TOPIC 1: SETS
Definition: A set is a collection of well-defined objects.
1. Given that Set A has 5 elements and B has 128 subsets.
(a) Find the number of subsets of A
(b) Find the number of elements of Set B.
Solutions
a) Number of subsets i s given by 2n where n is the number of elements in a given set.
Number of subset
= 2n
A has 5 elements
= 25
= 32, A has 32 subsets
b) No. of subset = 2n
128 = 2n
27 = 2n
n = 7, B has 7 elements
2. If E = { Natural numbers less than 13}
P = {x: x is a prime number}
O = {x: x is an old number}
S = x: x is a square number}
List Sets E, P, O and S and hence find the following:
(a) P′
(b) (P ∩ O)′
(c) (P ∪ S)′
(d) (P ∪ S ∪ O)′
Solutions
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
P = {2, 3, 5, 7, 11}
O = {1, 3, 5, 7, 9, 11}
S = {1, 4, 9}
a) P′ elements in the universal set that are not in P ∴ P′ = {1, 4, 6, 8, 9, 10, 12}
b) (P ∩ O)′ elements in the universal set that are not in P ∩ O = {3, 5, 7, 11}
∴ (P ∩ O)′ = {1, 2, 4, 6, 8, 9, 10, 12}
c) P ∪ S elements of both Sets P ∪ S = {1, 2, 4, 5, 7, 9, 11} and common elements should be listed once.
d) First list P ∪ S ∪ O = {1, 2, 3, 4, 5, 7, 9, 11} then list elements of universal that are not P ∪ S ∪ O ∴
(P ∪ S ∪ O)′ = {6, 8, 10, 12}
3. The diagram below shows three intersecting sets A, B and C. It is given that n (A) = 50, n (B) = 42
and n (C) = 62
a) By considering Sets B and C, show that
a + b = 38 – x
(i)
E
B
a
+
2b
=
58=
x
(ii)
A
b) Hence or otherwise, find the value of b
S
a
b
c) Given that
S = b + 10 find the values of S, a and x.
4
x
a
b
2b
C
Solutions
a) From Sets B, a + b + 4 + x = 42 and from C,
a + 2b + 4 + 4 + x = 62
a + b = 42 – 4 – x
a + 2b = 62 – 4 – x
a + b = 38 – x
a + 2b = 58 – x
∴ a + b = 38 – x
(i)
a + 2b = 58 – x (ii) Hence shown
1
b) Value of b can be solved by showing the two equations above simultaneously.
π + π = 38 − π₯
−(
)
π + 2π = 58 − π₯
−π = −20
∴ π = 20
Value a from Set A
Value of x from Set B
S + 2a + 4 = 50
a + b + x + 4 = 42
30 + 4 + 2a = 50
8 + 20 + x + 4 = 42
34 + 2a = 50
x = 42 – 8 – 20 – 4
2a = 56 – 34
x = 10
2π 16
= π
π
a=8
∴ a = 8, s = 30 and x = 10
4. At Hillcrest Technical Secondary School, a group of 70 take optional subjects as illustrated in the Venn
diagram below.
c) S = b + 10
S = 20 + 10
S = 30
E
History
Commerce
RE
x
3x – 5
2x
9
(i)
(ii)
Calculate the value of x
Find the value of Girls who take
a) History only
b) Commerce
Solutions
(i)
3x – 5 + x + 2x + 9 = 70
3x + x + 2x – 5 + 9 = 70
6x + 4 = 70
6x = 70 – 4
6π₯ 66
=
6
6
Value of x = 11
(ii) (a) History only = 3x – 5
3 (11) – 5
33 – 5
28 History only
(b) Commerce = x + 2x + 9
= 11 + 2(11) + 9
= 11 + 22 + 9
= 42
5. In the diagram below shows three Sets A, B and C
C
A
10
B
X
3
Give that n (A ∪ B ∪ πΆ) = 50 ππππ
(i) The value of x
(ii) n (A ∪ B)
(iii) n (B ∪ πΆ)′
(iv) n (A′ ∩ C′)
C
3X
5 i.
4
Solution
(i)
10 + x + 3 + 3x + 5 + 4 = 50
10 + 3 + 5 + 4 + x + 3x = 50
2
22 + 4x = 50
4x = 50 – 22
4π₯ 28
= 4
4
x=7
(ii)
∩ (A∪ π΅) = 10 + π₯ + 3 + 3π₯
= 10 + 7 + 3 +3(7)
= 10 + 7 + 3 + 21
= 41
(iii)
(B ∪ πΆ)′ = 10 + 4
= 10 + 4
= 14
(iv)
∩ (A′ ∩ C′) = 3 + 4
=7
6. On the Venn diagram below, shade the Set (A′ ∩ B) ∩ C
Solution
E
Hint
A
B
Shade A′ and B to give the region common to (A′ and
B)
Shade C to give the region common to (A′ ∩ B) ∩ C
C
7. A survey was conducted on 60 women connecting the types of SIM cards used in their cell phones for the
past 2 years. Their responses are given in the diagram below.
E
Cell Z
a) Given that 23 women have used Cell Z SIM
cards, find the values of a and b
b) How many women have used only two different
SIM cards?
c)
Airtel
a
4
14
2
3
b
10
MTN
d) If a woman is selected at random from the group, what is the probability that
i.
She has no cell phone
ii.
She used only type of a SIM card
e) How many women did not use MTN and Cell Z SIM Cards?
f) How many women used either Airtel or MTN SIM Cards but not Cell Z.?
Solutions
a) a + 4 + 2 + 3 = 23
a + 9 = 23
a = 23 – 9
a = 14
b) a + b + 4 + 2 + 3 + 10 + 14 + 8 = 60
3
14 + b + 4 + 2 + 3 + 10 + 14 + 8 = 60
b + 55
= 60
b
= 60 – 55
b=5
∴ π = 14 πππ π = 5
8
2
c) (i) No cell phone =
60 15
(iii)
Only one type of SIM card = a + 14 + 10
= 14 + 14 + 10
= 38
38
19
∴ ππππππππππ‘π¦ = 60 = 30
d) (Cell Z ∪ πππ)′ = 14 + 8
= 22
e) 10 + 5 + 2 + 3 + 14 + 4 = 38 used either Airtel or MTN but not Cell Z
8. If P = {-2, -1, 0, 1, 2,3, 4 …….}, express
a) P in set builder notation
P = {x: x ≥ - 2; x ∈ π) (The inequality ≥ is used since elements are building
from 2 (inclusive) to the positive side).
b) Write Set A in listed form,
A = (x : 2 ≤ x < 8
x < 8, π₯ ≥ 2
(By splitting the set builder notation)
(Using the number line)
2
3
4
5
6
7
8
∴ π΄ = {2, 3, 4, 5, 6,7}
(2 is included in the solution set when 8 is not included)
9. Using the information given in the Venn diagram below
E
A
.k
B
.f
.b
.m
.c
.g
a) List the Set (A ∩ B)
(A∩B) = {f, m}
b) List (A ∩B)′
(A∩B)′ = {k, b, c, g}
(Complement of a set are elements outside the given
Sets (A∩B)
10. E (Universal Set) = {2, 4, 6, 8, 10, 12, 14, 18, 20}
A = {2, 4, 6. 8}
B = {4, 8, 12, 14} C = {2, 4, 12, 18, 20}
i.
Illustrate the above information on a Venn diagram.
E
A
.6
B
.8
.2
4
.14
12
4
18
ii.
20
List the Set
(A ∩ B ∩ C) = {4}
iii.
C
(The intersection of three sets)
(A ∩ B′) ∪ πΆ
A ∩ B′ = {2, 4, 6, 8} ∩ {2, 6, 10, 16, 18, 20)
(Dealing with what is inside the
brackets by listing Set A and Set B′)
(A ∩ B′) = {2, 6}
Set C = {2, 4, 12, 18, 20}
∴ (π΄ ∩ π΅′ ) ∪ πΆ = {2, 6} ∪ {2, 4, 12, 18, 20}
(A ∩ B′) ∪ πΆ = {2, 4, 6, 12, 18, 20}
(Do not repeat elements in the union of sets)
iv.
Find the value of ∩ ( A ∪ π΅ ∪ πΆ)′
(A∪ π΅ ∪ πΆ)′ = {10, 16}
∴ ∩ (A∪ π΅ ∪ πΆ)′ = 2
v.
By listing
(There are two elements in that Set)
In the Venn diagram, shade the region
(A′∪ π΅) ∩ πΆ
(A′ ∪ π΅) = {12, 14, 18, 20, 10, 16) ∪ {4, 8, 12, 14}
(A′ ∪ π΅) = {4, 8, 10, 12, 14, 16, 18, 20}
(A′ ∪ π΅) ∩ πΆ = {4, 8, 10, 12, 14, 16, 18, 20} ∩ {2, 4, 12, 18, 20}
(A′ ∪ π΅) ∩ πΆ = {4, 12, 18, 20}
Shade the region where these elements are lying.
11. 70 learners at Mansa Secondary School were asked to mention their favorite subjects between
Mathematics and Science.
The results are shown in a Venn diagram below
E
Maths
25
i.
ii.
Science
30
10
How many learners like Mathematics only? 25 learners
How many learners do not like Mathematics nor Science?
70 – (25 + 30 + 10)
70 – (65)
70 – 65
= 5 learners do not like Mathematics nor Science
5
TOPC 2: INDEX NOTATION
Important Rules in indices
1.ππ × ππ = ππ+π
2. ππ ÷ ππ = ππ−π
π
3. π−π = ππ
4.ππ = π
5.(ππ )π = πππ
π
π
6. π π = √ππ or ( π√π)π
π
π
7.( π )−π = ( π )π
Solutions
1
Q1 a) Evaluate
5 −2
(3)
1
2
9
b) Simplify (π‘ 6 )
b)
(3)2 2
=[(π‘ 3 )2 ]
c)
2π₯ 3 π¦
6π₯π¦ 2
=
2π₯ 3−1
3π¦ 2−1
a)
b)
9π₯π¦ 6 2
[π₯ 3 π¦ 2 ]
=
3 2
(5)
1
=
b) Simplify
π
ππ
ππ
ππ
=
9π¦ 4 2
[ π₯2 ]
9π₯π¦ 6 2
[π₯ 3 π¦ 2 ]
Solution
π
ππ
1
π
a) 4−2 = 42 = ππ
1
1
=
=
1
Solution
5 −2
(3)
3 2 2
=[( 3 ) ]
π‘
Q2 a) Evaluate 4−2
2π₯ 3 π¦
6π₯π¦ 2
c) Simplify
1
1
9 2
( 6)
π‘
=
2 2
3π¦ 2
[( π₯ ) ]
=
c)6−2 =
πππ
π
1
62
2 −2
=
π
ππ
1
a) 30
Q4 a) Evaluate 50 − 5−1
b)5−1
b) Simplify(5π₯ 3 )2
c)6−2
=
5 2
( )
2
=
ππ
π
Q3 Work out the value of
d)(5)
16
1
c) Simplify [π16 ]2
2 −2
d)(5)
Solution
0
a) 3 = 1
b) 5−1 =
=
π
b) (5π₯ 3 )2 = ππππ
1
1
16
c) [π16 ]2
1
a) 50 − 5−1 = 1 − 5 = π
π
π
4 2 2
⌈(π8 ) ⌉
whole population are children? Give your answer
in its simplest form
b) Simplify 25π₯ 2 ÷ 5π₯ −4
Solution
π
= ππ
Q5a) the population of a country is 3.2 X 106.
There are 8 x 105 children. What fraction of the
8×105
a) 3.2×106 =
b) 25π₯ 2 ÷ 5π₯ −4 = 5π₯ 2−(−4) = πππ
Q6 Evaluate
a) 170
5
8×105
32×105
8
π
= 32 = π
5
b)42 = (22 )2 = (2)5 = ππ
2 −2
c)(0.2)−2 = (10)
5
2
b)4
c)(0.2)−2
Solution
a) 170 = 1
1 −2
Q7 a) Evaluate (4)
2
b) Simplify 643
6
10 2
= ( 2 ) = (5)2 = ππ
1
c)
a) π2 (π3 − 3π−2 ) = π2+3 − 3π−2+2 = ππ − π
4π₯ 2 π¦ 9 2
Simplify[ π₯ 4 π¦ ]
Solution
1 −2
a) (4)
2
1
4 2
= (1) = ππ
2
b) 643 = (43 )3 = 42 = ππ
1
4π₯ 2 π¦ 9 2
c) [ π₯ 4 π¦ ]
1
=
1
1
8×
4π¦ 8 2
42 π¦ 2
πππ
[ π₯ 2 ] = 2×1 = π
π₯ 2
2 (π3
−2 )
Q8 a) simplify π
1
3
1
1
b) (27π₯ 6 )3 = [33 (π₯ 2 )3 ]3 =[(3π₯ 2 )3 ]3 = πππ
Q9 a) Find the value of a when 3π ÷ 34 = 32
b) Find the value of b when 82 = 2
Solution
3 π ÷ 34 = 32
3 π = 34 × 32
∴π =2+4=π
b) 8π = 2
π
(2)3b = (2)1 ∴ 3b = 1π =
π
− 3π
b) Simplify (27π₯ 6 )
Solution
Q10
Q 11
7
PRACTICE QUESTIONS:
(1) Evaluate 42 + 41 + 40.
(2) (a) Find the value of 2n− π2 .
(i)
when n = 0,
(ii)
When n = 3.
1
(b) Find the value of 3 -2
(3) Find a, b and c when
(a) 3π ÷ 35 = 27,
(b) 125π = 5,
(c) 10π = 0.001.
ANSWERS:
(1)
21
(2) (a) (i) 1
(b) 9
(3) (a) 8
1
(b) 3
(c) -3
(ii) -1
8
TOPIC 3: ALGEBRA
1. Solve the equation
7
1
+
=4
x + 2 x −1
2. Peter cuts a square out of a rectangular piece of metal.
2x + 3
Diagram NOT
accurately drawn
x+2
x+4
x+2
The length of the rectangle is 2x + 3.
The width of the rectangle is x + 4.
The length of the side of the square is x + 2.
All measurements are in centimeters.
The shaded shape in the diagram shows the metal remaining.
The area of the shaded shape is 20 cm2.
(a)
Show that x2 + 7x – 12 = 0
(b)
(i)
Solve the equation x2 + 7x – 12 = 0
Give your answers correct to 4 significant figures.
(ii)
Hence, find the perimeter of the square.
Give your answer correct to 3 significant figures.
7r + 2 = 5(r – 4)
3. Solve
4. Simplify fully
(i)
(ii)
(p3)3
3q 4 ο΄ 2q 5
q3
5. The force, F, between two magnets is inversely proportional to the square of the distance, x,
between them.
When x = 3, F = 4.
9
6.
(a)
Find an expression for F in terms of x.
(b)
Calculate F when x = 2.
(c)
Calculate x when F = 64.
40 – x
=4+x
3
(a)
Solve
(b)
Simplify fully
4x 2 – 6x
4x 2 – 9
7. A van can carry a maximum load of 400 kg. It carries boxes weighing 20 kg and 40 kg. It
carries at least 7 boxes weighing 40 kg. The number of boxes weighing 40 kg is not more
than twice the number of 20 kg boxes.
Let x represent the number of 20 kg boxes and y the number of 40 kg boxes.
a) Write down three inequalities involving x and y.
b) Illustrate the three inequalities by a suitable diagram on graph paper. Let 2 cm represent 1
box on both axes.
c) From the diagram determine the least weight the van carries.
d) What combinations give the greatest weight?
8. The dimensions of a rectangle are such that its perimeter is greater than 20 metres and less
than 30 metres. One side must be greater than the other. The larger side must be less than
twice the size of the smaller side.
Let x represent the length of the smaller side and y the length of the larger one.
a) Write down four inequalities involving x and y.
b) On graph paper, illustrate these inequalities using a scale of 2 cm to represent 2 metres on
each axis, clearly showing the area containing the solution.
c) What whole number dimensions will satisfy these three inequalities?
9. Given that 3π =
5π−4π
2π−3π
, express q in terms of p and r. Find the value of q when p =2 and
r = -5.
10. The surface area of a solid cone is given by the formula
A = ππ 2 + πππ.
(i) Factorize fully the expression A = ππ 2 + πππ.
(ii) Rearrange the formula to express l in terms of π, r and A.
11. The formula used in connection with the mirror is
1
π’
(i)
(ii)
1
1
+π£ = π
Given that v = 9, and f = 5, find u.
Express v in terms of u and f.
10
TOPIC 4: MATRICES
1. Given that 936
, find
(a)
(b)
Solutions
(a)
(b)
2. Given that
(i)
(ii)
(iii)
(i)
, and
, find
The inverse of matrix A.
3A – B
AB
Solutions
Determinant of A = ( 5 x 0) – (2 x 1 )
= 0–2
=-2
=
(ii)
3A –B
=
=
=
(iii)
AB =
=
2
3. If P= (1
8
4 1
1 0
3 1)and Q =(2 1
2 6
1 0
4
1), evaluate PQ.
1
11
2 4 1 1
PQ = (1 3 1) (2
8 2 6 1
0 4
1 1)
0 1
2+8+1 0+4+0 8+4+1
=(1 + 6 + 1 0 + 3 + 0 4 + 3 + 1 )
8 + 4 + 4 0 + 2 + 0 32 + 3 + 6
11 4 13
=( 8 3 8 )
16 2 41
4. Solve the following simultaneous equations using the matrix method.
SOLUTION
3 2 π₯
3
(
) (π¦) = ( )
2 3
7
Find the determinant of the matrix of coefficient of
Det = (3 x 3) – (2x2)
=9–4
=5
Find inverse of matrix of coefficient of
Inverse =
)
( )
5.
If matrix A
, is a singular matrix , find the value of
Solution
A singular matrix have the determinant equal to zero
(
=0
- 24 = 0
12
2 0
π 0
6. If P= (
) and Q= (
), find
6 1
1 π
(a) PQ
(b) The values of π and π, given that PQ = P – Q
SOLUTION
2 0 π
(a) PQ = (
)(
6 1 1
2π + 0
=(
6π + 1
0
)
π
0+0
)
0+π
2π
0
=(
)
6π + 1 π
(b) PQ = P - Q
2π
(
6π + 1
0
2
)=(
π
6
2π
(
6π + 1
0
2−π
)=(
π
5
0
π
)−(
1
1
0
)
π
0
)
1−π
2π = 2 − π
6π + 1 = 5
π =1−π
3π = 2
6π = 4
2π = 1
3π
3
2
=3
2
π=3
6π
6
4
=6
2
π=3
2π
2
1
=2
1
π=2
ACTIVITY
π₯−2 π₯
1. Given that the determinant of matrix A = (
) is 4,
2
1
(a) Find the value of π₯,
(b) Write the inverse of matrix A
3 7
2. Given that A= (
), find the
2 5
(a) The determinant of A
(b) Inverse of A,
−2
(c) The value of π΄−1 ( )
1
1 π₯
3. Given that matrix A= (
),
−1 2
(a) Write an expression in terms of π₯, for the determinant of A.
(b) Find the value ofπ₯, given that the determinant of A is 5.
(c) Write π΄−1
−1 3
4. Given that M=(
), find
1 2
(a) π2
(b) π−1
π 2
5. Given that matrix Q= (
),
3 −2
(a) Write an expression in terms of π for the determinant of Q,
13
(b) Find the value of π, given that the determinant of Q is 2,
(c) Write π −1
SOLUTIONS FOR THE ACTIVITY
1. (a) π₯ = −6
−1
=
(b)π΄
−1
( 61
3
1
1)
13
2. (a) |π΄| = 1
5 −7
(b) π΄−1 = (
)
−2 3
3. (a) |π΄| = 2 + π₯
(b) π₯ = 3
−1
(c) π΄
=
2
(51
5
−3
5
1)
5
4 3
4. (a) π2 = (
)
1 7
(b) π
−1
=
−2
( 51
5
3
5
1)
5
5. (a) −2π − 6
(b) π = −4
(c) π
−1
=
1
(23
4
1
2
1
)
14
TOPIC 5: SIMILARITY AND CONGRUENCE
1. SIMILARITY
Two objects are said to be similar if:
i.
The corresponding angles are equal
ii.
The ratio of the corresponding sides is the same or equal.
1.1 SIMILARITY IN TRIANGLES
For two triangles to be similar, they need to satisfy any of the three cases:
i.
Three pairs of corresponding angles are equal (AAA)
ii.
The ratio of corresponding sides is the same (SSS)
iii.
Two pairs of corresponding sides are proportional and the included angles are equal (SAS).
2. CONGRUENCY
Two objects are congruent if they have the same shape and size.
QUESTIONS
1) State the two triangles in the diagram below which are similar. Give the reason why.
Answer: ΔABC and ΔADE are similar.
Since DE and BC are parallel, Λπ΄π΅πΆ = Λπ΄π·πΈ πππ Λπ΄πΆπ΅ = Λπ΄πΈπ·
(Corresponding Angles). ΛA is common to both triangles, hence satisfying AAA.
2) Determine whether or not the two rectangles below are similar
Solution
2:3 ≠ 3:5
Therefore XYWZ and RSTQ are not similar.
a) In the diagram, DE is parallel to AB, DE = 3cm, AB = 9cm and CD = 4cm
b) Find the ratio of corresponding sides.
c) Find the value of x
Solution
a) 3: 9 = 1: 3
3
4
b) =
9 4+π₯
3(4+x) = 9×4
12 + 3x = 36
3x = 24
15
X = 8cm
3) Name all the pairs of congruent triangles in the figure below:
Solution
ΔEOF = ΔGOF
ΔEOH = ΔGOH
ΔFEH = ΔFGH
4) Show that ΔABC and ΔPQR below are congruent
Answer:
ΛABC = ΛPQR,
ΛBAC = ΛRPQ = 30Λ and
ΛBCA = ΛPRQ = 40Λ
5) Find the length of YU in the diagram below
Solution:
ΛUVY = ΛXVZ (Vertically opposite angles)
UV = XV
YV = VZ
Therefore UY = XZ = 7cm.
6) A wall, which is 4m high, is built next to a street light that is 8m high. The shadow of the wall is 5m long.
How far is the wall from the street light?
16
Solution
4
8
5
= 5+π₯
4(5+x) = 8×5
20+ 4x = 40
X = 5m
NOTE: If two figures are similar and the lengths of their corresponding sides are in the ratio
π,
then the ratio of their area is ππ βΆ ππ ,
and the ratio of their volumes is ππ βΆ ππ .
PRACTICE QUESTIONS:
(1) A model of a tanker is made using a scale of 1: 20.
(a) The length of the tanker is 15 m. Calculate the length, in centimetres, of the model.
(b) The model holds 12 litres of liquid. Calculate the number of litres the tanker will hold.
(2) (a)
Are triangles ABC and DEF similar? Explain your answer clearly.
(b)
Triangle LMN and PQR are similar. Calculate the value of x.
(3) In the diagram, ABCD is a quadrilateral with BA parallel to CD.
AC and BD meet at X where CX = 8 cm and XA = 10 cm.
17
πβΆ
(a) Given that BD =27 cm , find the length BX.
(b) Find the ratio area of triangle BXC: area of triangle AXD.
(4) Two pots are geometrically similar. The height of the smaller pot is 5 cm.
The height of the bigger pot is 15 cm.
(a) The diameter of the base of the larger pot is 7 cm. Find the diameter of the base of the smaller pot.
(b) Find the ratio of the volume of the smaller pot to that of the larger. Give your answer in the form 1:
n.
(5) The ratio of the areas of the bases of two geometrically similar buckets is 4: 9.
(a) The area of the top of the smaller bucket is 480 cm2.
What is the area of the top of the larger bucket?
(b) Write down the ratio of the heights of the two buckets.
(c) Both buckets are filled with sand. The mass of sand in the larger bucket is 36 kg.
Find the mass of sand in the smaller bucket.
ANSWERS:
(1) (a) 75cm ,
(b) 96000 litres
(2) (a) not similar, because not all the 3 corresponding sides are proportional.
(b) 7
(3) (a) 15 cm,
(b) 1: 1
1
(4) (a) 23 cm,
(b) 1: 27
(5) (a) 1080 cm2,
(b)
2: 3,
(c)
18
2
3
10 kg.
TOPIC 6: KINEMATICS- TRAVEL GRAPHS
Problems involving distance, time, speed (velocity) and acceleration are given the name of
kinematics (kinema= motion)
UNITS USED:
• Distance travelled- (metres) or (m)
• Time taken- (seconds) or (s)
• Velocity- (metres/second) or (m/s)
• Acceleration- (metres/second/second) or (m/s2 )
NOTE:
πππ π‘ππππ π‘πππ£πππππ
• Velocity(speed)=
m/s
π‘πππ π‘ππππ
π£ππππππ‘π¦
• Acceleration =
m/s2
π‘πππ
GRADIENT GRAPHS
If motion of an object is given in a graph by a straight line then the object travels at
constant speed or uniform speed determined by the gradient of the line.
If the graph is a curve, then the object concerned has different speeds at each instant.
The gradient of the tangent to the curve at that point gives the speed of the object.
TRAVEL GRAPHS
(i)
DISTANCE TIME GRAPH
Constant rate of change = constant speed.
Varying rate of change = varying speed
19
EXAMPLE 1:
The distance time graph a shows an object starting from a point O, travelling 15m in
2 s, is stationary for another 2s and finally travels back to O in 1s.
Graph can be interpreted as follows:
The total distance travelled is 30 m (going + returning)
15π
From O to A: constant speed = 2π = 7.5 m/s
From A to B: The car is stationary, speed = 0 m/s
From B to O: The speed is constant =
15π
1π
= 15 m/s
(ii) SPEED (VELOCITY) TIME GRAPH
Hints:
• If the graph is a straight line parallel to the time axis (x-axis), then the object is said to travel
at constant speed, i.e. there is no acceleration.
• The total distance travelled in t sec is given by the area under speed time graph for that time.
NOTE:
• From O to A, Speed is constantly changing, hence there is constant (uniform) acceleration.
• From A to B, Speed is constant (not changing), hence there is no acceleration.
• From B to C, Speed is decreasing uniformly, hence there is deceleration.
EXAMPLE 2
The diagram shows the velocity - time graph of a particle during a period of t seconds.
20
Calculate
a) the acceleration of the particle in the first 10 seconds,
b) the value of t, if it travelled 50m from the 20π‘β second,
c) the average speed of the particle for the whole journey.
EXPECTED ANSWER:
(a) π =
π£−π’
π‘
, π΅π’π‘ π’ = 30π/π , π£ = 10π/π , π‘πππ π‘ = 10π
10 − 30
π=
2
π = −ππ/ππ
(b) Distance traveled is equal to the area under the graph.
1
π΄πππ = × π × β
2
1
50 = × (π‘ − 20) × 10
2
50 = 5π‘ − 100
5π‘ = 50 + 100
5π‘ 150
=
5
5
∴ π‘ = πππ
π‘ππ‘ππ πππ π‘ππππ π‘πππ£πππππ
(c) ππ£πππππ π ππππ =
π‘ππ‘ππ π‘πππ π‘ππππ
π·ππ π‘ππππ = ππππ π’ππππ π‘βπ ππππβ
∴ π· = ππππ π΄ + ππππ π΅ + ππππ πΆ.
1
1
π· = × 10 × 20 + 10 × 20 + × 10 × 10
2
2
π· = 100 + 200 + 50
π· = 350π πππ π‘πππ π‘ = 30π
350π
35
π
∴ π΄π£. πππππ = 30π = 3 π/π = ππ π π/π.
EXAMPLE 3
The diagram below is the speed-time graph of a bus which leaves a bus stop and accelerates uniformly for 10
seconds over a distance of 100m.
It then maintains the speed it has attained for 30 seconds and finally retards uniformly to rest at the next bus
stop. The whole jouney takes t seconds.
21
If the two bus stops are 1 kilometre apart, find
(i)
the value of V,
(ii)
the acceleration in the first 10 seconds,
(iii)
the total time (t) taken for the whole journey.
EXPECTED ANSWER
(i)
distance= ππππ π’ππππ π‘βπ ππππβ
1
= × πππ π × βπππβπ‘
2
1
100π = × 10π × π π/π
2
100
∴ π = 5 m/s
∴ π = ππm/s
π ππππ
(ππ) πππππππππ‘πππ =
m/s2
π‘πππ
20π/π
∴ πππππππππ‘πππ =
10π
∴ πππππππππ‘πππ = π m/s2
(πππ) πππ‘π: 1ππ = 1000π,
1
∴ π‘ππ‘ππ πππ π‘ππππ = (π + π)β
2
∴ π‘ =100−40
ππ’π‘ πππ π‘ = 1000π; π = 40 − 10 = 30π ; π = π‘; β =
1
∴ 1000 = (40 + π‘)20
2
∴ 100 = (40 + π‘)
20π
,
π
∴ π‘ = ππππππππ
π
∴ π‘ππ‘ππ π‘πππ π‘ππππ = ππππππππ
π
LEARNER’S ACTIVITY:
The diagram below is the speed-time graph of a particle. The particle accelerates uniformly from a speed of
vm/s to a speed of 5v m/s in 20 seconds.
(a) Find the expression in terms of v, for acceleration.
(b) The distance travelled by the object from 0 seconds to 20 seconds is 80m. Find the value of v.
(c) Find the speed at t = 15seconds.
22
EXPECTED ANSWERS
π£−π’
(a) acceleration= π‘
but u=v and v= 5v
∴Acceleration=
ππ−π
ππ
m/s2=
π
ππ
π/π2
ππ
∴Acceleration= π π π/π2
1
(π)π‘ππ‘ππ πππ π‘ππππ = (π + π)β
2
But π=v, b=5v, h=20 and dist=80m
1
∴ 80 = (π£ + 5π£)20
2
∴ 6v=8
8
∴V= 6 π/π
π
∴ π = π/π
π
π
(π )
8
4
∴acceleration= π × 6 π/π 2 = 15 π/π 2 ,
π−π
π
4
πππ‘ π ππππ ππ‘ π‘ = 15 π ππ = π¦= final velocity, πππ π£ = π/π be initial velocity.
∴ π‘π ππππ π ππππ ππ‘ π‘ = 15π , π€π π’π π ππππππππππππ =
3
4
∴
π−
4
3
2
π/π
=
15
ππ
4
∴ 15 × 4 = 15 × π¦ − 15 ×
3
4
∴4=π¦−
3
4
∴π¦ =4+
3
ππ
∴ π¦ = m/s = speed at t =15s
π
LEARNER ACTIVITY:
The diagram represents the speed-time graph of a moving object.
(i)
(ii)
(iii)
Calculate the speed of the object when t= 4s.
Calculate the distance travelled in the first 15 seconds.
