solutions to csec maths p2 january 2011
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Solutions to CSEC Maths P2 January 2011
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Question 1a part (i)
Calculate the exact value of
(5.82 + 1.02) Γ 2.5
= 34.66 Γ 2.5
= 86.65 to 2 decimal places
Question 1a part (ii)
Calculate the exact value of
πππ
π ππ
β π
π
=
229
143
β 3
7
= (22
9 Γ
3
14) β
3
7
= (66
126) β
3
7
=11
21β
3
7
=11 β 3(3)
21
=π
ππ
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Question 1b part (i)
Data Given
Employee is paid a basic wage of $9.50 per hour for a 40 hour week and one and a half times
this rate for overtime.
Calculate the basic weekly wage of one employee for a 40-hour work week
π΅ππ ππ π€πππππ¦ π€πππ = βππ’πππ¦ πππ‘π Γ
ππ’ππππ ππ βππ’ππ ππ π πππ ππ π€πππ π€πππ
= $9.50 Γ 40
= $380 πππ π€πππ
Question 1b part (ii)
Calculate the overtime wage for an employee who worked 6 hours overtime
ππ£πππ‘πππ π ππ‘π = 1.5 Γ $9.50
= $14.25 πππ ππ£πππ‘πππ βππ’π
ππ£πππ‘πππ π€πππ πππ 6 βππ’ππ = $14.25 Γ 6
= $85.50
Question 1b part (iii)
Data Given
A total of $12 084.00 in basic and overtime wages were paid to 30 employees by the company in
a certain week.
Calculate the total paid in overtime wages
π΅ππ ππ π€πππ ππ 30 ππππππ¦πππ πππ π 40 βππ’π π€πππ π€πππ
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= ππ’ππππ ππ ππππππ¦πππ Γ πππ ππ π€πππππ¦ π€πππ
= 30 Γ $380
= $11, 400
πππ‘ππ π€ππππ ππππ = $12,084 (πΊππ£ππ π·ππ‘π)
πππ‘ππ ππππ ππ ππ£πππ‘πππ π€ππππ
= πππ‘ππ π€ππππ ππππ β π΅ππ ππ π€πππ ππ 30 ππππππ¦πππ πππ π 40 βππ’π π€πππ π€πππ
= $12,084 β $11, 400
= $12,084 β $11, 400
= $684.00
Question 1b part (iv)
Calculate the total number of overtime hours worked by employees
πππ‘ππ ππ’ππππ ππ ππ£πππ‘πππ βππ’ππ π€πππππ ππ¦ ππππππ¦πππ = πππ‘ππ ππππ ππ ππ£πππ‘πππ π€ππππ
ππ£πππ‘πππ π ππ‘π
= $684.00
$14.25 πππ βππ’π
= 48 βππ’ππ
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Question 2a
Simplify 2π₯
5β
π₯
3
2π₯
5β
π₯
3
= 3(2π₯) β 5(π₯)
15
= 6π₯ β 5π₯
15
= π₯
15
Question 2b
Factorise π2π + 2ππ
π2π + 2ππ
= ππ(π + 2)
Question 2c
Express p as the subject of the formula π =π2βπ
π‘
π
1 =
π2βπ
π‘
ππ‘ = (π2 β π) (1)
π2 = ππ‘ + π
π = βππ‘ + π
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Question 2d part (i)
Data Given
Type of Box Number of Donuts per Box
Small box x
Large Box 2x + 3
In total 8 small boxes and 5 large boxes were sold.
Write an expression in terms of x to represent the Total number of donuts sold.
πππ‘ππ ππ’ππππ ππ ππππ’π‘π π πππ = ππ’ππππ ππ π ππππ πππ₯ππ π πππ + ππ’ππππ ππ πππ πππ₯ππ π πππ
= 8(π₯) + 5(2π₯ + 3)
= 8π₯ + 10π₯ + 15
= 18π₯ + 15
Question 2d part (ii)
Data Given
The total number of donuts sold was 195.
