Page 1 of 192
COPYRIGHT
All rights reserved. No part of this material may be reproduced, transmitted, stored, or used in any form or by any
means graphic, electronic, or mechanical, including photocopying, recording, scanning, web distribution, or any
information storage and retrieval systems, without permission from the author.
© Frederick Wamweene Mulenga 2020
DEDICATION
This study is dedicated to my dear wife Siabasimbi Bertha Mulenga for her understanding and support during my
absence from home for many months. To my only daughters Chineva, and my only sisters: Vumbi, Lillian, Agatha,
Annie, Mercy, Chawana just to mention a few thanks for the love you showed me during moments I was spending
much of my time on this material and to my mother.
ACKNOWLEDGEMENT
To my God, who continuously favours me even when I am not expecting it. You have always carried me high.
Thank you my Lord. My un-measured gratitude goes to my supervisor Mr. Banda (HOD, Maths Dept @ Choma
Day) for his guidance and assistance rendered to make this work successful and his understanding of my difficult
situation during the ailment of my grandma. For your Kindness and patience, thank you so much.
To all coursemates thank you, you were all brothers in need and indeed. Mukulu Wisely, you deserve special thanks
too for your expertise in computer skills which made this work organized. Shadreck Chisenga, thank you for the
valued corrections you made in this work.
Compiled & Solved by Dr. Frendo
Promoting Excellence In Mathematics © 2020
Page 2 of 192
COPYRIGHT………………………………………………………………………………………………
1
DEDICATION……………………………………………………………………………………………… 1
ACKNOWLEDGEMENT…………………………………………………………………………………
1
PREFACE…………………………………………………………………………………………………..
4
MATHEMATICAL FORMULAES……………………………………………………………………… 5
TOPICS………………………………………………………………………………………………………. 6
TOPIC 1 – SETS…………………………………………………………………………………………… 6
TOPIC 2 - ALGEBRA……………………………………………………………………………………… 9
TOPIC 3 - MATRICES……………………………………………………………………………………… 11
TOPIC 4 - PROBABILITY………………………………………………………………………………… 13
TOPIC 5 – SEQUENCES AND SERIES……………………………………………………………………15
TOPIC 6 – ALGORITHMS (PSEUDO CODE & FLOW CHART) …………………………………… 17
TOPIC 7 – LOCI AND CONSTRUCTION……………………………………………………………… 22
TOPIC 8 – VECTORS……………………………………………………………………………………… 26
TOPIC 9 – TRIGONOMETRY…………………………………………………………………………… 29
TOPIC 10 – STATISTICS………………………………………………………………………………….. 35
TOPIC 11 – GRAPHS OF POLYNOMIAL (QUADRATIC FUNCTIONS) …………………………… 45
TOPIC 12 – EARTH GEOMETRY………………………………………………………………………… 45
TOPIC 13 – LINEAR PROGRAMMING………………………………………………………………… 48
TOPIC 14 – MENSURATION……………………………………………………………………………
54
TOPIC 15 – CALCULUS…………………………………………………………………………………… 61
TOPIC 16 – GEOMETRICAL TRANSFORMATION…………………………………………………… 65
TOPIC 17 – FUNCTIONS………………………………………………………………………………… 73
TOPIC 18 – COORDINATE GEOMETRY……………………………………………………………… 74
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TOPIC 19 – ANGLES IN A CIRCLE……………………………………………………………………
77
TOPIC 20 – BEARING & SCALE MAP DRAWING…………………………………………………… 81
TOPIC 21 – VARIATION………………………………………………………………………………… 85
TOPIC 22 – SPEED & DISTANCE-TIME………………………………………………………………… 87
TOPIC 23 – INDICES & STANDARD FORM…………………………………………………………… 90
TOPIC 24 – SIMILARITIES……………………………………………………………………………… 92
TOPIC 25 – APPROXIMATION & LIMIT OF ACCURACY………………………………………… 95
ANSWERS TO ALL TOPICS………………………………………………………………………………. 96
Compiled & Solved by Dr. Frendo
Promoting Excellence In Mathematics © 2020
Page 4 of 192
is specifically designed to
assist students in their revision for tests and examinations. It is small and handy so that students can carry it around
conveniently. The materials written follow closely the latest paper two examination ‘O’ Level Syllabus D issued by
the Ministry of General Education and aims to provide students with maximum assistance by way of explanation,
advice, revision, practice, models, thinking process and answers.
I have gone to great lengths to make this text both pedagogically sound and error-free. If you have any suggestions,
or find potential errors, please contact Dr. Frendo at frendomulex4@gmail.com or call on: +260
976416759/954441708.
This book is therefore designed to help students to:

Acquire the basic skills and understanding which is vital to examination success.

Appreciate the use of mathematics as a tool for analysis and effective thinking.

Discover order, patterns and relations.

Communicate their thoughts through symbolic expressions and graphs.

Develop mathematical abilities useful in commerce, industry and public service.
Features of this booklet:

Concise Revision Notes,

Specimen Examination Paper,

Essential workings and Answers.
How to use this book:
1.
Use the revision notes to revise a particular topic after you have learnt that topic. This will refresh your
understanding of that topic.
2.
Use the revision exercises to test your understanding and your ability to recall important facts and concepts.
3.
Use this booklet to do a quick revision when preparing yourself for a test of for the examination.
4.
Use the specimen examination paper to have an overall practice that you have finished revising all the topics.
NO PART OF THIS PAMPHLET MAY BE REPRODUCED WITHOUT A PERMISSION OF THE
AUTHOR – DR. FRENDO…………………………………………!
Compiled & Solved by Dr. Frendo
Promoting Excellence In Mathematics © 2020
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Compiled & Solved by Dr. Frendo
Promoting Excellence In Mathematics © 2020
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Concise Revision Notes

A set is a clearly defined collection (finite or infinite) of objects. These objects are called memebers or elements
of the set.

Sets are denoted by capital letters (letters such as A, B, C), while { } are used to enclose the elements.

The symbol ∈ means ‘is an element of’. e.g. 𝒂 ∈ 𝑨 means ‘𝒂 is an element of the set A’. The symbol ∉ means ‘is
not an element of’.

Subset - set P is said to be the subset of the set Q if all the elements of P belong to the set Q. The symbol ⊂ is
used to denote the phrase ‘subset of’. P is a subset of Q is therefore written as P ⊂ Q.

The complement of a set A is defined as the set of all elements of the universal set u, which are not elements of
A. The complement of A is written as 𝑨′ .

The intersection of two sets A and B is defined as the set of all elements that belong to both A and B. The
operation ⋂ is used to define the intersection between two sets. Intersection of A and B is written as A ⋂ B.

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Page 7 of 192
Specimen Examination Questions:
1.
2018 Oct/Nov Exams - Q2 (b)
At Sambilileni College, 20 students study at least one of the three subjects; Mathematics (M), Chemistry (C)
and Physics (P). All those who study chemistry also study mathematics. 3 students study all the three subjects. 4
students study mathematics only, 8 students study chemistry and 14 students study mathematics.
i)
Draw a Venn diagram to illustrate the information.
ii)
How many students study
a) Physics only
b) two types of subjects only,
c)
2.
Mathematics and physics but not chemistry.
2018 Jul/Aug Exams - Q3 (a)
The diagram below shows how learners at Twatenda School travel to school. The learners use either buses (B),
Buses
Cars
cars (C) or walk (W) to school.
2
7
14
𝝐
4
i)
If 22 learners walk to school, find the value of 𝑥.
ii)
How many learners use
𝑥
3
7
Walk
a) Only one mode of transport,
b) Two different mode of transport.
3.
2019 Aug Exams Q2 (b).
The Venn diagram below shows the optional subjects that all the Grade 10 learners at Kusambilila Secondary
School took, in a particular year.
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(i)
Given that 12 learners took Music, find the value of 𝑥.
(ii)
How many learners were in Grade 10 this particular year?
(iii)
Find the number of learners who took:
(a) One optional subject only,
(b) Two optional subjects only.
4.
2017 Oct/Nov Exams - Q1 (b).
A survey carried out at Kamulima Farming Block showed that 44 farmers planted maize, 32 planted sweet
potatoes, 37 planted cassava, 14 planted both maize and sweet potatoes, 24 planted both sweet potatoes and
cassava, 20 planted both maize and cassava, 9 planted all the three crops and 6 did not plant any of these crops.
i)
Illustrate this information on a Venn diagram.
ii)
How many farmers:
a) Where at this farming block,
b) Planted maize only,
c)
5.
Planted two different crops.
2017 July/Aug Exams - Q3 (b)
The Venn diagram below shows tourist attractions visited by certain students in a certain week.
i)
Find the value of 𝑦 if 7 students visited Mambilima Falls only.
ii)
How many students visited?
a) Victoria falls but not Gonya Falls,
b) Two tourist attractions only,
c)
6.
One tourist attraction only?
2016 October Exams - Q2 (a)
Of the 50 villagers who can tune in to Kambani Radio Station, 29 listen to news, 25 listen to sports, 22 listen to
music, 11 listen to both news and sports, 9 listen to both sports and music, 12 listen to both news and music, 4
listen to all the three programmes and 2 do not listen to any programme.
i)
Draw a Venn diagram to illustrate this information.
ii)
How many villagers
a) Listen to music only,
b) Listen to one type of programme only,
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c)
7.
Listen to two types of programs only.
Miscellaneous Question
142 women were interviewed to find out what they used for cooking. The following results were obtained:
58 used gas cookers,
58 used microwave ovens,
63 used electric cookers,
4 used both gas cookers and electric cookers,
17 used both microwave ovens and gas cookers,
19 used both microwave ovens and electric cookers,
1 used gas cookers, microwave ovens and electric cookers,
2 cooked with solar energy only.
a) Draw a Venn diagram to illustrate this information.
b) Using your Venn diagram, how many women used:
8.
i)
Gas cookers only?
ii)
Microwave ovens and electric cookers only?
iii)
Microwave ovens or electric cookers?
iv)
Microwave ovens or electric cookers, but did not use gas cookers?
v)
Microwave ovens or solar energy?
100 members of a community were asked to state the activities they undertake during the day.
38 go to School.
18 go to School and also Trade.
54 go for Fishing.
22 go to Fishing and also Trade.
50 engage in Trading.
Each of these members undertakes at least one of the activities. The number of people who go to school only is
the same as the number who engages in Trading only.
Use the information to find the number of people who:
a) Undertake all the three activities,
b) Go to school only.
Concise Revision Notes:

1
Expressions such as 2x + 3y, 𝑎2 which contain letters put together using +, –, × 𝑎𝑛𝑑 ÷ are known as
2
Algebraic Expressions. Letters or symbols are used to represent quantities, and signs are used to represent
relations between them. Symbols or letters which have fixed values are called constants. Symbols or letters
which represent changing or changeable quantities are called variables.
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Page 10 of 192

When simplifying algebraic fractions, always fully factorize the numerators and denominators. It may be
possible therefore, to divide the numerators and deminators by any factors which they have in common.

× 𝑎𝑛𝑑 ÷ signs are not usually written in algebraic expressions. a × b is usually written as ab (so as not to
𝑎
confuse the multiplication sign with the letter x) and a ÷ b is also written as .
𝑏

Numbers are written first, followed by letters in alphabetical order. For example: 6𝑥 2 𝑦𝑧, 9𝑎𝑏𝑐 3 .

Only like terms can be added or subtracted

To add or subtract like terms of algebraic expressions with fractional coefficients, we first determine the
LCM of the denominators before performing the operation.

When an expression is written as the product of two or more factors, we say that is factorized.

A perfect square minus another perfect square is called difference of two squares. For example, 49 – 25 is
the difference between two perfect squares and this can be written as: 72 − 52 . If 𝑥 2 𝑎𝑛𝑑 𝑦 2 are two perfect
squares, then their difference can be expressed as: 𝑥 2 − 𝑦 2 = (𝑥 + 𝑦)(𝑥 − 𝑦).

For any quadratic equation in the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 can be solved using the formulae:
𝒙=
−𝐛 ± √𝐛 𝟐 − 𝟒𝐚𝐜
𝟐𝐚
Specimen Examination Questions:
1.
2018 Oct/Nov Exams
a) Simplify 𝑏2 − 𝑎2
𝑎 − 𝑏
b) Simplify
c)
2.
15𝑐𝑑
÷
3
9𝑐3 𝑛
10𝑐2 𝑑
2
3
4
−𝑥− 1
𝑥+ 1
Express
as a single fraction in its lowest terms.
2018 July/Aug Exams
a) Simplify
b) Express
c)
3.
12𝑑𝑛3
7𝑠𝑡 3
15𝑢3 𝑣 2
3
2𝑥 − 5
×
5𝑢3 𝑣
28𝑠 3 𝑡 2
4
−
𝑥− 3
as a single fraction in its lowest terms.
Factorise 𝑥 2 − 7𝑥 + 12 completely.
2017 Oct/Nov Exams
3
4
a) Simplify 14𝑥2 ÷ 7𝑥 3
9𝑦
18𝑦
2𝑥2 −
b) Simplify 𝑥 + 28
c)
4.
Express
1
𝑥− 4
−
2
5𝑥 − 1
as a single fraction in its lowest terms.
2017 July/ Aug Exams
a) Simplify
𝑚2 − 1
𝑚2 − 𝑚
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b) Simplify
c)
5.
𝑝2 𝑞 3
4
×
3
Express
8
𝑝𝑞
−
5𝑥 − 2
÷ 2𝑝2 𝑞
2
𝑥+3
as a single fraction in its lowest terms.
2016 October Exams
a) Simplify 𝑥2− 1
𝑥 −1
b) Simplify
c)
6.
Express
2
2
17𝑘
÷ 51𝑘
5𝑎
20𝑎2
2
2𝑥 − 1
−
1
3𝑥 + 1
as a single fraction in its lowest terms.
Solve the following equations, giving your answers correct to two decimal places:
a) 1 − 5𝑝 − 2𝑝2 = 0
(2005)
b) 7 − 5𝑥 − 𝑥 2 = 0
(2010)
c)
(2𝑥 − 1)(3𝑥 − 2) = 3
(2011)
2
d)
𝑥 + 2𝑥 = 7
(2016)
e)
𝑥 2 + 2𝑥 = 5
(2016)
“A bird sitting on a tree is never afraid of the branch breaking, because her trust is not on the branch but on its
wings. Always believe in yourself”. Mathematics is very simple…………!!!!!
Concise Revision Notes:

A matrix is an array of numbers (called elements) arranged in a rectangular pattern of rows and columns and
enclosed in brackets. A matrix is usually denoted by a capital letter. i.e. 𝐴 = (

2
1
3
).
−4
The order of a matrix defines the number of rows and the number of columns in the matrix. A matrix with 𝑚
rows and 𝑛 columns has an order of (𝒎 × 𝒏). E.g
3 columns
𝑃=(
1
5
2
4
3
)
−1
2 rows
Therefore, it is a (2 × 3 𝑚𝑎𝑡𝑟𝑖𝑥)

Equal matrices are matrices which are identical. They have the same order and the same corresponding
elements.

Matrices can only be added or subtracted if only they are of the same order. Therefore, to add or subtract
matrices, simply add or subtract the corresponding elements. The outcome will be a matrix of the same order.

Multiplying a matrix by a number (a scalar), simply multiply each element by that number.

Two matrices can only be multiplied if the number of the columns in the first matrixis the sam as the number of
rows in the second matrix. We then say that the two matrices are compatible or conformable. NB: we multiply
matrices row by column, i.e. (𝒓 × 𝒄).
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
The determinant of a matrix is the difference between the product of the elements in the major diagonal and the
product of the elements in the minor diagonal. Sometimes it is denoted by |𝑨|.

The inverse of a matrix is the reciprocal of the determinant multiplied by its adjoint. It is usually denoted by
𝐴−1 .

Identity matrix is a matrix when a matrix is multiplied by its inverse, i.e.
1
(
0
0
)
1
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q1 (a)
Given that 𝐴 = (
4
1
8
−5
) 𝑎𝑛𝑑 𝐵 = (
2
3
𝑦
),
5
a) Find the value of 𝑦𝑦, for which the determinants of A and B are equal,
b) Hence, find the inverse of B.
2.
2018 Jul/August Exams: Q1 (a)
Given that A = (
2𝑥
3
2
),
𝑥
a) find the positive value of 𝑥 for the determinant of A is 12,
b) hence or otherwise find 𝐴−1 .
3.
2019 Aug Exams Q2 (a).
The determine of matrix Q = (
8
𝑥−4
12
) is 8, find
𝑥
(a) The value 𝑥
(b) The inverse of Q.
4.
2017 Oct/Nov Exams: Q1 (a)
10
Given that K = (
11
−2
),
−2
a) the determinant of K,
b) The inverse of K.
5.
2016 Oct/Nov Exams: Q1 (a)
3
Given that (
𝑥
−2
),
4
a) The value of 𝑥, given that the determinant of Q is 2,
b) The inverse of Q.
6.
Miscellaneous Questions
a) Express each of the following as single matrices:
i)
(1
3
2) ( )
4
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ii)
(−1
0
2
2) ( 0
−3
5
3)
1
Promoting Excellence In Mathematics © 2020
Page 13 of 192
Concise Revision Notes:

Probability is the study of chances, or the likelihood of an event happening.

A favourable outcome refers to the event in question actually happening.

The total number of possible outcomes refers to all the different types of outcome one can get in a particular
situation. In general:
𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒂𝒏 𝒆𝒗𝒆𝒏𝒕 =
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒇𝒂𝒗𝒐𝒖𝒓𝒂𝒃𝒍𝒆 𝒐𝒖𝒕𝒄𝒐𝒎𝒆
𝒕𝒐𝒕𝒂𝒍 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒆𝒒𝒖𝒂𝒍𝒍𝒚 𝒍𝒊𝒌𝒆𝒍𝒚 𝒐𝒖𝒕𝒄𝒐𝒎𝒆𝒔

If the probability is = 0, then the event is impossible.

If the probability is = 1, then the event is certain to happen.

Probability of the event not occurring = 𝟏 − 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒐𝒇 𝒕𝒉𝒆 𝒆𝒗𝒆𝒏𝒕 𝒐𝒄𝒄𝒖𝒓𝒓𝒊𝒏𝒈

Combined events look at the probability of two or more events, i.e. a die or coin.

Exclusive events – A and B are exclusive events if they cannot both happen at the same. The rule for the
probabilities of exclusive events A and B
𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨) + 𝑷(𝑩)
NB: we add their separate probabilities to get the probability of either of them happening.

Independent events – if the happening of event A has no event on the possible happening of another event B,
then A and B are independent events. The rule for their probabilities is:
𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝑨) × 𝑷(𝑩)
NB: we multiply their separate probabilities to get the probability of both happening.

Tree diagram – when more than two combined events are being considered, then two-way tables cannot be
used and therefore, another method of representing information diagramatically is needed. Tree diagrams are a
good way of doing this.
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q1 (b)
A small bag contains 6 black and 9 red pens of the same type. Two pens are taken at random one after the other
without replacement.
Calculate the probability that both pens:
a) One is black,
b) Are of different colour.
2.
2018 Jul/Aug Exams, Q5 (a)
A box contains identical buttons of different colours. There are 20 black, 12 red and 4 white buttons in the box.
Two buttons are picked at random one after the other and not replaced in the box.
a) Draw a tree diagram, showing all the possible outcomes.
b) What is the probability that both buttons are white?
3.
2017 Oct/Nov Exams, Q2 (a)
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A box of chalk contains 5 white, 4 blue and 3 yellow pieces of chalk. A piece of chalk is selected at random
from the box and not replaced. A second piece of chalk is then selected.
a) Draw a tree diagram to show all the possible outcomes.
b) Find the probability of selecting pieces of chalk of the same colour.
4.
2017 July Exams, Q3 (a)
In a box of 10 bulbs, 3 are faulty. If two bulbs are drawn at random one after the other, find the probability that
(a) Both are good.
(b) One is faulty and the other one is good.
5.
2016 Oct/Nov Exams, Q2 (b)
A survey was carried out at certain hospital indicated that the probability that patient tested positive for malaria
is 0.6. What is the probability that two patients selected at random
(a) One tested negative while the other positive,
(b) Both patients tested negative.
6.
2016 July Exam Q9 (b)
A bag contains 3 black balls, and 2 white balls. Two balls are taken from the bag at random, one after another,
without replacement.
a) Draw a tree diagram to represent this information.
b) Calculate the probability that the two balls taken at random are of the same colour.
7.
2011 Exam Q5
A box contains 3 green apples and 5 red apples. An apple is picked from the box and not replaced then a second
apple is picked. Expressing the answer as a fraction in its simplest form, calculate:
a) The probability that both apples picked are green.
b) The probability that the two apples picked are of different colours.
8.
2012 Exam Q11 (b)
A box has 7 identical sweets. 3 of these are green and the rest are red. Chineva picks one sweet at random and
eats it. After sometime, she picks another one and eats it.
a) Construct a tree diagram to illustrate the outcomes of the two sweets taken.
b) Calculate the probability that the first sweet was red and the second was green.
9.
2013 Exam Q10 (b)
Dr. Frendo bought three oranges and two apples which he put in a bag. Later on he picked one fruit at random
from the bag and ate it. After sometimes he picked one fruit at random and ate it.
a) Construct a tree diagram to represent this information.
b) Hence or otherwise, find the probability that the two fruits picked were of different types.
10. 2015 Exam Q6 (b)
A box has 14 identical balls, three of which are blue. Two balls are drawn at random from the box, one after the
other without replacement.
Calculate the probability that:
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Page 15 of 192
a) The two balls are both blue,
b) At least one ball drawn is blue.
11. Micellaneous Question
Concise Revision Notes:

A sequence is a collection of terms arranged in a specific order, where each term is obtained according to a
rule. Examples of some simple sequence are given below;
2, 4, 6, 8, 10…
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1, 2, 4, 6, 8,…
1, 1, 2, 3,.
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
The terms in a sequence can be expressed as a1, a2, a3, a4, a5,………….an where:
a1 is the first term
a2 is the second term
an is the nth term

Arithmetic Progression sequences (AP)
In an arithmetic sequence there is a common difference (d) between successive terms. Exampples of some
arithmetic sequences are given below;
3
6
9
12
15…d = 3
7
2
-3
-8
-13…d = -5
NB: to find the common difference, add or subtract the first term from the second term.

