcs, sectors, segments
𝜃𝜃
× 2𝜋𝜋𝜋𝜋
360
𝜃𝜃
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 =
× 𝜋𝜋𝑟𝑟 2
360
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 − 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇
𝐴𝐴𝐴𝐴𝐴𝐴 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙ℎ =
(b) Circle theorems
(i) Angle subtended at the centre = 2 × angle subtended at the circumference by the same arc
(ii) Angles in the same segment are equal.
(iii) Angle in a semicircle = 90⁰
(iv) Opposite angles of a cyclic quadrilateral add up to 180⁰ .
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(v) Exterior angle of a cyclic quadrilateral = opposite interior angle
(iv) Alternate segment theorem
32 Vectors
Translation is described using column vector.
A column vector describes the sliding movement of object in both x direction and the y direction.
�����⃗ = �1�
AB
3
�����⃗ = �2�
BC
0
Negative vectors
𝐚𝐚 = �25�
Addition of vectors
0
�����⃗
CD = �−2
�
𝐛𝐛 = �−3
�
−2
�����⃗
DA = �−3
�
−1
−𝐚𝐚 = �−2
�
−5
−𝐛𝐛 = �32�
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Cambridge IGCSE Extended Mathematics Study Guide
𝐚𝐚 = �25�
𝐛𝐛 = �−3
�
−2
Subtraction of vectors
𝐚𝐚 + 𝐛𝐛 = �25� + �−3
� = �−1
�
−2
3
𝐚𝐚 = �25�
−𝐛𝐛 = �32�
𝐚𝐚 − 𝐛𝐛 = 𝐚𝐚 + (−𝐛𝐛) = �25� + �32� = �57�
Multiplying a vector by a scalar
𝐚𝐚 = �12�
2𝐚𝐚 = 2�12� = �24�
The magnitude of a vector
The greater the magnitude of a vector, the longer the length.
𝐚𝐚 = �34�
�����⃗ = �−6�
BC
8
|𝐚𝐚| = √32 + 42 = √25 = 5
�����⃗� = �(−6)2 + 82 = √100 = 10
�BC
Position vectors
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Cambridge IGCSE Extended Mathematics Study Guide
�����⃗ = �2� .
The position vector of A relative to O is OA
6
Vector geometry
In general vectors are not fixed in position.
If 𝐚𝐚 = �32� then all the vectors with the same magnitude and direction can also be labelled a .
33 Matrices
A matrix represents the information written in a rectangular array.
For example, two students Tom and Jenny sit a maths exam, a sience exam and an English exam.
Tom scores 70%, 65% and 80% respectively, whilst Jenny scores 60%, 75% and 45% respectively.
70 65 80
�
�
60 75 45
The order of the matrix is written as m × n where m is the number of rows and n is the number of columns.
• If m = n, it is a square matrix.
• If m = 1, it is a row matrix.
• If n = 1, it is a column matrix.
Addition and subtraction of matrices
Corresponding elements can added or subtracted.
8 2
6 6
14 8
�
�+ �
�= �
�
7 9
0 7
7 16
20
−5
15
�6�− �6�= � 0 �
4
8
4
Multiplying matrices by a scalar quantity
16 12
8 6
2�
�=�
�
4 2
8
4
1 8 6
4 3
�
�=�
�
2 1
2 4 2
Multiplying a matrix by another matrix
12 + 3 24 + 9
6 3
15 33
2 4
�2 4� �
� = � 4 + 4 8 + 12� = � 8 20�
1 3
0+1
0+3
0 1
1
3
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In general if matrix A is multiplied by matrix B then this is not the same as matrix B multiplied by matrix A .
𝐀𝐀 × 𝐁𝐁 ≠ 𝐁𝐁 × 𝐀𝐀
The identity matrix , I
1 0
The matrix 𝐈𝐈 = �
� is known as the identity matrix of order 2.
0 1
𝐀𝐀𝐀𝐀 = 𝐈𝐈𝐈𝐈 = 𝐈𝐈
The zero matrix
A matrix in which all the elements are zero is called a zero matrix.
Multiplying a matrix by a zero matrix gives a zero matrix.
4 2 0 0
0 0
�
��
�= �
�
−3 0 0 0
0 0
The determinant of a 2 × 2 matrix
5
6
If A = �
� , |𝐀𝐀| = det 𝐀𝐀 = (5 × −2) − (6 × −4) = −10 + 24 = 14
−4 −2
Finding the inverse of a matrix (by using the formula)
𝑎𝑎 𝑏𝑏
� , then det A = 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏
If A = �
𝑐𝑐 𝑑𝑑
1
𝑑𝑑
�
𝐀𝐀−1 =
det 𝐀𝐀 −𝑐𝑐
−𝑏𝑏
�
𝑎𝑎
Finding the inverse of a matrix (by using the definition)
2 3
If A = �
� then A × 𝐀𝐀−1 = 𝐈𝐈
3 5
𝑒𝑒 𝑓𝑓
Let 𝐀𝐀−1 be �
�
𝑔𝑔 ℎ
2 3 𝑒𝑒 𝑓𝑓
1 0
�
��
�= �
�
3 5 𝑔𝑔 ℎ
0 1
2𝑒𝑒 + 3𝑔𝑔 2𝑓𝑓 + 3ℎ
1 0
�= �
�
�
3𝑒𝑒 + 5𝑔𝑔 3𝑓𝑓 + 5ℎ
0 1
2𝑓𝑓 + 3ℎ = 0
2𝑒𝑒 + 3𝑔𝑔 = 1
and
3𝑒𝑒 + 5𝑔𝑔 = 0
3𝑓𝑓 + 5ℎ = 1
Solving simultaneous equations gives
𝑒𝑒 𝑓𝑓
5 −3
�= �
�
𝐀𝐀−1 = �
𝑔𝑔 ℎ
−3 2
Reflection
An object is reflected about the mirror line.
