ECZ 2011 · O-Level Paper 2 · Question 8 — Vectors In Two Dimensions
Question from ECZ 2011 · O-Level Paper 2 · Question 8
(a) The diagram below is a trapezium OABC. M is the midpoint of AB, OM and CA meet at X. OA = 4p, OC = 2q and CB = 2p.
(a(i)) Express as simply as possible, in terms of p and/or q
(a(i)(a)) CA,[1]
Verified working — step 1
Route C to A through O.(a(i)(b)) BA,[1]
Verified working — step 1
OB = OC + CB = 2q + 2p.(a(i)(c)) OM.[1]
Verified working — step 1
M is the midpoint of AB, so OM is the average of OA and OB.(a(ii)) Given that CX = hCA, express CX in terms of p, q and h.[1]
Verified working — step 1
CX = h CA, and CA = 4p - 2q from part (a)(i)(a).(a(iii)) Hence, show that OX = 4hp + 2(1 - h)q.[2]
Verified working — step 1
OX = OC + CX.(b) The variables x and y are connected by the equation y = 6 + 3x - 2x2. Some corresponding values of x and y are given in the table below. x-2-1012345y-81674-3-14p
(b(i)) Calculate the value of p.[1]
Verified working — step 1
p is the y-value at x = 5 on y = 6 + 3x - 2x2.(b(ii)) Using a scale of 2 cm to represent 1 unit on the x-axis and 2 cm to represent 5 units on the y-axis for -2 ≤ x ≤ 5 and -30 ≤ y ≤ 10, draw the graph of y = 6 + 3x - 2x2.[3]
Verified working — step 1
Scale: x-axis 2 cm to 1 unit (-2 to 5), y-axis 2 cm to 5 units (-30 to 10).(b(iii)) Showing your method clearly, use your graph to solve the equation -2x2 + 3x = -2.[2]
Verified working — step 1
Add 6 to both sides so the left matches the drawn curve.The full step-by-step working and final answer are included with Premium — checked against ECZ marking standards.
Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.
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