Given that the rate at which the object slows down after t=15 is equal to half the rate
at which the object accelerates during the first 6 seconds, calculate the time at which
It stops.
EXPECTED ANSWERS:
(i)
(ii)
speed at t= 4s is equal to 32m/s,
distance travelled= 528m,
23
(iii)
time at which it stops= t=35s.
PRACTICE QUESTION:
(Q) The diagram is the speed- time graph of a car which is uniformly retarded from u m/s to 20 m/s in 10
seconds. The car is then uniformly retarded at a different rate until it finally comes to rest after a further 40
seconds.
Calculate
(a) the speed of the car after 20 seconds,
(b) the retardation during the final 40 seconds of its motion,
(c) the value of u, if the distance travelled in the first 10 seconds is 275 metres.
ANSWERS
(a) 15 m/s,
(b)
1⁄ m/s2,
2
(c)
35 m/s.
TOPIC 7: SOCIAL AND COMMERCIAL ARITHMETIC
UNIT
NO.
3.1
SUB-TOPIC
SPECIFIC OUTCOMES
Profit and Loss
3.2
Simple and Compound Interest
3.3
Commission
3.4
3.5
Discount
Hire Purchase
βΊCalculate profit and loss in a
given business transaction
βΊCalculate simple and
compound interest.
βΊFind the total earnings by basic
salary plus straight commission.
βΊCalculate the discount
βΊDefine hire purchase.
βΊCalculate hire purchase price
3.6
Investments
3.6.1Shares and Dividends
3.6.2 Bonds
3.7
βΊ Calculate the value of the
shares.
βΊ Calculate dividends as per
investments
βΊCalculate the value of the bond
βΊGiven the exchange rates,
calculate the amount requested.
Foreign Exchange
3.1 PROFIT AND LOSS
Business transactions can either be in form of goods or services. A trader may buy goods either directly from
the manufacturer or a wholesaler. He or she thereafter retails (sells) them to his or her customers.
24
During the transactions, a business person may incur either profits or losses for the goods sold. If a business
person sells his or her goods at a higher price than the price he ordered at, he or she is said to have made a
profit. However, if for any reason, the cost price of goods is higher than the selling price then a loss is said to
have been incurred on the transaction.
βΊ Income – Cost Price = +(ππππππ‘)
βΊπΌπππππ − πΆππ π‘ πππππ = −(πΏππ π )
Example 3.1.1
Isaac bought 200 mathematical sets at K8.00 each. He sold half of them at K10.00 and the rest at K9.00. Did
he make a profit or loss?
Solution
πΆππ π‘ πππππ = πΎ8.00 × 200 = πΎ1,600.00
πππ‘ππ πΌπππππ = (100 × πΎ10.00) + (100 × πΎ9.00)
πΎ1000.00 + πΎ900.00
πΎ1900.00 (π΄ πππ ππ‘ππ£π π πππ’π‘πππ). πβπππππππ, πΌπ πππ ππππ π ππππππ‘ ππ πΎ300.00.
Furthermore, it is also important to note that:
π΄ππ‘π’ππ πππ π
× 100%
πΆππ π‘ πππππ
π΄ππ‘π’ππ ππππππ‘
πππππππ‘πππ ππππππ‘ =
× 100%
πΆππ π‘ πππππ
πππππππ‘πππ πΏππ π =
Example 3.1.2
A businessman buys a house at πΎ150,000 = 00 and later sells it at πΎ175,000 = 00. Find his percentage
profit.
SOLUTION
π΄ππ‘π’ππ ππππππ‘ = πΎ175,000 − πΎ150,000 = πΎ25,000.
π»ππππ, πππππππ‘πππ ππππππ‘ =
25,000
× 100%
150,000
= 16.6666667%
≅ ππ. π% (π πππππππππππ πππππππ)
3.2 Simple and Compound Interest
Interest is a charge for borrowing a sum of money called the principal. For a simple interest, the lender earns it
at once while compound interest is accrued and accumulated each time the period of maturity passes. It is an
interest charged on an interest.
ππ
π
βΊ ππππππ πΌππ‘ππππ π‘ (π. πΌ) = 100 π€βπππ π = πππππππππ, π
= πππ‘π πππ π = π‘πππ
βΊ Compound Interest calculated as
π΄ = π (1 +
π
π
)
100
π€βπππ π΄ ππ π‘βπ ππππ’ππ‘ ππ‘ π‘βπ πππ ππ π π¦ππππ , π ππ π‘βπ πππππππππ, π
ππ π‘βπ πππ‘π
πππππππ‘ πππ π¦πππ πππ π ππ π‘βπ ππππππ ππ π¦ππππ .
Example 3.2.1
25
1
Mr. Kaloba borrowed πΎ12,000.00 from a Micro-finance lending institution at a rate of 9 % for four months.
2
How much does he have to repay?
SOLUTION
π΄πππ’ππ‘ πππππ¦ππππ = πππππππππ + πΌππ‘ππππ π‘
9.5
4 12,000
πΎ12,000 + (
×
×
) = πΎ12, 380.00
100 12
1
Example 3.2.2
Mrs. Kobili invested a sum of πΎ400.00 at 10% compounded annually. How much will she get at the end of
three years?
SOLUTION
ππππ 0: πΎ400 ππ πππ‘ππππ π‘
ππππ 1: πΎ400 π€ππ‘β
10
× 400 = πΎ40.00 π»ππππ ππ‘ π‘βπ πππ ππ π¦πππ 1: πΎ400 + πΎ40 = πΎ440.00
100
10
× πΎ440 = πΎ44. π»ππππ ππ‘ π‘βπ πππ ππ π¦πππ 2. π΄πππ’ππ‘ = πΎ440 + πΎ44
100
= πΎ484.00
10
Yπππ3: πΎ484 π€ππ‘β
× πΎ484 = πΎ48.40. π»ππππ ππ‘ π‘βπ πππ ππ π¦πππ 3. π΄πππ’ππ‘ = πΎ484 + 48.4
100
= πΎ532.40
ππππ 2: πΎ440 π€ππ‘β
Therefore, Mrs Kobili will get π²πππ. ππ ππ‘ π‘βπ πππ ππ π¦πππ 3.
Alternatively, we can use the formula:
π΄ = π (1 +
π
π
)
100
= 400 (1 +
10 3
)
100
= 400 (1.1)3
π²πππ. ππ
3.3 Commission
An annual amount of money paid to a salesman or an agent of a business as a reward for services rendered is
called a commission. The most common method used by companies for calculating commission payment is
basic salary plus straight commission. This method involves a basic wage with an additional commission for
every unit sold.
Example 3.3.1
Mr Ndalama gets a monthly salary of πΎ4600. plus a 3% commission on sales. In one month, he sold goods
valued at πΎ63,000.00. Calculate his total earnings for that month.
SOLUTION
πππ‘ππ ππππππππ = ππππππ¦ + πΆπππππ π πππ
1
3
= πΎ4,600 + (
× πΎ63,000)
100
= πΎ4,600 + πΎ1,890
= πΎ6,490.00
3.4 DISCOUNT
26
Due to various market or economic factors, competition inclusive, traders resort to price reductions. Price
reduction is referred to as discount. It is usually given as a percentage of the selling price and usually offered
to as customers who wish to obtain goods on cash.
Example 3.4.1
A house was initially priced at π²πππ, πππ. ππ in a certain town. Mr. Malata requested for a 10% discount
since he was ready to pay by cash. Assuming the request was successful, how much did he pay for it?
SOLUTION
10
× 550,000 = πΎ55,000.
100
π»ππππ, πΎ550,000 − πΎ55,000 = πΎ495,000.00
Alternatively, we use the method of proportion as shown below:
100% = πΎ550,000.00
90% = π₯
π»ππππ, 100%π₯ = πΎ550,000 × 90%
πΎ550,000 × 90%
π₯=
100%
= π²πππ, πππ. ππ
3.5 Hire Purchase
Most often such properties as houses, cars and farms are quite expensive for a number of people to buy on
cash. They would rather buy them in instalments; a part payment. Usually, a buyer pays a deposit and the
balance is agreed upon to be paid in weekly or monthly amounts. This method is usually more expensive than
buying an item on cash.
Example 3.5.1
A plot of land can be bought for K40. 000=00 cash. It can also be bought on hire purchase by paying a
deposit of K29, 000 and then 12 monthly instalments of K1200.00 each. Calculate the hire purchase
price.
SOLUTION
π»πππ ππ’ππβππ π πππππ = πππππ ππ‘ + π‘ππ‘ππ πππ π‘ππππππ‘π
= πΎ29,000 + (12 π₯ πΎ1200)
πΎ29,000 + 14,400
π²ππ, πππ. ππ
3.6 INVESTMENTS
3.6.1 Shares and Dividends
Shares are owned by a large number of individuals or institutions in a public company. Once a share is bought,
the sum raised goes to the company as working capital. Depending on the profits raised by the company, the
shareholders are paid dividends on each share. Shares can also be bought or sold through middlemen called
brokers. There are two types of shares: preferred stock (share) or ordinary stock (share). The former carry
fixed interest rates and the latter are paid out after.
Dividends are calculated as a percentage of the nominal value of the shares:
πππ£πππππ πππ π βπππ
• πππππ % = πππππ ππππ πππ π βπππ × 100%
π·ππ πππ’ππ‘ =
•
π·ππ£ππππ πππ π βπππ =
π‘ππ‘ππ πππ£ππππ ππππ’ππ‘
ππ’ππππ ππ π βππππ ππ π π’ππ
Example 3.6.1
Mr. Bonanza bought 800 shares of a company at K1500=00 per share. If the nominal value of a share was
K1000=00,
(a) What did he pay for the 800 shares?
(b) Find the total nominal value of the shares.
SOLUTION
(π)πππ π‘ ππ π βππππ = 800 × πΎ1500 = πΎ1,200,000.00
(π)πππ‘ππ πππππππ π£πππ’π = πΎ1000 × 800 = πΎ800,000.00
Example 3.6.2
Kontolola owns 60% of the shares in his company. The company made a profit of K3, 000,000.00. An amount
of K1, 500, 00 was paid out as dividends to the shareholders who had bought 100,000 shares. Calculate:
27
(a) the amount paid out as dividends on one share.
(b) the amount paid out in dividends to the shareholder.
SOLUTION
π‘ππ‘ππ πππ£πππππ ππππ’ππ‘
(π)πππ£πππππ πππ π βπππ =
ππ’ππππ ππ π βππππ ππ π π’ππ
1,500,000
=
100,000
= πΎ15.
(π)π΄πππ’ππ‘ ππππ = 40% ππ 100,000 × πΎ15
= π²πππ, πππ. ππ
3.6.2 Value of the Bonds
A bond is like a loan. Government or a municipality is a borrower that needs larger sums of money than an
average bank can supply. Debts issued by government are often known as treasuries. Interest rate is sometimes
called a coupon.
Example 3.6.2.1
Mr. Khoswe buys a bond worth K200, 000 that pays an interest of 8% per annum. The interest is paid to him
semi-annually.
(a) Calculate the amount he will receive on maturity of the bond.
(b) Find the annual interest.
SOLUTION
(π) ππ. πΎβππ π€π π€πππ ππππ¦ ππππππ£π π‘βπ ππππππππ ππππ’ππ‘, π‘βππ‘ ππ πΎ200,000.00
(π)π΄πππ’ππ πππ‘ππππ π‘ = 8% ππ πΎ100,000 = πΎ8,000.00
3.7.0 FOREIGN EXCHANGE
Most countries work on a centesimal system. This is a system with a unit of currency which is equivalent to
100 smaller coins or denominations. For Zambia, the currency is kwacha and smaller denomination is called
ngwee.
3.7.1 RATES OF EXCHANGE
Currencies nowadays ‘float’ and change from day to day according to the laws of supply and demand. Rates of
exchange help the business operations, as they import or export raw materials and finished products.
Example 3.7.1
The bank displays the exchange rate at 1$= K10.2. If Zemba orders goods worth $50,000 from America. Find
how much kwacha is required for her to purchase $50,000=00.
SOLUTION
1$ = πΎ10.2
$50,000 = π₯
π₯ = πΎ10.2 × 50,000
= πΎ509999.999999
≅ π²πππ, πππ. ππ
Example 3.7.2
Mrs. Maulendo exchanges K20, 000.00 for British pounds. Calculate the amount in pounds that she receives
given that 1 pound = K20.
SOLUTION
£1 = πΎ20
π₯ = πΎ20,000.00
20,000.00
πΆπππ π ππ’ππ‘ππππ¦πππ π‘βπ ππππ£π π¦ππππ , π₯ =
20.00
π = £ππππ. ππ
1. Mr. Baldwin bought 500 shares of a company at k3100 per share. The nominal value of a share was K1000.
(a) What did he pay for the 500 shares?
(b) What is the total nominal value of the shares?
Solutions
Cost of shares = number of shares x cost per share
= 500 x K3100
= K1550000
(b) Total nominal value=number of shares x nominal value of a share
= 500x K1000
28
= K500000
Q2. The director of Mukuba pensions decide to pay a total dividend of K978000 on 1250000 shares
(a)Calculate the dividend per share.
(b)Mukuyu Trust holds 15000 shares in the company how much is paid out in dividends to the company?
Solutions
(a) dividend = total dividend amount
Number of shares issued
K978, 000
1250,000 = K0.78
=
(b) Total dividend = dividend x number of shares
= K0.78 x 15000
= K11, 700
Q3. The value of shares that Mr. ZIBA bought increased .When the market value per share was
K30.52, he
sold his 7500 share.
(a)If the dividend per share was k1.05 when he sold his shares, what was the total dividend he received when
he sold all his shares?
7500 shares x K1.05
=K 7875
(b)What would Mr Ziba have received for his shares if he had sold them at K23.23 per share?
23.23 X 7500
=K 174225
(c)How much profit did he make when he sold his shares at k28.23?
28.23 X 7500
=K211725 therefore profit will be equal to
= K211725 – K7875
= K203850
TOPIC 8: BEARING AND SCALE DRAWING
i). BEARING: Bearing refers to the direction of a movement.
We use three methods to show direction.
i. COMPAS BEARING (Nautical bearing)
βͺ There are four cardinal points on the compass.
N
W
E
S
βͺ
βͺ
There are other points half way between the cardinal points e.g. North West (NW), South East
(SE) etc.
Nautical bearings are measured as acute angles from the North or South to the East or West.
29
N
N
A
45°
O
ii. THREE FIGURE BEARING
βͺ These are given in three figures
βͺ They are always measured in the clockwise direction starting from the North.
N
Bearing of F from O is 064°
F
064°
O
2). SCALE DRAWING
When making a scale drawing;
i. Make a rough drawing (sketch diagram) to give a general idea of what the final diagram
should look like.
ii. Choose a suitable scale to give a large diagram.
βͺ The larger the diagram the more accurately you can read the lengths and angles.
βͺ Draw parallel lines from North to South (North lines) if you are making a scale
diagram to show bearings at the points from where you are reading the bearings.
7.
N
A
B
30°
C
In the diagram above, B is due East of A and due North of C. Angle BAC is 30°, find the bearing of B from C.
Solution
Find <ACB
<ABC+<BAC+<ACB=180°
(angles in a triangle)
90°+30°+<ACB=180°
Therefore, the bearing of B from C =360°-60°
<ACB=180°-120°
=300°
<ACB=60°
N
N
B
A
300
C
30
8).
N
N
A
110°
B
C
D
In the diagram above, πΆ is due south of π΄, π΄π΅ = π΅πΆ and the bearing of π΅ fro π΄ is 110°. Find the bearing of
πΆ from π΅.
Solution
< π΅π΄πΆ =< π΄πΆπ΅ = 70° (Base angles on an isosceles triangle)
< π΄πΆπ΅ =< πΆπ΅π· = 70°
∴the bearing of πΆ from π΅
= 180° + 70°
= 250°
9). Three towns π, π and π
are such that Q is 45ππ from P on a bearing of 030°. Make an accurate scale
drawing to show the positions of P, Q and R. From the scale drawing, find the distance and bearing of R from
P.
Solution
N
Sketch
N
Q 145°
45km
60km
030°
P
R
Accurate drawing:
Q
4.5cm
030°
6cm
P
From the drawing, ππ
= 7.6ππ
7.6 × 10 = 76ππ
The bearing of R from P is
R
1. B. Show this information in a diagram. SW is 225° from N, therefore: A is on a bearing of 225°.
Solution
At B, draw a line showing N and measure an angle of 225° clockwise from N. A is south west of point
31
N
B 2250
A
2. Figure 2 shows the bearing of P from Q.
Find the bearing of Q from P.
N
P
Q
300°
The bearing of Q from P is the angle between North and the direction
from Q to P as parallel to NX shown in figure 3,
It can be found in the following way:
At P, draw line NY.
Solution
N
N
< πππ = πππ – πππ‘πππππ‘π ππππππ
π
< πππ = 360β° − 300β° = 60β°
Hence < YPQ = 60°
Q
πβπππππππ, < πππ = 180° – 60° = 120°
Y
300°
πβπ πππππππ ππ π ππππ π ππ 120β°.
X
3. Points A, B, C and D lie on level ground.
The pint D is due north of A.
DAΜC = 140°, CAΜB = 90° and ABΜC = 75°. Find the bearing of:
North
D
140° A
75°
a) A from C
b) B from A
c) C from B
B
C
Solution
D
140° A
a) Y = 180° - 140° (interior opposite <s)
= 40°
∴ Bearing of A from C is 040°
b) X =360° - (140° +90°)=130°
∴ Bearing B from A is 130°
c) Bearing of C from B
<x+<z=180° (Interior opposite angles)
130°+<z=180°
x
B
C
750
32
<z=50°
∴ Bearing of C from B
=360° - (75°+ 50°)
=235°
4.
B
A
2.2m
1.9m
42° C
D
The diagram shows a frame work π΄π΅πΆπ·, π΄π· = 2.2π
π΅π· = 1.9π πππ π΅πΆΜπ· = 45β°. π΄π·ΜπΆ = 90β°
Calculate i) π΄π·Μπ΅ ii) π΅πΆ
b) A vertical flagpole, 18m high, stands on horizontal ground,
Calculate the angle of elevation of the top of the flagpole
from a point, on the ground, 25m from its base.
Solutions
a) i) πΌπ β π΄π·π΅,
1.9
πΆππ < π΄π·Μπ΅ = 2.2
1.9
< π΄π·Μπ΅ = πΆππ −1
2.2
= 30.27°
= 30.3° (1 π. π)
ππ) πΌπ β π΅π·πΆ,
1.9
πππ 42° =
π΅πΆ
π΅πΆ = 2.839
= 2.84π (3π . π).
b)
1.8m
Q
18
25m
πππ π =
25
π = 35.8β° (1 π. π)
∴ < ππ ππππ£ππ‘πππ ππ 35.8β°.
5. The diagram shows the position of a harbour, H, and three islands A, B and C.
C is due north of H.
C
A
128°
31km
54km
B
62°
H
33
The bearing of A from H is 062° and HAΜB = 128° HA = 54km and AB = 31 km.
a) Calculate the distance HB
b) Find the bearing of B from A
c) The bearing of A from C is 133°. Calculate the distance AC.
Solution
a) Using Cosine Rule
π»π΅ = 54 + 31 – 2(54)(31) πΆππ 128
= 77.059
= 77.1π (3π . π)
b).
N
A
π° = 62° (πππ‘ < π )
π₯
128° − 62° = 66°
128°
180° − 66° = 114°
62°
31km
∴Bearing of B from A is 114°
B
c). < π»πΆπ΄ = 180β° − 133β°
Using Sine Rule;
ππππ47β°
54
=
πππ 62β°
π΄πΆ
54 πππ 62β°
πππ 47β°
π΄πΆ =
= 65.19
= 65.2π (3π . π)
1. The diagram below shows three points P, Q and R on the map. Given that the bearing of Q from P is
035°, the bearing R from Q is 110° and < QRP = 33°
N
Q 110°
π₯ π¦
N
N
w
R
35°
P
Find:
(a)
(b)
(c)
(d)
the bearing of P from Q
the bearing of Q from R
the bearing of P from R
the bearing of R from P
Solution:
a) P from Q
π = 35° (πππ‘πππππ‘π ππππππ )
π = 180° − 110
π = 70°
The bearing of P from Q = X + Y + 110°
= 35° + 70° + 110°
= 215°
b) Q from R
34
110° + W + 180 (interior angles)
W = 180° - 110°
W = 70°
The bearing of Q from R = 360° - 70°
= 290°
c) P from R
360° − (70° + 33°) π ππππ π = 70
= 257°
∴ The bearing of P from R is 257°
d) R from P
π = 180° − (π₯ + π¦ + 33) (Interior angles)
π = 180° − (35° + 70° + 33°)
π = 180° − 138° = 42°
∴ The bearing of R from P is 42° + 35° = 077°
PRACTICE QUESTIONS:
(1) The diagram below shows an equilateral triangle ABC. A is due north of B and CN is
parallel to BA.
Find (a)
(b)
BCN,
the bearing of C from A.
Answer: (a)………………….
(b)………………….
(2) The diagram below shows triangle PQR in which R is due east of Q, angle PQR
=135° and angle QPR=33°.
Calculate (a) angle PRQ,
(b) the bearing of P from R.
35
Answer: (a)………………….
(b)…………………...
(3) The bearing of a point B from A is 129°. What is the bearing of A from B?
(4) The bearing of B from A is 072°.
(a) Find the bearing of A from B.
(b) C is due South of B and BA = BC. Find the bearing of A from C.
(5) A, B and C are three towns. C is equidistant from A and B.
The bearing of C from A is 132° and angle BAC = 75°.
Find
(a) (i) the acute angle ACB,
(ii) the reflex angle ACB,
(b) the bearing of A from C,
(c) the bearing of A from B.
(1)
(2)
(3)
(4)
(5)
ANSWERS
(a) angle BCN = 120° ,
(b) bearing of C from A = 120°
(a) 12°,
(b) bearing of P from R =256°
bearing of A from B = 309°
(a) bearing of A from B = 252° (b) bearing of A from C =306°
(a) (i) angle ACB = 30°
(ii) the reflex angle ACB = 330°
(c) bearing of A from C = 312°
(d) bearing of A from B = 27°
36
TOPIC 9: SYMMETRY
1. THE ORDER OF ROTATIONAL SYMMETRY
βͺ A shape has rotational symmetry if the image looks like the original shape.
βͺ The order of rotational symmetry is the number of times the image of the shape coincides
with the original shape within a full turn of 360°.
βͺ A shape or object has a point of rotational symmetry if is unchanged by a rotation through
an angle of 360°.
2. SYMMETRY IN SOLIDS
βͺ A 2D shape may have a centre of rotation but a 3D object would have an axis of rotation.
βͺ The order of rotational symmetry for a 3D object is the number of times it can be rotated
through an angle not greater than 360°, so that the image looks the same as the original
object before rotation.
3. PLANE OF SYMMETRY
βͺ A 2D shape may have a line of symmetry but a 3D object would have a plane of symmetry
along which you could see the reflection of the object.
βͺ An object can have more than one plane of symmetry.
The diagram shows a squared floor tile with centre o.
O
Describe the rotational symmetry about O of this floor tile.
Solution
360°
= 90°
4
∴ The floor tile has a rotational symmetry of 90° about O with the order of 4.
4. The diagram shows a simplified view of a water wheel.
O
Describe fully the symmetry of the water wheel about the centre O.
Solution
360°
Angle of rotational symmetry= π
360°
=
8
=45°
∴ The water wheel has an angle of rotational symmetry of 45° about center O and with an order of 8.
In the diagram below, shade any two small triangle such that the figure has rotational symmetry of order two.
37
Solution
1. (a) State the order of rotational symmetry of a regular decagon
(b) Write down those letters of the work AMBULANCE which have a vertical axis of symmetry.
Solution
(a) A regular decagon has a rotational symmetry of order 10.
(b) A, M, U
2. A prism has a cross section which is a regular hexagon. How many planes of symmetry does this
prism have?
Solution
(a) 7 planes of symmetry
3. On the regular hexagon below, draw all the lines of symmetry
4.
Solution
5. The 7 sided polygon in the diagram has 6 angles of x° and one of y°
x°
x°
y°
x°
x0
x°
(a) Draw the line of symmetry on the diagram.
38
x
x°
y°
x°
x0
x
x0
x°
(b)If Y = 126, calculate the value of x
Total interior angles = (7 – 2) × 180
= 5 × 180
= 900°
6π₯ + π¦ = 900
6π₯ + 126 = 900
6π₯ = 900 – 126
6π₯ = 774
π₯ = 129
6. How many planes of symmetry has a prism with a regular pentagon as the cross section
Solution
6 planes of symmetry
39
TOPIC 10: COMPUTERS
What is a computer?
A computer is an Electronic device that can:
- Accept data, as input
- Process the data
- Store data and information
- Produce information, as output.
BASIC ELEMENTS OF A COMPUTER
INPUT
PROCESS
OUTPUT
STORAGE
FLOW CHARTS
- A computer carries out all its tasks in a logical way.
- A set of logical steps that need to be followed in order to solve a problem are also referred to as
FLOW CHART.
How does one construct a flow chart?
-
To construct a flow chart one needs to firstly master the symbols used and their meaning.
BASIC FLOWCHART SYMBOLS
SYMBOL
MEANING
BEGIN/END OR
START/ STOP
INPUT/OUTPUT ORENTRY/DISPLAY
PROCESS
DECISION
PROGRAM FLOW
40
BASIC FOUR OPERATORS
SYMBOL
MEANING
+
ADDITION
β
SUBTRACTION
*
MULTIPLICATION
/
DIVISION
EXAMPLE:
- Construct a flow chart program to calculate the perimeter of a square, given its length.
- Since the formula is:
Perimeter = 4l
- Data needed for input is the length
BEGIN
ENTER length
IS length
≥ 0?
ERROR: length
MUST BE ≥ 0
NO
YES
PERIMETER = 4*length
DISPLAY PERIMETER
END
Carry out the following activities:
1.
Construct a flow chart program on how to find Mr. Mwansa’s age given his year of
birth as y.
2.
Prepare two questions using the well-known formulae or mathematical concepts on
which a flow chart can be used to arrive at its solutions.
41
Topic Study –Mathematics Computer & Calculator
FUNCTIONS ON A CALCULATOR
A relevant component of the syllabus but with most challenges when it comes to
teaching arising from the fact that:
1.
there is no standard or specific type of a calculator to be used in our
Schools.
2.
the subtopic is perceived as obvious by both the learners and teachers.
3.
the supposition that the subtopic is taken care of by the manual the
calculators come with.
- Hence, this subtopic goes un attended to in most cases.
- Thus far it is advised that we try to find resolutions to these
- Challenges and start teaching this subtopic.
-
BASIC COMPONENTS OF A COMPUTER
A computer is an Electronic device that can:
- Accept data, as input
- Process the data
- Store data and information
- Produce information, as output.
- Parts or devices that serve in each of the processes below:
INPUT
PROCESS
OUTPUT
STORAGE
-
Input parts/devices
Examples include Keyboard, mouse etc.
Process part/device
For example CPU or Systems Unit.
Output parts/ devices
Examples include Monitor, Printer etc.
Storage parts/devices
For example Flash disk, Memory card etc.
ALGORITHMS.
A flow chart as understood from the previous presentation can also be considered as an example of an
Algorithm
Definition
- An algorithm is generally a set of logical steps that need to be followed in order to solve a problem.
- For example the solving of a malaria problem using any anti-malaria drug.
METHODS OF IMPLEMENTING AN ALGORITHM.
- There are two basic methods of implementing an algorithm and these are:
•
-
(i)
FLOW CHARTS
(ii)
PSEUDO CODE
FLOW CHARTS
The underlying factors of any flow chart are the use of the correct symbols for each step and the
correct operation symbols.
For example; calculate how far point A is from Mufulira, given point A.
42
BEGIN
ENTER Point A
Is Point A
within
Mufulira
ERROR: Point A
MUST BE OUTSIDE
MUFULIRA
YES
NO
Measure the distance
between them.
•
DISPLAY
DISTANCE
PSEUDO CODE
END
-
This is a derivative of a flow chart and its underlying factors are the correct extraction of statements
inside a flow chart symbol, listing them vertically and preservation of the logical steps also known as
dentation.
- For example; An equivalent Pseudo code to the flow chart above is:
EQUIVALENT PSEUDO CODE
BEGIN
ENTER Point A,
IF Point A is within Mufulira THEN
DISPLAY Error message,
ENTER Point A,
ELSE
Measure Distance between them,
ENDIF
DISPLAY Distance,
END.
43
GROUP ACTIVITIES
Working in pairs
(i)
Identify a problem whose solution is dependent on one parameter.
(ii)
Demonstrate how you would get to this solution by using both methods of implementing an
algorithm.
(iii)
Each one of the members in a group to present each of the methods.
CONCLUSION
This topic on computer and calculator in mathematics does not replace the need for one to be computer literate
via the learning of computer studies but among others:
Provide for the knowledge on stages of problem solving (define a problem, analysis method of solution) and
knowledge on how to write a computer program.
44
TOPIC 11: APPROXIMATION
Specific outcome: work with relative and absolute error
π΄ππ πππ’π‘π πππππ
Relative error= πππ’π ππππ π’ππππππ‘
1) Find the absolute error, the upper limit and the lower limit for 8h
Solution
Least unit of measurement = 1h
1
Absolute error = ππ1β = 0.5β
2
Upper limit = 8 + 0.5 = 8.5β
Lower limit = 8 − 0.5 = 7.5β
2) Find the relative error of 5.5l
Solution
Absolute error = 0.05
0.05
5
1
Relative error= 5.5 = 550 = 110
3) Find the limits between which the areas of the following shapes must lie.
a) A square of side 3cm
b) A right-angled triangle with hypotenuse 5m and the other sides 3m and 4m long
Solution
a). least unit of measurement = 1cm
Absolute error = 0.5
Upper limit is 3 + 0.5 = 3.5ππ
Lower limit is 3 − 0.5 = 2.5ππ
Maximum area= (3.5 × 3.5)ππ = 12.25ππ2
Minimum area = (2.5 × 2.5)ππ = 6.25ππ2
Therefore, the limits are 6.25ππ2 πππ 12.25ππ2
b). least unit of measurement = 1m
Absolute error = 0.5
Upper limit for the height=4+0.5=4.5m
Lower limit for the height= 4-0.5=3.5m
Upper limit for the base= 3+0.5=3.5m
Lower limit for the base= 3-0.5= 2.5m
1
Area of a triangle = 2 × π × β
1
Therefore, maximum area= 2 × 4.5 × 3.5 =
1
1
15.5
2
= 7.875π2
Minimum area = 2 × 3.5 × 2.5 = 2 × 8.75 = 4.375π2
The limits are 4.375π2 πππ 7.875π2
1. The length and breadth of a rectangle, given to the nearest centimeter, are 15cm and 10cm
respectively. Find :
a) The shortest possible length and shortest possible breadth of the rectangle.
b) The longest possible length and the longest possible breadth of the rectangle
c) The limits between which the area must lie.
