# DK & JC's Maths P2 Pamphlet

```A QUICK LOOK AT MATHEMATICS PAPER 2 SYLLABUS D (4024/2)
FINAL EXAMINATION QUESTIONS (2016β 2018) WITH ANSWERS
MATHEMATICAL FORMULAE
οΏ½ πππ₯π₯ ππ ππππ =
ππππ ππ +1
+ ππ
ππ + 1
1
Conical Frustum ππ = ππ(ππ2 π»π» β ππ 2 β)
3
1
Pyramid frustum with a square base and top ππ = β(πΏπΏ2 + ππ2 + πΏπΏπΏπΏ)
Mean = π₯π₯Μ =
β ππππ
β ππ
, ππππ = οΏ½οΏ½
β ππ(π₯π₯βπ₯π₯Μ )2
β ππ
3
οΏ½ = οΏ½οΏ½
πππ‘π‘β π‘π‘π‘π‘π‘π‘π‘π‘ ππππππ ππππ: ππππ = ππ + (ππ β 1)ππ,
ππ
ππππ: πΊπΊππ = [2ππ + (ππ β 1)ππ], ππππ: πΊπΊππ =
2
β πππ₯π₯ 2
β ππ
β (π₯π₯Μ )2 οΏ½
ππππ: ππππ = ππππ ππβ1
(1βππ ππ )
1βππ
,r<1
Cosine Rule for βππππππ: ππππ = ππππ + ππππ β ππππππππππππππ
Compiled & Solved by Kachama Dickson. C & Chansa John
Contacts: 0966295655/0955295655/0976221226
Email: [email protected]
: [email protected]
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Both Mr. Kachama. D.C and Mr. Chansa J are graduates from the Copperbelt University
(CBU). Kachama D.C is a holder of bachelorβs degree in Mathematics Ed (BSc MA.Ed)
while Mr. Chansa. J is a holder of bachelorβ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.
βWe believe, this pamphlet will be of great help to you even as you prepare for your final
examinations:β
Topic
page
1. Algebra β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦........ 3
2. Matrices β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...β¦β¦β¦.β¦... 4
3. Setsβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 5
4. Probability β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 7
5. Sequences & Series β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 8
6. Pseudo code & Flow chartβ¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.... 9
7. Loci & Construction β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦β¦.. 12
8. Calculus β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦...β¦β¦β¦β¦β¦ 14
9. Vector Geometry β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦ 15
10. Trigonometry β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦..β¦...β¦ 17
11. Mensuration β¦β¦β¦β¦β¦β¦..β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.β¦ 20
12. Earth Geometry β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 22
14. Linear Programming β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦.. 28
15. Statistics β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦. 31
16. Transformation β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦β¦... 34
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
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
ππβππ
(a) Simplify
ππ 2 βππ 2
12ππππ 3
(b) Simplify
(c) Express
3
π₯π₯+1
β
9ππ 3 ππ
÷
15ππππ 3
4
π₯π₯β1
10ππ 2 ππ 2
as a single fraction in its lowest terms.
2. 2018 July/Aug Exams
(a) Simplify
(b) Express
7π π π‘π‘ 3
3
β
2π₯π₯β5
5π’π’ 3 π£π£
×
15π’π’ 3 π£π£ 2
28π π  3 π‘π‘ 2
4
as a single fraction in its lowest terms.
π₯π₯β3
3. 2017 Oct/Nov Exams
(a) Simplify
(b) Simplify
(c) Express
14π₯π₯ 3
7π₯π₯ 4
÷
9π¦π¦ 2
18π¦π¦ 3
2π₯π₯2 β8
1
π₯π₯β4
π₯π₯+2
β
2
5π₯π₯β1
as a single fraction in its lowest terms.
4. 2017 July/ Aug Exams
(a) Simplify
(b) Simplify
(c) Express
ππ 2 β1
ππ 2 βππ
ππ 2 ππ 3
3
4
5π₯π₯β2
β
×
8
ππππ
2
π₯π₯+3
÷ 2ππ2 ππ
as a single fraction in its lowest terms.
5. 2016 October Exams
(a) Simplify
(b) Simplify
(c) Express
π₯π₯β1
π₯π₯ 2 β1
17ππ 2
20ππ 2
2
2π₯π₯β1
÷
β
51ππ 2
5ππ
1
3π₯π₯+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 π΄π΄ = οΏ½
8
β5
οΏ½ and π΅π΅ = οΏ½
2
3
4
1
π¦π¦
οΏ½,
5
(a) find the value of π¦π¦, for which the determinants of A and B are equal,
(b) hence find the inverse of B.
2. 2018 Jul/August Exams: Q1(a)
2π₯π₯
3
Given that A = οΏ½
2
οΏ½,
π₯π₯
(a) find the positive value of π₯π₯ for the determinant of A is 12,
(b) hence or otherwise find Aβ1 .
3. 2017 Oct/Nov Exams: Q1(a)
3
Given that matrix M = οΏ½
5
β2
οΏ½
π₯π₯
(a) find the value of π₯π₯ 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 = οΏ½
β2
οΏ½, find
β2
(a) the determinant of K,
(b) the inverse of K.
5. 2016 Oct/Nov Exams: Q1(a)
Given that ππ = οΏ½
3
π₯π₯
β2
οΏ½, find
4
(a) the value of π₯π₯, 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
π₯π₯
7
(i)
(ii)
3
W
If 22 learners walk to school, find the value of π₯π₯.
How many learners use
(a) only one mode of transport,
(b) two different mode of transport.
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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 π¦π¦ 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.
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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 ππ + 4, ππ ππππππ 2ππ β 15 where ππ a
positive integer.
(a) find the value of ππ,
(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 + ππ, 10 + ππ and 15 + ππ. Find
(a) the value of n,
(b) the common ratio,
(c) the sum of the first 6 terms of this sequence.
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/ 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 π₯π₯ + 1, π₯π₯ β 3 and π₯π₯ β 1.
(a)
(b)
the value of π₯π₯,
(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 π₯π₯, π¦π¦
Let M = square root(π₯π₯ squared + π¦π¦ squared)
IF M< 0
THEN display error message β M must be positiveβ
ELSE
END IF
Display M
Stop
Draw the corresponding flow chart for the information given above.
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 10 of 84
2. 2018 July/Aug Exams, Q5(b)
Study the pseudo code below.
Start
Enter ππ, ππ, ππ
R= 1 β ππ
If R = 0 THEN
Print βthe value of r is not validβ
Else Sn =
End if
ππ(1βππ ππ )
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 βr must be positiveβ
No
1
π΄π΄ = β ππ β ππ β π π π π π π π π
2
Display Area
Stop
Write a pseudo code corresponding to the flow chart program above
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/ Together we can do Mathematics /0966-295655
Page 11 of 84
4. 2017 July/Aug Exams, Q6(b)
The diagram below is given in the form of a flow chart
Start
Enter a, r
Is
|ππ| < 1?
No
Yes
πΊπΊβ =
ππ
ππβ ππ
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
The display βerror messageβ and re-enter positive radius
Else enter height
If height < 0
The display βerror messageβ and re-enter positive height
1
Else Volume = β ππ β square radius β height
3
End if
Display volume
Stop
Draw the corresponding flowchart for the information given above.
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 12 of 84
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)
οΏ½ R = 50°
Construct triangle PQR in which PQ = 10cm, QR = 8cm and ππππ
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
(b)
Measure and write the length of CD
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/ Together we can do Mathematics /0966-295655
Page 13 of 84
(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 β 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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Page 14 of 84
TOPIC 8: CALCULUS
1. 2018 Oct/Nov Exams, Q3
2
(a) Evaluate β«β1(2 + π₯π₯ β π₯π₯ 2 )ππππ
(b) Find the equation of the normal to the curve π¦π¦ = π₯π₯ +
2. 2018 Jul/Aug Exams, Q7(b)
1
(a) Evaluate β«0 (π₯π₯ 2 β 2π₯π₯ + 3)ππππ
4
π₯π₯
at the point where π₯π₯ = 4
(b) Determine the equation of the normal to the curve π¦π¦ = 2π₯π₯ 2 β 3π₯π₯ β 2 that passes
through (3, 7)
3. 2017 Oct/Nov Exams, Q9 (b & c)
(a) Find the coordinates of the points on the curve π¦π¦ = 2π₯π₯ 3 β 3π₯π₯ 2 β 36π₯π₯ β 3 where
3
(b) Evaluate β«β1(3π₯π₯ 2 β 2π₯π₯)ππππ
4. 2017 July/Aug Exams, Q4b & 9c
5
(a) Evaluate β«2 (3π₯π₯ 2 + 2)ππππ
(b) Find the equation of the tangent to the curve π¦π¦ = π₯π₯ 2 β 3π₯π₯ β 4 at a point where
π₯π₯ = 2
5. 2016 Oct / Nov Exams, Q6
3
The equation of the curve isπ¦π¦ = π₯π₯ 3 β π₯π₯ 2 . Find
2
(a) equation of the normal where π₯π₯ = 2,
(b) the coordinates of the stationary points.
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TOPIC 9: VECTORS
1. 2018 Oct/Nov Exams, Q3
οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ and οΏ½οΏ½οΏ½οΏ½οΏ½β
BC = 2ππ and AE: AC = 1: 3
U
(i)
(ii)
Find in terms of ππ and/or ππ
(a) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄
(b) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅
U
(c) οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅
Hence or otherwise show that the points B, D and E are collinear
2. 2017 Oct/Nov Exams, Q2(b)
οΏ½οΏ½οΏ½οΏ½οΏ½β = 2p, οΏ½οΏ½οΏ½οΏ½οΏ½β
In the diagram below, OP
OQ = 4q and PX βΆ XQ = 1: 2
(i)
(ii)
Express in terms of p and / or q.
