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CSEC MATHEMATICS JANUARY 2020 PAPER 2
SECTION I
1. (a) Using a calculator, or otherwise, calculate the exact value of the following:
(i) 4 "#Γ "%β 1 "
(
SOLUTION:
)415 Γ
13, β 1
14
We first evaluate: 4 "#Γ "%
=215 Γ
13 =
75 = 1
25
Nowwehave, 1 9#β 1 "
(. Since 1 β 1 = 0, we now have:
25 β
14 =
8 β 520
= %9<
(in exact form)
(ii)4.1 β 1.259
0.005 SOLUTION:
4.1 β 1.259
0.005 4.1 β (1.25 Γ 1.25)
0.005 Using the calculator,
=4.1 β 1.5625
0.005
=2.53750.005
= 507.5 (in exact form)
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1(b) A stadium currently has a seating capacity of 15 400 seats.
(i) Calculate the number of people in the stadium when 75% of the seats are occupied.
SOLUTION:
75% of 15 400 seats
=75100 Γ 15400
=11 550 persons
(ii) The stadium is to be renovated with a new seating capacity of 20 790 seats. After the renovation, what will be the percentage increase in the number of seats?
SOLUTION:
Increase in the number of seats
= 20 790 β 15 400
= 5 390
So, percentage increase = BCDEFGHFICJKFCLMNFEOPHFGJHQEIRICGSCLMNFEOPHFGJH
Γ 100%
= #%U<"#(<<
Γ 100%
= 35%
1(c) A neon light flashes five times every 10 seconds. Show that the light flashes 43 200 times in one day.
SOLUTION:
1 day = 24 hours = 24 Γ 60 minutes = 24 Γ 60 Γ 60 seconds = 86 400 seconds
The light flashes five times in each 10-second interval.
The number of 10-second intervals in 86 400 seconds = VW(<<"<
= 8640
If there are 5 flashes in every 10 seconds, then the number of times the light flashes in one day = 8640Γ 5
= 43200
Q.E.D.
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2. (a) Factorise the following expressions completely.
(i) 5β9 β 12βπ
SOLUTION:
5β9 β 12βπ
= 5β. β β 12βπ
= β(5β β 12π)
(ii)2π₯9 β π₯ β 6
SOLUTION:
2π₯9 β π₯ β 6
(2π₯ + 3)(π₯ β 2)
2(b) Solve the equation
π + 3 = 3(π β 5)
SOLUTION:
π + 3 = 3π β 15
π β 3π = β15 β 3
β2π = β18
π = ]"V]9
π = 9
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2(c) Make π the subject of the formula
2π΄ = ππ9 + 3π‘
SOLUTION:
2π΄ = ππ9 + 3π‘
ππ9 = 2π΄ β 3π‘
π9 = 9c]%Jd
π = e9c]%Jd
2(d) A farmer plants two crops, potatoes and corn, on a ten-hectare piece of land. The number of hectares of corn planted, π, must be at least twice the number of hectares of potatoes, π.
Write two inequalities to represent the scenario above.
SOLUTION:
The area of the entire plot of land = 10 hectares
The number of hectares of corn = c
The number of hectares of potatoes = p
Clearly, the total area planted cannot exceed the area of the plot
Hence, π + π β€ 10
The number of hectares of corn is at least twice the number of hectares of potatoes
Hence, π β₯ 2π
The two inequalities are:
π + π β€ 10β¦β¦β¦β¦(1)
π β₯ 2π β¦β¦β¦β¦β¦.(2)
c 2p β₯
π + π β― 10
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3 (a) The diagram below shows a hexagonal prism.
Complete the following statement.
The prism has
______________ faces.
______________ edges and
______________ vertices.
SOLUTION:
Faces
The prism has 6 rectangular faces and 2 hexagonal faces.
Total number of faces = 6 + 2 = 8
Edges
The prism has 6 edges bounding each of the two hexagonal faces and 6 edges bounding the rectangular faces. Total number of edges = (6 Γ 2) + 6 = 18
Vertices
The prism has 6 vertices on each of the two hexagonal face.