Hence,
the
general
formula
for
the
nth
term
for
AP
is 𝑻𝒏 = 𝒂 + (𝒏 − 𝟏)𝒅,
where
𝒂 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚, 𝒏 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 𝑎𝑛𝑑 𝒅 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒.

Geometric Progression sequences (GP)
A sequence where there is a common ratio (r) between successive terms is known as a geometric sequence. For
example;
2
4
27
8
9
1
32….. r = 2
16
1
3
…………. r =
1
3
NB: to find the common ratio (r), divide the first term into the second term.

Hence,
the
general
formula
for
the
nth
term
for
GP
is:
𝑻𝒏 = 𝒂𝒓𝒏 − 𝟏 ,
𝒂 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑡𝑒𝑟𝑚, 𝒏 𝑖𝑠 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑒𝑟𝑚𝑠 𝑎𝑛𝑑 𝒓 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑜𝑚𝑚𝑜𝑛 𝑟𝑎𝑡𝑖𝑜
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q5 (b)
The first three terms of a geometric progression are k + 4, k, 2k − 15 where 𝑘 a positive integer.
a) Find the value of 𝑘,
b) List the first three terms of the geometric progression,
c)
2.
Find the sum to infinity.
2018 Jul/Aug Exams, Q2 (b)
In a geometric progression, the third term is
2
9
and the fourth term is
2
. Find:
27
a) The first term and the common ratio,
b) The sum of the first 5 terms of the geometric progression,
c)
3.
The sum to infinity.
2019 Aug Exams Q1 (b).
In a geometric progression, the third term is 16 and the fifth term is 4. Calculate
(i)
The first term an the common ratio,
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4.
(ii)
The tenth term,
(iii)
The sum to infinity
2017 Oct/Nov Exams, Q5 (a)
For the geometric progression 20, 5,
1 14. . . find
a) The common ratio,
b) The nth term,
c)
5.
The sum of the first 8 terms.
2017 July/ Aug Exams, Q2 (b)
The first three terms of a geometric progression are 6 + 𝑛, 10 + 𝑛, 𝑎𝑛𝑑 15 + 𝑛, Find
a) the value of n,
b) the common ratio,
c)
6.
the sum of the first 6 terms of this sequence.
2016 Oct/Nov Exams, 5 (b)
The first three terms of a geometric progression are 𝑥 + 1, 𝑥 − 3 and 𝑥 − 1. Calculate:
a) the value of 𝑥,
b) the first term,
c)
7.
the sum to infinity.
Miscellaneous Questions
i)
The first three terms of an AP are 𝑥 + 3, 2𝑥 + 6, 𝑎𝑛𝑑 8. Find
a) The value of 𝑥
b) The sum of the first 12 terms.
ii)
The first three terms of an AP are 𝑥, 2𝑥 + 1, 𝑎𝑛𝑑 5𝑥 − 1. Find
a) The value of x
b) The sum of the first 10 terms.
ii)
The terms 5 + 𝑥, 8 𝑎𝑛𝑑 1 + 2𝑥 are consecutive terms in an arithmetic sequence. Find
a) The value of x
b) State the three terms
Concise Revision Notes:

An algorithm is procedure consisting of a finite set of unambiguous rules (instructions) which specify a finite
sequence of operations that provides the solution to a problem, or to a specific class of problems for any
allowable set of input quantities (if there are inputs). In other word, an algorithm is a step-by-step procedure to
solve a given problem.
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
Pseudo code is one of the tools that can be used to write a preliminary plan that can be developed into a
computer program. Pseudocode is a generic way of describing an algorithm without use of any specific
programming language syntax. It is, as the name suggests, pseudo code — it cannot be executed on a real
computer, but it models and resembles real programming code, and is written at roughly the same level of
detail.

There are 6 basic symbols commonly used in flowcharting of assembly language Programs: Terminal, Process,
and input/output, Decision, Connector and Predefined Process. This is not a complete list of all the possible
flowcharting symbols; it is the ones used most often in the structure of Assembly language programming.
Symbol
Name
Function
Process (Rectangle)
Indicates any type of internal
operation inside the Processor or
Memory. It also Denotes a
process to be carried out e.g.
addition, subtraction, division
etc.
Used for any Input / Output
(I/O) operation. Indicates that
input/output (Parallelogram)
the computer is to obtain data or
output results. Also it denotes an
input operation.
Used to ask a question that can
be answered in a binary format
Decision (Diamond)
(Yes/No, True/False). Denotes a
decision (or branch) to be made.
The program should continue
along one of two routes. (e.g.
IF/THEN/ELSE).
Connector (Oval)
Allows the flowchart to be drawn
without intersecting lines or
without a reverse flow. Denotes
the beginning or end of the
program.
Predefined Process
Used to invoke a subroutine or
an Interrupt program.
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Indicates the starting or ending
of the program, process, or
Terminal
interrupt program.
Hybrid
Denotes an output operation.
Flow Lines
Shows direction of flow. Denotes
the direction of logic flow in the
program.
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q6 (b)
The program below is given in a form of a Pseudo code
Start
Enter 𝑥, 𝑦
Let 𝑀 = √𝑥 2 + 𝑦 2
IF 𝑀 < 0
THEN display error message “M must be positive”
ELSE
END IF
Display M
Stop
Draw the corresponding flow chart for the information given above.
2.
2018 July/Aug Exams, Q5(b)
Study the pseudo code below.
Start
Enter 𝑎, 𝑟, 𝑛
𝑅 =1−𝑟
If 𝑅 = 0, THEN
Print “the value of r is not valid”
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Else 𝑆𝑛 =
𝑎(1 − 𝑟 𝑛 )
𝑅
End if
Print Sn
Stop
Construct a flow chart corresponding to the Pseudo code above.
3.
2017 Oct/Nov Exams, Q6
Start
Enter r
Is
Yes
Error “r must be positive”
𝑟 < 0?
No
𝐴=
1
∗ 𝑟 ∗ 𝑟 ∗ sin 𝜃
2
Display Area
Stop
Study the flow chart above, write a pseudo code corresponding to the flow chart program above
4.
Micellaneous Question
The program below is given in the form of pseudocode. Read the two sides of a rectangle and calculate its area.
Start
Input the width (W) and Length (L) of a rectangle
Calculate the area (A) by multiplying L with W
Print A
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Stop
Draw the corresponding flowchart for the information given above.
5.
2017 July/Aug Exams, Q6 (b)
The diagram below is given in the form of a flow chart.
Start
Enter a, r
Is
|𝒓| < 𝟏?
𝑺∞ =
𝒂
𝟏−𝒓
Display sum to infinity
Stop
Write a pseudo code corresponding to the flow chart program above.
6.
2016 Oct/Nov Exams, Q3(b)
The program below is given in the form of a pseudo code.
Start
Enter radius
If 𝑟𝑎𝑑𝑖𝑢𝑠 < 0
The display “error message” and re-enter positive radius
Else enter height
If ℎ𝑒𝑖𝑔ℎ𝑡 < 0
The display “error message” and re-enter positive height
1
Else 𝑉 = ∗ 𝜋 ∗ 𝑠𝑞𝑢𝑎𝑟𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 ∗ ℎ𝑒𝑖𝑔ℎ𝑡
3
End if
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Display volume
Stop
Draw the corresponding flowchart for the information given above.
7.
Miscellaneous Question
The flowchart for an algorithm below shows the information to find the area of a circle of radius r.
Start
Enter 𝝅, 𝒓
𝑨 = 𝟑. 𝟏𝟒𝟐 ∗ 𝒓 ∗ 𝒓
Display Area
Stop
Write down the pseudocode for the flowchart for an algorithm
Concise Revision Notes:

The locus ( plural, loci) of a moveable point is the set of its possible positions as it moves under certain
conditions. Therefore, a locus is the set of all possible positions occupied by an object which varies its position
according to some given law.
Common locus:
i)
Locus of points at a given distance from a fixed point.
-
ii)
Locus of points at a given distance from a straight line.
-
iii)
iv)
v)
The locus of points is a circle.
The locus of points is the two straight parallel lines.
Locus of points equidistance from two given points.
The locus of points is the perpendicular bisector.
Locus of points equidistant from two straight lines.
The locus of points is the pair of bisectors of the angles between the lines.
Locus of points which subtend a given angle on a given line seqment.
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
The locus of points is the arc.
Circum-circle of a triangle.
-
A circle which passes through the three vertices of a triangle is called circum-circle of the triangle.
-
The construction of the circum-circle of a triangle uses the fact that the perpendicular bisector of a chord
of a circle passes through its centre. The three sides of the given ∆ form three chords of its circumcircle.
-
If the perpendicular bisectors of the three sides of a ∆ are constructed they will meet at a single point, the
circum-centre. The radius of the circumcircle is the distance of the circum-centre from any one of the
vertices of the given ∆.

Triangles anod other polygons can be drawn accurately by using a ruler, and a pair of compass or protractor.
This is called construction.
Specimen Examination Questions:
Answer the whole of these questions on a sheet of plain papers.
1.
2018 Oct/Nov Exams, Q4
a)
i)
Construct triangle XYZ in which XY = 9cm, YZ = 7cm and angle XYZ = 38°
ii)
Measure and write the length of XZ
b) On your diagram within triangle XYZ, construct the locus of points which are
c)
i)
6cm from Y
ii)
equidistant from XY and XZ
Mark clearly with letter P, within triangle XYZ, a point which is 6cm from Y and equidistance from XY
and XZ.
d) A point Q within triangle XYZ is such that its distance from Y is less than or equal to 6cm and its nearer to
XY than XZ. Indicate clearly by shading the region in which Q must lie.
2.
2018 Jul/Aug Exams, Q4
a)
i)
Construct triangle PQR in which PQ = 10cm, QR = 8cm and 𝑃𝑄̂ 𝑅 = 50° .
ii)
Measure and write the length of PR
b) On your diagram, within triangle PQR, construct the locus of points which are
c)
i)
equidistant from P and Q
ii)
equidistant from PR and PQ
iii)
5cm from R
A point T within triangle PQR is such that it is 5cm from R and equidistant from P and Q. Label point T.
d) Another point X is such that it is less than or equal to 5cm from R, nearer to Q than P and nearer to PQ
than PR. Indicate by shading, the region in which X must lie.
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3.
2017 Oct/Nov Exams, Q3
a) Construct a quadrilateral ABCD in which AB = 10cm, and angle ABC = 120°, angle BAD = 60°, BC =
7cm and AD = 11cm.
b) Measure and write the length of CD.
c)
Within the quadrilateral ABCD, draw the locus of points which are
i)
8cm from A
ii)
Equidistant from BC and CD
d) A point P, within the quadrilateral ABCD, is such that it 8cm from A and equidistant from BC and CD.
Label point P.
e)
Another point Q, within the quadrilateral ABCD, is such that, it is nearer to CD than BC and greater than
or equal to 8cm from A. indicate, by shading, the region in which Q must lie.
4.
2017 July/Aug Exams, Q4
a)
i)
Construct triangle PQR in which PQ is 9cm, angle PQR = 60° and QR = 10cm
ii)
Measure and write the length of PR.
b) On your diagram, draw the locus of points with triangle PQR which are
c)
i)
3cm from PQ,
ii)
7cm from R,
iii)
Equidistant from P and R.
A point M, within triangle PQR, is such that it is nearer to R than P, less than or equal to 7cm from R and
less than or equal to 3cm from PQ. Shade the region in which M must lie.
5.
2016 Oct/Nov Exams, Q4
a)
i)
Construct triangle ABC where AB = BC = CA = 7cm
ii)
Measure and write the size of ∠CAB.
b) Within triangle ABC construct the locus of points which are
c)
i)
Equidistant from AB and BC
ii)
4cm from B
iii)
3cm from AB
A point R, within triangle ABC, is such that it is nearer to BC than AB, less that 3cm from AB and less
than 4cm from B. Shade the region in which R must lie.
6.
2002 Oct/Nov Exam Q4
The floor of a sharp pen is in the shape of a quadrilateral ABCD. Each side of the quadrilateral is 3m long
and 𝐴𝐵̂ 𝐷 = 60°.
a) Using a scale of 2cm to represent 1m, draw quadrilateral ABCD.
b) Draw the axes of symmetry of quadrilateral ABCD
c)
What is the special name of the quadrilateral ABCD
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d) Measure the diagonal AC, giving your answer in meters, correct to 1 decimal place
e)
7.
Calculate the area of the floor in meters giving your answer correct to 2 decimal places.
Miscellaneous Question
a)
construct ∆𝑷𝑸𝑹 in which PQ = 8cm, QR = 7cm and PR = 3cm
b) Measure and write down the size of the largest angle in the triangle
c)
On your diagram, draw the locus of points which are:
i)
Equidistant from P and Q
ii)
Equidistant from Q and R
d) Draw the circle which touches P, Q, and R
e)
The point X inside ∆𝑷𝑸𝑹 is nearer to Q than to P and nearer to R than Q. indicate clearly by shading the
region in which X must lie.
Concise Revision Notes:

A vector is quantity having both magnitude (size) and direction.

A vector can be represented in magnitude and direction by a directed line segment with a length proportional to
the magnitude and an arrow showing the direction.
B
i.e.
𝒂
A
NB: the notation ⃗⃗⃗⃗⃗
𝐴𝐵 𝑜𝑟 𝑨𝑩 is used to indicate the endpoints and the direction (from A to B). The vector can
also denoted by the single letter i.e. 𝑎 𝑜𝑟 𝒂 can be used.

The
magnitude
of
a
vector
is
the
size
represented
by
a
segmented
line.
It
is
denoted
⃗⃗⃗⃗⃗ | 𝑜𝑟 |𝑎| 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦 𝑎𝑛𝑑 𝑖𝑡 𝑖𝑠 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑝𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑖. 𝑒. |𝑎| = √𝑥 2 + 𝑦 2 .
|𝐴𝐵

A vector can start from any point, as any line segment of the correct length and direction will represent the
vector. However, a vector ⃗⃗⃗⃗⃗
𝑂𝐴 starting from the origin O is called the position vector.

Vectors are combined or added using the parallelogram rule. For example, vector a is to be added with vector
⃗⃗⃗⃗⃗ = 𝑎 𝑎𝑛𝑑 𝑂𝐵
⃗⃗⃗⃗⃗ = 𝑏. Complete the parallelogram OACB.
b. place them to start from the same point O. Then, 𝑂𝐴
⃗⃗⃗⃗⃗ + 𝑂𝐵
⃗⃗⃗⃗⃗ . i.e.
The parallelogram rule states that⃗⃗⃗⃗⃗⃗
𝑂𝐶 = 𝑂𝐴
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by
Page 26 of 192

However, the simplest method of adding vectors is the triangular or end-on rule. This states that to combine or
add vectors, place the second to start from the “end” of the first and complete the triangle. The third side is the
sum of the two vectors. i.e.
C
O
A
⃗⃗⃗⃗⃗⃗ + ⃗⃗⃗⃗⃗
∴ ⃗⃗⃗⃗⃗⃗
𝑶𝑪 = 𝑶𝑨
𝑨𝑪
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q3
In the quadrilateral ABCD, ⃗⃗⃗⃗⃗
AB = 𝑎 𝑎𝑛𝑑 ⃗⃗⃗⃗⃗
AD = 𝑏, ⃗⃗⃗⃗⃗
BC = 2𝑏 and AE: AC = 1: 3.
i)
Find in terms of 𝒂 and/or 𝒃
ii)
a) ⃗⃗⃗⃗⃗
AE
b) ⃗⃗⃗⃗⃗
BE
c) ⃗⃗⃗⃗⃗
BD
Hence or otherwise show that the points B, D and E are collinear.
2.
2017 Oct/Nov Exams, Q2 (b)
In the diagram below, ⃗⃗⃗⃗⃗
OP = 2𝑝, ⃗⃗⃗⃗⃗
OQ = 4𝑞 and PX : XQ = 1 : 2.
i)
ii)
3.
Express in terms of 𝑝 and/or 𝑞
a) ⃗⃗⃗⃗⃗
PQ
b) ⃗⃗⃗⃗
PX
c) ⃗⃗⃗⃗⃗
OX
1
4
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗
Given that OC = ℎOX, show that CQ = 4 (1 − ℎ) 𝑞 − ℎ𝑝.
3
3
2017 July/Aug Exams, Q6 (a)
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In the diagram below, OABC is a parallelogram in which ⃗⃗⃗⃗⃗
OA = 𝑎 and ⃗⃗⃗⃗⃗
AB = 2𝑏 and AC intersect at D. E is the
midpoint of CD. E is the mid - point of CD.
Express in terms of 𝑎 and/or 𝑏
4.
𝐚) ⃗⃗⃗⃗⃗
OB
2017 Oct/Nov Exams, Q6(a)
b)
⃗⃗⃗⃗⃗
OE
⃗⃗⃗⃗⃗
c) CD
⃗⃗⃗⃗⃗ = 3𝑎 and ⃗⃗⃗⃗⃗
In the diagram below, OAB is a triangle in which OA
OB = 6𝑏. OC : CA = 2 : 3 and AD : DB = 1 : 2.
OD meets CB at E.
i)
Express each of the following in terms of 𝑎 and / or 𝑏.
𝐚) ⃗⃗⃗⃗⃗
AB
ii)
5.
b) ⃗⃗⃗⃗⃗
OD
c) ⃗⃗⃗⃗⃗
BC
⃗⃗⃗⃗⃗ , express BE
⃗⃗⃗⃗⃗ = ℎBC
⃗⃗⃗⃗⃗ in terms of ℎ, 𝑎 𝑎𝑛𝑑 𝑏.
Given that BE
2005 Oct/Nov Exam
In the diagram below, M is the midpoint of BP and P is on OP produced such that OA : AP = 1 : 2. ⃗⃗⃗⃗⃗
OA =
P
𝑎 𝑎𝑛𝑑 ⃗⃗⃗⃗⃗
OB = 𝑏.
M
N
A
𝒂
O
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𝒃
B
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Express each of the following in terms of 𝑎 and / or 𝑏
i)
a) ⃗⃗⃗⃗⃗
OP
c) ⃗⃗⃗⃗⃗
BP
d) ⃗⃗⃗⃗⃗⃗
BM
3
1
⃗⃗⃗⃗⃗ , show that ⃗⃗⃗⃗⃗⃗
Given that ⃗⃗⃗⃗⃗
BN = 𝑘BA
MN = (𝑘 − ) 𝑎 + ( − 𝑘) 𝑏
ii)
6.
b) ⃗⃗⃗⃗⃗
BA
2
2
2007 Oct/Nov Exam
1
In the diagram below, ⃗⃗⃗⃗⃗
OP = 𝑝, ⃗⃗⃗⃗⃗
OQ = 𝑞, ⃗⃗⃗⃗⃗⃗
OM = ⃗⃗⃗⃗⃗
OQ and PN : NQ = 3 : 1
3
P
N
𝒑
Q
X
𝒒
M
i)
O
Express in terms of 𝑝 and/or 𝑞
b) ⃗⃗⃗⃗⃗
PQ
7.
b)
⃗⃗⃗⃗⃗
ON
c) ⃗⃗⃗⃗⃗⃗
PM
ii)
⃗⃗⃗⃗⃗ , express OX in terms of p, q and h.
Given that ⃗⃗⃗⃗⃗
OX = ℎON
iii)
1
⃗⃗⃗⃗ = 𝑘PM
⃗⃗⃗⃗⃗⃗ , show that ⃗⃗⃗⃗⃗
Given that PX
OX = (1 − 𝑘)𝑝 + 𝑘𝑞.
3
2009 Oct/Nov Exam
In the diagram below, ABC and ADE are straight lines. CD and BE intersect at F so that BF : FE = 1 : 2. B is
C
the midpoint of AC and D is the midpoint of AE.
B
F
𝒒
A
𝒑
E
D
Given that ⃗⃗⃗⃗⃗
AB = 𝑞 and ⃗⃗⃗⃗⃗
AD = 𝑝
i)
Express in terms of 𝑝 and/or 𝑞
⃗⃗⃗⃗⃗
a) BE
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b)
⃗⃗⃗⃗
FE
⃗⃗⃗⃗⃗
c) DF
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ii)
⃗⃗⃗⃗⃗ , write down an expression for DC, and hence or otherwise show that ⃗⃗⃗⃗⃗
Given that ⃗⃗⃗⃗⃗
DC = ℎDF
AC =
1
2
3
3
(1 − ℎ) 𝑝 + ℎ𝑞.
Concise Revision Notes:

Trigonometry deals with the relations between the sides and angles of a triangle.