Ans: 15cm, given to the nearest centimeter, lies between 14.5cm and 15.5cm. 10cm lies between
9.5cm and 10.5cm
a) The shortest possible length = 14.5 and the shortest possible breadth = 9.5cm
b) The longest possible length =15.5cm and the longest possible breadth is 10.5cm
c) The smallest possible Area= 14.5 × 9.5
45
= 137.75cm2
The longest possible Area= 15.5 × 10.5
=162.75cm2
So, the area lies between 137.75cm2 and 162.75cm2 or (137.75cm2≤ π΄ ≤ 162.75cm2)
2. A car is driven a distance of 30km, measured to the nearest km, in 20 minutes, measured to the nearest
min. between what limits will the average speed lie?
Ans: 30km to the nearest km means.
29.5km ≤ distance ≤ 30.5km
20min to the nearest min means
19.5min ≤ time ≤20.5min
πππππ‘ππ π‘ πππ π‘ππππ
ππππ π‘ π‘πππ
Greatest average speed=
Least possible speed =
ππππ π‘ πππ π‘ππππ
πππππ‘ππ π‘
=
30.5ππ
19.5πππ
=
1.56ππ
πππ
= 94ππ/β
29.5ππ
= 20.5πππ = 1.44km/min = 86km/h
Hence, the average speed = (90± 4) km/h
3. The true value of the length or a rectangle is 5.5m. If this is recorded as 5m, find
A] The absolute error
B] The relative error
C] The percentage error
SOLUTIONS.
True value is 5.5m recorded value is 5m.
A] Absolute error = [recorded value –true value]
= [5-5.5] m = [-0.5] = 0.5m.
B] Relative error=
πππ πππ’π‘π πππππ
π‘ππ’π π£πππ’π
0.5π
=5.5π =0.09
C] The percentage error = relative error × 100%
= 0.09 × 100% = 9%
4. The length of a square is given as 10cm correct to the nearest centimetre calculate
A)
B)
C)
D)
E)
The lower bound length.
The upper bound length.
The smallest possible perimeter.
The largest possible perimeter.
The smallest possible area.
46
F) The largest possible area.
Solutions
The length of a square is 10cm.
n
1ππ
Error = 2 = 2 = 0.5cm
a) The lower bound length = actual length +error
=10cm − 0.5cm
=9.5cm
b) The upper bound length = actual length + error
= 10cm + 0.5cm = 10.5cm
c) Perimeter of a square = 4× length
Therefore, smallest possible perimeter = 4× lower breadth length
= 4× 9.5cm = 38cm
d) Largest possible perimeter = 4× upper bound length
= 4× 10.5cm = 42cm
e) Area of a square = L2
Therefore the smallest possible area = (lower bound length) 2
= (9.5cm) 2 = 90.25cm2
f) Largest possible area = (upper bound length)2
= (10.5cm) 2 = 110.25cm2
47
TOPIC 12: SEQUENCES SERIES
-53 = 3 - 4n + 4
4n = 53 + 7
Sequence
Sequence is a set of numbers listed in a welldefined order with a specific rule that can be used
to state the next numbers in that set.
1
1
ο΄ 4 n = 60 ο΄
4
4
n = 15
1. Write the next three terms of each of the
sequences below
4. The 10th term of an AP is 37 and the 16th term is
61, for this AP find:
(i)
The common difference
Answers
Tn = a + (n - 1) d
T10 = a + 9d
37 = a + 9d……………eqtn 1
and
T16 = a + (16 - 1) d
16 = a+15d…………..eqtn 2
and solve the equations
simultaneously.
a + 9d = 37
- (a+15d = 61)
(a) 1, 2, 4, 8…
Solutions
16, 32, 64
(b)-4,-1, 2, 5,8,11
Solutions
14, 17, 20
Series
A series is the sum of all the terms of a sequence
e.g.
−
(i) 1+2+4+8+16+,…….
d
Solutions
6
4
=
=
− 24
4
4
(ii) The first term
Solutions
First term
14, 17, 20
1. For the AP, 2+5+8+………… find
a + 9d = 37
th
(i)
The 10 term
(ii)
51st term
Solutions
a=2, first term
d=5-2
D=3
T10 = a+ (n-1) d
= 2+ (10-1)3
= 29
(iii)
The 51 st term
Solutions
T51 = 2+ (51 - 1)3
- = 2+ 50 x 3
= 152
(iv)
The nth term
Solutions
Tn = a + (n - 1) d
= 2+ (n - 1)3
= 2 + 3n - 3
= 3n - 1
a + 9(4) = 37
a + 36 = 37
a = 37 - 36
a=1
(iii) the 30th term
Solutions
Tn = a + (n - 1)
T30 = 1 + (30 - 1)4
T30 = 117
Practicing Questions
5. The nth term (Tn) of an AP is given by Tn =
1/2(4n - 3).
(a) State (i) the 5th term (ii) the 10th term
(iii) the 6th term
(b) the common difference
6. If x + 1, 2x - 1 and x + 5 are three
consecutive terms, find the value of x.
3. Find the number of terms in the AP 3 + (-1) + (5) +…. + (-53).
Solutions
a = 3, d = -4, Tn = -53
Tn = a + (n - 1) d
-53 = 3 + (n - 1) -4
Solutions
48
X+1
,
2x - 1
,
x
+5
T1
T2
T3
For an AP,
Common difference, d = T2 - T1 = T3 - T1
(2x - 1) - (x + 1) = (x + 5) - (2x - 1)
2x - 1 - x - 1 = x + 5 - 2x
2x - x - 1 - 1 = x - 2x + 5 + 1
X - 2 = -x + 6
X+x=2+6
2x = 8
X=4
7. (i) if the numbers 3,m,n and 8 are three
consecutive terms of an AP, find the values of m
and n.
Solutions
T6 = 1/2(4 x 6 - 3)
The 5th term
T10 = 1/2(4 x 10 30)
T6 = 1/2(24 - 3)
T5 = 1/2(4n - 3)
T10 = 1/2(40 3)
T6 = 1/2(21)
T5 = 1/2(4 x 5n - 3
T10 = 18.5
T6 = 10.5
T5 = 1/2(17)
T5 = 8.5
m=
16
2
m=
Therefore, m = 8 and n = 13
(ii)
The numbers m - 1, 4m + 1
and 5m - 1 are three
consecutive terms of an AP,
find the numbers.
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
2
b=
2n - 18…………….eq2
c
m − 1 + 5m − 1
2
8m - 6m = -2 - 2
2m = -4
M = -2
Substitute for m = 1 in the
series we get, -3, -7 and -11
8. (i) Find the arithmetic mean of the first 6 terms
of 3 + 8 +………
form
m=
T10 =
8m+2=6m - 2
n+3
= 2(2n - 18)
n + 3 = 4n - 36
n - 4n = -36 -3
-3n = -39
− 3n =
− 39
−3
− 3
n = 13
m=
+
2
2(4m + 1)
=m + 5m-1-1
= 2n - 18/1
3
2
a
4m + 1 =
Equate m = m
n+
T6 = 1/2(4 x 6 - 3)
The 5th term
1/2(4 x 10 - 30)
T6 = 1/2(24 - 3)
T5 = 1/2(4n - 3)
T10 = 1/2(40 - 3)
T6 = 1/2(21)
T5 = 1/2(4 x 5n - 3
T10 = 18.5
T6 = 10.5
T5 = 1/2(17)
T5 = 8.5
m - 1, 4m + 1, 5m - 1 and 4m
+ 1 is an arithmetic mean
between m - 1 and 5m - 1
M-3=n-m
and
n - m = 18 - n
M+m=n+3
n + n = 18 + m
2m = n + 3
2n = 18 + m
m=
…………..eq1
n + 3
m=
8
Answers
n+3
2
First term
a=3
Common difference d = 8-3
d=5
13 + 3
2
Therefore, the 6 terms are 3,8,13,18,23,28.
49
formula for the sum of the first n - 1
terms
Arithmetic mean = 3 + 8 + 13 + 18 + 23 +
28
6
93
Arithmetic mean =
Solutions
Sn = n2 + 2n
To find the first term
we put n = 1 in the
given sum
S1 = 12 + 2(1)
a=3
The common
difference
d = S2 - 2S1
=8-6
=2
Sn = n2 + 2n
Sn-1 = (n - 1)2 + 2(n 1)
= n2 - n - n + 1 + 2n 2
= n2 - 2n + 2n + 1 - 2
= n2 - 1
(i)
6
Arithmetic mean =
Or
15.5
Arithmetic mean = median
=
(ii)
=
15.5
(ii) Find the arithmetic mean and the
geometric mean of 4 and 64.
(iii)
13 + 18
2
Solutions
Given 4 and 64
Arithmetic mean =
=
4 + 64
2
68
2
Geometric Progression (GP)
= 34
Geometric mean = square root of 4 and 64
=2x8
= 16
9. An arithmetic progression has a 1st term to be 2
and common difference of 2, show that the sum of
the first nth terms of the AP is given by Sn = n2 + n.
hence find the sum of the 21st terms of an AP.
Answers
A = 2, d = 2
Sn = n
(2a + (n − 1)d )
2
n
=
=
=
A geometric progression (GP) is a sequence in
which each term is formed by multiplying the
previous term by a constant amount.
nth term of a Geometric Progression
The nth term of a GP with first term a and common
ratio r is: Tn = arn - 1
1. For a GP, 2 + 6 + 18 + …………, find (i) the
tenth term (ii) the 17th term
Solution
First term (a) = 2
Common ratio r =
(2 x 2 + (n − 1)2)
2
n
( 4 + 2n − 2)
2
T
T
n = 10
T10 = 2 ο΄ 310−1
T10 = 39366
= n + n2
=
1
6
2
r = 3
Tn = ar n −1
n
( 2 + 2n)
2
2
So the 10th term is 39 366
(ii) T17 = 2 X 317 - 1
16
=2x3
= 86 093 442
Sn = n2 + n is required
The sum of the first 21st terms
2. The third term of a GP is 9 and the tenth term is
19 683, find;
(i) the common ratio
(ii) the 8th term
S21 = 212 + 21
S21 = 441+ 21
= 462
Solutions
(i) T3 = ar2
ar2 = 9………….eqn (i)
10. The sum Sn of the first n terms of an
AP is given by Sn = n2 + n, find (i) the first
term (ii) common difference (iii) the
50
T10 = ar9
Ar9 = 19 683……..eqn (ii)
Dividing equation (i) by equation (ii)
x+3
x+2
Common ratio (r) =
ar 9 19683
=
ar 2
9
7
r 7 = 7 2187
r =3
=
The first term is 1
(iii) Tn = arn -1
T8 = 1 x 38 - 1
= 37
T8 = 2187 the 8th term is 2 187
3. Given that x +2, x + 3 and x + 6 are the first
three terms of a GP, find
(a) the value of x
(ii) the 5th term of the GP.
T5 = ar n −1
1 4
ο΄3
2
81
= or 40.5
2
=
The nth term
(Tn) of a GP is
given by Tn = 29 - n. Find
(i) the first term
(ii) the common ratio
(iii) the sum of the first 9 terms.
Solutions
(i) Common ratio (r) =
T
T
2
1
=
T
T
3
2
x+3 x+6
=
x+2 x+3
Solutions
(x + 3)(x
(i)
+ 3) =(x + 2) (x + 6)
X2 + 3x + 3x + 9 = X2 +
6x + 2x + 12
6x + 9 = 8x + 12
8x - 6x = 9 - 12
T n = 29 - n
T 1 = 29 - 1
T 1 = 28
T1 = 256
a = 256
(ii) To find the common ratio, first calculate the
second term (T2)
2 x
− 3
=
2
2
x = − 1 13
(ii) First term (a) =
+ 13
+ 12
ο¦ −3 3οΆ ο¦ −3 2οΆ
=ο§
+ ο· οΈο§
+ ο·
ο¨ 2 1οΈ ο¨ 2 1οΈ
3 1
= οΈ
2 2
3
= ο΄2
2
r =3
Therefore the common ratio is 3
(ii) ar2 = 9
A x (3)2 = 9
9a
9
=
9
9
a =1
−3
2
−3
2
T2 = 29-2
= 27
= 128
x+2
−3
+ 2
2
−3+ 4
=
2
1
a =
2
=
Common ratio (r)
=
T2
T1
128
256
1
=
2
=
51
(iii) Sum
a (1 − r n )
=
1− r
256[1 − ( 12 ) 9 ]
=
1 − 12
=
Common ratio (r) =
1 1
οΈ
4 8
1
r = ο΄8
4
1
8
n
a (r − 1)
S10 =
r −1
10
1
(2 − 1)
= 8
2 −1
1
(
1024
− 1)
= 8
1
1
= (1023)
8
S10 = 127.875
256(0.998046875)
r = 2anda =
1
2
255.5
0 .5
Sum = 511
=
Sum of a GP
6. Calculate, correct to three significant figures, the
sum of the 5 1 .......... first 8 terms of the
3
GP 12, 8,
Solutions
8
r=
12
3
r = or 0.75
4
First term a =
12
Geometric Mean
8. Find the geometric
of 4 and 64.
Solution
a (1 − r n )
1− r
12 (1 − ( 34 )8
=
1 − 34
S8 =
ο
=
ο
=
43.2
significant figures
=
4 ο΄ 64
=
4 ο΄
64
= 2 ο΄ 8
= 16
12(0.899837085)
1 − 34
10.79864502
0.25
= 43.19458008
Mean
9. The sum of infinity of a certain GP is 28. if the
first term is 37, find r
Sum to infinity
=
correct to 3
a
= 28
1− r
a = 37
Sο₯ =
1 1 1
, , ,......... ..
8 4 2
7. Work out the sum of the first 10 terms of
a
= 28
1− r
37
= 28
1− r
28(1 − r ) = 37
Solution
28 − 28r = 37
− 28r = 37 − 28
10. Write down the
number of terms in the
following GPs
28r
9
=
or − 032
28 − 28
(i) 2 + 4 + 8 + …………..+512
52
(ii) 81 + 27 + 9 + ………..
+
1
L = ar n −1
27
Solution
512 2 ο΄ 2n −1
=
2
2
n −1
256 = 2
First term a = 2, r =2
Last term = 512
28 = 2n - 1
8=nn=8+1
n=9
Factorising 256
2 128
2
1
64
The GP has 9 terms
2 32
2 16
2
8
2
4
2
2
a = 81, r =
2
(ii) 1 + 27 + 9 + ………..
1
1
n =1+
+
log L
a
log r
27
=1+
=1+
1
1
, last =
3
27
log
1
27
81
log 13
log( 217 ο΄
log 13
1
81
)
log( 13 ) 3 ο΄ ( 13 ) 4
n =1+
log( 13 )
log( 13 ) 7
n =1+
log( 13 )
n =1+
The GP has 8 terms
7 log( 13 )
log( 13 )
n =1+ 7
n=8
ARITHMETIC PROGRESSION
53
PRACTICE QUESTIONS
1. The fourteenth term of an AP is 5.2 and the twenty fifth term of the AP is 105.find the sum to the first 79
term of AP?
T14=a+ (14-1) d=58.2
therefore the value of a+24d=105
T25=a+ (25-1) d=105
a+24(4.25) =105
a +13d=58.2
a=105-102
a + 24d=105
a=3
Subtract = -11d=-46.8
-11
therefore the sum is S79=79/2[2(3) + (79-1)4.25]
-11
S79=13,331.25
D=4.25
2. The twenty first of an AP is 184.9 and the twelfth is 104.8 calculate the sum of 81.
T21=a+ (21-1) d=184.9
therefore the value of a is a+11d=104.8
T12=a + (12-1) d=104.8
a+11(8.9) =104.8
=a+20d =184.9
Subtract
a=6.9
a+11d=104
therefore the sum is S81=81/2[2(6.9) + (89-1)8.9]
=9d/9=80.1/9
=35,466.5
D=8.9
3. The ninth term of AP is 70.7 and the sixteenth term of the AP is 59.3 .find the T39
T19=a+18d=70.7
therefore a+18d=70.7
T16=a+15d=59.3
T39=2.3+ (39-1)3.8
a+18(3.8) =70.7
3d/3=11.4/3
a=70.7-68.4
D=3.8
T39=2.3+144.4
T39=146.7
a=2.3
4. The twenty third term of an AP is 159.5 and the eighteenth term of the AP is 123.5.find the sum of 18.
T23= a + (23-1) d=159.5 a+22d=159.5
T18=a + (18-1) d=123.5 a+22(7.2) =159.5
a +22d=159.5
a +17d=123.5
S18=18/2[2(1.1) + (18-1)7.2]
=1,121.4
a=159-158.4
a=1.1
5d/5=36/5
D=7.2
5. The twenty first term of an AP is 42.6 and the nineteenth term of the AP is 39.find T46
T21=a + (21-1) d=42.6
T19= a + (19-1) =39
2d/2=3.6/2
D=1.8
a+20d=42.6
T46=a + (n-1)
a+20(1.8) = 42.6
= 6.6+ (46.1)1.8
a=42.6-81
=87.6
a=6.6
54
GEOMETRICAL PROGRESSION
1. The third term of a G.P is 4096 and the fifth term is 1024. Calculate
T3=ar3-1
T5= ar5-1
S12= 16384×4095/4096/1/2
T3=ar2= 4096
T5= ar4= 1024
S12= 32760
Tr/T3= 1024/4096
ar3= 4096
Ar2= ¼
a (1/2)2= 4096
√r2=√1/4
a= 4096/ (1/2)2
r= ½
a=16384
2. The seventh term of a G.P is 729 and the fourth term is 19683. Find the fifth term.
T7= ar7-1
T4=ar4-1
ar6= 729
=ar3= 19683
T7/T4=r6/r3
r=
=1/27
r3=729/19683
TOPIC 13: CO-ORDINATE GEOMETRY
1. Find the length of AB if A (2, 3) and B (4, 5)
Y
B (4, 5)
5
A(2,3
)
3
0
2
4
A (2, 3) and B (4, 5)
55
X
Using Pythagoras theorem
AQ =2 units
QB = 2 units
Therefore, AB2 = (QA)2 = (BQ)2
AB2 = (4-2)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
AB2 = 8
AB = √8
From above, we can say that the distance along the x-axis = (x2 – x1) and distance along the y-axis = (y2 – y1)
AB2 = (X2 – X1)2 + (Y2 –Y1)2
AB2 = (4-1)2 + (5-3)2
AB2 = (2)2 + (2)2
AB2 = 4+4
AB = √8
Therefore, distance between two points = √ (π2 – π1)2 + (π2 – π1)2
2. Find the midpoint of A and B
ο© x1 + x2 y1 + y 2 οΉ
οͺ 2 + 2 οΊ
ο«
ο»
(Halfway x1 and x2 and halfway y1 and y2)
x1, y1
x2, y2
A (2, 3) and B (4, 5)
ο© 2 + 4 3 + 5οΉ
+
2 οΊο»
ο« 2
=οͺ
6
8
Midpoint= (2 + 2)
Midpoint = (3, 4)
3. Calculate the gradient of the line C (-5,-3) and D (-2, 6) and E (1, 2) and (3,-1)
56
.D (-2, 6)
6
5
4
3
.E(1,2)
2
1
-5
-4
-3
0 00
-2
0
-1
1
2
3
4
2
. F (3, -1)
-1
-2
-3
. C (-5,-3)
M = βy = M = Y1-Y2
βx
x1-x2
C = (-5, -3) and D (-2, 6)
MCD = 6-(-3)
-2-(-5)
MCD = 6+3
-2+5
MCD = 9
3
MCD = 3
TIP - since the line slops to the right, the gradient should be positive
X1,Y1, X2 , Y2
(ii) MEF =β Y = MEF = Y2 -Y1
EF = (1, 2) and (3,-1)
βX
X2 –X1
MEF = -1-2
3-1
MEF = -3
2
Since the line is sloping to the left, the gradient should be negative
4. (i) Find the equation of the straight line through (-2,-3) with a gradient 2.
Y – Y1 = M (X – X1)
Y-(-3) =2(X-(-2))
Y+3=2 (X+2)
Y+3=2X +4
Y=2X +4 -3
Y=2X+1
(equation to use when a point on the line
is given and gradient) M=Y2-Y1
X2-X1
57
(ii) Find the equation of the straight line through (2, 4) and (-2, -4)
M = Y-Y1
X-X1
Y-Y1 = M (X-X1)
since we do not have M then we find M as y2 –y1
Y – Y1 = y2 –y1
x2 –x1
x2 –x1
Y -4 = -4-4
-2-2 (X-2)
using the point (2, 4) to substitute in to the equation
Y – 4= 2 (X-2)
Y=2X-4+4
Y = 2X
(iii) Find the equation of the straight line through the point (4,-2) and the origin
i.e. (4,-2) and (0, 0)
origin has co-ordinates (0, 0)
Y- Y1 = M (X-X1)
Y-Y1 = Y2-Y1, (X-X1)
X2-X1
Y-0 = 0-(-2) (X-0)
using the point (0, 0)
0-4
Y = -2X OR 4Y = -2X
4
(iv) Find the equation of the straight line with the gradient 3 and y – intercept of 6
Y = MX +C , M is the gradient, C is the y-intercept, the point
Where the line cuts the y-axis
Therefore, Y = 3X + 6
by substituting the values of M and C
(V) Find the equation of the straight line with the gradient -2/9 and x-intercept 3
Y = MX+C
−2
Y = 9 (3) +C
Y=
0=
−2
3
−2
3
C=
X-intercept is the point where the line cuts the x-axis. Along
x-axis all the y- co-ordinates are zero (0)
+C
therefore, the point is (3, 0)
+C
2
3
The equation of the straight line is
−2
2
y = 9 x + 3 or 9y = -2x +6
, by multiplying through by 9.
(vi) Find the equation of the straight line with an x- intercept 4 and y- intercept 6.
i.e. (4, 0) and (0, 6)
M = Y2+Y1
X2+Y1
M = 6-0
0-4
−6
M= 4 =
−3
2
Using the point (0, 6) i.e. C =6
Y = MX +C
−3
Y = 2 x+6. Or
2y = -3x = 12
multiplying throughout by 2.
58
5. (i) Find the equation of the straight line that passes through the point (3, 8) and is
parallel to y = 2x -9
- The gradient of the new line is 2 since parallel lines have the same gradient
i.e. M1 = M2.
Using the point (3, 8)
Y = MX +C
8 = 2(3) + C
8 = 6 +C
2=C
Y = 2X +2
(ii) Find the equation of the line that is perpendicular to 3y – 2x = 4 and passes through the point (-7, 4)
3y = 2x +4
making y the subject of
y = 2x + 4
the formula
3
3
Therefore, M = 2
3
- Gradient of parallel lines M2× M2 = -1
2⁄
3 × π2 = −1
2M2 = -3
−3
M2 = 2
Using the point (-7, 4) to substitute into the equation
Y = MX + C,
−3 (−7)
4=
+C
2
4 = +C
2
4 -21 = C
1 2
8-21 = C
2
-13 =C
2
The equation perpendicular to 3y-2x=4 which passes through point (-7, 4) is
−3π₯ 13
Y=
or
2
2
2y = -3x - 13
PRACTICE QUESTIONS:
(1) Find the gradient of the straight line whose equation is 3y + x = 5.
(2)
Find the equation of the straight line passing through (-4, 4) and is
π₯
perpendicular to the straight line whose equation is π¦ + 7 = 1.
(3)
In the diagram below, the points A and B are (4, 0) and (0, 8) respectively.
59
Find the equation of AB.
(4) In the diagram, B is the point (0, 16) and C is the point (0, 6). The sloping line through B and the
horizontal line through C meet at the point A.
(a)
(b)
(c)
(d)
Write down the equation of the line AC.
Given that the gradient of the line AB is 2, find the equation of the line AB.
Calculate the coordinates of the point A.
Calculate the area of the triangle ABC.
ANSWERS:
1
(1) m = − 3
(2) y = 7x + 32
(3) y = -2x + 8
(4) (a) y = 6
(b) y = 2x + 1
(c) (-5, 6)
(d) area = 25 units2
TOPIC 14: RELATIONS, MAPPINGS AND FUNCTIONS
1. RELATIONS
A relation from set X to a set Y is a pairing of elements from set X to elements of set Y.
For example, given the sets X = {1, 3, 5, 0} and Y = {2, 4, 6, 7} such that x ο X and y ο Y ,
the relation “x is greater than y” is illustrated below by means of an arrow diagram (fig. 1)
and a Cartesian graph (fig. 2).
The same relation can be defined by the set of ordered pairs {(3, 2), (5, 2), (5, 4)}.
60
The set X is the domain and the set Y is the codomain. The element 2 in set Y is the
image of the elements 3 and 5 in set X. 4 is also the image of 5. The image set {2, 4} is
known as the range. Note that set Y is not the range. In this case, the range is a subset
of the codomain.
Examples
1. Two sets X = {0, 2, 4, 6, 8} and Y = {0, 1, 2, 3, 4, 5} are such that x ο X and y ο Y .
Show the relation “x is double y”
(i) by an arrow diagram
(ii) on the Cartesian plane
(iii) by a set of ordered pairs
2. The diagram below shows a relationship from the set A to the set B, where a ο A and
b ο B.
(i) Write down the domain, the codomain and the range.
(ii) Write down the set of ordered pairs defined by this arrow diagram.
(iii) Show this relationship on a Cartesian plane.
(iv) Write down a possible relationship from the set A to the set B.
Solutions
1.
(iii) The required set of ordered pairs is {(2, 1), (4, 2), (6, 3), (8, 4)}.
2. (i) Domain: A = {0, 1, 3, 5}, Codomain: B = {0, 2, 4, 5, 6}, Range: {0, 2, 4, 5}
(ii) {(0, 5), (1, 4), (3, 2), (5, 0)}
(iii)
61
(iv) The sum of a and b is 5 or a + b = 5 .
Exercise 1.1
1. Two sets F = {5, 3, 2} and G = {1, 2, 3, 4, 5, 6} are given.
(i) Use an arrow diagram to show the relation “is a factor of” from set F to set G.
Write down this relation
(ii) as a set of ordered pairs
(iii) on a Cartesian plane
ο»
1
1
ο½
2. A relation R is defined by ( 2 , 8) , (1, 4) , ( 2 , 2) , (4 , 1) , (8, 2 ) . The set D is the domain
and the set C is the range.
(i) List the sets C and D
(ii) Show this relation on a Cartesian graph.
(iii) Describe in words a possible relation from set D to set C. `
Answers to Exercise 1.1
1. (i)
(ii) {(5, 5), (3, 3), (3, 6), (2, 2), (2, 4) (2, 6)}
(iii)
1
1
2. (i) C = { 2 , 1, 2, 4, 8}, D = { 2 , 1, 2, 4, 8}
(ii)
62
(iii) The product of x and y is 4 or x y is equal to 4.
2. MAPPINGS
A mapping is a relation between the elements of one set (domain) and the elements of another
set (range). A mapping from set X to set Y links each element of X to exactly one element of set Y. If, in
addition, every element of Y is linked to exactly one element
of set X, then there is a one-to-one correspondence between the elements of set X and the
elements of set Y. In this case, we say that X maps onto Y.
If the number of elements in X and Y are not equal or if an element of Y (the codomain)
does not have a corresponding element in X (the domain), we say that X maps into Y.
Examples1. State whether each of the following relations from X to Y is a mapping or not. Give a reason for
your answer.
Solutions
(a) This is not a mapping because the element 1 from set X is mapped to two elements 4
and 5 in set Y.
(b) This is a mapping because each element from set X is mapped onto exactly one element
in set Y.
(c) This is a mapping because each element from set X is mapped onto exactly one element
in set Y.
(d) This is not a mapping because the element 2 from set X is not mapped to any elements
in set Y.
63
Exercise 2.1
1. State whether each of the following relations from R to S is a mapping or not.
Give a reason for your answer.
2. State whether each of the following relations from A to B is a mapping or not.
Give a reason for your answer.
3. x and y are variables on the set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}.
A relation on R is defined by the open sentence “x is three times y”.
(i) Write an arrow diagram for this relation.
(ii) Write down the set of ordered pairs (x, y) which define this relation.
(iii) Is this relation a mapping? Give a reason for your answer.
Answers to Exercise 2.1
1. (i) mapping; each element from set R is mapped onto exactly one element
in set S.
(ii) mapping; each element from set R is mapped onto exactly one element
in set S.
(iii) not a mapping; element a and b in set R not mapped to any in set S.
64
(iv) Mapping; each element from set R is mapped onto exactly one element
in set S.
(v) not a mapping; element b in set R not mapped to any in set S.
(vi) not a mapping; element c in set R not mapped to two elements (1 and 3) in set S.
2. (i) mapping; each element from set A is mapped onto exactly one element
in set B.
(ii) not a mapping; element ο¬ in set A not mapped to any in set B.
(iii) mapping; each element from set A is mapped onto exactly one element
in set B.
(iv) not a mapping; element ο in set A not mapped to any in set B.
(v) not a mapping; element ο in set A is mapped to two elements (2 and 9) in set B.
(vi) Mapping; each element from set A is mapped onto exactly one element
in set B.
3. (i)
(ii) { (3, 1), (6, 2), (9, 3), (12, 4), (15, 5)}.
(iii) a mapping; each element of the domain has exactly one element in the range.
3. FUNCTIONS
3.1 Definition
Let x ο X and y ο Y , where X = {1, 2, 3, 5} and Y = {2, 4, 6, 10, 12, 16}. This relation
is shown by the arrow diagram in fig. 7 below.
A function f is a relation mapping a set X into a set Y such that every element x in set X
has exactly one element y in set Y as shown in fig. 7. We say the function maps X into Y.
If we remove the element 16 in set Y we have the situation shown in fig. 8. Since the
elements are in one-to-one correspondence, the function maps X onto Y.
3.2 Notation
The symbol f (x) denotes the image of x under the function f. Since the image is y, this
is written as f ( x) = y . This means x is mapped onto y or symbolically x → y .
There are five main ways in which you can show the function f ( x) = 2 x − 1 , namely
(i)
y = 2x − 1
(ii) f ( x) = 2 x − 1
65
(iii) f : x → 2 x − 1
(iv) an arrow diagram
(v) on a Cartesian graph
3.3 Evaluating Functions
f (x) is the value of the function f at x? Thus, f (2) is the value of f at x = 2.
