JUNE 2017 CSEC MATHEMATICS PAPER 2
SECTION 1
1. (a)
Using a calculator, or otherwise, calculate the EXACT value of
æ 1 2ö 4
ç 4 -1 ÷ ÷
è 3 5 ø 15
(i)
SOLUTION:
sp
as
s
m
at
hs
.
co
m
æ 1 2ö 4
Required to calculate: ç 4 - 1 ÷ ÷ in exact form
è 3 5 ø 15
Calculation:
1 2 13 7
4 -1 = 3 5 3 5
5 (13) - 3 ( 7 )
=
15
65 - 21
=
15
44
=
15
So, now we have
æ 1 2 ö 4 44 4
÷
ç 4 -1 ÷ ÷ =
è 3 5 ø 15 15 15
.fa
=
w
( 3.1 - 1.15)
w
w
(ii)
11
44 15
´
15
41
= 11 (in exact form)
2
0.005
SOLUTION:
Required to calculate:
Calculation:
( 3.1 - 1.15)
0.005
2
(1.95)
=
( 3.1 - 1.15)
0.005
2
in exact form
2
(By calculator)
0.005
3.8025
=
0.005
= 760.5 (in exact form)
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Pg 1 of 40
(b)
A store is promoting a new mobile phone under two plans. Plan A and Plan B.
The plans are advertised as shown in the table below.
Deposit
Monthly installment
Number of months to repay
Tax on ALL payments
Plan B
$600
$80
6
5%
Calculate the TOTAL cost of a phone under Plan A.
co
m
(i)
Plan A
$400
$65
12
0%
SOLUTION:
Data: Table showing two mobile phone plans as advertised by a store.
at
hs
.
Required to calculate: The total cost of a phone under Plan A
Calculation:
Total cost under Plan A
= The deposit + ( Monthly installments ´ No. of months to repay ) + Tax
= $1180
sp
Determine which of the two plans, A or B, is the better deal. Justify your
answer.
SOLUTION:
Required to find: The better deal of the two plans
Solution:
Total cost under Plan B
= The deposit + ( Monthly installments ´ No. of months to repay ) + Tax
w
w
w
.fa
(ii)
as
s
= $ ( 400 + 780 )
m
= $400 + ( $65 ´12 ) + $0
= $600 + ( $80 ´ 6 ) +
5
( $600 + ( $80 ´ 6 ) )
100
5
( $600 + $480 )
100
= 1.05 ´ $ ( 600 + 480 )
= $ ( 600 + 480 ) +
= $1134
If the better deal is supposed to mean the plan that has a lesser cost, then
Plan B is the better deal as there is a savings of $1180 - $1134 = $46 .
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Pg 2 of 40
(c)
John’s monthly electricity bill is based on the number of kWh of electricity that
he consumes for that month. He is charged $5.10 per kWh of electricity
consumed. For the month of March 2016, two meter readings are displayed in the
table below.
Beginning 01 March
Ending 31 March
Calculate the TOTAL amount that John pays for electricity consumption
for the month of March 2016.
co
m
(i)
Meter Readings (kWh)
0 3 0 1 1
0 3 3 0 7
at
hs
.
SOLUTION:
Data: Table showing John’s electricity meter readings, in kWh, at the
beginning and end of March 2016. John is charged $5.10 per kWh for
electricity used.
Calculation:
Number of kWh used
m
Required to calculate: The total amount John pays for electricity for
March 2016
as
s
= End reading on 31 March - Beginning reading on 01 March
sp
=0330 7 0 3 0 11
296
.fa
The cost at $5.10 per kWh = 296 ´ $5.10 = $1509.60
w
w
\ If John pays the full amount that is required, then he would pay
$1 509.60.
w
(ii)
For the next month, April 2016, John pays $2 351.10 for electricity
consumption. Determine his meter reading at the end of April 2016.
SOLUTION:
Data: John pays $2 351.10 for electricity in April 2016.
Required to calculate: John’s meter reading at the end of April 2016
Calculation:
For the month of April John pays $2 351.10.
$2 351.10
At $5.10 per kWh, the number of kWh used in April =
$5.10
= 461 kWh
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Pg 3 of 40
Hence, the meter reading at the end of April should be the reading at the
end of March + 461
=0330 7 +
4 61
03768
Factorise the following expressions completely.
6 y 2 - 18xy
(i)
= 6 y ( y - 3x )
4m 2 - 1
m
(ii)
at
hs
.
SOLUTION:
Required to factorise: 6 y 2 - 18xy
Solution:
6 y 2 - 18 xy = 6 y ´ y - 6 y ´ 3x
co
m
2. (a)
as
s
SOLUTION:
Required to factorise: 4m 2 - 1
Solution:
4m2 - 1 = ( 2m) - (1)
This is now in the form of the difference of two squares.
