Date | May 2010 | Marks available | 11 | Reference code | 10M.2.hl.TZ1.13 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Determine, Hence, and Show that | Question number | 13 | Adapted from | N/A |
Question
Points A, B and C are on the circumference of a circle, centre O and radius r . A trapezium OABC is formed such that AB is parallel to OC, and the angle AˆOC is θ , π2⩽θ⩽π .
(a) Show that angle BˆOC is π−θ.
(b) Show that the area, T, of the trapezium can be expressed as
T=12r2sinθ−12r2sin2θ.
(c) (i) Show that when the area is maximum, the value of θ satisfies
cosθ=2cos2θ.
(ii) Hence determine the maximum area of the trapezium when r = 1.
(Note: It is not required to prove that it is a maximum.)
Markscheme
(a) OˆAB=π−θ(allied) A1
recognizing OAB as an isosceles triangle M1
so AˆBO=π−θ
BˆOC=π−θ(alternate) AG
Note: This can be done in many ways, including a clear diagram.
[3 marks]
(b) area of trapezium is T=areaΔBOC+areaΔAOB (M1)
=12r2sin(π−θ)+12r2sin(2θ−π) M1A1
=12r2sinθ−12r2sin2θ AG
[3 marks]
(c) (i) dTdθ=12r2cosθ−r2cos2θ M1A1
for maximum area 12r2cosθ−r2cos2θ=0 M1
cosθ=2cos2θ AG
(ii) θmax=2.205… (A1)
12sinθmax−12sin2θmax=0.880 A1
[5 marks]
Total [11 marks]
Examiners report
In part (a) students had difficulties supporting their statements and were consequently unable to gain all the marks here. There were some good attempts at parts (b) and (c) although many students failed to recognise r as a constant and hence differentiated it, often incorrectly.