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DK & JC's Maths P2 Pamphlet

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A QUICK LOOK AT MATHEMATICS PAPER 2 SYLLABUS D (4024/2)
FINAL EXAMINATION QUESTIONS (2016โ€“ 2018) WITH ANSWERS
MATHEMATICAL FORMULAE
๏ฟฝ ๐‘Ž๐‘Ž๐‘ฅ๐‘ฅ ๐‘›๐‘› ๐‘‘๐‘‘๐‘‘๐‘‘ =
๐‘Ž๐‘Ž๐‘Ž๐‘Ž ๐‘›๐‘› +1
+ ๐‘๐‘
๐‘›๐‘› + 1
1
Conical Frustum ๐•๐• = ๐œ‹๐œ‹(๐‘…๐‘…2 ๐ป๐ป โˆ’ ๐‘Ÿ๐‘Ÿ 2 โ„Ž)
3
1
Pyramid frustum with a square base and top ๐•๐• = โ„Ž(๐ฟ๐ฟ2 + ๐‘™๐‘™2 + ๐ฟ๐ฟ๐ฟ๐ฟ)
Mean = ๐‘ฅ๐‘ฅฬ… =
โˆ‘ ๐‘“๐‘“๐‘“๐‘“
โˆ‘ ๐‘“๐‘“
, ๐‘†๐‘†๐‘†๐‘† = ๏ฟฝ๏ฟฝ
โˆ‘ ๐‘“๐‘“(๐‘ฅ๐‘ฅโˆ’๐‘ฅ๐‘ฅฬ… )2
โˆ‘ ๐‘“๐‘“
3
๏ฟฝ = ๏ฟฝ๏ฟฝ
๐‘›๐‘›๐‘ก๐‘กโ„Ž ๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก ๐‘“๐‘“๐‘“๐‘“๐‘“๐‘“ ๐€๐€๐€๐€: ๐‘‡๐‘‡๐‘›๐‘› = ๐‘Ž๐‘Ž + (๐‘›๐‘› โˆ’ 1)๐‘‘๐‘‘,
๐‘›๐‘›
๐€๐€๐€๐€: ๐‘บ๐‘บ๐’๐’ = [2๐‘Ž๐‘Ž + (๐‘›๐‘› โˆ’ 1)๐‘‘๐‘‘], ๐†๐†๐†๐†: ๐‘บ๐‘บ๐’๐’ =
2
โˆ‘ ๐‘“๐‘“๐‘ฅ๐‘ฅ 2
โˆ‘ ๐‘“๐‘“
โˆ’ (๐‘ฅ๐‘ฅฬ… )2 ๏ฟฝ
๐†๐†๐†๐†: ๐‘‡๐‘‡๐‘›๐‘› = ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ ๐‘›๐‘›โˆ’1
(1โˆ’๐‘Ÿ๐‘Ÿ ๐‘›๐‘› )
1โˆ’๐‘Ÿ๐‘Ÿ
,r<1
Cosine Rule for โˆ†๐€๐€๐€๐€๐€๐€: ๐’‚๐’‚๐Ÿ๐Ÿ = ๐’ƒ๐’ƒ๐Ÿ๐Ÿ + ๐’„๐’„๐Ÿ๐Ÿ โˆ’ ๐Ÿ๐Ÿ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ๐’ƒ
Compiled & Solved by Kachama Dickson. C & Chansa John
Contacts: 0966295655/0955295655/0976221226
Email: [email protected]
: [email protected]
© 2019
Page 2 of 84
Both Mr. Kachama. D.C and Mr. Chansa J are graduates from the Copperbelt University
(CBU). Kachama D.C is a holder of bachelorโ€™s degree in Mathematics Ed (BSc MA.Ed)
while Mr. Chansa. J is a holder of bachelorโ€™s degree in Pure and Applied mathematics. To
show their competence and eligibility in mathematics, the two individuals have produced
many mathematics pamphlets and this is one of their best joint pamphlet.
This pamphlet consists of final examination questions from 2016 - 2018 for both grade
twelve (12) and G.C.E ordinary levels. Answers with all necessary working methods are
shown at the end of questions. All the Questions are copied directly from the examination
past papers for both July/August and October/ November exams.
โ€œWe believe, this pamphlet will be of great help to you even as you prepare for your final
examinations:โ€
TABLE OF CONTENTS
Topic
page
1. Algebra โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ........ 3
2. Matrices โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ...โ€ฆโ€ฆโ€ฆ.โ€ฆ... 4
3. Setsโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ 5
4. Probability โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ 7
5. Sequences & Series โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.. 8
6. Pseudo code & Flow chartโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.... 9
7. Loci & Construction โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.โ€ฆโ€ฆ.. 12
8. Calculus โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ...โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ 14
9. Vector Geometry โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ 15
10. Trigonometry โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ..โ€ฆ...โ€ฆ 17
11. Mensuration โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ..โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.โ€ฆ 20
12. Earth Geometry โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.. 22
13. Quadratic Function โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ 24
14. Linear Programming โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.. 28
15. Statistics โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ. 31
16. Transformation โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ... 34
17. Answers โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ. 38
CAUTIONS: NO PART OF THIS PAMPHLET MAY BE REPRODUCED WITHOUT A PERSSION OF THE
AUTHORS
ANY ERRORS IN THIS PAMPHLET ARE THE RESPONSIBILTY OF THE AUTHOR
© 2019 All rights Reserved
Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
Page 3 of 84
TOPIC 1: ALGEBRA
1. 2018 Oct/Nov Exams
๐‘๐‘โˆ’๐‘Ž๐‘Ž
(a) Simplify
๐‘Ž๐‘Ž 2 โˆ’๐‘๐‘ 2
12๐‘‘๐‘‘๐‘‘๐‘‘ 3
(b) Simplify
(c) Express
3
๐‘ฅ๐‘ฅ+1
โˆ’
9๐‘๐‘ 3 ๐‘›๐‘›
÷
15๐‘๐‘๐‘๐‘ 3
4
๐‘ฅ๐‘ฅโˆ’1
10๐‘๐‘ 2 ๐‘‘๐‘‘ 2
as a single fraction in its lowest terms.
2. 2018 July/Aug Exams
(a) Simplify
(b) Express
7๐‘ ๐‘ ๐‘ก๐‘ก 3
3
โˆ’
2๐‘ฅ๐‘ฅโˆ’5
5๐‘ข๐‘ข 3 ๐‘ฃ๐‘ฃ
×
15๐‘ข๐‘ข 3 ๐‘ฃ๐‘ฃ 2
28๐‘ ๐‘  3 ๐‘ก๐‘ก 2
4
as a single fraction in its lowest terms.
๐‘ฅ๐‘ฅโˆ’3
3. 2017 Oct/Nov Exams
(a) Simplify
(b) Simplify
(c) Express
14๐‘ฅ๐‘ฅ 3
7๐‘ฅ๐‘ฅ 4
÷
9๐‘ฆ๐‘ฆ 2
18๐‘ฆ๐‘ฆ 3
2๐‘ฅ๐‘ฅ2 โˆ’8
1
๐‘ฅ๐‘ฅโˆ’4
๐‘ฅ๐‘ฅ+2
โˆ’
2
5๐‘ฅ๐‘ฅโˆ’1
as a single fraction in its lowest terms.
4. 2017 July/ Aug Exams
(a) Simplify
(b) Simplify
(c) Express
๐‘š๐‘š 2 โˆ’1
๐‘š๐‘š 2 โˆ’๐‘š๐‘š
๐‘๐‘ 2 ๐‘ž๐‘ž 3
3
4
5๐‘ฅ๐‘ฅโˆ’2
โˆ’
×
8
๐‘๐‘๐‘๐‘
2
๐‘ฅ๐‘ฅ+3
÷ 2๐‘๐‘2 ๐‘ž๐‘ž
as a single fraction in its lowest terms.
5. 2016 October Exams
(a) Simplify
(b) Simplify
(c) Express
๐‘ฅ๐‘ฅโˆ’1
๐‘ฅ๐‘ฅ 2 โˆ’1
17๐‘˜๐‘˜ 2
20๐‘Ž๐‘Ž 2
2
2๐‘ฅ๐‘ฅโˆ’1
÷
โˆ’
51๐‘˜๐‘˜ 2
5๐‘Ž๐‘Ž
1
3๐‘ฅ๐‘ฅ+1
as a single fraction in its lowest terms.
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 4 of 84
TOPIC 2: MATRICES
1. 2018 Oct/Nov Exams, Q1(a)
Given that ๐ด๐ด = ๏ฟฝ
8
โˆ’5
๏ฟฝ and ๐ต๐ต = ๏ฟฝ
2
3
4
1
๐‘ฆ๐‘ฆ
๏ฟฝ,
5
(a) find the value of ๐‘ฆ๐‘ฆ, for which the determinants of A and B are equal,
(b) hence find the inverse of B.
2. 2018 Jul/August Exams: Q1(a)
2๐‘ฅ๐‘ฅ
3
Given that A = ๏ฟฝ
2
๏ฟฝ,
๐‘ฅ๐‘ฅ
(a) find the positive value of ๐‘ฅ๐‘ฅ for the determinant of A is 12,
(b) hence or otherwise find Aโˆ’1 .
3. 2017 Oct/Nov Exams: Q1(a)
3
Given that matrix M = ๏ฟฝ
5
โˆ’2
๏ฟฝ
๐‘ฅ๐‘ฅ
(a) find the value of ๐‘ฅ๐‘ฅ for which the determinant of M is 22,
(b) hence find the inverse of M.
4. 2017 July/ Aug Exams: Q1(a)
10
11
Given that K = ๏ฟฝ
โˆ’2
๏ฟฝ, find
โˆ’2
(a) the determinant of K,
(b) the inverse of K.
5. 2016 Oct/Nov Exams: Q1(a)
Given that ๐‘„๐‘„ = ๏ฟฝ
3
๐‘ฅ๐‘ฅ
โˆ’2
๏ฟฝ, find
4
(a) the value of ๐‘ฅ๐‘ฅ, given that the determinant of Q is 2,
(b) the inverse of Q.
Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
Page 5 of 84
TOPIC3: SETS
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) Mathematics and physics but not chemistry.
2. 2018 Jul/Aug Exams: Q3(a)
The diagram below shows how learners at Twatenda School travel to school. The
learners use either buses (B), cars (C) or walk (W) to school.
B
E
C
7
2
14
4
๐‘ฅ๐‘ฅ
7
(i)
(ii)
3
W
If 22 learners walk to school, find the value of ๐‘ฅ๐‘ฅ.
How many learners use
(a) only one mode of transport,
(b) two different mode of transport.
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 6 of 84
3. 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) Planted two different crops
4. 2017 July/Aug Exams: Q3 (b)
The Venn diagram below shows tourist attractions visited by certain students in a
certain week.
(i)
(ii)
Find the value of ๐‘ฆ๐‘ฆ if 7 students visited Mambilima Falls only.
How many students visited
(a) Victoria falls but not Gonya Falls,
(b) Two tourist attractions only,
(c) One tourist attraction only?
5. 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.
Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
Page 7 of 84
(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,
(c) Listen to two types of programs only.
TOPIC 4: PROBABILITY
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 to show all the possible outcomes.
(b) What is the probability that both buttons are white?
3. 2017 Oct/Nov Exams, Q2 (a)
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.
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 8 of 84
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.
TOPIC 5: SEQUENCES AND SERIES
1. 2018 Oct/Nov Exams, Q5(b)
The first three terms of a geometric progression are ๐‘˜๐‘˜ + 4, ๐‘˜๐‘˜ ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž 2๐‘˜๐‘˜ โˆ’ 15 where ๐‘˜๐‘˜ a
positive integer.
(a) find the value of ๐‘˜๐‘˜,
(b) list the first three terms of the geometric progression,
(c) find the sum to infinity.
2. 2018 Jul/Aug Exams, Q2(b)
2
In a geometric progression, the third term is and the fourth term is
(a) The first term and the common ratio,
9
2
27
. Find
(b) The sum of the first 5 terms of the geometric progression,
(c) The sum to infinity.
3. 2017 Oct/Nov Exams, Q5(a)
1
For the geometric progression 20, 5,1 , . . . , find
(a) the common ratio,
4
(b) the nth term,
(c) the sum of the first 8 terms.
4. 2017 July/ Aug Exams, Q2(b)
The first three terms of a geometric progression are 6 + ๐‘›๐‘›, 10 + ๐‘›๐‘› and 15 + ๐‘›๐‘›. Find
(a) the value of n,
(b) the common ratio,
(c) the sum of the first 6 terms of this sequence.
Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
Page 9 of 84
5. 2016 Oct/Nov Exams, 5(b)
The first three terms of a geometric progression are ๐‘ฅ๐‘ฅ + 1, ๐‘ฅ๐‘ฅ โˆ’ 3 and ๐‘ฅ๐‘ฅ โˆ’ 1.
(a)
(b)
the value of ๐‘ฅ๐‘ฅ,
(c)
the sum to infinity.
the first term,
TOPIC 6: PSEUDO CODE AND FLOW CHARTS
1. 2018 Oct/Nov Exams, Q6(b)
The program below is given in a form of a Pseudo code
Start
Enter ๐‘ฅ๐‘ฅ, ๐‘ฆ๐‘ฆ
Let M = square root(๐‘ฅ๐‘ฅ squared + ๐‘ฆ๐‘ฆ squared)
IF M< 0
THEN display error message โ€œ M must be positiveโ€
ELSE
END IF
Display M
Stop
Draw the corresponding flow chart for the information given above.
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2. 2018 July/Aug Exams, Q5(b)
Study the pseudo code below.
Start
Enter ๐‘Ž๐‘Ž, ๐‘Ÿ๐‘Ÿ, ๐‘›๐‘›
R= 1 โˆ’ ๐‘Ÿ๐‘Ÿ
If R = 0 THEN
Print โ€œthe value of r is not validโ€
Else Sn =
End if
๐‘Ž๐‘Ž(1โˆ’๐‘Ÿ๐‘Ÿ ๐‘›๐‘› )
R
Print Sn
Stop
Construct a flow chart corresponding to the Pseudo code above.
3. 2017 Oct/Nov Exams, Q6
Study the flow chart below
Start
Enter r
Is
r < 0?
Yes
Error โ€œr must be positiveโ€
No
1
๐ด๐ด = โˆ— ๐‘Ÿ๐‘Ÿ โˆ— ๐‘Ÿ๐‘Ÿ โˆ— ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
2
Display Area
Stop
Write a pseudo code corresponding to the flow chart program above
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4. 2017 July/Aug Exams, Q6(b)
The diagram below is given in the form of a flow chart
Start
Enter a, r
Is
|๐’“๐’“| < 1?
No
Yes
๐‘บ๐‘บโˆž =
๐’‚๐’‚
๐Ÿ๐Ÿโˆ’ ๐’“๐’“
Display sum to infinity
Stop
Write a pseudo code corresponding to the flow chart program above
5. 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 โ€œerror messageโ€ and re-enter positive radius
Else enter height
If height < 0
The display โ€œerror messageโ€ and re-enter positive height
1
Else Volume = โˆ— ๐œ‹๐œ‹ โˆ— square radius โˆ— height
3
End if
Display volume
Stop
Draw the corresponding flowchart for the information given above.
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TOPIC 7: LOCI & CONSTRUCTION
Answer the whole of these questions on a sheet of plain papers
1. 2018 Oct/Nov Exams, Q4
(a)
(i)
(ii)
(b)
(c)
Construct triangle XYZ in which XY = 9cm, YZ = 7cm and angle XYZ = 38°
Measure and write the length of XZ
On your diagram within triangle XYZ, construct the locus of points which are
(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)
(ii)
(b)
(c)
๏ฟฝ R = 50°
Construct triangle PQR in which PQ = 10cm, QR = 8cm and ๐๐๐๐
Measure and write the length of PR
On your diagram, within triangle PQR, construct the locus of points which are
(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.
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
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(c)
(d)
Within the quadrilateral ABCD, draw the locus of points which are
(i)
8cm from A
(ii)
Equidistant from BC and CD
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)
(ii)
(b)
(c)
Construct triangle PQR in which PQ is 9cm, angle PQR = 60° and QR = 10cm
Measure and write the length of PR.
On your diagram, draw the locus of points with triangle PQR which are
(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)
(b)
(c)
(i)
Construct triangle ABC where AB = BC = CA = 7cm
(ii)
Measure and write the size of โˆ CAB.
Within triangle ABC construct the locus of points which are
(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.
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TOPIC 8: CALCULUS
1. 2018 Oct/Nov Exams, Q3
2
(a) Evaluate โˆซโˆ’1(2 + ๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ 2 )๐‘‘๐‘‘๐‘‘๐‘‘
(b) Find the equation of the normal to the curve ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ +
2. 2018 Jul/Aug Exams, Q7(b)
1
(a) Evaluate โˆซ0 (๐‘ฅ๐‘ฅ 2 โˆ’ 2๐‘ฅ๐‘ฅ + 3)๐‘‘๐‘‘๐‘‘๐‘‘
4
๐‘ฅ๐‘ฅ
at the point where ๐‘ฅ๐‘ฅ = 4
(b) Determine the equation of the normal to the curve ๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 2 โˆ’ 3๐‘ฅ๐‘ฅ โˆ’ 2 that passes
through (3, 7)
3. 2017 Oct/Nov Exams, Q9 (b & c)
(a) Find the coordinates of the points on the curve ๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 โˆ’ 36๐‘ฅ๐‘ฅ โˆ’ 3 where
the gradient is zero.