(a) Calculate the number of donuts in a small box
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πππ‘ππ ππ’ππππ ππ ππππ’π‘π π πππ = 18π₯ + 15
195 = 18π₯ + 15
18π₯ = 195 β 15
18π₯ = 180
π₯ = 10
π΅ππ ππ ππ πππ£ππ πππ‘π: ππ’ππππ ππ ππππ’π‘π ππ π ππππ πππ₯ = π₯
β΄ ππ’ππππ ππ ππππ’π‘π ππ π ππππ πππ₯ = 10
(b) Calculate the number of donuts in a large box
π΅ππ ππ ππ πππ£ππ πππ‘π: ππ’ππππ ππ ππππ’π‘π ππ πππππ πππ₯ = 2π₯ + 3
β΄ ππ’ππππ ππ ππππ’π‘π ππ πππππ πππ₯ = 2(10) + 3
= 23
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Simplify 7 π5π3 Γ 2 π2π
= 7 Γ 2 Γ π5+2 Γ π3+1
= 14 π7π4
Data Given
One carton of milk measures 6 cm by 4 cm by 10 cm.
Calculate, in cm3, the volume of milk in a carton
ππππ’ππ ππ ππππ ππ π ππππ‘ππ = πΏ Γ π Γ π»
= 6 Γ 4 Γ 10
= 240 ππ3
Data Given
For a recipe to make ice-cream 3 litres of milk are required.
Calculate the number of cartons of milk that should be bought to make the ice-cream
1 πππ‘ππ = 1000 ππ3
3 πππ‘πππ ππ ππππ = 3 Γ 1000
Question 3a
Question 3b part (i)
Question 3b part (ii)
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= 3000 ππ3 ππ ππππ
ππ’ππππ ππ ππππ‘πππ ππ ππππ ππππ’ππππ = ππππ’ππ ππ ππππ ππππ’ππππ
ππππ’ππ ππ π ππππ‘ππ
= 3000 ππ3
240 ππ3
= 12.5
Therefore 13 cartons of milk should be purchased make the ice β cream according to its recipe.
Data Given
One carton of milk is poured into a cylindrical cup of internal diameter 5 cm.
π = 3.14
Calculate the height of milk in the cup
ππππ’ππ ππ ππππ ππ πππ ππππ‘ππ = ππππ’ππ ππ ππππ ππ π‘βπ ππ’π
ππππ’ππ ππ ππππ ππ π‘βπ ππ’π = ππ2β
Where π = πππππ’π ππ π‘βπ ππ’π
β = βπππβπ‘ ππ π‘βπ ππππ ππ π‘βπ ππ’π
π = πππ‘πππππ ππππππ‘ππ ππ π‘βπ ππ’π
2
= 5 ππ
2
= 2.5 ππ
Question 3b part (iii)
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ππππ’ππ ππ ππππ ππ π‘βπ ππ’π = ππ2β
240 ππ3 = (3.14) Γ (2.5 ππ)2 Γ β
β =240
(3.14) Γ (2.5 )2
β = 12.23 ππ
π = ππ. π ππ ππ π πππππππππππ πππππππ
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Question 4a part (i)
Data Given
π = {πβπππ ππ’πππππ ππππ 1 π‘π 12}
π» = {πππ ππ’πππππ πππ‘π€πππ 4 πππ 12}
π½ = {πππππ ππ’πππππ ππππ 1 π‘π 12}
List the members of π»
π» = {πππ ππ’πππππ πππ‘π€πππ 4 πππ 12}
= {5, 7, 9, 11}
Question 4a part (ii)
List the members of π½
π½ = { πππππ ππ’πππππ ππππ 1 π‘π 12}
= {2, 3, 5, 7, 11}
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Question 4a part (iii)
Draw a Venn diagram to represent π,π» and π½
Question 4b part (i)
Construct a triangle LMN with angle πΏππ = 60Β° ,ππ = 9 ππ πππ πΏπ = 7 ππ
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Question 4b part (ii)
Measure and state the size of ποΏ½ΜοΏ½πΏ
Using a protractor ποΏ½ΜοΏ½πΏ was found to be 48Β°.