(a) οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
PQ
οΏ½οΏ½οΏ½οΏ½β
(b) PX
οΏ½οΏ½οΏ½οΏ½οΏ½β
(c) OX
οΏ½οΏ½οΏ½οΏ½οΏ½β, show that CQ
οΏ½οΏ½οΏ½οΏ½οΏ½β = 4 οΏ½1 β β οΏ½q β 4β p.
Given that οΏ½οΏ½οΏ½οΏ½οΏ½β
OC = βOX
3
3
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3. 2017 July/Aug Exams, Q6(a)
οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ andοΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
In the diagram below, OABC is a parallelogram in which OA
AB = 2ππ. 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.
οΏ½οΏ½οΏ½οΏ½οΏ½β,
OB
οΏ½οΏ½οΏ½οΏ½οΏ½β
OE,
(a)
(b)
οΏ½οΏ½οΏ½οΏ½οΏ½β.
CD
(c)
4. 2017 Oct/Nov Exams, Q6(a)
οΏ½οΏ½οΏ½οΏ½οΏ½β = 6ππ .
In the diagram below, OAB is a triangle in which οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = 3ππ and ππππ
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 ππ and / or ππ
U
(a) οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
AB
(b) οΏ½οΏ½οΏ½οΏ½οΏ½β
OD
(c)
(ii)
οΏ½οΏ½οΏ½οΏ½οΏ½β
BC
οΏ½οΏ½οΏ½οΏ½οΏ½β, express BE
οΏ½οΏ½οΏ½οΏ½οΏ½β = hBC
οΏ½οΏ½οΏ½οΏ½οΏ½β in terms of β, ππ and ππ
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,
οΏ½ 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 260ππ2 , calculate the shortest
distance from R to KN.
(c) Sketch the graph of π¦π¦ = cos ππ ππππππ 0° β€ ππ β€ 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 ππ = 0.937 for ππππππ 0° β€ ππ β€ 360°
(c) Sketch the graph of π¦π¦ = sin ππ ππππππ 0° β€ ππ β€ 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
οΏ½ 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 ππ =
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 ΞΈ = 0.6792 for 0° β€ ΞΈ β€ 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 ΞΈ = 0.7 for 0° β€ ΞΈ β€ 180°.
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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. (Takeππ = ππ. ππππππ)
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π = 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. (Takeππ = 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 ππ = ππ. ππππππ ππππππ πΉπΉ = ππππππππππππ]
(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. [ ππ = 3.142 and R = 3437nm].
(a) Calculate the distance BC along the latitude 60°πΈπΈ 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 ππ = ππ. ππππππ ππππππ ππ = ππππππππ)
(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), ππ(80°N, 70°E), ππ(85°S, 70°E)and ππ(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 ππ = ππ. ππππππ ππππππ ππ = ππππππππππππ]
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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 Ο = 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.
1. 2018 Oct/Nov Exams, Q7(a)
The values π₯π₯ and π¦π¦ are connected by the equationπ¦π¦ = 2π₯π₯ 3 β 3π₯π₯ 2 + 5. Some
corresponding values of π₯π₯ and π¦π¦ are given in the table below.
π₯π₯
β2
β1.5
β1
β0.5
0
0.5
1
1.5
2
π¦π¦
ππ
β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 π₯π₯ βaxis for β2 β€ π₯π₯ β€ 2 and a scale
of 2cm to represent 5 units on the π¦π¦ βaxis forβ25 β€ π¦π¦ β€ 10, draw the graph
of π¦π¦ = 2π₯π₯ 3 β 3π₯π₯ 2 + 5.
(c) Use your graph to solve the equation 2π₯π₯ 3 β 3π₯π₯ 2 + 5 = π₯π₯
(d) Calculate an estimate of the gradient of the curve at the point where π₯π₯ = 1.5
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2. 2018 July/ Aug Exams, Q12a)
The diagram below shows the graph of π¦π¦ = π₯π₯ 3 + π₯π₯ 2 β 12π₯π₯
(a) Use the graph to solve the equation
(i)
π₯π₯ 3 + π₯π₯ 2 β 12π₯π₯ = 0
(ii)
π₯π₯ 3 + π₯π₯ 2 β 12π₯π₯ = π₯π₯ + 10
(b) Calculate an estimate of the
(i)
Gradient of the curve at the point where π₯π₯ = β3
(ii)
Area bounded by the curve, π₯π₯ = β3, π₯π₯ = β1 and π¦π¦ = β10
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3. 2017 Oct / Nov Exams, Q10(a)
The diagram below shows the graph of π¦π¦ = π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3
(a) Use the graph to find the solutions of the equations
(i)
π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 = 0
(ii)
π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 = 5
(b) Calculate an estimate of
(i)
the gradient of the curve at the point (β3,0)
(ii)
the area bounded by the curve, π₯π₯ = 0, π¦π¦ = 0 and π¦π¦ = 20
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4. 2017 July/Aug Exams, Q9(a)
The diagram below shows the graph of π¦π¦ = π₯π₯ 3 + π₯π₯ 2 β 5π₯π₯ + 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)
π₯π₯ 3 + π₯π₯ 2 β 5π₯π₯ + 3 = 0
(ii)
π₯π₯ 3 + π₯π₯ 2 β 5π₯π₯ + 3 = 5π₯π₯
(c) to calculate an estimate of the area bounded by the curve π₯π₯ = 0, π¦π¦ = 0 and π₯π₯ = β2
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5. 2016 Oct/ Nov Exams, Q9(a)
The values of x and y are connected by the equation π¦π¦ = π₯π₯(π₯π₯ β 2)(π₯π₯ + 2). Some
corresponding values of x and y are given in the table below
π₯π₯
π¦π¦
β3
β15
β2
(a) Calculate value of ππ
0
β1
3
0
0
1
β3
2
0
3
ππ
(b) Using a scale of 2cm to represent 1 unit on the x β axis for β3 β€ π₯π₯ β€ 3 and 2cm to
represent 5 units on the y- axis forβ16 β€ π¦π¦ β€ 16. Draw the graph of
π¦π¦ = π₯π₯(π₯π₯ β 2)(π₯π₯ + 2)
(c) Use your graph to solve the equations
(i)
(ii)
π₯π₯(π₯π₯ β 2)(π₯π₯ + 2) = 0
π₯π₯(π₯π₯ β 2)(π₯π₯ + 2) = π₯π₯ + 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 π₯π₯ be the number of learners and π¦π¦ 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 β€ π₯π₯ β€ 70 ππππππ 0 β€ π¦π¦ β€ 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 = 30π₯π₯ + 50π¦π¦
2. 2018 July/ Aug Exams, Q12a)
A tailor at a certain market intends to make dresses and suits for sale.
(a) Let π₯π₯ represent the number of dresses and π¦π¦ 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 π₯π₯ and π¦π¦ axes for
0 β€ π₯π₯ β€ 60 and 0 β€ π¦π¦ β€ 80. Shade the unwanted region to indicate clearly the
region where (π₯π₯, π¦π¦) 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 π₯π₯ and π¦π¦ 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 β€ π₯π₯ β€ 70 ππππππ 0 β€ π¦π¦ β€ 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 β€ π₯π₯ β€ 160 and 0 β€ π¦π¦ β€ 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 π₯π₯ represent the number of videos produced and π¦π¦ 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 β€ π₯π₯ β€ 800 ππππππ 0 β€ π¦π¦ β€ 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 < π₯π₯ β€ 20
20 < π₯π₯ β€ 30
30 < π₯π₯ β€ 40
40 < π₯π₯ β€ 50
50 < π₯π₯ β€ 60
60 < π₯π₯ β€ 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)
β€ 10
0
0
β€ 20
2
0.02
β€ 30
12
0.13
β€ 40
27
0.3
β€ 50
50
β€ 60
80
β€ 70
90
Using a scale of 2cm to represent 10 units on the x- axis for 0 β€ π₯π₯ β€ 70 and a
scale of 2cm to represent 0.1 units on the y β axis for 0 β€ π¦π¦ β€ 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 β€ π₯π₯ β€ 8 and a
scale of 2cm to represent 0. 1 unit on the y β axis for 0 β€ π¦π¦ β€ 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 < π₯π₯ β€ 5
13
5 < π₯π₯ β€ 10
27
0 < π₯π₯ β€ 15
35
15 < π₯π₯ β€ 20
16
20 < π₯π₯ β€ 25 2 25 β€ π₯π₯ β€ 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)
β€0
0
β€5
13
β€ 10
β€ 15
β€ 20
40
β€ 25
β€ 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)
semiβ 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 < π₯π₯ β€ 30
6
30 < π₯π₯ β€ 35
4
35 < π₯π₯ β€ 40
40 < π₯π₯ β€ 45
7
11
45 < π₯π₯ β€ 50
12
50 < π₯π₯ β€ 55
8
55 < π₯π₯ β€ 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)
β€ 25
β€ 30
0
5
β€ 35
9
β€ 40
16
β€ 45
27
β€ 50
β€ 55
β€ 60
Using a horizontal scale of 2cm to represent 10 newspapers on the x βaxis for
0 β€ π₯π₯ β€ 60 and a vertical scale of 4cm to represent 10 vendors on the y β
(iii)
axis for 0 β€ π¦π¦ β€ 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 < π₯π₯ β€ 10 10 < π₯π₯ β€ 20
20 < π₯π₯ β€ 30
30 < π₯π₯ β€ 40
40 < π₯π₯ β€ 50
50 < π₯π₯ β€ 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)
β€ 10
7
β€ 20
29
β€ 30
β€ 40
β€ 50
β€ 60
100
Using the scale of 2cm to represent 10 units on both axes, draw a smooth
cumulative frequency curve where 0 β€ π₯π₯ β€ 60 and 0 β€ π¦π¦ β€ 100.