3
(b) Evaluate โˆซโˆ’1(3๐‘ฅ๐‘ฅ 2 โˆ’ 2๐‘ฅ๐‘ฅ)๐‘‘๐‘‘๐‘‘๐‘‘
4. 2017 July/Aug Exams, Q4b & 9c
5
(a) Evaluate โˆซ2 (3๐‘ฅ๐‘ฅ 2 + 2)๐‘‘๐‘‘๐‘‘๐‘‘
(b) Find the equation of the tangent to the curve ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 2 โˆ’ 3๐‘ฅ๐‘ฅ โˆ’ 4 at a point where
๐‘ฅ๐‘ฅ = 2
5. 2016 Oct / Nov Exams, Q6
3
The equation of the curve is๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 โˆ’ ๐‘ฅ๐‘ฅ 2 . Find
2
(a) equation of the normal where ๐‘ฅ๐‘ฅ = 2,
(b) the coordinates of the stationary points.
Continue working hard, Rome was not built in one day
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TOPIC 9: VECTORS
1. 2018 Oct/Nov Exams, Q3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐‘Ž๐‘Ž and ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
In the quadrilateral ABCD, AB
AD = b, ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
BC = 2๐‘๐‘ and AE: AC = 1: 3
U
(i)
(ii)
Find in terms of ๐‘Ž๐‘Ž and/or ๐‘๐‘
(a) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด
(b) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต
U
(c) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต
Hence or otherwise show that the points B, D and E are collinear
2. 2017 Oct/Nov Exams, Q2(b)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘ = 2p, ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘
In the diagram below, OP
OQ = 4q and PX โˆถ XQ = 1: 2
(i)
(ii)
Express in terms of p and / or q.
(a) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
PQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
(b) PX
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
(c) OX
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—, show that CQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 4 ๏ฟฝ1 โˆ’ โ„Ž ๏ฟฝq โˆ’ 4โ„Ž p.
Given that ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OC = โ„ŽOX
3
3
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3. 2017 July/Aug Exams, Q6(a)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘ = ๐‘Ž๐‘Ž and๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘
In the diagram below, OABC is a parallelogram in which OA
AB = 2๐‘๐‘. OB and
AC intersect at D. E is the midpoint of CD. E is the mid - point of CD.
U
Express in terms of a and / or b.
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘,
OB
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘
OE,
(a)
(b)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ‘.
CD
(c)
4. 2017 Oct/Nov Exams, Q6(a)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 6๐‘๐‘ .
In the diagram below, OAB is a triangle in which ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = 3๐‘Ž๐‘Ž and ๐‘‚๐‘‚๐‘‚๐‘‚
U
U
OC : CA = 2 : 3 and AD : DB = 1 : 2. OD meets CB at E.
(i)
Express each of the following in terms of ๐‘Ž๐‘Ž and / or ๐‘๐‘
U
(a) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
AB
(b) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OD
(c)
(ii)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
BC
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—, express BE
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = hBC
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— in terms of โ„Ž, ๐‘Ž๐‘Ž and ๐‘๐‘
Given that BE
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TOPIC10: TRIGONOMETRY
1. 2018 Oct/Nov Exams, Q8
In the diagram below, K, N, B and R are places on horizontal surface. KN = 80m,
๏ฟฝ N = 52°
NB = 50m and KR
80m
K
60°
N
50m
B
52°
(a) Calculate
R
(i)
KR
(ii)
the area of triangle KNB
(b) Given that the area of triangle KNR is equal to 3 260๐‘š๐‘š2 , calculate the shortest
distance from R to KN.
(c) Sketch the graph of ๐‘ฆ๐‘ฆ = cos ๐œƒ๐œƒ ๐‘“๐‘“๐‘“๐‘“๐‘“๐‘“ 0° โ‰ค ๐œƒ๐œƒ โ‰ค 360°
2. 2018 July/Aug Exams, Q8
(a) Three villages A, B and C are connected by straight paths as shown in the diagram
below.
A
15km
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 ๐œƒ๐œƒ = 0.937 for ๐‘“๐‘“๐‘“๐‘“๐‘“๐‘“ 0° โ‰ค ๐œƒ๐œƒ โ‰ค 360°
(c) Sketch the graph of ๐‘ฆ๐‘ฆ = sin ๐œƒ๐œƒ ๐‘“๐‘“๐‘“๐‘“๐‘“๐‘“ 0° โ‰ค ๐œƒ๐œƒ โ‰ค 360°
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3. 2017 Oct/Nov Exams, Q7
(a) The diagram below shows the Location of houses for a village Headman (H), his
Secretary (S) and a Trustee (T). H is 1.3 km from S, T is 1.9 km from H and
๏ฟฝ S = 130°
TH
Calculate
(i)
the area of triangle THS
(ii)
the distance TS
(iii)
the shortest distance from H to TS
(b) Find the angle between 0° and 90° which satisfies the equation cos ๐œƒ๐œƒ =
4. 2017 July/ Aug Exams, Q10
2
3
(a) In Triangle PQR below, QR = 36.5, angle PQR = 36° and angle QPR = 46°.
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Calculate
(i)
PQ
(ii)
the area of triangle PQR
(iii)
the shortest distance from R to PQ
(b) Solve the equation sin ฮธ = 0.6792 for 0° โ‰ค ฮธ โ‰ค 360°
5. 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
(i)
MI
(ii)
the area of triangle MKI
(iii)
the shortest distance from K to MI
(b) Solve the equation tan ฮธ = 0.7 for 0° โ‰ค ฮธ โ‰ค 180°.
Donโ€™t wait for somebody to tell you that you can do it, just believe in yourself that you can do it
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TOPIC 11: MENSURATION
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.
10cm
9cm
4cm
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. (Take๐…๐… = ๐Ÿ‘๐Ÿ‘. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ)
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 ๐œ‹๐œ‹ = 3.142]
Given that EX = 4cm, DB = 12cm and DE = 15cm, calculate
(i)
the height AE, of the smaller con AXY.
(ii)
the volume of XBCY, the shape that remained.
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5. 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 = 1000litres)
(b) A cone has a perpendicular height of 12 cm and slant height of 13 cm, calculate its
total surface area. (Take๐œ‹๐œ‹ = 3.142).
TOPIC 12: EARTH GEOMETRY
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 ๐…๐… = ๐Ÿ‘๐Ÿ‘. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ ๐š๐š๐š๐š๐š๐š ๐‘น๐‘น = ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”]
(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.
(i)
Calculate the difference in longitudes between B and Q.
(ii)
Find the position of Q.
2. 2018 July/ Aug Exams, Q7(a)
In the diagram below, A and B are points on latitude 60°N while C is a point on latitude
60°S. [ ๐œ‹๐œ‹ = 3.142 and R = 3437nm].
(a) Calculate the distance BC along the latitude 60°๐ธ๐ธ in nautical miles.
(b) A ship sails from C to D in 12 hours. Find its speed in notes.
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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 ๐›‘๐›‘ = ๐Ÿ‘๐Ÿ‘. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ ๐š๐š๐š๐š๐š๐š ๐‘๐‘ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘)
(a) Calculate the difference in latitudes between W and Y.
(b) Calculate the distance in nautical miles between
(i)
X and Z along the longitudes 105°E
(ii)
Y and Z along the circle of latitude 30°S
4. 2017 July/Aug Exams, Q12 (a)
P (80°N, 10°E), ๐๐(80°N, 70°E), ๐‘๐‘(85°S, 70°E)and ๐’๐’(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.
[Take ๐›‘๐›‘ = ๐Ÿ‘๐Ÿ‘. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ ๐š๐š๐š๐š๐š๐š ๐‘๐‘ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘]
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5. 2016 Oct/ Nov Exams, Q9(a)
The points A, B, C and D are on the surface of the earth.
(Take ฯ€ = 3.142 and R = 3437nm)
(a)
(b)
(c)
Find the difference in latitude between points C and B.
Calculate the length of the circle of latitude 50°N in nautical miles.
Find the distance AD in nautical miles.
TOPIC 13: QUADRATIC FUNCTIONS
1. 2018 Oct/Nov Exams, Q7(a)
The values ๐‘ฅ๐‘ฅ and ๐‘ฆ๐‘ฆ are connected by the equation๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 + 5. Some
corresponding values of ๐‘ฅ๐‘ฅ and ๐‘ฆ๐‘ฆ are given in the table below.
๐‘ฅ๐‘ฅ
โˆ’2
โˆ’1.5
โˆ’1
โˆ’0.5
0
0.5
1
1.5
2
๐‘ฆ๐‘ฆ
๐‘๐‘
โˆ’8.5
(a) Calculate the value of P
0
4
5
4.5
4
5
9
(b) Using a scale of 4cm to represent 1 unit on the ๐‘ฅ๐‘ฅ โˆ’axis for โˆ’2 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 2 and a scale
of 2cm to represent 5 units on the ๐‘ฆ๐‘ฆ โˆ’axis forโˆ’25 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 10, draw the graph
of ๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 + 5.
(c) Use your graph to solve the equation 2๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 + 5 = ๐‘ฅ๐‘ฅ
(d) Calculate an estimate of the gradient of the curve at the point where ๐‘ฅ๐‘ฅ = 1.5
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2. 2018 July/ Aug Exams, Q12a)
The diagram below shows the graph of ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 12๐‘ฅ๐‘ฅ
(a) Use the graph to solve the equation
(i)
๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 12๐‘ฅ๐‘ฅ = 0
(ii)
๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 12๐‘ฅ๐‘ฅ = ๐‘ฅ๐‘ฅ + 10
(b) Calculate an estimate of the
(i)
Gradient of the curve at the point where ๐‘ฅ๐‘ฅ = โˆ’3
(ii)
Area bounded by the curve, ๐‘ฅ๐‘ฅ = โˆ’3, ๐‘ฅ๐‘ฅ = โˆ’1 and ๐‘ฆ๐‘ฆ = โˆ’10
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3. 2017 Oct / Nov Exams, Q10(a)
The diagram below shows the graph of ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3
(a) Use the graph to find the solutions of the equations
(i)
๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 = 0
(ii)
๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 = 5
(b) Calculate an estimate of
(i)
the gradient of the curve at the point (โˆ’3,0)
(ii)
the area bounded by the curve, ๐‘ฅ๐‘ฅ = 0, ๐‘ฆ๐‘ฆ = 0 and ๐‘ฆ๐‘ฆ = 20
I was not born a mathematician, but I am determined to
Be a mathematician (Young Prof)
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4. 2017 July/Aug Exams, Q9(a)
The diagram below shows the graph of ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 5๐‘ฅ๐‘ฅ + 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)
๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 5๐‘ฅ๐‘ฅ + 3 = 0
(ii)
๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 5๐‘ฅ๐‘ฅ + 3 = 5๐‘ฅ๐‘ฅ
(c) to calculate an estimate of the area bounded by the curve ๐‘ฅ๐‘ฅ = 0, ๐‘ฆ๐‘ฆ = 0 and ๐‘ฅ๐‘ฅ = โˆ’2
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5. 2016 Oct/ Nov Exams, Q9(a)
The values of x and y are connected by the equation ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ(๐‘ฅ๐‘ฅ โˆ’ 2)(๐‘ฅ๐‘ฅ + 2). Some
corresponding values of x and y are given in the table below
๐‘ฅ๐‘ฅ
๐‘ฆ๐‘ฆ
โˆ’3
โˆ’15
โˆ’2
(a) Calculate value of ๐‘˜๐‘˜
0
โˆ’1
3
0
0
1
โˆ’3
2
0
3
๐‘˜๐‘˜
(b) Using a scale of 2cm to represent 1 unit on the x โ€“ axis for โˆ’3 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 3 and 2cm to
represent 5 units on the y- axis forโˆ’16 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 16. Draw the graph of
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ(๐‘ฅ๐‘ฅ โˆ’ 2)(๐‘ฅ๐‘ฅ + 2)
(c) Use your graph to solve the equations
(i)
(ii)
๐‘ฅ๐‘ฅ(๐‘ฅ๐‘ฅ โˆ’ 2)(๐‘ฅ๐‘ฅ + 2) = 0
๐‘ฅ๐‘ฅ(๐‘ฅ๐‘ฅ โˆ’ 2)(๐‘ฅ๐‘ฅ + 2) = ๐‘ฅ๐‘ฅ + 2
TOPIC 14: LINEAR PROGRAMING
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 ๐‘ฅ๐‘ฅ be the number of learners and ๐‘ฆ๐‘ฆ 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 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 70 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 70 respectively and shade the unwanted region to
indicate clearly where the solution of the inequalities lie.
(c) (i)
If the group has 25 learners, what is the minimum number of teachers that
must accompany them?
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(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 = 30๐‘ฅ๐‘ฅ + 50๐‘ฆ๐‘ฆ
2. 2018 July/ Aug Exams, Q12a)
A tailor at a certain market intends to make dresses and suits for sale.
(a) Let ๐‘ฅ๐‘ฅ represent the number of dresses and ๐‘ฆ๐‘ฆ 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 ๐‘ฅ๐‘ฅ and ๐‘ฆ๐‘ฆ axes for
0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 60 and 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 80. Shade the unwanted region to indicate clearly the
region where (๐‘ฅ๐‘ฅ, ๐‘ฆ๐‘ฆ) must lie.
(c) (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) Calculate this maximum profit.
3. 2017 Oct / Nov Exams, Q10(a)
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 ๐‘ฅ๐‘ฅ and ๐‘ฆ๐‘ฆ 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 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 70 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 70 respectively and shade the unwanted region to
show clearly the region where the solution of the inequalities lie.
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(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 o 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 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 160 and 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 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 ๐‘ฅ๐‘ฅ represent the number of videos produced and ๐‘ฆ๐‘ฆ 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.
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(b)
Using a scale of 2cm to represent 100 copies on both axes, draw the x and y
axes for 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 800 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 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
TOPIC 15: STATISTICS
1. 2018 Oct/Nov Exams, Q9
The frequency table below shows the distribution of marks obtained by 90 learners on a
test.
10 < ๐‘ฅ๐‘ฅ โ‰ค 20
20 < ๐‘ฅ๐‘ฅ โ‰ค 30
30 < ๐‘ฅ๐‘ฅ โ‰ค 40
40 < ๐‘ฅ๐‘ฅ โ‰ค 50
50 < ๐‘ฅ๐‘ฅ โ‰ค 60
60 < ๐‘ฅ๐‘ฅ โ‰ค 70
Marks(x)
frequency
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
Marks(x)
Cumulative frequency
Relative Cumulative frequency
(ii)
(iii)
โ‰ค 10
0
0
โ‰ค 20
2
0.02
โ‰ค 30
12
0.13
โ‰ค 40
27
0.3
โ‰ค 50
50
โ‰ค 60
80
โ‰ค 70
90
Using a scale of 2cm to represent 10 units on the x- axis for 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 70 and a
scale of 2cm to represent 0.1 units on the y โ€“ axis for 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 1, draw a
smooth relative cumulative frequency curve.
Showing your method clearly, use your graph to estimate the 65th Percentile.
2. 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
frequency
2
1
3
4
5
6
7
8
5
4
6
10
16
18
(a) Calculate the standard deviation
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(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
(ii)
Fruits per tree
2
3
4
5
6
7
8
Cumulative frequency
1
6
10
16
26
42
60
Relative cumulative frequency
0.02
0.1
0.17
0.27
Using a scale of 1cm to represent 1 unit on the x-axis for 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 8 and a
scale of 2cm to represent 0. 1 unit on the y โ€“ axis for 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 1, draw a
smooth relative frequency curve.
(iii)
Showing your method clearly, use your graph to estimate the 70th Percentile.
3. 2017 Oct / Nov Exams, Q8
The table below shows the amount of money spent by 100 learners at school on a
particular day.
Amount in Kwacha
Frequency
0 < ๐‘ฅ๐‘ฅ โ‰ค 5
13
5 < ๐‘ฅ๐‘ฅ โ‰ค 10
27
0 < ๐‘ฅ๐‘ฅ โ‰ค 15
35
15 < ๐‘ฅ๐‘ฅ โ‰ค 20
16
20 < ๐‘ฅ๐‘ฅ โ‰ค 25 2 25 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 30
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
Cumulative frequency
(ii)
โ‰ค0
0
โ‰ค5
13
โ‰ค 10
โ‰ค 15
โ‰ค 20
40
โ‰ค 25
โ‰ค 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)
Showing your method clearly, use your graph to estimate the
semiโˆ’ interquartile range.