Examples
1. The functions f ( x) = 2 x + 3 , g ( x) = 1 − 4 x and
(a) Evaluate each of the following:
(i) f (0)
(ii) g (−3)
1
(iv) f ( 4 )
(b) Find the value of x if
(i) f ( x) = 1
Solutions
1. (a) (i) f ( x) = 2 x + 3
(iii) h (6)
h ( 103 )
(vi)
(ii) g ( x) = 9
(iii) f ( x) = 2 g ( x)
(ii) g ( x) = 1 − 4 x
(iii)
g (−3) = 1 − 4(−3)
= 1 + 12
= 13
(vi) h ( x) = − 13 x
g ( 52 ) = 1 − 4( 52 )
= 12 + 3
= 72
h ( x) = − 13 x
h (6) = − 13 (6)
= −2
(v) g ( x) = 1 − 4 x
f ( 14 ) = 2( 14 ) + 3
(b)(i) f ( x) = 1
2x + 3 = 1
2x = − 2
x = −1
are given.
(v) g (1)
f (0) = 2(0) + 3
=0+3
=3
(iv) f ( x) = 2 x + 3
h ( x) = − 13 x
h (103 ) = − 13 (103 )
= 1 − 85
= − 53
= − 101
(ii) g ( x) = 9
1 − 4x = 9
− 4x = 8
x = −2
(iii) f ( x) = 2 g ( x)
2 x + 3 = 2(1 − 4 x)
2 x + 3 = 2 − 8x
10 x = − 1
x = − 101
Exercise 3.1
1. Given the functions f ( x) = x − 13 , g ( x) = 1 −
(a) Evaluate each of the following:
(i) f (31)
(ii) g (− 9)
1
(iv) f ( 2 )
(b) Find the value of x if
(i) g ( x) = 3
(v)
2
3
x
and h
( x) = 12 ( 5x + 1) ,
(iii) h (2)
g ( 15 )
(vi)
(ii) f ( x) = − 1
Answers to Exercise 3.1
11
1. (a) (i) 18
(ii) 7
(iii) 2
(iv)
(b) (i) x = − 3 (ii) x = 12 (iii) x = 79
− 252
h ( 103 )
(iii) 3 f ( x) = h ( x)
(v)
13
15
(vi)
5
4
3.4 Composite functions
If f and g are two functions, f g (x) is the image of g (x) under f. f g (x) is a composite
Function where g (x) is performed first and then f is performed on the result of g.
Examples
66
h ( x) = 2 ( x + 1) , find
1. Given the functions f ( x) = 4 x − 3 , g ( x) = 1 − 6 x ,
(i) f g (12)
(ii) f h (5)
(iii) g f (0)
(iv) h f (1)
(v) g h (−13)
(vi) f g (x)
(vii) h f (x)
(viii) f h g (6)
(ix) f h g (x)
1
2. The functions f ( x) =
1
a
and g ( x) = b x + 4 .
2x − 1
(i) Given that f (−1) = − 2 and g (2) = 10 , find the value of a and the value of b.
(ii) Find the value of x for which f (x) is not defined.
(iii) Evaluate g f (0) .
Solutions
1. (i)
f g (12) = f [1 − 16 (12) ]
(ii)
= f (−1)
= 4(−1) − 3
= −7
(iii) g f (0) = g [4(0) − 3 ]
= g (−3)
= f (3)
= 4(3) − 3
=9
(iv) h f (1) = h [4(1) − 3]
= h (1)
= 1 − 16 (−3)
= 32
(v)
= 12 (1+ 1)
=1
g h (−13) = g [ 12 (−13 + 1) ]
(vi)
= g (−6)
= 1 − 16 (−6)
f g ( x) = f (1 − 16 x )
= 4(1 − 16 x) − 3
= 4 − 64 x − 3
= 2
(vii)
f h (5) = f [ 12 (5 + 1) ]
h f ( x) = h (4 x − 3)
(viii)
= 1 − 23 x
f h g (6) = f h[1 − 16 (6) ]
= 12 [ (4 x − 3) + 1]
= f h (0)
= (4 x − 3) +
= f [ 12 (0 + 1) ]
1
2
= 2 x − 32 +
1
2
= f ( 12 )
= 4( 12 ) − 3
1
2
= 2x − 1
= −1
(ix)
f h g ( x) = f h (1 − 16 x)
= f [ 12 ο»(1 − 16 x ) + 1ο½ ]
= f [ 12 (2 − 16 x ) ]
= f (1 − 121 x )
= 4(1 − 121 x ) − 3
= 4 − 13 x − 3
= 1 − 13 x
67
f (−1) = − 2
2. (i)
a
= −2
2(−1) − 1
a
= −2
−3
a=6
g (2) = 10
b (2) + 4 = 10
2b = 6
b=3
(ii) f (x) is not defined when the denominator is zero.
2x − 1 = 0
x=
f ( x) =
(iii)
1
2
.
6
and g ( x) = 3 x + 4 .
2x − 1
g f (0) = g (− 6)
= 3 (− 6) + 4
= − 14 .
Exercise 3.2
1. Given the functions f ( x) = x + 8 , g ( x) = 2 x − 5 and h ( x) = 2( x − 1) , find
(i) f g (−2)
(ii) h f (1)
(iii) g h (1)
(iv) h f (−8)
(v) g f (6)
(vii) f g (x)
(vii) h f (x)
(viii) f h g (2)
(ix) f h g (x)
1
2. The functions f ( x) =
12
and g ( x) = m x − 4 .
ax + 1
(i) Given that f (−2) = − 4 and g (−3) = 2 , find the value of a and the value of m.
(ii) Find the value of x for which f (x) is not defined.
(iii) Evaluate f g (6) .
Answers to Exercise 3.2
1.
(i) 2
(viii)
−2
(ii) 16
(iii) − 5
(iv)
−2
(v) 2 (vi)
1
2
x+3
(vii) 2 x + 14
(ix) x − 4
2. (i) a = 2 , m = − 2
(ii)
x = − 12
(iii) 6
3.3 The Inverse of a Function
−1
−1
The inverse of a function f (x) is written as f ( x) . If f ( x) = y , then x = f ( y) .
In the original function f ( x) = y , x→ y (x maps onto y). In the inverse function
f −1 ( y) = x , y → x
(y maps onto x). This is shown in the arrow diagrams below.
68
This means that the Domain in the original function becomes the Range in the inverse
function, and the Range in the original function becomes the Domain in the inverse
function. In other words x and y interchange.
3.3.1 Finding the Inverse function
A one-to-one function has an inverse function.
Examples
1. Find the inverse of the f unction f ( x) = 3x − 4 .
2. given that f ( x) =
x
, find f
2x − 1
−1
( x) .
Solutions
1. Method 1: Interchanging x and y and making y the subject
f ( x) = 3 x − 4
y = 3x − 4
x = 3 y − 4 , interchanging x and y
3y = x + 4
x + 4
, making y the subject
y =
3
x+4
.
ο f −1 ( x) =
3
Method 2: making x the subject
f ( x) = 3 x − 4
y = 3x − 4
3x = y + 4
y + 4
, making x the subject
x =
3
−1
y +4
, since x = f ( y)
f −1 ( y ) =
3
x
+ 4
, writing the inverse in terms of x.
ο f −1 ( x) =
3
2.
3x
2x − 1
3x
y=
2x − 1
3y
x=
2y − 1
f ( x) =
2 xy − x = 3 y
2 xy − 3 y = x
69
y (2 x − 3 ) = x
x
2x − 3
x
ο f −1 ( x) =
.
2x − 3
y=
Remark
−1
It is a good idea to check if you have found the correct inverse function f ( x) . You do
this by taking any suitable value of x within the domain and then find the value f (x) .
Replace this value in the inverse
f −1 ( x)
to see if you get the original value of x.
Let’s take function f ( x) = 3x − 4 in example 1, whose inverse function is f
Taking x = 3, f (3) = 3(3) − 4 = 5 andο 3→ 5 . Now, f
This shows that f
−1
( x) =
−1
(5) =
7.
x+4
.
3
x+4
is the correct inverse function of f (x) .
3
f ( x) = 6 x − 7
4. g ( x) =
( x) =
5+4
= 3 andο 5→ 3 .
3
Exercise 3.3
Find the inverse of each of the following functions:
1.
−1
2. f ( x) = 1 − x
x
2x − 3
5. h ( x) =
3.
1 − 4x
x
g ( x) = 12 x − 4
6. f ( x) =
12 − 5 x
3
h ( x) = 13 ( x + 4 )
Answers to Exercise 3.3
1.
f −1 ( x) = 16 ( x + 7 )
4. f
7.
−1
( x) =
2.
3x
2x − 1
f −1 ( x) = 1 − x
5. h
−1
( x) =
3.
1
x+4
g −1 ( x) = 2 x + 8
6. f
−1
( x) =
12 − 3x
5
f −1 ( x) = 3x − 4
Miscellaneous Exercises
1. Given the functions f : x → 2 − x , g : x → 7 x − 3 and h
1
(i) h (−1) (ii) f (−2) (iii) g ( 7 )
(vii) Find the value of x if g ( x) = 4
(iv) g f (1)
: x → x 2 , evaluate
(v) f h (1)
(vi) h g (x)
2. The functions f : x → x + 1 and g : x → a x + b are given.
(i) If f (0) = g (0) and g (2) = 15 , find the value of a and the value of b.
(ii) Find g f (x) and g f (−1)
(iii) Find the inverse of g (x) .
2
3. The function g ( x) = c x + d , where c and d are constants. Given that g (2) = 3
and g (−3) = 13 , find
(i) the value of c and the value of d.
(ii) g f (x) , if f ( x) = 2 x .
(iii) the value of x if 2 g ( x) = 6
2
(iv) f −1 (4) − g (0)
70
4. Given the functions f ( x) = 2 x + 1 , g ( x) = 3x − 1 and h ( x) =
(a) Evaluate
(i) f g (4)
(b) Find
(ii) g f (4)
1 − 4x
,
6+ x
(iii) h f g (−1)
2
(i) g h (x)
(ii) f g ( x)
(c) Find the value of x if
(i) f ( x) = g ( x) (ii) g ( x) = g h ( x) (iii) f g ( x) = 4 x
(d) Find
−1
−1
−1
(i) f ( x)
(ii) g ( x)
(iii) h ( x)
(e) Find the value of x for which h (x) is not defined.
-------------------------------------------------------------------------------------------------------
71
TOPIC 15: QUADRATIC EQUATIONS
A quadratic equation is expressed in the form of ππ₯ 2 + ππ₯ + π = 0 where π ≠ 0. We can solve quadratic
equations using factorization method, completing the square method or using the formula.
Examples.
1. Solve the following equations.
(a) π₯ 2 = −2π₯ + 3
(b) 2π₯ 2 + π₯ − 3 = 0
4
1
(c) 5 π₯ 2 + 5 π₯ − 1 = 0
Solutions
1. (a) π₯ 2 = −2π₯ + 3
π₯ 2 + 2π₯ − 3 = 0
π₯ 2 + 3π₯ − π₯ − 3 = 0
π₯(π₯ + 3) − 1(π₯ + 3) = 0
(π₯ + 3)(π₯ − 1) = 0
π₯ + 3 = 0, π₯ − 1 = 0
π₯ = −3, π₯ = 1
(b) 2π₯ 2 + π₯ − 3 = 0
2π₯ 2 + 3π₯ − 2π₯ − 3 = 0
π₯(2π₯ + 3) − 1(2π₯ + 3) = 0
(2π₯ + 3)(π₯ − 1) = 0
2π₯ + 3 = 0, π₯ − 1 = 0
−3
π₯= 2 , π₯=1
1
π₯ = −1 2 , π₯ = 1
(c)
4 2
π₯
5
2
1
+ 5 π₯ − 1 = 0 multiply all the terms by 5 we get
4π₯ + π₯ − 5 = 0
4π₯ 2 − 4π₯ + 5π₯ − 5 = 0
4π₯(π₯ − 1) + 5(π₯ − 1) = 0
π₯ − 1 = 0 , 4π₯ + 5 = 0
1
π₯ = 1 , π₯ = −1 2
2. Solve the equations below and give your answers correct to 2 decimal places.
(a) π₯ 2 + 4π₯ − 3 = 0
(b) 2π₯ 2 − 5π₯ − 6 = 0
Solutions
2. (a) π₯ 2 + 4π₯ − 3 = 0
π₯ 2 + 4π₯ = 3
π₯ 2 + 4π₯ + 22 = 3 + 22
( π₯ + 2)2 = 3 + 4
(π₯ + 2)2 = 7
π₯ + 2 = β√7
72
π₯ = β√7 − 2
π₯ = −2.645751311 − 2, π₯ = 2.645751311 − 2
π₯ = −4.65 , π₯ = 0.65
(b).2π₯ 2 − 5π₯ − 6 = 0
Divide all the terms by 2
5
π₯2 − 2 π₯ − 3 = 0
5
2
π₯2 − π₯ = 3
5
Now add the square of half of (− 2)
5
−5
−5
π₯ 2 − 2 π₯ + ( 4 )2 = 3 + ( 4 )2
5
25
(π₯ − 4)2 = 3 + 16
5
73
(π₯ − )2 =
4
16
5
4
73
16
π₯ − = β√
π₯=β
√73
5
+4
4
π₯ = −0.89, π₯ = 3.39
3. Solve the equation π₯ 2 + 2π₯ = 5, giving your answer correct to 2 decimal places.
Solution
Using the formula
π₯ 2 + 2π₯ − 5 = 0
π₯=
−π±√π2 −4ππ
2π
π₯=
−2 ± √22 − 4(1)(−5)
2(1)
π₯=
−2 ± √4 + 20
2
π₯=
−2 ± √24
2
π₯=
−2 ± 4.8989794856
2
π = 1, π = 2 π = −5
73
π₯ = 1.45, π₯ = −3.45
Activity
1. Solve the following equations
(a) π₯ 2 − 2π₯ = 0
(b) −3π₯ 2 − 8 = 10π₯
2. Solve the equation π₯ 2 + 4π₯ − 10 = 0, giving your answer correct to two decimal places.
3. Solve the equation 1 − 2π − 5π2 = 0, giving your answers correct to two decimal places.
4. Solve the equation 7 − 5π₯ − π₯ 2 = 0, giving your answers correct to 2 decimal places.
5. Solve the equation π₯ 2 + π₯ − 1 = 0, giving your answers correct to 2 decimal places.
6. (2π₯ − 1)(3π₯ − 2) = 3
Solutions
1. (a). π₯ = 0 , π₯ = 2
−2
(b) π₯ = 4, π₯ =
3
2.
π₯ = 1.74, π₯ = −5.74
3.
π₯ = −0.69, π₯ = 0.29
4.
π₯ = −4.64, π₯ = 1.14
5.
π₯ = 0.62, π₯ = −1.62
TOPIC 16: VARIATION
1) Direct Variation
π¦
a) π¦ π£πππππ ππππππ‘ππ¦ ππ π₯ ππ π‘βπ πππ‘ππ = π. πβπππππππ, π¦ = ππ₯
π₯
b) πβπ π π¦ππππ ∝ πππππ‘ππ ′πππππππ‘πππππ π‘π ′ . πβπππππππ, π¦ ∝ π₯
2) Inverse Variation
π¦
π
a) π¦ π£πππππ πππ£πππ πππ¦ ππ π₯ ππ π‘βπ πππ‘ππ 1⁄π₯ = π. πβπππππππ, π¦ = π₯
1
b) πβπ π π‘ππ‘πππππ‘ π¦ π£πππππ πππ£πππ πππ¦ ππ π₯ ππ π€πππ‘π‘ππ ππ π¦ ∝ π₯
c) If two quantities π₯ and π¦ vary in such a way that when one increases the other decreases, or vice
versa, then the two are said to be showing inverse variation.
3) Joint Variation: A variation in which one variable depends on two or more other variables.
4) Partial Variation: Variation as the sum of parts.
Direct Variation
1
1) It is given that y is directly proportional to the square of x and that y = 1 when x = 2. Find
a) π‘βπ πππππ’ππ πππ π¦ ππ π‘ππππ ππ π₯,
b) π‘βπ π£πππ’ππ ππ π₯ π€βππ π¦ = 9.
π΄ππ .
a. π¦ ∝ π₯ 2
π¦ = ππ₯ 2
π
π) 9 = 4π₯ 2
9
3
π₯2 = 4 ∴ π₯ = ± 2
1=4
π = 4 ∴ π¦ = 4π₯ 2
2) π¦ ππ ππππππ‘ππ¦ πππππππ‘πππππ π‘π π‘βπ π ππ’πππ ππππ‘ ππ π₯. πΊππ£ππ π‘βππ‘ π¦ = 12 π€βππ π₯ = 36, ππππ
a) π‘βπ πππππ’ππ πππ π¦ ππ π‘ππππ ππ π₯,
b) π‘βπ π£πππ’π ππ π₯ π€βππ π¦ = 10.
74
π΄ππ .
a) π¦ ∝ √π₯
π¦ = π √π₯
π) 10 = 2√π₯
√π₯ = 5
1
12 = π√36
π₯2 = 5
12 = 6π
(π₯ 2 ) = 52
1
π = 2 ∴ π¦ = 2√π₯
2
π₯ = 25
Inverse Variation:
π
1) A formula connecting x and y is π¦ = 3 , where k is a constant. Given that
π₯
π¦ = −1 π€βππ π₯ = 2, πππππ’πππ‘π π‘βπ π£πππ’π ππ
a) π,
b) π₯ π€βππ π¦ = 64.
π΄ππ .
π
a) −1 = 23 ∴ π = −8
−8
b) 64 = π₯ 3
64π₯ 3 = −8
−8
π₯ 3 = 64
π₯3 =
1
(π₯ 3 )3
−1
8
=
2) It is given that π
1
−1 3
(8 )
12
= π
√
∴π₯=
−1
2
a. π·ππ πππππ π‘βπ πππππ‘ππππ βππ πππ‘π€πππ π πππ π ππ π€ππππ ππ¦ πππππππ‘πππ π‘βπ π πππ‘ππππ ππ π‘βπ
πππ π€ππ π ππππ.
b. πΆππππ’πππ‘π π π€βππ π = 4.
π΄ππ .
a) π π£πππππ πππ£πππ πππ¦ ππ π π ππ’πππ ππππ‘ ππ π.
12
b) 4 = π
√
4√π = 12
1
1
2
√π = 3, π 2 = 3, (π 2 ) = 32 , π = 32 ∴ π = 9
3) It is given that π¦ varies inversely as π₯. Some corresponding values of π₯ and π¦ are given in the table
below.
0.6
0.9
π₯
π
30
9
π¦
π
a) Find the equation connecting π₯ and π¦,
b) Find the values of π and π.
π΄ππ .
1
18
a) π¦ ∝ π₯
π) π = 0.9
π
π¦=π₯
30 =
0.9π = 18 ∴ π = 20
π
0.6
π = 18 ∴ π¦ =
9=
18
π₯
18
π
9π = 18 ∴ π = 2
75
Joint Variation
1
2
1) Given that y varies directly as x and z and that y = 9 when x = 6 πππ π§ = , find;
π΄ππ .
π) π¦ ∝ π₯π§
a) k (the constant of variation ),
b) the value of y when x=4 and z = 3,
1
c) the value of x when y = 4 2 and z = 5.
π) π¦ = 3π₯π§
π¦ = ππ₯π§
π) π¦ = 3π₯π§
=3×4×3
1
9 = π(6) (2)
∴ π¦ = 36
9
2
=3×π₯×5
9
2
= 15π₯
3
9=3∴π=3
30π₯ = 9 ∴ π₯ = 10
Partial Variation:
1) The resistance (R Newton’s) to the motion of the motor vehicle is partly constant and partly varies
directly as the square of the velocity (v m/s). Write down a formula for R in terms of v.
π΄ππ .
R = a + bv2
2) The velocity v m/s of a body moving with constant acceleration is given by v = u + at where t seconds
is the time the body is in motion and then u and a are constants. If v = 30 when t = 10 and v = 40 when
t =15, find the values of u and a.
π΄ππ .
30 = u + 10a ……………………………………………………………………………………….. (i)
40 = u + 15a ……………………………………………………………..………………………… (ii)
Solving these equations simultaneously we get, a= 2 and u = 10.
TUTORIAL
(1) Given that y is proportional to π₯ 3 and that y =250 when π₯ = 10, find
(a) the value of the constant π,
(b) y when π₯ = 4,
(c) π₯ when y = 54.
(2) It is given that π€ varies directly as the square of π₯ and inversely as y.
(a) Write an equation for π€, in terms of π₯, π¦ and a constant π.
1
π΄nπ .
4
π΄ππ . 16
Aππ . 6
π΄ππ . π€ =
ππ₯ 2
π¦
(b) If π₯ = −6, π¦ = 12 and π€ = 15, find π.
π΄ππ . 5
(c) Find the value of π¦ when π₯ = 8 and π€ = 20.
π΄ππ . 16
(3) Two variables π and π have corresponding values as shown in the table below.
π
π
3
5
6
5
7
14
5
Given that π varies directly as π, find
(a) the constant of variation, π
(b) the value of π when π = 5,
(c) the value of π when π = 6.
6
2
π΄ππ . 5
π΄ππ . 2
π΄ππ . 15
76
(4) Given that π₯ varies as π¦ and inversely as π§ 2 and that π₯ = 12 when π¦ = 3 and π§ = 2, ππππ
16π¦
(a) the equation connecting π₯, π¦ and z
π΄ππ . π₯ = π§2
(b) the value of π₯ when π¦ = 3 and π§ = 4,
π΄ππ . 3
(c) the values of z when x =4 and π¦ = 25.
π΄ππ . 10
TOPIC 17: CIRCLE THEOREM
Hint:
- Discuss circle properties such as Radius, Diameter, Circumference, Sector, Chords, Segment, and Tangent
properties. Then move on to angles in a circle.
LENGTH ELEMENTS IN A CIRCLE
A circle may be defined as a set of points in a plane equidistant from a given point called the centre.
AREA ELEMENTS IN A CIRCLE
CHORD PROPERTIES
77
A chord is any straight line joining two points on the circumference. It divides the circumference into two arcs:
The longer arc is called the Major arc while the shorter arc is called the Minor arc.
TANGENT PROPERTIES
Theorems of angles in a Circle.
1. Angle at the centre theorem:
- Angle subtended by an arc at the centre is twice the angle subtended at the circumference.
2. Angle in the same segment:
- Angles subtended by the same segment of a circle are equal.
78
3. Angle in a semi-circle:
- Angle in a semi-circle= 90°.
4. Cyclic-quadrilateral:
5. Angles associated with a cyclic-quadrilateral:
-
The opposite angles of a cyclic-quadrilateral are supplementary.
The exterior angle of a cyclic-quadrilateral is equal to the opposite interior angle.
6. Alternate Segment theorem:
- The angle between a chord and a tangent at the point of contact is equal to any angle in the Alternate
Segment.
ACTIVITY 1
79
In the diagram, A, B, C, D and E lie on the circumference of a circle.
DE is parallel to CB. Angle ACE = 48°, angle CED = 65° and angle CBE = 73°. Calculate;
a) ππππππ΄π΅πΈ b) ππππππ΄πΆπ΅,
c) ππππππ΅πΈπΆ, d) ππππππΆπ·πΈ
EXPECTED ANSWERS:
1)
π·πΈ β¦ πΆπ΅
< π΄πΆπΈ = 48°
< πΆπΈπ· = 65°
< πΆπ΅πΈ = 73°
(a)
(b)
(πΊππ£ππ )
(Given)
(πΊππ£ππ )
(Given)
< π΄π΅πΈ =< π΄πΆπΈ = 48°
∴<ABE=48°
(Angles in the same segment)
< π΄πΆπ΅ =< π΅πΆπΈ−< π΄πΆπΈ
π»πππ‘:
π΅π’π‘ < π΅πΆπΈ =< π·πΆπΈ = 65°
∴< π΄πΆπ΅ = 65° − 48°
∴< π¨πͺπ© = ππ°
(c)
< π΅πΈπΆ+< π΅πΆπΈ+< πΆπ΅πΈ = 108°
∴ < π΅πΈπΆ = 180° − (< π΅πΆπΈ+< πΆπ΅πΈ)
= 180° − (65° + 73° )
= 180° − 138
∴< π©π¬πͺ = ππ°
(d)
< πΆπ·πΈ+< πΆπ΅πΈ = 180°
∴ < πΆπ·πΈ = 180° − (< πΆπ΅πΈ)
(πππ‘πππππ‘π ππππππ πππ πππ’ππ)
(π π’π ππ ππππππ ππ π π‘πππππππ)
(Opposite angles of a cyclic quadrilateral)
< πΆπ·πΈ = 180° − 73°
∴ < πͺπ«π¬ = πππ°
ACTIVITY 2
In the diagram below, ABC is a tangent to the circle BDEF at B, angle DFB= 40°,
angle EBF=15°
and angle DBE= 45°.
80
Find (a) angle DEB,
(b) Angle DEF,
(c) Angle ABF.
EXPECTED ANSWERS:
(a)
angle DEB= angle DFB= 40°
(angles in the same segment)
(b)
angle DEF+ angle DBF=180°
∴ πππππ π·πΈπΉ + 60° =180°
(opposite angles of a cyclic quadrilateral)
∴ πππlπ π·πΈπΉ =180° − 60°
∴ πππππ π«π¬π = πππ°
(c)
angle ABF= angle BDF
(alternate segment theory)
∴ πππππ π΄π΅πΉ =180° − (45° + 15° + 40°)
∴ πππππ π¨π©π = ππ°.
ACTIVITY 3.
In the diagram below, TAP and TBQ are tangents to the circle, centre O, and TO meet the circle at S and R. If
angle ATR= 34°,
find,
(i ) angle ARB,
(ii)
angle ASB,
(iii) angle PAR.
EXPECTED ANSWERS:
81
ACTIVITY 4.
In the diagram below, A, B, C and D lie on the circumference of the circle, center o.
BO is parallel to CD, angle BAD=62° πππ π΅πΆπΈ ππ π π π‘ππππβπ‘ ππππ.
Calculate:
(a)
angle t,
(b)
angle BCD,
(c)
angle OBC,
(d)
angle DCE.
EXPECTED ANSWERS:
(a)
angle t= 2× ππππππ΅π΄π·
∴angle t= 2× 62°
∴angle t=124°.
(b)
(c)
(d)
angle BCD + angle BAD= 180°
∴ angle BCD =180° − 62°
∴ angle BCD = 118°.
angle OBC= 180° − 118°
∴ angle OBC= ππ°
angle DCE= ππππππ΅π΄π· = ππ°
(angle at the centre = 2×angle on the
circumference)
(opposite angles of a cyclic quad.)
(angles associated with parallel lines)
(exterior angle of cyclic quad =
opposite interior angle)
PRACTICE QUESTIONS:
(1) O is the centre of the circle through A, B, C and D. Angle BOC= 100° and angle OBA=62°.
82
Calculate
(i)
BAC,
(ii)
OCB,
(iii)
ADC.
(2) In the diagram, AB is a diameter of the circle, Centre O. P and Q are two points on the circle and
APR is a straight line.
Given that angle QBA = 67° and angle PAQ = 32° ,
Calculate
(a) angle QAB,
(b) angle RPQ,
(c) angle POB.
(3) BT is a diameter of a circle and A and C are points on the circumference. The tangent to the circle at
the point T meets AC produced at P.
Given that angle ATB = 42° and angle CAT = 26°,
Calculate
(i)
angle CBT,
(ii)
angle ABT,
(iii)
angle APT.
(4) The circle BCDE has centre O. AB and AC are tangents to the circle. COFE is a straight
line.
83
(i) Give a brief reason why angle OBA is a right angle.
(ii) Given that angle CED = 56°,
Calculate,
(a) angle BOC,
(b) angle BDC,
(c) angle OFD.
(5) In the diagram below, A, B, C, and D are points on the circumference of a circle. TCM and
MDE are tangents to the circle at C and D respectively.
ANSWERS:
(1) (i)
50°,
(ii) 40° ,
(iii)
78° .
(2) (a)
23° ,
(b) 67° ,
(c)
110° .
(3) (i)
26° ,
(ii) 48° ,
(iii)
22°.
(4) (ii)
BOC = 124° ,
(b) BDC = 62° ,
(c)
OFD = 93° .
(5) (i) angle BCT = 23° , (ii) angle BAD = 70° , (iii) angle ABD = 55°, (iv) angle DMC = 86°.
84
TOPIC 18: CONSTRUCTION AND LOCI
85
86
87
TOPIC 19: TRIGONOMETRY QUESTIONS
Sine, cosine, tangent
The trigonometric ratios sine, cosine and tangent are defined in terms of the hypotenuse, opposite and
adjacent side of a right-angled triangle.
πππππ ππ‘π
ππππ π = π»π¦πππ‘πππ’π π
Opposite
Hypotenuse
π΄πππππππ‘
Cosine= π»π¦πππ‘πππ’π π
π
πππππππ‘ π =
Adjacent
πππππ ππ‘π
π΄πππππππ‘
Sine rule
Sine rule is used when you are given
i.
two angle and one side
ii.
two side and a non-included angle
In obtuse-angled triangle π ππ π = π ππ (180° − π)
Cosine rule
Cosine rule is used when you are given;
i.
two side and an included angle
ii.
three side only
In obtuse-angled triangle πππ π = −πππ (180° − π)
Pythagoras
In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the two adjacent
sides.
TRIGONOMETRY ON THE CARTESIAN PLANE
1. Determining the signs of the three trig ratios in the quadrants.
2
2 quadrant
nd
90°
1
1 quadrant
st
180°
360°
3
3rd quadrant
4
4th quadrant
270°
The quadrants are numbered 1 to 4 in an anti-clock wise direction.
i.
angles between 0 and 90 fall in the first quadrant
ii.
angles between 90 and 180 fall in the second quadrant
iii.
angles between 180 and 270 fall in the third quadrant
iv.
angle between 270 and 360 fall in the fourth quadrant
We can find trig ratios on Cartesian plane given a point
y
P
i)
π
0
r
y
x
X
88
y
P
r
ii) π¦
π
π₯
π
π
y
y
iii)
π₯
π
X
0
y
r
P
y
iv)
π
π₯
0
X
r
y
P
• P(x, y) is a point in each of the four quadrants of the Cartesian plane.
The length ππ = π, where π > π and < πππ = π
π¦
π
πππ π = , πΆππ π =
•
π₯
π
and πππ π =
π¦
π
The signs of x and y differ from quadrant to
quadrant so the values of sin, cos and tan will differ as well.
•
π is the angle between the line OP and the positive π₯ − ππ₯ππ
TRIGONOMETRIC EQUATIONS
Sinπ, cosπ and tanπ can be positive or negative depending on the quadrant within which π fall.