Ratio of sides a right – angled triangle i.e.
NB: the longest side is the hypotenuse (hyp). The other two sides are considered as opposite (opp) and
adjacent (adj) in relation to one of the acute angles as shown above.

Let ∠𝐴 = 𝜃, there are three important ratios of the sides, which depends on the value of θ. They are 𝒔𝒊𝒏𝒆 (sin),
𝒄𝒐𝒔𝒊𝒏𝒆 (cos) and 𝒕𝒂𝒏𝒈𝒆𝒏𝒕 (tan), defined as follows:
𝐬𝐢𝐧 𝜽 =

𝒐𝒑𝒑
𝐜𝐨𝐬 𝜽 =
𝒉𝒚𝒑
𝒂𝒅𝒋
𝐭𝐚𝐧 𝜽 =
𝒉𝒚𝒑
𝒐𝒑𝒑
𝒂𝒅𝒋
The area of a triangle is defined as:
𝟏
Area of ∆𝑨𝑪𝑩 = 𝒂𝒃 ∙ 𝐬𝐢𝐧 𝑪
𝟐

The sine rule for any triangle – it is usual to label the sides as 𝒂, 𝒃 𝒂𝒏𝒅 𝒄 opposite the angles A, B and C
respectively. Therefore, this is defined as:
Sine Rule:

𝒂
𝐬𝐢𝐧 𝑨
𝒃
𝒄
= 𝐬𝐢𝐧 𝑩 = 𝐬𝐢𝐧 𝑪 𝒐𝒓
𝐬𝐢𝐧 𝑨
𝒂
=
𝐬𝐢𝐧 𝑩
𝒃
=
𝐬𝐢𝐧 𝑪
𝒄
The cosine rule for any triangle – if two sides and an included angle are known, then sine rule is
inconformable. For this, cosine rule is used. This is defined as:
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
The rule begins and ends with the same letter, starting with a side and end with the angle opposite that side.

NB: if ∠𝐶 is obtuse, 𝑐𝑜𝑠 𝐶 will be negative.

Positive, Negative and Basic Angles:

With reference to the origin O, an angle measured anticlockwise from the 𝑥 − 𝑎𝑥𝑖𝑠 is positive and an angle
measured clockwise from the the 𝑥 − 𝑎𝑥𝑖𝑠 is negative. i.e.

The basic angle is the acute angle between a straight line from the origin, O and 𝑥 − 𝑎𝑥𝑖𝑠𝑥 − 𝑎𝑥𝑖𝑠. E.g.
𝜶𝟏 , 𝜶𝟐 , 𝜶𝟑 , 𝜶𝟒 are the basic angles.

Signs of trugonometrical ratios in a CAST CIRCLE:

For 𝜃 in the first quadrant (𝑄1 ), all 𝑠𝑖𝑛 𝜃, 𝑐𝑜𝑠 𝜃 and 𝑡𝑎𝑛 𝜃 are positive. Therefore, the range is:
𝟎° ≤ 𝜽 ≤ 𝟗𝟎°.

For 𝜃 in the second quadrant (𝑄2 ) only 𝑠𝑖𝑛 𝜃 is positive. Therefore, the range is: 𝟎° ≤ 𝜽 ≤ 𝟏𝟖𝟎°.

For 𝜃 in the third quadrant (𝑄3 ) only 𝑡𝑎𝑛 𝜃 is positive. Therefore, the range is: 𝟎° ≤ 𝜽 ≤ 𝟐𝟕𝟎°.
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
For 𝜃 in the fourth quadrant (𝑄4 ) only 𝑐𝑜𝑠 𝜃 is positive. Therefore, the range is: 𝟎° ≤ 𝜽 ≤ 𝟑𝟔𝟎°.
NB: the sentence “All Science Teachers are Crazy” will help you remember the positive ratios.

Basic Trigonometrical Equations:
To solve trigonometrical equations of the form 𝒔𝒊𝒏 𝜽 = 𝒌, 𝒄𝒐𝒔 𝜽 = 𝒌 and 𝒕𝒂𝒏 𝜽 = 𝒌, carry out the following
steps, using 𝑠𝑖𝑛 𝜃 = 𝑘 as an illustration:
To solve 𝒔𝒊𝒏 𝜽 = 𝒌:
Step 1: find the basic angle, 𝜶 such that 𝑠𝑖𝑛 𝛼 = |𝑘|, |𝑘| means take the positive value of 𝑘.
Step 2: find the quadrants in which 𝜽 must lie from the sign (+ ve or – ve) of 𝑠𝑖𝑛 𝜃.
Step 3: find the relevant angles in these quadrants and solve for 𝜽. A basic equation usually has two (2)
solutions for 𝟎° ≤ 𝜽 ≤ 𝟑𝟔𝟎° (i.e. one complete turn of 360° .
Note:
a) For 𝑸𝟏 − 𝜽 = 𝜶
b) For 𝑸𝟐 − 𝜽 = 𝟏𝟖𝟎° − 𝜶
c)
For 𝑸𝟑 − 𝜽 = 𝟏𝟖𝟎° + 𝜶
d) For 𝑸𝟒 − 𝜽 = 𝟑𝟔𝟎° − 𝜶
Specimen Examination Questions:
1.
2019 Aug Exam Q8 (a).
(a) In the triangle ABC below, AC = 275𝑘𝑚, angle BAC = 125° and angle ACB = 40° .
Calculate
(i)
The distance BC,
(ii)
The area of triangle ABC,
(iii)
The shortest distance from A to BC.
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(b) Solve equation 13cos θ = 5 for 0 ≤ θ ≤ 360°
2.
2018 Oct/Nov Exams, Q8
In the diagram below, K, N, B and R are places on horizontal surface. KN = 80m,
̂ N = 52°.
NB = 50m and KR
80 m
K
60°
N
50 m
B
a) Calculate:
i)
KR
52°
R
b) Given that the area of triangle KNR is equal to 3 260𝑚2 , calculate the shortest distance from R to KN.
ii)
c)
the area of triangle KNB
Sketch the graph of 𝑦 = cos 𝜃 for 0° ≤ θ ≤ 360°.
d) Solve sin 𝜃 = −0.5 for 0° ≤ θ ≤ 360°.
3.
2018 July/Aug Exams, Q8
a) Three villages A, B and C are connected by straight paths as shown in the diagram below.
A
B
79°
40°
C
Given that AB = 15km, angle ABC = 79° and angle ACB = 40°, calculate the:
i)
Distance AC
ii)
area of triangle ABC
iii)
Shortest distance from B to AC.
b) Solve the equation cos 𝜃 = 0.937 for 0° ≤ θ ≤ 360°.
c)
4.
Sketch the graph of 𝑦 = sin 𝜃 for 0° ≤ θ ≤ 360°.
2017 Oct/Nov Exams, Q7
a) The diagram below shows the Location of houses for a village Headman (H), his Secretary (S) and a
̂ S = 130°.
Trustee (T). H is 1.3 km from S, T is 1.9 km from H and TH
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Calculate:
i)
the area of triangle THS.
ii)
the distance TS
iii)
the shortest distance from H to TS
2
b) Find the angle between 0° and 90° which satisfies the equation cos 𝜃 = .
3
5.
2017 July/ Aug Exams, Q10
a) In Triangle PQR below, QR = 36.5, angle PQR = 36° and angle QPR = 46°.
Calculate:
i)
PQ
ii)
the area of triangle PQR
iii)
the shortest distance from R to PQ
b) Solve the equation sin θ = 0.6792 for 0° ≤ θ ≤ 360°.
6.
2016 Oct/Nov Exams, Q10
a) The diagram below shows the location of three secondary schools, namely Mufulira (M), Kantanshi (K)
and Ipusukilo (I) in Mufulira district. M is 5km from K, I is 3km from K and angle MKI is 110° .
Calculate:
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i)
MI
ii)
the area of triangle MKI
iii)
the shortest distance from K to MI.
b) Solve the equation tan 𝜃 = 0.7 for 0° ≤ θ ≤ 180°.
7.
i)
Miscellaneous Question:
A, B and C are Food Reserve Agency Maize buying points in a given district. B is 32km from A on a
bearing of 060° and C is 27km on a bearing of 130° as shown in the diagram below.
N
B
60°
A
Calculate:
130°
(a) Angle CAB,
(b) The area of triangle ABC,
C
(c) BC to the nearest kilometer,
(d) AX, given that there is a shopping centre X along BC such that AX is the shortest distance from A.
ii)
During a soccer training session, Frendo (F) was standing at the centre of the goal posts, Mukulu (M) was
standing 21 metres from Frendo’s position, Chitalu (C) was 19 metres from Frendo and angle CFM = 110°,
as shown in the diagram below.
M
C
110°
F
a) Calculate the area of FMC to the nearest square meter.
b) Chitalu (C) rolls a ball along CM for Mukulu (M) to kick to the goal mouth. Calculate the distance
CM.
c)
Frendo (F) is free to intercept the ball at any point along CM before it reaches Mukulu at M. find the
shortest distance which Frendo could run in order to intercept it.
iii)
Three towns Choma (C), Monze (M) and Namwala (N) are such that the distance from Monze to Choma is
100km and Monze to Namwala 80km. the bearing of Namwala from Monze is 060° and the bearing of
Choma from Monze is 290° .
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N
C
N
°
𝑥 ° 60
(a) Calculate the
M
290°
(i)
Value of 𝑥
(ii)
Distance from Choma to Namwala correct to 2 decimal places
(iii)
Area covered by triangle CMN correct to 2 decimal places
(iv)
Given that Nikoh (H) is a bus station on the Choma – Namwala route such that MH is the shortest
distance from Monze to Nikoh, calculate this shortest distance MH, correct to 2 decimal places
(b) Hence, find how far Nikoh is from Namwala, giving your answer correct to 2 decimal places.
(c) sin2 θ + 2 sin θ cos θ = 0 for 0° ≤ θ ≤ 360°.
Concise Revision Notes:

The mode is the value occurring the most often.

The median is the middle most value when all the data is arranged in order of size.

The mean is the average value of data set. It is found by adding together all the values of the data and then
dividing that total by the number of data values.

The simplest measure of spread is the range. The range is simply the difference between the largest and
smallest values in the data.

The mean of ungrouped data is found by adding together all the values of the data and then dividing that total
by the number of data values. Therefore, it is defined as:
̅=
𝒎𝒆𝒂𝒏, 𝒙

∑𝒙
𝒏
The mean of grouped data can only be an estimate as the position of the data within a group is not known. An
estimate is made by calculating the the mid – interval value for a group and then assigning all of the data
within the group that mid – interval value. Thus, it is calculated as:
̅=
𝒎𝒆𝒂𝒏, 𝒙
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∑ 𝒇𝒙
∑𝒇
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
When the data are grouped, their individual values are lost. Hence, we cannot arrange them in order to find the
median. We assume that they are spread evenly through each class so we can only estimate the median. We look
at a graphical method of estimating the median, using a cumulative frequency polgon.

Variance measures the distances or spread of data about the mean.

Standard deviation is the square root of the variance.

The cumulative frequency range is divided into two equal parts by the median. Sometimes it is useful to divide
the frequency range into four equal parts called the quartiles. The second quartile is the median and the other
two dividing lines are called the lower quartile (𝑸𝟏 ) and upper quartile (𝑸𝟐 ).
1

The lower quartile is way along the cumulative frequency axis 𝑦 − 𝑎𝑥𝑖𝑠 and then reciprocate to 𝑥 − 𝑎𝑥𝑖𝑠.

The upper quartile is way along the cumulative frequency 𝑦 − 𝑎𝑥𝑖𝑠 and then reciprocate to 𝑥 − 𝑎𝑥𝑖𝑠.

The difference between the values of the variable at the upper quartile and lower quartile is called the
4
3
4
interquartile range. (𝒊. 𝒆. 𝒊𝒏𝒕𝒆𝒓𝒒𝒖𝒂𝒓𝒕𝒊𝒍𝒆 𝒓𝒂𝒏𝒈𝒆 = 𝑸𝟐 − 𝑸𝟏 )
𝑸𝟐 −𝑸𝟏

Half of the interquartile range is called the semi – interquartile range (𝑖. 𝑒.

The relative cumulative frequency is the quotient between the cumulative frequency of a particular value and
2
).
the total number of data. Therefore, the relative cumulative frequency helps in determining the percentile. To
find the relative cumulative frequency.
𝒊. 𝒆. 𝑹𝑪𝑭 =
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𝒄𝒍𝒂𝒔𝒔 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒄𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒗𝒆 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
𝒕𝒐𝒕𝒂𝒍 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚
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Page 37 of 192
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q9
The frequency table below shows the distribution of marks obtained by 90 learners on a test.
𝒎𝒂𝒓𝒌𝒔 (𝒙)
10 < 𝑥 ≤ 20
20 < 𝑥 ≤ 30
30 < 𝑥 ≤ 40
40 < 𝑥 ≤ 50
50 < 𝑥 ≤ 60
60 < 𝑥 ≤ 70
𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 (𝒇)
2
10
15
23
30
10
a) Calculate the standard deviation
b) Answer this part of this question on the sheet of graph paper.
i)
copy and complete the cumulative frequency table
Mark (𝑥)
≤ 10
≤ 20
≤ 30
≤ 40
≤ 50
≤ 60
≤ 70
Cumulative frequency
0
2
12
27
50
80
90
Relative Cumulative frequency
0
0.02
0.13
0.3
ii)
Using a scale of 2cm to represent 10 units on the x – axis for 0 ≤ 𝑥 ≤ 70 and a scale of 2cm to
represent
0.1
units
on
the
y-axis
for
0 ≤ 𝑦 ≤ 1, draw a smooth relative cumulative frequency curve.
iii)
2.
Showing your method clearly, use your graph to estimate the 65 th Percentile.
2018 July/ Aug Exams, Q11
A farmer planted 60 fruit trees. In a certain month, the number of fruits per tree was recorded and the results
were as shown in the table below.
Fruits per tree
2
3
4
5
6
7
8
Frequency
1
5
4
6
10
16
18
a) Calculate the standard deviation
b) Answer this part of the question on the sheet of graph paper
i)
Using the table above, copy and complete the relative cumulative frequency table below.
Fruits per tree
2
3
4
5
6
7
8
Cumulative frequency
1
6
10
16
26
42
60
0.02
0.1
0.17
0.27
Relative cumulative frequency
ii)
Using a scale of 1cm to represent 1 unit on the x-axis for 0 ≤ 𝑥 ≤ 8 and a scale of 2cm to
represent 0. 1 unit on the y – axis for 0 ≤ 𝑥 ≤ 1, draw a smooth relative frequency curve.
iii)
3.
Showing your method clearly, use your graph to estimate the 70 th Percentile.
2017 Oct / Nov Exams, Q8
The table below shows the amount of money spent by 100 learners at school on a particular day.
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𝒂𝒎𝒐𝒖𝒏𝒕 𝒊𝒏 𝒌𝒘𝒂𝒄𝒉𝒂
0<𝑥≤5
5 < 𝑥 ≤ 10
10 < 𝑥 ≤ 15
15 < 𝑥 ≤ 20
20 < 𝑥 ≤ 25
25 < 𝑥 ≤ 30
𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 (𝒇)
13
27
35
16
7
2
a) Calculate the standard deviation.
b) Answer this part of the question on a sheet of graph paper.
i)
Using the table above, copy and complete the cumulative frequency table below.
Amount (𝐢𝐧 𝐤𝐰𝐚𝐜𝐡𝐚)
Frequency
ii)
≤0
≤5
≤ 10
0
13
40
≤ 15
≤ 20
≤ 25
≤ 30
100
Using a scale of 2cm to represent 5 units on the horizontal axis and 2cm to represent 10 units on
the vertical axis, draw a smooth cumulative frequency curve.
iii)
4.
Showing your method clearly, use your graph to estimate the semi − interquartile range.
2017 July/ Aug Exams, Q8
The frequency table below shows the number of copies of newspapers allocated to 48 newspaper vendors.
Copies of Newspaper
Number of vendors
25 < 𝑥 ≤ 30
30 < 𝑥 ≤ 35
35 < 𝑥 ≤ 40
40 < 𝑥 ≤ 45
45 < 𝑥 ≤ 50
50 < 𝑥 ≤ 55
55 < 𝑥 ≤ 60
6
4
7
11
12
8
1
a) Calculate the standard deviation.
b) Answer this part of the question on a sheet of graph paper.
i)
Using the table above, copy and complete the cumulative frequency table below.
Copies of Newspaper
≤ 25
≤ 30
≤ 35
≤ 40
≤ 45
0
5
9
16
27
Number of vendors
ii)
≤ 50
≤ 55
≤ 60
Using a horizontal scale of 2cm to represent 10 newspapers on the x – axis for
0 ≤ 𝑥 ≤ 60 and a vertical scale of 4cm to represent 10 vendors on the y – axis for 0 ≤ 𝑥 ≤ 50,
draw a smooth cumulative frequency curve.
iii)
5.
Showing your method clearly, use your graph to estimate the 50 th Percentile.
2016 Oct/ Nov Exams, Q7
The ages of people living at Pamodzi Village are recorded in the frequency table below.
Age
0 < 𝑥 ≤ 10
10 < 𝑥 ≤ 20
20 < 𝑥 ≤ 30
30 < 𝑥 ≤ 40
40 < 𝑥 ≤ 50
50 < 𝑥 ≤ 60
7
22
28
23
15
5
Number of people
a) Calculate the standard deviation
b) Answer this part of the question on a sheet of graph paper
i)
Using the table above, copy and complete the cumulative frequency table below.
Age
Number of people
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≤ 10
≤ 20
7
29
≤ 30
≤ 40
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≤ 50
≤ 60
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ii)
Using the scale of 2cm to represent 10 units on both axes, draw a smooth cumulative frequency
curve where 0 ≤ 𝑥 ≤ 60 and 0 ≤ 𝑦 ≤ 100.
Showing your method clearly, use your graph to estimate the Semi− interquartile range.
iii)
Concise Revision Notes:

The general expression for a quadratic function takes the form 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 where 𝒂, 𝒃 𝑎𝑛𝑑 𝒄 are constants.

If the graph of a quadratic function is plotted, the smooth curve produced is called a parabola.

If 𝑎 > 0, the graph of 𝒚 = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 has the
with a lowest point P called the minimum point.

If 𝑎 < 0, the graph of 𝒚 = 𝒂𝒙 + 𝒃𝒙 + 𝒄 has the

To find the maximum or minimum value of 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐.
𝟐
𝑏𝑥
𝑐
i)
Rewrite 𝑓(𝑥) as a [𝑥 2 +
ii)
Complete the square on 𝑥 2 +
iii)
The maximum/minimum occurs at 𝑥 = −
iv)
If 𝑎 < 0, the value of 𝑓(𝑥) = 𝑓 (−
𝑎
with a highest point Q called the maximum point.
+ ].
𝑎
𝑏𝑥
𝑎
to convert 𝑓(𝑥) to 𝑎 [(𝑥 +
If 𝑎 > 0, the value of 𝑓(𝑥) = 𝑓 (−
𝑏
2𝑎
𝑏
2𝑎
𝑏
2𝑎
𝑐
𝑏2
𝑎
4𝑎2
) + −
].
and the maximum/minimum value of 𝑓(𝑥) is (−
𝑏
2𝑎
).
) is maximum.
) is minimum.
Note: If 𝑎 < 0, the coordinates of the maximum point are [(−
If 𝑎 > 0, the coordinates of the maximum point are [(−

2
𝑏
2𝑎
𝑏
2𝑎
𝑏
2𝑎
) , 𝑓 (−
) , 𝑓 (−
𝑏
2𝑎
𝑏
2𝑎
)].
)].
The gradient of a curve, however, is not constant: its slope changes. To calculate the gradient of a curve at
a specific point, the following steps need to be taken:
i)
Draw a suitable tangent to the curve at that point.
ii)
Calculate the gradient of the tangent.
NB: the gradient (m) of a straight line is constant and is calculated by considering the coordinates of
two of the points on the line and then carrying out the calculation 𝒎 =

𝒚𝟐 − 𝒚𝟏
𝒙𝟐 − 𝒙𝟏 .
If a graph of a function is plotted, then it can be used to solve equations.
NB: to solve the equation, 𝑦 = 0. Therefore, where the curve intersects the 𝑥 − 𝑎𝑥𝑖𝑠 gives the solution to
the equation.