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4. 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 < ๐‘ฅ๐‘ฅ โ‰ค 30
6
30 < ๐‘ฅ๐‘ฅ โ‰ค 35
4
35 < ๐‘ฅ๐‘ฅ โ‰ค 40
40 < ๐‘ฅ๐‘ฅ โ‰ค 45
7
11
45 < ๐‘ฅ๐‘ฅ โ‰ค 50
12
50 < ๐‘ฅ๐‘ฅ โ‰ค 55
8
55 < ๐‘ฅ๐‘ฅ โ‰ค 60
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
Number of
Vendors
(ii)
โ‰ค 25
โ‰ค 30
0
5
โ‰ค 35
9
โ‰ค 40
16
โ‰ค 45
27
โ‰ค 50
โ‰ค 55
โ‰ค 60
Using a horizontal scale of 2cm to represent 10 newspapers on the x โ€“axis for
0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 60 and a vertical scale of 4cm to represent 10 vendors on the y โ€“
(iii)
axis for 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 50, draw a smooth cumulative frequency curve.
Showing your method clearly, use your graph to estimate the 50th Percentile.
5. 2016 Oct/ Nov Exams, 7
The ages of people living at Pamodzi Village are recorded in the frequency table below.
0 < ๐‘ฅ๐‘ฅ โ‰ค 10 10 < ๐‘ฅ๐‘ฅ โ‰ค 20
20 < ๐‘ฅ๐‘ฅ โ‰ค 30
30 < ๐‘ฅ๐‘ฅ โ‰ค 40
40 < ๐‘ฅ๐‘ฅ โ‰ค 50
50 < ๐‘ฅ๐‘ฅ โ‰ค 60
Ages
Number of
7
22
28
23
15
5
People
(a) Calculate the standard deviation
(b) Answer part of this question on a sheet of a graph paper
(i)
Using the table above, copy and complete the cumulative frequency table
below.
Age
Number of people
(ii)
โ‰ค 10
7
โ‰ค 20
29
โ‰ค 30
โ‰ค 40
โ‰ค 50
โ‰ค 60
100
Using the scale of 2cm to represent 10 units on both axes, draw a smooth
cumulative frequency curve where 0 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 60 and 0 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 100.
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(iii)
Showing your method clearly, use your graph to estimate the
Semiโˆ’ interquartile range.
TOPIC 16: TRANSFORMATION
1. 2018 Oct/Nov Exams, Q10
Study the diagram below and answer the questions that follow.
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(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
1
โˆ’2
๏ฟฝ
0
๏ฟฝ, find the coordinates of S.
1
2. 2018 July/Aug Exams, Q10
Using a scale of 1cm to represent 1 unit, on both axes, draw x and y axes for
โˆ’8 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 12 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž โˆ’ 6 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 14.
(a)
(b)
Draw and label triangle X with vertices (2,4), (4,4) and (4,1).
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 โ€“axis as the invariant line and shear factor -2 maps triangle X onto
triangle S. Draw and label triangle S.
(e)
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
3. 2017 Oct/Nov Exams, Q12
Using a scale of 1cm to represent 1 unit on each axis, draw x and y axes for
โˆ’6 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 10 and โˆ’10 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 8.
(a) A quadrilateral ABCD has vertices A(โˆ’5,7), B(โˆ’4,8), C(โˆ’3,7) and
D(โˆ’4,4) while
its imagine has vertices A1 (โˆ’5, โˆ’3), B1 (โˆ’6, โˆ’2), C1 (โˆ’5, โˆ’1) and D1 (โˆ’2, โˆ’2).
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(i)
(ii)
(b)
Draw and label the quadrilateral ABCD and its image A1 B1 C1 D1 .
Describe fully the transformation which maps the quadrilateral ABCD
onto quadrilateral A1 B1 C1 D1 .
A2 B2 C2 D2
(i)
(ii)
(c)
โˆ’2 0
๏ฟฝ maps the quadrilateral ABCD on the quadrilateral
0 1
The matrix ๏ฟฝ
Find the coordinates of the vertices of the quadrilateral A2 B2 C2 D2
Draw and label the quadrilateral A2 B2 C2 D2 .
The quadrilateral ABCD is mapped onto quadrilateral A3 B3 C3 D3 where
A3 is (4, โˆ’8), B3 is (2, โˆ’10), is C3 (0, โˆ’8)and D3 is (2. โˆ’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
โˆ’6 โ‰ค ๐‘ฅ๐‘ฅ โ‰ค 10 and โˆ’6 โ‰ค ๐‘ฆ๐‘ฆ โ‰ค 12.
(a)
A quadrilateral ABCD has vertices A(1,1), B(2,1), C(3,2) and D(2,3) while
its imagine has vertices ๐ด๐ด1 (3,2), ๐ต๐ต1 (6,1), ๐ถ๐ถ1 (9,2)
(i)
(ii)
(b)
Draw and label the quadrilateral ABCD and its image A1 B1 C1 D1 .
Describe fully the transformation which maps the quadrilateral ABCD
onto quadrilateral A1 B1 C1 D1 .
The matrix ๏ฟฝ
A2 B2 C2 D2 .
(i)
(ii)
(c)
and ๐ท๐ท1 (6,3).
1
3
0
๏ฟฝ maps the quadrilateral ABCD on the quadrilateral
1
Find the coordinates of quadrilateral A2 B2 C2 D2 .
Draw and label quadrilateral A2 B2 C2 D2 .
Quadrilateral A3 B3 C3 D3 has vertices A3 (โˆ’2, โˆ’4), B3 (โˆ’4, โˆ’2), C3 (โˆ’6, โˆ’4) and
D3 (โˆ’4, โˆ’6). Describe fully the transformation which maps quadrilateral ABCD
onto A3 B3 C3 D3 .
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5. 2017 Oct/Nov Exams, Q12
Study the diagram below and answer questions that follow.
(a)
(b)
An enlargement maps triangle ABC onto triangle A1 B1 C1 . Find
(i)
the centre of enlargement
(ii)
the scale factor
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)
โˆ’3
A transformation with matrix ๏ฟฝ
0
not drawn on the diagram. Find
0
๏ฟฝ maps triangle ABC onto triangle A4 B4 C4
1
(i)
The scale factor of this transformation
(ii)
The coordinates of A4, B4 and C4 .
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
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Answers to All the Topic Questions
TOPIC1: ALGEBRA
๐‘๐‘โˆ’๐‘Ž๐‘Ž
1. (a)
๐‘Ž๐‘Ž 2 โˆ’๐‘๐‘ 2
=
=
2. (a)
โˆ’๐Ÿ๐Ÿ
7๐‘ ๐‘ ๐‘ก๐‘ก 3
15๐‘ข๐‘ข 3 ๐‘ฃ๐‘ฃ
=
=
4. (a)
=
=
=
7×๐‘ ๐‘ ×๐‘ก๐‘ก×๐‘ก๐‘ก×๐‘ก๐‘ก
×
15×๐‘ข๐‘ข×๐‘ข๐‘ข×๐‘ข๐‘ข
๐’•๐’•
Ans
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐’”๐’”๐Ÿ๐Ÿ
14๐‘ฅ๐‘ฅ 3
9๐‘ฆ๐‘ฆ 2
14๐‘ฅ๐‘ฅ 3
9๐‘ฆ๐‘ฆ 2
÷
×
3
28×๐‘ ๐‘ ×๐‘ ๐‘ ×๐‘ ๐‘ ×๐‘ก๐‘ก×๐‘ก๐‘ก
2๐‘ฅ๐‘ฅโˆ’5
7๐‘ฅ๐‘ฅ 4
×
=
7×๐‘ฅ๐‘ฅ×๐‘ฅ๐‘ฅ×๐‘ฅ๐‘ฅ×๐‘ฅ๐‘ฅ
๐‘š๐‘š 2 โˆ’1
๐‘š๐‘š 2 โˆ’12
๐‘š๐‘š (๐‘š๐‘š โˆ’1)
(๐‘š๐‘š +1)(๐‘š๐‘š โˆ’1)
๐‘š๐‘š (๐‘š๐‘š โˆ’1)
Ans
๐‘๐‘ 2 ๐‘ž๐‘ž 3
=
=
=
4
๐‘๐‘ 2 ๐‘ž๐‘ž 3
4
3๐‘ฅ๐‘ฅโˆ’9โˆ’8๐‘ฅ๐‘ฅ+20
(2๐‘ฅ๐‘ฅโˆ’5)(๐‘ฅ๐‘ฅโˆ’3)
3๐‘ฅ๐‘ฅโˆ’8๐‘ฅ๐‘ฅโˆ’9+20
(2๐‘ฅ๐‘ฅโˆ’5)(๐‘ฅ๐‘ฅโˆ’3)
๐’‘๐’‘
โˆ’๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“+๐Ÿ๐Ÿ๐Ÿ๐Ÿ
1
(c)
๐‘ฅ๐‘ฅโˆ’4
=
2(๐‘ฅ๐‘ฅ 2 โˆ’22 )
8
×
8
÷ 2๐‘๐‘2 ๐‘ž๐‘ž
4
Ans
×
8
๐‘๐‘×๐‘ž๐‘ž
5๐‘ฅ๐‘ฅโˆ’2๐‘ฅ๐‘ฅโˆ’1+8
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘+๐Ÿ•๐Ÿ•
Compiled and Solved by Kachama Dickson C & Chansa John
= (๐’™๐’™โˆ’๐Ÿ’๐Ÿ’)(๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“โˆ’๐Ÿ๐Ÿ) Ans
3
=
×
5๐‘ฅ๐‘ฅโˆ’1
1(5๐‘ฅ๐‘ฅโˆ’1)โˆ’2(๐‘ฅ๐‘ฅโˆ’4)
(๐‘ฅ๐‘ฅโˆ’4)(5๐‘ฅ๐‘ฅโˆ’1)
= (๐‘ฅ๐‘ฅโˆ’4)(5๐‘ฅ๐‘ฅโˆ’1)
(c)
1
2๐‘๐‘ 2 ๐‘ž๐‘ž
2
5๐‘ฅ๐‘ฅโˆ’1โˆ’2๐‘ฅ๐‘ฅ+8
๐‘ฅ๐‘ฅ+2
×
โˆ’
= (๐‘ฅ๐‘ฅโˆ’4)(5๐‘ฅ๐‘ฅโˆ’1)
๐‘ฅ๐‘ฅ+2
2(๐‘ฅ๐‘ฅ+2)(๐‘ฅ๐‘ฅโˆ’2)
๐‘๐‘๐‘๐‘
Ans
= (๐Ÿ๐Ÿ๐Ÿ๐Ÿโˆ’๐Ÿ“๐Ÿ“)(๐’™๐’™โˆ’๐Ÿ‘๐Ÿ‘) Ans
๐‘ฅ๐‘ฅ+2
๐‘๐‘×๐‘๐‘×๐‘ž๐‘ž×๐‘ž๐‘ž×๐‘ž๐‘ž
๐’’๐’’
4
๐‘ฅ๐‘ฅ+2
๐‘๐‘๐‘๐‘
โˆ’๐’™๐’™โˆ’๐Ÿ•๐Ÿ•
(๐’™๐’™+๐Ÿ๐Ÿ)(๐’™๐’™โˆ’๐Ÿ๐Ÿ)
๐‘ฅ๐‘ฅโˆ’3
= ๐Ÿ๐Ÿ(๐’™๐’™ โˆ’ ๐Ÿ๐Ÿ) Ans
×
3๐‘ฅ๐‘ฅโˆ’4๐‘ฅ๐‘ฅโˆ’3โˆ’4
(๐‘ฅ๐‘ฅ+1)(๐‘ฅ๐‘ฅโˆ’1)
=
2(๐‘ฅ๐‘ฅ 2 โˆ’4)
=
(b)
๐‘š๐‘š 2 โˆ’๐‘š๐‘š
โˆ’
2๐‘ฅ๐‘ฅ 2 โˆ’8
=
18×๐‘ฆ๐‘ฆ×๐‘ฆ๐‘ฆ×๐‘ฆ๐‘ฆ
3๐‘ฅ๐‘ฅโˆ’3โˆ’4๐‘ฅ๐‘ฅโˆ’4
(๐‘ฅ๐‘ฅ+1)(๐‘ฅ๐‘ฅโˆ’1)
=
9×๐‘๐‘×๐‘๐‘×๐‘๐‘×๐‘›๐‘›
4
๐‘ฅ๐‘ฅโˆ’1
3(๐‘ฅ๐‘ฅโˆ’1)โˆ’4(๐‘ฅ๐‘ฅ+1)
(๐‘ฅ๐‘ฅ+1)(๐‘ฅ๐‘ฅโˆ’1)
=
10×๐‘๐‘×๐‘๐‘×๐‘‘๐‘‘×๐‘‘๐‘‘
โˆ’
3(๐‘ฅ๐‘ฅโˆ’3)โˆ’4(2๐‘ฅ๐‘ฅโˆ’5)
(2๐‘ฅ๐‘ฅโˆ’5)(๐‘ฅ๐‘ฅโˆ’3)
=
(b)
18๐‘ฆ๐‘ฆ 3
×
๐‘ฅ๐‘ฅ+1
=
=
7๐‘ฅ๐‘ฅ 4
Ans
๐’Ž๐’Ž
Ans
๐Ÿ—๐Ÿ—๐’„๐’„๐Ÿ๐Ÿ
5×๐‘ข๐‘ข×๐‘ข๐‘ข×๐‘ข๐‘ข×๐‘ข๐‘ข×๐‘ฃ๐‘ฃ
18๐‘ฆ๐‘ฆ 3
9×๐‘ฆ๐‘ฆ×๐‘ฆ๐‘ฆ
๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’
๐’Ž๐’Ž+๐Ÿ๐Ÿ
๐Ÿ–๐Ÿ–๐’๐’๐Ÿ๐Ÿ
=
14×๐‘ฅ๐‘ฅ×๐‘ฅ๐‘ฅ×๐‘ฅ๐‘ฅ
๐’™๐’™
15×๐‘๐‘×๐‘‘๐‘‘×๐‘‘๐‘‘×๐‘‘๐‘‘
(b)
28๐‘ ๐‘  3 ๐‘ก๐‘ก 2
10๐‘๐‘ 2 ๐‘‘๐‘‘ 2
9๐‘๐‘ 3 ๐‘›๐‘›
3
(c)
10๐‘๐‘ 2 ๐‘‘๐‘‘ 2
12 ×๐‘‘๐‘‘×๐‘›๐‘›×๐‘›๐‘›×๐‘›๐‘›
=
5๐‘ข๐‘ข 3 ๐‘ฃ๐‘ฃ
9๐‘๐‘ 3 ๐‘›๐‘›
×
15๐‘๐‘๐‘๐‘ 3
=
2 ×
÷
12๐‘‘๐‘‘๐‘‘๐‘‘ 3
=
Ans
๐’‚๐’‚+๐’ƒ๐’ƒ
=
=
15๐‘๐‘๐‘๐‘ 3
โˆ’1(๐‘Ž๐‘Žโˆ’๐‘๐‘)
(๐‘Ž๐‘Ž+๐‘๐‘)(๐‘Ž๐‘Žโˆ’๐‘๐‘)
=
3. (a)
12๐‘‘๐‘‘๐‘‘๐‘‘ 3
(b)
1
2×๐‘๐‘×๐‘๐‘×๐‘ž๐‘ž
โˆ’
2
๐‘ฅ๐‘ฅ+3
3(๐‘ฅ๐‘ฅ+3)โˆ’2(5๐‘ฅ๐‘ฅโˆ’2)
=
=
=
5๐‘ฅ๐‘ฅโˆ’2
(5๐‘ฅ๐‘ฅโˆ’2)(๐‘ฅ๐‘ฅ+3)
3๐‘ฅ๐‘ฅ+9โˆ’10๐‘ฅ๐‘ฅ+4
(5๐‘ฅ๐‘ฅโˆ’2)(๐‘ฅ๐‘ฅ+3)
3๐‘ฅ๐‘ฅโˆ’10๐‘ฅ๐‘ฅ+9+4
(5๐‘ฅ๐‘ฅโˆ’2)(๐‘ฅ๐‘ฅ+3)
โˆ’๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ•+๐Ÿ๐Ÿ๐Ÿ๐Ÿ
(๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“โˆ’๐Ÿ๐Ÿ)(๐’™๐’™+๐Ÿ‘๐Ÿ‘)
Ans
/ Together we can do Mathematics /0966-295655
Page 39 of 84
5. (a)
=
๐‘ฅ๐‘ฅโˆ’1
(b)