Question 4b part (iii)
Show the point, K, such that KLMN is a parallelogram
Since the opposite sides of a parallelogram are equivalent in length;
ππΏ = ππΎ = 7 ππ
ππ = πΏπΎ = 9 ππ
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Question 5a (i)
Data Given
Equation of a straight line is 3π¦ = 2π₯ β 6
Determine the gradient of the line
3π¦ = 2π₯ β 6
In the form π¦ = ππ₯ + π, where the gradient of the line is given by βmβ.
π¦ =2
3π₯ β 2
πΊπππππππ‘ (π1) = 2
3
Question 5a (ii)
Determine the equation of the line perpendicular to 3π¦ = 2π₯ β 6 that passes through the point
(4, 7)
πΊπππππππ‘ ππ πΏπππ = π2
π1 Γ π2 = β1 ; π ππππ π‘βπ πππππ πππ ππππππππππ’πππ
π2 = β1
π1
= β1
23
= β3
2
πΈππ’ππ‘πππ ππ πΏπππ: (π¦ β π¦1) = π (π₯ β π₯1)
(π¦ β 7) = β3
2 (π₯ β 4)
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(π¦ β 7) = β3
2 π₯ +
12
2
2π¦ β 14 = β3 π₯ + 12
2π¦ = β3 π₯ + 26
Question 5b part (i)
Data Given: Diagram showing π: π₯ β π₯2 β π where π₯ π {3, 4, 5, 6, 7, 8, 9, 10}
Calculate the value of π
Using the data given where 4 is mapped onto 11.
π₯2 β π = 11
π = π₯2 β 11
π = (4)2 β 11
π = 16 β 11
π = 5
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Question 5b part (ii)
Calculate the value of π(3)
π(π₯) = π₯2 β π
Substituting π₯ = 3 and π = 5 gives
π(3) = (3)2 β 5
π(3) = 9 β 5
= 4
Question 5b part (iii)
Calculate the value of π₯ when π(π₯) = 95
π(π₯) = π₯2 β π
π₯2 β π = 95
π₯2 = 95 + π
π₯2 = 95 + 5
π₯2 = 100
π₯ = β100
π₯ = Β±10
Based on given data the value of x would be 10 when f(x) = 95.
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Question 6 part (i)
Data Given: Line graph showing the monthly sales, in thousands of dollars, at a school cafeteria
for the period January to May 2010.
Copy and complete the table
Month Jan Feb Mar Apr May
Sales in
$Thousands
38 35 27 15 10
Question 6 (ii)
Find the two months in between which there was the greatest decrease in sales
π·ππππππ π πππ‘π€πππ π½ππ πππ πΉππ = $38,000 β $35,000
= $3,000
π·ππππππ π πππ‘π€πππ πΉππ πππ πππ = $35,000 β $27,000
= $8,000
π·ππππππ π πππ‘π€πππ πππ πππ π΄ππ = $27,000 β $15,000
= $12,000
π·ππππππ π πππ‘π€πππ π΄ππ πππ πππ¦ = $15,000 β $10,000
= $5,000
β΄ πβπ ππππππ π‘ πππππππ π ππππ’πππ πππ‘π€πππ π‘βπ ππππ‘βπ ππ ππππβ πππ π΄ππππ.