So, 4m2 - 1 = ( 2m - 1)( 2m + 1)
2
.fa
sp
2
2t 2 - 3t - 2
w
w
w
(iii)
SOLUTION:
Required to factorise: 2t 2 - 3t - 2
Solution:
2t 2 - 3t - 2 = ( 2t + 1)( t - 2 )
é 2t 2 - 4t + t - 2 ù We may even expand to confirm our result
ê 2
ú
ë 2t - 3t - 2 û
(b)
Write as a single fraction and simplify
5 p + 2 3 p -1
3
4
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Pg 4 of 40
5 p + 2 3 p -1
as a simplified single fraction.
3
4
Solution:
5 p + 2 3 p -1
3
4
4 ( 5 p + 2 ) - 3 ( 3 p - 1)
12
(c)
20 p + 8 - 9 p + 3
12
20 p - 9 p + 8 + 3
=
12
11 p + 11
=
12
11( p + 1)
=
12
=
A formula is given as d =
4h
.
5
co
m
Required to write:
at
hs
.
SOLUTION:
Determine the value of d when h = 29 . Give your answer correct to 3
significant figures.
as
s
m
(i)
sp
SOLUTION:
4h
Data: d =
and h = 29
5
Required to calculate: d
w
w
w
.fa
Calculation:
When h = 29
4 ( 29 )
d=
5
=
116
5
= 23.2
= 4.816
(using the calculator)
= 4.82 (correct to 3 significant figures)
(ii)
Make h the subject of the formula.
SOLUTION:
Required to make: h the subject of the formula
Solution:
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Pg 5 of 40
4h
5
Squaring both sides to remove the hindrance of the root sign
4h
d2 =
5
2
\ 5 ´ d = 4h
d=
h=
The Universal set, U, is defined as follows:
U = {x : x Î N , 2 < x < 12}
at
hs
.
3. (a)
5d 2
4
co
m
4h = 5d 2
The sets M and R are subsets of U such that
M = {odd numbers}
as
s
m
R = {square numbers}
(i)
List the members of the subset M.
sp
SOLUTION:
Data: U = {x : x Î N , 2 < x < 12} and M and R are subsets of U where
.fa
M = {odd numbers} and R = {square numbers}.
Required to list: The members of the subset M.
Solution:
w
w
U = {3, 4, 5, 6, 7, 8, 9, 10, 11}
w
(ii)
\ M = {3, 5, 7, 9, 11}
List the members of the subset R.
SOLUTION:
Required to list: The members of the subset R
Solution:
R = {4, 9}
(iii)
Draw a Venn diagram that represents the relationship among the defined
subsets of U.
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Pg 6 of 40
Using a ruler, a pencil and a pair of compasses,
at
hs
.
(b)
co
m
SOLUTION:
Required to draw: A Venn diagram to illustrate the information given
about the sets U, M and R
Solution:
(i)
construct accurately, the square ABCD, with sides 6 cm.
as
s
w
w
w
.fa
sp
Solution:
m
SOLUTION:
Required to construct: A square ABCD with sides 6 cm.
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Pg 7 of 40
co
m
at
hs
.
m
as
s
sp
.fa
w
w
w
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Pg 8 of 40
co
m
at
hs
.
construct, as an extension of your diagram in (b) (i), the trapezium DABQ
so that ÐABQ = 120° .
m
(ii)
as
s
[Note: Credit will be given for clearly drawn construction lines.]
w
w
w
.fa
sp
SOLUTION:
Required To Construct: An extension of the diagram previously down
trapezium DABQ with the ÐABQ = 120° .
Solution:
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Pg 9 of 40
(iii)
Hence, measure and state the length of BQ.
4. (a)
1
The function f is defined as f ( x ) = x - 2 .
3
(i)
Find the value of f ( 3) + f ( -3) .
SOLUTION:
1
( -3) - 2
3
= -1 - 2
= -3
m
1
( 3) - 2
3
= 1- 2
= -1
f ( -3) =
as
s
f ( 3) =
at
hs
.
1
Data: f ( x ) = x - 2
3
Required to find: f ( 3) + f ( -3)
Solution:
co
m
SOLUTION:
Required to state: The length of BQ, by measurement
Solution:
BQ = 6.9 cm (by measurement with a ruler)
= -1 - 3
= -4
w
.fa
sp
So,
f ( 3) + f ( -3) = -1 + ( -3)
w
w
(ii)
Calculate the value of x for which f ( x ) = 5.
SOLUTION:
Required to calculate: x when f ( x ) = 5
Calculation:
𝑓(𝑥) = 5
1
∴ 𝑥−2 =5
3
1
𝑥 =5+2=7
3
𝑥 =7×3
(× 3)
𝑥 = 21
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Pg 10 of 40
Determine the inverse function, f -1 ( x )
x = 3( y + 2)
Replace y by x to obtain:
f -1 ( x ) = 3 ( x + 2)
m
The graph below shows two straight lines, ℓ1 and ℓ2 . Line ℓ1 intercepts the y –
axis at ( 0, 1) . Line ℓ2 intercepts the x and y axes at (12, 0 ) and ( 0, 6 )
respectively.
w
w
w
.fa
sp
as
s
(b)
at
hs
.