๐‘ฅ๐‘ฅ 2 โˆ’1
๐‘ฅ๐‘ฅโˆ’1
๐‘ฅ๐‘ฅ 2 โˆ’12
๐‘ฅ๐‘ฅโˆ’1
= (๐‘ฅ๐‘ฅ+1)(๐‘ฅ๐‘ฅโˆ’1)
=
๐Ÿ๐Ÿ
๐’™๐’™+๐Ÿ๐Ÿ
Ans
17๐‘˜๐‘˜ 2
20๐‘Ž๐‘Ž 2
=
=
÷
17๐‘˜๐‘˜ 2
51๐‘˜๐‘˜ 2
5๐‘Ž๐‘Ž
5๐‘Ž๐‘Ž
= 20๐‘Ž๐‘Ž 2 × 51๐‘˜๐‘˜ 2
17×๐‘˜๐‘˜×๐‘˜๐‘˜
20×๐‘Ž๐‘Ž×๐‘Ž๐‘Ž
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
×
Ans
2
(c)
=
5×๐‘Ž๐‘Ž
=
51×๐‘˜๐‘˜×๐‘˜๐‘˜
โˆ’
1
2๐‘ฅ๐‘ฅโˆ’1
3๐‘ฅ๐‘ฅ+1
2(3๐‘ฅ๐‘ฅ+1)โˆ’1(2๐‘ฅ๐‘ฅโˆ’1)
(2๐‘ฅ๐‘ฅโˆ’1)(3๐‘ฅ๐‘ฅ+1)
=
=
6๐‘ฅ๐‘ฅ+2โˆ’2๐‘ฅ๐‘ฅ+1
(2๐‘ฅ๐‘ฅโˆ’1)(3๐‘ฅ๐‘ฅ+1)
6๐‘ฅ๐‘ฅโˆ’2๐‘ฅ๐‘ฅ+2+1
(2๐‘ฅ๐‘ฅโˆ’1)(3๐‘ฅ๐‘ฅ+1)
๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’+๐Ÿ‘๐Ÿ‘
(๐Ÿ๐Ÿ๐Ÿ๐Ÿโˆ’๐Ÿ๐Ÿ)(๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘+๐Ÿ๐Ÿ)
Ans
TOPIC 2: MATRICES
1. (a) deter A = (4 × 2) โˆ’ (1 × โˆ’5)
(b)
= 8 โˆ’ (โˆ’5)
=8+5
โˆด ๐‘ซ๐‘ซ๐‘ซ๐‘ซ๐‘ซ๐‘ซ ๐‘จ๐‘จ = 13 Ans
deter ๐ต๐ต = (8 × 5) โˆ’ (3 × ๐‘ฆ๐‘ฆ)
13 = 40 โˆ’ 3๐‘ฆ๐‘ฆ
13 โˆ’ 40 = โˆ’3๐‘ฆ๐‘ฆ
โˆ’27 = โˆ’3๐‘ฆ๐‘ฆ dividing both sides by 3 we have,
๐’š๐’š = ๐Ÿ—๐Ÿ— Ans
2. (a) Deter A = (2๐‘ฅ๐‘ฅ × ๐‘ฅ๐‘ฅ) โˆ’ (2 × 3)
2๐‘ฅ๐‘ฅ 2 โˆ’ 6 = 12
2๐‘ฅ๐‘ฅ 2 = 18
๐‘ฅ๐‘ฅ 2 = 9
๐‘ฅ๐‘ฅ = โˆš9
๐‘ฅ๐‘ฅ = ±3
โˆด ๐’™๐’™ = ๐Ÿ‘๐Ÿ‘ Ans
๐Ÿ‘๐Ÿ‘. (a) Deter M= (3× ๐‘ฅ๐‘ฅ) โˆ’ (5 × โˆ’2)
22 = 3๐‘ฅ๐‘ฅ โˆ’ (โˆ’10)
22 = 3๐‘ฅ๐‘ฅ + 10
22 โˆ’ 10 = 3๐‘ฅ๐‘ฅ
12 = 3๐‘ฅ๐‘ฅ
๐’™๐’™ = ๐Ÿ’๐Ÿ’ Ans
8 ๐‘ฆ๐‘ฆ
๏ฟฝ,
3 5
8 9
๏ฟฝ
B=๏ฟฝ
3 5
๐Ÿ๐Ÿ
๐Ÿ“๐Ÿ“ โˆ’๐Ÿ—๐Ÿ—
๏ฟฝ Ans
๐๐ โˆ’๐Ÿ๐Ÿ = ๏ฟฝ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ โˆ’๐Ÿ‘๐Ÿ‘
๐Ÿ–๐Ÿ–
๐ต๐ต = ๏ฟฝ
(b) A = ๏ฟฝ
2×3
3
๐‘จ๐‘จโˆ’๐Ÿ๐Ÿ
2
6 2
๏ฟฝ=๏ฟฝ
๏ฟฝ
3
3 3
๐Ÿ๐Ÿ
๐Ÿ‘๐Ÿ‘ โˆ’๐Ÿ๐Ÿ
๏ฟฝ Ans
= ๏ฟฝ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ โˆ’๐Ÿ‘๐Ÿ‘
๐Ÿ”๐Ÿ”
โˆ’2
๏ฟฝ
4
๐Ÿ๐Ÿ
๐Ÿ’๐Ÿ’ ๐Ÿ๐Ÿ
๏ฟฝ Ans
= ๏ฟฝ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ โˆ’๐Ÿ“๐Ÿ“ ๐Ÿ‘๐Ÿ‘
(b) ๐‘€๐‘€ = ๏ฟฝ
๐‘ด๐‘ดโˆ’๐Ÿ๐Ÿ
3
5
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 40 of 84
๐Ÿ๐Ÿ
4. (a) |K|= (10 × โˆ’2) โˆ’ (11 × โˆ’2)
= โˆ’20 โˆ’ (โˆ’22)
= โˆ’20 + 22
โˆด ๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘๐‘‘ ๐พ๐พ = ๐Ÿ๐Ÿ Ans
(b) ๐Š๐Š โˆ’๐Ÿ๐Ÿ = ๏ฟฝ
๐Ÿ๐Ÿ
5. (a) deter Q = (3 × 4) โˆ’ (๐‘ฅ๐‘ฅ × โˆ’2)
โˆ’๐Ÿ๐Ÿ
โˆ’๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐Ÿ๐Ÿ
๏ฟฝ Ans
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
3 โˆ’2
๏ฟฝ
โˆ’5 4
๐Ÿ๐Ÿ ๐Ÿ’๐Ÿ’ โˆ’๐Ÿ๐Ÿ
๏ฟฝ Ans
= ๏ฟฝ
๐Ÿ๐Ÿ ๐Ÿ“๐Ÿ“
๐Ÿ‘๐Ÿ‘
(b) Q = ๏ฟฝ
๐‘ธ๐‘ธโˆ’๐Ÿ๐Ÿ
2 = 12 โˆ’ (โˆ’2๐‘ฅ๐‘ฅ)
2 = 12 + 2๐‘ฅ๐‘ฅ
2 โˆ’ 12 = 2๐‘ฅ๐‘ฅ
โˆ’10 = 2๐‘ฅ๐‘ฅ
๐’™๐’™ = โˆ’๐Ÿ“๐Ÿ“ Ans
TOPIC 3: SETS
1. (i)
(ii) (a) Physics only = 6 students Ans
E Maths
(b) two subject only = 2 + 5
= 7 students Ans
(c) Maths and Physics not chem = 2 Ans
Chemistry
5
4
2
0
3
0
6
Physics
2. (i) 4 + ๐‘ฅ๐‘ฅ + 3 + 7 = 22
๐‘ฅ๐‘ฅ + 14 = 22
๐‘ฅ๐‘ฅ = 22 โˆ’ 14
๐’™๐’™ = ๐Ÿ–๐Ÿ– Ans
3. (i)
E
Ans
(ii) (a) only 1 mode of transport = 7 + 14 + 7 = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
(b) 2 different modes of transport
=4+2+3+8
= 17 Ans
Maize
(ii) (a) total = 6 +19 + 5+ 9+ 3+ 15+ 2+ 11
11
19
2
Cassava
(b) maize only = 19 farmers
9
5
6
15
3
4. (i) 2๐‘ฆ๐‘ฆ + 1 = 7
2๐‘ฆ๐‘ฆ = 7 โˆ’ 1
= 70 farmers
2๐‘ฆ๐‘ฆ = 6 โ†’ ๐’š๐’š = 3 Ans
(c) two different crops = 11+5+15+9
Sweet potatoes
= 40 farmers
(ii) (a) Victoria falls but not Gonya = 6+2 = 8 students
(b) two tourist attraction only = 4 +1 + 2 = 7 students
(c) one tourists attraction only = 6 + 8 + 7 = 21 students
Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
Page 41 of 84
5. (i) E
(ii) (a) Music only = 5 villagers
News
Sports
7
(b) one type only = 10 + 9 + 5
= 24 villagers
(c) two types of programs only = 8 + 7 + 5
= 20 Villagers
9
10
4
8
5
Music
5
TOPIC 4: PROBABILITY
1. Total 6 + 9 = 15 (hint: use the tree diagram for easy calculations of probabilities)
B
BB
R
BR
B
RB
5/14
B
6๏ฟฝ
15
9/14
6/14
9/15
R
8/14
R
(a) P( one is black) = P(BB) + P(BR) + P(RB)
=๏ฟฝ
=
=
=
6
15
30
210
138
210
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
×
+
5
14
54
๏ฟฝ+๏ฟฝ
210
Ans
+
6
15
54
210
×
9
14
RR
(b) P(different colour) = P(BR)+P(RB)
๏ฟฝ+๏ฟฝ
9
15
×
6
14
๏ฟฝ
= ๏ฟฝ
=
=
=
6
15
54
210
108
210
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
×
+
9
๏ฟฝ+๏ฟฝ
14
54
210
9
15
×
6
14
Ans
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๏ฟฝ
Page 42 of 84
2. (a) Total ๐Ÿ๐Ÿ๐Ÿ๐Ÿ + ๐Ÿ๐Ÿ๐Ÿ๐Ÿ + ๐Ÿ’๐Ÿ’ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
B
BB
R
BR
W
BW
RB
19/35
12/35
B
20/36
4/35
20/35 B
12/36
11/35
R
R
W
B
4/35
4/36
RR
RW
WB
W 20/35
12/35
R
WB
W
WW
3/35
(b) P( both white) = P(WW)
=
=
=
4
36
12
×
1260
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
3
35
Ans
3. (a) Total ๐Ÿ“๐Ÿ“ + ๐Ÿ’๐Ÿ’ + ๐Ÿ‘๐Ÿ‘ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
(b) P(Same colour) = P(WW) + P( BB) + (YY)
= ๏ฟฝ
=
5
12
20
132
×
+
4
11
12
๏ฟฝ+๏ฟฝ
132
+
4
12
6
132
×
=
3
11
38
๏ฟฝ+๏ฟฝ
132
=
3
12
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”
×
Ans
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2
11
๏ฟฝ
/ Together we can do Mathematics /0966-295655
Page 43 of 84
4. Faulty = 3 and good = 10 โ€“ 3 = 7
Use a tree diagram below for easy calculations
(a) P( both good) = P(G, G)
=๏ฟฝ
=
=
7
10
56
90
๐Ÿ•๐Ÿ•
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
6
× ๏ฟฝ
9
Ans
(b) P (one is faulty and the one is good) = ๐‘ƒ๐‘ƒ(๐บ๐บ, ๐น๐น) + ๐‘ƒ๐‘ƒ(๐น๐น, ๐บ๐บ)
=
=๏ฟฝ
7
3
× ๏ฟฝ+๏ฟฝ
10
21 21
+
90 90
=
=
9
3
10
7
× ๏ฟฝ
9
42
90
๐Ÿ•๐Ÿ•
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
Ans
5. P( negative) = 1- P(positive)
= 1-0.6
= 0.4
(a) P (1 negative & other positive) = (0.6× 0.4) + (0.4 × 0.6)
(b) P(both positive) = (0.4× 0.4)
= ๐ŸŽ๐ŸŽ. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
= 0.24 + 0.24
= 0.48
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Page 44 of 84
TOPIC 5: SEQUENCES AND SERIES
1. (a) ๐‘˜๐‘˜ + 4, ๐‘˜๐‘˜, 2๐‘˜๐‘˜ โˆ’ 15
๐‘˜๐‘˜
๐‘˜๐‘˜+4
(b) ๐‘˜๐‘˜ + 4, ๐‘˜๐‘˜, 2๐‘˜๐‘˜ โˆ’ 15
2๐‘˜๐‘˜โˆ’15
=
๐‘˜๐‘˜
๐‘˜๐‘˜ 2 = (๐‘˜๐‘˜ + 4)(2๐‘˜๐‘˜ โˆ’ 15)
(c) ๐‘†๐‘†โˆž =
12 + 4, 12, 2(12) โˆ’ 15
๐‘†๐‘†โˆž =
16, 12, 9,โ€ฆ Ans
๐‘†๐‘†โˆž =
๐‘˜๐‘˜ 2 = 2๐‘˜๐‘˜ 2 โˆ’ 15๐‘˜๐‘˜ + 8๐‘˜๐‘˜ โˆ’ 60
๐‘†๐‘†โˆž =
2๐‘˜๐‘˜ 2 โˆ’ ๐‘˜๐‘˜ 2 โˆ’ 7๐‘˜๐‘˜ โˆ’ 60 = 0
๐‘˜๐‘˜ 2 โˆ’ 7๐‘˜๐‘˜ โˆ’ 60 = 0
๐‘˜๐‘˜ 2 + 5๐‘˜๐‘˜ โˆ’ 12๐‘˜๐‘˜ โˆ’ 60 = 0
๐‘˜๐‘˜(๐‘˜๐‘˜ + 5) โˆ’ 12(๐‘˜๐‘˜ + 5) = 0
(๐‘˜๐‘˜ + 5)(๐‘˜๐‘˜ โˆ’ 12) = 0
๐‘˜๐‘˜ = โˆ’5 ๐‘œ๐‘œ๐‘œ๐‘œ ๐‘˜๐‘˜ = 12
โˆด ๐’Œ๐’Œ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
2
๐‘Ž๐‘Ž =
2
3โˆ’1
๐‘†๐‘†5 =
27
3
2
27
2
27
=
=
2
9๐‘Ÿ๐‘Ÿ 2
2๐‘Ÿ๐‘Ÿ
9
× ๐‘Ÿ๐‘Ÿ 3
2
9๐‘Ÿ๐‘Ÿ 2
๐‘†๐‘†5 = 3(
in (ii) we have
๐‘†๐‘†5 =
๐‘Ÿ๐‘Ÿ =
๐‘Ž๐‘Ž =
3
2
1 2
9×๏ฟฝ ๏ฟฝ
3
๐‘Ž๐‘Ž = 2
243
242
81
)
1
4
4โˆ’3
4
16
1
4
)÷
)×
=2
๐‘Ž๐‘Ž
1โˆ’๐‘Ÿ๐‘Ÿ
2
๐‘†๐‘†โˆž =
2
80
81
2
= ๐Ÿ๐Ÿ. ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ— Ans
1
3
1โˆ’
๐‘†๐‘†โˆž =
18
54
1
3
243
3
243โˆ’1
3
243
242
(c) ๐‘†๐‘†โˆž =
54๐‘Ÿ๐‘Ÿ = 18
๐‘Ÿ๐‘Ÿ =
1 5
3
1
1โˆ’
3
1
2(1โˆ’ 5 )
3
2
3
๐‘†๐‘†5 = 2(
= ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ. Equation (ii)
Replacing ๐‘Ž๐‘Ž by
=
1โˆ’๐‘Ÿ๐‘Ÿ
๐‘†๐‘†5 = 2(1 โˆ’
๐‘‡๐‘‡4 = ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ 4โˆ’1
2
16
16
2(1โˆ’๏ฟฝ ๏ฟฝ )
๐‘†๐‘†5 =
โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.. equation (i)
9๐‘Ÿ๐‘Ÿ 2
3
4
1โˆ’
12
๐‘Ž๐‘Ž(1โˆ’๐‘Ÿ๐‘Ÿ ๐‘›๐‘› )
(b) ๐‘†๐‘†๐‘›๐‘› =
= ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ 2
9
,r=
๐‘บ๐‘บโˆž = ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” Ans
2. (a) ๐‘‡๐‘‡๐‘›๐‘› = ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ ๐‘›๐‘›โˆ’1
๐‘‡๐‘‡3 = ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ
๐‘Ž๐‘Ž
1โˆ’๐‘Ÿ๐‘Ÿ
16
2
2
3
๐‘บ๐‘บโˆž = ๐Ÿ‘๐Ÿ‘ Ans
=
2
9×
1
9
๐Ÿ๐Ÿ
โˆด the first term is 2 and common ratio is Ans
๐Ÿ‘๐Ÿ‘
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Page 45 of 84
3. (a) ๐‘Ÿ๐‘Ÿ =
5
1
(c) ๐‘†๐‘†๐‘›๐‘› =
= = 0.25
20
4
(b) ๐‘‡๐‘‡๐‘›๐‘› = ๐‘Ž๐‘Ž๐‘Ÿ๐‘Ÿ ๐‘›๐‘›โˆ’1
๐‘†๐‘†8 =
1 ๐‘›๐‘›โˆ’1
๐‘‡๐‘‡๐‘›๐‘› = 20 ๏ฟฝ ๏ฟฝ
๐‘‡๐‘‡๐‘›๐‘› = 20
โˆด ๐‘ป๐‘ป๐’๐’
=
๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐Ÿ’๐Ÿ’๐’๐’โˆ’๐Ÿ๐Ÿ
๐‘†๐‘†8 =
Ans
10+๐‘›๐‘›
6+๐‘›๐‘›
=
๐‘‡๐‘‡2
=
๐‘‡๐‘‡1
15+๐‘›๐‘›
10+๐‘›๐‘›
๐‘‡๐‘‡3
๐‘‡๐‘‡2
=โ‹ฏ
(c)
๐‘‡๐‘‡๐‘›๐‘›
100 + 20๐‘›๐‘› + ๐‘›๐‘› = 90 + 21n + ๐‘›๐‘›
100 โˆ’ 90 = 21๐‘›๐‘› โ€“ 20๐‘›๐‘›
10 = ๐‘›๐‘›
โˆด ๐’๐’ =10
The GP is: 16, 20, 25 . . .