Total number of vertices = 6 Γ 2 = 12
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3(b) A Sports Club owns a field, PQRS, in the shape of a quadrilateral. A scale diagram of this field is shown below. (1 centimetre represents 10 metres)
In the following parts show all your construction lines.
The field is to be divided with a fence from π to the side π π so that different sports can be played at the same time.
Each point on the fence is the same distance fromππ as from ππ
(i) Using a straight edge and compasses only, construct the line representing the fence.
(ii) Write down the length of the fence , in metres.
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SOLUTION:
(i) Points equidistant from the line ππ and ππ will lie on the line bisecting the angle πππ.
The bisector of the angle QPS meets RS at F.
(ii) The line BF is 7.4 cm by measurement.
Since 1 centimetre represents 10 metres, the length of the fence is
7.4 Γ 10 = 74πππ‘πππ
PF is the line representing the fence.
T
For any point, T, on PF, the two right-angled triangles are congruent
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3 (c) A quadrilateral πππ π and its image πβ²πβ²π β²πβ² are shown on the grid below.
(i) Write down the mathematical name for the quadrilateral πππ π (ii) πππ π is mapped onto πβ²πβ²π β²πβ² by an enlargement with scale factor, π, about the
centre, πΆ(π, π). Using the diagram above, determine the values of π, π and π.
SOLUTION:
(i) The quadrilateral πππ π has only one pair of parallel sides and is therefore a trapezium.
(ii) To locate the center of enlargement, join image points to their corresponding object points by straight lines. Produce them to meet at the center of enlargement.
We join ππβ² and ππy and the point of intersection, πΆ(β4,1)is shown below. It is not necessary to join more than two such lines since all such lines are concurrent.
Hence, π = β4 and π = 1.
The scale factor of the enlargement is the ratio of image length to object length.
Hence, π = zy{yz{
= %"= 3
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4 (a) The function π is defined as
π(π₯) =2π₯ + 75
(i) Find the value of π(4) + π(β4).
SOLUTION:
π(4) =2(4) + 7
5 =8 + 75 =
155 = 3.
π(β4) =2(β4) + 7
5 =β8+ 75 = β
15
Hence, π(4) + π(β4) = 3 + )β15, = 2
45
C
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(ii) a) Calculate the value of π₯ for which π(π₯) = 9.
SOLUTION:
π(π₯) = 9
Therefore,2π₯ + 75 = 9
2π₯ + 7 = 45
2π₯ = 45 β 7
2π₯ = 38
π₯ = 19
b) Hence, or otherwise, determine the value of π]"(9)
SOLUTION:
From above, when π₯ = 19, π(π₯) = 9.
This means that, for the function π(π₯), an input of 19 produces an output of 9. We can also say that the function, πmaps 19 onto 9.
Hence, the inverse function, π]"(π₯),will map 9 onto 19. So, π]"(19) = 9.
Alternatively, we could find π]"(π₯), and substitute π₯ = 19 in this inverse function.
Let.π¦ =2π₯ + 75
Interchange π₯ andπ¦:
π₯ =2π¦ + 75
5π₯ = 2π¦ + 7
2π¦ = 5π₯ β 7
π¦ =5π₯ β 72
π]"(π₯) =5π₯ β 72
π]"(9) =5(9) β 7
2 =382 = 19
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4(b) The graph below shows two straight lines, πΏ"andπΏ9. πΏ" intercepts the π₯ and π¦ axes at (4, 0) and (0, 2) respectively. πΏ9 intercepts the π₯ and π¦ axes at (1.5, 0) and (0,β3) respectively.
(i) Determine the equation of the line πΏ".
SOLUTION:
The gradient of πΏ" can be found by inspection as 9](= β "
9
OR by using the points (0,2)and(4,0) and the gradient formula, οΏ½οΏ½]οΏ½οΏ½οΏ½οΏ½]οΏ½οΏ½
.