Approximation of Area enclosed under the curve:
In general the area of a shape enclosed by a curve can only be found approximately. One method is to use the
Trapezium rule.
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i)

Trapezium Method Rule
Divide the area under the curve into different shapes trapeziums, rectangles, squares or triangles and use
their appropriate formulas for their areas:
𝟏
a) Area of trapezium = (𝒂 + 𝒃) × 𝒉
𝟐
𝟏
b) Area of triangle = 𝒃𝒉
𝟐
c)
Area of rectangle = 𝒍𝒃
d) Area of square = 𝒍𝟐
ii)

Counting the Square Method
Divide the area under the curve into convient shapes, based on the squares of the graph paper in each
shape, the number of “small squares” that it contains. NB: the small squares have sides 2mm.

Estimate the remaining area by counting anything more than half a small square as one square, ignoring
those bits which are less than half a small square.
Specimen Examination Questions:
All the questions in this topic are to be answered on the sheet of graph papers.
1.
2018 Oct/Nov Exams, Q7(a)
The values 𝑥 and 𝑦 are connected by the equation 𝒚 = 𝟐𝒙𝟑 − 𝟑𝒙𝟐 + 𝟓.
Some corresponding values of 𝑥 and 𝑦 are given in the table below.
𝒙
−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
𝒚
𝑝
−8.5
0
4
5
4.5
4
5
9
a) Calculate the value of P
b) Using a scale of 4cm to represent 1 unit on the 𝑥 − axis for −2 ≤ 𝑥 ≤ 2 and a scale of 2cm to represent 5
units on the 𝑦 − axis for −25 ≤ 𝑦 ≤ 10, draw the graph of
𝒚 = 𝟐𝒙𝟑 − 𝟑𝒙𝟐 + 𝟓.
c)
Use your graph to solve the equation 2𝑥 3 − 3𝑥 2 + 5 = 𝑥
d) Calculate an estimate of the gradient of the curve at the point where 𝑥 = 1.5.
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2.
2018 July/ Aug Exams, Q12 (a)
The diagram below shows the graph of 𝒚 = 𝒙𝟑 + 𝒙𝟐 + 𝟏𝟐𝒙 .
a) Use the graph to solve the equation:
i)
𝑥 3 + 𝑥 2 + 12𝑥 = 0
ii)
𝑥 3 + 𝑥 2 + 12𝑥 = 𝑥 + 10
b) Calculate an estimate of the:
3.
i)
Gradient of the curve at the point where 𝑥 = −3.
ii)
Area bounded by the curve, 𝑥 = −3, 𝑥 = −1 and 𝑦 = −10.
2017 Oct / Nov Exams, Q10 (a).
The diagram below shows the graph of 𝒚 = 𝒙𝟑 + 𝟑𝒙𝟐 − 𝒙 − 𝟑.
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a) Use the graph to find the solutions of the equations:
i)
𝑥 3 + 3𝑥 2 − 𝑥 − 3 = 0
ii)
𝑥 3 + 3𝑥 2 − 𝑥 − 3 = 5
b) Calculate an estimate of:
i)
the gradient of the curve at the point (−3, 0).
ii)
the area bounded by the curve, 𝑥 = 0, 𝑦 = 0 and 𝑦 = 20.
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4.
2017 July/Aug Exams, Q9 (a)
The diagram below shows the graph of 𝒚 = 𝒙𝟑 + 𝒙𝟐 − 𝟓𝒙 + 𝟑.
Use the graph:
a) to calculate an estimate of the gradient of the curve at the point (2, 5).
b) to solve the equations:
i)
𝑥 3 + 𝑥 2 − 5𝑥 + 3 = 0
ii)
𝑥 3 + 𝑥 2 − 5𝑥 + 3 = 5𝑥
c)
Calculate an estimate of the area bounded by the curve 𝑥 = 0, 𝑦 = 0 and 𝑥 = −2.
5.
2016 Oct/ Nov Exams, Q9 (a)
The values of x and y are connected by the equation 𝑦 = 𝑥(𝑥 − 2)(𝑥 + 2). Some corresponding values of x and
y are given in the table below;
𝒙
−3
−2
−1
0
1
2
3
𝒚
−15
0
3
0
−3
0
𝑘
a) Calculate value of 𝑘
b) Using a scale of 2cm to represent 1 unit on the x – axis for −3 ≤ 𝑥 ≤ 3 and 2cm to represent 5 units on the
y - axis for −16 ≤ 𝑥 ≤ 16. Draw the graph of
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𝑦 = 𝑥(𝑥 − 2)(𝑥 + 2).
c)
6.
Use your graph to solve the equations:
i)
𝑥(𝑥 − 2)(𝑥 + 2) = 0
ii)
𝑥(𝑥 − 2)(𝑥 + 2) = 𝑥 + 2
2006 Oct/Nov Exam Q7.
The variable 𝑥 and 𝑦 are connected by the equation 𝒚 = 𝟖 − 𝒙 − 𝟐𝒙𝟐 .
Some of the corresponding values of 𝑥 and 𝑦, correct to 1 decimal place where necessary are given in the table
below.
𝒙
−3
−2
−1
−0.5
0
0.5
1
2
3
𝒚
−7
2
7
8
8
𝒒
5
−2
−13
a) Calculate the value of q.
b) Taking 2cm to represent 1 unit on the 𝑥 − 𝑎𝑥𝑖𝑠, and 1cm to represent 2 units on the 𝑦 − 𝑎𝑥𝑖𝑠, draw
the graph of 𝒚 = 𝟖 − 𝒙 − 𝟐𝒙𝟐 .
c)
Use your graph to solve:
i)
𝟖 − 𝒙 − 𝟐𝒙𝟐 = 𝟎
ii)
𝟖 − 𝒙 − 𝟐𝒙𝟐 = 𝒙 + 𝟐
d) Estimate the area bounded by the curve, the 𝑥 − 𝑎𝑥𝑖𝑠, 𝒙 = −𝟏 and 𝒙 = 𝟏.
e)
7.
Find the maximum value of 𝒚 = 𝟖 − 𝒙 − 𝟐𝒙𝟐 .
2010 Oct/Nov Exam Q9.
𝟏
The variable x and y are connected by the equation 𝒚 = 𝒙(𝟏𝟎 − 𝒙𝟐 ).
𝟓
The table below shows some corresponding values of x and y. the values of y are given to one decimal place
where necessary.
𝒙
−1
0
1
2
3
3.5
4
𝒚
−1.8
0
1.8
2.4
0.6
−1.6
𝒑
(a) Calculate the value of p.
(b) Using a scale of 2cm to 1 unit on both axes for −1 ≤ 𝑥 ≤ 5 and −5 ≤ 𝑦 ≤ 3, draw the graph of 𝒚 =
𝟏
𝟓
𝒙(𝟏𝟎 − 𝒙𝟐 ).
(c) By drawing a tangent to the curve, estimate the gradient of the curve at the point (1, 1.8).
(d) On the same graph, draw the line whose equation is 5𝑦 + 4𝑥 = 4.
𝟏
𝟒
𝟒
𝟓
𝟓
𝟓
(e) Use your graphs to find the solutions of 𝒚 = 𝒙(𝟏𝟎 − 𝒙𝟐 ) = − 𝒙 + .
(f) Estimate the area bounded by the curve between 𝑥 = 1, 𝑥 = 3 and the line 𝑦 = 0.
8.
Miscellaneous Question.
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Fig. 8.21 shows the graphs of 𝑦 = 2 − 𝑥 − 𝑥 2 𝑎𝑛𝑑 𝑥 + 2𝑦 = 2 intersecting at A and B.
(a) From the graph, find the solutions of the equations
(i)
𝑥2 + 𝑥 − 2 = 0
(ii)
𝑥2 + 𝑥 − 1 = 0
(b) Write down and simplify an equation in 𝑥 whose roots are the 𝑥 − coordinates of A and B.
(c) By drawing a suitable tangent, estimate the gradient of the curve at the point where 𝑥 = −1
.
Concise Revision Notes:

Earth Geometry deals with the location of places, calculation of time of distances, speed in knots, determining
the positions of points and determining time differences between places on the earth’s surface etc.

Terminologies used:
 Latitude – these are an imaginary line that runs from West to East on the earth’s surface and they are
parallel to each other.
 Equator – is the line of latitude which cuts the earth into two equal parts called hemispheres.
 Longitude - these are an imaginary line that runs from North to South on the earth’s surface and they are
also known as meridians.
 Great circles – any imaginary circle around the earth’s surface whose centre is at the centre of earth is
called the Great Circle.
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
When describing positions on the earth’s surface, start first with the latitudes then followed by longitudes
i.e. (𝒍𝒂𝒕𝒊𝒕𝒖𝒅𝒆, 𝒍𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒆), include the direction, i.e. N, S, E or W.

The radius of the circle of latitude is defined by 𝒓 = 𝑹 𝒄𝒐𝒔 𝒙° . NB: the radius (R) of the earth is approximately
6370km or 3437 nautical miles (nm).
Note that: 𝟏° = 𝟔𝟎𝒏𝒎.

The circumference of a circle is the distance around the circle. Therefore, it is defined by the formula: 𝑪 =
𝟐𝝅𝑹 𝒄𝒐𝒔 𝒙° , where 𝑥 ° is the latitude.

When two points are passing through on the same latitude, but different longitudes, then the distance between
them is defined by: 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 =
𝜽
𝟑𝟔𝟎°
× 𝟐𝝅 𝒄𝒐𝒔 𝒙° , where θ is the difference in longitudes and 𝑥 ° is the latitude.
This is called the distance along a circle of latitude.

When two points are passing through on the same longitude, but different latitudes, then the distance between
them is defined by: 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 =
𝑥°
𝟑𝟔𝟎°
× 𝟐𝝅𝑹, where 𝑥 ° is the difference in latitudes. This is called the distance
along a circle of longitude.

Speed in nautical miles per hour is measured in knots.

Time calculation

The earth rotates on its axis from west to east making a complete turn/rotation of 360° in 24 hours.
Therefore, it rotates 15° in 1 hour.

The world is divided into time zones which are determined from the universal line at the Greenwich
meridian. The universal time is called Greenwich Mean Time (GMT).
Note that:
i) For every 15° moved to the east of the Greenwich meridian, one hour is gained in time against GMT.
ii) For every 15° moved to the west of the Greenwich meridian, one hour is lost in time against GMT.

Time is always calculated along the lines of longitudes only.
Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q12 (b)
The points A(15° N, 40° E), B (35° N, 70° E) and C (35° S, 40° E) are on the surface of the Earth. [use 𝛑 =
𝟑. 𝟏𝟒𝟐, 𝐑 = 𝟔𝟑𝟕𝟎𝐤𝐦].
a) Calculate the distance AC in kilometers.
b) An aero plane takes off from point B and flies due west on the same latitude covering a distance of 900km
to point Q.
2.
i)
Calculate the difference in longitudes between B and Q.
ii)
Find the position of Q.
2018 July/ Aug Exams, Q7(a)
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In the diagram below, A and B are points on latitude 60° N while C is a point on latitude 60° S. [𝜋 = 3.142 and R
= 3437nm].
a) Calculate the distance BC along the latitude 60° E in nautical miles.
b) A ship sails from C to D in 12 hours. Find its speed in notes.
3.
2017 Oct/ Nov Exams, Q9 (a)
W, X, Y and Z are four points on the surface of the earth as shown in the diagram below. (Take 𝛑 = 𝟑𝟑.𝟏42
and 𝐑 = 𝟑437nm).
a) Calculate the difference in latitudes between W and Y.
b) Calculate the distance in nautical miles between
4.
i)
X and Z along the longitudes 105° E
ii)
Y and Z along the circle of latitude 30° S
2017 July/Aug Exams, Q12 (a)
P (80° N, 10° °E), 𝐐 (80° N, 70° E), 𝐑 (85° S, 70° E) and 𝐒 (85° S, 10° E) are the points on the surface of the earth.
i)
Show the points on a clearly labeled sketch of the surface of the earth
ii)
Find in nautical miles:
a) The distance QR along the longitude,
b) The circumference of latitude 85° S.
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(Take 𝛑 = 𝟑𝟑.𝟏42 and 𝐑 = 𝟑437nm).
5.
2016 Oct/ Nov Exams, Q9 (a)
The points A, B, C and D are on the surface of the earth.
(Take π = 3.142 and R = 3437nm).
a) Find the difference in latitude between points C and B.
b) Calculate the length of the circle of latitude 50° N in nautical miles.
c)
6.
Find the distance AD in nautical miles.
Miscellaneous Question.
Two aircraft tracking stations are located at (120° W, 40° N) and at (100° W, 40° N). Each has a range of 500
nautical miles.
i)
By what distance do they overlap?
ii) An aeroplane, flying east along the 40° N parallel of latitude, is picked up by the first station at 07 00 hours
and is out of range of the second by 12 30 hours. What was the average speed of the aeroplane?
iii) Find the longitude of the aeroplane when it is first picked up.
Concise Revision Notes:

Linear programming is a method of solving problems involving two variables that are subject to certain
conditions using the graphical representation of inequalities. (The names come from – programming: a
method, linear: using straight lines).

Linear programming is a way of finding a number of possible solutions to a problem given a number of
constraints or conditions. But it is more than this – it is also a method for minimizing or maximizing a linear
function in two or more variables i.e. 𝒚 = 𝒂 + 𝒃𝒙.
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
The function of linear programming is to either maximize or minimize a linear function subject to certain
conditions/constraints represented by linear inequations.
For instance, solving the inequation 3𝑥 − 1 < 2𝑥 + 5, 𝑥 𝜖 𝑅 and draw the graph of its solution set.
𝒚
𝒙<𝟔
Solution:
𝒙
3𝑥 − 1 < 2𝑥 + 5
2
4
6
3𝑥 − 2𝑥 < 1 + 5
𝑥 < 6 𝑎𝑛𝑠:
NB: in the graph above, the line 𝑥 = 6 is the boundary between the wanted region (the solution set) and the
unwanted region. Since the points on the line 𝑥 = 6 are not included in the solution set the line 𝑥 = 6 is broken
or dotted. Since 𝑥 = 6, the points to the left of the line 𝑥 = 6 are in the required region. Thus, the region to the
right of the line is shaded to show that it is not wanted.

Therefore, if an inequality has the sign < 𝒐𝒓 > draw a broken or dotted line on a graph. And if it is ≤ 𝒐𝒓 ≥
draw undotted or just solid straight line. And always shade opposite the inequality sign, unless the one that
passes through the origin shade according to the given sign.

In linear programing, a mathematical model is sometimes used as a system of inequalities which are formed
from the conditions in a problem.

To find the maximum (or minimum) value of the objective function, we consider the vertex of the feasible
region. If the point does not give a practical solution, then the nearest points should be considered.

To get the best solution called the optimal solution, we must find points whose coordinates satisfy the given
conditions and which give the maximum value of 𝑥 + 𝑦. The other alternative is to solve any two inequalities
simultaneously to confirm picked points.

However, an understanding of the following symbols is necessary under linear programming:
>
Means is “greater than”
≥
Means is “greater than or equal to”
<
Means is “less than”
≤
means is less than or equal to”
≥
can mean “at least or more than”
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≤
can mean “not more than”
Specimen Examination Questions:
All the questions in this topic are to be answered on the sheet of graph papers.
1.
2018 Oct/Nov Exams, Q7 (a)
A hired bus is used to take learners and teachers on a trip. The number of learners and teachers must be more
than 60. There must be at least 35 people on the trip. There must be at least 6 teachers on the trip. The number
of teachers on the trip should not be more than 14.
Let 𝑥 be the number of learners and 𝑦 be the number of teachers.
a) Write four inequalities which present the information above.
b) Using a scale of 2cm to represent 10 units on both axes, draw the x and y axes for 𝟎 ≤ 𝒙 ≤ 𝟕𝟎 and
𝟎 ≤ 𝒚 ≤ 𝟕𝟎 respectively and shade the unwanted region to indicate clearly where the solution of the
inequalities lie.
c)
Using your graph:
i)
If the group has 25 learners, what is the minimum number of teachers that must accompany them?
ii)
If 8 teachers go on the trip, what is the maximum number of learners that can be accommodated
on the bus?
d) If T is the amount in Kwacha paid by the whole group, what is the cost per learner if T = 30𝑥 + 50𝑦.
2.
2018 July/ Aug Exams, Q12 (a).
A tailor at a certain market intends to make dresses and suits for sale.
a) Let 𝑥 represent the number of dresses and 𝑦 the number of suits. Write the inequalities which represent
each of the conditions below.
i)
The number of dresses should not exceed 50.
ii)
The number of dresses should not be more than the number of suits.
iii)
The cost of making a dress is K140.00 and that o a suit is K210.00. The total should be at least K10
500.00.
b) Using a scale of 2cm to represent 10 units on both axes, draw 𝑥 - axes for 𝟎 ≤ 𝒙 ≤ 𝟔𝟎 and 𝟎 ≤ 𝒚 ≤ 𝟖𝟎.
Shade the unwanted region to indicate clearly the region where (𝑥, 𝑦) must lie.
c)
Using your grapg:
i)
The profit on a dress is K160.00 and on a suit is K270.00. Find the number of dresses and suits the
tailor must make for maximum profit.
ii)
3.
Calculate this maximum profit.
2017 Oct / Nov Exams, Q10 (a)
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Himakwebo orders maize and groundnuts for sale. The order price for a bag of maize is K75.00 and that of a
bag of groundnuts is K150.00. He is ready to spend up to K7 500.00 altogether. He intends to order at least 5
bags of maize and at least 10 bags of groundnuts. He does not want to order more than 70 bags altogether.
a) If 𝑥 and 𝑦 are the number of bags of maize and groundnuts respectively, write four inequalities which
represent these conditions.
b) Using a scale of 2cm to represent 10 bags on each axis, draw the 𝒙 and 𝒚 axes for 𝟎 ≤ 𝒙 ≤ 𝟕𝟎 and
𝟎 ≤ 𝒚 ≤ 𝟕𝟎 respectively and shade the unwanted region to show clearly the region where the solution of
the inequalities lie.
c)
Given that a profit on the bag of maize is K25.00 and on the bag of groundnuts is K50.00, how many bags
of each type should he order to have the maximum profit?
d) What is this estimate of the maximum profit?
4.
2017 July/Aug Exams, Q9 (a)
Makwebo prepares two types of sausages, Hungarian and Beef, daily for sale. She prepares at least 40
Hungarian and at least 10 Beef sausages. She prepares not more than 160 sausages altogether. The number of
Beef sausages prepared are not more than the number of Hungarian sausages.
a)
Given that 𝒙 represents the number of Hungarian sausages and 𝒚 the number of Beef sausages, write four
inequalities which represent these conditions.
b) Using a scale of 2cm to present 20cm sausages on both axes, draw the x and y axes for 𝟎 ≤ 𝒙 ≤ 𝟏𝟔𝟎 and
𝟎 ≤ 𝒚 ≤ 𝟏𝟔𝟎 respectively and shade the unwanted region to show clearly the region where the solution of
the inequalities lie.
c)
The profit on the sale of each Hungarian sausage is K3.00 and on each beef sausage is K2.00. How many of
each type of sausages are required to make maximum profit?
d) Calculate this maximum profit.
5.
2016 Oct/ Nov Exams, 11 (a)
A Health Lobby group produced a guide to encourage healthy living among local community. The group
produced the guide in two formats: a short video and a printed book. The group needs to decide the number of
each format to produce for sale to maximize profit.
Let 𝑥 represent the number of videos produced and 𝑦 the number of printed books produced.
a) Write the inequalities which represent each of the following conditions:
i)
the total number of copies produced should not be more than 800,
ii)
the number of video copies to be at least 100
iii)
the number of printed books to be at least 100.
b) Using a scale of 2cm to represent 100 copies on both axes, draw the x and y axes for 𝟎 ≤ 𝒙 ≤ 𝟖𝟎𝟎 and
𝟎 ≤ 𝒙 ≤ 𝟖𝟎𝟎 respectively and shade the unwanted region to indicate clearly the region where the solution
of the inequalities lie.
c)
The profit on the sale of each video copy is K15.00 while the profit on each printed book is K8.00. How
many of each type were produced to make maximum profit.
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6.
2006 Oct/Nov Exams Q11.
Dr. Frendo decides to buy two types of chickens; broilers and layers. Broilers cost K 15 000 per chicken and
layers cost K 20 000 per chicken. He has K 600 000 available and decides to buy at least 30 chickens. He also
decides that at least one third of the chickens should be layers. He buys 𝒙 broilers and 𝒚 layers and also makes
sure that there are at least 15 broilers.
a) Write down four inequalities which correspond to the above conditions.
b) Illustrate these inequalities on a graph using a scale of 2cm to represent 5 units on both axes.
c)
He makes a profit of K 20 000 on each broiler and K 10 000 on each layer. Assuming he sells all his
chickens, find how many chickens of each type he should buy to maximize his profit.
d) Hence, or otherwise calculate his profit.
7.
Miscellaneous Question
8.
2012 Oct/Nov Exam Q11.
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𝒚
30
25
20
15
𝐑
10
5
0
𝒙
10
5
15
20
The graph above shows three inequalities that satisfy Dr. Frendo’s intentions to purchase Science and
Mathematics text books for his daughter.
i)
Given that 𝒙 represents the number of Science text books and 𝒚 represents Mathematics text books,
write the three inequalities that represent the unshaded region R.
ii)
Given that a Science textbook costs K25 000 and a Mathematics textbook K10 000, find the largest
number of Science and Mathematics textbooks that can be bought. Hence, calculate the total cost of the
textbooks.
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Concise Revision Notes:

Mensuration means the measurement and calculation of areas and volumes. The basic formulae for common
shapes and solids are given in fig. 3.1 and 3.2. Therefore, these should be memorized. (NB: A = area or surface
area of a solid and V = volume).
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r
H
R
𝟏

Conical Frustum: 𝑽 = 𝝅(𝑹𝟐 𝑯 − 𝒓𝟐 𝒉)

Pyramid frustum with a square base and top: 𝑽 = 𝒉(𝑳𝟐 + 𝒍𝟐 + 𝑳𝒍)
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𝟑
𝟏
𝟑
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SI Units
Units
Abbreviation
Basic Units
Mass
1 kilogram
1 kg
1000 g
1 hectogram
1 hg
100 g
1 gram
1g
1g
1 milligram
1 mg
0.001 g
1 decigram
1 dg
0.1 g
1 centigram
1 cg
0.01 g
1 gram
1g
1000 mg
1 000 gram
1000 g
1 000 000 mg
1 tonne
1t
1 kilometre
1 km
1000 m
1 millimetre
1 mm
0.001 m
1 centimetre
1 cm
0.01 m
1 metre
1m
1000 mm
1 kilometre
1 km
1000000 mm
1 second
1s
1s
1 minute
1 min
60 s
1 hour
1h
3600 s
Square millimetre
mm2
Square centimetre
cm2
Length
Time
Area
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1000 kg ≈ 1000000 g
1 cm2 ≈ 100 mm2
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Volume
Capacity
Square metre
m2
1 m2 ≈ 10 000 m2
Square kilometre
km2
1 km2 ≈ 1000000 m2
Cubic millimetre
mm3
Cubic centimetre
cm3
1 cm3 ≈ 1000 mm3
Cubic metre
m3
1 m3 ≈ 1000000 cm3
Millilitre
mℓ
1 mℓ = 1cm3
Litre
ℓ 1ℓ = 1000 mℓ
1ℓ = 1000cm3
Specimen Examination Questions:
1.
2018 Oct/ Nov Exams, Q12(a)
The diagram below is a frustum of a rectangular pyramid with a base 14cm long and 10cm wide. The top of a
frustum is 8cm long and 4cm wide.
8cm
4cm
10cm
14cm
Given that the height of the frustum is 11.4cm, calculate its volume.
2.
2018 July/August Exams, Q6
The diagram below shows a bin in the form of a frustum with square base ends of t sides 4cm and 10cm
respectively. The height of the bin is 9cm.
Find the volume of the bin.
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3.
2017 Oct/ Nov Exams, Q4 (b)
The figure below is a frustum of a cone. The base diameter and top diameter are 42cm and 14 cm respectively,
while the height is 20cm. (Take 𝝅 = 𝟑.𝟏42).
Calculate its volume.
4.
2017 July/Aug Exams, Q12 (a)
The figure below is a cone ABC form which BCXY remained after the small cone AXY was cut of [Take 𝜋 =
3.142].
Given that EX = 4cm, DB = 12cm and DE = 15cm, calculate
5.
i)
the height AE, of the smaller con AXY.
ii)
the volume of XBCY, the shape that remained.
2016 Oct/ Nov Exams; Q 9 (b and c)
a) The cross section of a rectangular tank measures 1.2m by 0.9. If it contains fuel to a depth of 10m, find the
number of litres of fuel in the tank. (1m3 = 1000 litres).
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b) A cone has a perpendicular height of 12cm and slant height of 13cm, calculate its total surface area. (Take
𝜋 = 3.142).
6.
Miscellaneous Question.
i)
The bucket is in the shape of a frustum of a cone as shown below. Find the capacity in litres of a
bucket 24cm in diameter at the top, 16cm in diameter at the bottom and 18cm deep.
24
18
16
ii)
iii)
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7.
2019 Aug Exams Q10 (a).
The diagram below shows a frustum TQRS of a cone. [Take 𝜋 𝑎𝑠 3.142].
Given that US = 3cm, UV = 10cm and RV = 8cm, calculate its volume.
Concise Revision Notes:

The word Calculus comes from a Latin word meaning "small stone", because it is like understanding something
by looking at small pieces. In simplicity, calculus is the study of change with the basic focus being on rate of
change and accumulation. Therefore, it is divided into two categories namely; Differential calculus (rate of
change) and integral calculus (accumulation).

Differential Calculus cuts something into small pieces to find how it changes.

Integral Calculus joins (integrates) the small pieces together to find how much there is.

The essence of calculus is the derivative. The derivative is the instantaneous rate of change of a function with
respect to one of its variables. This is equivalent to finding the slope of the tangent line to the function at a
point. Thus, the process of finding the derivative of a given function is termed as Differentiation.

The notation
𝑑𝑦
𝑑𝑥
is the differential notation. We call this differentiation of the derivative or differential
coefficient of 𝑦 with respect to 𝑥.
𝑑𝑦
𝑑𝑥
is sometimes written as 𝑦 ′ . One type of notation for derivatives is
sometimes called prime notation (y').

A derivative is always the derivative of a function with respect to a variable. We mean the derivative of the
function f(x) with respect to the variable x. For instance, given that

𝑦 = 𝑎𝑥 𝑛 , 𝑡ℎ𝑒𝑛
𝑑𝑦
𝑑𝑥
= 𝑛𝑎𝑥 𝑛−1 . This is called the General rule for differentiation. NB: The derivative of a
constant is equal to zero.
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The
𝑑𝑦
𝑑𝑥
gives the gradient function for a curve and the value of
𝑑𝑦
𝑑𝑥
at a given point is the gradient of the curve at
that point and therefore, the gradient of the tangent there. Gradient is usually denoted by 𝑚. Therefore, 𝑚 =
𝑑𝑦
𝑑𝑥
at a given point.

Differentiation from the first principle is defined by:
𝑑𝑦
𝑑𝑦
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
= lim ( ) ⇒ lim [
]
ℎ→0
𝑑𝑥 ℎ→0 𝑑𝑥
ℎ

Equation Of A Tangent and Normal To A Curve

The tangent is a straight line which just touches the curve at a given point.

The normal is a straight line which is perpendicular to the tangent.

To calculate the equations of these lines we shall make use of the fact that the equation of a straight line
passing through the point with coordinates (𝑥1 , 𝑦1 ) and having gradient m is given by 𝑦 − 𝑦1 = 𝑚(𝑥 −
𝑥1 ).

We also make use of the fact that if two lines with gradients 𝑚1 𝑎𝑛𝑑 𝑚2 respectively are perpendicular,
then 𝑚1 𝑚2 = −1.

Indefinite Integral and Definite Integral
𝑑𝑦

If

∫𝑎 𝑔(𝑥)𝑑𝑥 = [𝑓(𝑥)]𝑏𝑎 ⇒ 𝑓(𝑏) − 𝑓(𝑎), where f(x) is the indefinite integral of g(x).
𝑑𝑥
= g(x), then y = ∫ 𝑔(𝑥)𝑑𝑥 + 𝑐[c = arbitrary constant]
𝑏
𝑏
𝑎
Note: ∫𝑎 𝑔(𝑥)𝑑𝑥 = − ∫𝑏 𝑔(𝑥)𝑑𝑥


Rules on integration
𝑥 𝑛+1

∫ 𝑥 𝑛 𝑑𝑥 =

∫ 𝑎𝑥 𝑛 𝑑𝑥 = a ∙ ∫ 𝑥 𝑛 𝑑𝑥 ⇒

∫ 1 𝑑𝑥 = x + c

∫ 𝑎 𝑑𝑥 = 𝑎 ∙ ∫ 1 𝑑𝑥 ⇒ 𝑎𝑥 + 𝑐

∫(𝑎𝑥 + 𝑏)𝑛 ) 𝑑𝑥 =

∫[𝑔1 (𝑥) ± 𝑔2 (𝑥)]𝑑𝑥 = ∫ 𝑔1 (𝑥)𝑑𝑥 ± ∫ 𝑔2 (𝑥)𝑑𝑥
𝑛+1
+ c, where n ≠ −1
𝑎 ∙ 𝑥 𝑛+1
𝑛+1
((𝑎𝑥 + 𝑏)𝑛+1 )
𝑎(𝑛+1)
+𝑐
+𝑐
Stationary Points
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Maximum, Minimum and Point of Ineflection
𝑑𝑦

A point on a curve at which the gradient is zero, where

At a stationary point, the tangent to the curve is horizontal and the curve is “flat”.

Basically, there are three (3) types of stationary points. These are:
𝑑𝑥
= 0, is called a stationary point.
A) Minimum Point
In this case, the gradient of the curve is negative to the left of point P. To the right of point P, the gradient
of the curve is positive.
To the left of P
𝒅𝒚
𝒅𝒙
At point P
𝒅𝒚
<𝟎
𝒅𝒙
To the right of P
𝒅𝒚
=𝟎
𝒅𝒙
>𝟎
B) Maximum Point
In this case, the gradient of the curve is positive to the left of point P. To the right of point P, the gradient of
the curve is negative.
To the left of P
𝒅𝒚
𝒅𝒙
At point P
𝒅𝒚
>𝟎
𝒅𝒙
To the right of P
=𝟎
𝒅𝒚
𝒅𝒙
<𝟎
C) Point of Inflection
In this case, the gradient has the same sign each side of the stationary point. A point of inflection which has
zero gradients (such as this one) is called a horizontal point of inflection or saddle point.
NB: Maximum and minimum points are also called turning points as this is where the graph “turns” and the
gradient changes from positive to negative or vice versa.
Thus, to find the stationary points on a curve 𝑦 = 𝑓(𝑥) and to distinguish between them, we proceed as follows:
𝑑𝑦
1.
Find the gradient function
2.
Equate to zero the expression for
3.
Find the values of 𝑥 (𝑖. 𝑒. 𝑥1 , 𝑥2 , 𝑥3 … ) which satisfy this equation.
4.
Consider the sign of
5.
Finally, find the values 𝑦1 , 𝑦2 , 𝑦3 … which correspond to 𝑥1 , 𝑥2 , 𝑥3 …
𝑑𝑦
𝑑𝑥
𝑑𝑥
of the curve.
𝑑𝑦
𝑑𝑥
.
on either side of these points.
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Specimen Examination Questions:
1.
2018 Oct/Nov Exams, Q3
2
a) Evaluate ∫−1(2 + 𝑥 − 𝑥 2 ) 𝑑𝑥
b) Find the equation of the normal to the curve 𝑦 = 𝑥 +
2.
4
𝑥
at the point where 𝑥 = 4.
2018 Jul/Aug Exams, Q7(b)
1
a) Evaluate ∫0 (𝑥 2 − 2𝑥 + 3) 𝑑𝑥
b) Determine the equation of the normal to the curve 𝑦 = 2𝑥 2 − 3𝑥 − 2 that passes through (3, 7).
3.
2017 Oct/Nov Exams, Q9 (b & c)
a) Find the coordinates of the points on the curve 𝑦 = 2𝑥 3 − 3𝑥 2 − 36𝑥 − 3 where the gradient is zero.
3
b) Evaluate ∫−1(3𝑥 2 − 2𝑥) 𝑑𝑥
4.
2017 July/Aug Exams, Q4b & 9c
5
a) Evaluate ∫2 (3𝑥 2 − 2) 𝑑𝑥
b) Find the equation of the tangent to the curve 𝑦 = 𝑥 2 − 3𝑥 − 4 at a point where 𝑥 = 2.
5.
2016 Oct / Nov Exams, Q6
3
The equation of the curve is 𝑦 = 𝑥 3 − 𝑥 2 . Find
2
a) equation of the normal where 𝑥 = 2,
b) the coordinates of the stationary points.
6.
Miscellaneous Questions:
a) The line 𝑦 = 4𝑥 − 5cuts the curve 𝑦 = 𝑥 2 − 2𝑥 at two points.
Find the gradient of 𝑦 = 𝑥 2 − 2𝑥 at these two points.
b) Differentiate each of the following from first principle to find
𝑑𝑦
𝑑𝑥
.
(i) 𝑦 = 𝑥 2 + 3𝑥
(ii) 𝑦 = 3𝑥 2
c)
Find the following:
(i) ∫ 3𝑥 4 𝑑𝑥
(ii) ∫ 4𝑥 3 + 3𝑥 + 2 𝑑𝑥
(iii) ∫ 3𝑥 2 +
2
𝑥2
− 1 𝑑𝑥
d) Evaluate each of the following:
1
(i) ∫0 8𝑥 3 + 3 𝑑𝑥
2 𝑥4 + 2
(ii) ∫1
𝑥2
𝑑𝑥
2
(iii) ∫0 (2𝑥 − 3)4 dx
(iv) ∫(3𝑥 − 2)5 𝑑𝑥
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e)
Find the coordinates of the stationary points on the curve and determine their nature. Sketch the curve
where necessary.
(i) 𝑦 = 8𝑥 − 2𝑥 2
(ii) 𝑦 = 𝑥 3 + 3𝑥 2 + 1
Concise Revision Notes:

Geometrical transformation of plane figures deals with the movement of points (figures) on a surface. In the
process of changing positions, the figures may be changed in size, shape or orientation. However, this change
may be partial or complete or none at all.

In order to determine the type of transformation that has occurred, it is usually necessary to consider the
changes of that have taken place on the origin figure.

The six (6) geometrical transformations includes;
i)
Translation
ii)
Reflection
iii)
Rotation
iv)
Enlargement
v)
Shear, and
vi)
Stretch
A) TRANSLATION
A translation is a geometrical transformation in which all the points in original figure move through the same
distance in a specific direction. Therefore, a column vector is used as a movement of points in a translation
𝑥
represented by a translation matrix that is expressed as (𝑦), it means that each of the points in the object will
move through
𝑥 units in the horizontal direction and 𝑦 units in the vertical direction. NB:
𝑻𝒓𝒂𝒏𝒔𝒍𝒂𝒕𝒊𝒐𝒏 𝒗𝒆𝒄𝒕𝒐𝒓 = 𝒊𝒎𝒂𝒈𝒆 − 𝒐𝒃𝒋𝒆𝒄𝒕.
Properties of translation
The following points are true for all translations:


No point is invariant, i.e. all points move.

Every point moves the same distance in the same specific direction

The dimensions of the object (angles & lengths) are maintained.

The direction or parallelism is maintained.
A translation is described fully by stating the column vector. Hence, we say; is a translation with a column
vector of………………..?
B) REFLECTION
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For reflection to take place there must exist a line of reflection through which it should occur. The line of
reflection is usually called the mirror line. A reflection is a geometrical transformation in which all points in
the object move such that the distance from object to the mirror line is equal to the distance from the image to
the mirror line.

Finding the Image of a Reflection:
If the line 𝑙 is a mirror line and A is a point on the object, the image of A can be found by dropping a
perpendicular from A to the line 𝑙. The image of point A will be a point on the other side of mirror line such that
the distance from A to the mirror line 𝑙 is the same as that from the image of A to the mirror line 𝑙.

Finding the Mirror Line of a Reflection:
When the object and the image of reflection are given, the mirror line can be found by drawing the
perpendicular bisector of the line joining any one of the points on the object to its image, i.e. If ABC is a
triangle that is mapped onto 𝐴1 𝐵1 𝐶1 by a reflection in the line 𝑙, then the line 𝑙 is the perpendicular bisector of
𝐴𝐴1 𝐵𝐵1 𝐶𝐶1 .

Properties of Reflection:
The following points are true for all reflections:


The distance of the object to the mirror line is equal to that of the image to the mirror line.

The mirror line is the line of symmetry between the object and the image.

The mirror line is a perpendicular bisector of lines joining objects to their images.
Describing a Reflection:
A reflection is fully decribed by stating the equation of the line of reflection (the mirror line).
C) ROTATION:
A rotation is a transformation in which all the points in the object move in a circular direction. For rotation to
take place there must exist a centre of rotation, and a mention should be made of the direction and the angle
through which the rotation must take place.

Finding the Image of a Rotation:
Let ∆ABC be an object and let the point 𝑋 be the centre of rotation. Suppose you are asked to find the image of
∆ABC when it is rotated through 90° at X in the clockwise direction.
To find the image of A, we follow the steps below:

o
Step 1 – join the point A to the centre X with a straight line.
o
Step 2 – with the protrator at the point X, measure the angle such that angle 𝐴𝑋𝐴1 = 90° .
o
Step 3 – make sure that the length 𝐴𝑋 = 𝐴1 𝑋.
o
Repeat the process in order to find the image of 𝐵1 𝑎𝑛𝑑 𝐶1 .
Finding the Centre, Angle and Direction of a Rotation:
Let the object, ∆ABC be given and let the ∆ 𝐴1 𝐵1 𝐶1 be the image of ∆ABC under a rotation. Suppose you are
wanted to find the centre, angle and direction of a rotation. The following steps can be followed to do so:
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o
Step 1 – join one point on the object to its image and construct a perpendicular bisector of the resulting
straight line, i.e. join A to 𝐴1 and draw the mediator of 𝐴𝐴1 .
o
Step 2 – join another point on the object to its image and construct a perpendicular bisector of the resulting
straight line, e.g. join B to 𝐵1 and draw the mediator of 𝐵𝐵1 .
o
Step 3 – let the mediator of 𝐴𝐴1 meet the mediator of 𝐵𝐵1 at the point X.
o
Step 4 – the point X is the centre of rotation.
o
Step 5 – join one point and its image to the centre of rotation with straight lines. The angle created is the
angle of rotation.
o

Step 6 – the direction of the rotation can be determined from the orientation of the diagram.
Properties of a Rotation:
The following points are true for all rotations:


The centre of rotation is invariant

A point and its image are at the same distance from the centre of rotation.
Describing a Rotation:
A rotation is described fully by stating the centre of rotation, the angle and the direction of the rotation.
D) ENLARGEMENT
An enlargement is a transformation in which all the points of the object move in such a way that the image is
geometrically similar to the origin shape. For an enlargement to take place there must exist a centre of
enlargement and the enlargement factor or ratio of enlargement.
NB: all the objects and their images lie in the same straight line with the centre of enlargement.

Enlargement Factor
If a ∆ 𝐴1 𝐵1 𝐶1 is the image of the ∆ABC under an enlargement with centre at X and enlargement factor 𝑘, we
have the following:

𝒌=
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒊𝒎𝒂𝒈𝒆 𝒇𝒓𝒐𝒎 𝒄𝒆𝒏𝒕𝒓𝒆 𝑿
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒐𝒇 𝒐𝒃𝒋𝒆𝒄𝒕 𝒇𝒓𝒐𝒎 𝒄𝒆𝒏𝒕𝒓𝒆 𝑿
𝒌=
𝑿𝑨𝟏 𝑿𝑩𝟏 𝑿𝑪𝟏
=
=
𝑿𝑨
𝑿𝑩
𝑿𝑪
Finding the Image of an Enlargement
Let ∆ABC be an object and let the point X be the centre of enlargement. Suppose you are asked to find the
image of ∆ABC when it is enlaged at X with the enlargement factor 𝑘.
o
Join the point A to the centre X and extend to a point 𝐴1 such that 𝑋𝐴1 = 𝑘 ∙ 𝑋𝐴.
o
Join the point B to the centre X and extend to a point 𝐵1 such that 𝑋𝐵1 = 𝑘 ∙ 𝑋𝐵.
o
Join the point C to the centre X and extend to a point 𝐶1 such that 𝑋𝐶1 = 𝑘 ∙ 𝑋𝐶.
The ∆ A1 B1 C1 is the image of ∆ABC under an enlargement with centre at X and enlargement factor 𝑘.
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NB: when the enlargement factor is negative, the ratio remains the same BUT the image is on the opposite
side of the centre of enlargement and the image is oppositely inverted.