SOLUTION:
Required to find: f -1 ( x )
Solution:
1
f ( x) = x - 2
3
1
Let y = x - 2
3
Make x the subject of the formula:
1
y+2= x
3
3( y + 2) = x
co
m
(iii)
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Pg 11 of 40
Calculate the gradient of the lines ℓ1 and ℓ2 .
(i)
SOLUTION:
Data: Diagram showing two lines ℓ1 and ℓ2 . The y – intercept of ℓ1 is
( 0, 1) . The y – intercept of ℓ2 is ( 0, 6) and the x – intercept is (12, 0) .
Required to calculate: The gradient of ℓ1 and ℓ2
Calculation:
Two points on ℓ1 are ( 0, 1) and ( 2, 5) which is the point of intersection of
at
hs
.
co
m
ℓ1 and ℓ2 . Each was obtained by a read-off.
5−1
∴ Gradient of ℓ1 =
2−0
4
=
2
=2
Determine the equation of the line ℓ1
as
s
(ii)
m
Two points on ℓ2 are (12, 0 ) and ( 0, 6 ) .
6−0
∴ Gradient of ℓ2 =
0 − 12
1
=−
2
w
.fa
sp
SOLUTION:
Required to find: The equation of the line ℓ1
Solution:
The general equation of a straight line is of the form y = mx + c , where m
is the gradient and c is the intercept on the y – axis. In this case, we have
already found m = 2 and noted that c = 1.
\ The equation of the line ℓ1 is y = 2 x + 1.
w
w
(iii)
What is the relationship between ℓ1 and ℓ2 ? Give a reason for your
answer.
SOLUTION:
Required to find: The relationship between ℓ1 and ℓ2 .
Solution:
Gradient of ℓ1 = 2
1
Gradient of ℓ2 = − 2
1
Gradient of ℓ1 ×Gradient of ℓ2 = 2 ´ - = -1
2
Hence, ℓ1 is perpendicular to ℓ2 since the product of the gradients of
perpendicular lines is -1 .
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Pg 12 of 40
PTRS, not drawn to scale, is a quadrilateral. Q is a point on PT such that
QT = QR = QP . Angle QRT = 76° .
co
m
5. (a)
Determine, given a reason for each step of your answer, the measure of
angle RQT
at
hs
.
(i)
w
w
w
.fa
sp
as
s
m
SOLUTION:
Data: Diagram showing a quadrilateral PTRS such that QT = QR = QP
and angle QRT = 76° .
Required to find:
ˆ
RQT
Solution:
Angle RTQ = 76°
(The base angles of the isosceles triangle RQT are equal)
\ Angle RQT = 180° - ( 76° + 76° )
= 28° (The sum of interior angles in a triangle = 180° )
(ii)
angle PRT
SOLUTION:
Required to find: angle PRT
Solution:
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Pg 13 of 40
co
m
Angle QPR = Angle QRP
(The base angles of the isosceles triangle RQP are equal)
\ Angle QRP =
28°
= 14°
2
at
hs
.
Angle QPR + Angle QRP = 28°
(The exterior angle of a triangle is equal to the sum of the interior opposite angles)
angle SPT, given that angle SRT = 145° and angle PSR = 100° .
as
s
(iii)
m
Hence, angle PRT = 76° + 14° = 90°
w
w
w
.fa
sp
SOLUTION:
Data: Angle SRT = 145° and angle PSR = 100°
Required to find: angle SPT
Solution:
Angle SRP = 145° - 90°
= 55°
Angle SPR = 180° - (100° + 55°)
= 25°
(Sum of the interior angles in a triangle = 180° )
Hence, angle SPT = 25° + 14°
= 39°
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Pg 14 of 40
The diagram below shows a triangle ABC and its image, A¢B¢C ¢ , under a single
transformation.
at
hs
.
co
m
(b)
Describe completely the transformation that maps DABC to DA¢B¢C ¢ .
m
(i)
w
w
w
.fa
sp
as
s
SOLUTION:
Data: Diagram showing triangle ABC and its image A¢B¢C ¢ after a single
transformation.
Required To Describe: The transformation that maps DABC unto
DA¢B¢C ¢ .
Solution:
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Pg 15 of 40
A¢B¢C ¢ is congruent to ABC and is re-oriented. Hence, the transformation
is a rotation. The perpendicular bisectors of BB¢ and AA¢ meet at O.
ˆ ¢ = 90° and the ‘sweep’ from OB
Hence, O is the center of rotation. BOB
to OB¢ is clockwise. The transformation is therefore a 90° clockwise
rotation about O.
æ 4ö
The translation vector T = ç ÷ maps DA¢B¢C ¢ to DA¢¢B¢¢C ¢¢ . On the
è -5 ø
diagram above, drawn the DA¢¢B¢¢C ¢¢ .
co
m
(ii)
SOLUTION:
w
w
w
.fa
sp
as
s
m
at
hs
.