16
=
5
4
๐‘œ๐‘œ๐‘œ๐‘œ ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
๐‘‡๐‘‡
๐‘ฅ๐‘ฅโˆ’3
=
1
๐‘ฅ๐‘ฅโˆ’1
๐‘ฅ๐‘ฅโˆ’3
S6
๐‘‡๐‘‡
๐‘‡๐‘‡
2
๐‘›๐‘› โˆ’1
๐‘ฅ๐‘ฅ 2 โˆ’ 6๐‘ฅ๐‘ฅ + 9 = ๐‘ฅ๐‘ฅ 2 โˆ’ 1
โˆ’6๐‘ฅ๐‘ฅ = โˆ’1 โˆ’ 9
๐‘ฅ๐‘ฅ =
3
๐Ÿ๐Ÿ
๐Ÿ‘๐Ÿ‘
Hence the GP is;
8 โˆ’4 2
3
,
3
, ,โ€ฆ
3
1.25โˆ’1
16(3.814697266 โˆ’1)
=
(C)
๐‘†๐‘†โˆž =
๐‘†๐‘†โˆž =
๐‘†๐‘†โˆž =
0.25
16(2.814697266 )
0.25
45.03515625
0.25
๐‘Ž๐‘Ž
1โˆ’๐‘Ÿ๐‘Ÿ
8/3
8/3
16
9
3
2
8
3
8
3
โˆด ๐‘บ๐‘บโˆž = ๐Ÿ๐Ÿ
5
3
5
1
2
1โˆ’(โˆ’ )
๐‘†๐‘†โˆž =
๐‘†๐‘†โˆž =
10
๐’™๐’™ = ๐Ÿ๐Ÿ Ans
for r > 1
๐‘Ÿ๐‘Ÿโˆ’1
16((1.25)6 โˆ’1)
÷
๐‘†๐‘†โˆž = ×
โˆ’6๐‘ฅ๐‘ฅ = โˆ’10
6
5
S6
=
๐‘Ž๐‘Ž(๐‘Ÿ๐‘Ÿ ๐‘›๐‘› โˆ’1)
S 6 = 180.140625
โˆด S6 = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐ŸAns
(๐‘ฅ๐‘ฅ โˆ’ 3) (๐‘ฅ๐‘ฅ โˆ’ 3) = ( ๐‘ฅ๐‘ฅ โˆ’ 1) ( ๐‘ฅ๐‘ฅ + 1)
๐‘ฅ๐‘ฅ =
๐‘†๐‘†๐‘›๐‘› =
S6 =
2
5. (a) We know that r = ๐‘‡๐‘‡2 = ๐‘‡๐‘‡3 = โ‹ฏ ๐‘‡๐‘‡ ๐‘›๐‘›
๐‘ฅ๐‘ฅ+1
0.75
๐‘†๐‘†6 =
๐‘‡๐‘‡๐‘›๐‘› โˆ’1
2
20
0.75
20(0.9999847412 )
๐‘บ๐‘บ๐Ÿ–๐Ÿ– = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ•๐Ÿ• Ans
(10 + ๐‘›๐‘›) (10 + ๐‘›๐‘›) = (6 + ๐‘›๐‘›) (15 + ๐‘›๐‘›)
(b) ๐’“๐’“ =
1โˆ’0.25
20 (1โˆ’0.00001558906 )
๐‘†๐‘†8 = 26.66625977
4. (a) to find n, we use the common ratio formula
That is r=
for r < 1
1โˆ’๐‘Ÿ๐‘Ÿ
20(1โˆ’(0.25)8 )
๐‘†๐‘†8 =
4
1๐‘›๐‘› โˆ’1
4 ๐‘›๐‘› โˆ’1
๐‘Ž๐‘Ž(1โˆ’๐‘Ÿ๐‘Ÿ ๐‘›๐‘› )
๐Ÿ•๐Ÿ•
๐Ÿ—๐Ÿ—
3
2
2
3
Ans
5
+ 1. โˆ’ 3, โˆ’ 1 โ€ฆ โ€ฆ ..
3
3
๐Ÿ–๐Ÿ–
(b) the first term โ€œ๐‘Ž๐‘Žโ€ = Ans
๐Ÿ‘๐Ÿ‘
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Page 46 of 84
TOPIC 6: PSEUDO CODE AND FLOWCHART
1.
Start
Enter ๐’™๐’™, ๐’š๐’š
M = (๐’™๐’™^2+๐’š๐’š^2)
Is
M< 0
yes
Error โ€˜Mโ€™ must be
positive
No
Display
M
Stop
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Page 47 of 84
2.
Start
Enter ๐’‚๐’‚, ๐’“๐’“, ๐’๐’
๐‘น๐‘น = ๐Ÿ๐Ÿ โˆ’ ๐’“๐’“
Is
๐‘น๐‘น = ๐ŸŽ๐ŸŽ?
๐๐๐๐
๐‘บ๐‘บ๐’๐’ =
yes
value of r
is not valid
๐’‚๐’‚(๐Ÿ๐Ÿโˆ’๐’“๐’“๐’๐’ )
๐Ÿ๐Ÿโˆ’๐’“๐’“
Display ๐‘บ๐‘บ๐’๐’
Stop
3. Start
Enter radius
If radius < 0
Then display โ€œerror message โ€œ
1
Else Area = โˆ— ๐‘Ÿ๐‘Ÿ โˆ— ๐‘Ÿ๐‘Ÿ โˆ— ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
2
End if
Display Area
Stop
4. Start
Enter a and r
If |๐‘Ÿ๐‘Ÿ| < 1
Then sum to infinity =
Else
End if
Display sum to infinity
Stop
๐‘Ž๐‘Ž
1โˆ’๐‘Ÿ๐‘Ÿ
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Page 48 of 84
5.
Start
Enter, r
Is
yes
๐’“๐’“ < 0?
Error
โ€œr must be positiveโ€
No
Enter h
Is h < 0?
Yes
Error
โ€œh must be Positiveโ€
No
๐Ÿ๐Ÿ
๐’—๐’— = ๐…๐… โˆ— ๐’“๐’“ โˆ— ๐’“๐’“ โˆ— ๐’‰๐’‰
๐Ÿ‘๐Ÿ‘
Display Volume
Stop
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/ Together we can do Mathematics /0966-295655
Page 49 of 84
TOPIC 7: LOCI AND CONSTRUCTION
1.
b (i)
Z
b (ii)
5.5 cm
T
7 cm
P (c)
38 °
9 cm
X
Y
R
2.
b(ii
(i)
7.8 cm
8 cm
b(iii
X
T
P
๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“°
10 cm
b(i)
Q
D
3.
8.9cm
C
11cm
P
QP
PPPPP
7 cm
AA
60°
10cm
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ°
B
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Page 50 of 84
R
4.
b(iii)
b(ii)
10 cm
9.5cm
b(i)
P
60°
9 cm
Q
5.
C
b (i)
b (ii)
7 cm
b (iii)
7 cm
R
A
7 cm
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TOPIC 8: CALCULUS
2
1. (a) โˆซโˆ’1(2 + ๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ 2 )๐‘‘๐‘‘๐‘‘๐‘‘
= ๏ฟฝ2๐‘ฅ๐‘ฅ +
=๏ฟฝ
2(2)
+
1
๐‘ฅ๐‘ฅ 2
2
22
2
โˆ’
โˆ’
8
๐‘ฅ๐‘ฅ 3
3
23
3
2
๏ฟฝ โˆ’1
m1 =
๏ฟฝ โˆ’ ๏ฟฝ2(โˆ’1) โˆ’
1
1
= ๏ฟฝ4 + 2 โˆ’ ๏ฟฝ โ€” ๏ฟฝ2 โˆ’ + ๏ฟฝ
10
3
โˆ’7
2
=๏ฟฝ ๏ฟฝโˆ’๏ฟฝ ๏ฟฝ
=
=
10
3
+
3
27
6
6
7
6
4
(b) ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ + โ†’
(โˆ’1)2
3
2
โˆ’
(โˆ’1)3
3
๐‘ฅ๐‘ฅ
๐‘‘๐‘‘๐‘‘๐‘‘
=1โˆ’
๐‘‘๐‘‘๐‘‘๐‘‘
R
3
=๏ฟฝ โˆ’
3
1
2(1)2
2
3
0
3
0
2
1
= โˆ’4โˆ’0
=
16
3
1โˆ’12
=
3
4
4
3
๐Ÿ‘๐Ÿ‘
๐‘‘๐‘‘๐‘‘๐‘‘
๐‘‘๐‘‘๐‘‘๐‘‘
P
Multiplying through by
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘ = โˆ’๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’ + ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘ Ans
= 4๐‘ฅ๐‘ฅ โˆ’ 3 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž (3,7) โ†’
1
๐‘š๐‘š2 = โˆ’1 × = โˆ’
9
1
9
๐‘‘๐‘‘๐‘‘๐‘‘
๐‘‘๐‘‘๐‘‘๐‘‘
= ๐‘š๐‘š1 = 4(3) โˆ’ 3 = 9
โˆด equation of the normal to the curve is given by
๐’š๐’š โˆ’ ๐’š๐’š๐Ÿ๐Ÿ = ๐‘š๐‘š2 (๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ1 )
1
๐‘ฆ๐‘ฆ โˆ’ 7 = โˆ’ (๐‘ฅ๐‘ฅ โˆ’ 3)
3
11
=โˆ’
16
+5
3
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
(b) ๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 2 โˆ’ 3๐‘ฅ๐‘ฅ โˆ’ 2
โˆ’ (3)๏ฟฝ โˆ’ ๏ฟฝ โˆ’ + 3(0)๏ฟฝ
= ๏ฟฝ โˆ’ 1 โˆ’ 3๏ฟฝ โˆ’ 0
12
๐’š๐’š โˆ’ ๐’š๐’š๐Ÿ๐Ÿ = ๐‘š๐‘š2 (๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ1 )
4
๐‘ฆ๐‘ฆ โˆ’ 5 = โˆ’ (๐‘ฅ๐‘ฅ โˆ’ 4)
3
2. (a) โˆซ0 (๐‘ฅ๐‘ฅ 2 โˆ’ 2๐‘ฅ๐‘ฅ โˆ’ 3)๐‘‘๐‘‘๐‘‘๐‘‘
13
=
4
P
โˆ’ 3๐‘ฅ๐‘ฅ๏ฟฝ 10
16
โˆด Equation of the normal is given by
3
2
16โˆ’4
4
๐Ÿ‘๐Ÿ‘
2๐‘ฅ๐‘ฅ 2
=
at ๐‘ฅ๐‘ฅ = 4
๐‘ฆ๐‘ฆ = 4 + = 4 + 1 = 5
๐’š๐’š = โˆ’ ๐’™๐’™ +
โˆ’
4
16
4
๐‘ฅ๐‘ฅ 2
To find y replace ๐‘ฅ๐‘ฅ = 4 in the original equat
3
๐Ÿ’๐Ÿ’
3
=1โˆ’
4
4
๐‘ฅ๐‘ฅ 3
4
42
=1โˆ’
m 2 = โˆ’1 × = โˆ’
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ +
=๏ฟฝ
๐‘‘๐‘‘๐‘‘๐‘‘
m 1 × m 2 = โˆ’1 (tangent โŠฅ to normal)
๏ฟฝ
= 4. ๐Ÿ“๐Ÿ“ Ans
1
๐‘‘๐‘‘๐‘‘๐‘‘
9
1
๐‘ฆ๐‘ฆ โˆ’ 7 = โˆ’ ๐‘ฅ๐‘ฅ +
3
๐Ÿ๐Ÿ
= โˆ’๐Ÿ‘๐Ÿ‘ Ans
1
9
3
9
1
9
66
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ + +
๐Ÿ‘๐Ÿ‘
9
1
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ +
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ +
9
7
9
1
3+63
3
9
9
๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ— = โˆ’๐’™๐’™ + ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” Ans
A journey of thousand miles
starts with one step
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 52 of 84
3. (a) ๐‘ฆ๐‘ฆ = 2๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 โˆ’ 36๐‘ฅ๐‘ฅ โˆ’ 3
๐‘‘๐‘‘๐‘‘๐‘‘
= 6๐‘ฅ๐‘ฅ 2 โˆ’ 6๐‘ฅ๐‘ฅ โˆ’ 36
๐‘‘๐‘‘๐‘‘๐‘‘
3
(b) โˆซโˆ’1(3๐‘ฅ๐‘ฅ 2 โˆ’ 2๐‘ฅ๐‘ฅ)๐‘‘๐‘‘๐‘‘๐‘‘
=๏ฟฝ
0 = 6๐‘ฅ๐‘ฅ 2 โˆ’ 6๐‘ฅ๐‘ฅ โˆ’ 36 dividing through by 6
3๐‘ฅ๐‘ฅ 3
3
3
2๐‘ฅ๐‘ฅ 2
๏ฟฝ 3
2 โˆ’1
3
โˆ’ ๐‘ฅ๐‘ฅ 2 ] โˆ’1
3
2]
โˆ’
= [๐‘ฅ๐‘ฅ
= [(3) โˆ’ (3)
๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 6 = 0
โˆ’ [(โˆ’1)3 โˆ’ (โˆ’1)2 ]
= (27 โˆ’ 9) โ€“ (โˆ’1 โˆ’ 1)
=18 โ€“ (โˆ’2)
= 18+2 = 20
(๐‘ฅ๐‘ฅ + 2) (๐‘ฅ๐‘ฅ โˆ’ 3) = 0
๐‘ฅ๐‘ฅ = โˆ’2 ๐‘œ๐‘œ๐‘œ๐‘œ ๐‘ฅ๐‘ฅ = 3
When ๐‘ฅ๐‘ฅ = โˆ’2
3
โˆด โˆซโˆ’1(3๐‘ฅ๐‘ฅ 2 โˆ’ 2๐‘ฅ๐‘ฅ)๐‘‘๐‘‘๐‘‘๐‘‘ = 20 Ans
๐‘ฆ๐‘ฆ = โˆ’16 โˆ’ 12 + 72 โˆ’ 3
๐‘ฆ๐‘ฆ = 41
When ๐‘ฅ๐‘ฅ = 3
๐‘ฆ๐‘ฆ = 2(โˆ’2)3 โˆ’ 3(โˆ’2)2 โˆ’ 36(โˆ’2) โˆ’ 3
๐‘ฆ๐‘ฆ = 2(3)3 โ€“ 3(3)2 โˆ’ 36 (3) โˆ’ 3
๐‘ฆ๐‘ฆ = โˆ’84
โˆด the coordinates on curve are (โˆ’ ๐Ÿ๐Ÿ, ๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’) and (๐Ÿ‘๐Ÿ‘, โˆ’๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–) Ans
5
4. (a) โˆซ2 (3๐‘ฅ๐‘ฅ 2 + 2)๐‘‘๐‘‘๐‘‘๐‘‘
3๐‘ฅ๐‘ฅ 2+1
=๏ฟฝ
2+1
3๐‘ฅ๐‘ฅ 3
+
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ2 โˆ’ 3๐‘ฅ๐‘ฅ โˆ’ 4
(b)
2๐‘ฅ๐‘ฅ 0+1 5
๏ฟฝ
0+1 2
2๐‘ฅ๐‘ฅ
= ๏ฟฝ + ๏ฟฝ 52
3
1
= [๐‘ฅ๐‘ฅ 3 + 2๐‘ฅ๐‘ฅ] 52
3
๐‘š๐‘š =
๐‘ฆ๐‘ฆ โˆ’ ๐‘ฆ๐‘ฆ1 = ๐‘š๐‘š(๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ1 )
๐‘ฆ๐‘ฆ โˆ’ (โˆ’6) = 1(๐‘ฅ๐‘ฅ โˆ’ 2)
= (135) โ€“ (12)
= ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐ŸAns
๐‘‘๐‘‘๐‘‘๐‘‘
๐‘‘๐‘‘๐‘‘๐‘‘
๐‘ฆ๐‘ฆ + 6 = ๐‘ฅ๐‘ฅ โˆ’ 2
๐’š๐’š = ๐’™๐’™ + ๐Ÿ–๐Ÿ– Ans
(b) At the stationary points,
2
2
= 3๐‘ฅ๐‘ฅ 2 โˆ’ 3๐‘ฅ๐‘ฅ = 3(2) โ€“ 3(2)= 6
1
6
Which is the gradient of the normal?