Finding the centre, and scale factor of an enlargement
Let ∆ABC be an object and let the ∆ 𝐴1 𝐵1 𝐶1 be the image of ∆ABC under an enlargement. The centre and scale
factor of this enlargement can be found as follows:
o
Step 1 – join any two points to their image and object and project the lines. i.e. join say, A to 𝐴1 and
𝐵 𝑡𝑜 𝐵1 and extend the lines 𝐴𝐴1 and 𝐵𝐵1 .
o
Step 2 – let the projections meet at a point. i.e. let the projections 𝐴𝐴1 and 𝐵𝐵1 meet at a point, say X.
o
Step 3 – the point X is the centre of the enlargement.
o
Step 4 – the scale factor of the enlargement is given by the ratio.
𝑲=
𝑿𝑨𝟏 𝑿𝑩𝟏 𝑿𝑪𝟏
=
=
𝑿𝑨
𝑿𝑩
𝑿𝑪
The ∆ 𝐴1 𝐵1 𝐶1 is the image of ∆ABC under an enlargement with centre at X and enlargement factor 𝑘.
E) SHEAR
A shear is a transformation in which points move parallel to the invariant line in such a way that the ratio of
distance moved by a point to that of its distance from the invariant line is constant. Note that an invariant line is
one on which the points on the object maintain their positions even after being transformed.
For a shear, the shear (scale) factor is given by the followinng relationship:
Shear factor 𝒌
=
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒎𝒐𝒗𝒆𝒅 𝒃𝒚 𝒂 𝒑𝒐𝒊𝒏𝒕 𝒑𝒂𝒓𝒂𝒍𝒍𝒆𝒍 𝒕𝒐 𝒕𝒉𝒆 𝒊𝒏𝒗𝒂𝒓𝒊𝒂𝒏𝒕 𝒍𝒊𝒏𝒆
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒑𝒐𝒊𝒏𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒊𝒏𝒗𝒂𝒓𝒊𝒂𝒏𝒕 𝒍𝒊𝒏𝒆
F) STRETCH
A stretch is a transformation in which points points move perpendicular to the invariant line in such a way that
the ratio of a distance of the image from the invariant line to that of its object from the invariant line is
constant.
For a stretch, the stretch (scale) factor is given by the following relationship:
Stretch factor 𝒌
=
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒎𝒐𝒗𝒆𝒅 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒊𝒎𝒂𝒈𝒆 𝒕𝒐 𝒕𝒉𝒆 𝒊𝒏𝒗𝒂𝒓𝒊𝒂𝒏𝒕 𝒍𝒊𝒏𝒆
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒐𝒃𝒋𝒆𝒄𝒕 𝒕𝒐 𝒕𝒉𝒆 𝒊𝒏𝒗𝒂𝒓𝒊𝒂𝒏𝒕 𝒍𝒊𝒏𝒆
G) MATRICES ASSOCIATED WITH TRANSFORMATION
Apar from the translation, all the transformation discussed above can be represented by some square matrices
of order 2. When the coordinates of the vertices of the original shape or object are given and a 2 × 2 matrix
representing a standard transformation, the coordinates of the vertices of the image can be found by
multiplying the matrix with the coordinates.
𝑎
Thus, if (𝑥, 𝑦)is the object, (𝑥1 , 𝑦1 ) is the image and 𝑀 = (
𝑏
𝑥1
𝑎 𝑐 𝑥
(𝑦 ) = (
) ( ).
𝑏 𝑑 𝑦
1
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𝑐
) is the matrix, then;
𝑑
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NOTE: to find the matrix describing a given transformation,
 Find the image of (1, 0) say, (𝑎, 𝑏)
 Find the image of (0, 1) say, (𝑐, 𝑑)
𝑎
𝑐
 The required matrix is (𝑏 𝑑 ).
The following are some of the matrices for some selected standard transformations:
(i)
Isometries
(a) Translation:
𝑎
If A(𝑥, 𝑦) is an object, A(𝑥1 , 𝑦1 ) is an image and ( 𝑏 ) is a column vector, then;
𝑥1
𝑥
𝑎
(𝑦 ) = ( ) + (𝑦).
𝑏
1
(b) Reflection:
1
0
0
)
−1
 Reflection in the line 𝑦 = 0 (𝑥 − 𝑎𝑥𝑖𝑠): (
−1 0
)
0 1
0 1
(
)
1 0
 Reflection in the line 𝑥 = 0 (𝑦 − 𝑎𝑥𝑖𝑠): (
 Reflection in the line 𝑦 = 𝑥:
 Reflection in the 𝑦 = −𝑥:
0
−1
−1
)
0
(
(c) Rotation:
 Clockwise rotation through 90° at (0, 0): (
0
−1
1
)
0
0
1
 Anticlockwise rotation through 90° at (0, 0): (
0
−1
 Half turn rotation through 180° at (0, 0): (
(ii)
−1
)
0
−1
)
0
Similarities
(a) Enlargement
 Enlargement with scale factor 𝑘 at (0, 0): (𝑘 0)
0 𝑘
 Enlargement with scale factor – 𝑘 at (0, 0): (−𝑘 0 )
0 −𝑘
(iii)
Affines
(a) Shear:
 Invariant line 𝑦 = 0 with shear factor 𝑘: (1 𝑘)
0
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1
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 Invariant line 𝑥 = 0 with shear factor −𝑘: (1 0)
𝑘
1
(b) Stretch:
 Invariant line 𝑦 = 0 with stretch factor 𝑘: (1 0)
0
𝑘
 Invariant line 𝑥 = 0 with stretch factor 𝑘: (𝑘 0)
0
1
Specimen Examination Questions:
1.
2018 Oct/Nov Exam Q10
Study the diagram below and answer the questions that follow.
a) Triangle R is the image of triangle P under a rotation. Find the coordinates of the centre, angle and the direction
of the rotation.
b) A single transformation maps triangle P onto triangle M. describe fully this transformation.
c)
Triangle P maps onto triangle V by a stretch. Find the matrix of this transformation
d) If triangle P is mapped onto triangle S by a shear represented by the matrix (
Compiled & Solved by Dr. Frendo
1
−2
0
), find the coordinates of S.
1
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2.
2018 July/Aug Exams, Q10
Using a scale of 1cm to represent 1 unit, on both axes, draw x and y axes for −8 ≤ 𝑥 ≤ 12 and −6 ≤ 𝑦 ≤ 14.
a) Draw and label triangle X with vertices (2, 4), (4, 4) and (4, 1).
b) Triangle X is mapped onto triangle U with vertices (6, 12), (12, 12) and (12, 3) by a single transformation.
c)
i)
Draw and label triangle U.
ii)
Describe fully this transformation.
A 90° clockwise rotation about the origin maps triangle X onto triangle W. Draw and label triangle W.
d) A shear with X – axis as the invariant line and shear factor -2 maps triangle X onto triangle S. Draw and
label triangle S.
e)
3.
Triangle X is mapped onto triangle M with vertices (4, 4), (8, 4) and (8, 1).
i)
Draw and label triangle M.
ii)
Find the matrix which represents this transformation.
2017 Oct/Nov Exams, Q12
Using a scale of 1cm to represent 1 unit on each axis, draw x and y axes for −6 ≤ 𝑥 ≤ 10
and −10 ≤ 𝑦 ≤ 8.
a) A quadrilateral ABCD has vertices A (−5, 7), B (−4, 8), C (−3, 7) and D (−4, 4) while its imagine has
vertices A1 (−5, −3), B1 (−6, −2), C1 (−5, −1) and D1 (−2, −2).
i)
Draw and label the quadrilateral ABCD and its image A1 B1 C1 D1 .
ii)
Describe
fully
the
transformation
which
maps
the
quadrilateral
ABCD
onto
quadrilateral A1 B1 C1 D1 .
b) The matrix (
c)
−2
0
0
) maps the quadrilateral ABCD on the A2 B2 C2 D2 .
1
i)
Find the coordinates of the vertices of the quadrilateral A2 B2 C2 D2 .
ii)
Draw and label the quadrilateral A2 B2 C2 D2 .
The quadrilateral ABCD is mapped onto quadrilateral A3 B3 C3 D3 . where A3 is (4, −8), B3 is (2, −10), C3
is (0, −8) and D3 is (2, −2). Describe fully this transformation.
4.
2017 July/Aug Exams, Q7
Using a scale of 1cm to represent 1 unit on each axis, draw 𝑥 and 𝑦 axes for −6 ≤ 𝑥 ≤ 10 and −6 ≤ 𝑦 ≤ 12.
a)
A quadrilateral ABCD has vertices A (1, 1), B (2, 1), C (3, 2) and D (2, 3) while it’s imagine has
vertices A1 (3, 2), B1 (6, 1), C1 (9, 2), and D1 (6, 3).
i)
Draw and label the quadrilateral ABCD and its image A1 B1 C1 D1 .
ii)
Describe
fully
the
transformation
which
maps
the
quadrilateral
quadrilateral A1 B1 C1 D1 .
b) The matrix (
1
3
0
) maps the quadrilateral ABCD on the quadrilateral A2 B2 C2 D2 .
1
i)
Find the coordinates of quadrilateral A2 B2 C2 D2 .
ii)
Draw and label quadrilateral A2 B2 C2 D2 .
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ABCD
onto
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c)
Quadrilateral
A 3 B 3 C 3 D3
has
vertices
A3 (−2, −4), B3 (−4, −2), C3 (−6, −4), and D3 (−4,
−6).Describe fully the transformation which maps quadrilateral ABCD onto A3 B3 C3 D3 .
5.
2017 Oct/Nov Exams, Q12
Study the diagram below and answer questions that follow.
a) An enlargement maps triangle ABC onto triangle A1 B1 C1 . Find
i)
the centre of enlargement
ii)
the scale factor
b) Triangle ABC is mapped onto triangle A2 B2 C2 by a shear. Find the matrix which presents this
transformation.
c)
Triangle ABC is mapped onto triangle A3 B3 C3 by a single transformation. Describe this transformation
fully.
d) A transformation with matrix (
−3
0
0
) maps triangle ABC onto triangle A4 B4 C4 not drawn on the
1
diagram. Find
i)
The scale factor of this transformation
ii)
The coordinates of A4 , B4 and C4 .
iii)
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Concise Notes
(a) Definition:
A function f is a process which takes an input x and maps it onto a unique output f(x). f(x) is also called the image
of x. it is written as f : x ⟼ f(x) or f(x) = output. E.g. f: x ⟼ 2x + 3.

When the input x = 1. f : 1 ⟼ 2(1) + 3 = 5. I.e. the output or image is 5.

Note: the letter x can be replaced by any other letter.
E.g. f : t ⟼ 2t +3 or f : 𝜃 ⟼ 2𝜃 + 3.
The letter used is called a dummy variable.

The set of inputs for a function is called the domain.

The set of outputs (or images) for the function is called the range.
(b) Composite functions or combined functions

Two functions f and g can be combined to give another function called a combined or composite function.

The composite function fg is obtained by first finding g(x) followed by finding f [g(x)].
𝑔
𝑓
i.e. f g : x → g(x) → f [g(x)].
Note:
(i)
The second function g is dealt first before the first function f.
(ii)
Generally g f ≠ f g, although there could be some value(s) of x for which g f = f g.
(iii)
𝑓𝑓 −1 (𝑥) = x = 𝑓 −1 𝑓(x).
(iv)
𝑓 2 Means f f. 𝑓 3 means f (𝑓 2 ) = f f f and so on.
(v)
(𝑓𝑔)−1 = 𝑔−1 𝑓 −1
(c) How to find the inverse function
Step 1: let the function f(x) = y
Step 2: find x in terms of y
E.g. f : x ⟼ 2x +3
𝑦 = 2𝑥 + 3
2𝑥 = 𝑦 − 3
𝑥=
𝑦−3
2
Hence, the inverse function is𝑓 −1 : y ⟼
𝑦−3
2
or 𝑓 −1 (𝑦)=
𝑦−3
2
Step 3: note that it is usual to use the dummy variable x in place of y. therefore, replace y with x in inverse
function. Thus, 𝑓 −1 : x ⟼
𝑥 −3
2
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or 𝑓 −1 (𝑥)=
𝑥−3
2
.
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Sample Questions:
1.
Functions f and g are defined as follows:
𝒇: 𝒙 ⟼ 𝟐𝒙 + 𝟏 and 𝒈: 𝒙 ⟼
𝟏
𝒙 − 𝟏
x≠1
Find:
a) f (3)
b) x, when f(x) = −11
c)
gf(x)
d) 𝑓 −1 (𝑥)
2.
Functions f and g are defined by;
𝒇(𝒙) ⟼
𝒙 + 𝟑
𝟒−𝒙
, x ≠ 4 and 𝒈(𝒙) ⟼ 𝟑𝒙 − 𝟒.
Find:
a) gf (x)
b)
gg(−2)
c)
𝑓 −1 (−1)
d)
𝑔−1 𝑓(𝑥)
Concise Revision Notes
Basic Concepts:
𝒙𝟏 + 𝒙𝟐 𝒚𝟏 + 𝒚𝟐

Midpoint between two points (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is defined by: (

Distance between (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is √(𝒙𝟐 − 𝒙𝟏 )𝟐 + (𝒚𝟐 − 𝒚𝟏 )𝟐 .

Gradient of a line through (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is 𝑚 =

if a line 𝑙1 has a gradient 𝑚1 and another line 𝑙2 has a gradient 𝑚2 , then:
𝒚𝟐 − 𝒚𝟏
𝒙𝟐 − 𝒙𝟏
𝟐
,
𝟐
).
.
(a) 𝑙1 ⊥ 𝑙2 ⟺ 𝑚1 𝑚2 = −1,
(b) 𝑙1 ∥ 𝑙2 ⟺ 𝑚1 = 𝑚2 . NB: All parallel lines have equal gradient.

Collinear points are points lying on a straight line. Three points A, B and C are collinear ⟺ gradient of AB =
gradient of BC = gradient of AC.

Equation of a straight line:
(a) Equation of a straight line passing through (𝑥1 , 𝑦1 ) with gradient 𝑚 is defined by:
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𝒚 − 𝒚𝟏
=𝒎
𝒙 − 𝒙𝟏
(b) Equation of a straight line passing through (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is defined by:
𝒚 − 𝒚𝟏
𝒙 − 𝒙𝟏
=
𝒚𝟐 − 𝒚𝟏
𝒙𝟐 − 𝒙𝟏
Note: Alternatively, find the gradient 𝑚 =
𝒚𝟐 − 𝒚𝟏
𝒙𝟐 − 𝒙𝟏
first, then use the equation in (a).
(c) Equation of a straight line with gradient 𝑚 and 𝑦 − intercept 𝑐 is 𝒚 = 𝒎𝒙 + 𝒄.

Points of intersection of two straight lines:
The coordinates of the point of intersection of 2 non-parallel straight lines are obtained by solving the
equations of the two lines simultaneously.
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Specimen Examination Questions:
1.
2019 June/July Exams Q4 & Q20 (b)
(a) Find the gradient of a line which passes through (−5, 3) and (−4, 1).
(b) The equation of a line A is 3𝑥 + 2𝑦 = 10. Line B is parallel to line A and passes through the point (4, 6).
Find the equation of the line B.
2.
2018 June/July Exams Q3 & Q20 (a)
(a) A straight line passing through A(3, 2) and B(5, 𝑦) has gradient −2. Find the value of 𝑦.
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(b) Find the equation of a line which is parallel to 2𝑥 + 𝑦 = 3 passing through (−2, 3).
3.
2017 August/Sept Exams Q20
The diagrambelow shows a cartesian plane with points A(6, 6), B(0, −2), C(0, 6) and D(6, 0).
Find the
(a) Equation of the line CD
(b) Distance AB
4.
2016 Oct/Nov Exams Q14 (b)
Find the equation of the straight line passing through (−4, 4) and is perpendicular to the straight line whose
𝑥
equation is 𝑦 + = 1
7
5.
2009 Oct/Nov Exams Q3 (b & c)
(a) Find the equation of a line I passing through a point (0, 5) whose gradient is 3.
(b) If B is a point (6, −3) and C is (−2, 1), find
(i)
6.
The gradient of line BC
(ii)
the equation of line BC.
2003 Oct/Nov Exams Q3 & 18 (a & b)
(a) P is a point (−4, 3) and O is the origin on the coordinate plane. Find the coodinates of the midpoint of OP
(b)
In the diagram ABCD is a square. A is a point (0, 10), B is the point (3, 6) and C is a point (7, 9). Find
(a) (i) the coordinates of D
(ii) The area of the square ABCD
(b) The equation of the side AB.
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Concise Revision Notes
Basic Concepts:


Parts of a circle:

A chord – is a line drawn across a circle.

A diameter – is a line dividing a circle into two equal parts.

An arc – is a piece of the circumference.

The angle 𝜃 – is the angle subtended by the arc at the centre, between the two radii.

A segment – is a portion of a circle cutt off by the chord.

A sector – it is a part of a circle formed by two radii and an arc.

Circumference – is the distance around the circle.
Angle properties of the circle:
Consider the following five (5) properties of the circle
1.
The angle at the centre of a circle
The angle subtended at the centre of a circle by an arc is twice the size of the angle on the circumference of a circle
subtended by the same arc. Both diagrams below illustrate this theorem.
𝑥
2.
The angle in a semi-circle
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The angle in a semi-circle is always equal to 90° . For example, in the diagram below, if AB represents the diameter
of the circle, then the angle at C and D is 90° . Similarly, ∠ADB = 𝟗𝟎° , this is because ∠AOB = 2∠ADB and
∠𝑨𝑶𝑩 = 𝟏𝟖𝟎° .
3.
Angles in the same segment
Angles in the same segment of a circle are equal. This can be explained simply by using the theorem that the angle
subtended at the centre is twice the angle on the circumference. Consider the diagram below, if the angle at the
centre is 2𝑥 ° , and then each of then each of the angles at the circumference must be equal to 𝑥 ° .
Similarly, ∠AD1 B = ∠AD2 B = ∠AD3 B as each is
half the angle at the centre.
4.
Angles in opposite segments ( angles in a cyclic quadrilateral)
Points P, Q, R and S all lie on the circumference of the circle (below). They are called concyclic points. Joining the
points P, Q, R and S produces a cyclic quadrilateral.

The opposite angles are supplementary, i.e. they add up to 180° . Since 𝑝° + 𝑟 ° = 180° (supplementary angles) and
𝑟 ° + 𝑡 ° = 180° (angles on a straight line) it follows that 𝑝° = 𝑡 ° .

Therefore, the exterior opposite angle of a cyclic quadrilateral is equal to the interior opposite angle.
5.
The angle between a tangent at a point and the radius of a circle
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The angle between a tangent at a point and the radius to the same point on the circle is a right angle (90° ). i.e.
∠OAC and ∠OBC = 90° .
Specimen Examination Questions:
1.
2015 Aug/Sept Exams Q3 (a)
̂ B = 30° .
̂ C = 81° and PA
In the diagram below, PAQ is a tangent to the circle at point A. AC bisects BĈD. AB
Calculate:
2.
(i)
∠ACB
(ii)
∠ADC
(iii)
∠DAQ
(iv)
∠CAQ
2019 Jun/Jul Exams Q16
In the diagram below, A, B, C, D and E are points on a circumference of the circle with centre O. DE = AD,
∠ACB = 20° and ∠AOE = 80° .
Find:
3.
(a)
∠ADE
(b)
∠DAE
(c)
∠BAD
2017 Jul/Aug Exams Q12
In the diagram below, points A, B, C and D are on a circle. BD is the diameter of the circle. ∠ACB = 𝟒𝟐°,
∠CAD = 𝟑𝟑° and the lines AC and BD intersects at X.
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Find:
(a) ∠CBD
(b) ∠ACD
(c) ∠AXB
4.
2018 Jul/Aug Exam Q13
The diagram below, shows a circle with a tangent RWS. The points V, W, X and Y are on the circle such that
̂𝐖 = 𝟒𝟒° , 𝐕𝐖
̂ 𝐘 = 𝟓𝟒° and 𝐒𝐖
̂ 𝐕 = 𝟑𝟗° .
𝐗𝐘
Calculate:
5.
(i)
∠RWX
(ii)
∠XVW
(iii)
∠YXW
2008 Oct/Nov Exams Q3
The diagram below, A, B, C and D are points on the points on the circumference of a circle. Line DC produced
meets BF at E. The point O is the centre, DF and BF are tangents to the circle.
D
A 75°
O 𝑥°
C
Given that ∠BAD = 75° , find:
(i)
angle 𝑥.
(ii)
∠BCE
(iii)
∠ODF
(iv)
∠BFD
F
6.
B
2011 Oct/Nov Exams Q8
E
In the diagram below, A, B and C are points on the circumference of the circle with centre O. AT and TB are
tangents to the circle, C𝐴̂O = 30° and C𝐵̂O = 29° .
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Find:
(a) 𝐴𝐶̂ 𝐵
(b) 𝐴𝑇̂𝐵
Concise Revision Notes
Basic Concepts:

Bearings are a measure of direction, with north taken as a reference. If you are travelling north, your bearing
is 𝟎𝟎𝟎° .

There are many ways of giving bearings or directions. The most common method is the one in which North is
reckoned to be Zero and angles are measured from the north in a clockwise direction.