æ 4ö
Data: DA¢B¢C ¢ is mapped to DA¢¢B¢¢C ¢¢ by T = ç ÷ .
è -5 ø
Required To Draw: DA¢¢B¢¢C ¢¢
Solution:
See graph, 𝐴7 = (1, 4), 𝐵7 = (1, 1) and 𝐶 7 = (3, 1)
æ 4ö
T =ç ÷
è -5 ø
DA¢B¢C ¢ ¾¾¾¾
®DA¢¢B¢¢C ¢¢
æ 4ö
ç ÷
è -5 ø
A¢ (1, 4 ) ¾¾¾
® A¢¢ ( 5, - 1)
æ 4ö
ç ÷
è -5 ø
B ¢ (1, 1) ¾¾¾
® B ¢¢ ( 5, - 4 )
æ 4ö
ç ÷
è -5 ø
C ¢ ( 3, 1) ¾¾¾
® C ¢¢ ( 7, - 4 )
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Pg 16 of 40
22
.
7
The diagram below, not drawn to scale, shows a field in the shape of a sector of
a circle with center O and diameter 28 m. Angle POZ is 90°.
In this problem take π to be
co
m
6. (a)
(i)
at
hs
.
Calculate:
the area of the field
w
.fa
sp
as
s
m
SOLUTION:
Data: Diagram showing a field POZ in the shape of a sector of a circle
with center O and diameter 20 m. Angle POZ is 90°.
Required to calculate: The area of the field
Calculation:
Diameter 28 m
=
= 14 m
Radius =
2
2
90°
2
Area of the field =
´ p (14 )
360°
1 22
= ´ ´ 14 ´ 14
4 7
= 154 m 2
w
w
(ii)
the perimeter of the field
SOLUTION:
Required to calculate: The perimeter of the field
Calculation:
The perimeter of the field
= Length of radius OP + Length of arc PZ + Length of radius ZO
90°
= 14 +
( 2 ´ p ´14 ) + 14
360°
22
æ1
ö
= 14 + ç ´ 2 ´ ´14 ÷ + 14
7
è4
ø
= 50 m
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Pg 17 of 40
The diagram below, not drawn to scale, shows a triangular prism ABCDEF. The
cross-section is the right-angled triangle, ABC, where AB = 6 cm and BC = 10 cm.
at
hs
.
co
m
(b)
Calculate
the area of the triangle ABC
m
(i)
w
w
w
.fa
sp
as
s
SOLUTION:
Data: Diagram showing the triangular prism ABCDEF which crosssection is a right-angled triangle. AB = 6 cm and BC = 10 cm.
Required to calculate: The area of triangle ABC
Calculation:
BC 2 = AC 2 + AB 2
(10 )
2
= AC 2 + ( 6 )
(Pythagoras' Theorem)
2
6´8
cm 2
2
48
=
cm 2
2
= 24 cm 2
Area of DABC =
AC 2 = 100 - 36
AC 2 = 64
AC = 64
= 8 cm (taking the positive value of the root)
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Pg 18 of 40
(ii)
the length of the prism, if the volume is 540 cm3
\ 540 = 24 ´ Length of prism
540
Length of prism =
24
= 22.5 cm
(iii)
the surface area of the prism.
co
m
SOLUTION:
Data: The volume of the prism is 540 cm3.
Required to calculate: The length of the prism.
Calculation:
Volume of prism = Area of cross-section ´ Length of prism
at
hs
.
SOLUTION:
Required to calculate: The surface area of the prism.
Calculation:
Surface area of the prism = Sum of the area of all five sides
6´8
= 24 cm2
2
6´8
Area of DDEF =
= 24 cm2
2
Area of rectangle ABED = 6 ´ 22.5 = 135 cm2
Area of rectangle ADFE = 8 ´ 22.5 = 180 cm2
Area of rectangle BEFC = 10 ´ 22.5 = 225 cm2
Hence, the surface area of the prism = 24 + 24 + 135 + 180 + 225
= 588 cm2
w
.fa
sp
as
s
m
Area of DABC =
w
w
7. The table below shows the speeds, to the nearest kmh-1, of 90 vehicles that pass a
checkpoint.
Speed (in kmh-1)
0 – 19
20 – 39
40 – 59
60 – 79
80 – 99
100 – 119
Frequency
5
11
26
37
9
2
Cumulative Frequency
5
16
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Pg 19 of 40
(a)
For the class interval 20 – 39, as written in the table above, complete the
following sentences.
(i)
The upper class limit is …………………………………………………….
(ii)
The class width is …………………………………………………………..
(iii)
Sixteen vehicles passed a checkpoint at no more than ……………....kmh-1
co
m
SOLUTION:
Data: Cumulative frequency table showing the speeds, in kmh-1, of 90 vehicles
passing a certain checkpoint.
The class width is 39.5 - 19.5 = 20 .
(iii)
Sixteen vehicles passed a checkpoint at no more than 39.5 kmh-1.
m
(ii)
Complete the table shown above by inserting the missing values for the
cumulative frequency column.
as
s
(b)
at
hs
.