๐‘ฆ๐‘ฆ โˆ’ ๐‘ฆ๐‘ฆ1 = ๐‘š๐‘š2 (๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ1 ) (2,2)
1
1
6
1
6
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ +
6
1
๐‘ฆ๐‘ฆ = โˆ’ ๐‘ฅ๐‘ฅ +
6
2+12
6
14
6
Compiled and Solved by Kachama Dickson C & Chansa John
2
For ๐‘ฅ๐‘ฅ = 1
2
=0
1
2
โˆด the stationary points are;
โ†’ Multiplying throughout by 6, we get
๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” = โˆ’๐’™๐’™ + ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
๐‘‘๐‘‘๐‘‘๐‘‘
3
๐‘ฆ๐‘ฆ = (0) 3 - (0)2 = 0
3
๐‘ฆ๐‘ฆ =โˆ’ ๐‘ฅ๐‘ฅ + + 2
๐‘‘๐‘‘๐‘‘๐‘‘
3๐‘ฅ๐‘ฅ(๐‘ฅ๐‘ฅ โˆ’ 1) = 0
๐‘ฅ๐‘ฅ = 0 ๐‘œ๐‘œ๐‘œ๐‘œ ๐‘ฅ๐‘ฅ = 1
For ๐‘ฅ๐‘ฅ = 0.
๐‘ฆ๐‘ฆ = (1)3 โˆ’ (1)2 = โˆ’
๐‘ฆ๐‘ฆ โˆ’ 2 = โˆ’ (๐‘ฅ๐‘ฅ โˆ’ 2)
6
2
3๐‘ฅ๐‘ฅ โˆ’3๐‘ฅ๐‘ฅ = 0
P
โˆด ๐‘š๐‘š1 = 6 which is the gradient of tangent to the curve
๐‘ค๐‘ค๐‘ค๐‘ค know that tangent is perpendicular to the normal,
๐‘š๐‘š1 ๐‘š๐‘š2 = โˆ’1, at ๐‘ฅ๐‘ฅ = 2, ๐‘ฆ๐‘ฆ = 2
๐‘š๐‘š2 = โˆ’
= 2๐‘ฅ๐‘ฅ โˆ’ 3 at ๐‘ฅ๐‘ฅ = 2, ๐‘š๐‘š = 2(2) โˆ’ 3 = 1
โˆด equation of the tangent is given by
3
= (125 + 10) โ€“ (8 + 4)
3
๐‘‘๐‘‘๐‘‘๐‘‘
๐‘ฆ๐‘ฆ = (2)2 โˆ’ 3(2) โˆ’ 4 = โˆ’6
= ( (5) + 2(5) ) โˆ’ (2 + 2(2))
5. (a) ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 โˆ’ 2 ๐‘ฅ๐‘ฅ 2
๐‘‘๐‘‘๐‘‘๐‘‘
๐Ÿ๐Ÿ
(0, 0) and (1, โˆ’ ) Ans
๐Ÿ๐Ÿ
/ Together we can do Mathematics /0966-295655
Page 53 of 84
TOPIC 9: VECTORS
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 1 ๐ด๐ด๐ด๐ด
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
1. (i) (a) ๐ด๐ด๐ด๐ด
๐Ÿ๐Ÿ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ( ๐’ƒ๐’ƒ โˆ’ ๐’‚๐’‚ )
(ii) Since ๐ต๐ต๐ต๐ต
3
๐Ÿ‘๐Ÿ‘
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐ด๐ด๐ด๐ด
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + ๐ต๐ต๐ต๐ต
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด = ๐‘Ž๐‘Ž + 2๐‘๐‘
U
U
๐Ÿ‘๐Ÿ‘
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ( ๐’‚๐’‚ + 2๐’ƒ๐’ƒ ) Ans
โˆด ๐‘จ๐‘จ๐‘จ๐‘จ
๐Ÿ‘๐Ÿ‘
U
U
(b) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด
1
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = -๐‘Ž๐‘Ž + ( ๐‘Ž๐‘Ž + 2๐‘๐‘ )
๐ต๐ต๐ต๐ต
U
3
U
1
U
2
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โˆ’๐‘Ž๐‘Ž + ๐‘Ž๐‘Ž + ๐‘๐‘
3
โˆ’3๐‘Ž๐‘Ž+๐‘Ž๐‘Ž
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต =
3
2
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐’ƒ๐’ƒ โˆ’ ๐’‚๐’‚
and ๐ต๐ต๐ต๐ต
๐Ÿ๐Ÿ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ
U
๐Ÿ๐Ÿ
U
3
U
2
U
+ ๐‘๐‘
3
2
โˆด the points ๐‘ฉ๐‘ฉ, ๐‘ฌ๐‘ฌ and ๐‘ซ๐‘ซ are collinear
(c)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โˆ’๐ด๐ด๐ด๐ด
๐ด๐ด๐ด๐ด
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โˆ’ ๐‘Ž๐‘Ž + ๐‘๐‘
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ = ๐’ƒ๐’ƒ โ€“ ๐’‚๐’‚ Ans
U
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โˆ’ ๐‘Ž๐‘Ž + ๐‘๐‘
๐ต๐ต๐ต๐ต
2
3
2
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = b โ€“ ๐‘Ž๐‘Ž
โˆด ๐ต๐ต๐ต๐ต
3
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐Ÿ๐Ÿ ( ๐’ƒ๐’ƒ โˆ’ ๐’‚๐’‚ ) Ans
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ
๐Ÿ‘๐Ÿ‘
U
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = PO
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + OQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
2. (i) (a) PQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โˆ’ 2p + 4q
PQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 4q โ€“ 2p
PQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 2 (2q โ€“p) Ans
โˆด ๐๐๐๐
(b)
(c)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 1 PQ
PX
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐Ÿ๐Ÿ (4q โ€“2p) Ans
PX
๐Ÿ‘๐Ÿ‘
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OX = OP
PX
4
2
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OX = 2p + q โˆ’ p
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OC = hOX
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = h (4p + 4q )
OC
โˆด
3
3
4h
4h
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OC = p + q
3
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = CO
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + OQ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
CQ
4h
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โˆ’ ( P + 4h q) + 4q
CQ
3
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 4q โˆ’ 4h q โ€“ 4h p
CQ
3
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐Ÿ’๐Ÿ’(1โˆ’ ๐ก๐ก) q โ€“ ๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’ p hence shown
๐‚๐‚๐‚๐‚
๐Ÿ‘๐Ÿ‘
๐Ÿ‘๐Ÿ‘
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 2p โˆ’ 2 pโˆ’ 4 q
OX
3
6pโˆ’2p
4q
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OX =
โˆ’
3
3
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐Ÿ’๐Ÿ’ p โˆ’ ๐Ÿ’๐Ÿ’ q
๐Ž๐Ž๐Ž๐Ž
๐Ÿ‘๐Ÿ‘
๐Ÿ‘๐Ÿ‘
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3.
(a)
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 1 CA
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
(b) CD
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = OA
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + AB
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OB
2
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 1 (โˆ’2๐‘๐‘ +๐‘Ž๐‘Ž )
CD
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐’‚๐’‚ + 2๐’ƒ๐’ƒ
๐Ž๐Ž๐Ž๐Ž
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— and
(b) To find ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OE, first find CA
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ถ๐ถ๐ถ๐ถ
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = CO
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + OA
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
CA
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โˆ’2b + a
๐ถ๐ถ๐ถ๐ถ
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด = โˆ’ ๐ถ๐ถ๐ถ๐ถ
4
2
U
U
1
= โˆ’ ๐‘๐‘ + ๐‘Ž๐‘Ž
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— =
๐‘ช๐‘ช๐‘ช๐‘ช
๐Ÿ๐Ÿ
๐Ÿ๐Ÿ
U
2
U
๐’‚๐’‚ โ€“ ๐’ƒ๐’ƒ Ans
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โˆ’ 3 ( ๐‘Ž๐‘Ž โˆ’2๐‘๐‘ )
๐ด๐ด๐ด๐ด
4
U
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— + AE
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
โˆด ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
OE = OA
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = ๐‘Ž๐‘Ž โˆ’ ( ๐‘Ž๐‘Ž โˆ’2๐‘๐‘ )
4
U
U
3
U
6
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = ๐‘Ž๐‘Ž โˆ’ ๐‘Ž๐‘Ž + ๐‘๐‘
4
4๐‘Ž๐‘Žโˆ’3๐‘Ž๐‘Ž
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ =
1
4
3
+ ๐‘๐‘
4
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = ๐‘Ž๐‘Ž + ๐‘๐‘
4
๐Ÿ๐Ÿ
U
4
U
4
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = (๐’‚๐’‚ + 3๐’ƒ๐’ƒ ) Ans
๐‘ถ๐‘ถ๐‘ถ๐‘ถ
๐Ÿ’๐Ÿ’
U
U
4. (i) (a) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด = โˆ’3a + 6b
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด = 6b โˆ’3a
โˆด ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘จ๐‘จ๐‘จ๐‘จ = 3 (2b โˆ’a) Ans
(b) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด
1
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ด๐ด๐ด๐ด
3
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = 3a + 2b โ€“ a
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 2a + 2b
๐‘‚๐‘‚๐‘‚๐‘‚
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = 2(๐’‚๐’‚ + ๐’ƒ๐’ƒ ) Ans
๐‘ถ๐‘ถ๐‘ถ๐‘ถ
U
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
(ii) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โ„Ž๐ต๐ต๐ต๐ต
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = โ„Ž(6 a โ€“ 6b)
๐ต๐ต๐ต๐ต
3
1
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚ = 3a + (6b โˆ’3a)
U
(c) ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚
2
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
โˆ’๐‘‚๐‘‚๐‘‚๐‘‚ + ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐‘‚๐‘‚๐‘‚๐‘‚
5
2
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โˆ’ 6b + (3a)
5
6๐‘Ž๐‘Ž
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ—
๐ต๐ต๐ต๐ต = โˆ’ 6b +
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— =
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ
๐Ÿ”๐Ÿ”
๐Ÿ“๐Ÿ“
5
a โˆ’ 6 b Ans
5
๐’‰๐’‰
๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝโƒ— = ๐Ÿ”๐Ÿ”( a โ€“ hb) Ans
๐‘ฉ๐‘ฉ๐‘ฉ๐‘ฉ
๐Ÿ“๐Ÿ“
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Compiled and Solved by Kachama Dickson C & Chansa John
/ Together we can do Mathematics /0966-295655
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TOPIC 10: TRIGONOMETRY
๐‘›๐‘›
1. (a)(i)
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
๐‘›๐‘›
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  60
=
๐‘›๐‘› =
=
๐‘Ÿ๐‘Ÿ
(ii) Area of โˆ†๐พ๐พ๐พ๐พ๐พ๐พ
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
80
1
A = ๐‘˜๐‘˜๐‘˜๐‘˜ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  52
80๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  60
2
(b) Shortest distance from R to KN
S .d =
2
A = 2000 ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 60°
A = 1732.050808
โˆด ๐‘จ๐‘จ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐’Ž๐’Ž๐Ÿ๐Ÿ Ans
๐‘›๐‘› = 87.92016097
โˆด ๐Š๐Š๐Š๐Š = ๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–. ๐Ÿ—๐Ÿ— ๐ฆ๐ฆ Ans
Sd =
1
A = (50)(80) sin 60°
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  52
2๐ด๐ด
๐‘๐‘
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ–
= ๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’. ๐Ÿ–๐Ÿ–๐Ÿ–๐Ÿ– Ans
(c) Graph of ๐‘ฆ๐‘ฆ = cos ๐œƒ๐œƒ
0°
1
๐œƒ๐œƒ
cos ๐œƒ๐œƒ
90°
0
1
0°
90°
180°
โˆ’1
270°
0
360°
1
๐’š๐’š = ๐œ๐œ๐œ๐œ๐œ๐œ ๐œฝ๐œฝ
180°
270°
360°
โˆ’1
2. (a) (i)
๐‘๐‘
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
๐‘๐‘
=
=
๐‘๐‘
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
15
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  79
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  40
15 ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  79
๐‘๐‘ =
๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘  40
๐‘๐‘ = 22.9
AC = 22.9km Ans
(b) ๐‘๐‘๐‘๐‘๐‘๐‘๐‘๐‘ = 0.937
๐œƒ๐œƒ = (๐‘๐‘๐‘๐‘๐‘๐‘ โˆ’1 (0.937)
๐œƒ๐œƒ = 20.4°
(ii) first find angle BAC
Angle BAC = 180 โˆ’ (79 + 40) = 61
1
๐ด๐ด = ๐‘๐‘๐‘๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
2
1
(iii) S.d =
2×150
22.9
= 13.100
= 13.1 Ans
๐ด๐ด = (15)(22.9)๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 61°
2
๐ด๐ด = 150.2159349
๐‘จ๐‘จ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐’Ž๐’Ž๐Ÿ๐Ÿ Ans
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๐œƒ๐œƒ = 360° โˆ’ 20.4) = 339.4
โˆด ๐œฝ๐œฝ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ’๐Ÿ’ ๐š๐š๐š๐š๐š๐š ๐œฝ๐œฝ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘. ๐Ÿ’๐Ÿ’ Ans
(c) ๐’š๐’š = ๐’”๐’”๐’”๐’”๐’”๐’” ๐œฝ๐œฝ
0°
0
๐œƒ๐œƒ
Sin ๐œƒ๐œƒ
90°
1
180°
0
270°
-1
360°
0
1
0
90
180
270
360
๐‘ฆ๐‘ฆ = sin ๐œƒ๐œƒ
โˆ’1
1
3. (a) (i) A = ๐‘ก๐‘ก๐‘ก๐‘ก sin ๐ป๐ป
2
(iii) Shortest distance =
1
=
๐ด๐ด = × 1.3 × 1.9 ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 130°
2
A = 1.235 sin 130°
A = 0.946
A = ๐ŸŽ๐ŸŽ.946 km2 Ans
(ii) โ„Ž2 = ๐‘ก๐‘ก 2 + ๐‘ ๐‘  2 โˆ’ 2๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก๐‘ก
h = 5.3 โ€“ ( โˆ’3.175370792)
๐œƒ๐œƒ =
2
h = 5.3 + 3.17537
h2 = 8.475370792
P
(ii)
๏ฟฝ = 180° โˆ’ (46° + 36°)
R
r
=
=
sinP
p
sin 46°
36.5
rsin 46° = 36.5 sin98°
r=
36.5 sin 98°
sin 46°
1
A = × p × r × sin Q
2
1
= × 36.5 × 50.2 × sin 36°
2
= 98°
r
sin 98°
3
2
cos โˆ’1 ๏ฟฝ ๏ฟฝ
3
P
2
sin R
2
๐œƒ๐œƒ = 48.1896851
โˆด ๐œƒ๐œƒ = 48.2° Ans
P
โˆด
2.9
(b) cos ๐œƒ๐œƒ =
2
4. (a) (i) to find PQ, first find angle R
๐‘๐‘
2×0.95
= 0.65517
S.d = ๐ŸŽ๐ŸŽ. ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” ๐ค๐ค๐ค๐ค Ans
h2 = (1.3)2 + (1.9)2 โˆ’ 2(1.3)(1.9)cos130°
h = โˆš8.475370792
h = 2.911249009
โˆด ๐“๐“๐“๐“ = ๐Ÿ๐Ÿ. ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ— Ans
2๐ด๐ด
= 916.15 sin 36°
โˆด A = 538.4994589 โ‰ˆ ๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“. ๐Ÿ“๐Ÿ“ ๐ค๐ค๐ฆ๐ฆ๐Ÿ๐Ÿ Ans
(iii) Shortest distance =
(b) sin๐œƒ๐œƒ = 0.6792
2๐ด๐ด
๐‘๐‘
=
2×538..2
50.2
= ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ•๐Ÿ• ๐ค๐ค๐ค๐ค A
๐œƒ๐œƒ = sinโˆ’1 (0.6792) โ†’ press shift / 2nd f and then sin
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r = 50.24716343
โˆด ๐๐๐๐ = ๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
5. (a) (i) k2 = i2 + m2 โ€“ 2imcos K
2
2
2
k = 5 + 3 โˆ’2(5)(3) cos110°
2=
k 34 โ€“ (- 10.2606043)
k2 = 44.26060643
๐œƒ๐œƒ = 42.8° and ๐œƒ๐œƒ = 180 โˆ’ 42.8 = 137.2°
โˆด ๐œฝ๐œฝ = ๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’. ๐Ÿ–๐Ÿ–°, ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ° Ans
1
(ii) A = × ๐‘–๐‘– × ๐‘š๐‘š ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 
1
2
(iii) shortest d =
A = (5)(3 )๐‘ ๐‘ ๐‘ ๐‘ ๐‘ ๐‘ 110°
2
A = 7.5 sin110°
A = 7.047694656
A = ๐Ÿ•๐Ÿ•. ๐ŸŽ๐ŸŽ๐ŸŽ๐ŸŽ๐ŸŽ๐ŸŽ๐ฆ๐ฆ๐Ÿ๐Ÿ Ans
k = โˆš44.26060643
k = 6.656864368
โˆด ๐Œ๐Œ๐Œ๐Œ = ๐Ÿ”๐Ÿ”. ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” Ans
=
2๐ด๐ด
๐‘๐‘
2×7.05
6.65
= ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
(b) tan๐œƒ๐œƒ = 0.7
๐œƒ๐œƒ = tanโˆ’1 (0.7)
๐œƒ๐œƒ = 34.9920202
๐œƒ๐œƒ = 34.9920202
๐›‰๐›‰ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘° Ans
TOPIC 11: MENSURATION
1.