A bearing can have any value from 0° to 360°. It is usual to give all bearings as three figures. This is known as
a three-figure bearing. So, in the above example, the bearing is to be written as 060°, using three figures. Here
are two more examples, i.e. F is on a bearing of 110° and H is on a bearing of 330° from G.
Maps and Scale Models:
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
Scale drawings are used when an accurate diagram, drawn in proportion, is needed. Common uses of scale
drawings include maps and plans. The use of scale drawings involves understanding how to scale
measurements.

For instance;
(a) a map is drawn to a scale of 1 : 10 000. If two objects are 1 cm apart on the map, how far apart are they in
real life? Give your answer in metres.
Explanation: A scale of 1 : 10 000 means that 1 cm on the map represents 10 000 cm in real life. Therefore, the
distance = 10 000 𝑐𝑚 ⟹ 100 𝑚 .
(b) A model boat is built to a scale of 1 : 50. If the length of the real boat is 12 m, calculate the length of the
model boat in cm.
Explanation: A scale of 1 : 50 means that 50 cm on the real boat is 1 cm on the model boat.
12𝑚 = 1200𝑐𝑚
Therefore, the length of the model boat = 1200 ÷ 50 ⟹ 24 𝑐𝑚.

Area and Volume of Maps:

If a map has a scale 1 ∶ 𝑛, then:
Length has a scale of 1 ∶ 𝑛
Area has a scale of 1 ∶ 𝑛2

Note on Units:
If a model has a scale of 1 ∶ 𝑛
Length has a scale of 1 ∶ 𝑛
1 km = 1 000 m
1 km = 100 000 cm
1 m2 = 10 000 cm2
Area has a scale of 1 ∶ 𝑛2
Volume has a scale of 1 ∶ 𝑛3
1 m3 = 1 000 000 cm3
Specimen Examination Questions:
1.
2019 Jul/Aug Exams Q14
The diagram below shows Mr Moenda’s trip. He travels on a bearing of 141° from A to B. He then decides to
continue with his trip from B on a bearing of 255° to C. The angle BCA = 35° .
Find the bearing of
(a) A from B
(b) A from C
2.
2018 Jul/Aug Exams Q17
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In the diagram below, a cyclist starts from town W and cycles on a bearing of 216° to town P. She then leaves
town P and cycles on a bearing of 296° to Z. Z is west of W.
Calculate:
3.
(i)
̂Z
PW
(ii)
̂Z
WP
2017 Jul/Aug Exams Q18
A, B and C are the three points level ground. B is on the bearing 070° from A and C is on a bearing of 130°
from B.
Ccalculate the bearing of:
(a) A from B
(b) B from C
4.
2016 Jul/Aug Exams Q20 (a)
The diagram below shows the positions of three towns Kime (K), Teswa (T) and Luwaya (L). Angle KTL =
110° and the bearing of L from T is 123° .
Find the bearing of:
5.
(i)
T from L
(ii)
T from K
2016 Oct/Nov Exams Q15
The diagram below shows an equilateral triangle ABC. A is due North of B and CN is parallel to BA.
Find:
(a) 𝐵𝐶̂ 𝑁
(b) The bearing of C from A
6.
2018 Jul/Aug Exams Q9 (b)
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The scale of a map is 1 : 20 000. The actual are of residential plots is 60km2 . Calculate the area of residential
plots on the map in square centimetres.
7.
2016 Jul/Aug Exams Q19 (a)
A map is drawn to a scale of 1 to 50 000. Calculate:
8.
(i)
The length of a road, in kilometres, which is 5cm long on the map,
(ii)
The actual area of a small town, in square kilometres, represented by an area of 10cm2 on the map.
Miscellaneous Questions:
(a) On a map with a scale of 1 ∶ 20 000, a garden has an area of 5 cm2 . Calculate the actual area of the garden.
(b) A map has a scale of 1 ∶ 500. A small public garden on the map has an area of 14 cm2 . Calculate the actual
area of this garden.
(c) A model car is made on a scale of 1 ∶ 20. The length of the model is 24 cm. the area of the windscreen of
the model is 32 cm2 . The volume of of the boot of the model is 90 cm3 . Calculate the actual:
(i)
Length of the car
(ii)
Area of the windscreen
(iii)
Volume of the boot
Concise Revision Notes
Basic Concepts:

When two or more variables 𝑥 𝑎𝑛𝑑 𝑦 are connected by a mathematical relationship, there are some important
relations describing the variation of 𝑦 with respect to 𝑥. It is usually denoted by the sign ∝.

Basically, there are three types of variations. These are:

Direct variation:

𝑦 is proportional to 𝑥, or 𝑦 varies directly as 𝑥. We write 𝒚 ∝ 𝒙 (𝑦 varies as 𝑥). Then 𝒚 = 𝒌𝒙, where 𝑘 is a
constant (the constant of proportionality). i.e. 𝒚 ∝ 𝒙 ⟹ 𝒚 = 𝒌𝒙.
NB: the constant, 𝒌 can be found if one pair of the corresponding values are known.

Indirect/Inverse Variation:

If 𝒚 ∝
𝟏
𝒙
i.e. 𝒚 ∝
then, 𝑦 is proportional to the reciprocal of 𝑥 or 𝑦 varies inversely as 𝑥.
𝟏
𝒙
⟹ 𝒚 = 𝒌𝒙 𝒐𝒓 𝒙𝒚 = 𝒌.
 Square Law Variation:
𝑦 is proportional to the square of 𝑥, or 𝑦 varies as 𝑥 2 . i.e. 𝒚 ∝ 𝒙𝟐

Inverse Square:
𝑦 is inversely proportional to the square of 𝑥. i.e. 𝒚 ∝
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𝟏
𝒙𝟐
⟹ 𝒚 = 𝒙𝒌𝟐
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
Cube Law Variation:
𝑦 varies as the cube of 𝑥. i.e. 𝒚 ∝ 𝒙𝟑

Joint Variation:
The volume 𝑉 of a cylinder is given by 𝑽 = 𝝅𝒓𝟐 𝒉. This is an example of joint variation: 𝑉 varies directly
as the square of the radius (r) and directly as the height (h).
NB: when dealing with any form of variation, convert the statement into an equation using a constant 𝒌,
and find 𝒌 from a given pair of values of the variables.
Specimen Examination Questions:
1.
2019 Jul/Aug Exams Q18.
Given that 𝒚 varies directly as 𝒙 and inversely as the square of 𝒛, and that 𝒚 = 𝟏𝟎 when 𝒙 = 𝟑𝟐 and 𝒛 = 𝟒,
find:
(a) The value of 𝑘, the constant of variation,
(b) 𝑦 when 𝑥 = 20 and 𝑧 = 5,
(c) 𝑧 when 𝑥 = 9 and 𝑦 = 5.
2.
2018 Jul/Aug Exams Q15.
Two variables 𝑥 and 𝑦 have corresponding values as shown in the table below.
𝒙
𝒚
2
20
3
40
𝑎
104
Given that 𝑦 varies directly as (𝒙𝟐 + 𝟏), find the:
(a) Constant of variation, 𝑘,
(b) Equation connecting 𝑦 and 𝑥,
(c) Values of 𝑎.
3.
2017 Jul/Aug Exams Q14.
The table below shows the relationship between two variables 𝑥 and 𝑦. It is given that 𝑦 varies inversely as the
square root of 𝑥, where 𝑥 is positive.
𝒚
2
8
𝒙
16
1
8
9
𝑎
(a) Write an expression for 𝑦 in terms of 𝑥 and the constant of variation, 𝑘.
(b) Find the value of
4.
(i)
𝑘
(ii)
𝑎
2016 Oct/Nov Exams Q20.
It is given that 𝒘 varies directly as the square of 𝒙 and inversely as 𝒚.
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(a) Write an expression for 𝑤 in terms of 𝑥, 𝑦 and a constant 𝑘.
(b) If 𝑥 = −6, 𝑦 = 12 and 𝑤 = 15, find 𝑘.
(c) Find the value of 𝑦 when 𝑥 = 8 and 𝑤 = 20.
5.
2014 oct/Nov Exams Q18.
Given that 𝑦 = 𝒌𝒙𝟐 − 𝟏, where 𝒌 is a constant and that 𝒚 = 𝟏𝟕 when 𝒙 = 𝟑, find
(a) The value of 𝑘,
(b) The value of 𝑦 when 𝑥 = −5,
(c) The values of 𝑥when 𝑦 = 7.
Concise Revision Notes
Basic Concepts:

Distance Time Graph:

The graph above shows the relationship between the distance of a moving object (measured from 0 in m)
and the corresponding times (in minutes). Starting from 0 the object travels 15m in 2 min, is then stationary
for another 2 min and finally travels back to 0 in 1 min. the total distance is 30m.

Therefore, the speed of the object will be the gradient of the line for that part, as this represents the rate of
change of distance with respect to time (for a curved part of the graph , the gradient at any point will give
the speed at that time).
i.e. from 0 to A, the speed is constant (straight line) =

From A to B, the speed is zero, as no distance is travelled. The car is not moving. From B back to 0, the
speed is constant =


15

15
1
2
≅ 7.5𝑚/𝑚𝑖𝑛.
≅ 15𝑚/𝑚𝑖𝑛.
Thus, the average speed for the whole journey =
30
5
≅ 6𝑚/𝑚𝑖𝑛.
Speed Time Graph:
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Note that: ∗ 𝒂 =
𝒗− 𝒖
𝒕
∗ 𝒗 = 𝒖 + 𝒂𝒕
𝟏
𝒗𝟐 − 𝒖 𝟐
∗ 𝒙 = 𝒖𝒕 + 𝒂𝒕𝟐 𝒐𝒓 𝒙 =
𝟐
𝟐𝒂𝒕
Where a: acceleration, v: final velocity, u: initial
velocity, t: time and x: distance travelled.

The graph above shows the relationship between speed and time for a journey. NB: the two important
features of a speed-time graph are:
(i) The gradient gives the rate of change of speed, i.e. the acceleration (if +) or the deceleration (if −), and
(ii) The area under the graph represents the distance travelled (in the correct units).
Specimen Examination Questions:
1.
2019 Jul/Aug Exams Q23.
The diagram below shows a speed time graph of an object. It starts from rest and accelerates uniformly for 2
seconds until it reaches a speed of 10m/s. it moves at this constant speed for 6 seconds then accelerates until it
reaches a speed of V m/s after 5 seconds. Finally, it retards for the next 8 seconds until it comes to halt.
Calculate the
(a) acceleration during the first 2 seconds,
(b) value of V if the retardation in the last 8 seconds is 3m/s 2 ,
(c) average speed for the whole journey.
2.
2018 Jul/Aug Exams Q23.
The diagram below shows the speed-time graph of a particle. The particle started off from rest and accelerated
uniformly for 10 seconds. It then travelled at a constant speed for 20 seconds and then decelerated to rest.
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(a) Find the speed V the particle reached if its aceleration was 2m/s 2 in
the first 10 seconds.
(b) Given that the total distance covered was 750m, find the value of 𝒕 in the diagram.
(c) What was the speed at 40 seconds?
3.
2017 Jul/Aug Exams Q23.
The diagram below shows the speed-time graph of a 100m sprint who accelerates uniformly for 3 secods until
he reaches a speed of 12m/s. he maintains the speed for 7 seconds and then uniformly retards for a further 4
seconds and comes to a stop.
Calculate the
(a) acceleration during the first 3 seconds.
(b) retardation at the end of his race.
(c) distance he covered in the first 10 seconds.
4.
2016 Jul/Aug Exams Q23.
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The diagram below shows the speed-time graph of a car which starts from rest and accelerates uniformly for 20
seconds till it reaches a speed of 30m/s. it then moves at a constant speed for some time before it starts
decelerating. It comes to rest after 100 seconds.
Given that the total distance travelled is 2 400 metres, calculate;
(a) the value of 𝑡,
(b) the retardation in the last part of the journey,
(c) the speed of the car at the ninety fifth seconds.
5.
2015 Oct/Nov Exams Q23.
The figure below shows the speed-time graph of cars driven by Sulanji and Lumbwa.
(a) Find the acceleration of Sulanji’s car during the first 10 seconds.
(b) Calculte how far Lumbwa’s car travelled before coming to rest
(c) What is Sulanji’s acceleration between 55𝑡ℎ 𝑎𝑛𝑑 30𝑡ℎ seconds?
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Concise Revision Notes
Basic Concepts:

Index Notation:
𝒂𝒏 , where 𝑎 is the base and 𝑛 is the index (or power)

Root Notation:
𝒏
𝟐
𝟑
√𝒂 is the 𝑛𝑡ℎ root of 𝑎. E.g. √𝒂 is the square root of 𝑎, (for square roots, √ with the “2” ommitted), √𝒂 is
𝟒
the cube root of 𝑎, √𝒂 is the fourth root of 𝑎, and so on……..
𝒏
NB: √
sign always represents the positive root. Thus, √𝟓 is the positive square root of 5 and −√5 is a
negative root of 5.

Laws of Indices:

𝒂𝒎 × 𝒂𝒏 – multiplication law:
When multiplying powers of the same base, ADD the indices. i.e. 𝒂𝒎 × 𝒂𝒏 = 𝒂

𝒎+ 𝒏
𝒂𝒎 ÷ 𝒂𝒏 - division law:
When dividing powers of the same base, SUBTRACT the indices. i.e. 𝒂𝒎 ÷ 𝒂𝒏 = 𝒂

(𝒂𝒎 )𝒏 – power of a power:
When taking the power of a power, MULTIPLY the indices. i.e. (𝒂𝒎 )𝒏 = 𝒂

𝒂−𝒎
𝒎𝒏
– negative index:
A negative power means the reciprocal of the base to the positive power. i.e. 𝒂−𝒎 =

𝒎− 𝒏
𝟏
𝒂𝒎
𝒂𝟎 – zero index:
When the base is raised to the power of zero, the outcome is always equal to 1. i.e. 𝑎0 = 1, for all values of
𝑎 except 0.
𝒎

𝒂 𝒏 – fractional index:
When a base is raised to a fraction power, is the root of the denominator raised to the power of the
𝒎
𝒏
𝒎
𝒏
numerator. i.e. 𝒂 𝒏 = ( √𝒂) 𝒐𝒓 √𝒂𝒎 .


(𝒂 × 𝒃)𝒎 = 𝒂𝒎 × 𝒃𝒎

( ) =
𝒂 𝒎
𝒂𝒎
𝒃
𝒂𝒎
Standard Form:
A number with one digit before the decimal point (e.g. 3.142) is in standard form.

Standard index form:
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Any number written in the form 𝑨 × 𝟏𝟎𝒏 where 𝟏 ≤ 𝑨 < 𝟏𝟎 and 𝑛 is an integer, is in standard index form
(sometimes called scientific notation).
Specimen Examination Questions:
1.
2019 Jul/Aug Exams Q2 & 12 (a).
3
4
(a) Evaluate ( √81)
(b) Evaluate 30 × 33 + 31 .
2.
2018 Jul/Aug Exams Q2 & 11 (b).
2
(a) Evaluate 273
(b) Solve the equation 22𝑥 −
3.
1
= 16−2𝑥
2017 Jul/Aug Exams Q9 (b) & 11 (b).
(a) Solve the equation 25𝑥 = 5
𝑥
(b) Given that 𝑥 = 3.2 × 34 and 𝑦 = 4 × 32 , evaluate .
𝑦
4.
2016 Jul/Aug Exams Q2 (b).
Evaluate 32 + 23 × 20
5.
2012 Oct/Nov Exams Q3.
The population of an African country in 2010 was 13 046 508. Express this population in standard form, correct
to 3 significant figures.
6.
2013 Oct/Aug Exams Q3 & 4 (b).
(a) Express 0.002304 in standard form, correct to one decimal place.
3 −2
(b) Evaluate ( )
4
7.
2003 Oct/Nov Exams Q15.
Given that 𝑥 =
27
8
, find the value of
(a) 𝑥 0
1
(b) 𝑥 3
2
(c) 𝑥 −3
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Concise Revision Notes
Basic Concepts:
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Page 94 of 192
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Page 95 of 192
Specimen Examination Questions:
1.
2019 Jul Exams Q19 (b).
The ratio of the volumes of two similar solids is 64 : 27. The surface area of the smaller solid is 180cm2 . What
is the surface area of the bigger solid?
2.
2018 Jul Exams Q14 (a).
The ratio of the heights of two contains that are geometrically similar is 2 : 3. If the surface area of the smaller
container is 80cm2 , fiind the surface area of the larger container.
3.
2017 Jul Exams Q4.
Two tins are geometrically similar. If the ratio of their volumes is 27 : 64, find the ratio of their curved surface
areas.
4.
2016 Jul Exams Q22 (b).
Two similar solids have surface areas in the ratio 4 : 25. If the volume of the bigger solid is 62.5cm3 , calculate
the volume of the smaller solid.
5.
2016 Oct/Nov Exams Q8 (b).
The base areas of two containers that are geometrically similar are 80cm2 and 180cm2 respectively. If the
capacity of the larger container is 54 litres, calculate the capacity of the smaller.
6.
2015 Oct/Nov Exams Q22 (b).
Two cylindrical mugs are geometrically similar. The smaller mug has diameter 8cm and holds 125cm3 of a
liquid. If the larger mug holds 1000cm3 of a liquid, find its diameter.
7.
2013 Oct/Nov Exams Q22 (a).
Two containers are geometrically similar. The ratio of their base areas is 9 : 16. If the volume of the smaller
container is 135cm3 , what would be the volume of the larger container?
8.
2010 Oct/Nov Exams Q16.
The two tins are geometrically similar. The ratio of their volumes is 64 : 27.
(a) Calculate the ratio of their curved surface areas.
(b) Given that the height of the larger tin is 28cm, calculate the height of the smaller tin.
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Concise Revision Notes
Basic Concepts:

Approximation & Estimation:
Approximations are very useful when calculating. Numbers can be approximated to obtain rough estimates of
calculations, thus, rough estimates are not accurate. Ways of approximating include;

Rounding off numbers:
Numbers can be rounded off to the nearest hundred, ten, whole numbers, etc, or to a given number of
decimal places. NB: the digits 1, 2, 3 and 4 are rounded down and the digits 5, 6, 7, 8 and 9 are
rounded up.