Required to complete: The sentences given for the class interval 20 – 39
Solution:
(See the modified table done)
(i)
The upper class limit is 39.
Class Boundaries
L.C.B
U.C.B.
Frequency
Cumulative
Frequency
w
w
w
Speed, x,
(in kmh-1)
L.C.L U.C.L
.fa
sp
SOLUTION:
Required to complete: The cumulative frequency give.
Solution:
L.C.L-lower class limit, L.C.B-lower class boundary
U.C.L-upper class limit, U.C.B-upper class boundary
Points to Plot
(U.C.B., CF)
( 0, 0)
0 – 19
0 £ x < 19.5
5
5
20 – 39
19.5 £ x < 39.5
11
16
40 – 59
39.5 £ x < 59.5
26
26 + 16 = 42
60 – 79
59.5 £ x < 79.5
37
37 + 42 = 79
80 – 99
79.5 £ x < 99.5
9
9 + 79 = 88
100 – 119
99.5 £ x < 119.5
2
2 + 88 = 90
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(19.5, 5)
( 39.5, 16)
(59.5, 42)
( 79.5, 79)
( 99.5, 88)
(119.5, 90)
Pg 20 of 40
(c)
On the grid provided, using a scale of 2 cm to represent 20 kmh-1 on the x – axis
and 2 cm to represent 10 vehicles on the y – axis, draw the cumulative frequency
curve to represent the information in the table.
w
w
w
.fa
sp
as
s
m
at
hs
.
co
m
SOLUTION:
Required to draw: The cumulative frequency curve to illustrate the information
given
Solution:
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Pg 21 of 40
(d)
(i)
On your graph, draw reference lines to estimate the speed at which no
more than 50% of the vehicles drove as they passed the check point.
w
w
w
.fa
sp
as
s
m
at
hs
.
co
m
SOLUTION:
Required to show: The speed at which no more than 50% of the vehicles
drove past the checkpoint using reference lines
Solution:
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Pg 22 of 40
(ii)
What is the estimated speed?
SOLUTION:
Required to state: The estimated speed at which no more than 50% of the
vehicles passed the checkpoint
Solution:
The estimated speed is 62 kmh-1
Draw Figure 5 of the sequence in the space below.
as
s
(a)
m
at
hs
.
co
m
8. The first four figures in a sequence are shown below. Figure 1 is a single black dot, while
each of the others consist of black dots arranged in an equilateral manner.
w
w
w
.fa
Solution:
sp
SOLUTION:
Data: Figures showing a sequence of black dots arranged in an equilateral
manner.
Required to draw: The fifth figure in the sequence
(b)
How many dots would be in Figure 6?
SOLUTION:
Required to state: The number of dots in Figure 6
Solution:
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Pg 23 of 40
Figure
1
2
3
4
5
Number of Dots
1
3
6
10
15
co
m
By observation, the number of dots in the figure appear to be,
½ × (the number of the figure) × (1 added to the number of the figure)
1
\ Figure 6 should have 6 ( 6 + 1) = 21 dots
2
at
hs
.
The table below refers to the figures and the number of dots in each figure. Study the
patterns below.
Number of Dots, d, in terms n
Number of Dots Used, d
1
1
´1´ (1 + 1)
2
1
2
1
´ 2 ´ ( 2 + 1)
2
3
1
´ 3 ´ ( 3 + 1)
2
6
as
s
m
Figure, n
sp
3
.fa
11
Complete the row which corresponds to Figure 11 in the table above.
w
w
(c)
w
n
SOLUTION:
Data: Table showing the pattern of the number of dots used in the sequence of
figures.
Required to complete: The row in the table that corresponds to Figure 11
Solution:
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Pg 24 of 40
Number of Dots, d, in terms n
Number of Dots Used, d
1
1
´1´ (1 + 1)
2
1
2
1
´ 2 ´ ( 2 + 1)
2
3
3
1
´ 3 ´ ( 3 + 1)
2
6
11
1
´11´ (11 + 1)
2
66
at
hs
.
n
(d)
co
m
Figure, n
Determine which figure in the sequence has 210 dots.
sp
as
s
m
SOLUTION:
Required to find: The figure that has 210 dots in the sequence
Solution:
1
n ( n + 1) = 210
2
n ( n + 1) = 420
= 20 ´ ( 20 + 1)
.fa
\ n = 20
Write a simplified algebraic expression for the number of dots, d, in the Figure n.
w
(e)
w
\ Figure 20 has 210 dots.
w
SOLUTION:
Required to state: An algebraic expression for the number of dots, d, in Figure n
Solution:
1
1
´ n ´ ( n + 1) = n ( n + 1)
2
2
The completed table looks like:
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Pg 25 of 40
Number of Dots Used, d
1
1
´1´ (1 + 1)
2
1
2
1
´ 2 ´ ( 2 + 1)
2
3
3
1
´ 3 ´ ( 3 + 1)
2
6
11
1
´11´ (11 + 1)
2
n
1
´ n ´ ( n + 1)
2
co
m
Number of Dots, d, in terms n
at
hs
.