The Cartesian diagram
•
Sine +ve
S
all positive
A
Tan +ve
T
Cosine +ve
C
The cast diagram is very useful when solving trig equations
89
•
•
A trigonometric equations will usually have two solutions
To solve equations of the form π πππ = π, πππ π = π ππ π‘ππ π = π
i.
find the reference angle ∝ in the 1st quadrant
ii.
determine in which two quadrants π will lie
iii.
find the corresponding angle in the two quadrant
In the diagram below P has coordinates (12, 5)
y
P (12, 5)
r
0
y
π₯
Find the value of;
a) < sin <XOP
b) cos <XOP
c) tan XOP
Solutions
First find side ‘r’ by using the Pythagoras theorem.
π2 = π¦2 + π₯2
π 2 = 52 + 122
π 2 = 25 + 144
π 2 = 169
√π 2 = √169
π = 13π’πππ‘π .
π
π΄
π
∴ (a) π ππ < πππ = π»
(π) πππ < πππ = π»
(π) π‘ππ < πππ = π΄
π¦
π₯
π¦
=
=
=
π
π
π₯
5
12
5
=
=
=
13
13
12
5. Solve the equation sin π=0.766 for 0° ≤ π ≤ 180°.
Solution
πππ π = 0.766
∝= π ππ−1 0.766
= 49.99603866°
= 50°
Sin is positive in the 1st and 2nd quadrant
(i) 1st quadrant
π = 50°
(ii) 2nd quadrant
π = 180° − 50°
π = 30°
∴ π = 50° πππ π = 130°
6. Solve the equation π‘πππ = −5.67 πππ 0° ≤ π ≤ 360°.
Solution
π‘πππ = −5.67
∝ = π‘ππ−1 5.67
= 79.99°
= 80°
Tan is negative in the 2nd and 4th quadrants.
2nd quadrant
4th quadrant
π = 180° − 80°
π = 360° − 80°
90
= 100°
∴ π = 100° πππ π = 280°
= 280°
A kite (k) being flown is such that its vertical height HK is 6m and the angle formed between the vertical
height and the string Skit is attached to is 60° as shown in the diagram below.
K
60°
6m
S
H
If π ππ 60° = 0.866, πππ 60° = 0.5 and π‘ππ 60° = 1.73, calculate the length of the string SK.
Solution
π΄
πΆππ π =
π»
6
πΆππ 60° =
ππΎ
6 = ππΎπππ 60°
6
ππΎ =
= 12π.
0.5
In the diagram below, AC=10cm, BC=5cm and <ACB=60°. Given that
π ππ60° = 0.866, πππ 60° = 0.5 πππ π‘ππ60° = 1.73. Calculate the value of (π΄π΅)2 .
B
5cm
A
60°
10cm
Solution
(π΄π΅)2 = (π΄πΆ)2 + (π΅πΆ)2 − 2π΄π΅ × π΅πΆ × πππ πΆ
= 102 + 52 − 2(10)(5)(0.5)
= 100 + 25 − 50
2
(π΄π΅) = 75ππ
The figure below shows triangle ABC in which AC=5cm. Given that
π ππ π΅ = 0.5, πππ π΅ = 0.9 πππ π‘πππ΅ = 0.6. Calculate the length of BC;
C
5cm
/
A
B
Solution
π
πππ π = π»
π΄πΆ
πππ π΅ = π΅πΆ
5
0.5 = π΅πΆ
0.5π΅πΆ = 5
π΅πΆ = 5 ÷ 0.5
∴ π΅πΆ = 10ππ
91
C
2. PQ and R are fishing camps along the banks of Lake Kariba joined by straight paths PQ, QR and RP. P is
7.6km from Q and Q is 13.2km from R and <PQR=120°.
Q
120°
13.2km
R
P
7.6km
a) Calculate;
i. The distance PR
ii. The area of triangle PQR
iii. Find the shortest distance from Q to PR
b) A fisherman takes 30 minutes to move from R to P. calculate his average speed in km/h.
Solution
a) (i) (ππ
)2 = (7.6)2 + (13.2)2 − 2(7.6 × 13.2 × πππ 120°)
= 232 − 2(−50.16)
2
(ππ
) = 332.32
ππ
= 18.2ππ (3 π . π)
1
(ii) Area of trianglePQR=2 × 7.6 × 13.2 π ππ 120°
= 43.4ππ2
1
(iii) In the formula, π΄ = π β ( β is a shortest distance)
2
A=43.4ππ2, π = ππ
= 18.2ππ
1
43.4= × 18.2 × β
2
43.4
β =
9.1
β = 4.773ππ
β = 4.8ππ
∴ π‘βπ π βπππ‘ππ π‘ πππ π‘ππππ ππ 4.8ππ
π·ππ π‘ππππ
30
b) π΄π£πππππ π ππππ = ππππ
π· = 18.2ππ , π = 60 = 0.5βππ
18.2ππ
= 0.5βππ
= 36.4ππ/β
4) π΄π΅πΆ is a triangle in which πππππ π΄πΆπ΅ = 90°, D is a point on π΄πΆ,
π΄π΅ = 20ππ, π΅πΆ = 12ππ, πΆπ· = 5ππ, π΅π· = 13ππ and π·π΄ = 11ππ. Giving each answer as a fraction,
find;
a) Tan <CDB
b) Cos <CAB
c) Sin <ADB
C
C
5cm
D
12cm
11cm
A
13cm
20 cm
B
Solution
92
π΅πΆ
a) π‘ππ < πΆπ·π΅ = π·πΆ
c) π ππ < π΄π·π΅ =?
12
π·πΆ 2 = 132 − 122
π ππ < π΅π·πΆ = 13
= 169 − 144
2
π·πΆ = 25
π·πΆ = 5ππ
12
∴ π‘ππ < πΆπ·π΅ = 5
π ππ < π΄π·π΅ = − 13
12
16
b) πππ < πΆπ΄π΅ = 20
4
=5
5) In the diagram below QP = 1.8 π’πππ‘π , ππ
= 2.5 π’πππ‘π , find the size of the angle marked π.
P
1.8units
Q
R
Solution
π
2.5units
R
πππππ ππ‘π
π»π¦πππ‘πππ’π π
1.8
πππ π =
2.5
1.8
π = π ππ−1( )
2.5
π = 46°
πππ π =
6)
C
A
4cm
5cm
120°
B
D
ABC is a triangle with AB=5cm. BC=4cm and angle ABC=120°. AB is produced to D and angle BCD=90°.
Using as much information in the table below as necessary;
sin
cos
tan
0.87
-0.5
-1.73
120°
Calculate,
a) The area of the triangle ABC
b) The length of BD
Solution
1
a) Area of triangle π΄π΅πΆ = 2 × 5 × 4 × π ππ 120°
1
= 2 × 20 × 0.87
= 8.7ππ2
π΅πΆ
b) π΅π· =? πππ < πΆπ΅π· = π΅π·
4
πππ 60° = π΅π·
4
0.5 = π΅π·
∴ π΅π· = 8ππ.
93
PRACTICE QUESTIONS:
(1) In the right angled triangle ABC, P is a point on the side AB. Given that AP= 4cm,
PB = 5cm, BC = 12cm and PC = 13cm, calculate
(a)
(b)
(c)
(d)
AC,
cos BPC,
tan PAC,
sin APC.
(2) ABCD represents a building with a vertical flagpole AP on the roof.
The point O is on the same level as C and D.
The angle of elevation of A from O is 15°, OA = 60 metres and POA =7°
(i)
(ii)
Calculate
(a) the height AD of the building,
(b) the height of the flagpole, AP.
Given also that AB =10 metres, calculate the angle of elevation of P from B.
(3) In the diagram, ABC represents a horizontal triangular field and AD represents
a vertical tree in the corner of the field. A path runs along the edge BC of the field.
AB =83m, AC= 46m and angle BAC= 67°.
(a) The angle of elevation of the top of the tree when viewed from B is 14°. Calculate the
height of the tree.
(b) Calculate the length of the path BC.
(c) Calculate the area of the field ABC.
(d) Calculate the shortest distance from A to the path BC.
(e) Calculate the greatest angle of elevation of the top of the tree when viewed from any point
on the path.
(4) The diagram shows a triangular field PQR.
94
Calculate (a) angle QPR,
(b) The area of triangle QPR.
ANSWERS:
(1) (a) AC = 15cm,
(2) (i) (a) 15.53m
(ii)
(b) cos BPC=
(b) 7.89m
5
13
,
(c) tan PAC =
4
3
,
(d) sin APC=
12
13
.
38.3° .
(3) (a) 20.7m ,
(b) 77.6m ,
(4) (a) 101.7°,
(b)
(c )
1760m2 ,
(d) 45.3m ,
(e)
24.6° .
1660.76m2.
TOPIC 20: MENSURATION
Specific outcome:
• Calculate the area of a sector.
• Calculate surface area of three dimensional figures.
• Calculate volume of prisms.
• Solve problems involving area and volume.
1
1. [Volume of a cone= 3 × πππ π ππππ × βπππβπ‘]
The diagram shows a plant pot. The open end of the plant pot is a circle of radius 10cm. The closed end is a
circle of radius 5cm. The height of the pot is 12cm. The plant pot is part of the right circular cone of height
24cm.
I.
II.
Calculate the volume of the plant pot. Give your answer in litres.
How many of these plant pots can be completely filled from a 75litre bag of compost?
95
Solutions
i.
1
3
Volume of larger cone= × ππ 2 × β
1
= 3 × 3.142 × 102 × 24
=2513.6πππ
1
Volume of the smaller cone= × 3.142 × 52 × 12
3
=314.2πππ
Volume of plant pot= 2513.6 − 314.
= 2199.4ππ3
1 litre → 1000ππ3
π₯ → 2199.4ππ3
π₯=
2199.4
1000
= 2.1994
=2.20 litres
75
ii.
The number of pots =2.1994 = 34.100
Therefore the number of completely filled = 34
2.
Diagram 1
Diagram 2
Diagram 1 shows a hollow cone whose sloping edge is of length t cm. The radius of the circular top is r cm.
The cone is cut along its sloping edge and laid flat to form the sector OPQ of a circle of radius t cm as shown
in Diagram 2.
1). Find an expression in terms of r, for the length of the arc PQ
2). It is given that t = 5r
a). calculate POQ
b). given also that π‘ = √40, calculate the area of the sector OPQ, expressing your answer as a multiple of π.
Solutions
1). Length of arc PQ= circumference of circular top
Arc PQ= 2πr
π
2).a).length of arc= 360 × 2ππ
2ππ =
πππ
360
× 2π × 5π, since t= 5r
720 = πππ
πππ = πππ
π
b). Area of sector = 360 × ππ 2
72
= 360 × π × (√40)2
96
=
72
×
36∅
π × 4∅
= ππ
3. The base of a pyramid is a square with diagonals of length 6cm. The sloping faces are isosceles
triangles with equal sides of length 7cm. the height of the pyramid is √π. calculate π
Solutions
Hint
Construct a right-angled triangle and use the Pythagoras theorem to find the height.
Applying the Pythagoras theorem
72 = β2 +32
72 − 32 = β2
49 − 9 = β2
(√40)2 = (√β)2
40 = β
β = 40
Therefore, π = 40
4. A solid cuboid measures 7cm by 5cm by 3cm
7cm
5cm
3cm
7
I.
Calculate the total surface area of the cuboid.
II.
A cube has the same volume as the cuboid. Calculate the length of an edge of this cube
Solutions
i).The total surface area =2(7x5) + 2(3x5) + 2(7x3)
= 70 + 30 + 42
= 142cm2
ii). Volume of the cuboid = 7 x 5 x 3
= 105ππ3
Length of edge = √105
= 4.72cm (to 3s.f)
5. A closed container is made by joining together a cylinder of radius 9cm and a hemisphere of radius
9cm as shown in diagram 1
The length of the cylinder is 18cm.
The container rests on a horizontal surface and is exactly half full of water.
97
a. Calculate the surface area of the inside of the container that is in contact with the
water.
b. Show that the volume of the water is 972πππ3
Solutions
1
2
a. Surface area = ππ 2 +
1
2
81
= π
2
1
2
1
4
= π(9)2 + [2π(9)(18)] + [4π(9)2
1
1
(2ππβ) + [4π(9)2 ]
2
4
+ 162π + 81π
81
= π ( + 162 + 81)
2
= 283.5 x 3.142
= 891πππ π‘π 3π π.
1 2
1 2
b. Volume of water = 2 [3 ππ 3 + π(81)(18)]
= 2 [3 π(9)3 + π(81)(18)]
1
= 2 (486π + 1458π)
1
= 2 (1944π)
=972π
πππ βππππ π βππ€π.
5. A car’s windscreen wiper left a part of a windscreen unwiped, as shown in the diagram below
a) Calculate the length of arc BB1
b) Calculate the length of arc AA1
c) Calculate the Area of the shaded part of the windscreen.
Ans; a) BB1 =
2
=
2
= 54.977 = 54.98cm
b) AA1 =
=
2
2
98
= 36.6519
= 36.65cm
c) A = area of sector OBB1 – area of sector OAA1
=
2
=
2
2
-
2
= 577.267 – 256.563
= 320.704
= 320.70cm
6.
O
In the diagram, O is the centre of a circle of radius 8cm. PQ is a chord and POQ= 150 . The minor segment of
the circle formed by the chord PQ is shaded. (Take as 3.142), calculate
i) the length of triangle the minor arc PQ
ii) the area of triangle OPQ
iii) the shaded area
Ans: i) minor arc PQ=
2
=
2
=20.9cm
ii) area of triangle OPQ = ab sin
=
8
8 sin 150
2
= 16cm
iii) area of sector =
=
2
3.142
= 83.79cm
Shaded area = area of sector – area of triangle
= 83.29cm2 – 16cm2
= 67.8cm2
7. OAB is the sector of a circle of radius r cm, AOB = 60
A
O
B
99
Find, in its simplest form, an expression in terms of r and
a) The area of the sector
b) The perimeter of the sector
a) Area of sector =
for
=
=
b) Arc length AB =
=
x2 r
=
r
x2 r
Perimeter of the sector = OA + OB + AB
=r+r+
r
= 2r +
=r(
r
)
8. In the diagram below, MN is an arc of a circle whose centre is O and radius 21cm
Given that
MON = 120 , calculate the area of the sector MON
[Take
to be
]
Ans:
x
x 212 = 462cm2
8. The net of a square based pyramid is shown below
40cm
Calculate the total surface area of the pyramid
Ans: TSA= S (2L +S)
= 40(2 x 32.02 + 40)
= 4161.6cm2
9. The area of the base of a square- based pyramid is 100mm2 and its slant height is 25mm
a) Calculate the total surface area of the pyramid
b) Express the relationship between the total surface area of the pyramid and the
100
area of its base as a ratio in its simplest form
ans: S2 = 100mm2
s= 10mm
a) TSA= s(2l + s)
= 10(50 + 10)
= 10(60)
= 600mm2
b) Ratio
=600: 100
= 6:1
10. The volume of a cone is x base area x height. The area of the curved surface of a cone of radius r and
slant L πrl.
A solid cone has a base radius of 8cm and a height of 15cm.
Calculate
(i). its volume
(ii).its slant height
(iii). its curved surface area
(iv).its total surface area
Solutions
(i)
Volume = πr2h
= x
x 82 x 15
=1005.3cm2
(ii)
Slant height =
= 17cm
(iii)
Curved surface area = πrl
= π x 8 x 17
= 427.256
= 427.26 cm2
(iv)
Total surface area =
+
2
= π (8) + 427.26
= 201.09 + 427.26
= 628.348 cm2
12. The volume of a cone = x base area x height. The diagram shows a plant pot. The open end of the plant
pot is a circle of radius 10cm. the closed end is a circle of 5cm. The height of the pot is 12cm. The plant
pot is part of a right circular cone of height 24cm
101
(i). calculate the volume of the plant pot. Give your in litres
(ii). A smaller plant pot is geometrically similar to the original plant pot. The open end of this plant pot is a
circle of radius 5cm.
How many of these plants pots can be completely filled from a 75 litre bag of compost?
Solution
(i) Volume of larger cone
=β
× πr2 × h
=2513.6 cm3
Volume of smaller cone
=β
× 3.142 × (5)2 × 12
=314.2cm2
Volume of plant pot
=2513.6 314.2
=2199.4cm3
=1 litre = 1000cm3
=2.1994 litres
=2.20 litres
(i)
number of pots =
= 34.100
number of pots completely filled = 34
(ii)
=
3
3
=
=
=
Volume of small pot = ×2.1994
No of pots =
=0.2749 litres
= 272.826
No of pots completely filled= 272
The base of a pyramid is a square with diagonals of length 6cm.
The sloping faces are isosceles triangles with equal sides of length 7cm
102
The height of the pyramid is
cm. Calculate L
Solution
Applying Pythagoras theorem
72=h2+32
49=h2+9
h2=40
h=
2
=40
Or L=40cm.
TOPIC 21:PROBABILITY
Probability is a measure of the likelihood of an event happening. Let E be an event, the probability of E
happening P (E) is given by
P (E) =
πππ πππ π ππππ ππ’π‘πππππ π‘βππ‘ πππ£ππ’π (πΈ)
π‘ππ‘ππ πππ π ππππ ππ’π‘πππππ
Example
In a family of five children, two are sons and three are daughters. If parents wish to take a child to China to
study for a first degree, what is the probability of sponsoring a girl.
β Two sons
β Three daughters
β There are a total of 5 possibilities of sponsorship.
2
P(son) = 5
3
P(daughter) = 5
3
Therefore, probability of sponsorship a girl is 5.
β The probability of an event not taking place = 0.
β The probability of an event guaranteed to take place = 1.
103
Examples
In a chicken pen there are 6 geese and 10 chickens. A bird is picked from the pen for slaughter, what is the
probability that it is
a) Geese
6 geese
total = 6 + 10 = 16
10 chickens
6
3
P (geese) = 16 = 8
b) A chicken
β chicken are 10
10 5
P(chicken) = 16 = 8
c) A cow
0
P(cow) = 16 = 0 since there is no cow in the pen.
d) P(bird). There are 16 birds.
16
P(bird) = 16 = 1, since every animal in the pen is a bird.
Probability Tools
β The common tools used to build probability theory are coins (unbiased), dice, packs of cards and other
familiar items such as balls and beads.
DIE
An unbiased die is a perfect cube in which all the faces have the same chance landing on top. The faces are
marked with dots 1 to 6.
A PACKED OF CARDS
A set of cards has 52 cards. 26 cards are black which are 13 spades and 13 clovers. It has 26 red cards which
are 13 diamonds and 13 hearts. They are distributed as knights (K), Queen (Q) , Jacks (J), Ace (A) one heart,
one diamond, one clover, one spade.
Example
A card is drawn at random, what is the probability that it is
β A jack
4
1
P (J) = 52 = 13
β A diamond
13 1
P (D) = 52 = 4
β Black
26 1
P (B) = 52 = 2
MUTUALLY EXCLUSIVE EVENTS
β Exclusive means separated, mutually exclusive events are events that cannot occur together. It A and
B are exclusive events the we can talk of P(A or B).
β The probability of A, B .... that are mutually exclusive events P(A or B) = P(A) + P(B).
INDEPENTENT EVENTS
β These are event that do not affect one another and could happen together. For example, if we toss a
coin twice then the outcomes of the first does not affect the outcomes of the second toss.
104
β If an unbiased die and unbiased coin are tossed the possibilities (probability) space is shown as
follows.
ACTIVITY
1
2
π»π
ππ
H
T
Toss of a coin
Toss of a coin
3
4
π»3
π»4
ππ
ππ
π»2
π2
β There are 12 possibilities
1
P(H and even number) = P(H) × P(even number) = 2 ×
If A, B, C ............ are independent then
P( A and B and C) = P(A) × P(B) × P(C) × … … ….
3
6
3
5
6
π»5
ππ
π»6
ππ
1
= 12 = 4 .
EXAMPLES
7
1. The probability that Chakupaleza will go for remedial lesson on a particular day is 10 what is the
probability that she will not go for her remedial lesson on that particular day.
[2017 MATHEMATICS PAPER 1 QUESTION 10a]
Solution
ο P (Not Going) = 1-
7
10
=
10−7
10
=
3
10
2. A die and a com are rolled and tossed respectively. What is the probability of getting a five and a
tail.
[ 2017 JULY MATHEMATICS PAPER 1 QUESTION 11(a)]
Solution
The above events are independent events.
1
1
ο P (5) = 5 , P (T) = 2
P (5 and Y) = P (5) X P (T)
=
1
6
=
1
12
X
1
2
3. A bag contains 3 black balls and two white balls. Two balls are taken from the bag at random, one
after another, without replacement.
[JULY 2016 MATH P2 Q 9b]
i.
Draw a try diagram to represent this information
ii.
Calculate the probability that two balls taken at random are of the same color.
105
Solution
Black=3, White=2
Total Ball=5,
1ST
Without Replacement
2ND
2
4
3
5
2
4
2
5
B
B
BB
(1/5)(2/5)=3/10
3
4
W
BW
(3/5)(2/4)=3/10
WB
(2/5)(3/4)=3/10
1
4
P (balls of the same colour) = P (BB) + P (WW)
3
1
= 10 + 10
B
=
4
10
W
=
π
π
W
WW
(2/5)(1/4)=1/10
4. A box contains 3 green apples and 5 red apples. An apple is picked from the box and not replaced then
a second apple is picked. Expressing the answer as a fraction in its simplest form calculate.
i.
The probability that both apples picked are green.
ii.
The probability that the two apples picked are of different colours.
Solution
β’ first draw the tree diagram
G= Green Apples (3),
R = Red Apple (5)
Total= 8
G=3,
R=5
3
8
G
5
7
5
8
R
GG
2
7
3
7
4
7
G(GR)
G(RG)
R(RR)
106
=
3
28
=
15
56
=
15
56
=
5
14
i.
P (both Green)
=
ii.
P (different colors) =
3
8
x
2
7
=
3
28
15
56
x
15
56
=
30
56
15
= 26
5. The Venn diagram below shows the number of students who took business studies (B), Human
Resource (H) and community Development(C) at Mafundisho College !00 students took these three
courses.
[OCTOBER 2015 MATHS PAPER 2 QUESTION 4]
i.
ii.
iii.
Find the value of x,
If a student is chosen at random. What is the probability that the student took.
One course
At least two courses.
Solution
i.
ii.
X+15+15=10+10+5+15=100
X+70=100
X=100-70
X=30
15+15+30
P (one course) =
=
iii.
P(at least two course)
60
6
=
100
100
π‘π€π ππ ππππ πππ’ππ π
=
100
10+10+5+15
=
100
40
= 100
2
= 5
100
=
3
5
1
6. In a certain community, the probability that a person chosen at random is of left handed is 10. Find the
expected number of left handed people in a sample of 2000 people randomly chosen from the
community.
Solution
Expected number = N x P (A)
=2000 x
1
10
= 200
ο So the expected number of left handed people in a sample of 2000 is 200.
7. A survey carried out at a certain hospital indicates that the probability that a patient tested positive for
malaria is 0.6. What is the probability that two patients selected at random.
107
i.
ii.
One tested negative while the other positive
Both patients tested negative.
Solution
β’ If 0.6 tested positive then 0.4 tested negative
β’ This information can be shown using the tree diagram.
+
0.6
0.4
P (++) = 0.36
−
P(+-) = 0.24
+0.6
0.4
−
i.
+
0.6
0.4
−
P (-+) = 0.24
P (- -) = 0.16
P ( one- while the other +) = 0.4 x 0.6 + 0.4 x 0.6
= 0.24 + 0.24
=0.48
ii.
P (both patient tested -) = 0.4 x 0.6
=0.16
8. 60% of people in a certain community can eat caterpillars. For a person randomly chosen from the
community. Find In its simple term, the probability that he/she
a) Eat caterpillars
b) Does not eat caterpillars
Solution
If 60% eat, then 40% do not eat caterpillars
a) P ( eat) =
60
100
=
40
6
10
=
4
3
5
2
b) P (do not) = 100 = 10 = 5
1. A box of chalk contains 5white, blue 4 and 3 yellow pieces of chalk. A piece is selected at random
from the box and not replaced. A second piece of chalk is then selected.
i.
Draw the tree diagram all the possible outcomes
ii.
Find the probability of selecting pieces of the same color.
108
Solution
W= 5 B=4 Y=3 Total=5+4+3 =12
Results
W
5
12
4
11
4
11
3
11
4
12
5
11
3
12
B
Y
ii). P ( same color)
3
11
3
11
5
11
4
11
2
11
Probability=12
W
WW
B
WB
Y
WY
W
BW
B
BB
Y
BY
W
YW
B
YB
Y
YY
P (WW) = 5/12 X
4/11 =5/33
P (WB) =5/12 X 4/11
= 5/33
P (WY) = 5/12 X
3/11 = 5/44
P (BW) = 4/12 X
5/11 = 5/33
P (BB) =4/12 X 3/11
=1/11
P (BY) = 4/12 X
3/11 =1/11
P (YW) = 3/12 X
5/11 = 5/44
P (YB) = 3/12 X
4/11 =1/11
P (YY) = 3/12
X2/11 = 1/22
= P (WW) + P (BB) + P(YY)
=
5
33
=
19
66
1
1
+ 11 + 22
3
2. In a test, probability that Mubita is number 1 is 14 and the probability that John is number 1 is
the probability that Mubita or John is number one, assuming no tie.
1
,
5
find
Solutions
The events, Mubita or John are mutually exclusive, therefore,
P (Mubita or John is ο£1) = p(m) + p(J)
=
=
3
14
+
1
5
29
.
70
QUESTIONS TO ANSWER.
1. A bag contains 5 while, 3 yellow and 2 green balls all identical except the color. A ball is drawn and set
aside. A second ball is drawn. What is the probability that;
4
25
12
white………………………………………………………………
25
a) The balls are of different colors…………………………………………..………..
b) One of the balls
2. Draw the probability (possibility) space for three tosses of an unbalanced coin and determine the
probality that the result are;
109
3
8
1
tail………………………………………………………8
Two heads and a tail……………………………………………………………….
i.
ii.
First two head and then a
3. During a tournament the probabilities that Mphundu School wins volleyball, netball and hockey are
2 1
3
, πππ 5 respectively. At the end of the tournament. What is the probability that Mphundu girls
3 5
2
a) Wins all the three games:………………………………………………………..25
67
b) Wins at least one game:…………………………………………………………. 75
31
c) Wins two games:………………………………………………………………… 75
4. The probabilities that Kenya, Tanzania and Uganda will be overall winners at the next all Africa
1 41
1
games are 7,20, and 10 respectively. What is the probability that
6
7
41
One of the three countries will be the overall winner ……………………….……140
513
None of the three countries will win?..................................................................... 700
a) Kenya will not be the winner?..................................................................................
b)
c)
Hint: a tree diagram not suitable for this question.
TOPIC 22: STATISTICS
1. (a) The number of people living in six houses is 3, 8, 4, x, y and z, if
The median is 7.5.
The mode is 8.
The mean is 7.
Find the value of x, y and z.
[5]
(b) Calculate the standard deviation
[3]
(c) The grouped frequency table below shows the amount (KA) spent on travel by a number of
students.
(i)
(ii)
Write down an estimate for the total amount in terms of m and n.
[2]
The calculated estimate of the mean amount is K13 exactly. Write down an equation containing m
and n, and show that it simplifies to
2m + 17n = 120.
[3]
A student drew a histogram to represent this data. The area of the rectangle representing the 0 <
π΄ ≤ 10 group was equal to the sum of the areas of the other two rectangles. Write an equations
in π and π for this relationship
[1]
(iv)
Find the values of m and n by solving the simultaneous equations
2m + 17n = 120,
m + n = 15.
[3]
2. Answer the whole of this question on a sheet of graph paper.
In a survey, 200 shoppers were asked how much they had just spent in a supermarket. The results are shown
in the table.
(iii)
(a) (i)
(ii)
Write down the modal class.
[1]
Calculate an estimate of the mean amount, giving your answer correct to 2 decimal places.
110
(b) (i)
(ii)
Make a cumulative frequency table for these 200 shoppers.
[2]
Using a scale of 2 cm to represent K20 on the horizontal axis and 2 cm to represent 20
shoppers on the vertical axis, draw a cumulative frequency diagram for this data. [4]
(c)
Use your cumulative frequency diagram to find
(i)
The median amount,
[1]
(ii)
The upper quartile,
[1]
(iii)
the inter-quartile range,
[1]
(iv)
How many shoppers spent at least K75?
[2]
3. A group of children were asked how much money they had saved. The histogram and table show the
results.
Use the histogram to calculate the values of p, q and r.
[4]
4. Answer the whole of this question on a sheet of graph paper.
120 passengers on an aircraft had their baggage weighed. The results are shown in the table.
(a) (i) Write down the modal class.
[1]
(ii) Calculate an estimate of the mean mass of baggage for the 120 passengers. Show all your
working.
[4]
(iii) Sophia draws a pie chart to show the data. What angle should she have in the 0 < π ≤ 10
sector?
[1]
(b) Calculate the Standard deviation for the data in the table above.
[7]
π¨π΅πΊπΎπ¬πΉπΊ
1. (a) For median to be 7.5, the numbers must be arranged such as
3, 4, π₯ , 8, π¦ πππ π§
π₯+ 8
∴ 7.5 = 2
π₯ = 15 − 8
π₯ = 7
If the mode is 8, y can be 8 as well
The mean is given as 7
βΉ 3+4 +7+ 8 + 8+π§ = 6 ×7
111
π§ = 42 − 30
π₯ = 7, π¦ = 8
(b)
and π§ = 12
∑ π₯2
√
SD =
π
2
− π₯Μ
2
Where ∑ π₯ = 32 + 42 + 72 + 82 + 82 + 122
= 9 + 16 + 49 + 64 + 64 + 144
∑ π₯ 2 = 346
346
−
6
ππ· = √
(i)
(ii)
49
ππ· = 2.94
15(5) + 15π + 30π
πΎπ΄ = 75 + 15π + 30π
75+15π+30π
= 13
15+π+π
2π + 17π = 120 ππΈπ·
π + π = 15
(iii)
2. (a)
(i) Modal Class is 60 < π₯ ≤ 80
(ii) ππππ =
10(10)+ 32(30)+ 48(50)+54(70)+36(90)+20(120)
200
12880
π₯Μ
= 200 = πΎ64.80
(b) (i)
Amount (KA)
Number of shoppers
≤ 20
10
≤ 40
42
≤ 60
90
≤ 80
144
≤ 100
180
≤ 140
200
(b) (i) ππππππ πΎ63 − πΎ64
(ii) π3 = πΎ82 – πΎ84
(iii) π3 − π1 = πΎ38 − πΎ41
(iv) ππ πππ $75 πππππππ ππ πΆπ’π. πΉπππ. πΊπππβ – 67 ππ 68 ππ 69 ππ 70
ππ 71 ππ 72
The Cumulative Frequency Curve
A cumulative frequency curve, also called the Orgive curve can be used to find the mean, quartiles (lower, upper,
interquartile, semi-interquartile) of a given distribution. To find the cumulative frequency, find the accumulated
totals and plot them against the data or score values. The cumulative frequency is formed by joining the points
with a smooth curve.