Significant figures:
Numbers are sometimes rounded off to a given number of significant figures. The significance of a digit
depends on its position in the number. NB: the first significant figures in a decimal fraction is the first nonzero digit in the fraction
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SOLUTIONS TO ALL THE TOPIC QUESTIONS
Question One (1) - 2018 Oct/Nov Exams: Q2 (b)
i)
E M
M aths
Chemistry
0
5
4
2
ii)
a) physics only = 6 students ans:
3
b) two types of subjects only = 𝟕 𝐬𝐭𝐮𝐝𝐞𝐧𝐭𝐬 𝐚𝐧𝐬: (2 + 5).
c)
6
Physics
Mathematics and physics but not chemistry
= 𝟏𝟒 𝐬𝐭𝐮𝐝𝐞𝐧𝐭𝐬 𝐚𝐧𝐬: (4 + 2 + 6)
Question Two (2) - 2018 Jul/Aug Exams: Q3 (a)
𝑥 + 3 + 4 + 7 = 22
i)
ii)
𝑥 + 14 = 22
𝑥 = 22 − 14
a) only one mode of transport = 14 + 7 + 7 ⟹ 𝟐𝟖 𝒍𝒆𝒂𝒓𝒏𝒆𝒓𝒔 𝒂𝒏𝒔:
𝑥 = 𝟖 𝒂𝒏𝒔:
b) 2 different modes of transport = 2 + 3 + 4 + 8 ⇒ 𝟏𝟕 𝒍𝒆𝒂𝒓𝒏𝒆𝒓𝒔 𝒂𝒏𝒔:
Question three (3) - 2017 Oct/Nov Exams: Q1 (b).
i)
E
M
5
19
ii) Number of farmers:
S
3
9
11
15
2
6
a) At the block = 𝟕𝟎 𝒇𝒂𝒓𝒎𝒆𝒓𝒔 𝒂𝒏𝒔:
b) Planted maize only = 𝟏𝟗 𝒇𝒂𝒓𝒎𝒆𝒓𝒔 𝒂𝒏𝒔:
c) Planted two different crops = 5 + 9 + 11 + 15 = 𝟒𝟎 𝒂𝒏𝒔:
C
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Question four (4) - 2017 July/Aug Exams: Q3 (b)
i)
2𝑦 + 1 = 7
ii) How many students visited?
2𝑦 = 7 − 1
2𝑦 = 6
a) Victoria falls but not Gonya Falls = 2 + 6 ⇒ 𝟖 𝒔𝒕𝒖𝒅𝒆𝒏𝒕𝒔 𝒂𝒏𝒔:
𝒚 = 𝟑 𝒂𝒏𝒔:
b) Two tourist attractions only = 2 + 1 + 4 ⇒ 𝟕 𝒔𝒕𝒖𝒅𝒆𝒏𝒕𝒔 𝒂𝒏𝒔:
c) One tourist attraction only = 6 + 8 + 7 ⇒ 𝟐𝟏 𝒔𝒕𝒖𝒅𝒆𝒏𝒕𝒔 𝒂𝒏𝒔:
Question Five (5) - 2016 October Exams Q2 (a)
i)
E
ii) How many villagers:
Sports
News
7
10
4
8
9
a) Listen to music only = 𝟓 𝒗𝒊𝒍𝒍𝒂𝒈𝒆𝒓𝒔 𝒂𝒏𝒔:
5
b) Listen to one type of programme only = 10 + 9 + 5
5
2
= 𝟐𝟒 𝒗𝒊𝒍𝒍𝒂𝒈𝒆𝒓𝒔 𝒂𝒏𝒔:
Music
c) Listen to two types of programs only = 7 + 8 + 5
= 𝟐𝟎 𝒗𝒊𝒍𝒍𝒂𝒈𝒆𝒓𝒔 𝒂𝒏𝒔:
Question Six (6) – Miscellaneous Question
a)
E
b) How many women used:
M
G
16
38
3
1
23
18
41
E
i)
Gas cookers only = 𝟑𝟖 𝒘𝒐𝒎𝒆𝒏 𝐚𝐧𝐬:
ii)
Microwave ovens and electric cookers only = 18 + 1 ⇒
𝟏𝟗 𝒘𝒐𝒎𝒆𝒏 𝐚𝐧𝐬:
2
iii)
Microwave ovens or electric cookers = 23 + 18 + 16 + 1 + 41 +
3 ⇒ 𝟏𝟎𝟐 𝒘𝒐𝒎𝒆𝒏 𝐚𝐧𝐬:
iv)
Microwave ovens or electric cookers, but did not use gas cookers
= 23 + 18 + 41 ⇒ 𝟖𝟐 𝒘𝒐𝒎𝒆𝒏 𝐚𝐧𝐬:
v)
Microwave ovens or solar energy = 23 + 18 + 1 + 16 + 2 ⇒
𝟔𝟎 𝒘𝒐𝒎𝒆𝒏 𝐚𝐧𝐬:
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Question Seven (7) – Miscellaneous Question
1.
1.
2018 Jul/Nov Exams
2018 Oct/Nov Exams
a)
⇒
⇒
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2.
2018 July/Aug Exams
(c) 𝑥 2 − 7𝑥 + 12
𝑥 2 − 3𝑥 − 4𝑥 + 12
(𝑥 2 − 3𝑥) − (4𝑥 + 12)
𝑥(𝑥 − 3) − 4(𝑥 − 3)
(𝒙 − 𝟑)(𝒙 − 𝟒) 𝐀𝒏𝒔:
2.
2017 Oct/Nov Exams
3.
2017 July/ Aug Exams
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−𝟓𝒙 + 𝟏𝟏
𝐀𝐧𝐬:
(𝟐𝒙 − 𝟓)(𝒙 − 𝟑)
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4.
5.
2016 October Exams
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Oct/Nov Exam
a)
𝟏 − 𝟓𝒑 − 𝟐𝒑𝟐 = 𝟎
(2005)
2
2
2
Arrange the expression 1 − 5𝑝 − 2𝑝 = 0 in the form: 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0, we have: 2𝑝 + 5𝑝 − 1 = 0.
⇒𝑝
=
⇒𝑝
=
⇒𝑝
=
⇒𝑝
=
∴ 𝑝1 =
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
−(5)±√(5)2 −4(2)(−1)
2(2)
−5±√25+8
4
−5±√33
4
−5 + √33
−5 − √33
or 𝑝2 =
4
4
𝐇𝐞𝐧𝐜𝐞, 𝐩𝟏 = 𝟎. 𝟏𝟗 𝐨𝐫 𝐩𝟐 = −𝟐. 𝟔𝟗 𝐀𝐧𝐬:
b) 𝟕 − 𝟓𝒙 − 𝒙𝟐 = 𝟎
(2010)
2
2
2
Arrange the expression 7 − 5𝑥 − 𝑥 = 0 in the form: 𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0, we have: 𝑥 + 5𝑥 − 7 = 0.
⇒𝑥
⇒𝑥
⇒𝑥
=
=
=
−𝑏±√𝑏2 −4𝑎𝑐
2𝑎
⇒𝑥
=
−(5)±√(5)2 −4(1)(−7)
2(1)
−5±√25+28
2
∴ 𝑥1 =
−5±√53
2
−5 + √53
−5 − √53
or 𝑥2 =
2
2
𝐇𝐞𝐧𝐜𝐞, 𝒙𝟏 = 𝟏. 𝟏𝟒 𝐨𝐫 𝒙𝟐 = −𝟔. 𝟏𝟒 𝐀𝐧𝐬:
x=(-(5)±√((5)^2-4(1)(-7) ))/2(1)
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c) (𝟐𝒙 − 𝟏)(𝟑𝒙 − 𝟐) = 𝟑
(2011)
Equating each term i.e. (2𝑥 − 1) 𝑎𝑛𝑑 (3𝑥 − 2) to 3, we have:
⇒ (2𝑥 − 1) = 3 or (3𝑥 − 2) = 3
⇒ 2𝑥 − 1 = 3 or 3𝑥 − 2 = 3
∴ 𝒙𝟏 = 𝟐
𝒐𝒓
𝒙𝟐 = 𝟏
𝟐
𝐀𝒏𝒔:
𝟑
d) 𝒙𝟐 + 𝟐𝒙 = 𝟕
(2016)
Arrange the expression 𝑥 2 + 2𝑥 = 7 in the form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, we have: 𝑥 2 + 2𝑥 − 7 = 0.
⇒𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
⇒𝑥=
−(2) ± √(2)2 − 4(1)(−7)
2(1)
⇒𝑥=
−2 ± √4 + 28
2
⇒𝑥=
−2 ± √32
2
∴ 𝑥1 =
−2 + √32
−2 − √32
or 𝑥2 =
2
2
𝑯𝒆𝒏𝒄𝒆,
𝒙𝟏 = 𝟏. 𝟖𝟑 𝒐𝒓 𝒙𝟐 = −𝟑. 𝟖𝟑 𝐀𝐧𝐬:
𝟐
𝒙 + 𝟐𝒙 = 𝟓
e)
(2016)
Arrange the expression 𝑥 2 + 2𝑥 = 5 in the form: 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, we have: 𝑥 2 + 2𝑥 − 5 = 0.
𝑥=
−𝑏 ± √𝑏 2 − 4𝑎𝑐
2𝑎
𝑥=
−(2) ± √(2)2 − 4(1)(−5)
2(1)
∴ 𝑥1 =
𝑥=
−2 ± √4 + 20
2
𝑯𝒆𝒏𝒄𝒆,
𝑥=
−2 ± √24
2
Compiled & Solved by Dr. Frendo
−2 + √24
−2 − √24
or 𝑥2 =
2
2
𝒙𝟏 = 𝟏. 𝟒𝟓 𝒐𝒓 𝒙𝟐 = 𝟑. 𝟑𝟓 𝐀𝐧𝐬:
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1.
2018 Oct/Nov Exams, Q1 (a)
⇒
⇒
⇒
⇒
2.
2018 Jul/August Exams: Q1 (a)
A−1 =
1
3
(
12 −3
−2
)
6
⇒
𝟏
⇒
−𝟏
∴𝐀
=(
⇒
⇒
⇒
⇒
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−
𝟒
𝟏
𝟒
−
𝟏
𝟔
𝟏)
𝟐
Ans:
Page 104 of 192
3.
2019 Aug Exams Q2 (a).
4.
2019 Aug Exams Q2 (a).
8
Q=(
𝑥−4
12
)
𝑥
(a) (8 × 𝑥) − (𝑥 − 4 × 12) = 8
(b) Q−1 =
Q−1 =
8𝑥 − 12𝑥 + 48 = 8
1
1 10
(
8 −6
−4𝑥 = −40
𝒙 = 𝟏𝟎 Ans:
5.
(𝑎𝑑𝑗𝑜𝑖𝑛𝑡)
𝑑𝑒𝑡 𝑄
𝟓
𝐐
−𝟏
=
(
−
𝟒
𝟑
𝟒
−12
)
8
−
𝟑
𝟐
𝟏
) Ans:
2017 Oct/Nov Exams: Q1 (a)
−𝟏 𝟏
(− 𝟏𝟏 𝟓) Ans:
𝟐
6.
2016 Oct/Nov Exams: Q1 (a)
7.
Miscellaneous Questions
(i)
(1
3
2) ( )
4
(ii) (−1
0
2
2) ( 0
−3
5
3)
1
(1 × 3 + 2 × 4)
(3 + 8)
(𝟏𝟏) Ans:
Compiled & Solved by Dr. Frendo
−1 × 2 + 0 × 0 + 2 × −3
)
−1 × 5 + 0 × 3 + 2 × 1
−2 + 0 − 6
⇒(
)
−5 + 0 + 2
−𝟖
⇒ ( ) Ans:
−𝟑
⇒(
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1.
2018 Oct/Nov Exams, Q1 (b)
2.
2018 Jul/Aug Exams, Q5 (a)
(a)
19
35
B
20
36
12
36
12
35
4
35
20
35
R
4
36
11
35
4
35
20
35
W
12
35
B BB
R BR
W BW
B RB
R RR
W RW
B WB
R WR
3
35
W WW
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3.
2017 Oct/Nov Exams, Q2 (a)
4.
2017 July Exams, Q3 (a)
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5.
2016 Oct/Nov Exams, Q2 (b)
6.
2016 July Exam Q9 (b)
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7.
2011 Exam Q5
8.
2012 Exam Q11 (b)
9.
2013 Exam Q10 (b)
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10. 2015 Exam Q6 (b)
11. Micellaneous Question
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1.
2018 Oct/Nov Exams, Q5 (b)
2.
2018 Jul/Aug Exams, Q2 (b)
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3.
2019 Aug Exams Q1 (b).
(i)
2
𝑎𝑟 = 16………(i)
4
Taking equation (iii), we
have:
= 4……...(ii)
𝑎𝑟
𝑎=
16
𝑟2
………….(iii)
16 4
∙𝑟 =4
𝑟2
𝑎=
4
16𝑟
𝑟2
=4
16𝑟 4 −
2
=4
𝑎=
16𝑟 2 = 4
𝑟2 =
𝑎=
4
16
16
𝑟2
16
1 2
( )
2
16
1
4
𝑎 = 16 ÷
4
𝑟 = ±√
16
𝑎 = 16 × 4
𝟏
𝒓 = ± Ans: (common ratio)
𝟐
(ii)
𝒂 = 𝟔𝟒 Ans: (first term)
T10 = 𝑎𝑟9
T10 = 64 (
T10 = 64 ∙
𝐓𝟏𝟎 =
(iii)
1
4
S∞ =
𝟏
𝟖
1
2
9
)
1
512
Ans:
𝑎
𝑟 −1
, 𝑓𝑜𝑟 𝑟 < 1
S∞ =
64
1
−1
2
S∞ =
64
1
−
2
S∞ = 64 ÷ −
1
2
S∞ = 64 × −2
𝐒∞ = −𝟏𝟐𝟖 Ans:
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4.
2017 Oct/Nov Exams, Q5 (a)
5.
2017 July/ Aug Exams, Q2 (b)
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6.
2016 Oct/Nov Exams, 5 (b)
8.
Miscellaneous Questions
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Page 114 of 192
1.
2018 Oct/Nov Exams, Q6 (b)
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2.
2018 July/Aug Exams, Q5(b)
3.
2017 Oct/Nov Exams, Q6
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4.
Micellaneous Question
5.
2017 July/Aug Exams, Q6 (b)
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6.
2016 Oct/Nov Exams Q3 (b).
7.
Miscellaneous Question
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Compiled & Solved by Dr. Frendo
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1.
2018 Oct/Nov Exams, Q3
2.
2017 Oct/Nov Exams, Q2 (b)
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3.
2017 July/Aug Exams, Q6 (a)
4.
2017 Oct/Nov Exams, Q6(a)
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5.
2005 Oct/Nov Exams.
6.
2007 Oct/Nov Exam
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7.
2009 Oct/Nov Exam.
1.
2019 Aug Exam Q8 (a).
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2.
2018 Oct/Nov Exams, Q8
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3.
2018 July/Aug Exams, Q8
4.
2017 Oct/Nov Exams, Q7
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5.
2017 July/ Aug Exams, Q10
6.
2016 Oct/Nov Exams, Q10
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7.
Miscellaneous Question:
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1.
2018 Oct/Nov Exams, Q9
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2.
2018 July/ Aug Exams, Q11
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(b)
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3.
2017 Oct / Nov Exams, Q8
(a)
(b)
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4.
2017 July/ Aug Exams, Q8
(a)
(b)
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5.
2016 Oct/ Nov Exams, Q7
(a)
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(b)
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1.
2018 Oct/Nov Exams, Q7(a)
2.
2018 July/ Aug Exams, Q12 (a)
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3.
2017 Oct / Nov Exams, Q10 (a).
4.
2017 July/Aug Exams, Q9 (a)
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5.
2016 Oct/ Nov Exams, Q9 (a)
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6.
2006 Oct/Nov Exam Q7.
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7.
2010 Oct/Nov Exam Q9.
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8.
Miscellaneous Question
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1.
2018 Oct/Nov Exams, Q12 (b)
B
A
C
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2.
2018 July/ Aug Exams, Q7(a)
3.
2017 Oct/ Nov Exams, Q9 (a)
4.
2017 July/Aug Exams, Q12 (a)
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5.
2016 Oct/ Nov Exams, Q9 (a)
6.
Miscellaneous Question.
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1.
2018 Oct/Nov Exams, Q7 (a)
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2.
2018 July/ Aug Exams, Q12 (a).
3.
2017 Oct / Nov Exams, Q10 (a)
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4.
2017 July/Aug Exams, Q9 (a)
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5.
2016 Oct/ Nov Exams, 11 (a)
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6.
2006 Oct/Nov Exams Q11.
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7.
Miscellaneous Question
8.
2012 Oct/Nov Exam Q11.
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1.
2018 Oct/ Nov Exams, Q12(a)
2.
2018 July/August Exams, Q6
3.
2017 Oct/ Nov Exams, Q4 (b)
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4.
2017 July/Aug Exams, Q12 (a)
5.
2016 Oct/ Nov Exams; Q 9 (b and c)
6.
Miscellaneous Question.
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7.
2019 Oct/Nov Exams
h = 6cm
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1.
2018 Oct/Nov Exams, Q3
2.
2018 Jul/Aug Exams, Q7(b)
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3.
2017 Oct/Nov Exams, Q9 (b & c)
4.
2017 July/Aug Exams, Q4b & 9c
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5.
2016 Oct / Nov Exams, Q6
6.
Miscellaneous Questions:
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1.
2018 Oct/Nov Exam Q10
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2.
2018 July/Aug Exams, Q10
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3.
2017 Oct/Nov Exams, Q12
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4.
2017 July/Aug Exams, Q7
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5.
2017 Oct/Nov Exams, Q12
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TOPIC 17 – FUNCTIONS
1.
𝒇: 𝒙 ⟼ 𝟐𝒙 + 𝟏 and 𝒈: 𝒙 ⟼
𝟏
𝒙 − 𝟏
x≠1
2.
f (x) ⟼
𝑥 + 3
a) gf (x) = 3x – 4
(a) 𝑓(3) = 2𝑥 + 1
𝑓(3) = 2(3) + 1
gf (x) = 3(
𝒇(𝟑) = 𝟕 Ans:
(b) 𝑓: 𝑥 ⟼ 2𝑥 + 1
gf (x) =
−11 = 2𝑥 + 1
2𝑥 + 1 = −11
gf (x) =
2𝑥 = −11 − 1
gf (x) =
𝒙 = −𝟔 Ans:
(d) 𝑓
−1 (𝑥)
𝟐𝒙
Ans:
𝑦 = 2𝑥 + 1
∴𝒇
=
𝟐
4 − 𝑥
𝟕𝒙 − 𝟕
𝟒 − 𝒙
Ans:
𝒈 𝒇(𝒙) =
19 − 3𝑥
4 − 𝑥
𝟏𝟗 − 𝟑𝒙
𝟏𝟐 − 𝟑𝒙
gg(−2) = 3(−10) – 4
gg(−2) = −34 Ans:
𝑦=
𝑥 + 3
4−𝑥
x + xy = 4y – 3
Ans:
x(1 + y) = 4y − 3
x=
4𝑦 − 3
(1 + 𝑦)
𝑓 −1 (−2) =
𝑓 −1 (−2) =
4𝑥 − 3
(1 + 𝑥)
−8 − 3
(1 −2)
⇒ 𝑓 −1 (−1) = 11 Ans:
I’ve always enjoyed mathematics. It is the most precise and
concise way of expressing an idea.
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3
𝑔 𝑓(𝑥) =
−𝟏
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3
𝑥 + 3 − 4𝑥 + 16
4−𝑥
−1
𝑔−1 𝑓(𝑥) =
𝑥+ 4
𝑥 + 3
+4
4−𝑥
𝑔−1 𝑓(𝑥) =
−4
4 − 𝑥
3𝑥 + 9 − 16 +4𝑥
4𝑦 − 3 = x + xy
𝑦 − 1
2
𝒙 − 𝟏
3𝑥 + 9
𝑔−1 𝑓(𝑥) =
)– 4
4𝑦 – 𝑥𝑦 = x + 3
𝑦 − 1 = 2𝑥
−𝟏 (𝒙)
4−𝑥
c) 𝑓 −1 (−2)
= 2𝑥 + 1
𝑥=
𝑥 + 3
gg(−2) = −10
1
𝑔𝑓(𝑥) =
2𝑥 + 1 − 1
𝟏
d) 𝑔−1 𝑓(𝑥) =
b) gg(−2) = 3(−2) – 4
(c) 𝑔𝑓(𝑥) = 𝑥 −1 1
𝒈𝒇(𝒙) =
g (x) ⟼ 3x – 4
&
4−𝑥
×
3
19 − 3𝑥
4 − 𝑥
1
3
Ans:
÷3
Page 174 of 192
1.
2019 June/July Exams Q4 & Q20 (b)
2.
2018 June/July Exams Q3 & Q20 (a)
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3.
2017 August/Sept Exams Q20
4.
2016 Oct/Nov Exams Q14 (b)
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5.
2009 Oct/Nov Exams Q3 (b & c)
6.
2003 Oct/Nov Exams Q3 & 18 (a & b)
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6.
2015 Aug/Sept Exams Q3 (a)
7.
2019 Jun/Jul Exams Q16
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8.
2017 Jul/Aug Exams Q12
9.
2018 Jul/Aug Exam Q13
10. 2008 Oct/Nov Exams Q3
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11. 2011 Oct/Nov Exams Q8
1.
2019 Jul/Aug Exams Q14
2.
2018 Jul/Aug Exams Q17
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3.
2017 Jul/Aug Exams Q18
4.
2016 Jul/Aug Exams Q20 (a)
5.
2016 Oct/Nov Exams Q15
6.
2018 Jul/Aug Exams Q9 (b)
7.
2016 Jul/Aug Exams Q19 (a)
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8.
Miscellaneous Questions:
1.
2019 Jul/Aug Exams Q18.
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2.
2018 Jul/Aug Exams Q15.
3.
2017 Jul/Aug Exams Q14.
4.
2016 Oct/Nov Exams Q20.
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5.
2014 oct/Nov Exams Q18.
1.
2019 Jul/Aug Exams Q23.
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2.
2018 Jul/Aug Exams Q23.
3.
2017 Jul/Aug Exams Q23.
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4.
2016 Jul/Aug Exams Q23.
5.
2015 Oct/Nov Exams Q23.
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Promoting Excellence In Mathematics © 2020
Page 187 of 192
1.
2019 Jul/Aug Exams Q2 & 12 (a).
2.
2018 Jul/Aug Exams Q2 & 11 (b).
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Promoting Excellence In Mathematics © 2020
Page 188 of 192
3.
2017 Jul/Aug Exams Q9 (b) & 11 (b).
4.
2016 Jul/Aug Exams Q2 (b).
5.
2012 Oct/Nov Exams Q3.
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Promoting Excellence In Mathematics © 2020
Page 189 of 192
6.
2013 Oct/Aug Exams Q3 & 4 (b).
7.
2003 Oct/Nov Exams Q15.
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Promoting Excellence In Mathematics © 2020
Page 190 of 192
1.
2019 Jul Exams Q19 (b).
2.
2018 Jul Exams Q14 (a).
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Promoting Excellence In Mathematics © 2020
Page 191 of 192
3.
2017 Jul Exams Q4.
4.
2016 Jul Exams Q22 (b).
5.
2016 Oct/Nov Exams Q8 (b).
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Promoting Excellence In Mathematics © 2020
Page 192 of 192
6.
2015 Oct/Nov Exams Q22 (b).
7.
2013 Oct/Nov Exams Q22 (a).
8.
2010 Oct/Nov Exams Q16.
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Promoting Excellence In Mathematics © 2020