66
1
n ( n + 1)
2
Show that there is no diagram that has exactly 1 000 dots.
m
(f)
Figure, n
sp
as
s
SOLUTION:
Required To Show: No diagram in the sequence has 1 000 dots.
Solution:
1
Let n ( n + 1) = 1000
2
.fa
n ( n + 1) = 2000
w
w
w
There are no two consecutive integers, n and n +1 being consecutive integers,
whose product is exactly 2 000. Therefore, no diagram has 1 000 dots.
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Pg 26 of 40
SECTION II
Answer TWO questions in this section.
ALGEBRA AND RELATIONS, FUNCTIONS AND GRAPHS
The velocity – time graph below shows the motion of a cyclist over a period of 40
seconds.
m
at
hs
.
co
m
9. (a)
Calculate the gradient of
as
s
(i)
OA
sp
a)
w
w
w
.fa
SOLUTION:
Data: A velocity – time graph showing the motion of a cyclist for
a 40 second period.
Required to find: The gradient of OA.
Solution:
O = ( 0, 0) and A = ( 25, 10)
b)
Gradient of OA =
10 - 0 2
=
25 - 0 5
AB
SOLUTION:
Required to find: The gradient of AB.
Solution:
A = ( 25, 10) and B = ( 40, 10 )
Gradient of AB =
10 - 10
= 0 (as expected for a horizontal line)
40 - 25
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Pg 27 of 40
(ii)
Complete the following statements.
The cyclist started from rest, where his velocity was ……………….. ms-1,
and steadily increased his velocity by ……………….. ms-1 each second
during the first 25 seconds. During the next 15 seconds, his velocity
remained constant, that is, his acceleration was …………….. ms-2.
(iii)
at
hs
.
co
m
SOLUTION:
Required to complete: The sentences given.
Solution:
The cyclist started from rest, where his velocity was 0 ms-1, and steadily
2
increased his velocity by ms-1 each second during the first 25 seconds.
5
During the next 15 seconds, his velocity remained constant, that is, his
acceleration was 0 ms-2.
Determine the average speed of the cyclist over the 40-second period.
as
s
m
SOLUTION:
Required to find: Average speed of the cyclist over the 40 second period.
Solution:
Total distance covered
Average speed =
Total time taken
w
.fa
sp
Area under the graph = Distance covered
1
= {( 40 - 25) + ( 40 - 0 )} ´10
2
1
= (15 + 40 ) ´10
2
= 275 m
w
\ Average speed =
= 6.875 ms-1 or 6
w
(b)
275 m
40 s
7
ms -1
8
Consider the following pair of simultaneous equations:
x 2 + 2 xy = 5
x+ y =3
(i)
WITHOUT solving, show that (1, 2 ) is a solution for the pair of
simultaneous equations.
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Pg 28 of 40
SOLUTION:
Data: The pair of simultaneous equations x 2 + 2 xy = 5 and x + y = 3 .
Required to show: (1, 2 ) is a solution for the pair of simultaneous
equations without solving
Solution:
Substitute x = 1 and y = 2 into both equations to test for ‘truth or falsity’
When x = 1 and y = 2
x 2 + 2 xy = 5
2
+ 2 (1)( 2 ) = 5
co
m
(1)
at
hs
.
1+ 4 = 5
5 = 5 (True)
(ii)
as
s
m
When x = 1 and y = 2
x+y=3
1+2=3
3 = 3 (True)
x = 1 and y = 2 satisfy both equations
\(1, 2 ) is a solution for the pair of simultaneous equations.
Solve the pair of simultaneous equations above to determine the other
solution.
w
w
w
.fa
sp
SOLUTION:
Required to find: The other solution for the pair of simultaneous
equations.
Solution:
Let x 2 + 2 xy = 5
…Œ
x+ y =3
…
From equation 
y = 3- x
…Ž
Substitute equation Ž into equation Œ
x2 + 2 x (3 - x ) = 5
x2 + 6x - 2x2 = 5
- x2 + 6x - 5 = 0
´ -1
x2 - 6 x + 5 = 0
( x - 1)( x - 5) = 0
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Pg 29 of 40
x -1 = 0
x -5 = 0
x =1
x=5
We already know that x = 1 is part of one set of solution. We consider the
other.
When x = 5
y = 3- x
= 3-5
\ The other solution is ( 5, - 2 ) .
co
m
= -2
P, Q, R and S are four points on the circumference of the circle shown below.
Angle QRS = 58° .
w
w
.fa
sp
as
s
m
10. (a)
at
hs
.
MEASUREMENT, GEOMETRY AND TRIGONOMETRY
w
Using the geometrical properties of a circle to give reasons for each step of your
answer, determine the measure of
(i)
ÐSPQ
SOLUTION:
Data: Diagram of a circle where P, Q, R and S lie on its circumference
and angle QRS = 58° .