Question
Answer the whole of this question on a sheet of graph paper.
The waiting time for 55 passengers at the power tools bus station in Kitwe for them to board a Lusaka bound
bus on a particular day were as follows:Waiting time
(in minutes)
Number of
Passengers
1≤ π₯ ≤ 3
4≤π₯ ≤6
7≤ π₯ ≤ 9
10≤ π₯ ≤ 12
13 ≤ π₯ ≤ 15
6
11
20
13
5
≤ 12
≤ 15
a) Calculate the estimate of the mean waiting time.
b) Copy and complete the cumulative frequency table below.
Waiting time
(in minutes)
Number of
Passengers
≤3
≤6
6
17
≤9
55
c) Using a horizontal scale of 2cm to represent 2 minutes for times from 0 to 15 minutes and a vertical scale
of 2cm to represent 10 passengers.
Draw a smooth cumulative frequency curve.
112
d) Showing your method clearly, use your graph to estimate the
i.
Median
ii.
Lower quartile
iii.
Upper quartile
iv.
Interquartile range
v.
Semi-interquartile range
vi.
60th percentile
e) Find the number of passengers who waited for more than 6 minutes.
f) (i) If a passenger was chosen at random, find the probability that he waited for less than 9 minutes.
g) If two passengers were chosen at random. Find the probability that they both waited for more than 12
minutes.
Relative Cumulative Frequency Table
The Table below shows a frequency table of the marks obtained by 120 pupils in a Mathematics Test.
Marks
0 -4
5–9
10 – 14 15 – 19 20 – 24 25 – 29 30 – 34 35 – 39
40 – 44 45 – 49
Frequency
0
4
6
9
i.
ii.
10
14
24
28
19
Construct the relative cumulative frequency curve for the above mentioned data.
From the curve, estimate the 74th percentile.
a) Relative cumulative frequency =
Mark
πΆπππ π πππ‘πππ£ππ ππ’ππ’πππ‘ππ£π πππππ’ππππ¦
πππ‘ππ πππππ’ππππ¦
Frequency
Cumulative
Frequency
0
Relative Cumulative
Frequency
0
=0
120
0–4
0
5–9
4
4
4
= 0.03
120
10 – 14
6
10
10
= 0.08
120
15 – 19
10
20
20
= 0.17
120
20 – 24
14
34
34
= 0.28
120
25 – 29
24
58
58
= 0.48
120
30 – 34
28
86
86
= 0.72
120
35 – 39
19
105
105
= 0.88
120
40 – 44
9
114
114
= 0.95
120
45 – 49
6
120
120
= 1.00
120
113
6
1.00
0.9
0.8
74th Percent
Frequency
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
5
10
15
20
25
30
35
40
45
50
Marks
b) 74th Percentile
=74 (1.00)
100
= 0.74
= 34.5 Marks
Variance and Standard Deviation
1.1
Ungrouped Data
Variance
This is a measure of spread that measures the distances or spread of data about the mean. It is found by
the formula.
Variance = ∑ππ=1
(π₯ − π₯)
π
114
Where x is the mean of x1, x2, x3, …..., xi and R is the number of observations.
An alternative method that can be used to calculate the variance for ungrouped is
π₯2
Variance = ∑ π − (π₯)
1.2
2
Standard Deviation
The standard deviation is the positive square root of the variance.
Standard deviation = √ππππππππ
(S.D)
It can be calculated by the formula
Standard deviation = √∑(π₯ − π₯) 2
π
Alternatively, the standard deviation of the ungrouped data can be found by the formula
Where π₯ ππ π‘βπ ππππ
S.D = √∑ π₯2 − (π₯) 2
π
Question 1
Calculate the variance and standard deviation of the following set of data.
8, 5, 10, 25 and 32.
Solution
Mean π₯ = ∑π₯
π
8+5+10+25+32
5
80
x= 5
Mean π₯ =
x = 16
Method 1
Using
(π₯−π₯) 2
π
Variance = ∑
Score
x
8
5
10
25
32
Ζ© π₯ = 80
Deviation
x–x
-8
-11
-6
9
16
Ζ© (π₯ − π₯) = 0
(π − π)²
64
121
36
81
256
Ζ© (π₯ − π₯)² = 558
2.0. GROUPED DATA
2.1 THE MEAN OF GROUPED DATA
The mean of grouped data also called the estimate of the mean is denoted by x is calculated using the formula
Mean (π₯) = ∑ππ₯
∑π
115
2.2 THE STANDARD DEVIATION OF GROUPED DATA
The formula used to calculate the standard deviation of the grouped data is
Standard deviation = √∑ π(π₯ − π₯) 2
π
Where
π = ∑π
π₯ = ∑ππ₯
∑π
The alternative formula for calculating the standard deviation is
2
Standard deviation = √∑ ππ₯ − π₯ 2
∑π
where
Variance ==Mean (π₯) = ∑(π₯ − π₯)2
π
= 558
5
=111.6 (Id.p)
S.D = √ππππππππ
= √∑(π₯ π
− π₯)2
= √558
5
=10.56 (2d.p)
Standard deviation = √∑ π₯ 2 − (π₯) 2
π
π₯
π₯2
8
64
5
25
10
100
25 π = ππ
625
32
1024
π₯ = ∑π₯
π
= 80
5
∑π₯ = 80 ∑π₯2=1838
Standard deviation = √∑ π₯2 − (π₯)2
π
= √1838 − 16 2
5
= √111.6
= ππ. ππ
116
π₯ = ∑ππ₯
∑π
Question
An intelligence quotient test that was taken by pupils at Kitwe Boys Secondary School in Kitwe Showed the
following results.
Time taken
(Minutes)
Frequency
0<x<1
1<x<2
2<x<3
3<x<5
5<x<10
10
15
25
40
25
Find
(a) The estimate of the mean
(b) The standard deviation
Solution
To calculate the estimate of the mean and standard deviation, we use the mid-internal or midpoint x
Time
0 < π₯ <1
1 < π₯ <2
..
2 < π₯ <3
3 < π₯ <5
5 < π₯ <10
π
0.5
1.5
2.5
4
7.5
π
10
15
25
40
25
ππ₯
5
22.5
62.5
160
187.5
∑π = 115 ∑ππ₯ = 437.5
ππ₯ 2
2.5
33.75
156.25
640
1406.25
∑ππ₯ = 2238.75
(a) Estimate of the mean
π₯ = ∑ππ₯
∑π
π₯ = 437.5
115
π = π. ππ
(b)S.D
S.D= √∑ ππ₯ 2 − (π₯)2
∑π
S.D= √2238.75 − 3.80 2
115
π. π· = 2.2 (πΌπ. π)
(c) (i) Median = ½ Q
Q2 = ½ x 55
=27.5
Q2 =7.8 Minutes
(ii) Lower Quartile = ¼ Q
Q1 = ¼ x55
= 13.75
Q1 = 5.2 Minutes
(iii) Upper quartile = ¾ Q
Q3 =3/4 x55
= 41.25
=9.7 Minutes
(iv) Interquartile range = Upper quartile – Lower quartile
= Q3 - Q1
9.7- 5.2
4.5 Minutes
117
(v) Semi – interquartile range = 1.2 x interquartile range
½ x 4.5 Minutes
2.25 Minutes
(Vii) 60th Percentile = 60 X 56
100
60% of Q = 33.6
= 8.45 Minutes
(d) 55-17
=38 Pupils
P (> 6 Minutes) = 38
55
(e) (i) P(<9 Minutes ) = 37
55
(ii) P (< 12 Minutes and < 12 Minutes)
P (> 2Minutes) x P(< 12 Minutes )
5 X 4 =2
55
54 297
a) Mean (π₯) = ∑ππ₯
π₯ = 2x 6 + 5x11 + 8x20 + 11x12 + 14+5
∑π
55
π₯ = 12+55+160+132+70
= 429
55
55
Waiting time (mm)
Frequency
<3
16
π₯ = 7.8 Minutes (I.d.p)
<6
<9
17
37
<12
50
<15
55
Number of Passengers
60
50
Upper Quartile
(Q )
40
60th Percentile
Median (Q2)
30
20
Lower Quartile
10
0
2
4
6
8
10
Waiting time in
minutes
118
12
14
16
TOPIC 23: GRAPHS OF FUNCTIONS
QUADRATIC FUNCTIONS
1. (a) Make a table of values of the function π(π₯) = π₯ 2 − 2π₯, with the domain −2 ≤ π₯ ≤ 4,
ππΈπ
, and sketch the graph. Use a scale of 1cm to 1 unit on both axes.
(b) Write down the range
(c) Find the turning point and state whether it is maximum or minimum
(d) Write down the equation of the line of symmetry.
SOLUTION
(a) π(π₯) = π₯ 2 − 2π₯
-2
-1
0
1
2
3
π₯
4
1
0
1
4
9
π₯2
4
2
0
-2
-4
-6
−2π₯
8
3
0
-1
0
3
π(π₯)
Points are: (-2,8), (-1,3), (0,0), (1,-1), (2,0), (3,3), (4,8).
4
16
-8
8
(b) From table: the smallest value π(π₯) is -1 and largest value is 8.
Therefore the range−1 ≤ π(π₯) ≤ 8, π(π₯)πΈπ
(c) Turning point (1,-1), minimum.
(d) Line of symmetry π₯ = 1
2. Answer the whole of the question on a sheet of graph paper.
The valuables π₯ and π¦ are connected by the equation π¦ = π₯ 2 − 4π₯ + 3
Some of the corresponding values of π₯ and π¦ correct to one decimal place where necessary are given
in the table below.
0
0.2
0.5
0.8
1
1.5
2
2.3
2.5
2.8
3
π₯
3
2.2
r
0.4
0
-0.8
-1
-0.9
-0.8
-0.4
0
π¦
(a) Calculate the value of r, correct to one decimal place.
(b) Using a scale of 2cm to represent 1 unit on the horizontal axis and 2cm to represent 1 unit on the vertical
axis, draw the graph of π¦ = π₯ 2 − 4π₯ + 3 for 0 ≤ π₯ ≤ 3.
(c) By drawing a suitable straight line on the same axes, use your graph to find the values of π₯ which satisfy
the equation π₯ 2 − 4π₯ + 3 = π₯ + 1
(d) By drawing a suitable tangent, find the gradient of the curve at the point where π₯ = 1.5
SOLUTION
2. (a) π = 1.0
(b)
119
(c) By drawing the line π¦ = π₯ + 1 on the graph then the answer for the equation π₯ 2 − 4π₯ + 3 = π₯ + 1 is
π₯ = 0.4
(d) Taking two points within the tangent, the gradient of the curve at π₯ = 1.5 is
π¦ −π¦
π= 2 1
π₯2 −π₯1
−1.3−(−0.5)
=
2−1
−0.8
= 1
= −0.8
3. Answer the whole of this question on a sheet of graph paper.
The variables π₯ and π¦ are connected by the equation
π₯2
12
π¦=
+ − 6,
6
π₯
The table below shows some corresponding values of π₯ andπ¦. The values of π¦ are given correct to one
decimal place where appropriate.
1
1.5
2
3
4
5
6
7
π₯
6.2
2.4
0.7
0.6
2
π¦
−0.5
−0.3
π
(a) Calculate the value of π, correct to one decimal place.
(b) Using a scale of 2 cm to 1 unit on each axis, draw a horizontal π₯-axis for 0 ≤ π₯ ≤ 8 and a vertical π¦axis for−1 ≤ π¦ ≤ 7.
On your axes, plot the points given in the table and join them with a smooth curve.
(c) By drawing a tangent, find the gradient of the curve at the point (1.5, 2.4).
(d) Showing your method clearly, use your graph to find the values of π₯ in the range 1 ≤ π₯ ≤ 7 for
which
π₯2
6
+
12
π₯
=7
π₯
4.
(e) (i) On the same axes, draw the graph of the straight line π¦ = .
(ii) Using your graphs, find the values of π₯ in the range 1 ≤ π₯ ≤ 7 for which
π₯2
12
π₯
(iii) By using your graphs, solve the equation 6 + π₯ − 6 = 4.
(f) Estimate the area bounded by the curve, π₯ = 3, π₯ = 5 and π¦ = 2
SOLUTION
3. (a) π = 3.9
(b)
120
π₯2
6
+
12
π₯
π₯
− 6 ≤ 4.
(c) Taking two points within the tangent, the gradient of the curve at (1.5, 2.4) is
π¦2 −π¦1
π₯2 −π₯1
4.7−1
= 1−1.9
3.7
= − 0.9
π=
= −4.1
(d) π₯ = 5.4, π₯ = 1.9
(e) (i) 2.1 ≤ π₯ ≤ 5.6
(ii) π₯ = 2.1 , π₯ = 5.6
(f) Area = 4.2 square unit
4. Answer the whole of this question on a sheet of graph paper.
1
The variable x and y are connected by the equation π¦ = 2 π₯ 2 (3 − π₯).
Some corresponding values, corrected to 1 decimal place where necessary, are given in the following
table.
π₯
-1.5
-1.3
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.3
3.5
π¦
5.1
3.6
2
0.4
0
0.3
1
1.7
2
1.6
0
-1.6
-3.1
1
a. Taking 2 cm to represent 1 unit on each axis, draw the graph of π¦ = 2 π₯ 2 (3 − π₯), for values of π₯
in the range -1.5≤ π₯ ≤ 3.5.
1
b. Use your graph to write down the value of the largest solution of the equation 2 π₯ 2 (3 − π₯) = 1.
c. By drawing a tangent, find the gradient of the curve at the point where π₯ = 2.5.
121
SOLUTION
(a)
½ x2(3 – x) = 1 i.e. y=1
The gradient of the curve = gradient of tangent
6.3
= - 3.4
= - 1.85
ACTIVITY
1. Answer the whole of this question on a sheet of graph paper.
1
The variable π₯ and π¦ are connected by the equation π¦ = 5 π₯(10 − π₯ 2 ).
The table below shows some corresponding value of π₯ and π¦. The value of π¦ are given correct to
decimal place where necessary.
(b)
(c)
π₯
π¦
−1
−1.8
0
1
2
3
3.5
4
0
1.8
2.4
0.6
−1.6
π
one
(a) Calculate the value of π.
(b) Using a scale of 2 cm to 1 unit on both axes for −1 ≤ π₯ ≤ 5 and −5 ≤ π¦ ≤ 3, draw the graph of π¦ =
1
π₯(10 − π₯ 2 ).
5
(c) By drawing a tangent to the curve, estimate the gradient of the curve at the point (1, 1.8).
(d) On the same graph, draw the line whose equation is 5π¦ + 4π₯ = 4.
1
4
4
(e) Use your graphs to find the solutions of 5 π₯(10 − π₯ 2 ) = − 5 π₯ + 5.
2. Answer the whole of this question on a sheet of graph paper.
The variables π₯ and π¦ are connected by the equationπ¦ = π₯ 2. − 2π₯ + 1.
Some of the corresponding values of π₯ and π¦ correct to 1 decimal place are given in the table
π₯
−1.5
−1
−0.5
π¦
π
4
2.3
(a) Calculate the value of π.
0
1
0.5
0.3
122
1
0
1.5
0.3
2
1
2.5
2.3
below.
3
4
(b) Using a scale of 2 cm to represent 1 unit on both axes, draw the graph of π¦ = π₯ 2 − 2π₯ + 1 for −2 ≤
π₯ ≤ 3 and 0 ≤ π¦ ≤ 10.
(c) Calculate an estimate of the gradient of the curve at the point (0, 1).
(d) Showing your method clearly, use your graph to solve the equationπ₯ 2 − 2π₯ + 3 = 1.5.
(e) Estimate the area bounded by the curve π₯ = 0, π₯ = 2 and π¦ = 3.
3. Answer the whole of this question on a sheet of graph paper.
The table below gives some values of π₯ and the corresponding values of π¦, where π¦ = 30 − 18π₯ + π₯ 3 .
π₯
-4
-3
-2
-1
0
1
2
3
4
π¦
38
57
58
47
30
13
2
3
22
a. Using a scale of 2 cm to represent 1 unit, draw a horizontal π₯-axis for −4 ≤ π₯ ≤ 4.
Using a scale of 2 cm to represent 10 units, draw a vertical π¦-axis for 0 ≤ π¦ ≤ 60.
On your axes, plot the points given in the table and join them with a smooth curve.
b. π΄ is the point (0, 27) and π΅ is the point (3, 3).
(i)
Draw , on the axes the line which passes through π΄ and π΅
(ii)
Find the equation ofπ΄π΅
c. By drawing a tangent, find the gradient of the curve π₯ = 1.
TOPIC 24: LINEAR PROGRMMING
LINEAR EQUATIONS
β’ A linear inequation is an algebraic expression containing one of the signs:<, >, ≤ ππ ≥ instead of an
= sign.
β’ An inequation is a mathematical statement which relates algebraic expressions using either of the
following signs.
a) < (πππ π π‘βππ)
b) > (πππππ‘ππ π‘βππ)
c) ≤ (πππ π π‘βππ ππ πππ’ππ π‘π)
d) ≥ (πππππ‘ππ π‘βππ ππ πππ’ππ π‘π)
NOTE:
Linear inequations are solved (worked out) in the same way as linear equations .However when an
inequality is divided by or multiplied by a negative the sign change.
Examples
1. Solve each of the following inequalities
a) 3π₯ + 2 ≤ 8
b) 1− 3π₯⁄5 < 4
Solutions
a) 3π₯ + 2 ≤ 8
3π₯ ≤ 8 − 2
3π₯ ≤ 6
3π₯
6
≤3
3
π₯=2
3π₯
b) 1 − 5 < 4
−3π₯
4−1
5
−3π₯
<3
5
−5
−3π₯
× 5
3
<3×
−5
3
π₯ > −5
2. Solve the inequation 7 − 2π₯ < 9 and show it’s solution set on the number line.
123
7 − 2π₯ < 9
−2π₯ < 9 − 7
−2π₯ < 2
−2π₯
2
< −2
−2
π₯ > −1
-1 0 1 2 3 4
5 6 7
8
3. Find the range of π₯ for which −1 < 5 − 3π₯ ≤ 11 and illustrate its solution set on the number line.
−1 < 5 − 3π₯ ≤ 11
−1 < 5 − 3π₯ ≤ 11
−1 < 5 − 3π₯
5 − 3π₯ ≤ 11
3π₯ < 5 + 1
−3π₯ ≤ 11 − 5
3π₯ < 6
−3π₯ ≤ 6
3π₯⁄ < 6⁄
−3π₯⁄ ≤ 6⁄
3
3
3
3
π₯<2
π₯ ≥ −2
-3
-2
-1 0
1
2 3
APPLICATION OF INTERGERS ON INEQUATIONS
Examples
1. find the smallest integer π₯, for which 3π₯ > 28
3π₯ > 28
3π₯⁄ > 28⁄
3
3
1
π₯ > 93
2. Given that 0.5 ≤ π₯ ≤ 0.9 and 4 ≤ π¦ ≤ 8. find
a) The largest possible value of π₯π¦.
b) The largest possible value of π¦ − π₯
Solutions
Note: identify the smallest and the largest values in each in solution set and then work out the
problem
a) π₯ = (0.5,0.6,0.7,0.8,0.9)
π ππππππ π‘ = 0.5, ππππππ π‘ = 0.9
π¦ = (4,5,6,7,8,9)
π ππππππ π‘ = 4, ππππππ π‘ = 8
π₯π¦
(s,s)
(s,l)
(l, s)
(l, l)
0.5× 4
0.5× 8
0.9× 4 0.9× 8
2
4
3.6
7.2
The largest value of π₯π¦ = 7.2
b) π¦ − π₯
π −π
π −π
π−π
π−π
4-0.5
4-0.9
8-0.5
8-0.9
3.5
3.1
7.5
7.1
The smallest value of π¦ − π₯ = 3.1
ILLUSTRATION OF INEQUATIONS ON THE CARTESIAN GRAPH (XOY-PLANE)
β’ Inequalities are presented on the xoy-plane in the same as linear equations. However; two types
of lines are used depending on the sign of the inequality given.
β’ The following are the lines used to show the inequalities
a) Solid (continuous) lines: these lines are associated with inequality signs ≤ or ≥ which means that the
solution set of the given inequalities lies on the boundary drawn.
124
b) Broken (dotted) lines: these lines are associated with inequality signs < or > which means that the
solution set of the inequality does not lie on the given boundary.
β’ In order to draw the given inequality on the Cartesian graph, the following steps
are
followed.
STEP 1: Convert an inequality into an equation
STEP 2: Make a table of values especially with those inequalities which involves two
Variables
NOTE: For those inequalities which have one variables. There is no need to make a table of values because
the lines will either be parallel to the x-axis or y-axis.
1) Any inequality with x variable will be drawn parallel to y-axis passing through the xaxis at a
given number.
2) Any inequality with y variable will be drawn parallel to x-axis passing through the y-axis at a given number.
STEP 3: Plot the points and draw the line with respect to the sign of the inequality given.
STEP 4: Determine the unwanted region by using the test point.
NOTE: The test point should be a point away from the given line because if you take a point
along the given line the decision will not be arrived at.
STEPT 5: Shade the unwanted region and indicate the inequality on the line drawn
Examples
Illustrate the solution set of the following inequalities on the Cartesian graph.
a) π₯ ≤ 2
b) π¦ > 1
c) π¦ + π₯ ≤ 2
d) π₯ ≥ −1, π¦ ≥ 0, π¦ + π₯ < 1
Solution
a) π₯ ≤ 2
π₯=2
-1
1
2
3
Test point (3, 1)
π₯≤2
π₯=2
3=2
But 3>2
b)
π¦>1
2
1
-2
-1
0
1 2
Test point (1,2)
π¦>1
π¦=1
2=1
2≠1
but 2 > 1
125
c)π₯ + π¦ ≤ 2
π₯+π¦ =2
2
1
-2
-1
0
1
2
3
FINDING INEQUALITIES FROM THE XOY PLANE
β’ When finding inequalities from the Cartesian plane or graph, the following steps are
followed.
STEP 1: Find the equation of the given line whose inequality is to be determined.
NOTE: i) For the line which is parallel to y-axis, their equations are given by x= n, where n
is x- intercept (n is a point at which the line passes through the x-axis)
ii) For the lines which are parallel to the x-axis their equations are given by y= n,
where n is the value of y at y- intercept. (n is a point at which the line crosses
the y-axis)
iii) For non-vertical lines or horizontal lines (inclined lines), their equations are
given by the formula.
(a) π¦ − π¦1 = π(π₯ − π₯1)
( π) π¦ = ππ₯ + π
π¦−π¦1
π₯−π₯1
(c) π¦2−π¦1 = π₯2−π₯1
STEP 2: Determine the equation sign by using test point
Note: A test point must be taken from the wanted region
Caution: Never take a test point along the line
STEP 3: Replace the equal sign the equation found in step 1 with the sign found
in step 2 to form an inequation
Example
1. In the diagram below R is the unshaded region
6
4
3
R
2
1
1 2 3 4 5 6
126
•
Write down the three inequalities which describe the shaded region R.
Solutions
(i)
π¦=π
π¦=2
Test points (2, 3)
π¦=2
(ii)
3≠2
But 3 > 2
π¦≥2
π₯=π
π₯=1
Test points (2, 3)
2≠1
2>1
π₯≥1
(iii) (6, 0)
π=
(0, 6)
π¦2−π¦1
π₯2−π₯1
6−0
π = 0−6
6
π=
−6
π = −1
π¦ − π¦1 = π(π₯ − π₯1)
π¦ − 0 = −1(π₯ − 6)
π¦ = −π₯ + 6
Test points (2, 3)
π¦ = −π₯ + 6
3 = −2 + 6
3<4
π¦ ≤ −π₯ + 6
Example 2
The diagram below shows triangle PQR with vertices p (-1, 2), Q (2, 2) and R (-1, -1).
P (-1, 2)
Q (2, 2)
R (-1,-1)
127
Write down the inequalities which describes the region inside the triangle PQR
Solutions
(i ) Q(2,2)
π¦=π
π¦−2
Test points (1,1)
π¦=2
1≠2
But 1 < 2
= π¦≤2
(ii) p (-1,2 )
π₯=π
π₯ = −1
Test points (2,3)
2 ≠ −1
But 2 > −1
π₯ ≥ −1
(iii) R(-1,-1) , Q (2,2)
π¦ −π¦
π = π₯2 −π₯1
2
1
2+1
π = 2+1
3
π=3
π=1
π¦ − π¦1 = π(π₯ − π₯1)
π¦ − 2 = 1(π₯ − 2)
π¦−2 = π₯−2
π¦ =π₯−2+2
π¦=π₯
Test points (4, 3)
π¦=π₯
3=4
But 3 < 4
π¦≤π₯
128
Example 3
In the diagram below R is the unwanted region.
8
7
6
5
4
3
2
R
1
-4 -3 -2 -1 0 1 2
3
4
5
6
7
a) Write three inequalities which describe the region R.
b) Find the maximum value of 2π₯ + π¦ within the region R.
Solutions
a)
(i) π₯ = π
π₯=3
Testing point (1, 1)
π₯=3
1≠3
But 1 < 3
∴π₯ ≤3
(ii) (-2, 0) and (0, 2)
π¦ −π¦
π = π₯1 −π₯2
1
π=
2
2−0
0+2
π=1
π¦ − π¦1 = π(π₯ − π₯1 )
π¦ − 0 = 1(π₯ + 2)
π¦ = π₯+2
129
Test point (1, 1)
π¦ = π₯+2
1 =1+2
1≠3
But 1 < 3
∴π¦ ≤π₯+2
(iii) (0, 1) (0, 2)
π=
π¦2 − π¦1
π₯2 − π₯1
π=
0−2
2−0
π = −1
π¦ − π¦1 = π(π₯ − π₯1 )
π¦ − 2 = −1(π₯ − 0)
π¦ = −π₯ + 2
Test point (2, 2)
π¦ = −π₯ + 2
2=π
2≠0
But 2 > 0
∴ π¦ ≤ −π₯ + 2
b). 2π₯ + π¦
(0,2)
2π₯ + π¦
2(0) + 2
=2
(3,5)
2π₯ + π¦
2(3) + 5
=11
(2,0)
2π₯ + π¦
2(2) + 0
=4
(3,0)
2π₯ + π¦
2(3) + 0
=6
∴ πππ₯πππ’π π£πππ’π ππ 2π₯ + π¦ = 11
LINEAR PRORAMMING
β’ Linear indicates relationship arising from conditions represented by a straight line. The
relationship is of the form π¦ = π + ππ₯
β’ Programming involves that decisions are made systematically. Therefore, linear programming
is the branch of mathematics that deals with decision making under the conditions of
uncertainty.
β’ The function of linear programming is either to maximize profit or minimize costs subject to
certain constraints represented by inequation.
MATHEMATIC MODEL
β’ A mathematical model is a system of in equations which are formed from the conditions specified in a
mathematical problem.
β’ The conditions to which inequation are formed are called constrains.
β’ The conditions which gives rise to inequalities in a given mathematical are.
a) Maximum means less than or equal to (≤)
130
At most means less than or equal to (≤)
Available means less than or equal to(≤)
Not more than means less than or equal to(≤)
Not exceeds means less than or equal to(≤)
Less than(<)
At least means greater than or equal to.
Maximum means greater than or equal to.
Not less than means greater than or equal to.
As many as means greater than or equal to.
More than means greater than(>)
Exceeds means greater than(>)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
l)
NOTE:
•
•
Inequation are formulated based on the variables and figures given in a mathematical problem.
Wherever the above stated condition appears in the mathematical problem, then it means that there
is an inequality that lies in that sentence.
Example:
1. Joshua has K4. He buys a book at K1.5 and pens at K0.5 each. He intends to have not more than
2books .The money spent on pens must be at least one-third of the money
spent on books. Let x be the number of books and y be the number of pens. Formulate 3
inequalities which satisfy the given conditions.
Solutions
(i).1.5π₯ πππππ + 0.5π₯ πππππππ ≤ 4
1.5π₯ + 0.5π¦ ≤ 4
1.5π₯ + 0.5π¦ ≤ 4 / 10
15π₯
5
+
5π¦
5
≤
40
5
3π₯ + π¦ ≤ 8
(ii).πππππ ≤ 2
π₯≤2
1
(iii). 0.5 × πππππππ ≥ 3 × 1.5 × πππππ
0.5π¦ ≥
1
× 1.5π₯
3
0.5π¦
0.5π₯
≥
0.5
0.5
π¦≥π₯
131
2. Abraham buys sweets and sales them to his classmates at school. He buys two type of sweets type ‘A’
and type ‘B’. Type ‘A’ sweets costs K2 each and type ‘B’ sweets costs K3 each. He has K30 to spend
on sweets. His sweet pouch can carry 12 sweets. He decided to buy three at least three of each type of
sweet. Let ‘A’ sweets be X and type ‘B’ sweets be Y. Write four inequalities which satisfy the above
conditions.
Solutions
(i)
2(π₯) + 3(π¦) ≤ πΎ30
2π₯ + 3π¦ ≤ 30
(ii)
π₯ + π¦ ≤ 12
(iii)
π₯≥3
π¦≥3
3. Mr Mbangu intends to start a transport company; he intends to buy Taxis and Mini Buses. Let ‘X’ be
number of taxis and ‘Y’ number of Mini Buses. Write an equality which represents each of the
following conditions.
• The total number of vehicles should be at least 10.
• A taxi takes up three units of parking space while a mini bus takes up six units of parking
space. He has a maximum of sixty units of parking space.
• The number of mini bus should be at least two third the number of taxi.
Solutions
(i)
(ii)
(iii)
π₯ + π¦ ≥ 10
π₯(3π’πππ‘π ) + π¦(6π’πππ‘π ) ≤ 60π’πππ‘π
3π₯ 6π¦
60
+
≤
3
3
3
π₯ + 2π¦ ≤ 20
2
π¦≥3 × π₯
2π₯
π¦≥
3
4. Mrs Kawena bakes two types of cakes for sale. Type ‘A’ and ‘B’.
(i). To satisfy her regular customers daily, she must bake:
• At least ten cakes of type A.
• At least twenty cakes of type B.
Taking x to represent the number of cakes of type A and ‘y’ to represent cakes of type B.
Write two inequalities which satisfies the above conditions.
(ii). To avoid wastage, the total number of cakes she should bake per day must not exceed 70.
Write another inequality which satisfies this condition.