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(iii)
Semiβ 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
β2
οΏ½
0
οΏ½, 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
β8 β€ π₯π₯ β€ 12 ππππππ β 6 β€ π¦π¦ β€ 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 β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
β6 β€ π₯π₯ β€ 10 and β10 β€ π¦π¦ β€ 8.
(a) A quadrilateral ABCD has vertices A(β5,7), B(β4,8), C(β3,7) and
D(β4,4) while
its imagine has vertices A1 (β5, β3), B1 (β6, β2), C1 (β5, β1) and D1 (β2, β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)
β2 0
0 1
The matrix οΏ½
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, β8), B3 is (2, β10), is C3 (0, β8)and D3 is (2. β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
β6 β€ π₯π₯ β€ 10 and β6 β€ π¦π¦ β€ 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 π΄π΄1 (3,2), π΅π΅1 (6,1), πΆπΆ1 (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 οΏ½
A2 B2 C2 D2 .
(i)
(ii)
(c)
and π·π·1 (6,3).
1
3
0
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 (β2, β4), B3 (β4, β2), C3 (β6, β4) and
D3 (β4, β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)
β3
A transformation with matrix οΏ½
0
not drawn on the diagram. Find
0
οΏ½ maps triangle ABC onto triangle A4 B4 C4
1
(i)
The scale factor of this transformation
(ii)
The coordinates of A4, B4 and C4 .
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Answers to All the Topic Questions
TOPIC1: ALGEBRA
ππβππ
1. (a)
ππ 2 βππ 2
=
=
2. (a)
βππ
7π π π‘π‘ 3
15π’π’ 3 π£π£
=
=
4. (a)
=
=
=
7×π π ×π‘π‘×π‘π‘×π‘π‘
×
15×π’π’×π’π’×π’π’
ππ
Ans
ππππππππππ
14π₯π₯ 3
9π¦π¦ 2
14π₯π₯ 3
9π¦π¦ 2
÷
×
3
28×π π ×π π ×π π ×π‘π‘×π‘π‘
2π₯π₯β5
7π₯π₯ 4
×
=
7×π₯π₯×π₯π₯×π₯π₯×π₯π₯
ππ 2 β1
ππ 2 β12
ππ (ππ β1)
(ππ +1)(ππ β1)
ππ (ππ β1)
Ans
ππ 2 ππ 3
=
=
=
4
ππ 2 ππ 3
4
3π₯π₯β9β8π₯π₯+20
(2π₯π₯β5)(π₯π₯β3)
3π₯π₯β8π₯π₯β9+20
(2π₯π₯β5)(π₯π₯β3)
ππ
βππππ+ππππ
1
(c)
π₯π₯β4
=
2(π₯π₯ 2 β22 )
8
×
8
÷ 2ππ2 ππ
4
Ans
×
8
ππ×ππ
5π₯π₯β2π₯π₯β1+8
ππππ+ππ
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= (ππβππ)(ππππβππ) Ans
3
=
×
5π₯π₯β1
1(5π₯π₯β1)β2(π₯π₯β4)
(π₯π₯β4)(5π₯π₯β1)
= (π₯π₯β4)(5π₯π₯β1)
(c)
1
2ππ 2 ππ
2
5π₯π₯β1β2π₯π₯+8
π₯π₯+2
×
β
= (π₯π₯β4)(5π₯π₯β1)
π₯π₯+2
2(π₯π₯+2)(π₯π₯β2)
ππππ
Ans
= (ππππβππ)(ππβππ) Ans
π₯π₯+2
ππ×ππ×ππ×ππ×ππ
ππ
4
π₯π₯+2
ππππ
βππβππ
(ππ+ππ)(ππβππ)
π₯π₯β3
= ππ(ππ β ππ) Ans
×
3π₯π₯β4π₯π₯β3β4
(π₯π₯+1)(π₯π₯β1)
=
2(π₯π₯ 2 β4)
=
(b)
ππ 2 βππ
β
2π₯π₯ 2 β8
=
18×π¦π¦×π¦π¦×π¦π¦
3π₯π₯β3β4π₯π₯β4
(π₯π₯+1)(π₯π₯β1)
=
9×ππ×ππ×ππ×ππ
4
π₯π₯β1
3(π₯π₯β1)β4(π₯π₯+1)
(π₯π₯+1)(π₯π₯β1)
=
10×ππ×ππ×ππ×ππ
β
3(π₯π₯β3)β4(2π₯π₯β5)
(2π₯π₯β5)(π₯π₯β3)
=
(b)
18π¦π¦ 3
×
π₯π₯+1
=
=
7π₯π₯ 4
Ans
ππ
Ans
ππππππ
5×π’π’×π’π’×π’π’×π’π’×π£π£
18π¦π¦ 3
9×π¦π¦×π¦π¦
ππππ
ππ+ππ
ππππππ
=
14×π₯π₯×π₯π₯×π₯π₯
ππ
15×ππ×ππ×ππ×ππ
(b)
28π π  3 π‘π‘ 2
10ππ 2 ππ 2
9ππ 3 ππ
3
(c)
10ππ 2 ππ 2
12 ×ππ×ππ×ππ×ππ
=
5π’π’ 3 π£π£
9ππ 3 ππ
×
15ππππ 3
=
2 ×
÷
12ππππ 3
=
Ans
ππ+ππ
=
=
15ππππ 3
β1(ππβππ)
(ππ+ππ)(ππβππ)
=
3. (a)
12ππππ 3
(b)
1
2×ππ×ππ×ππ
β
2
π₯π₯+3
3(π₯π₯+3)β2(5π₯π₯β2)
=
=
=
5π₯π₯β2
(5π₯π₯β2)(π₯π₯+3)
3π₯π₯+9β10π₯π₯+4
(5π₯π₯β2)(π₯π₯+3)
3π₯π₯β10π₯π₯+9+4
(5π₯π₯β2)(π₯π₯+3)
βππππ+ππππ
(ππππβππ)(ππ+ππ)
Ans
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5. (a)
=
π₯π₯β1
(b)
π₯π₯ 2 β1
π₯π₯β1
π₯π₯ 2 β12
π₯π₯β1
= (π₯π₯+1)(π₯π₯β1)
=
ππ
ππ+ππ
Ans
17ππ 2
20ππ 2
=
=
÷
17ππ 2
51ππ 2
5ππ
5ππ
= 20ππ 2 × 51ππ 2
17×ππ×ππ
20×ππ×ππ
ππ
ππππππ
×
Ans
2
(c)
=
5×ππ
=
51×ππ×ππ
β
1
2π₯π₯β1
3π₯π₯+1
2(3π₯π₯+1)β1(2π₯π₯β1)
(2π₯π₯β1)(3π₯π₯+1)
=
=
6π₯π₯+2β2π₯π₯+1
(2π₯π₯β1)(3π₯π₯+1)
6π₯π₯β2π₯π₯+2+1
(2π₯π₯β1)(3π₯π₯+1)
ππππ+ππ
(ππππβππ)(ππππ+ππ)
Ans
TOPIC 2: MATRICES
1. (a) deter A = (4 × 2) β (1 × β5)
(b)
= 8 β (β5)
=8+5
β΄ π«π«π«π«π«π« π¨π¨ = 13 Ans
deter π΅π΅ = (8 × 5) β (3 × π¦π¦)
13 = 40 β 3π¦π¦
13 β 40 = β3π¦π¦
β27 = β3π¦π¦ dividing both sides by 3 we have,
ππ = ππ Ans
2. (a) Deter A = (2π₯π₯ × π₯π₯) β (2 × 3)
2π₯π₯ 2 β 6 = 12
2π₯π₯ 2 = 18
π₯π₯ 2 = 9
π₯π₯ = β9
π₯π₯ = ±3
β΄ ππ = ππ Ans
ππ. (a) Deter M= (3× π₯π₯) β (5 × β2)
22 = 3π₯π₯ β (β10)
22 = 3π₯π₯ + 10
22 β 10 = 3π₯π₯
12 = 3π₯π₯
ππ = ππ Ans
8 π¦π¦
οΏ½,
3 5
8 9
οΏ½
B=οΏ½
3 5
ππ
ππ βππ
οΏ½ Ans
ππ βππ = οΏ½
ππππ βππ
ππ
π΅π΅ = οΏ½
(b) A = οΏ½
2×3
3
π¨π¨βππ
2
6 2
οΏ½=οΏ½
οΏ½
3
3 3
ππ
ππ βππ
οΏ½ Ans
= οΏ½
ππππ βππ
ππ
β2
οΏ½
4
ππ
ππ ππ
οΏ½ Ans
= οΏ½
ππππ βππ ππ
(b) ππ = οΏ½
π΄π΄βππ
3
5
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ππ
4. (a) |K|= (10 × β2) β (11 × β2)
= β20 β (β22)
= β20 + 22
β΄ ππππππππππ πΎπΎ = ππ Ans
(b) ππ βππ = οΏ½
ππ
5. (a) deter Q = (3 × 4) β (π₯π₯ × β2)
βππ
βππππ
ππ
οΏ½ Ans
ππππ
3 β2
οΏ½
β5 4
ππ ππ βππ
οΏ½ Ans
= οΏ½
ππ ππ
ππ
(b) Q = οΏ½
πΈπΈβππ
2 = 12 β (β2π₯π₯)
2 = 12 + 2π₯π₯
2 β 12 = 2π₯π₯
β10 = 2π₯π₯
ππ = βππ 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 + π₯π₯ + 3 + 7 = 22
π₯π₯ + 14 = 22
π₯π₯ = 22 β 14
ππ = ππ Ans
3. (i)
E
Ans
(ii) (a) only 1 mode of transport = 7 + 14 + 7 = ππππ
