(a) Copy and complete the table of values for the relation V = -X² + X + 2 for -3 ≤ x ≥ 3.
(b) Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the v-axis, draw a graph of the relation
y = -X² + X + 2.
(c) From the graph, find the:
(i)Minimum value of y;
(ii)Roots of the equation X² - x -2 = 0;
(iii)Gradient of the curve at x = -0.5.
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Observation
This question was reported to have been attempted by majority of the candidates who were able to complete the table of values and plotted the graph correctly. However, majority ofthe candidates were reported not to draw the tangent to the curve as required. Hence they were unable to determine the gradient of the curve at
x = -0.5.
Question 11
In the diagram, L.PTQ = L.PSR = 900, /PQ/ = 10 ern, /PS/ = 14.4 cm and /TQ/ = 6 cm.
Calculate the area of quadrilateral QRST.
(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.
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Observation
This question was reported not to be very popular and candidates' performance was described as not satisfactory. Majority of the candidates were reported not to apply the concept of similar triangles correctly. Others did not recognize the quadrilateral as a trapezium and so failed to use the correct formula when finding its area. Part (b) of the question was reported not to have been done satisfactorily either. Candidates were expected to show that:
/PT/ = ..../102 - 62 = 8 cm. m: =!J2L i.e ~ = 14.4.Hence, /SR/ = 10.8 cm.
/TO/ /SR/ 6 /SR/
Area of quadrilateral QRST = Yz (6 + 10.8) x 6.4 = 53.76 cm2. Some candidates were reported to have subtracted the area of triangle PQT from triangle PRS. This was also in order. In part (b) if the side of the square was y, then new breadth = 90 x y = 0.9y.
100
New length = 115 x Y = 1.15y. New area = 1.15y x 0.9y = 1.035/.
100
Hence, ratio = 1.035y² : y² = 1.035 : 1 or 207:200 .
Question 12
The frequency distribution of the weight of 100 participants in a high jump competition is as
shown below:
Weight (kg)
|
20-29 30 - 39 40 - 49 50 - 59 60 - 69 70-79 |
Number of participants
|
10 18 22 25 16 9 |
(b) Draw the cumulative frequency curve.
(c) From the curve, estimate the:
(i) median;
(ii) semi-interquartile range;
(iii) probability that a participant chosen at random weighs at least 60 kg.
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Observation
This question was reported to be attempted by majority of the candidates. According to the report, candidates' performance in this question was fair. Majority of them were reported not to have read from their Ogives correctly. Others did not draw the Ogive using class boundaries.
The median was 49.5 while the first quartile(Ql) was 37.8 and the third Quartile (~)59.5.
Hence, the semi-interquartile range Q1-Q1 = 59.5 - 32.8 = 10.85 ± 1.
2 2
Number of participants who weighed at least 60kg = 25. Therefore probability of choosing a
participant who weighed at least 60kg = 25= -.1 .
100 4
Question 13
(a)The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4(20/27) .Find Its irst term.
(b)Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.
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Observation
This question was reported to have been attempted by majority of the candidates and their performance was commended. However in part (a), a few of them divided the indices instead of subtracting them. A few others were not able to manipulate the fraction involved. Part (b) was also well attempted and majority of them obtained full marks.
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