34 Transformations
−1 0
The transformation matrix for reflection about y-axis is �
�.
0 1
1 0
The transformation matrix for reflection about x-axis is �
�.
0 −1
0 1
The transformation matrix for reflection about 𝑦𝑦 = 𝑥𝑥 is �
�.
1 0
0 −1
�.
The transformation matrix for reflection about 𝑦𝑦 = −𝑥𝑥 is �
−1 0
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Rotation
An object is rotated about a specific point.
Finding the centre and angle of rotation
Consider the triangle ABC is rotated and its new position AˈBˈCˈ
cos θ − sin θ
The transformation matrix for rotation θ in a anticlockwise direction centre at the origin is �
�.
sin θ cos θ
Enlargement
If an object is enlarged, the result is an image which is mathematically similar to the object but of a different
size.
1
The rectangle ABCD undergoes an enlargement with scale factor centre at the point O .
2
3
1
AˈBˈ
= =
scale factor =
6
2
AB
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Negative enlargement
In negative enlargement, each point and its image are on opposite sides of the centre of enlargement.
scale factor of enlargement is −2
Transformations and inverse matrices
If an object is transformed by a matrix A , then the inverse matrix 𝐀𝐀−𝟏𝟏 gives the inverse transformation, i.e. it
maps the image back onto the object.
Combinations of transformations
An object can be transformed by a series of transformation matrices.
These matrices can be replaced by a single matrix which maps the original object onto the final image.
The transformation matrix ABC to AˈBˈCˈ is rotation 180⁰ centre at the origin.
cos 180 − sin 180
−1 0
�
�=�
�
sin 180 cos 180
0 −1
The transformation matrix AˈBˈCˈ to AˈˈBˈˈCˈˈ is enlargement with scale factor 2 centre
at the origin.
1 0
2 0
2�
�= �
�
0 1
0 2
The single transformation matrix which maps ABC onto AˈˈBˈˈCˈˈ is
2 0 −1 0
−2 0
�
��
�= �
�
0 2
0 −1
0 −2
35 Probability
Probability is the study of chance, or likelihood of an event happening.
In general,
number of favourable outcomes
P(the event occurring) =
number of equally likely outcomes
A probability scale goes from 0 to 1.
P(the event not occurring) = 1 − P(the event occurring)
Example: An ordinary fair dice is rolled.
(a) Calculate the probability of getting a six.
Number of favourable outcomes = 1
(i.e. getting a 6)
Number of possible outcomes = 6
(i.e. getting a 1, 2, 3, 4, 5 or 6)
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P(getting a six) =
(b) Calculate the probability of not getting a six.
1
6
P(not getting a six) = 1 −
1
5
=
6
6
(c) Calculate the probability of getting an even number.
Number of favourable outcomes = 3
(i.e. getting a 2, 4 or 6)
Number of possible outcomes = 6
(i.e. getting a 1, 2, 3, 4, 5 or 6)
3
1
P(getting an even number) = =
6
2
Relative frequency
A football referee always used a special coin to toss for ends. He made a simple experiment by spinning the
coin lots of times. His results are shown below.
Number of trials
100
200
300
400
500
600
700
800
900
1000
Number of heads
40
90
142
210
260
290
345
404
451
499
Relative frequency
0.4
0.45
0.47…
0.525
0.52
0.48
0.49
0.505
0.50…
0.499
number of successful trials
total number of trials
The greater the number of trials, the better the estimated probability or relative frequency is likely to be.
Relative frequency =
Example: There is a group of 250 people in a hall. A girl calculates that the probability of randomly picking
someone that she knows from the group is 0.032. Calculate the number of people that the girl knows.
number of favourable results(F)
Probability =
number of possible results
F
0.032 =
250
F = 8 people
The girl knows 8 people in the group.
36 Further probability
Combined events look at the probability of two or more events.
Example: Two coins are tossed. All the possible outcomes are shown in a two-way table.
(i) Calculate the probability of getting two heads.
1
4
(ii) Calculate the probability of getting a head and a tail in any order.
P(getting two heads) =
P(getting a head and a tail in any order) =
2
1
=
4
2
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Tree diagrams
Tree diagrams can show more than two combined events.
Example: A coin is tossed three times.
(i) What is the probability of getting three heads?
The favourable outcome is HHH.
1
8
(ii) What is the probability of getting two heads and one tail in any order?
The favourable outcomes are HHT, HTH, THH.
P(getting three heads) =
P(getting two heads and one tail in any order) =
3
8
(iii) What is the probability of getting at least one head?
The favourable outcomes are with either one, two or three heads, i.e. all of them except TTT.
7
P(getting at least one head) =
8
(iv) What is the probability of getting no heads?
The favourable outcome is TTT.
1
P(getting no heads) =
8
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