(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) 2π¦π¦ + 1 = 7
2π¦π¦ = 7 β 1
= 70 farmers
2π¦π¦ = 6 β ππ = 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
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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
6οΏ½
15
9/14
6/14
9/15
R
8/14
R
(a) P( one is black) = P(BB) + P(BR) + P(RB)
=οΏ½
=
=
=
6
15
30
210
138
210
ππππ
ππππ
×
+
5
14
54
οΏ½+οΏ½
210
Ans
+
6
15
54
210
×
9
14
RR
(b) P(different colour) = P(BR)+P(RB)
οΏ½+οΏ½
9
15
×
6
14
οΏ½
= οΏ½
=
=
=
6
15
54
210
108
210
ππππ
ππππ
×
+
9
οΏ½+οΏ½
14
54
210
9
15
×
6
14
Ans
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οΏ½
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2. (a) Total ππππ + ππππ + ππ = ππππ
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
ππ
ππππππ
3
35
Ans
3. (a) Total ππ + ππ + ππ = ππππ
(b) P(Same colour) = P(WW) + P( BB) + (YY)
= οΏ½
=
5
12
20
132
×
+
4
11
12
οΏ½+οΏ½
132
+
4
12
6
132
×
=
3
11
38
οΏ½+οΏ½
132
=
3
12
ππππ
ππππ
×
Ans
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2
11
οΏ½
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4. Faulty = 3 and good = 10 β 3 = 7
Use a tree diagram below for easy calculations
(a) P( both good) = P(G, G)
=οΏ½
=
=
7
10
56
90
ππ
ππππ
6
× οΏ½
9
Ans
(b) P (one is faulty and the one is good) = ππ(πΊπΊ, πΉπΉ) + ππ(πΉπΉ, πΊπΊ)
=
=οΏ½
7
3
× οΏ½+οΏ½
10
21 21
+
90 90
=
=
9
3
10
7
× οΏ½
9
42
90
ππ
ππππ
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)
= ππ. ππππ Ans
= 0.24 + 0.24
= 0.48
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TOPIC 5: SEQUENCES AND SERIES
1. (a) ππ + 4, ππ, 2ππ β 15
ππ
ππ+4
(b) ππ + 4, ππ, 2ππ β 15
2ππβ15
=
ππ
ππ 2 = (ππ + 4)(2ππ β 15)
(c) ππβ =
12 + 4, 12, 2(12) β 15
ππβ =
16, 12, 9,β¦ Ans
ππβ =
ππ 2 = 2ππ 2 β 15ππ + 8ππ β 60
ππβ =
2ππ 2 β ππ 2 β 7ππ β 60 = 0
ππ 2 β 7ππ β 60 = 0
ππ 2 + 5ππ β 12ππ β 60 = 0
ππ(ππ + 5) β 12(ππ + 5) = 0
(ππ + 5)(ππ β 12) = 0
ππ = β5 ππππ ππ = 12
β΄ ππ = ππππ Ans
2
ππ =
2
3β1
ππ5 =
27
3
2
27
2
27
=
=
2
9ππ 2
2ππ
9
× ππ 3
2
9ππ 2
ππ5 = 3(
in (ii) we have
ππ5 =
ππ =
ππ =
3
2
1 2
9×οΏ½ οΏ½
3
ππ = 2
243
242
81
)
1
4
4β3
4
16
1
4
)÷
)×
=2
ππ
1βππ
2
ππβ =
2
80
81
2
= ππ. ππππ Ans
1
3
1β
ππβ =
18
54
1
3
243
3
243β1
3
243
242
(c) ππβ =
54ππ = 18
ππ =
1 5
3
1
1β
3
1
2(1β 5 )
3
2
3
ππ5 = 2(
= ππππ β¦β¦β¦β¦β¦β¦β¦. Equation (ii)
Replacing ππ by
=
1βππ
ππ5 = 2(1 β
ππ4 = ππππ 4β1
2
16
16
2(1βοΏ½ οΏ½ )
ππ5 =
β¦β¦β¦β¦β¦.. equation (i)
9ππ 2
3
4
1β
12
ππ(1βππ ππ )
(b) ππππ =
= ππππ 2
9
,r=
πΊπΊβ = ππππ Ans
2. (a) ππππ = ππππ ππβ1
ππ3 = ππππ
ππ
1βππ
16
2
2
3
πΊπΊβ = ππ Ans
=
2
9×
1
9
ππ
β΄ the first term is 2 and common ratio is Ans
ππ
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3. (a) ππ =
5
1
(c) ππππ =
= = 0.25
20
4
(b) ππππ = ππππ ππβ1
ππ8 =
1 ππβ1
ππππ = 20 οΏ½ οΏ½
ππππ = 20
β΄ π»π»ππ
=
ππππ
ππππβππ
ππ8 =
Ans
10+ππ
6+ππ
=
ππ2
=
ππ1
15+ππ
10+ππ
ππ3
ππ2
=β―
(c)
ππππ
100 + 20ππ + ππ = 90 + 21n + ππ
100 β 90 = 21ππ β 20ππ
10 = ππ
β΄ ππ =10
The GP is: 16, 20, 25 . . .
16
=
5
4
ππππ ππ. ππππ
ππ
π₯π₯β3
=
1
π₯π₯β1
π₯π₯β3
S6
ππ
ππ
2
ππ β1
π₯π₯ 2 β 6π₯π₯ + 9 = π₯π₯ 2 β 1
β6π₯π₯ = β1 β 9
π₯π₯ =
3
ππ
ππ
Hence the GP is;
8 β4 2
3
,
3
, ,β¦
3
1.25β1
16(3.814697266 β1)
=
(C)
ππβ =
ππβ =
ππβ =
0.25
16(2.814697266 )
0.25
45.03515625
0.25
ππ
1βππ
8/3
8/3
16
9
3
2
8
3
8
3
β΄ πΊπΊβ = ππ
5
3
5
1
2
1β(β )
ππβ =
ππβ =
10
ππ = ππ Ans
for r > 1
ππβ1
16((1.25)6 β1)
÷
ππβ = ×
β6π₯π₯ = β10
6
5
S6
=
ππ(ππ ππ β1)
S 6 = 180.140625
β΄ S6 = ππππππAns
(π₯π₯ β 3) (π₯π₯ β 3) = ( π₯π₯ β 1) ( π₯π₯ + 1)
π₯π₯ =
ππππ =
S6 =
2
5. (a) We know that r = ππ2 = ππ3 = β― ππ ππ
π₯π₯+1
0.75
ππ6 =
ππππ β1
2
20
0.75
20(0.9999847412 )
πΊπΊππ = ππππ. ππ Ans
(10 + ππ) (10 + ππ) = (6 + ππ) (15 + ππ)
(b) ππ =
1β0.25
20 (1β0.00001558906 )
ππ8 = 26.66625977
4. (a) to find n, we use the common ratio formula
That is r=
for r < 1
1βππ
20(1β(0.25)8 )
ππ8 =
4
1ππ β1
4 ππ β1
ππ(1βππ ππ )
ππ
ππ
3
2
2
3
Ans
5
+ 1. β 3, β 1 β¦ β¦ ..
3
3
ππ
(b) the first term βππβ = Ans
ππ
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TOPIC 6: PSEUDO CODE AND FLOWCHART
1.
Start
Enter ππ, ππ
M = (ππ^2+ππ^2)
Is
M< 0
yes
Error βMβ must be
positive
No
Display
M
Stop
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2.
Start
Enter ππ, ππ, ππ
πΉπΉ = ππ β ππ
Is
πΉπΉ = ππ?
ππππ
πΊπΊππ =
yes
value of r
is not valid
ππ(ππβππππ )
ππβππ
Display πΊπΊππ
Stop
3. Start
Then display βerror message β
1
Else Area = β ππ β ππ β π π π π π π π π
2
End if
Display Area
Stop
4. Start
Enter a and r
If |ππ| < 1
Then sum to infinity =
Else
End if
Display sum to infinity
Stop
ππ
1βππ
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5.
Start
Enter, r
Is
yes
ππ < 0?
Error
βr must be positiveβ
No
Enter h
Is h < 0?
Yes
Error
βh must be Positiveβ
No
ππ
ππ = ππ β ππ β ππ β ππ
ππ
Display Volume
Stop
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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
ππππ°
10 cm
b(i)
Q
D
3.