Gradient of πΏ" =<]9(]<
= β 9(= β "
9
The general equation of any straight line is π¦ = ππ₯ + π,where π is the gradient and π is theπ¦-intercept. πΏ" has a π¦-intercept of 2, so π = 2. Hence, the equation of πΏ" is:
π¦ = β12π₯ + 2
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OR
Using the point (π₯", π¦") = (2,0) and π = β"9,the equation of a line is given as:
π¦ β π¦"π₯ β π₯"
= π
π¦ β 2π₯ β 0 = β
12
2(π¦ β 2) = βπ₯
2π¦ β 4 = βπ₯
2π¦ = βπ₯ + 4
π¦ = β12π₯ + 2
(ii) What is the gradient of the line πΏ9, given that πΏ" and πΏ9 are perpendicular?
SOLUTION:
The product of the gradients of perpendicular lines is β1.
Gradient of πΏ" Γ Gradient of πΏ9 = β1
β "9Γ Gradient of πΏ9 = β1
GradientofπΏ9 = β1
β12
GradientofπΏ9 = 2
OR
By using the points (1.5, 0) and (0,β3),
Gradient of πΏ9:
=β3 β 00 β 1.5 =
β3β1.5 = 2
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5. A group of 100 students estimated the mass, π(grams) of a seed. The cumulative frequency curve below shows the results.
(i) (a) Using the cumulative frequency curve, estimate the (i) median (ii) upper quartile
SOLUTION:
(i) The median is the middle score in a distribution and will occur at the 50th percentile. The estimate of the median is approximately 3.2 grams.
(ii) The upper quartile is the 75th percentile. The estimate of the upper quartile approximately 4.3 grams.
[These values are obtained from the curve and shown on the diagram.]
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(iii) Semi-interquartile range
To obtain the semi-interquartile range, we use the formula: {οΏ½]{οΏ½9
, where π% is the upper quartile and π"the lower quartile.
The lower quartile is at the 25th percentile and shown on the graph.
π" β 2.3πππππ% β 4.3π
π% β π"2 =
4.3 β 2.32 =
22 = 1
The semi-interquartile range is 1.
Medianβ 3.2π
Upper Quartile β 4.3π
Lower Quartile β 2.3π
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(iv) Number of students whose estimate is 2.8 grams or less.
SOLUTION:
The cumulative frequency that corresponds to a mass of 2.8 grams is 38.
Hence 38 students had estimates that were less than 2.8 grams.
(b) (i) Use the cumulative frequency curve given to complete the frequency table below.
38
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SOLUTION:
The cumulative frequency that corresponds to π β€ 4 is 68. Hence, 68 students had estimates that were less than or equal to 4 grams.
Since 20 students had estimates that were less than or equal to 2 grams, the number having estimates between 2 and 4 grams is 68 β 20 = 48.
The cumulative frequency that corresponds to π β€ 6 is 93. Hence, 93 students had estimates that were less than or equal to 6 grams.
Since 68 students had estimates that were less than or equal to 4 grams, the number having estimates between 4 and 6 grams is 93 β 68 = 25
The completed table is shown below:
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(b) (ii) A student is chosen at random. Find the probability that the student estimated the mass to be greater than 6 grams.
From the frequency table, the number of students who estimated the mass to be greater than 6 grams is 6 + 1 = 7.
The total number of students is 100.
Probability (estimate is greater than 6 grams)
=ππ’πππππππ π‘π’ππππ‘π π€βπππ π‘ππππ‘πππ‘βππππ π π‘ππππππππ‘πππ‘βππ6π
πππ‘ππππ’πππππππ π‘π’ππππ‘π
=7100
48
25
βπ = 100
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6. (a) The radius of EACH circle in the rectangle ππππshown below is 7 cm. The circles fit exactly into the rectangle
(ii) Show that the area of the rectangle is 2 352 cm2.
SOLUTION:
The diameter of the circle = 7ππ Γ 2 = 14ππ
The length of the rectangle = 14ππ Γ 4 = 56ππ
The width of the rectangle = 14ππ Γ 3 = 42ππ
The area of the rectangle = 56 Γ 42ππ9
= 2352ππ9
(iii) Calculate the area of the shaded region.