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Question 6 part (iii)
Calculate the mean monthly sales for the period January to May 2011
πππ‘ππ ππππ‘βππ¦ π ππππ ππππ Jan to May = $ (38,000 + 35,000 + 27,000 + 15,000 + 10,000)
= $125,000
ππππ ππππ‘βππ¦ π ππππ ππππ Jan to May = πππ‘ππ ππππ‘βππ¦ π ππππ ππππ Jan to May
ππ’ππππ ππ ππππ‘βπ
= $125,000
5
= $25,000
Question 6 part(iv)
Data Given: Total sales for the period January to June was $150,000.
Calculate the sales for the month of June
πππππ πππ π‘βπ ππππ‘β ππ π½π’ππ = πππ‘ππ π ππππ πππ π½ππ π‘π π½π’ππ β πππ‘ππ π ππππ ππππ π½ππ π‘π πππ¦
= $ (150,000 β 125,000)
= $25,000
Question 6 part(v)
Comment on the sales in June compared to the sales for the months January to May
The sales showed an increase when compared to the two previous months; April and May.
The sales were equivalent to the mean monthly sales for the period January to May 2010.
The month of June sales was the only month that displayed an increase since January.
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Question 7 (i)
Data Given: Diagram showing the triangle RST and its transformation RβSβTβ after undergoing a
transformation.
Write the coordinates of R and Rβ
πΆππππππππ‘ππ ππ π = (2, 4)
πΆππππππππ‘ππ ππ π β² = (2, 0)
Question 7 (ii)
Describe the transformation which maps triangle RST onto triangle R'S'T'
If the line y = 2 is drawn, it can be seen the vertices of RST and RβSβTβ are located the same
distance away from the line on opposite sides.
RST and RβSβTβ are also identical and RβSβTβ is a lateral inversion of RST.
Therefore the triangle RST was reflected along the line y = 2 to give the triangle RβSβTβ.
Question 7 (iii) part (a)
Data Given: RST undergoes an enlargement, centre (0, 4), scale factor, 3.
Draw triangle RββSββTββ, the image of triangle RST under the enlargement
For the enlargement; centre C = (0, 4)
πΆππππππππ‘ππ ππ π β²β² = 3 Γ πΆπ
= [(3 Γ 2), (3 Γ 0 + 4)]
= (6, 4)
πΆππππππππ‘ππ ππ πβ²β² = 3 Γ πΆπ
= [(3 Γ 3), (3(7 β 4) + 4)]
= (9, 13)
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πΆππππππππ‘ππ ππ πβ²β² = 3 Γ πΆπ
= [(3 Γ 5), (3(1) + 4)]
= (15, 7)
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Question 7 (iii) part b
Data Given: Area of triangle RST = 4 square units
Calculate the area of triangle RββSββTββ
π΄πππ ππ π‘πππππππ π ββπββπββ = (π ππππ ππππ‘ππ )2 Γ ππππ ππ π‘πππππππ π ππ
= (3 )2 Γ 4
= 36 π ππ’πππ π’πππ‘π
Question 7 (iii) part c
State two geometrical relationships between triangles RST and RββSββTββ
Triangles RST and RββSββTββ similar where RST is the object and RββSββTββ is an enlargement of the
object.
Regarding the sides of RST and RββSββTββ:
π β²β²πβ²β²
π π=
πβ²β²πβ²β²
ππ=
π β²β²πβ²β²
π π= 3 which is the scale factor
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Question 8a part (i)
Data Given: An answer sheet showing a rectangle, A, of area 20 square units and perimeter 24
units.