๐ป๐ป
๐ป๐ปโˆ’11.4
=
14
8
14( H โ€“ 11.4) = 8H
14H โˆ’ 159.6 = 8H
14H โ€“ 8H = 159.6
2.
6H = 159.6
H = 26.6
h = 26.6 โ€“ 11.4 = 15.2 cm
๐ป๐ป
๐ป๐ปโˆ’9
=
10
4
10(H โˆ’ 9) = 4H
10Hโˆ’90 = 4H
10Hโˆ’4H = 90
6H = 90
H = 15
โˆด ๐ป๐ป = 15๐‘๐‘๐‘๐‘ ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž โ„Ž = 15 โˆ’ 9 = 6๐‘๐‘๐‘๐‘
1
โˆด ๐‘‰๐‘‰ = (๐ฟ๐ฟ๐ฟ๐ฟ๐ฟ๐ฟ โˆ’ ๐‘™๐‘™๐‘™๐‘™โ„Ž)
3
1
๐‘‰๐‘‰ = (14 × 10 × 26.6 โˆ’ 8 × 4 × 15.6)
3
1
๐‘‰๐‘‰ = (3724 โˆ’ 486.4)
๐‘‰๐‘‰ =
3
1
3
(3237.6)
V = 1079.2
V = 1080 ๐œ๐œ๐œ๐œ๐Ÿ‘๐Ÿ‘ Ans
1
โˆด V = [๐ฟ๐ฟ2 ๐ป๐ป โˆ’ ๐‘™๐‘™ 2 โ„Ž] (square base)
3
1
V = [102 × 15 โˆ’ 42 × 6]
3
1
V = (1500 โˆ’ 96)
1
3
V = (1404)
3
V = 468 ๐’„๐’„๐’Ž๐’Ž๐Ÿ‘๐Ÿ‘ Ans
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3. First find the height of the small cone that was
7
=
20+โ„Ž
V=
21โ„Ž = 140 + 7โ„Ž
(13230 โˆ’ 490)
3
3.142
3
(212 × 30 โˆ’ 72 × 10)
(12740)
1
1
V= ๐œ‹๐œ‹๐œ‹๐œ‹2 ๐ป๐ป โˆ’ ๐œ‹๐œ‹๐œ‹๐œ‹ 2 โ„Ž
12
3
1
3
V = ๐œ‹๐œ‹(๐‘…๐‘… ๐ป๐ป โˆ’ ๐‘Ÿ๐‘Ÿ 2 โ„Ž)
12๐‘ฅ๐‘ฅ = 4(15 +๐‘ฅ๐‘ฅ)
3
1
2
V = × 3.142(122 × 22.5 โˆ’ 42 × 7.5)
12๐‘ฅ๐‘ฅ = 60 +4๐‘ฅ๐‘ฅ
3
12๐‘ฅ๐‘ฅ โˆ’4๐‘ฅ๐‘ฅ =60
V=
8๐‘ฅ๐‘ฅ = 60
V=
๐‘ฅ๐‘ฅ = 7.5
โˆด ๐€๐€๐€๐€ = ๐Ÿ•๐Ÿ•. ๐Ÿ“๐Ÿ“ And AD = 7.5 +15 = 22.5cm
1cm3 โ†’ 1000๐‘™๐‘™
10.8cm3โ†’ ๐‘ฅ๐‘ฅ
๐‘ฅ๐‘ฅ = 10.8 × 1000๐‘™๐‘™
๐‘ฅ๐‘ฅ = 10800๐‘™๐‘™
โˆด ๐‘ฝ๐‘ฝ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ (3 sig fig)
3
3.142
(ii) Volume that remained is the that of the frustum
15+๐‘ฅ๐‘ฅ
5. (a) V = ๐‘™๐‘™๐‘™๐‘™โ„Ž
V= 1.2× 0.9 × 10
V= 10.8cm3
3
V = 13343.02667
V = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐’„๐’„๐’„๐’„๐Ÿ‘๐Ÿ‘ Ans
4. (i) Let height EA = ๐‘ฅ๐‘ฅ, then
=
1
3
3.142
V=
21โ„Ž โˆ’7โ„Ž = 140
14โ„Ž= 140
โ„Ž= 10
๐’‰๐’‰ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ ๐’„๐’„๐’„๐’„ and H = 10 + 20 = 30cm
4
1
V=
21
21โ„Ž = 7(20+โ„Ž)
๐‘ฅ๐‘ฅ
3
V = ๐œ‹๐œ‹(๐‘…๐‘… 2 ๐ป๐ป โˆ’ ๐‘Ÿ๐‘Ÿ 2 โ„Ž)
Cutoff.
โ„Ž
1
โˆด ๐‘‰๐‘‰ = ๐œ‹๐œ‹๐‘…๐‘…2 ๐ป๐ป โˆ’ ๐œ‹๐œ‹๐‘Ÿ๐‘Ÿ 2 โ„Ž
3.142
3
3.142
3
(3240 โˆ’ 120)
(3120)
V = 3267.68
V = 3270 ๐’„๐’„๐’Ž๐’Ž๐Ÿ‘๐Ÿ‘ (correct to 3 sig figures)
(b) first find the radius of a cone
r2= 132 โˆ’ 122
r2 = 25
r = โˆš25
r = 5cm
โˆด ๐‘‡๐‘‡. ๐‘†๐‘†. ๐ด๐ด = ๐œ‹๐œ‹๐‘Ÿ๐‘Ÿ 2 + ๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
T.S.A = ๐œ‹๐œ‹๐œ‹๐œ‹(๐‘Ÿ๐‘Ÿ + ๐‘™๐‘™)
T.S.A = 3.142 × 5(5 + 13)
T. S. A = 282.78c๐’Ž๐’Ž๐Ÿ๐Ÿ
12
Great works are performed not by strength but by perseverance
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13
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(Kachama Dickson.C)
TOPIC 9: EARRTH GEOMETRY
N
1. (a)
B
A
35° N
15°N
0°
70°E
C
35°๐‘†๐‘†
40°E
SA
S
๐œƒ๐œƒ
AC =
AC =
AC =
360
50°
× 2๐œ‹๐œ‹๐œ‹๐œ‹ where ๐œƒ๐œƒ = 15° + 35° = 50°
× 2 × 3.142 × 6370
360°
2001454
360
AC = 5,559.594444
AC = 5,560km Ans
(b) (i) BQ =
900 =
๐›ผ๐›ผ
(ii) longitude of Q = 70° โˆ’ 9.9° = 60.1°
× 2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
360
2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
โˆด position of Q (๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘°๐‘บ๐‘บ, ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”. ๐Ÿ๐Ÿ°๐‘ฌ๐‘ฌ) Ans
360
2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹ = 900 × 360°
๐›ผ๐›ผ =
๐›ผ๐›ผ =
32400
2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
32400
(2×3.142×6370 ×cos 35°)
๐›ผ๐›ผ = 9.88109
โˆด the difference in longitude is 9.9°
2. (a) D =
D=
D=
๐œƒ๐œƒ
× 2๐œ‹๐œ‹๐œ‹๐œ‹ where ๐œƒ๐œƒ = 60° + 60° = 120°
360
120
× 2 × 3.142 × 3437
360
2591772 .96
or D = ๐œƒ๐œƒ × 60
D = 120° × 60
D = ๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ•๐Ÿ• nm Ans
360
D = 7199.369333
โˆด ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐ = 7200nm Ans
(b)
To find speed, first find distance CD
D=
D=
D=
๐œถ๐œถ
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
120
× ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
× 2 × ๐œ‹๐œ‹ × 3437 cos 60
360
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ.๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
โˆด speed =
๐‘ซ๐‘ซ
speed =
๐‘ป๐‘ป
3600
12
speed = 300knots Ans
D = 3599.684667
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D = 3600nm
3. (a) Difference in latitudes between W and Y
๐œฝ๐œฝ = 80° + 30° = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ°
(b) (i) XZ =
๐œƒ๐œƒ
XZ =
(ii) YZ =
× 2๐œ‹๐œ‹๐œ‹๐œ‹
360
110°
360°
× 2 × 3.142 × 3437
XZ = 6599.421889
XZ = 6600nm Ans
QR =
165°
360°
๐œƒ๐œƒ
360°
YZ =
× 2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹ where ๐›ผ๐›ผ = 15° + 105 = 120°
× 2 × 3.142 × 3437 × ๐‘๐‘๐‘๐‘๐‘๐‘30°
360°
2244541 .224
360
YZ = 6234.836734
YZ =6230nm Ans
4. (i)
(ii)(a) Distance QR =
YZ =
๐›ผ๐›ผ
360°
120°
× 2๐œ‹๐œ‹๐œ‹๐œ‹ ๐œƒ๐œƒ = 80° + 85° = 165°
× 2 × 3.142 × 3437
QR = 9899.132833
QR =9900 nm Ans
5. (a) Difference in latitudes
ฮธ = 50° + 70° = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ°
(b) C = 2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
C = 21600 Cos 50°
C=13884.21237nm
C= 13900nm
(b) C = 2๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹๐œ‹
C = 21600 Cos ๐œƒ๐œƒ
C = 21600× ๐‘๐‘๐‘๐‘๐‘๐‘ 85°
C = 1882.564043
C =1900nm Ans
(c) AD =
๐œƒ๐œƒ
× 2๐œ‹๐œ‹๐œ‹๐œ‹
360°
120°
AD =
360°
× 2 × 3.142 × 3437
AD = 7199.369333
AD = 7200nm Ans
Produced by Sir DK & Sir JC
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/ Together we can do Mathematics /0966-295655
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TOPIC 13: QUADRATIC FUNCTIONS
1.
(ii) From part (i) we can see that
2. (a) (i) ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 12๐‘ฅ๐‘ฅ โˆ’ (๐‘ฅ๐‘ฅ 3 + ๐‘ฅ๐‘ฅ 2 โˆ’ 12๐‘ฅ๐‘ฅ)
3
2
3
2
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ + 10
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘ฅ โˆ’ 12๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅ + 12๐‘ฅ๐‘ฅ
๐‘ฆ๐‘ฆ = 0 โ†” ๐‘ฅ๐‘ฅ โˆ’ ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž
when we draw this line on the
โˆด ๐’™๐’™ = โˆ’๐Ÿ’๐Ÿ’, ๐’™๐’™ = ๐ŸŽ๐ŸŽ and ๐’™๐’™ = ๐Ÿ‘๐Ÿ‘
same graph using any points we have
Note that the values of ๐‘ฅ๐‘ฅ are the points where
๐’™๐’™ = โˆ’๐Ÿ‘๐Ÿ‘. ๐Ÿ•๐Ÿ• ± ๐Ÿ๐Ÿ, ๐’™๐’™ = โˆ’๐ŸŽ๐ŸŽ. ๐Ÿ•๐Ÿ• ± and
3
2
the curve ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ + ๐‘ฅ๐‘ฅ โˆ’ 12๐‘ฅ๐‘ฅ meets the line ๐‘ฆ๐‘ฆ = 0
๐’™๐’™ = ๐Ÿ‘๐Ÿ‘. ๐Ÿ“๐Ÿ“ ± ๐Ÿ๐Ÿ
๐’™๐’™
โˆ’3
0
2
3
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7
๐’š๐’š
2. (b)(i) Draw a straight line touching only
at ( -3, 18) and pick any two points
lying on the same line
๐‘š๐‘š =
๐‘š๐‘š =
12
13
(ii) From the graph, we can see that the
required area is simply the areas
of the two trapeziums
e. g (โˆ’3, 18 ) and (โˆ’4.5,4)
๐‘š๐‘š =
10
๐‘ฆ๐‘ฆ2 โˆ’๐‘ฆ๐‘ฆ1
๐‘ฅ๐‘ฅ 2 โˆ’๐‘ฅ๐‘ฅ 1
4โˆ’18
โˆ’4.5โˆ’(โˆ’3)
โˆ’14
โˆ’1.5
๐’Ž๐’Ž = ๐Ÿ—๐Ÿ—. ๐Ÿ‘๐Ÿ‘ ± ๐Ÿ๐Ÿ
1
A = (๐‘Ž๐‘Ž + ๐‘๐‘)โ„Ž
2
1
2
1
2
1
A = (58) + (52)
2
2
A = 29 + 26
๐€๐€ = ๐Ÿ“๐Ÿ“๐Ÿ“๐Ÿ“ ± ๐Ÿ๐Ÿ Square units
3. (a) (i) ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 โˆ’ (๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3)
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 โˆ’ ๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 + ๐‘ฅ๐‘ฅ + 3
๐‘ฆ๐‘ฆ = 0
On the graph, when ๐‘ฆ๐‘ฆ = 0, (๐‘ฅ๐‘ฅ โˆ’ ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž)
๐’™๐’™ = โˆ’๐Ÿ‘๐Ÿ‘, ๐’™๐’™ = โˆ’๐Ÿ๐Ÿ and ๐’™๐’™ = ๐Ÿ๐Ÿ
(ii) ๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 โˆ’ (๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 5)
๐‘ฆ๐‘ฆ = ๐‘ฅ๐‘ฅ 3 + 3๐‘ฅ๐‘ฅ 2 โˆ’ ๐‘ฅ๐‘ฅ โˆ’ 3 โˆ’ ๐‘ฅ๐‘ฅ 3 โˆ’ 3๐‘ฅ๐‘ฅ 2 + ๐‘ฅ๐‘ฅ + 5)
๐‘ฆ๐‘ฆ = โˆ’3 + 5
1
A = (28 + 30)1 + (30 + 22)1
(b) (i) to find the gradient, draw a
straight line touching the curve
only at (โˆ’3,0) and pick any
two points lying on this line
for Example (โˆ’3,0) and (โˆ’2,8)
๐‘š๐‘š =
๐‘š๐‘š =
๐‘ฆ๐‘ฆ2 โˆ’๐‘ฆ๐‘ฆ1
๐‘ฅ๐‘ฅ 2 โˆ’๐‘ฅ๐‘ฅ 1
8โˆ’0
โˆ’2โˆ’(โˆ’3)
๐Ÿ–๐Ÿ–
๐’Ž๐’Ž = = ๐Ÿ–๐Ÿ– Ans
๐Ÿ๐Ÿ
(ii) Area = A of rectangle + trapezium
๐‘ฆ๐‘ฆ = 2
On the graph when y=2, then
๐’™๐’™ = โˆ’๐Ÿ๐Ÿ. ๐Ÿ”๐Ÿ”, ๐’™๐’™ = โˆ’๐Ÿ๐Ÿ. ๐Ÿ“๐Ÿ“ and ๐’™๐’™ = ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ Ans
1
A = l× ๐‘๐‘ + (๐‘Ž๐‘Ž + ๐‘๐‘)โ„Ž
2
1
A = 1 × 20 + (20 + 5)
2
A = 20 + 12.5
A = 32.5 square units
4. (a) To find the gradient of the curve, simply draw a straight line touching the curve at (2,5)
and pick any two points lying on the line drawn, for example (2,5) and (1, โˆ’7) and use the
gradient formula to find the gradient
๐‘ฆ๐‘ฆ2 โˆ’ ๐‘ฆ๐‘ฆ1
๐‘š๐‘š =
๐‘ฅ๐‘ฅ2 โˆ’ ๐‘ฅ๐‘ฅ1
โˆ’7 โˆ’ 5
๐‘š๐‘š =
1โˆ’2
โˆ’12
๐‘š๐‘š =
โˆ’1
๐’Ž๐’Ž = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Ans
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4. (b) (i) ๐’š๐’š = ๐‘ฅ๐‘ฅ3 + ๐‘ฅ๐‘ฅ2 โˆ’ 5๐‘ฅ๐‘ฅ + 3 โˆ’ (๐‘ฅ๐‘ฅ3 + ๐‘ฅ๐‘ฅ2 โˆ’ 5๐‘ฅ๐‘ฅ + 3)
๐’š๐’š = ๐‘ฅ๐‘ฅ3 + ๐‘ฅ๐‘ฅ2 โˆ’ 5๐‘ฅ๐‘ฅ + 3 โˆ’ ๐‘ฅ๐‘ฅ3 โˆ’ ๐‘ฅ๐‘ฅ2 + 5๐‘ฅ๐‘ฅ โˆ’ 3
๐‘ฆ๐‘ฆ = 0
On the graph, when ๐‘ฆ๐‘ฆ = 0(๐‘ฅ๐‘ฅ โˆ’ ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž)
๐’™๐’™ = โˆ’๐Ÿ‘๐Ÿ‘ and ๐’™๐’™ = ๐Ÿ๐Ÿ Ans
(c) From the graph in the given bounds, we can
make two trapeziums.