SOLUTION:
Area of the 12 circles = 12 Γ ππ9 = 12 Γ 99οΏ½Γ 79
= 12 Γ 99οΏ½Γ 7 Γ 7 = 1848ππ9
Area of the shaded region = Area of the rectangle β Area of the 12 circles
= 2352 β 1848ππ9
= 504ππ9
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6 (b) The diagram below, not drawn to scale, shows triangle MNP in which angle MNP = angle PMN = 52< and MN = 12.5 cm.
(i) State the type of triangle shown above.
SOLUTION:
Since angle N = 180< β 2(52)< β 52<,only two angles of triangle MNP are equal.
Hence, triangle MNP is isosceles.
(ii) Determine the value of angle PNM.
SOLUTION:
Angle PNM = 180< β 2(52)< = 76< (sum of angles in a triangle is 180<)
(iii) Calculate the area of triangle MNP.
SOLUTION:
760
12.5 cm
Area of triangle MNP:
=12 Γ 12.5 Γ 12.5 Γ π ππ76
<
=12 Γ 12.5 Γ 12.5 Γ 0.970
= 75.80ππ9
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7. A sequence of figures is made up stars, using dots and sticks of different lengths. The first three figures in the sequence are shown below.
Study the pattern of numbers in each row of the table below. Each row relates to a figure in the sequence of figures stated above. Some rows have not been included in the table.
(a) Complete Rows (i), (ii) and (iii).
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SOLUTION:
Part (a)
(i) The number of sticks is 12 times the Figure Number. So, in Figure 5, there will be 5 Γ 12 = 60 sticks.
The number of dots is one more than the number of sticks. Since Figure 5 has 60 sticks it will have (60 + 1) = 61 dots
(ii) The Figure Number will be the number of dots divided by 12. If a figure has 156 sticks then, the Figure Number will be (156 Γ· 12) = 13
Hence, the number of dots for Figure 13 is 156 + 1 = 157.
(iii) The number of sticks is always 12 times the Figure Number. So, the πJKfigure will have 12π sticks.
The number of dots is always one more than the number of sticks. So, the πJK figure will have [(12 Γ π) + 1] = 12π + 1 sticks.
The completed table is shown below:
60 61
13 157
12π 12π + 1
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(b) The sum of the number of dots in two consecutive figures is recorded. This information for the first three pairs of consecutive figures is shown in the table below.
Determine the total number of dots in
(i) Figure 7 and 8 (ii) Figure πand Figure (π + 1)
SOLUTION:
Part (b)
(i) The total number of dots in Figures 7 and 8
The number of dots in Figure 7 = 12(7) + 1 = 85 (π = 7)
The number of dots in Figure 8 = 12(8) + 1 = 97 (π = 8)
Therefore the total number of dots in Figures 7 and 8 = 85 + 97 = 182
(iii) The total number of dots in Figure πand Figure (π + 1) The number of dots in Figure π is12π + 1 The number of dots in Figure π + 1 = 12(π + 1) + 1
Therefore, the total number of dots in Figure πand Figure (π + 1)
= [12π + 1] + [12(π + 1) + 1]
= 12π + 1 + 12π + 12 + 1
= 24π + 14
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SECTION II
ALGEBRA, RELATIONS, FUNCTIONS AND GRAPHS
8. (a) Solve the pair of equations:
π¦9 + 2π¦ + 11 = π₯
π₯ = 5 β 3π¦
SOLUTION:
Let
π¦9 + 2π¦ + 11 = π₯ Eq (1)
π₯ = 5 β 3π¦ Eq (2)
Substitute π₯ from Equation 2 in Equation 1:
π¦9 + 2π¦ + 11 = 5 β 3π¦
π¦9 + 2π¦ + 11 β 5 + 3π¦ = 0
π¦9 + 5π¦ + 6 = 0
(π¦ + 3)(π¦ + 2) = 0
π¦ + 3 = 0πππ¦ + 2 = 0
π¦ = β3πππ¦ = β2
When π¦ = β3, π₯ = 5 β 3(β3) = 5 + 9 = 14
When π¦ = β2, π₯ = 5 β 3(β2) = 5 + 6 = 11
Solutions are: π₯ = 14ππππ¦ = β3ππ π₯ = 11ππππ¦ = β2
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(b) The function f is defined as follows:
π(π₯) = 4π₯9 β 8π₯ β 2
(i) Express π(π₯) in the form π(π₯ + β)9 + π, where π, β and πare constants.