Required to calculate: ÐSPQ
Calculation:
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Pg 30 of 40
co
m
at
hs
.
ÐSPQ = 180° - 58°
= 122°
(The opposite angles of a cyclic quadrilateral are supplementary.)
ÐOQS
m
(ii)
w
w
w
.fa
sp
as
s
SOLUTION:
Required to calculate: ÐOQS
Calculation:
ÐSOQ = 2 ( 58° )
= 116°
(The angle subtended by a chord at the center of a circle is twice the angle
that the chord subtends at the circumference, standing on the same arc.)
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Pg 31 of 40
co
m
at
hs
.
ÐOSQ = ÐOQS (The base angles of an isosceles triangle are equal)
ÐOSQ + ÐOQS = 180° - 116°
1
( 64° )
2
= 32°
A ship leaves Port A and sails 52 km on a bearing of 044° to Port B. The ship
then changes course to sail to Port C, 72 km away, on a bearing of 105°.
sp
(b)
as
s
\ÐOQS =
m
= 64°
On the diagram below, not drawn to scale, label the known distances
travelled and the known angles.
w
w
w
.fa
(i)
SOLUTION:
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Pg 32 of 40
(ii)
at
hs
.
co
m
Data: A ship leaves Port A and sails on a bearing of 044° to Port B, 52 km
away. It then sails to Port C on a bearing of 105°, 72 km away.
Required to label: The sides and angles given on the diagram
Solution:
Determine the measure of ÐABC.
w
w
w
.fa
sp
as
s
m
SOLUTION:
Required to find: ÐABC
Solution: Let the line NB be extended to S, as shown.
ÐABS = 44° (Alternate angles are equal)
ÐCBS = 180° - 105°
= 75°
(two angles which make a straight line are supplementary)
\ÐABC = 44° + 75°
= 119°
(iii)
Calculate, to the nearest km, the distance AC.
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Pg 33 of 40
SOLUTION:
Required to calculate: AC
Calculation:
By the cosine rule:
AC 2 = ( 52 ) + ( 72 ) - 2 ( 52 )( 72 ) cos (119° )
2
2
= 2 704 + 5184 - ( -3630.25 )
= 11518.25
= 107.3 km
= 107 km (to the nearest km)
Show that the bearing of A from C, to the nearest degree, is 260°.
at
hs
.
(iv)
co
m
AC = 11518.25
.fa
sp
as
s
m
SOLUTION:
Required to show: The bearing of A from C is 260°
Solution:
w
w
w
ÐBCN = 75° (Co-interior angles are supplementary)
By the sine rule:
52
107.3
=
ˆ
sin ACB sin119°
ˆ = 52sin119°
sin ACB
107.3
= 0.4238
ˆ = sin -1 ( 0.4238 )
ACB
= 25.07°
The bearing of A from C = 360° - ( 75° + 25.07°) = 259.9>
= 260° (correct to the nearest degree
Q.E.D.
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Pg 34 of 40
VECTORS AND MATRICES
11. (a)
Matrices A and B are such that
æ3 2ö
æ 4 0ö
and
A=ç
B
=
÷
ç
÷.
è5 4ø
è 3 -1 ø
2´2 ´ 2´2 = 2´2
as
s
m
e12 ö
æe
= ç 11
÷
è e21 e22 ø
e11 = ( 3 ´ 4 ) + ( 2 ´ 3)
at
hs
.
SOLUTION:
æ3 2ö
æ 4 0ö
Data: A = ç
÷ and B = ç
÷
è5 4ø
è 3 -1 ø
Required to show: AB ¹ BA
Proof:
æ3 2öæ 4 0ö
A´ B = ç
֍
÷
è 5 4 ø è 3 -1 ø
co
m
Show by multiplying A and B, that AB ¹ BA .
(i)
= 12 + 6
sp
= 18
.fa
æ 4 0 öæ 3 2 ö
B´ A = ç
֍
÷
è 3 -1 øè 5 4 ø
2´2
´
2´2 = 2´2
w
w
w
e12 ö
æe
= ç 11
÷
è e21 e22 ø
e11 = ( 4 ´ 3) + ( 0 ´ 5 )
= 12 + 0
= 12
e11 of AB ¹ e11 of BA . There is no need for further working since,
regardless of whether the remaining three corresponding entries of AB are
equal or not to those of BA, all the entries of AB will never be equal to the
entries of BA. \ AB ¹ BA .
Q.E.D.
(ii)
Find A-1 , the inverse of A.
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Pg 35 of 40
SOLUTION:
Required to find: A-1
Solution:
det A = ( 3 ´ 4 ) - ( 2 ´ 5 )
(iii)
at
hs
.
1 æ 4 - ( 2) ö
A-1 = ç
÷
3 ø
2 è - ( 5)
æ 2 -1 ö
÷
=ç 5
3÷
çè 2
2ø
co
m
= 12 - 10
=2
Write down the 2 ´ 2 matrix representing the matrix product AA-1.