(iii). The point (x, y) represents X cakes of type A. Using a scale of 2 cm to represents 10
cakes. On each axis draw x and y axis for 0 ≤ π₯ ≤ 80 πππ 0 ≤ π¦ ≤ 80. present the three
inequalities above on your graph and shade the unwanted
region to indicate were x and y must lie.
Solutions
(i)
π πππππ ππππππ‘πππ
ππππ π‘ ππππππ‘πππ
π¦ ≥ 20
132
π₯ ≥ 10
(ii) π₯ + π¦ ≤ 70
x
0
y
70
π₯ + π¦ ≤ 70
70
0
0 + π¦ ≤ 70
π¦ ≤ 70
π₯ + π¦ ≤ 70
70 + π¦ ≤ 70
π¦≤0
70
20
10
APPLICATION OF LINEAR PROGRAMMING
•
•
The objective of linear programming is to minimize cost and maximize profit.
Objective function: Is a mathematically designed rule used to determine the feasibility of investment
e.g. 2π₯ + 3π¦.
133
•
•
Feasibility Region: This is the region of the xoy plane were the objective function holds goods, I.e a
region were all the inequalities (constraints) defining the region are satisfied.
Feasible points: these are points usually whole number value of the variables lying in the feasible
region.
QUESTIONS FOR PRACTICE
1. Answer this question on a sheet of graph paper
Makwebo prepares two types of sausages, Hungarian and beef, daily for sale. She prepares at least 40
Hungarian and at least 10 beef sausages. She prepares not more than 160 sausages altogether. The
number of beef sausages prepared are not more than the number of Hungarian sausages.
(a) Given that x represents the number of Hungarian sausages and y the number of beef sausages,
write four inequalities which represents these conditions.
[4]
(b) Using a scale of 2cm to represent 20 sausages on both axes, draw the x and y axes for 0 ≤ π₯ ≤
160 πππ 0 ≤ π¦ ≤ 160 respectively and shade the unwanted region to show clearly the region
where the solution of the inequalities lie.
[4]
(c) The profit on the sale of each Hungarian sausage is K2.00. How many of each type of sausages are
required to be prepared to make maximum profit?
[2]
(d) Calculate maximum profit.
[2]
(2) Answer the whole of this question on a sheet of graph paper.
Himakwebo orders maize and ground nuts for sale. The order price of the bag of maize is K75.00 and that of a
bag of groundnuts is K150.00. He is prepared to spend up to K7500.00 altogether. He intends to order at least
5 bags of maize and at least 10 bags of groundnuts. He does not want to order more than 70 bags altogether.
(a) If x and y are the number of bags of maize and groundnuts respectively, write four inequalities which
represents these conditions.
[4]
(b) Using a scale of 2cm to represent 10 bags on each axis, draw the x and y axes for 0 ≤ π₯ ≤
70 πππ 0 ≤ π¦ ≤ 70 respectively and shade the unwanted region to show clearly the region to show
clearly the region where the solution of the inequalities lie.
[4]
(c) Given that the profit on a bag of maize is K25.00 and on a bag of groundnuts is K50.00, ho many bags
of each type should he order to have maximum profit?
[2]
(d) What is this estimate of the maximum profit?
[2]
ANSWERS FOR THE ABOVE QUESTIONS
Q1.(a) [i] π₯ ≥ 40
[ii] π¦ ≥ 10
[iii] π₯ + π¦ ≤ 160
[iv] π₯ ≥ π¦
134
(b)
160
π₯ ≥ 40
x ≥ y
x + y ≤ 160
R
x ≥ 10
10
40
160
(c).150 Hungarians sausages
10 beef sausages
(d). Objective function;
3π₯ + 2π¦
3(150) + 2(10)
= K470.00
Q2. (a) [i] π₯ + 2π¦ ≤ 100
[ii] π₯ ≥ 5
[iii] π¦ ≥ 10
[iv] π₯ + π¦ ≤ 70
135
(b)
70
x≥5
x + 2y ≤ 100
50
R
y ≥ 10
10
5
(c). 40 bags of maize
70
100
x + y ≤ 70
30 bags of maize
(d). Maximum profit (Objective function)
25π₯ + 50π¦
25(40) + 50(30)
πΎ1000.00 + πΎ1500.00
= πΎ2500.00
TOPIC 25: VECTORS IN TWO DIMENSIONS
INTRODUCTION
A vector is a quantity that has both magnitude and direction, i.e. velocity 20 mπ −1 north east.
REPRESENTATION OF VECTORS
βββββ where the arrow head shows the sense of direction. We can
Vectors are represented by a line segment i.e. π΄π΅
also represent vectors using small letters printed in bold.
EQUALITY OF VECTORS
In the figure below, the line segments AB, CD and EF are parallel (in the same direction) and equal in length.
Since they are equal in length and parallel to each other, each line can represent the same vector and
βββββ =πΆπ·
βββββ =πΈπΉ
βββββ
π΄π΅
MAGNITUDE OF A VECTOR
The magnitude or modulus of a vector βββββ
π΄π΅ is the lenth of a line segment representing the vector to the scale
used. We denote this as|π΄π΅|. The magnitude of a vector π is written as |π|
ZERO OR NULL VECTOR
The vector which has no magnitude (and of course no direction) is the zero vector or null vector
SCALAR MULTIPLE OF A VECTOR
136
Given a vector π, we can make multiples of this vector
π
2π
π
π
π
i.e. PQ=2π. PQ has the same direction as π but twice it’s magnitude
βββββββ =2|π|=2a
|ππ|
EXAMPLE
What type of quadrilateral is ABCD if
a) βββββ
π΄π΅ = βββββ
π·πΆ
βββββ
ββββββββ
b) π΄π΅ = 3π·πΆ
SOLUTION
βββββ
a) AB=DC and AB//DC, then ABCD is a parallelogram. it follows therefore that βββββ
π΄π·=π΅πΆ
D
C
A
B
b) AB= 3DC and AB//DC. Then ABCD is a trapezium
D
C
A
B
ACTIVITY
Copy the figure below and draw vectors
a) 2p
b) –p
3
c) 4p
π
137
ADDITION AND SUBTRACTION OF VECTORS
Suppose that A, B and C are the points (1, 1), (8, 2) and (3, 5) respectively as shown in the diagram
C (3, 5)
B (8, 2)
A (1,1)
Consider the paths that can be taken to travel from A to C.
Travelling directly gives the vector
βββββ
π΄πΆ =(24)
Alternatively, it is possible to go from A to B and then from B to C
VECTOR ADDITION
Vector addition is defined to mean one displacement followed by another and so, in this case
βββββββ
βββββ +π΅πΆ
βββββ
π΄πΆ =π΄π΅
βββββ
π΄πΆ is called the resultant of the vectors βββββ
π΄π΅ and βββββ
π΅πΆ
Notice that this definition makes sense when the column vector are considered
2
7
−5
( ) = ( )+( )
4
1
3
Notice also that, it does not matter in what order two vectors are added. From the diagram
b
a
c
b
c=a+b
a
This is often called the triangle law for vector addition
However, if the dotted vectors are used the result obtained is
πΆ = π + π πππ π π
π+π =π+π
because of this important result the rule for adding vectors is sometimes called the parallelogram law for
vector addition.
VECTOR SUBTRACTION
Subtraction of vectors π − π can be thought of as π + (−π). Hence, if the vectors a and b are as shown, the
vector π − π is formed by adding the vector – π to the vector π
π
π
138
π
π−π
−π
WORKED EXAMPLES
1. (a) The position vector of a point A is (
(b)(i) Find the column vector m such that (
and AB = (
−8
−5
)−π =( )
6
2
, find the coordinates of B.
(ii) Hence find |π|
−1
(c).Given that PQ= ( ), find QP in component form.
9
Solutions
1. (a) π΄π΅ = π΄π + ππ΅
−3
−2
(
) = ( ) = ππ΅
2
1
−3
−2
ππ΅ = ( ) − ( )
2
1
−1
=( )
1
π©(−π, π)
−8
−5
(b)(i)( ) − π = ( )
6
2
−8
−5
(
)−( )= π
6
2
π
π=( )
−π
(ii)|π| = √(3)2 + (−4)2 = √9 + 16
= √25
=5
−1
(c).ππ = ( )
9
π
ππ = ( )
−π
15
6
−9
2. Given that π’ = ( ), π£ = ( ) andπ€ = ( ), find
π
−8
10
(a) (i) |π’|
(ii) 2π’ + π£
(b).Given that vector π€ is parallel to vector π’, calculate the value of π.
Solution
2(a) (i) |π’| = √(6)2 + (−8)2
= √36 + 64
= √100
=10 units
6
−9
(ii) 2π’ + π£ = 2 ( ) + ( )
−8
10
12
−9
=(
)+( )
−16
10
π
= (
)
−π
(iii) π’ = ππ€ π€βπππ π ππ π ππππ π‘πππ‘ then
15
6
(
) = π( )
π
−8
139
15π
6
( )=(
)
ππ
−8
6 = 15π
2
π=5,
2
−8 = ππ, ππ’π‘ π =
5
2
−8 = π
5
2π = −40
π = −ππ
3. In the diagram, ππ΄ = 2π, ππ΅ = 3π and π΅π = π − π. The lines OX and AB intersects at L.
(i)
Express as simply as possible in terms of π and / or π
(a) ππ
(b) AB
(ii)
Given that π΄πΏ =hπ΄π΅, express π΄πΏ in terms of π, π and h.
(iii)
Hence show that ππΏ = (2 − 2β)π + 3βπ
Solution
3. (i)(a) ππ = ππ΅ + π΅π
= 3π + π − π
= 3π − π + π
=ππ + π
(b) π΄π΅ = π΄π + ππ΅
= −2π + 3π
= 3π −2p
(ii).π΄πΏ =hπ΄π΅
π΄πΏ =h(3π − 2π)
=πππ − ππ
(iii) ππΏ = ππ΄ + π΄πΏ
=2π + 3βπ − 2βπ
=2π − 2βπ + 3βπ
= (−ππ)π + πππ
4. In the diagram, ππ΄ = 2π, ππΆ = 3π and π΄π΅ = 2π. The lines OB and AC intersects at X.
140
Express as simply as possible in terms of π and / or π.
(a) ππ΅
(b) BC
(ii)
Given that πΆπ = βπΆπ΄, express πΆπ in terms of π, π and β.
(iii)
Hence show that ππ = (3 − 3β)π + 2βπ
(i)
Solutions
4(i)(a) ππ΅ = ππ΄ + π΄π΅
(b) π΅πΆ = π΅π + ππΆ
= 2π + 2π
= −(2π + 2π) + 3π
= −2π − 2π + 3π
= 3π − 2π − 2π
(ii) πΆπ = βπΆπ΄ππ’π‘ πΆπ΄ = πΆπ + ππ΄
=2π − 3π
=πππ − πππ
(iii) ππ = ππΆ + πΆπ
= 3π + 2βπ − 3βπ
= (π − π)π + πππ shown.
= π − ππ
πΆπ = β(2π − 3π)
= 3π − 3βπ + 2βπ
PRACTICE QUESTIONS
(1)
OABC is a parallelogram.
The point X on AC is such that AX = 1⁄5 π΄πΆ. The point Y on AB is such that AY = 1⁄4 π΄π΅. Given
that OA = 20p and OC= 20q, express in terms of p and q
(i)
AC,
(ii) AX,
(iii) OX,
(iv) OY.
What do the results of (iii) and (iv) tell you about O, X and Y?
141
(2) In the diagram XVZ is a straight line, XY = 8p, XZ = 4p + 9q and YV =-6p + cq.
(i)
Express XV in terms of p, q and c.
(ii)Given that XV = h XZ, form an equation involving p, q, h and c.
(iii) The point K is outside triangle XYZ and is such that
XK = -4p + 3q.
Is XK parallel to YV? Justify your answer.
(3) In the triangle ORS, the point A on OR is such that OA = 2AR. B is the midpoint of OS, X is the midpoint
of AB and OX produced meets RS at Y.OA = 2p and OB =2q.
(a) Express in terms of p and/or q
(i)
AB,
(ii)
AX,
(b) Given that RY = k
RS,
(iii)
OX,
(iii)
RS
Express RY in terms of p, q and h.
(c) Hence show that OY = 3(1- h) p +4hq.
(d) Given also that OY =k OX, Express OY in terms of p, q and k.
(e) Using these two expressions for OY, find the value of h and the value of k.
(f) Find the ratio RY: YS.
(g) Express XY in terms of p and q.
(4) P =(−ππ), q = (−π
) , π = (ππ ) ands =(ππ).
π
(a) Find (i)
2p + 3q,
(ii)
P− q.
(c) Given that 2p = r + s calculate the value of π₯ and the value of y.
142
(5)
OA = (−2
)and OB = (34).
1
(a)
Given that OC = OA + OB, express OC as a column vector.
(b) Express AB as a column vector.
4
(c) If LM = (−2
), what is the special name given to the quadrilateral ALMO?
π
(6) p = (−π
) , q = (−π
)and r = (π
).
π
π
(a) Find q.
(b) Express 2p–q as a column vector.
(c) Given that p is parallel to r, find m.
ANSWERS:
(1) (i) 20(q–p),
(ii) 4( q – p)
(iii) 4(4p+ q) (iv) 5(4p+ q)
The points O, X and Y lie on the same line.
(2) (i) 2p + cq , (ii) h= 1⁄2 , c= 41⁄2 ,
(ii)
YV= 3⁄2 (−4π· + 3π) ,
XK is parallel to YV.
(3) (a) (i) 2p −2q ,
( b)
h(4q – 3p ) ,
(f)
3: 4 ,
(ii) q – p ,
(iii) p+ q ,
(d) k( p+ q ) ,
(g)
(e )
(iv) 4q – 3p
.
h= 3⁄7 , k = 1 5⁄7.
5⁄ (π· + π)
7
1
8
(4) (a) (i) (−1
(ii) (−3
(b)
x= 6 , y =−11.
) ,
),
π
5
(5) (a) OC = (π) ,
(b) AB =(3) ,
(c)
Trapezium.
5
(6) (a) 5,
(b)
(c)
- 1⁄2 .
(−12) ,
TOPIC 26: GEOMETRICAL TRANSFORMATIONS
1. DEFINITIONS
A geometrical transformation moves an object (a point or a shape) from one position to
another. During a transformation some points will move while others may not move.
The new position is known as the image of the original object.
Those points which do not move are known as invariant points and the line on which
these points lie is the line of invariant points. At this level, there are seven types of
transformations that you are expected to know in preparation for your examinations.
These are: translation, reflection, rotation, dilatation (enlargement or reduction), stretch
and shear.
143
2. TRANSLATION
A translation is represented by a
2ο΄1 column matrix or vector T = ο¦ο§ ba οΆο· .
ο¨ οΈ
A translation is completely described by stating the column vector of T.
Finding the image under a Translation
ο¦ x1 οΆ ο¦ x οΆ ο¦ a οΆ
ο· = ο§ ο· + ο§ ο·.
ο¨ y1 οΈ ο¨ y οΈ ο¨ b οΈ
The image of the point ( x, y) is ( x1 , y1 ) and is defined by ο§
Examples
(a) Find the image of the point (3, − 2) under the translation T = ο¦ο§ 6 οΆο· .
ο¨ 2οΈ
(b) The image of the point (9, − 1) is (4, 5) . Find the translation vector T.
(c) Under the translation T = ο¦ο§ − 7 οΆο· , the image of the point A is (2 , − 5) .
5
ο¨
οΈ
Find the coordinates of the point A.
Solutions
ο¦ x1 οΆ ο¦ x οΆ ο¦ a οΆ
ο·=ο§ ο·+ο§ ο·
ο¨ y1 οΈ ο¨ y οΈ ο¨ b οΈ
ο¦ x1 οΆ ο¦ 3 οΆ ο¦ 6 οΆ
ο§ y ο· = ο§ − 2ο· + ο§ 2ο·
ο¨ 1οΈ ο¨ οΈ ο¨ οΈ
ο¦ x1 οΆ ο¦ x οΆ ο¦ a οΆ
ο·=ο§ ο·+ο§ ο·
ο¨ y1 οΈ ο¨ y οΈ ο¨ b οΈ
(a) ο§
(b) ο§
ο¦ x1 οΆ ο¦ 9 οΆ
ο§ y ο· = ο§ 0ο·
ο¨ 1οΈ ο¨ οΈ
The image is (9, 0) .
ο¦ x1 οΆ ο¦ x οΆ ο¦ a οΆ
ο·=ο§ ο·+ο§ ο·
ο¨ y1 οΈ ο¨ y οΈ ο¨ b οΈ
(c) ο§
ο¦ο§ 4 οΆο· = ο¦ 9 οΆ + ο¦ο§ a οΆο·
ο§ ο·
ο¨ 5 οΈ ο¨ − 1οΈ ο¨ b οΈ
ο¦ 2 οΆ = ο¦ xοΆ + ο¦− 7οΆ
ο§ − 5ο· ο§ y ο· ο§ 5ο·
οΈ
ο¨ οΈ ο¨ οΈ ο¨
ο¦ο§ a οΆο· = ο¦ 4 οΆ − ο¦ 9 οΆ
ο§ ο· ο§ ο·
ο¨ b οΈ ο¨ 5 οΈ ο¨ − 1οΈ
ο¦ο§ a οΆο· = ο¦ − 5 οΆ
ο§
ο·
ο¨bοΈ ο¨ 6 οΈ
ο¦ο§ x οΆο· = ο¦ 2 οΆ − ο¦ο§ − 7 οΆο·
ο§
ο·
ο¨ y οΈ ο¨ − 5οΈ ο¨ 5 οΈ
ο¦ο§ x οΆο· = ο¦ 9 οΆ
ο§
ο·
ο¨ y οΈ ο¨ − 10 οΈ
ο T = ο¦ο§ − 65 οΆο·
ο¨ οΈ
ο
A is (9, − 10)
Exercise 1
(a) Find the image of the point (8, − 4) under the translation T = ο¦ο§ 5 οΆο· .
ο¨ 4οΈ
(b) The image of the point (−9, − 3) under a translation T is (7, 5) .
Find the translation vector T.
(c) Under the translation T = ο¦ο§ 2 οΆο· , the image of the point B is (2 , − 5) . Find the
ο¨ 3οΈ
coordinates of the point B.
(d) The image of the point (0, − 1) under a translation T is (8, 8) .
Find the translation vector T.
(e) Under the translation T = ο¦ο§ − 1οΆο· , the image of the point C is (11, − 5) . Find the
4
ο¨
οΈ
coordinates of the point C.
144
Answers
(a) (13, 0)
(b) (16, 8)
(c) (0, − 8)
(d) (8, 9)
(e) (12, − 9)
---------------------------------------------------------------------------------------------------------------3. REFLECTION
A reflection is completely described by giving the equation of the line of reflection.
The following table gives the matrices associated with the given line of reflection.
Line of Reflection
The line y = 0
(or the x – axis)
Matrix
1
0οΆ
M x = ο¦ο§
ο·
ο¨ 0 − 1οΈ
−1
M y = ο¦ο§
ο¨ 0
0
M y = x = ο¦ο§
ο¨1
The line x = 0
(or the y – axis)
The line y = x
The line
y = −x
0οΆ
1 ο·οΈ
1οΆ
0 ο·οΈ
− 1οΆ
0 ο·οΈ
0
M y = x = ο¦ο§
−
ο¨ 1
Finding the image under a Reflection
ο¦ x1 οΆ
ο¦ xοΆ
ο· = Mο§ yο· ,
y
ο¨ οΈ
ο¨ 1οΈ
The image ( x1 , y1 ) of the point ( x, y) under a reflection is defined by ο§
where M is the
2ο΄ 2 matrix associated with the given reflection.
Examples
(a) Find the image of the point (2 , − 5) under reflection in
(i) the line x = 0
(ii) the line y = x
(b) The point (−2 , − 8) is the image of the point Q under reflection in the line y = − x .
Find the coordinates of Q.
Solutions
ο¦ x1 οΆ ο¦ − 1 0 οΆ ο¦ 2 οΆ
ο·ο§ ο·
ο·=ο§
ο¨ y1 οΈ ο¨ 0 1 οΈ ο¨ − 5 οΈ
ο¦ x1 οΆ ο¦ − 2 οΆ
ο§ y ο· = ο§ − 5ο·
ο¨ 1οΈ ο¨ οΈ
ο¦ x1 οΆ ο¦ 0 1 οΆ ο¦ 2 οΆ
ο·ο§ ο·
ο·=ο§
ο¨ y1 οΈ ο¨ 1 0 οΈ ο¨ − 5 οΈ
ο¦ x1 οΆ ο¦ − 5 οΆ
ο§ y ο· = ο§ 2ο·
ο¨ 1οΈ ο¨ οΈ
(a) (i) ο§
(ii) ο§
The image is (−2 , − 5)
(b)
the image is (−5 , 2)
− 1οΆ ο¦ x οΆ
ο¦− yοΆ
−2
ο§ ο· ο ο¦ο§ − 8 οΆο· = ο§
ο·
0 ο·οΈ ο¨ y οΈ
ο¨
οΈ ο¨− xοΈ
−2 = − y ο y = 2
−8 = − x ο x = 8
The coordinates of Q are (8, 2).
ο¦ο§ − 2 οΆο· = ο¦ 0
ο§
ο¨ − 8 οΈ ο¨ −1
145
Exercise 2
(a) Find the image of the point (2 , − 5) under reflection in
(i) the line y = 0
(ii) the line y = − x
(iii) the y – axis
(b) The point (−3 , 12) is the image of the point R under reflection in the line y = x .
Find the coordinates of R.
(c) Under a reflection in the line x = 0, the point B is mapped onto the point (−6 , 15) .
Find the coordinates of B.
(d) Find the image of the point (4 , − 3) when it is reflected in the line y = 2.
Answers
(a)(i) (2 , 5)
(ii ) (5 , − 2)
(iii ) (−2 , − 5)
(b) (12 , − 3)
(c) (6 , 15)
(d) (4 , − 1)
4. ROTATION
A rotation is completely described by giving the centre, the angle and the
direction (clockwise or anticlockwise) of rotation.
The following table gives the matrices associated with Rotations.
Rotation, Centre (0, 0)
90° anticlockwise
(positive quarter turn)
Matrix
0
− 1οΆ
R +90 = ο¦ο§
1
0 ο·οΈ
ο¨
0
1οΆ
R +90 = ο¦ο§
0 ο·οΈ
ο¨ −1
−1
0οΆ
R180 = ο¦ο§
0
−
1ο·οΈ
ο¨
0
1οΆ
R 2700 = ο¦ο§
−
1
0 ο·οΈ
ο¨
90° clockwise
(negative quarter turn)
180°
(Half Turn)
270°
(three quarter turn)
Finding the image under a Rotation
(i) When the centre is the origin (0, 0)
The image ( x1 , y1 ) of the point ( x, y) under a rotation, centre (0, 0), is defined by
ο¦ x1 οΆ
ο¦ xοΆ
ο§ y ο· = R ο§ y ο· , where R is the 2ο΄ 2 matrix associated with the given rotation.
ο¨ οΈ
ο¨ 1οΈ
(ii) When the centre is (r, s), where r ≠ 0 and s ≠ 0
The image ( x1 , y1 ) of the point ( x, y) under a rotation, centre (r, s), is defined
ο¦ x1 οΆ
ο¦ x−rοΆ
ο· = Rο§ y − sο· +
y
ο¨
οΈ
ο¨ 1οΈ
by ο§
ο¦ο§ r οΆο· , where R is the 2ο΄ 2 matrix associated with the
ο¨sοΈ
given rotation.
146
Finding the Centre and Angle of a Rotation
Examples
(a) Find the image of the point (4 , 1) under a 90° negative quarter turn, centre (0, 0)
(b) Find the image of the point (2 , − 5) under a clockwise rotation, centre (2, 4).
(c) The point (−2 , − 8) is the image of the point Q under an anticlockwise rotation of 90°,
centre (1, 3). Find the coordinates of Q.
Solutions
(a) ο¦ο§ 0 1 οΆο· ο¦ο§ 4 οΆο· = ο¦ο§ 1 οΆο·
ο¨ −1 0οΈ ο¨ 1οΈ ο¨ − 4οΈ
ο¦ 0 − 1οΆ ο¦ 2 − 2 οΆ +
ο§1
0 ο·οΈ ο§ο¨ − 5 − 4 ο·οΈ
ο¨
0 − 1οΆ ο¦ 0 οΆ ο¦ 2 οΆ
= ο¦ο§
+ο§ ο·
0 ο·οΈ ο§ο¨ − 9 ο·οΈ ο¨ 4 οΈ
ο¨1
9
οΆο·
= ο¦ο§ οΆο· + ο¦ο§ 2
4
0
ο¨
οΈ
ο¨ οΈ
11 οΆ
= ο¦ο§
ο·
ο¨ 4 οΈ
The image is (11, 4)
(b)
The image is (1, − 4)
(c) ο¦ο§ − 1
ο¨ 0
0 οΆ ο¦ a − 1 οΆ ο¦1οΆ ο¦ − 2 οΆ
+ ο§ ο· =ο§
ο·
− 1ο·οΈ ο§ο¨ b − 3 ο·οΈ ο¨ 3 οΈ ο¨ − 8 οΈ
ο¦ − a + 1 οΆ + ο¦1οΆ = ο¦ − 2οΆ
ο§ − b + 3 ο· ο§ 3ο· ο§ − 8 ο·
οΈ
ο¨
οΈ ο¨ οΈ ο¨
− a +2=− 2 ο a=4
− b + 6 = − 8 ο b =14
ο Q is (4, 14)
ο¦ο§ 2 οΆο·
ο¨ 4οΈ
Exercise 3
(a) Find the image of the point (7 , − 1) under a 90° positive quarter turn, centre (0, 0).
(b) Find the image of the point (9 , − 3) under a half turn rotation, centre (2, 4).
(c) The point (−4 , 5) is the image of the point D under an anticlockwise rotation of 90°,
centre (1, 3). Find the coordinates of D.
5. DILATATION
A Dilatation can either be an Enlargement (where an object is enlarged in size) or
a Reduction (where an object is reduced in size).
Enlargement
147
An Enlargement is represented by the
2ο΄ 2 matrix E = ο¦ο§ k0
ο¨
0οΆ .
k ο·οΈ
The linear scale factor is k, where k Λ 1 and the area scale factor is k2, which is
2
the determinant of the matrix E. The object and its image are in the ratio 1 : k .
This means if the area of the object is A cm2, the area of the image will be k 2 A cm2.
AB
If the image of the line AB is A1B1, then the linear scale factor k = 1 1 .
AB
If k Λ 0 (negative), the image is turned round. The transformation is equivalent to
a combination of an enlargement followed by a 180° rotation or vice-versa.
If k = 1, the object will remain where it is since, in this case, we shall have
the identity matrix ο¦ο§ 1
ο¨0
0οΆ .
1 ο·οΈ
An enlargement is completely described by stating the centre of enlargement and
the linear scale factor.
Reduction
This has all the properties of Enlargement except that the lengths are reduced
by the same linear scale factor k, where − 1 Λ k Λ 1.
Finding the Centre of an Enlargement
Finding the image under an Enlargement
(i) When the Centre is the origin (0, 0)
The image ( x1 , y1 ) of the point ( x, y) under an enlargement, centre (0, 0), is defined
ο¦ x1 οΆ ο¦ k
ο·=ο§
ο¨ y1 οΈ ο¨ 0
by ο§
0οΆο¦ x οΆ
k ο·οΈ ο§ο¨ y ο·οΈ , where
ο¦ο§ k
ο¨0
0 οΆ is the matrix associated with the enlargement
k ο·οΈ
and k as the linear scale factor.
(ii) When the Centre is (r, s), where r ≠ 0 and s ≠ 0
The image ( x1 , y1 ) of the point ( x, y) under an enlargement, centre (r, s), is defined
ο¦ x1 οΆ ο¦ k
ο·=ο§
ο¨ y1 οΈ ο¨ 0
by ο§
0οΆο¦ x − r οΆ
k ο·οΈ ο§ο¨ y − s ο·οΈ +
ο¦ο§ r οΆο· , where ο¦ k
ο§0
ο¨sοΈ
ο¨
0 οΆ is the matrix associated with the
k ο·οΈ
enlargement and k as the linear scale factor.
Examples
(a) Find the image of the point (4 , 1) under an enlargement, scale factor 2, centre (0, 0).
(b) Find the image of the point (2 , − 5) , under an enlargement, scale factor 3, centre (2, 4).
(c) The point (−2 , − 8) is the image of the point Q under an enlargement scale factor − 3 ,
centre (1, 3). Find the coordinates of Q.
Solutions
148
(a) ο¦ο§ 2
ο¨0
0 οΆ ο¦ 4οΆ ο¦ 8οΆ
ο§ ο·=
2 ο·οΈ ο¨ 1 οΈ ο§ο¨ 2 ο·οΈ
The image is (8 , 2)
(c) ο¦ο§ − 3
ο¨ 0
(b)
0 οΆ ο¦ 2 − 2 οΆ ο¦ 2οΆ
+ο§ ο·
3 ο·οΈ ο§ο¨ − 5 − 4 ο·οΈ ο¨ 4 οΈ
0οΆ ο¦ 0 οΆ ο¦ 2οΆ
+
3 ο·οΈ ο§ο¨ − 9 ο·οΈ ο§ο¨ 4 ο·οΈ
0 οΆ ο¦ 2οΆ
= ο¦ο§
ο· + ο§ 4ο·
−
27
ο¨
οΈ ο¨ οΈ
2
οΆ
= ο¦ο§
ο·
ο¨ − 23 οΈ
The image is (2 , − 23)
ο¦3
ο§0
ο¨
3
= ο¦ο§
0
ο¨
0 οΆ ο¦ a − 1 οΆ ο¦1οΆ ο¦ − 2 οΆ
+ ο§ ο·=ο§
ο·
− 3 ο·οΈ ο§ο¨ b − 3 ο·οΈ ο¨ 3 οΈ ο¨ − 8 οΈ
ο¦ − 3a + 3 οΆ + ο¦ 1 οΆ = ο¦ − 2 οΆ
ο§ − 3b + 9 ο· ο§ 3 ο· ο§ − 8 ο·
οΈ
ο¨
οΈ ο¨ οΈ ο¨
− 3a + 4 = − 2 ο a = 2
− 3b +12 = − 8 ο b = 6 23
ο Q is (2, 6 23 )
Exercise 4
(a) Find the image of the point (1, − 2) , under an enlargement, scale factor 2, centre (3, 1).
(b) Find the image of the point (−4 , 0) under an enlargement, scale factor 4, centre (0, 0).
(c) The
point (−7 , 10) is the image of the point C under an enlargement scale factor − 2 ,
centre (5, 0). Find the coordinates of C.