Draw and label
(a) rectangle B of area 27 square units and perimeter 24 units
π΄πππ ππ ππππ‘πππππ = π Γ π€ = 27 π ππ’πππ π’πππ‘π
πππππππ‘ππ ππ ππππ‘πππππ = 2π + 2π€ = 24 π’πππ‘π
π Γ π€ = 27 β¦β¦β¦β¦β¦β¦. Equation 1
2π + 2π€ = 24 .β¦β¦β¦β¦.. Equation 2
From Equation 2
π = 12 β π€β¦β¦β¦β¦β¦β¦Equation 3
Substituting Equation 3 into Equation 1
(12 β π€)π€ = 27
π€2 β 12π€ + 27 = 0
(π€ β 9)(π€ β 3) = 0
β΄ π€ = 9 ππ π€ = 3
Taking w=3;
π = 12 β π€
π = 12 β 3 = 9
Therefore for Rectangle B; length = 9 units and width = 3 units
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(b) rectangle C of area 32 square units and perimeter 24 units
π΄πππ ππ ππππ‘πππππ = π Γ π€ = 32 π ππ’πππ π’πππ‘π
πππππππ‘ππ ππ ππππ‘πππππ = 2π + 2π€ = 24 π’πππ‘π
π Γ π€ = 32 β¦β¦β¦β¦β¦β¦. Equation 1
2π + 2π€ = 24 .β¦β¦β¦β¦.. Equation 2
From Equation 2
π = 12 β π€β¦β¦β¦β¦β¦β¦Equation 3
Substituting Equation 3 into Equation 1
(12 β π€)π€ = 32
π€2 β 12π€ + 32 = 0
(π€ β 8)(π€ β 4) = 0
β΄ π€ = 8 ππ π€ = 4
Taking w=4;
π = 12 β π€
π = 12 β 4 = 8
Therefore for Rectangle B; length = 8 units and width = 4 units
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Question 8a part (ii)
Complete the table to show the dimensions of Rectangles B and C
Rectangle Length Width Area
(square units)
Perimeter
(units)
A 10 2 20 24
B 9 3 27 24
C 8 4 32 24
D
E
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Data Given
Rectangle D has a perimeter of 24 cm, with length, π, and width, π€.
Determine the values of π and π€ for the area of the rectangle D to be as large as possible
π΄πππ ππ ππππ‘πππππ, π΄ = π Γ π€
πππππππ‘ππ ππ ππππ‘πππππ = 2π + 2π€ = 24 π’πππ‘π
π΄ = π Γ π€ β¦β¦β¦β¦β¦β¦. Equation 1
2π + 2π€ = 24 .β¦β¦β¦β¦.. Equation 2
From Equation 2
π = 12 β π€β¦β¦β¦β¦β¦β¦Equation 3
Substituting Equation 3 into Equation 1
π΄ = (12 β π€) Γ π€
π΄ = 12π€ β π€2
At Maximum
ππ΄
ππ€= 0
12 β 2π€ = 0
2π€ = 12
π€ = 6
When w = 6;
π = 12 β π€
= 12 β 6
Question 8b
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= 6
Since the maximum area occurs at l = 6 and w = 6 rectangle D is actually a square.
πππ₯πππ’π π΄πππ = π Γ π
= 6 Γ 6
= 36 π ππ’πππ π’πππ‘π
Question 8c
Data Given
Rectangle E has a perimeter of 36 cm, with length, π, and width, π€.