8.9cm
C
11cm
P
QP
PPPPP
7 cm
AA
60°
10cm
ππππππ°
B
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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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B
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TOPIC 8: CALCULUS
2
1. (a) β«β1(2 + π₯π₯ β π₯π₯ 2 )ππππ
= οΏ½2π₯π₯ +
=οΏ½
2(2)
+
1
π₯π₯ 2
2
22
2
β
β
8
π₯π₯ 3
3
23
3
2
οΏ½ β1
m1 =
οΏ½ β οΏ½2(β1) β
1
1
= οΏ½4 + 2 β οΏ½ β οΏ½2 β + οΏ½
10
3
β7
2
=οΏ½ οΏ½βοΏ½ οΏ½
=
=
10
3
+
3
27
6
6
7
6
4
(b) π¦π¦ = π₯π₯ + β
(β1)2
3
2
β
(β1)3
3
π₯π₯
ππππ
=1β
ππππ
R
3
=οΏ½ β
3
1
2(1)2
2
3
0
3
0
2
1
= β4β0
=
16
3
1β12
=
3
4
4
3
ππ
ππππ
ππππ
P
Multiplying through by
ππππ = βππππ + ππππ Ans
= 4π₯π₯ β 3 ππππ (3,7) β
1
ππ2 = β1 × = β
9
1
9
ππππ
ππππ
= ππ1 = 4(3) β 3 = 9
β΄ equation of the normal to the curve is given by
ππ β ππππ = ππ2 (π₯π₯ β π₯π₯1 )
1
π¦π¦ β 7 = β (π₯π₯ β 3)
3
11
=β
16
+5
3
ππππ
(b) π¦π¦ = 2π₯π₯ 2 β 3π₯π₯ β 2
β (3)οΏ½ β οΏ½ β + 3(0)οΏ½
= οΏ½ β 1 β 3οΏ½ β 0
12
ππ β ππππ = ππ2 (π₯π₯ β π₯π₯1 )
4
π¦π¦ β 5 = β (π₯π₯ β 4)
3
2. (a) β«0 (π₯π₯ 2 β 2π₯π₯ β 3)ππππ
13
=
4
P
β 3π₯π₯οΏ½ 10
16
β΄ Equation of the normal is given by
3
2
16β4
4
ππ
2π₯π₯ 2
=
at π₯π₯ = 4
π¦π¦ = 4 + = 4 + 1 = 5
ππ = β ππ +
β
4
16
4
π₯π₯ 2
To find y replace π₯π₯ = 4 in the original equat
3
ππ
3
=1β
4
4
π₯π₯ 3
4
42
=1β
m 2 = β1 × = β
π¦π¦ = β π₯π₯ +
=οΏ½
ππππ
m 1 × m 2 = β1 (tangent β₯ to normal)
οΏ½
= 4. ππ Ans
1
ππππ
9
1
π¦π¦ β 7 = β π₯π₯ +
3
ππ
= βππ Ans
1
9
3
9
1
9
66
π¦π¦ = β π₯π₯ + +
ππ
9
1
π¦π¦ = β π₯π₯ +
π¦π¦ = β π₯π₯ +
9
7
9
1
3+63
3
9
9
ππππ = βππ + ππππ Ans
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starts with one step
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3. (a) π¦π¦ = 2π₯π₯ 3 β 3π₯π₯ 2 β 36π₯π₯ β 3
ππππ
= 6π₯π₯ 2 β 6π₯π₯ β 36
ππππ
3
(b) β«β1(3π₯π₯ 2 β 2π₯π₯)ππππ
=οΏ½
0 = 6π₯π₯ 2 β 6π₯π₯ β 36 dividing through by 6
3π₯π₯ 3
3
3
2π₯π₯ 2
οΏ½ 3
2 β1
3
β π₯π₯ 2 ] β1
3
2]
β
= [π₯π₯
= [(3) β (3)
π₯π₯ 2 β π₯π₯ β 6 = 0
β [(β1)3 β (β1)2 ]
= (27 β 9) β (β1 β 1)
=18 β (β2)
= 18+2 = 20
(π₯π₯ + 2) (π₯π₯ β 3) = 0
π₯π₯ = β2 ππππ π₯π₯ = 3
When π₯π₯ = β2
3
β΄ β«β1(3π₯π₯ 2 β 2π₯π₯)ππππ = 20 Ans
π¦π¦ = β16 β 12 + 72 β 3
π¦π¦ = 41
When π₯π₯ = 3
π¦π¦ = 2(β2)3 β 3(β2)2 β 36(β2) β 3
π¦π¦ = 2(3)3 β 3(3)2 β 36 (3) β 3
π¦π¦ = β84
β΄ the coordinates on curve are (β ππ, ππππ) and (ππ, βππππ) Ans
5
4. (a) β«2 (3π₯π₯ 2 + 2)ππππ
3π₯π₯ 2+1
=οΏ½
2+1
3π₯π₯ 3
+
π¦π¦ = π₯π₯2 β 3π₯π₯ β 4
(b)
2π₯π₯ 0+1 5
οΏ½
0+1 2
2π₯π₯
= οΏ½ + οΏ½ 52
3
1
= [π₯π₯ 3 + 2π₯π₯] 52
3
ππ =
π¦π¦ β π¦π¦1 = ππ(π₯π₯ β π₯π₯1 )
π¦π¦ β (β6) = 1(π₯π₯ β 2)
= (135) β (12)
= ππππππAns
ππππ
ππππ
π¦π¦ + 6 = π₯π₯ β 2
ππ = ππ + ππ Ans
(b) At the stationary points,
2
2
= 3π₯π₯ 2 β 3π₯π₯ = 3(2) β 3(2)= 6
1
6
Which is the gradient of the normal?
π¦π¦ β π¦π¦1 = ππ2 (π₯π₯ β π₯π₯1 ) (2,2)
1
1
6
1
6
π¦π¦ = β π₯π₯ +
6
1
π¦π¦ = β π₯π₯ +
6
2+12
6
14
6
Compiled and Solved by Kachama Dickson C & Chansa John
2
For π₯π₯ = 1
2
=0
1
2
β΄ the stationary points are;
β Multiplying throughout by 6, we get
ππππ = βππ + ππππ Ans
ππππ
3
π¦π¦ = (0) 3 - (0)2 = 0
3
π¦π¦ =β π₯π₯ + + 2
ππππ
3π₯π₯(π₯π₯ β 1) = 0
π₯π₯ = 0 ππππ π₯π₯ = 1
For π₯π₯ = 0.
π¦π¦ = (1)3 β (1)2 = β
π¦π¦ β 2 = β (π₯π₯ β 2)
6
2
3π₯π₯ β3π₯π₯ = 0
P
β΄ ππ1 = 6 which is the gradient of tangent to the curve
π€π€π€π€ know that tangent is perpendicular to the normal,
ππ1 ππ2 = β1, at π₯π₯ = 2, π¦π¦ = 2
ππ2 = β
= 2π₯π₯ β 3 at π₯π₯ = 2, ππ = 2(2) β 3 = 1
β΄ equation of the tangent is given by
3
= (125 + 10) β (8 + 4)
3
ππππ
π¦π¦ = (2)2 β 3(2) β 4 = β6
= ( (5) + 2(5) ) β (2 + 2(2))
5. (a) π¦π¦ = π₯π₯ 3 β 2 π₯π₯ 2
ππππ
ππ
(0, 0) and (1, β ) Ans
ππ
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TOPIC 9: VECTORS
οΏ½οΏ½οΏ½οΏ½οΏ½β = 1 π΄π΄π΄π΄
οΏ½οΏ½οΏ½οΏ½οΏ½β
1. (i) (a) π΄π΄π΄π΄
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½β = ( ππ β ππ )
(ii) Since π΅π΅π΅π΅
3
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½β = π΄π΄π΄π΄
οΏ½οΏ½οΏ½οΏ½οΏ½β + π΅π΅π΅π΅
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ = ππ + 2ππ
U
U
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½β = ( ππ + 2ππ ) Ans
β΄ π¨π¨π¨π¨
ππ
U
U
(b) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ + οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄
1
οΏ½οΏ½οΏ½οΏ½οΏ½β = -ππ + ( ππ + 2ππ )
π΅π΅π΅π΅
U
3
U
1
U
2
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = βππ + ππ + ππ
3
β3ππ+ππ
U
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ =
3
2
U
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ β ππ
and π΅π΅π΅π΅
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
U
ππ
U
3
U
2
U
+ ππ
3
2
(c)
οΏ½οΏ½οΏ½οΏ½οΏ½β + οΏ½οΏ½οΏ½οΏ½οΏ½β
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = βπ΄π΄π΄π΄
π΄π΄π΄π΄
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = β ππ + ππ
U
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
U
U
οΏ½οΏ½οΏ½οΏ½οΏ½β = β ππ + ππ
π΅π΅π΅π΅
2
3
2
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = b β ππ
β΄ π΅π΅π΅π΅
3
3
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ ( ππ β ππ ) Ans
ππ
U
U
οΏ½οΏ½οΏ½οΏ½οΏ½β = PO
οΏ½οΏ½οΏ½οΏ½οΏ½β + OQ
οΏ½οΏ½οΏ½οΏ½οΏ½β
2. (i) (a) PQ
οΏ½οΏ½οΏ½οΏ½οΏ½β = β 2p + 4q
PQ
οΏ½οΏ½οΏ½οΏ½οΏ½β = 4q β 2p
PQ
οΏ½οΏ½οΏ½οΏ½οΏ½β = 2 (2q βp) Ans
β΄ ππππ
(b)
(c)
οΏ½οΏ½οΏ½οΏ½οΏ½β
οΏ½οΏ½οΏ½οΏ½β = 1 PQ
PX
3
οΏ½οΏ½οΏ½οΏ½β = ππ (4q β2p) Ans
PX
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½β
οΏ½οΏ½οΏ½οΏ½οΏ½β + οΏ½οΏ½οΏ½οΏ½β
OX = OP
PX
4
2
οΏ½οΏ½οΏ½οΏ½οΏ½β
OX = 2p + q β p
3
οΏ½οΏ½οΏ½οΏ½οΏ½β
οΏ½οΏ½οΏ½οΏ½οΏ½β
OC = hOX
οΏ½οΏ½οΏ½οΏ½οΏ½β = h (4p + 4q )
OC
β΄
3
3
4h
4h
οΏ½οΏ½οΏ½οΏ½οΏ½β
OC = p + q
3
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = CO
οΏ½οΏ½οΏ½οΏ½οΏ½β + OQ
οΏ½οΏ½οΏ½οΏ½οΏ½β
CQ
4h
οΏ½οΏ½οΏ½οΏ½οΏ½β = β ( P + 4h q) + 4q
CQ
3
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = 4q β 4h q β 4h p
CQ
3
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ(1β π‘π‘) q β ππππ p hence shown
ππππ
ππ
ππ
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = 2p β 2 pβ 4 q
OX
3
6pβ2p
4q
οΏ½οΏ½οΏ½οΏ½οΏ½β
OX =
β
3
3
3
οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ p β ππ q
ππππ
ππ
ππ
Compiled and Solved by Kachama Dickson C & Chansa John / Together we can do Mathematics /0966-295655
Page 54 of 84
Get your own copy direct from the Authors
3.