SOLUTION:
4π₯9 β 8π₯ β 2
= 4(π₯9 β 2π₯) β 2
= 4[(π₯ β 1)9 β 1] β 2
= 4(π₯ β 1)9 β 4 β 2
= 4(π₯ β 1)9 β 6
OR
4π₯9 β 8π₯ β 2
= 4(π₯9 β 2π₯) β 2
= 4(π₯ β 1)9+?
4(π₯ β 1)(π₯ β 1) = 4(π₯9 β 2π₯ + 1) = 4π₯9 β 8π₯ + 4
The constant in the original expression is β2. Hence, 4+?= β2 and ? = β6
So, 4π₯9 β 8π₯ β 2 = 4(π₯ β 1)9 β 6
OR
Expanding and equating coefficients;
π(π₯ + β)9 + π = π(π₯9 + 2π₯β + β9) + π
= ππ₯9 + 2π₯πβ + πβ9 + π
4π₯9 β 8π₯ β 2 = ππ₯9 + 2π₯πβ + πβ9 + π
Equating coefficients:
π = 4 2πβ = β8 πβ9 + π = β2
2(4) Γ β = β8 4(β1)9 + π = β2
β = β1 π = β6
So, π(π₯ + β)9 + π = 4(π₯ β 1)9 β 6
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(ii) State the minimum value of π(π₯)
π(π₯) = 4(π₯ β 1)9 β 6
SOLUTION:
4(π₯ β 1)9 β₯ 0, for all values of π₯.
π(π₯)MIC = 0 β 6 = β6
(iii) Determine the equation of the axis of symmetry.
The axis of symmetry passes through the minimum point.
This occurs when π₯ β 1 = 0, orπ₯ = β1
OR
The equation of the axis of symmetry is π₯ = ]N9G
, for a quadratic of the form π(π₯) = ππ₯9 + ππ₯ + π
For the function, π(π₯) = 4π₯9 β 8π₯ β 2, π = 4, π = β8
π₯ = ](]V)9(()
= β1
The equation of the axis of symmetry is π₯ = β1.
(c) The speed-time graph below, not drawn to scale, shows the three-stage journey of a car over a period of 40 seconds.
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Determine the acceleration of the car for EACH of the following stages of the journey
Stage 2
Stage 3
SOLUTION:
Acceleration of the car in Stage 2 of the journey
Stage 2 of the journey is represented by a horizontal line. This indicates that the velocity is constant.
Since the velocity is not changing, the rate of change of velocity is zero.
Therefore the acceleration, π = 0π/π 9.
OR
The acceleration can be calculated by finding the gradient of the line. A horizontal line has a gradient of zero. By calculation,
Gradient = "9]"9%9]"W
= 0.
(16, 12) (32, 12)
Acceleration of the car in Stage 3 of the journey
(32, 12) Gradient = "9]<%9](<
= "9]V= β %
9π/π 9
The car is decelerating (slowing down). The acceleration is β1.5π/π 9 or the deceleration is 1.5π/π 9.
(40, 0)
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9. (a) The circle shown below has center O and the points A, B, C and D lying on the circumference. A straight line passes through the points A and B. Angle πΆπ΅π· = 49< and angle ππ΄π΅ = 37<.
(i) Write down the mathematical names of the straight lines BC and OA.
SOLUTION: The straight line, BC, joins two points on the circle and it is therefore a chord of the circle. The straight line OA joins the center of the circle to a point on the circle and is therefore a radius of the circle.