Determine the values of π and π€ for the area of the rectangle E to be as large as possible For rectangle E to have a maximum value then it is also a square. Therefore π = π€ = π
πππππππ‘ππ ππ ππππ‘πππππ = 4π€ = 36 π’πππ‘π
π€ = 9 π’πππ‘π
π = 9 π’πππ‘π
πππ₯πππ’π π΄πππ = π Γ π
= 9 Γ 9
= 81 π ππ’πππ π’πππ‘π
Rectangle Length Width Area
(square units)
Perimeter
(units)
A 10 2 20 24
B 9 3 27 24
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C 8 4 32 24
D 6 6 36 24
E 9 9 81 36
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Question 9a part (i)
Data Given
π(π₯) = 2π₯ β 7
π₯ πππ π(π₯) = βπ₯ + 3
Evaluate π(5)
π(π₯) = 2π₯ β 7
π₯
π(5) = 2(5) β 7
5
= 3
5
Question 9a part (ii)
(a) Write an expression in terms of x for πβ1(π₯)
πΏππ‘ π¦ = π(π₯)
π¦ =2π₯ β 7
π₯
ππ€ππ‘πβ π₯ πππ π¦
π₯ = 2π¦ β 7
π¦
π₯π¦ = 2π¦ β 7
2π¦ β π₯π¦ = 7
π¦(2 β π₯) = 7
π¦ = 7
2 β π₯
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β΄ πβ1(π₯) = 7
2βπ₯, π₯ β 2
(b) Write an expression in terms of π₯ for ππ(π₯)
Replacing π₯ in π(π₯ ) with π(π₯)
ππ(π₯) = β2π₯ β 7
π₯+ 3
= β2π₯ β 7
π₯+
3
π₯
= β2π₯ β 7 + 3π₯
π₯
= β5π₯ β 7
π₯
= β5 β7
π₯ , π₯ β 0
Question 9b part (i)
Express the quadratic function 1 β 6π₯ β π₯2, in the form π β π(π₯ + β)2
1 β 6π₯ β π₯2 β‘ π β π(π₯ + β)2
1 β 6π₯ β π₯2 β‘ π β [π (π₯ + β)(π₯ + β)]
1 β 6π₯ β π₯2 β‘ π β [π (π₯2 + 2βπ₯ + β2)]
1 β 6π₯ β π₯2 β‘ π β ππ₯2 β 2πβπ₯ β πβ2
Equating coefficients
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π₯2; βπ = β1
π = 11
π₯; β2πβ = β6
β2(1)β = β6
β2β = β6
β = 3
ππππ π‘πππ‘; π β πβ2 = 1
π β (1)(3)2 = 1
π β 9 = 1
π = 10
β΄ 1 β 6π₯ β π₯2 β‘ 10 β (π₯ + 3)2 in the form π β π(π₯ + β)2 where π = 10, π = 1 πππ β = 3.
Question 9b part (ii)
(a) State the maximum value of 1 β 6π₯ β π₯2
1 β 6π₯ β π₯2 β‘ 10 β (π₯ + 3)2
πππ₯πππ’π π£πππ’π ππ 1 β 6π₯ β π₯2 = 10 β 0
= 10
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(b) State the equation of the axis of symmetry
πΏππ‘ π¦ = 1 β 6π₯ β π₯2
Rearranging gives
π¦ = βπ₯2 β 6π₯ + 1 π€βππβ ππ ππ π‘βπ ππππ π¦ = ππ₯2 + ππ₯ + π
πΈππ’ππ‘πππ ππ π‘βπ ππ₯ππ ππ π π¦ππππ‘ππ¦; π₯ = βπ
2π
= β(β6)
2(β1)
= 6
β2
= β3
πΈππ’ππ‘πππ ππ π‘βπ ππ₯ππ ππ π π¦ππππ‘ππ¦ ππ π₯ = β3
Question 9b part (iii)
Determine the roots of 1 β 6π₯ β π₯2
1 β 6π₯ β π₯2 = 0
π πππππ 1 β 6π₯ β π₯2 β‘ 10 β (π₯ + 3)2
10 β (π₯ + 3)2 = 0
(π₯ + 3)2 = 10
π₯ + 3 = Β±β10
π₯ = Β±β10 β 3
π₯ = +β10 β 3 ππ π₯ = ββ10 β 3
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π₯ = 3.162 β 3 ππ π₯ = β3.162 β 3
π₯ = 0.162 ππ π₯ = β6.162
π₯ = 0.16 ππ β 6.16 ππ π π ππππππ ππππππ
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Question 10a part (i)
Data Given
Calculate < ππΊπΉ
Triangle OGF is an isosceles triangle since the radius of a circle is constant;
πΏππππ‘β ππ ππΊ = πΏππππ‘β ππ ππΉ = πππππ’π ππ π‘βπ ππππππ
Therefore the base angles would be equivalent;
ποΏ½ΜοΏ½πΉ = ποΏ½ΜοΏ½πΊ
=180Β° β 118Β°
2 since the total sum of angles in a triangle = 180Β°
=62Β°
2
= 31Β°
Question 10a part (ii)
Calculate < π·πΈπΉ
SGH is a tangent to the circle so an angle of 90Β° is created where it comes into contact with the
circle.