(a)
οΏ½οΏ½οΏ½οΏ½οΏ½β = 1 CA
οΏ½οΏ½οΏ½οΏ½οΏ½β
(b) CD
οΏ½οΏ½οΏ½οΏ½οΏ½β = OA
οΏ½οΏ½οΏ½οΏ½οΏ½β + AB
οΏ½οΏ½οΏ½οΏ½οΏ½β
OB
2
οΏ½οΏ½οΏ½οΏ½οΏ½β = 1 (β2ππ +ππ )
CD
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ + 2ππ
ππππ
U
οΏ½οΏ½οΏ½οΏ½οΏ½β and
(b) To find οΏ½οΏ½οΏ½οΏ½οΏ½β
OE, first find CA
οΏ½οΏ½οΏ½οΏ½οΏ½β
πΆπΆπΆπΆ
οΏ½οΏ½οΏ½οΏ½οΏ½β = CO
οΏ½οΏ½οΏ½οΏ½οΏ½β + OA
οΏ½οΏ½οΏ½οΏ½οΏ½β
CA
οΏ½οΏ½οΏ½οΏ½οΏ½β = β2b + a
πΆπΆπΆπΆ
3
οΏ½οΏ½οΏ½οΏ½οΏ½β
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ = β πΆπΆπΆπΆ
4
2
U
U
1
= β ππ + ππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β =
πͺπͺπͺπͺ
ππ
ππ
U
2
U
ππ β ππ Ans
U
οΏ½οΏ½οΏ½οΏ½οΏ½β = β 3 ( ππ β2ππ )
π΄π΄π΄π΄
4
U
U
οΏ½οΏ½οΏ½οΏ½οΏ½β + AE
οΏ½οΏ½οΏ½οΏ½οΏ½β
β΄ οΏ½οΏ½οΏ½οΏ½οΏ½β
OE = OA
3
οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = ππ β ( ππ β2ππ )
4
U
U
3
U
6
οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = ππ β ππ + ππ
4
4ππβ3ππ
U
οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ =
1
4
3
+ ππ
4
3
οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = ππ + ππ
4
ππ
U
4
U
4
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = (ππ + 3ππ ) Ans
πΆπΆπΆπΆ
ππ
U
U
4. (i) (a) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ = οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ + οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ = β3a + 6b
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄ = 6b β3a
β΄ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
π¨π¨π¨π¨ = 3 (2b βa) Ans
(b) οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ + οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄
1
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ + οΏ½οΏ½οΏ½οΏ½οΏ½β
π΄π΄π΄π΄
3
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = 3a + 2b β a
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = 2a + 2b
ππππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = 2(ππ + ππ ) Ans
πΆπΆπΆπΆ
U
οΏ½οΏ½οΏ½οΏ½οΏ½β
(ii) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = βπ΅π΅π΅π΅
οΏ½οΏ½οΏ½οΏ½οΏ½β = β(6 a β 6b)
π΅π΅π΅π΅
3
1
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ = 3a + (6b β3a)
U
(c) οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ + οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ
2
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β
βππππ + οΏ½οΏ½οΏ½οΏ½οΏ½β
ππππ
5
2
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = β 6b + (3a)
5
6ππ
οΏ½οΏ½οΏ½οΏ½οΏ½β
π΅π΅π΅π΅ = β 6b +
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β =
ππ
ππ
5
a β 6 b Ans
5
ππ
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½β = ππ( a β hb) Ans
ππ
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TOPIC 10: TRIGONOMETRY
ππ
1. (a)(i)
π π π π π π π π
ππ
π π π π π π  60
=
ππ =
=
ππ
(ii) Area of βπΎπΎπΎπΎπΎπΎ
π π π π π π π π
80
1
A = ππππ π π π π π π π π
π π π π π π  52
80π π π π π π  60
2
(b) Shortest distance from R to KN
S .d =
2
A = 2000 π π π π π π 60°
A = 1732.050808
β΄ π¨π¨ = ππππππππππππ Ans
ππ = 87.92016097
β΄ ππππ = ππππ. ππ π¦π¦ Ans
Sd =
1
A = (50)(80) sin 60°
π π π π π π  52
2π΄π΄
ππ
ππππππππ
ππππ
= ππππ. ππππ Ans
(c) Graph of π¦π¦ = cos ππ
0°
1
ππ
cos ππ
90°
0
1
0°
90°
180°
β1
270°
0
360°
1
ππ = ππππππ π½π½
180°
270°
360°
β1
2. (a) (i)
ππ
π π π π π π π π
ππ
=
=
ππ
π π π π π π π π
15
π π π π π π  79
π π π π π π  40
15 π π π π π π  79
ππ =
π π π π π π  40
ππ = 22.9
AC = 22.9km Ans
(b) ππππππππ = 0.937
ππ = (ππππππ β1 (0.937)
ππ = 20.4°
(ii) first find angle BAC
Angle BAC = 180 β (79 + 40) = 61
1
π΄π΄ = ππππ π π π π π π π π
2
1
(iii) S.d =
2×150
22.9
= 13.100
= 13.1 Ans
π΄π΄ = (15)(22.9)π π π π π π 61°
2
π΄π΄ = 150.2159349
π¨π¨ = ππππππππππππ Ans
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ππ = 360° β 20.4) = 339.4
β΄ π½π½ = ππππ. ππ ππππππ π½π½ = ππππππ. ππ Ans
(c) ππ = ππππππ π½π½
0°
0
ππ
Sin ππ
90°
1
180°
0
270°
-1
360°
0
1
0
90
180
270
360
π¦π¦ = sin ππ
β1
1
3. (a) (i) A = π‘π‘π‘π‘ sin π»π»
2
(iii) Shortest distance =
1
=
π΄π΄ = × 1.3 × 1.9 π π π π π π 130°
2
A = 1.235 sin 130°
A = 0.946
A = ππ.946 km2 Ans
(ii) β2 = π‘π‘ 2 + π π  2 β 2π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘
h = 5.3 β ( β3.175370792)
ππ =
2
h = 5.3 + 3.17537
h2 = 8.475370792
P
(ii)
οΏ½ = 180° β (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 β1 οΏ½ οΏ½
3
P
2
sin R
2
ππ = 48.1896851
β΄ ππ = 48.2° Ans
P
β΄
2.9
(b) cos ππ =
2
4. (a) (i) to find PQ, first find angle R
ππ
2×0.95
= 0.65517
S.d = ππ. ππππππ π€π€π€π€ Ans
h2 = (1.3)2 + (1.9)2 β 2(1.3)(1.9)cos130°
h = β8.475370792
h = 2.911249009
β΄ ππππ = ππ. ππππππππ Ans
2π΄π΄
= 916.15 sin 36°
β΄ A = 538.4994589 β ππππππ. ππ π€π€π¦π¦ππ Ans
(iii) Shortest distance =
(b) sinππ = 0.6792
2π΄π΄
ππ
=
2×538..2
50.2
= ππππ. ππ π€π€π€π€ A
ππ = sinβ1 (0.6792) β press shift / 2nd f and then sin
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r = 50.24716343
β΄ ππππ = ππππ. ππππππ Ans
5. (a) (i) k2 = i2 + m2 β 2imcos K
2
2
2
k = 5 + 3 β2(5)(3) cos110°
2=
k 34 β (- 10.2606043)
k2 = 44.26060643
ππ = 42.8° and ππ = 180 β 42.8 = 137.2°
β΄ π½π½ = ππππ. ππ°, ππππππ. ππ° Ans
1
(ii) A = × ππ × ππ π π π π π π π π
1
2
(iii) shortest d =
A = (5)(3 )π π π π π π 110°
2
A = 7.5 sin110°
A = 7.047694656
A = ππ. πππππππ¦π¦ππ Ans
k = β44.26060643
k = 6.656864368
β΄ ππππ = ππ. ππππππππ Ans
=
2π΄π΄
ππ
2×7.05
6.65
= ππ. ππππππππ Ans
(b) tanππ = 0.7
ππ = tanβ1 (0.7)
ππ = 34.9920202
ππ = 34.9920202
ππ = ππππ° Ans
TOPIC 11: MENSURATION
1.