6. SHEAR
ο¦1 k οΆ
ο·ο· .
A Shear with the x- axis as the invariant line is represented by the matrix H = ο§ο§
ο¨0 1οΈ
ο¦ 1 0οΆ
ο·.
1 ο·οΈ
A Shear with the y- axis as the invariant line is represented by the matrix H = ο§ο§
ο¨k
The shear factor =
distance moved by point
.
distance of that point from the invariant line
When finding the shear factor from a given diagram, you should note the following:
(i) a horizontal movement to the right is positive and a horizontal movement to the left
is negative.
(ii) a vertical movement up is positive and a vertical movement down is negative.
In the diagram above, where the image is shaded, the x- axis is invariant and the movement
of points is horizontal right. Therefore, the shear factor k is given by
BB 3
distance moved by B
1 32 οΆ
k=
= 1 = and the matrix is then H = ο¦ο§ο§
ο·ο· .
distance of B from the x - axis BC 2
ο¨0 1οΈ
Remarks
(i) the x-axis is the invariant line
(ii) DC is the line of invariant points of the figure being transformed by this shear.
A Shear is completely described by stating the equation of the invariant line, the linear scale factor and the
direction.
149
Exercise 5
Each of the diagrams (a), (b), (c) and (d) below contains a unit square ABCD and its image
under a shear. Each image is shown shaded. For each diagram,
(i) State the invariant line and, where possible, the line of invariant points
(ii) Find the shear factor and write down the shear matrix
(iii) Describe each shear completely.
Answers to Exercise 5
(a) (i) invariant line: x – axis, no line of invariant points
ο¦ 1 32 οΆ
3
(ii) k = 2 , H = ο§ο§
ο·ο· .
ο¨0 1οΈ
3
(iii) Shear, x – axis invariant, factor 2 , horizontal right.
(b) (i) invariant line: x – axis, no line of invariant points
ο¦1 − 2οΆ
ο·.
1 ο·οΈ
(ii) k = − 2 , H = ο§ο§
ο¨0
(iii) Shear, x – axis invariant, factor − 2 , horizontal left.
(c) (i) invariant line: y – axis, line of invariant points: AD
ο¦ 1 − 3οΆ
ο·.
1 ο·οΈ
(ii) k = − 3 , H = ο§ο§
ο¨0
(iii) Shear, y – axis invariant, factor − 3 , vertical down.
(d) (i) invariant line: y – axis, no line of invariant points
ο¦1 2οΆ
ο·.
1 ο·οΈ
(ii) k = 2 , H = ο§ο§
ο¨0
(iii) Shear, y – axis invariant, factor 2 , vertical up.
150
7. STRETCH
A Stretch with the x- axis as the invariant line is represented by the matrix S = ο¦ο§ 1
0οΆ
k ο·οΈ .
A Stretch with the y- axis as the invariant line is represented by the matrix S = ο¦ο§ k
0οΆ
1 ο·οΈ .
ο¨0
ο¨0
The stretch factor =
distance of image from invariant line
.
distance of object from invariant line
When finding the stretch factor from a given diagram, you should note the following:
(i) A horizontal movement to the right is positive and a horizontal movement to the left
is negative.
(ii) A vertical movement up is positive and a vertical movement down is negative.
In the diagram above, the image of the square ABCD under a stretch is shaded. The
invariant line is the y- axis. There is no line of invariant points since all the points of the
square have moved. The movement of points is horizontal right. Therefore, the stretch
factor k is given by
k=
distance of B1 from the y − axis 4
ο¦ 2 0οΆ
= = 2 and the matrix is then S = ο§ο§
ο·ο· .
distance of B from the y - axis
2
ο¨0 1οΈ
A Stretch is completely described by stating the invariant line, the linear scale factor
and the direction.
Exercise 6
Each of the diagrams (a) – (b) below contains a unit square ABCD and its image under
a stretch. Each image is shown shaded. For each diagram,
(i) state the invariant line and, where possible, the line of invariant points
(ii) find the stretch factor and write down the stretch matrix
(iii) Describe each stretch completely.
151
Answers to Exercise 6
(a) (i) invariant line: y – axis, line of invariant points: AD
ο¦ 3 0οΆ
ο·.
1 ο·οΈ
(ii) k = 3 , S = ο§ο§
ο¨0
(iii) Stretch, y – axis invariant, factor 3 , horizontal right.
(b) (i) invariant line: x – axis, line of invariant points: DC
ο¦1
(ii) k = − 2 , S = ο§ο§
ο¨0
0οΆ
ο·.
− 2 ο·οΈ
(iii) Stretch, x – axis invariant, factor − 2 , vertical down.
(c) (i) invariant line: y – axis, line of invariant points: AD
ο¦ − 3 0οΆ
ο·.
1 ο·οΈ
(ii) k = − 3 , S = ο§ο§
ο¨ 0
(iii) Stretch, y – axis invariant, factor − 3 , horizontal left.
(d) (i) invariant line: x – axis, line of invariant points: DC
ο¦1 0οΆ
ο·.
3 ο·οΈ
(ii) k = 3 , S = ο§ο§
ο¨0
(iii) Stretch, x – axis invariant, factor 3 , vertical up.
----------------------------------------------------------------------------------------------------------------
152
Exercise 5: Miscellaneous
ο¦1 0οΆ
ο·ο· , B =
ο¨ 0 −1οΈ
1. (a) A = ο§ο§
ο¦ −1 0 οΆ
ο§ο§
ο·ο· , C =
ο¨ 0 − 1οΈ
ο¦− 6 οΆ
ο§ο§
ο·ο· , D =
ο¨ 1οΈ
ο¦ −1
ο§ο§
ο¨ 0
0οΆ
ο·, F =
1 ο·οΈ
ο¦ 0
ο§ο§
ο¨ −1
− 1οΆ
ο·.
0 ο·οΈ
(i) Name the transformation that each of matrix A, B, C, D and F represents.
(ii) Find the image of P (−2, − 3) under
(a) a reflection in the line y = 0 (b) a positive quarter turn
(c) the translation given above
0 οΆ
ο¦ 6− x
ο·ο· represents an enlargement.
ο¨8 − 2 y 4 + x οΈ
(b) The matrix L = ο§ο§
(i) Find the value of x and the value of y.
(ii) Write down the area scale factor.
-----------------------------------------------------------------------------------------------------------2. (a) Triangle V is the image of βXYZ, with vertices X(1, 2), Y(4, 2) and Z(1, 3),
ο¦−1 0 οΆ
ο·ο· .
under the transformation given by the matrix A = ο§ο§
ο¨ 0 2οΈ
(i) Draw and label triangle V using a scale of 1 unit to 1 cm on each axis.
Take − 5 ≤ x ≤ 5 and − 7 ≤ y ≤ 7.
(ii) Triangle W is the image of triangle V under a rotation of 180Λ about
the origin O. Draw and label triangle W.
(iii) Determine the matrix representing the single transformation which maps
triangle V onto triangle W.
(b) The point (x, y) is mapped onto the point ( x1 , y1 ) by the transformation D
ο¦ x1 οΆ
ο¦ xοΆ
ο¦1 − 2οΆ
ο¦ 2
ο·ο· → A ο§ο§ ο·ο· + B , where A = ο§ο§
ο·ο· and B = ο§ο§
1οΈ
ο¨ yοΈ
ο¨0
ο¨− 5
ο¨ y1 οΈ
(i) Find the coordinates of the point P1 , the image of P (3, 2) under D.
(ii) Q1 (−1, 6) is the image of Q(a, b) under the transformation D.?
described by ο§ο§
οΆ
ο·ο· .
οΈ
Find the value of a and the value of b.
---------------------------------------------------------------------------------------------------------------3. Answer the whole of this question on a sheet of graph paper
Triangle A has vertices (4, 1), (4, -1) and (5, 1)
Triangle B has vertices (1, 4), (-1, 4) and (1, 5)
(i) Using a scale of 1 cm to represent 1 unit on each axis, draw axes for values
of x and y in the range -6 ≤ x ≤ 6 and -6 ≤ y ≤ 6 . Draw and label the
triangles A and B.
(ii) Describe fully the single transformation which maps triangle A onto
triangle B
ο¦ 0
ο¨−1
(iii) The transformation represented by the matrix ο§ο§
1οΆ
ο· maps triangle A
0 ο·οΈ
onto triangle C. Draw and label triangle C.
(iv) Write down the matrix representing the transformation which maps
triangle B onto triangle C.
(v) Given also that triangle C is mapped onto triangle D by a translation given
ο¦− 5
ο¨ 3
by ο§ο§
οΆ
ο·ο· , draw and label triangle D.
οΈ
---------------------------------------------------------------------------------------------------------4. The points A (1,1) , B ( 2, 3 ) and C ( 3, 2 ) are vertices of οABC .
(a) Using a scale of 1 cm for 1 unit on each axis, draw and label οABC .
ο¦ 2 1οΆ
ο·ο· .
S = ο§ο§
ο¨ 0 1οΈ
153
οABC is transformed to οA1 B1C1 , where A1 , B1 and C1 are respectively the
images of A, B and C under the transformation with matrix S.
(b) (i) Find the coordinates of A1 , B1 and C1 .
(ii) Draw and label οA1 B1C1 .
(c)
ο¦ 0 2οΆ
ο·ο· .
T = ο§ο§
−
1
1
ο¨
οΈ
οA1 B1C1 is transformed to οA2 B2 C2 , where A2 , B2 and C 2 are respectively
the images of A1 , B1 and C1 under the transformation with matrix T.
Draw and label οA2 B2 C2 .
An enlargement, centre O, followed by a rotation about O transforms οABC
onto οA2 B2 C2 .
(d)
Find
(i) the scale factor of the enlargement,
(ii) the angle of rotation.
(e) Find the matrix of the transformation that maps οABC onto οA2 B2 C2 .
----------------------------------------------------------------------------------------------------------5. (a) Find the matrix for the stretch S parallel to the y – axis if the x – axis is invariant
and the point P (1, 2) is mapped onto PΚΉ (1, 6).
(b) (i) Plot the points A (2, 1), B(3, 5) and C(5, 1).
(ii) If S is a stretch such that the y – axis is invariant and the point (1, 0) is mapped
onto (3, 0), plot the image of triangle ABC under S.
(iii) Find the matrix S and state its determinant.
(iv) Find the ratio of the areas of the two triangles.
--------------------------------------------------------------------------------------------------------------
TOPIC 27: EARTH GEOMETRY
(i) THE EARTH
Hints:
• The Earth is very nearly spherical.
• Mean Radius of the Earth is 6 370 km or 3 437nm.
• Longitude: The meridian that passes through Greenwich is called the Prime meridian. All longitudes
run through the North Pole and the South Pole. All longitudes are measured west or east of the Prime
(Greenwich) Meridian.
•
•
•
Latitude: Latitude marks the distance (in degrees) given to a place north or south of the equator
All latitudes parallel to the equator.
Among the latitudes, the equator is the only great circle.
154
LEARNER’S ACTIVITY:
In the diagram below,
(i)
(ii)
(iii)
(iv)
(v)
state the longitude of the points A, B, C, T and R.
find the difference in longitude between A and T; A and R; E and N; B and C.
find the difference in longitude between A and B; B and T; T and C; C and R.
state the latitude of the points C, D, E, M, N and R.
find the difference in latitude between C and D; D and E; C and E; M and N; Rand N.
DISTANCE BETWEEN POINTS ALONG THE SAME CIRCLE OF LONGITUDE.
NOTE:
• If the two given points are in the same hemisphere,
π½ = π
πππππππππ ππ πππππππ
π.
• If the two given points are in different hemispheres,
π½ = πππ ππ πππππ ππππππ ππ πππππππ
ππ.
i) Distance in nautical miles
• along a Great circle, 1° of arc = 60 nm
∴distance between two points= π½ × ππππ.
where π is the difference in latitude between the given
points.
π½
Or distance between two points = πππ × ππ
πΉ ππ.
Where is the difference in latitude and R= 3 437nm
is the radius of the Earth?
ii) Distance in kilometres (km)
• Radius of the Earth in kilometres = 6 370km
π½
∴ π
πππππππ πππππππ πππ ππππππ =
× ππ
πΉ ππ
πππ
With R= 6370 km.
155
iii) distance between any two points on the equator:
NOTE: Distance is calculated as in i) and ii), but
π½ is the difference in Longitude between the
two given points for points on the same side
of the Prime Meridian, OR the sum of their
Longitudes, if they are on opposite sides of
the Prime Meridian.
iv) distance between points on the same circle of Latitude
other than the Equator.
• Formulae:
i) distance = π½ × ππππ × ππππΆ ππ, where π½is the difference OR sum in
longitude and πΆis the
angle of Latitude.
π½
ii) distance= πππ × ππ
π,but r=RcosπΆ
π½
∴Distance== πππ × ππ
πΉππππΆ
ππ
NOTE: In nm use R= 3437nm; in km use R= 6370km. and let π
= π or 3.142.
ACTIVITY 1.
Find the distance along a circle of latitude between P (40°N, 30°E) and
Q (40°N, 50°E).
EXPECTED ANSWERS:
- difference in longitude=50° − 30° = 20°
- latitude= 40°
Distance= π × 60 × πΆπππΌ
∴Distance= 20° × 60 × πΆππ40°
∴Distance=919 nm.
ACTIVITY 2.
The diagram below shows a wire model of the earth, the circle of latitude in the north is 80°π and the circle
of latitude0°. The meridian πππ
π is 60°πΈ and meridian ππππ is directly oppositeπππ
π.
156
i) State the position of the point π.
ii) Find the distance along the circle of latitude 80°π between π and π in ππ.
iii) Calculate the shortest distance between π and π
iv) If the time at π
is 20: 00βππ , what is the time at π
EXPECTED ANSWERS:
i.
ii.
∴ π(80π, 120π)
180
× 2 × 3.142 × 6370 × πππ 80
360
= 3475.5ππ
= ππππππ (3 π . π)
=
iii. SHORTEST DISTANCE BETWEEN P AND Q.
π
π·=
× 2ππ
360
20
× 2 × 3.142 × 6370
360
π· = 2224ππ
π· = ππππππ(3 π . π)
π·=
iv.
180
15
= 12 βππ’ππ
∴ 20: 00 − 12: 00 = ππ: ππ πππππ.
ACTIVITY 3.
The diagram below is a sketch of the earth and on it are the points P (20°N, 80°E), Q (40°S,
80°E) and R (40° S, 30° E). [Use π = 3.142 πππ π
= 6370ππ]
157
(i)
(ii)
Calculate the distance QR in kilometers.
An aeroplane starts from P and flies due west on the same latitude covering a distance
of 1 232km to point T.
(a) Calculate the difference in angles between P and T,
(b) Find the position of T.
EXPECTED ANSWERS:
(i)
Dist.ππ =
π
360°
DISTANCE OF PQ:
-difference in longitude= 80° − 30°
= 50° = π,
ππ ππ ππ πππ‘ππ‘π’ππ = 40°π = πΌ,
× 2ππ
πππ πΌ
50°
× 2 × 3.142 × 6370πππ 40°
360°
∴ Distance PQ = 4258.8964 km
≈ ππππ. πππ (1 π. π. )
π
(ii)
(a) Distance =
× 2ππ
πππ πΌ, but Dist.= 1232km,
360°
R= 6370km,
πΌ = 20°.
π
∴ 1232 =
× 2 × 3.142 × 6370πππ 20°
360°
104.4861975
∴ π=
1232
∴ π = 11.79103106
≈ ππ. π°
∴ Difference in angles between P and T≈ ππ. π°
∴ π·ππ π‘. ππ =
(π)Position of T is: 80° − ππ. ππππ° = ππ. πππ°
π»(ππ°π΅, ππ. π°πΊ)
ACTIVITY 4.
In the diagram below, A (65°N, 5°E), B (65°N, 45°W) and Care three
points on the surface of the model of the earth and O is the centre of the
model. The point C due south of A, is such that AOC = 82° .
[π = 3.142, πΉ = 3437nm]
158
∴
(i)
(ii)
(iii)
State the longitude of A,
Calculate the latitude of C,
Calculate, in nautical miles, the shortest distance
(a) between A and C and measured along the common longitude,
(b) between A and B measured along the circle of latitude.
EXPECTED ANSWERS:
(i)
(ii)
(iii)
Longitude of A = π° π¬
Latitude of C = 82° − 65° = 17° π
(a)π° ππ πππ = ππππ,
Difference in latitude= 82° = π,
∴ Shortest distance between A and C:
π × 60ππ = 82° × 60ππ = ππππ ππ.
(b) Difference in longitude between A and B
=45°π + 5°πΈ = ππ°.
∴ Distance along circle of latitude 65°π΅, between A and B
= π × 60ππ × πππ πΌ
= 50° × 60ππ × πππ 65°
= ππππ. ππππππ ≈ ππππ. πππ(1 d.p.)
ACTIVITY 5
In the diagram below, the points P and Q lie on the same latitude, O is the Centre of
the earth and angle NOQ = 60°. (ππππ π = 3.142 πππ π
= 6370ππ)
(i)
(ii)
(iii)
State the latitude where the points P and Q are lying.
Find the distance between P and T.
Given that the point P is on longitude 18°π and the time difference between P and Q is
5hours, calculate the longitude on which Q lies.
EXPECTED ANSWERS:
(i)
Latitude where P and Q lie = 90° − 60°
= ππ° N.
159
(ii)
(iii)
difference in latitude = 30° − 0°
= 30° = π.
π
∴ π·ππ π‘ππππ πππ‘π€πππ π· πππ πΈ =
× 2ππ
360°
30°
=
× 2 × 3.142 × 6370ππ
360°
= ππππ. ππππππππ
= ππππ. ππππ (2 π. π. )
Note: πππ = ππ°.
5hr = y°
∴y° = 5βππ π‘πππ ππππππππππ = ππ°
∴ ππππππππ
π ππ πΈ = 75° − 18°
∴ ππππππππ
π ππ πΈ = 57°.
ACTIVITY 6
The diagram below shows a wire model of the earth. The circle of latitude in the north
is 50°π΅and the circle of latitude in the south is 60°πΊ. A and C are on longitude 55°W
while B and D are on longitude 50°E.
(Take π = 3.142 πππ π
= 3437ππ)
(i)
(ii)
(iii)
(iv)
Write the positions, using longitudes and latitudes, of the points A and D.
Calculate the difference in longitudes between A and B.
Given that the time at town D is 09 20 hours, what would be the time at town C?
Calculate the distance BD along the longitude 50°E in nautical miles.
EXPECTED ANSWER:
(i)
A(50°N, 55°W) and D(60°S, 50°E)
(ii)
Difference in longitudes between A and B= 55° + 50° = πππ°
(iii)
TIME AT C=
Note: C (60°S, 55°W), D (60°S, 50°E)
Difference in longitudes between C and D=55° + 50° = πππ°
Note: 1hr = 15°
X hrs. = πππ°
∴ X hrs. =
(iv)
πππ°
15°
= 7βππ time difference,
∴ TIME AT C= 09 20hrs −7βππ (C is West of D, so subtract)
∴ TIME AT C= 02 20hrs.
Difference in latitudes= 50° +60° = 110°,
∴ π«πππππππ πππππππ π© πππ
π« = 110° ×60nm= ππππππ.
160
TOPIC 28: INTRODUCTION TO CALCULUS
Calculus is a branch of mathematics which was developed by Newton (1642-1727) and Leibnitz
(1646-1716) to deal with changing quantities.
EXPECTED OUTCOMES:
A: Differentiation.
1. Differentiate functions from first principles.
2. Differentiate functions using the formula
3. Calculate equations of tangents and normals
B: Integration
Find indefinite integrals
Evaluate simple definite integrals
Find the area under the curve
A: DIFFERENTIATION.
1. DIFFERENTIATING FUNCTIONS FROM FIRST PRINCIPLES.
EXAMPLES:
1. If f ( x ) = 2 x − 5 , f ( x) from first principle.
'
SOLUTION
f '( x) = lim
h →o
f ( x + h) − f ( x )
h
DATA:
f ( x) = 2 x − 5
f ( x + h) = 2( x + h) − 5 [Plug in these functions in the formula above]
f ( x + h) − f ( x )
h
2( x + h) − 5 − (2 x − 5)
f '( x) = lim
h →o
h
2 x + 2h − 5 − 2 x + 5
= lim
h →o
h
2h
= lim
h →o h
= lim 2
f '( x) = lim
h →o
h →o
f '( x) = 2
2. Find
dy
2
from first principle for the function y = 2 x .
dx
SOLUTION:
dy
f ( x + h) − f ( x )
= lim
dx h→o
h
DATA.
f ( x) = 2 x 2
f ( x + h) = 2( x + h) 2
[plug in these in the formula above]
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f '( x) = lim
h →o
f ( x + h) − f ( x )
h
dy
2( x + h) 2 − 2 x 2
= lim
dx h→o
h
2( x 2 + 2 xh + h 2 ) − 2 x 2
= lim +
h →o
h
2
2 x + 4 xh + 2h 2 − 2 x 2
= lim
h →o
h
2
4 xh + 2h
= lim
h →o
h
= lim 4 x + 2h
h →o
dy
= 4 x (note that as h → o, 2h = 0)
dx
EXERCISE:
1. Find f '( x) for each of the following functions by first principle.
(a ) f ( x) = 5 x + 4
(b) f ( x) = x 2 − 1
(c) f ( x) = 20 x 2 − 6 x + 7
(a) f '( x) = 5
Expected Answers:
(b) f '( x) = 2 x
(c) f '( x) = 40 x − 6
2. DIIFFERENTIATING FUNCTIONS USING THE FORMULA:
A. The Derivate of ax n
Given a function y = f ( x) = ax then it follows that
n
dy
= f '( x) = anx n −1 .
dx
Examples
1. Given that y = 7 , find
dy
.
dx
SOLUTION:
y = 7 is the same as y = 7 x0 sin ce x0 = 1
y = 7 x0
dy
= (0)7 x 0−1
dx
dy
=0
dx
NOTE: The derivate of any constant is Zero (0)
2. Given that y = 5 x , find
dy
.
dx
162
SOLUTION:
y = 5xis the same as y = 5x1
y = 5 x1
dy
= 1(5) x1−1
dx
dy
= 1(5) x 0
dx
dy
=5
dx
3. Find the derived function of y = 2 x + 5x − x + 2
SOLUTION:
4
3
2
y = 2 x 4 + 5 x3 − x 2 + 2
dy
= 4(2) x 4−1 + 3(5) x3−1 − 2 x 2−1 + 0
dx
dy
= 8 x3 + 15 x 2 − 2 x1
dx
dy
= 8 x3 + 15 x 2 − 2 x
dx
B. The Derivate of (ax + b)
n
The derivate of the function y = (ax + b) is given by the formula
n
dy
= n(ax + b)n −1 ο΄ a
dx
EXAMPLE:
1. If y = (3x + 5)4 , find
dy
.
dx
SOLUTION:
y = (3 x + 5) 4
dy
= 4(3x + 5) 4−1 ο΄ 3
dx
dy
= 12(3 x + 5)3
dx
C. The Derivate of a product. (Product Rule)
n
m
If y = (ax + b) (cx + d ) we can let u = (ax + b) and v = (cx + d ) . From it follows that, the derivative
of a product is given by the formula
dy
du
dv
=v
+u
dx
dx
dx
Example:
1. Given that y = (3x + 1) (2 x − 5) , find
2
3
dy
.
dx
SOLUTION:
y = (3x + 1)2 (2 x − 5)3
We let ---
163
u = (3x + 1) 2 and v = (2 x − 1)3
du
= 2(3x + 1) ο΄ 3 and
dx
du
= 6(3x + 1)
dx
dv
= 3(2 x − 1) 2 ο΄ 2
dx
dv
= 6(2 x − 1) 2
dx
From the above it follows that =
dy
du
dv
= v +u
dx
dx
dx
dy
= (2 x − 1)3 6(3x + 1) + (3 x + 1) 2 6(2 x − 1) 2
dx
= 6(2 x − 1)3 (3 x + 1) + 6(3 x + 1) 2 (2 x − 1) 2
= 6(2 x − 1) 2 (3 x + 1)[2 x − 1 + 3 x + 1]
= 6(2 x − 1) 2 (3 x + 1)(5 x)
= 30 x(2 x − 1) 2 (3x + 1)
C: THE DERIVATE OF A QUOTIENT
If y = f ( x) is a ratio of functions u and v where u and v are also functions of x , the derivative of
the function y with respect to x is given by the formula
du
dv
v −u
dy
vu '− uv '
= dx 2 dx =
dx
v
v2
Example:
( x − 3) 2
1. Differentiate
.
( x + 2) 2
SOLUTION:
SINCE
y=
( x − 3)2
( x + 2)2
We let u = ( x − 3) and v = ( x + 2)
2
2
u ' = 2( x − 3) and v ' = 2( x + 2)
dy vu '− uv '
=
dx
v2
dy ( x + 2) 2 2( x − 3) − ( x − 3) 2 2( x + 2)
=
2
dx
ο©( x + 2 ) 2 οΉ
ο«
ο»
2
2( x + 2) ( x − 3) − 2( x − 3) 2 ( x + 2)
=
( x + 2) 4
2( x + 2)( x − 3)[( x + 2) − ( x − 3)]
=
( x + 2) 4
2( x − 3)( x + 2 − x + 3)
=
( x + 2)3
2( x − 3)(5)
=
( x + 2)3
dy 10( x − 3)
=
dx ( x + 2)3
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EXERCISE:
−2
1. Differentiate y = 6 x + 3x − x + 9
2. Differentiate ( x − 4 x)
3
2
7
3. Differentiate f ( x) = ( x + 1) ( x + 2)
EXPECTED ANSWERS:
2
3
dy
= −12 x −3 + 6 x − 1
dx
dy
2.
= 7(3x 2 − 4)( x3 − 4 x)7
dx
1.
3. f '( x) = ( x + 1)(5x + 7)( x + 2)
2
TANGENTS AND NORMALS
If y = f ( x) is a curve, we can find the gradient at any point on the curve. This gradient is equal to the
gradient of the tangent to the curve at that point.
If the gradient of the tangent is m1 and that of the normal line is m2 it follows that m1 ο΄ m2 = −1
The tangent and the normal are perpendicular to each other at the point of contact. Example
(FINDING THE GRADIENT OF THE TANGENT AND THENORMAL)
1. Find the gradient of the tangent and the normal to the curve y = 3x + 4 x + 1 at the point where
x = 4.
SOLUTION
2
y = 3x 2 + 4 x + 1 is the equation of the curve.
The gradient of the tangent is
dy
= m1
dx
m1 = 6 x + 4 and since the value of x = 4
m1 = 6(4) + 4
m1 = 24 + 4
m1 = 28
To find the gradient of the normal, recall that
165
m1 ο΄ m2 = −1
28 ο΄ m2 = −1
m2 =
ο The gradient of the tangent, m1 = 28 and the gradient of the normal, m2 =
−1
28
−1
28
Example (FINDING THE EQUATION OF THE TANGENT AND THE NORMAL)
1. Find the equation of the tangent and the normal to the curve y = 3x + 4 x + 1 at the point where
2
x = 4.
SOLUTION
We have the value for x. So let’s find the corresponding value for y.
y = 3 x 2 + 4 x + 1, x = 4
y = 3(4) 2 + 4(4) + 1
y = 3(16) + 16 + 1
ο The point is (4, 65)
y = 48 + 16 + 1
y = 65
The gradient of the tangent is m1 = 28 and that of the normal is m2 =
Equation of the tangent
−1
at the point (4, 65)
28
Equation of the normal
y = m2 x + c
−1
(4) + c
28
−1
65 =
+c
7
1
65 + = c
7
456
c=
7
−1
456
y=
x+
28
7
65 =
y = m1 x + c
65 = 28(4) + c
65 = 112 + c
c = −47
y = 28 x − 47
EXERCISE
1. Find the equation of the tangent and the normal to the curve x = 4 y at the point (6,9).
2
Expected Answers: y = 3x − 9 for the tangent and y =
−1
x + 11 for the normal.
3
INTEGRATION
The inverse of differentiation or the reverse of differentiation is called integration.
Since integration is the reverse of differentiation the following steps must be taken:1. Increase the power of the variable by 1.
2. Divide the term (variable term) by the new power.
3. Then finally add the arbitrary term C
Example (INDEFINITE INTEGRALS)
An indefinite integral must contain an arbitrary constant (C). An integral of the form
ο² f ( x) dx is
called an indefinite integral.
1. Integrate the following gradient functions
(a)
dy
= 3x
dx
(b) f '( x) = 6 x + 2 x − x
3
SOLUTION:
166
2
(a)
f '( x) = 6 x3 + 2 x 2 − x
dy
= 3 x = 3 x1
dx
3 x1+1
y=
+c
2
3x 2
y=
+2
2
f '( x) = 6 x3 + 2 x 2 − x1
6 x3+1 2 x 2+1 x1+1
f ( x) =
+
−
+c
4
3
2
6 x 4 2 x3 x 2
f ( x) =
+
− +c
4
3
2
3
2
1
f ( x) = x 4 + x3 − x 2 + c
2
3
2
(b)
EXERCISE
1. Integrate the following gradient functions
(a) 5x 2 − x + 1 (b) x6 − 3x4 + 2 x2 + 1
EXPECTED ANSWERS:
1(a) y =
5 3 1
x − x+ x+c
3
2
(b) y =
1 7 3 5 2 3
x − x + x + x +c
7
5
3
DEFINITE INTEGRALS
b
A definite integral is an integral performed between the limits. Thus A =
ο² f ( x) dx is an integral
a
performed between the limiting values a and b for x.
NOTE that y =
dy
ο² dx = ο² f '( x) dx
Example
1. Evaluate the definite integral of 4 x3 − 1 between x = 1 and x = 3
f ( x) = 4 x3 − 1
3
ο²
1
3
f ( x) dx = ο² (4 x 3 − 1) dx
1
4
= [ x 4 − x]13
4
= [ x 4 − x]13
= (34 − 3) − (14 − 1)
= (81 − 3) − (1 − 1)
= 78 − 0
= 78
Integrals can be used to compute the area under a given curve.
Example
1. Find the area of the region bounded by the curve y = x + 1 , the ordinates x=1 and x=2 and the xaxis.
2
167
2
A = ο² ( x 2 + 1) dx
1
x3
+ x]12
3
23
13
= ( + 2) − ( + 1)
3
3
8
1
= ( + 2) − ( + 1)
2
3
1
A = 3 units 2
3
=[
EXERCISE:
2
3
1. Find the area under the curve y = x + x between x=1 and x=3 Expected Answer 12 units 2
2
2. Find the area enclosed by the x – axis, the curve y = 3x + 2 and the straight lines x=3 and x= 5.
Expected Answer. 102 squared units.
2
168