2´2 ´ 2´2
as
s
m
SOLUTION:
Required to write: The 2 ´ 2 matrix AA-1.
Solution:
A ´ A-1 = I
= 2´ 2
(i)
Write the following pair of simultaneous equations as a matrix equation.
w
w
w
.fa
(b)
sp
æ1 0ö
I =ç
÷
è0 1ø
(ii)
3x + 2 y = 1
5x + 4 y = 5
SOLUTION:
Data: 3x + 2 y = 1 and 5 x + 4 y = 5
Required to write: The given pair of simultaneous equations as a matrix
equation
Solution:
æ 3 2 öæ x ö æ 1 ö
ç
÷ç ÷ = ç ÷
è 5 4 øè y ø è 5 ø
Write the solution of your matrix equation in (b) (i) as a product of two
matrices.
SOLUTION:
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Pg 36 of 40
Required to write: The solution of the matrix equation as a product of
two matrices
Solution:
æ 3 2 öæ x ö æ 1 ö
ç
÷ç ÷ = ç ÷
è 5 4 øè y ø è 5 ø
æ x ö æ1ö
Aç ÷ = ç ÷
è y ø è 5ø
´ A-1
-1 ö
÷æ1ö
3 ÷ ç ÷ as a product of two matrices.
è5ø
2ø
at
hs
.
æ 2
æ xö ç
ç ÷=ç 5
è yø è 2
co
m
æ xö
æ1ö
A ´ A-1 ç ÷ = A-1 ç ÷
è yø
è 5ø
æ xö
æ1ö
I ç ÷ = A-1 ç ÷
è yø
è 5ø
sp
æ 4ö
The position vectors of the points P and Q relative to an origin O, are OP = ç ÷
è 3ø
æ5ö
and OQ = ç ÷ respectively.
è0ø
The diagram below shows that PR = 3 OP and QS = 3 OQ .
w
w
w
.fa
(c)
as
s
m
If calculated, (which was not required according to the question) the right
æ -3 ö
hand side would give ç ÷ and equating corresponding entries gives
è 5ø
x = -3 and y = 5 .
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Pg 37 of 40
æ xö
Express in the form ç ÷ , vector
è yø
(i)
•
OS
m
æ5ö
QS = 3 ç ÷
è0ø
æ15 ö
=ç ÷
è 0ø
at
hs
.
co
m
SOLUTION:
Data: Diagram showing position of vectors of the points P and Q,
æ 4ö
æ5ö
where OP = ç ÷ and OQ = ç ÷ and PR = 3 OP and QS = 3 OQ .
è 3ø
è0ø
æ xö
Required to express: OS in the form ç ÷ .
è yø
Solution:
æ5ö
OQ = ç ÷
è0ø
as
s
OS = OQ + QS
.fa
sp
æ 5 ö æ15 ö
= ç ÷+ç ÷
è0ø è 0ø
æ 20 ö
æ xö
= ç ÷ and is of the form ç ÷ , where x = 20 and y = 0.
è 0ø
è yø
w
w
w
•
PQ
SOLUTION:
æ xö
Required to express: PQ in the form ç ÷ .
è yø
Solution:
PQ = PO + OQ
æ 4ö æ5ö
= -ç ÷ + ç ÷
è 3ø è0ø
æ 1ö
= ç ÷ and is of the form
è -3 ø
æ xö
ç ÷ , where x = 1 and y = -3.
è yø
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Pg 38 of 40
•
RS
SOLUTION:
æ xö
Required to express: RS in the form ç ÷ .
è yø
Solution:
PR = 3OP
m
æ 4 ö æ12 ö
= ç ÷+ç ÷
è 3ø è 9ø
æ16 ö
=ç ÷
è12 ø
at
hs
.
OR = OP + PR
co
m
æ 4ö
= 3ç ÷
è 3ø
æ 12 ö
=ç ÷
è 9ø
as
s
RS = RO + OS
æ xö
ç ÷ , where x = 4 and y = -12.
è yø
w
.fa
sp
æ16 ö æ 20 ö
= -ç ÷ + ç ÷
è12 ø è 0 ø
æ 4ö
=ç
÷ and is of the form
è -12 ø
w
w
(ii)
State TWO geometrical relationships between PQ and RS.
SOLUTION:
Required to state: Two geometrical relationships between PQ and RS.
Solution:
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Pg 39 of 40
æ 1ö
PQ = ç ÷
è -3 ø
æ 4ö
RS = ç
÷
è -12 ø
æ 1ö
= 4ç ÷
è -3 ø
co
m
= 4 PQ
Hence, RS is a scalar multiple (which is 4) of PQ . Therefore, RS and
æ -1 ö
RS = 4 ç ÷
è 3ø
= 4 ´ PQ
RS = 4 PQ
at
hs
.
PQ are parallel.
w
w
w
.fa
sp
as
s
m
That is to say, the length of RS is 4 times the length of PQ .
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Pg 40 of 40