π»π»
π»π»β11.4
=
14
8
14( H β 11.4) = 8H
14H β 159.6 = 8H
14H β 8H = 159.6
2.
6H = 159.6
H = 26.6
h = 26.6 β 11.4 = 15.2 cm
π»π»
π»π»β9
=
10
4
10(H β 9) = 4H
10Hβ90 = 4H
10Hβ4H = 90
6H = 90
H = 15
β΄ π»π» = 15ππππ ππππππ β = 15 β 9 = 6ππππ
1
β΄ ππ = (πΏπΏπΏπΏπΏπΏ β ππππβ)
3
1
ππ = (14 × 10 × 26.6 β 8 × 4 × 15.6)
3
1
ππ = (3724 β 486.4)
ππ =
3
1
3
(3237.6)
V = 1079.2
V = 1080 ππππππ Ans
1
β΄ V = [πΏπΏ2 π»π» β ππ 2 β] (square base)
3
1
V = [102 × 15 β 42 × 6]
3
1
V = (1500 β 96)
1
3
V = (1404)
3
V = 468 ππππππ Ans
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3. First find the height of the small cone that was
7
=
20+β
V=
21β = 140 + 7β
(13230 β 490)
3
3.142
3
(212 × 30 β 72 × 10)
(12740)
1
1
V= ππππ2 π»π» β ππππ 2 β
12
3
1
3
V = ππ(ππ π»π» β ππ 2 β)
12π₯π₯ = 4(15 +π₯π₯)
3
1
2
V = × 3.142(122 × 22.5 β 42 × 7.5)
12π₯π₯ = 60 +4π₯π₯
3
12π₯π₯ β4π₯π₯ =60
V=
8π₯π₯ = 60
V=
π₯π₯ = 7.5
β΄ ππππ = ππ. ππ And AD = 7.5 +15 = 22.5cm
1cm3 β 1000ππ
10.8cm3β π₯π₯
π₯π₯ = 10.8 × 1000ππ
π₯π₯ = 10800ππ
β΄ π½π½ = ππππππππππππ (3 sig fig)
3
3.142
(ii) Volume that remained is the that of the frustum
15+π₯π₯
5. (a) V = ππππβ
V= 1.2× 0.9 × 10
V= 10.8cm3
3
V = 13343.02667
V = ππππππππππππππππ Ans
4. (i) Let height EA = π₯π₯, then
=
1
3
3.142
V=
21β β7β = 140
14β= 140
β= 10
ππ = ππππ ππππ and H = 10 + 20 = 30cm
4
1
V=
21
21β = 7(20+β)
π₯π₯
3
V = ππ(ππ 2 π»π» β ππ 2 β)
Cutoff.
β
1
β΄ ππ = ππππ2 π»π» β ππππ 2 β
3.142
3
3.142
3
(3240 β 120)
(3120)
V = 3267.68
V = 3270 ππππππ (correct to 3 sig figures)
(b) first find the radius of a cone
r2= 132 β 122
r2 = 25
r = β25
r = 5cm
β΄ ππ. ππ. π΄π΄ = ππππ 2 + ππππππ
T.S.A = ππππ(ππ + ππ)
T.S.A = 3.142 × 5(5 + 13)
T. S. A = 282.78cππππ
12
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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°ππ
40°E
SA
S
ππ
AC =
AC =
AC =
360
50°
× 2ππππ where ππ = 15° + 35° = 50°
× 2 × 3.142 × 6370
360°
2001454
360
AC = 5,559.594444
AC = 5,560km Ans
(b) (i) BQ =
900 =
πΌπΌ
(ii) longitude of Q = 70° β 9.9° = 60.1°
× 2ππππππππππππ
360
2ππππππππππππππ
β΄ position of Q (ππππ°πΊπΊ, ππππ. ππ°π¬π¬) Ans
360
2ππππππππππππππ = 900 × 360°
πΌπΌ =
πΌπΌ =
32400
2ππππππππππππ
32400
(2×3.142×6370 ×cos 35°)
πΌπΌ = 9.88109
β΄ the difference in longitude is 9.9°
2. (a) D =
D=
D=
ππ
× 2ππππ where ππ = 60° + 60° = 120°
360
120
× 2 × 3.142 × 3437
360
2591772 .96
or D = ππ × 60
D = 120° × 60
D = ππππππππ nm Ans
360
D = 7199.369333
β΄ ππππππππππππππππ ππππ = 7200nm Ans
(b)
To find speed, first find distance CD
D=
D=
D=
πΆπΆ
ππππππ
120
× ππππππππππππππ
× 2 × ππ × 3437 cos 60
360
ππππππππππππππ.ππππ
ππππππ
β΄ speed =
π«π«
speed =
π»π»
3600
12
speed = 300knots Ans
D = 3599.684667
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D = 3600nm
3. (a) Difference in latitudes between W and Y
π½π½ = 80° + 30° = ππππππ°
(b) (i) XZ =
ππ
XZ =
(ii) YZ =
× 2ππππ
360
110°
360°
× 2 × 3.142 × 3437
XZ = 6599.421889
XZ = 6600nm Ans
QR =
165°
360°
ππ
360°
YZ =
× 2ππππππππππππ where πΌπΌ = 15° + 105 = 120°
× 2 × 3.142 × 3437 × ππππππ30°
360°
2244541 .224
360
YZ = 6234.836734
YZ =6230nm Ans
4. (i)
(ii)(a) Distance QR =
YZ =
πΌπΌ
360°
120°
× 2ππππ ππ = 80° + 85° = 165°
× 2 × 3.142 × 3437
QR = 9899.132833
QR =9900 nm Ans
5. (a) Difference in latitudes
ΞΈ = 50° + 70° = ππππππ°
(b) C = 2ππππππππππππ
C = 21600 Cos 50°
C=13884.21237nm
C= 13900nm
(b) C = 2ππππππππππππ
C = 21600 Cos ππ
C = 21600× ππππππ 85°
C = 1882.564043
C =1900nm Ans
ππ
× 2ππππ
360°
120°
360°
× 2 × 3.142 × 3437
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1.
(ii) From part (i) we can see that
2. (a) (i) π¦π¦ = π₯π₯ 3 + π₯π₯ 2 β 12π₯π₯ β (π₯π₯ 3 + π₯π₯ 2 β 12π₯π₯)
3
2
3
2
π¦π¦ = π₯π₯ + 10
π¦π¦ = π₯π₯ + π₯π₯ β 12π₯π₯ β π₯π₯ β π₯π₯ + 12π₯π₯
π¦π¦ = 0 β π₯π₯ β ππππππππ
when we draw this line on the
β΄ ππ = βππ, ππ = ππ and ππ = ππ
same graph using any points we have
Note that the values of π₯π₯ are the points where
ππ = βππ. ππ ± ππ, ππ = βππ. ππ ± and
3
2
the curve π¦π¦ = π₯π₯ + π₯π₯ β 12π₯π₯ meets the line π¦π¦ = 0
ππ = ππ. ππ ± ππ
ππ
β3
0
2
3
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7
ππ
2. (b)(i) Draw a straight line touching only
at ( -3, 18) and pick any two points
lying on the same line
ππ =
ππ =
12
13
(ii) From the graph, we can see that the
required area is simply the areas
of the two trapeziums
e. g (β3, 18 ) and (β4.5,4)
ππ =
10
π¦π¦2 βπ¦π¦1
π₯π₯ 2 βπ₯π₯ 1
4β18
β4.5β(β3)
β14
β1.5
ππ = ππ. ππ ± ππ
1
A = (ππ + ππ)β
2
1
2
1
2
1
A = (58) + (52)
2
2
A = 29 + 26
ππ = ππππ ± ππ Square units
3. (a) (i) π¦π¦ = π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 β (π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3)
π¦π¦ = π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 β π₯π₯ 3 β 3π₯π₯ 2 + π₯π₯ + 3
π¦π¦ = 0
On the graph, when π¦π¦ = 0, (π₯π₯ β ππππππππ)
ππ = βππ, ππ = βππ and ππ = ππ
(ii) π¦π¦ = π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 β (π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 5)
π¦π¦ = π₯π₯ 3 + 3π₯π₯ 2 β π₯π₯ β 3 β π₯π₯ 3 β 3π₯π₯ 2 + π₯π₯ + 5)
π¦π¦ = β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 (β3,0) and pick any
two points lying on this line
for Example (β3,0) and (β2,8)
ππ =
ππ =
π¦π¦2 βπ¦π¦1
π₯π₯ 2 βπ₯π₯ 1
8β0
β2β(β3)
ππ
ππ = = ππ Ans
ππ
(ii) Area = A of rectangle + trapezium
π¦π¦ = 2
On the graph when y=2, then
ππ = βππ. ππ, ππ = βππ. ππ and ππ = ππ. ππ Ans
1
A = l× ππ + (ππ + ππ)β
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, β7) and use the
π¦π¦2 β π¦π¦1
ππ =
π₯π₯2 β π₯π₯1
β7 β 5
ππ =
1β2
β12
ππ =
β1
ππ = ππππ Ans
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4. (b) (i) ππ = π₯π₯3 + π₯π₯2 β 5π₯π₯ + 3 β (π₯π₯3 + π₯π₯2 β 5π₯π₯ + 3)
ππ = π₯π₯3 + π₯π₯2 β 5π₯π₯ + 3 β π₯π₯3 β π₯π₯2 + 5π₯π₯ β 3
π¦π¦ = 0
On the graph, when π¦π¦ = 0(π₯π₯ β ππππππππ)
ππ = βππ and ππ = ππ 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 = (ππ + ππ)ππ
2
1
(ii) similarly we have π¦π¦=5π₯π₯
Use any points to draw the
line π¦π¦ = 5π₯π₯ and find the
values of x where it meets
the curve for example use
the points in this table
below
π₯π₯ -1
0
1
5
π¦π¦ -5
0
5
10
β΄ ππ = βππ. ππ, ππ = 0.3 and ππ=ππ. ππ
1
A= (9 + 8)1 + (8 + 3)
2
1
1
2
A= (17) + (11)
2
2
A= 8.5 + 5.5
A= ππππ Square units
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TOPIC 14: LINEAR PROGRAMING
1.