(ii) Determine the value of EACH of the following angles. Show detailed working where necessary and give a reason to support your answer. a) x b) y SOLUTION: Consider triangle OAB ππ΄ = ππ΅ (radii)
Hence, triangle OAB is isosceles. Angle ππ΅π΄ = 37< (Base angles of an isosceles triangle are equal) π₯< = 180< β (37 + 37)< (Sum of angles in a triangle is 180<) π₯< = 180< β 74< π₯ = 104
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Consider triangle BCD Angle π΅πΆπ· = 90< (Angle in a semi-circle is a right angle) π¦< = 180< β (90< + 49<) (Sum of angles in a triangle is 180<) π¦< = 180< β 139< π¦ = 41
(b) The diagram below, not drawn to scale, shows the route of a ship cruising from
Palmcity (π) to Quayton (π) and then to Rivertown (R). The bearing of πfrom π is 133< and the angle πππ is 56<.
(i) Calculate the value of the angle π€. SOLUTION:
π₯
π₯
Let this angle be x. π₯ = 180< β 133< = 47<
(Angles on a straight line) The two North lines are parallel, hence, this angle is also equal to x (alternate angles). Therefore, π€< = 180< β (56 + 47)< π€ = 77
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(ii) Determine the bearing of P from Q.
The angle that represents the bearing of P from Q is show in the diagram. It is calculated as
360< β 47< = 313<
(iii) Calculate the distance RP.
Applying the cosine rule to triangle PQR we have:
π π9 = 2109 + 2909 β 2(210)(290)πππ 56< = 44100 + 84100 β 121800 Γ 0.559 = 128200β 68109.70 = 60090.30 π π = β60090.30 π π = 245.13ππ
47<
Bearing of P from Q = 3130
P
Q
N
N
P
Q
R
210 km
290 km
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10. (a) The transformation, π = )0 ππ 0, maps the point π onto π β² as shown in the diagram
below.
(i) Determine the value of p and of q.
SOLUTION:
The transformation, π = )0 ππ 0, maps π (2,β5) onto π y(5,2). Hence,
)0 ππ 0, Β©
2β5Βͺ = Β©52Βͺ
2Γ2 2Γ1 2Γ1
)0 Γ 2 + π Γ β5π Γ 2 + 0 Γ β5, = Β©52Βͺ
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)β5π2π , = Β©52Βͺ
Equating corresponding elements, β5π = 5and2π = 2 So, π = β1andπ = 1
(ii) Describe fully the transformation, M.
SOLUTION: The transformation, π = Β©0 β1
1 0 Βͺ, is an anticlockwise (or positive) rotation about the origin through 90<.
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(b) πππ π is a parallelogram in which ππ«««««β = π’and ππ««««β = π£.
M is a point on QS such that ππ:ππ = 1: 2
(i) Write in terms of πand π an expression for
a) ππ«««««β
SOLUTION: ππ«««««β = ππ«««««β + ππ««««β = β(π) + π
ππ«««««β = βπ + π
b) ππ««««««β
SOLUTION: ππ:ππ = 1: 2 Therefore, ππ = "
%ππ
ππ««««««β =13 (βπ + π)
ππ««««««β = β13π +
13π
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(ii) Show that ππ ««««««β = "%(π + 2π)
SOLUTION: ππ ««««««β = ππ««««««β + ππ «««««β ππ ««««««β = βππ««««««β + ππ «««««β = βΒ©β"
%π + "
%πΒͺ + π Β²ππ «««««β = ππ««««β = πΒ³
=13π β
13π + π
=13π +
23π
=13 (π + 2π)
Q.E.D.
(iii) T is the mid-point of PQ. Prove that R, M and T are collinear.
SOLUTION: To prove that R, M and T are collinear, consider the vectors ππ «««««β andππ ««««««β .
ππ «««««β = ππ«««««β + ππ «««««β
ππ «««««β =12π + π
ππ «««««β =12 (π + 2π)
T
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We know from part (ii) that
ππ ««««««β =13 (π + 2π)
ππ ««««««β =13(π + 2π) =
23Β΄12 (π + ππ)
ΒΆ
ππ ««««««β =23ππ «««««β
The vector ππ ««««««β is a scalar multiple of ππ «««««β (The scalar multiple being 9
%).
Hence, ππ ««««««β is parallel to ππ «««««β . In both line segments, R is a common point. Therefore, R, M and T are collinear.
Q.E.D.
13(π+
2π)
12(π+
2π)
T
R
M
R