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Based on the given diagram; ππΊπ = 90Β°
ππΊπ· = ππΊπ β 65Β°
ππΊπ· = 90Β° β 65Β°
= 25Β°
π·πΊπΉ = ππΊπ· + ππΊπΉ
π·πΊπΉ = 25Β° + 31Β°
= 56Β°
π·οΏ½ΜοΏ½πΉ = 180Β° β π·πΊπΉ
π·οΏ½ΜοΏ½πΉ = 180Β° β 56Β°
= 124Β°
Question 10b part (i)
Data Given
J, K and L are three sea ports.
A ship began its journey at L, sailed to K, then to L and returned to J.
The bearing of K from J is 54Β° and Z is due east of K.
π½πΎ = 122 ππ and πΎπΏ = 60 ππ
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Draw a clearly labelled diagram to represent the data given
Question 10b part (ii)
(a) Calculate the measure of angle JKL
π½οΏ½ΜοΏ½πΏ = 90Β° + 54Β°
= 144Β°
(b) Calculate the distance JL
Applying cosine rule π2 = π2 + π2 β 2ππ cosπ΄
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Where
π = π½πΏ, π = π½πΎ, π = πΎπΏ πππ π΄ = π½οΏ½ΜοΏ½πΏ
π2 = π2 + π2 β 2ππ cosπ΄
π½πΏ2 = π½πΎ2 + πΎπΏ2 β 2 (π½πΎ)(πΎπΏ) cos( π½οΏ½ΜοΏ½πΏ)
π½πΏ2 = (122)2 + (60)2 β 2 (122)(60) cos( 144Β°)
π½πΏ2 = 14884 + 3600 β (β11844)
π½πΏ2 = 30328
π½πΏ = β30328
π½πΏ = 174.149
Distance JL = 174. 15 to 2 decimal places
(c) Calculate the bearing of J from L
Applying sine rule to triangle JKL
π
sinπ΄=
π
sinπ΅
Where
π = π½πΎ, π΄ = πΎοΏ½ΜοΏ½π½, π = π½πΏ and π΅ = π½οΏ½ΜοΏ½πΏ
π
sinπ΄=
π
sinπ΅
π½πΎ
sin(πΎοΏ½ΜοΏ½π½)=
π½πΏ
sin( π½οΏ½ΜοΏ½πΏ )
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122
sin(πΎοΏ½ΜοΏ½π½)=
174.15
sin( 144Β° )
sin(πΎοΏ½ΜοΏ½π½) =122
174.15sin( 144Β° )
sin(πΎοΏ½ΜοΏ½π½) =122
296.282
πΎοΏ½ΜοΏ½π½ = sinβ1 (122
296.282)
πΎοΏ½ΜοΏ½π½ = 24.31Β°
π΅ππππππ ππ π½ ππππ πΏ = 270Β° β πΎοΏ½ΜοΏ½π½
= 270Β° β 24.31Β°
= 245.7Β°
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Question 11a part (i)
Data Given
Under a matrix transformation π = (π ππ π
), the points V and W were mapped onto Vβ and Wβ:
π (3, 5) β πβ²(5,β3)
π (7, 2) β πβ²(2,β7)
Determine the values of π, π , π πππ π
π Γ π = πβ²
(π ππ π
) Γ (35) = (
5β3
)
(3π + 5π3π + 5π
) = ( 5β3
)
Equating terms gives
3π + 5π = 5β¦β¦β¦β¦β¦β¦β¦β¦β¦ .πΈππ’ππ‘πππ 1
3π + 5π = β3β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 2
π Γ π = πβ²
(π ππ π
) Γ (72) = (
2β7
)
(7π + 2π7π + 2π
) = ( 2β7
)
Equating terms gives
7π + 2π = 2β¦β¦β¦β¦β¦β¦β¦β¦β¦ .πΈππ’ππ‘πππ 3
7π + 2π = β7β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 4
3π + 5π = 5β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 1
7π + 2π = 2β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 3