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3.
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4.
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5.
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TOPIC 15: STATISTICS
1. (a)
class
ππ
ππππ
ππ
10< π₯π₯ β€ 20
15
225
20 < π₯π₯ β€ 30
25
625
10
250
6250
30 < π₯π₯ β€ 40
35
1225
15
525
18375
40< π₯π₯ β€ 50
45
2025
23
1035
46575
50 < π₯π₯ β€ 60
55
3025
30
1650
90750
60 < π₯π₯ β€ 70
65
4225
10
1650
42250
Totals
Mean π₯π₯Μ =
=
=οΏ½
οΏ½ ππ = ππππ
30
οΏ½ ππππ = ππππππππ
450
ππππππ
οΏ½ ππππππ = ππππππππππππ
β ππππ
β ππ
4140
90
= 46
β ππππ 2
SD = οΏ½ β
2
ππππ
ππππ
β (π₯π₯)2
204650
90
β (46)2
= β2273.888889 β 2116
= β157.8888889
SD = 12.57 Ans
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2. (a)
2
ππ
ππππ
4
3
9
4
ππ
1
2
ππππ
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
οΏ½ ππ = ππππ
οΏ½) =
Mean ( ππ
β ππππ
SD = οΏ½ β
ππ
60
οΏ½ ππππππ = ππππππππ
β ππππ
β ππ
=
ππππππ
ππππ
= ππ. ππππ
β (π₯π₯Μ )2
2559
=οΏ½
οΏ½ ππππ = ππππππ
ππππππ
β (6.32)2
= β42.65 β 39.9424
= β2.7076
= 1.645478654
= ππ. ππππ Ans
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Class marks
0 < π₯π₯ β€ 5
5 < π₯π₯ β€ 10
10 < π₯π₯ β€ 15
15 < π₯π₯ β€ 20
20 < π₯π₯ β€ 25
25 < π₯π₯ β€ 30
οΏ½=
ππ
=
ππ
2.5
7.5
12.5
17.5
22.5
27.5
β ππππ
13
27
35
16
7
2
ππ
οΏ½ ππ = 100
ππππ
32.5
205.5
437.5
280
157.5
55
οΏ½ ππππ = 1168
οΏ½)ππ
(ππ β ππ
12.25
17.2225
0.7225
34.2225
117.7225
251.2225
οΏ½)ππ
ππ(ππ β ππ
159.25
465.0
25.288
547.58
824.058
502.445
οΏ½ ππ(π₯π₯ β π₯π₯Μ )2
= 2523.621
β ππ
1165
100
= ππππ. ππππ
SD = οΏ½
β ππ(ππβππ
οΏ½)ππ
β ππ
ππππππππ.ππππππ
SD = οΏ½
ππππππ
SD = βππππ. ππππππππππ
SD = 5 .023565467
SD = ππ Ans
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Class marks
Midpoints
(π₯π₯)
Frequency
(ππ)
30 < π₯π₯ β€ 35
32.5
4
40 < π₯π₯ β€ 45
42.5
25 < π₯π₯ β€ 30
35 < π₯π₯ β€ 40
45 < π₯π₯ β€ 50
55 < π₯π₯ β€ 55
55 < π₯π₯ β€ 60
Totals
οΏ½) =
Mean (ππ
47.5
52.5
57.5
ππππ 2
5
137.5
756.25
3781.25
7
262.5
1406.25
9843.75
11
12
8
1
οΏ½ ππ = 48
130
467.5
570
420
57.5
οΏ½ ππππ = 2045
1056.25
1806.25
2256.25
2256.25
3306.25
4225
19868.75
27075
22050
3306.25
οΏ½ πππ₯π₯ 2 = 90150
β ππππ
β ππ
ππππ
= 42 .6
β fx 2
βf
90150
=οΏ½
37.5
π₯π₯ 2
ππππππππ
=
SD = οΏ½οΏ½
27.5
ππππ
48
β (xοΏ½ )2 οΏ½
β (42.6)2
= β1878.125 β 1814.76
= β63.365
=ππ. ππππ ππππππ
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Class marks
0 < π₯π₯ β€ 10
5
30 < π₯π₯ β€ 40
35
10 < π₯π₯ β€ 20
20 < π₯π₯ β€ 30
40 < π₯π₯ β€ 50
50 < π₯π₯ β€ 60
Mean, π₯π₯Μ =
=
15
ππ
25
45
55
7
22
ππ
28
23
15
5
οΏ½ ππ = ππππππ
35
330
ππππ
700
805
675
275
οΏ½ ππππ = ππππππππ
25
ππππ
225
625
1225
2025
3025
fππππ
175
4950
17500
28175
30375
15125
οΏ½ ππ ππππ = ππππππππππ
β ππππ
β ππ
2820
100
= 28.2
β΄ ππππ = οΏ½
β ππ π₯π₯ 2
β (π₯π₯Μ )2
β ππ
96300
SD = οΏ½
100
β (28.2)2
SD = β963 β 795.24
SD = β167.76
SD = 12.95221989
SD = ππππ. 95 Ans
(b)
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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 οΏ½
ππ
ππ
ππ
οΏ½, then pick any two coordinates of P which corresponds to V and
ππ
form four equations as follows:
οΏ½
ππ
ππ
ππ 2
οΏ½οΏ½
ππ 1
4
β4
οΏ½=οΏ½
1
4
β8
οΏ½
1
ππ + 4ππ = β4 β¦β¦β¦β¦β¦β¦β¦β¦.(i)
ππ + ππ = β8 β¦β¦β¦β¦β¦β¦β¦β¦(ii)
2ππ + ππ = 1β¦β¦β¦β¦β¦β¦.(iii)
4ππ + ππ = 1β¦β¦β¦β¦β¦β¦β¦.(iv)
Solving these equations (i) and (ii) simultaneously, we have ππ = β2 ππππππ ππ = 0
Similarly solving equations (iii) and (iv) we have ππ = 0 and ππ = 1
β΄ the required matrix is
οΏ½
ππ
ππ
βππ
ππ
οΏ½=οΏ½
ππ
ππ
ππ
οΏ½
ππ
(d) To find the coordinates of S, we need multiply the given matrix by the coordinates of P.
1 0 2 2 4
οΏ½οΏ½
οΏ½
β2 1 1 4 1
2+0
2+0
4+0
οΏ½
οΏ½
β4 + 1 β4 + 4 β8 + 1
ππ ππ ππ
οΏ½
οΏ½
βππ ππ βππ
οΏ½
There the coordinates of S are (ππ, βππ), (ππ, ππ)and (ππ, βππ)
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2.
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3.
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4.
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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.
β΄ the centre of enlargement is (1, 2) Ans
(ii)
Scale factor =
ππ
ππ
ππ 1
οΏ½οΏ½
ππ 4
ππ
ππ
= ππ Ans
ππ
οΏ½, then
ππ
1 3
3
οΏ½=οΏ½
οΏ½
1 5
4
(b) Let the matrix be οΏ½
οΏ½
3.2
1.6
ππ + 4ππ = 1 β¦β¦β¦β¦β¦.(i)
ππ + 4ππ = 1 β¦β¦β¦β¦β¦β¦β¦..(iii)
3ππ + 4ππ = 3 β¦β¦β¦β¦β¦.(ii)
3ππ + 4ππ = 5β¦β¦β¦β¦β¦β¦β¦..(iv)
Solving the equations (i) and (ii) simultaneously yields ππ = 1 ππππππ ππ = 0.
Similarly solving equations (iii) and (iv) simultaneously yields ππ = 2 and ππ = β
β΄ the required matrix is οΏ½
ππ
ππ
ππ ππ
ππ
οΏ½ = οΏ½ππ β πποΏ½ Ans
ππ
ππ
1
4
(c) Triangle ABC is mapped onto triangle ππ ππ ππππ ππππ 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 οΏ½
ππ
0
β3
0
οΏ½ and οΏ½
0
1
0
οΏ½, we have ππ = β3
1
β΄ the scale factor of this transformation is ππ = βππ 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)
β3 0 1 3 1
οΏ½οΏ½
οΏ½
0 1 4 4 5
β3 + 0 β9 + 0 β3 + 0
οΏ½
οΏ½
0+4
0+4
0+5
β3 β9 β3
οΏ½
οΏ½
4
4
5
οΏ½
β΄ the coordinates of ππ ππ, ππππ and ππππ are (βππ, ππ), (βππ, ππ) and (βππ, ππ)
respectively
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