Hence to find the area, we find the total areas of these
two trapeziums:
1
A = (๐’‚๐’‚ + ๐’ƒ๐’ƒ)๐’‰๐’‰
2
1
(ii) similarly we have ๐‘ฆ๐‘ฆ=5๐‘ฅ๐‘ฅ
Use any points to draw the
line ๐‘ฆ๐‘ฆ = 5๐‘ฅ๐‘ฅ and find the
values of x where it meets
the curve for example use
the points in this table
below
๐‘ฅ๐‘ฅ -1
0
1
5
๐‘ฆ๐‘ฆ -5
0
5
10
โˆด ๐’™๐’™ = โˆ’๐Ÿ‘๐Ÿ‘. ๐Ÿ–๐Ÿ–, ๐’™๐’™ = 0.3 and ๐’™๐’™=๐Ÿ๐Ÿ. ๐Ÿ’๐Ÿ’
1
A= (9 + 8)1 + (8 + 3)
2
1
1
2
A= (17) + (11)
2
2
A= 8.5 + 5.5
A= ๐Ÿ๐Ÿ๐Ÿ๐Ÿ Square units
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TOPIC 14: LINEAR PROGRAMING
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TOPIC 15: STATISTICS
1. (a)
class
๐’™๐’™
๐’™๐’™๐Ÿ๐Ÿ
๐’‡๐’‡
10< ๐‘ฅ๐‘ฅ โ‰ค 20
15
225
20 < ๐‘ฅ๐‘ฅ โ‰ค 30
25
625
10
250
6250
30 < ๐‘ฅ๐‘ฅ โ‰ค 40
35
1225
15
525
18375
40< ๐‘ฅ๐‘ฅ โ‰ค 50
45
2025
23
1035
46575
50 < ๐‘ฅ๐‘ฅ โ‰ค 60
55
3025
30
1650
90750
60 < ๐‘ฅ๐‘ฅ โ‰ค 70
65
4225
10
1650
42250
Totals
Mean ๐‘ฅ๐‘ฅฬ… =
=
=๏ฟฝ
๏ฟฝ ๐’‡๐’‡ = ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—
30
๏ฟฝ ๐’‡๐’‡๐’‡๐’‡ = ๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’
450
๐’‡๐’‡๐’™๐’™๐Ÿ๐Ÿ
๏ฟฝ ๐’‡๐’‡๐’™๐’™๐Ÿ๐Ÿ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
โˆ‘ ๐‘“๐‘“๐‘“๐‘“
โˆ‘ ๐‘“๐‘“
4140
90
= 46
โˆ‘ ๐‘“๐‘“๐‘“๐‘“ 2
SD = ๏ฟฝ โˆ‘
2
๐’‡๐’‡๐’‡๐’‡
๐‘“๐‘“๐‘“๐‘“
โˆ’ (๐‘ฅ๐‘ฅ)2
204650
90
โˆ’ (46)2
= โˆš2273.888889 โˆ’ 2116
= โˆš157.8888889
SD = 12.57 Ans
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2. (a)
2
๐’™๐’™
๐’™๐’™๐Ÿ๐Ÿ
4
3
9
4
๐’‡๐’‡
1
2
๐’‡๐’‡๐’‡๐’‡
4
5
15
45
16
4
16
64
5
25
6
30
150
6
36
10
60
360
7
49
16
112
784
8
64
18
144
1152
๏ฟฝ ๐’‡๐’‡ = ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”
๏ฟฝ) =
Mean ( ๐’™๐’™
โˆ‘ ๐‘“๐‘“๐‘“๐‘“
SD = ๏ฟฝ โˆ‘
๐‘“๐‘“
60
๏ฟฝ ๐’‡๐’‡๐’™๐’™๐Ÿ๐Ÿ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
โˆ‘ ๐’‡๐’‡๐’‡๐’‡
โˆ‘ ๐’‡๐’‡
=
๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”
= ๐Ÿ”๐Ÿ”. ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
โˆ’ (๐‘ฅ๐‘ฅฬ… )2
2559
=๏ฟฝ
๏ฟฝ ๐’‡๐’‡๐’‡๐’‡ = ๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘๐Ÿ‘
๐’‡๐’‡๐’™๐’™๐Ÿ๐Ÿ
โˆ’ (6.32)2
= โˆš42.65 โˆ’ 39.9424
= โˆš2.7076
= 1.645478654
= ๐Ÿ๐Ÿ. ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ” Ans
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Class marks
0 < ๐‘ฅ๐‘ฅ โ‰ค 5
5 < ๐‘ฅ๐‘ฅ โ‰ค 10
10 < ๐‘ฅ๐‘ฅ โ‰ค 15
15 < ๐‘ฅ๐‘ฅ โ‰ค 20
20 < ๐‘ฅ๐‘ฅ โ‰ค 25
25 < ๐‘ฅ๐‘ฅ โ‰ค 30
๏ฟฝ=
๐’™๐’™
=
๐’™๐’™
2.5
7.5
12.5
17.5
22.5
27.5
โˆ‘ ๐‘“๐‘“๐‘“๐‘“
13
27
35
16
7
2
๐’‡๐’‡
๏ฟฝ ๐‘“๐‘“ = 100
๐’‡๐’‡๐’‡๐’‡
32.5
205.5
437.5
280
157.5
55
๏ฟฝ ๐‘“๐‘“๐‘“๐‘“ = 1168
๏ฟฝ)๐Ÿ๐Ÿ
(๐’™๐’™ โˆ’ ๐’™๐’™
12.25
17.2225
0.7225
34.2225
117.7225
251.2225
๏ฟฝ)๐Ÿ๐Ÿ
๐’‡๐’‡(๐’™๐’™ โˆ’ ๐’™๐’™
159.25
465.0
25.288
547.58
824.058
502.445
๏ฟฝ ๐‘“๐‘“(๐‘ฅ๐‘ฅ โˆ’ ๐‘ฅ๐‘ฅฬ… )2
= 2523.621
โˆ‘ ๐‘“๐‘“
1165
100
= ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”
SD = ๏ฟฝ
โˆ‘ ๐’‡๐’‡(๐’™๐’™โˆ’๐’™๐’™
๏ฟฝ)๐Ÿ๐Ÿ
โˆ‘ ๐’‡๐’‡
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ.๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”๐Ÿ”
SD = ๏ฟฝ
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
SD = โˆš๐Ÿ๐Ÿ๐Ÿ๐Ÿ. ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
SD = 5 .023565467
SD = ๐Ÿ“๐Ÿ“ Ans
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Class marks
Midpoints
(๐‘ฅ๐‘ฅ)
Frequency
(๐‘“๐‘“)
30 < ๐‘ฅ๐‘ฅ โ‰ค 35
32.5
4
40 < ๐‘ฅ๐‘ฅ โ‰ค 45
42.5
25 < ๐‘ฅ๐‘ฅ โ‰ค 30
35 < ๐‘ฅ๐‘ฅ โ‰ค 40
45 < ๐‘ฅ๐‘ฅ โ‰ค 50
55 < ๐‘ฅ๐‘ฅ โ‰ค 55
55 < ๐‘ฅ๐‘ฅ โ‰ค 60
Totals
๏ฟฝ) =
Mean (๐’™๐’™
47.5
52.5
57.5
๐‘“๐‘“๐‘“๐‘“ 2
5
137.5
756.25
3781.25
7
262.5
1406.25
9843.75
11
12
8
1
๏ฟฝ ๐‘“๐‘“ = 48
130
467.5
570
420
57.5
๏ฟฝ ๐‘“๐‘“๐‘“๐‘“ = 2045
1056.25
1806.25
2256.25
2256.25
3306.25
4225
19868.75
27075
22050
3306.25
๏ฟฝ ๐‘“๐‘“๐‘ฅ๐‘ฅ 2 = 90150
โˆ‘ ๐’‡๐’‡๐’‡๐’‡
โˆ‘ ๐’‡๐’‡
๐Ÿ’๐Ÿ’๐Ÿ’๐Ÿ’
= 42 .6
โˆ‘ fx 2
โˆ‘f
90150
=๏ฟฝ
37.5
๐‘ฅ๐‘ฅ 2
๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
=
SD = ๏ฟฝ๏ฟฝ
27.5
๐‘“๐‘“๐‘“๐‘“
48
โˆ’ (x๏ฟฝ )2 ๏ฟฝ
โˆ’ (42.6)2
= โˆš1878.125 โˆ’ 1814.76
= โˆš63.365
=๐Ÿ•๐Ÿ•. ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ— ๐€๐€๐€๐€๐€๐€
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Class marks
0 < ๐‘ฅ๐‘ฅ โ‰ค 10
5
30 < ๐‘ฅ๐‘ฅ โ‰ค 40
35
10 < ๐‘ฅ๐‘ฅ โ‰ค 20
20 < ๐‘ฅ๐‘ฅ โ‰ค 30
40 < ๐‘ฅ๐‘ฅ โ‰ค 50
50 < ๐‘ฅ๐‘ฅ โ‰ค 60
Mean, ๐‘ฅ๐‘ฅฬ… =
=
15
๐’™๐’™
25
45
55
7
22
๐’‡๐’‡
28
23
15
5
๏ฟฝ ๐’‡๐’‡ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
35
330
๐’‡๐’‡๐’‡๐’‡
700
805
675
275
๏ฟฝ ๐’‡๐’‡๐’‡๐’‡ = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ๐Ÿ
25
๐’™๐’™๐Ÿ๐Ÿ
225
625
1225
2025
3025
f๐’™๐’™๐Ÿ๐Ÿ
175
4950
17500
28175
30375
15125
๏ฟฝ ๐’‡๐’‡ ๐’™๐’™๐Ÿ๐Ÿ = ๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—๐Ÿ—
โˆ‘ ๐‘“๐‘“๐‘“๐‘“
โˆ‘ ๐‘“๐‘“
2820
100
= 28.2
โˆด ๐‘†๐‘†๐‘†๐‘† = ๏ฟฝ
โˆ‘ ๐‘“๐‘“ ๐‘ฅ๐‘ฅ 2
โˆ’ (๐‘ฅ๐‘ฅฬ… )2
โˆ‘ ๐‘“๐‘“
96300
SD = ๏ฟฝ
100
โˆ’ (28.2)2
SD = โˆš963 โˆ’ 795.24
SD = โˆš167.76
SD = 12.95221989
SD = ๐Ÿ๐Ÿ๐Ÿ๐Ÿ. 95 Ans
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TOPIC 16: TRANSFORMATION
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1. (a) It is a clockwise rotation of 90° about the origin.
(b) It is an enlargement, centre (0, 0) and scale factor 2
(c) Let the matrix be ๏ฟฝ
๐‘Ž๐‘Ž
๐‘๐‘
๐‘๐‘
๏ฟฝ, then pick any two coordinates of P which corresponds to V and
๐‘‘๐‘‘
form four equations as follows:
๏ฟฝ
๐‘Ž๐‘Ž
๐‘๐‘
๐‘๐‘ 2
๏ฟฝ๏ฟฝ
๐‘‘๐‘‘ 1
4
โˆ’4
๏ฟฝ=๏ฟฝ
1
4
โˆ’8
๏ฟฝ
1
๐‘Ž๐‘Ž + 4๐‘๐‘ = โˆ’4 โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.(i)
๐‘Ž๐‘Ž + ๐‘๐‘ = โˆ’8 โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ(ii)
2๐‘๐‘ + ๐‘‘๐‘‘ = 1โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.(iii)
4๐‘๐‘ + ๐‘‘๐‘‘ = 1โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.(iv)
Solving these equations (i) and (ii) simultaneously, we have ๐‘Ž๐‘Ž = โˆ’2 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž ๐‘๐‘ = 0
Similarly solving equations (iii) and (iv) we have ๐‘๐‘ = 0 and ๐‘‘๐‘‘ = 1
โˆด the required matrix is
๏ฟฝ
๐’‚๐’‚
๐’„๐’„
โˆ’๐Ÿ๐Ÿ
๐’ƒ๐’ƒ
๏ฟฝ=๏ฟฝ
๐ŸŽ๐ŸŽ
๐’…๐’…
๐ŸŽ๐ŸŽ
๏ฟฝ
๐Ÿ๐Ÿ
(d) To find the coordinates of S, we need multiply the given matrix by the coordinates of P.
1 0 2 2 4
๏ฟฝ๏ฟฝ
๏ฟฝ
โˆ’2 1 1 4 1
2+0
2+0
4+0
๏ฟฝ
๏ฟฝ
โˆ’4 + 1 โˆ’4 + 4 โˆ’8 + 1
๐Ÿ๐Ÿ ๐Ÿ๐Ÿ ๐Ÿ’๐Ÿ’
๏ฟฝ
๏ฟฝ
โˆ’๐Ÿ‘๐Ÿ‘ ๐ŸŽ๐ŸŽ โˆ’๐Ÿ•๐Ÿ•
๏ฟฝ
There the coordinates of S are (๐Ÿ๐Ÿ, โˆ’๐Ÿ‘๐Ÿ‘), (๐Ÿ๐Ÿ, ๐ŸŽ๐ŸŽ)and (๐Ÿ’๐Ÿ’, โˆ’๐Ÿ•๐Ÿ•)
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5. (a) (i) To find the centre of enlargement, join any corresponding two points of the
object and the image, the intersection point is the centre of enlargement.
โˆด the centre of enlargement is (1, 2) Ans
(ii)
Scale factor =
๐‘Ž๐‘Ž
๐‘๐‘
๐‘๐‘ 1
๏ฟฝ๏ฟฝ
๐‘‘๐‘‘ 4
๐‘Ž๐‘Ž
๐‘๐‘
= ๐Ÿ๐Ÿ Ans
๐‘๐‘
๏ฟฝ, then
๐‘‘๐‘‘
1 3
3
๏ฟฝ=๏ฟฝ
๏ฟฝ
1 5
4
(b) Let the matrix be ๏ฟฝ
๏ฟฝ
3.2
1.6
๐‘Ž๐‘Ž + 4๐‘๐‘ = 1 โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.(i)
๐‘๐‘ + 4๐‘‘๐‘‘ = 1 โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ..(iii)
3๐‘Ž๐‘Ž + 4๐‘๐‘ = 3 โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ.(ii)
3๐‘๐‘ + 4๐‘‘๐‘‘ = 5โ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆโ€ฆ..(iv)
Solving the equations (i) and (ii) simultaneously yields ๐‘Ž๐‘Ž = 1 ๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž๐‘Ž ๐‘๐‘ = 0.
Similarly solving equations (iii) and (iv) simultaneously yields ๐‘๐‘ = 2 and ๐‘‘๐‘‘ = โˆ’
โˆด the required matrix is ๏ฟฝ
๐‘Ž๐‘Ž
๐‘๐‘
๐Ÿ๐Ÿ ๐ŸŽ๐ŸŽ
๐‘๐‘
๏ฟฝ = ๏ฟฝ๐Ÿ๐Ÿ โˆ’ ๐Ÿ๐Ÿ๏ฟฝ Ans
๐‘‘๐‘‘
๐Ÿ’๐Ÿ’
1
4
(c) Triangle ABC is mapped onto triangle ๐€๐€ ๐Ÿ‘๐Ÿ‘ ๐๐๐Ÿ‘๐Ÿ‘ ๐‚๐‚๐Ÿ‘๐Ÿ‘ by an anticlockwise rotation of 90°
about (0, 0) or by a clockwise rotation of 270° about (0, 0).
(d) (i)
(ii)
Comparing the matrices ๏ฟฝ
๐‘˜๐‘˜
0
โˆ’3
0
๏ฟฝ and ๏ฟฝ
0
1
0
๏ฟฝ, we have ๐‘˜๐‘˜ = โˆ’3
1
โˆด the scale factor of this transformation is ๐’Œ๐’Œ = โˆ’๐Ÿ‘๐Ÿ‘ Ans
To find the coordinates of A4, B4 and C4 (image) we need to multiply the given
matrix with the coordinates of triangle ABC (object)
โˆ’3 0 1 3 1
๏ฟฝ๏ฟฝ
๏ฟฝ
0 1 4 4 5
โˆ’3 + 0 โˆ’9 + 0 โˆ’3 + 0
๏ฟฝ
๏ฟฝ
0+4
0+4
0+5
โˆ’3 โˆ’9 โˆ’3
๏ฟฝ
๏ฟฝ
4
4
5
๏ฟฝ
โˆด the coordinates of ๐€๐€ ๐Ÿ’๐Ÿ’, ๐๐๐Ÿ’๐Ÿ’ and ๐‚๐‚๐Ÿ’๐Ÿ’ are (โˆ’๐Ÿ‘๐Ÿ‘, ๐Ÿ’๐Ÿ’), (โˆ’๐Ÿ—๐Ÿ—, ๐Ÿ’๐Ÿ’) and (โˆ’๐Ÿ‘๐Ÿ‘, ๐Ÿ“๐Ÿ“)
respectively
THE END OF SOLUTIONS TO ALL QUESTIONS: ENJOYOUR REVISION & GOD BLESS YOU!!!!!!
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