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Solving simultaneously
From Equation 1
π =5β5π
3
Substituting into Equation 3
7 (5 β 5π
3 ) + 2π = 2
35 β 35π
3+ 2π = 2
Multiplying throughout by 3
35 β 35π + 6π = 6
β29π = β29
π = 1
π =5 β 5π
3=
5 β 5(1)
3= 0
3π + 5π = β3β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 2
7π + 2π = β7β¦β¦ .β¦ . . . . β¦β¦ .β¦πΈππ’ππ‘πππ 4
Solving simultaneously
From Equation 2
π =β3β5π
3
Substituting into Equation 4
7 (β3 β 5π
3 ) + 2π = β7
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β21 β 35π
3+ 2π = β7
Multiplying throughout by 3
β21 β 35π + 6π = β21
β29π = 0
π = 0
π =β3 β 5π
3=
β3 β 5(0)
3= β1
Therefore, π = (π ππ π
) = ( 0 1β1 0
)
Question 11a part (ii)
State the coordinates of Z such that π (π₯, π¦) β πβ²(5, 1) under the transformation, M
π (π₯, π¦) β πβ²(5, 1)
( 0 1β1 0
) (π₯π¦) = (
51)
( 0π₯ + 1π¦β1π₯ + 0π¦
) = (51)
( π¦βπ₯
) = (51)
Since the matrices are both 2 Γ 1, we can equate terms
π¦ = 5
βπ₯ = 1
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π₯ = β1
π (π₯, π¦) = (β1, 5).
Question 11a part (iii)
Describe the geometric transformation, M
The transformation matrix, π = ( 0 1β1 0
) rotates the matrix it is applied to in a clockwise
manner about the origin O at an angle of 90Β°.
Question 11b part (i)
Data Given
ππββββ β and ππ ββββ β are position vectors with respect to the origin,π.
Point, P = (2, 7) and ππ ββββ ββ = ( 4β3
)
(a) Write ππββ ββ ββ ππ π‘βπ ππππ (ππ)
Point, P = (2, 7)
Hence
ππββ ββ ββ = (27)
πβππβ ππ ππ π‘βπ ππππ (ππ) π€βπππ π = 2 πππ π = 7.
(b) Write ππ ββ ββ ββ ππ π‘βπ ππππ (ππ)
ππ ββ ββ ββ = ππββ ββ ββ + ππ ββββ ββ
ππ ββββ ββ ββ ββ = (27) + (
4β3
)
= (64)
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πβππβ ππ ππ π‘βπ ππππ (ππ) π€βπππ π = 6 πππ π = 4.
Question 11b part (ii)
Data Given
Point, π = (14,β2)
(a) Find π πββββββ
π = (14,β2)
ππββββ β = ( 14β2
)
π πββββββ = π πββββ β + ππββββββ
ππ ββ ββ β = β (64) + (
14β2
)
= ( 8β6
)
(b) Show P, R and S are collinear.
ππββ ββ = ππ ββββ ββ + π πββββββ
R is a common point as ππ ββββ β πππ π πββββ β are parallel to each other and will lie on the line PS.
Using ππ ββββ β πππ π πββββ β ; in order for the points to be collinear
ππ ββββ β = π€ (π πββββ β) π€βπππ π ππ π π πππππ ππ’ππ‘ππππ
ππ ββββ ββ = ( 4β3
)
π πββββββ = ( 8β6
) = 2 ( 4β3
) π€βπππ π = 2
Therefore P, R and S are collinear.
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