Date | May 2021 | Marks available | 2 | Reference code | 21M.2.SL.TZ1.2 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The diagram below shows a circular clockface with centre OO. The clock’s minute hand has a length of 10 cm10cm. The clock’s hour hand has a length of 6 cm6cm.
At 4:004:00 pm the endpoint of the minute hand is at point AA and the endpoint of the hour hand is at point BB.
Between 4:004:00 pm and 4:134:13 pm, the endpoint of the minute hand rotates through an angle, θθ, from point AA to point CC. This is illustrated in the diagram.
A second clock is illustrated in the diagram below. The clock face has radius 10 cm10cm with minute and hour hands both of length 10 cm10cm. The time shown is 6:006:00 am. The bottom of the clock face is located 3 cm3cm above a horizontal bookshelf.
The height, hh centimetres, of the endpoint of the minute hand above the bookshelf is modelled by the function
h(θ)=10 cos θ+13, θ≥0,h(θ)=10cosθ+13, θ≥0,
where θθ is the angle rotated by the minute hand from 6:006:00 am.
The height, gg centimetres, of the endpoint of the hour hand above the bookshelf is modelled by the function
g(θ)=-10 cos(θ12)+13, θ≥0,g(θ)=−10cos(θ12)+13, θ≥0,
where θθ is the angle in degrees rotated by the minute hand from 6:006:00 am.
Find the size of angle AˆOBAˆOB in degrees.
Find the distance between points AA and BB.
Find the size of angle θθ in degrees.
Calculate the length of arc ACAC.
Calculate the area of the shaded sector, AOCAOC.
Write down the height of the endpoint of the minute hand above the bookshelf at 6:006:00 am.
Find the value of hh when θ=160°θ=160°.
Write down the amplitude of g(θ)g(θ).
The endpoints of the minute hand and hour hand meet when θ=kθ=k.
Find the smallest possible value of kk.
Markscheme
4×360°124×360°12 OR 4×30°4×30° (M1)
120°120° A1
[2 marks]
substitution in cosine rule (M1)
AB2=102+62-2×10×6×cos(120°)AB2=102+62−2×10×6×cos(120°) (A1)
AB=14AB=14 cmcm A1
Note: Follow through marks in part (b) are contingent on working seen.
[3 marks]
θ=13×6θ=13×6 (M1)
=78°=78° A1
[2 marks]
substitution into the formula for arc length (M1)
l=78360×2×π×10l=78360×2×π×10 OR l=13π30×10l=13π30×10
=13.6 cm (13.6135…, 4.33π, 13π3)=13.6 cm (13.6135…, 4.33π, 13π3) A1
[2 marks]
substitution into the area of a sector (M1)
A=78360×π×102A=78360×π×102 OR l=12×13π30×102l=12×13π30×102
=68.1 cm2 (68.0678…, 21.7π, 65π3)=68.1 cm2 (68.0678…, 21.7π, 65π3) A1
[2 marks]
2323 A1
[1 mark]
correct substitution (M1)
h=10 cos(160°)+13h=10cos(160°)+13
=3.60 cm (3.60307…)=3.60 cm (3.60307…) A1
[2 marks]
1010 A1
[1 mark]
EITHER
10×cos(θ)+13=-10×cos(θ12)+1310×cos(θ)+13=−10×cos(θ12)+13 (M1)
OR
(M1)
Note: Award M1 for equating the functions. Accept a sketch of h(θ)h(θ) and g(θ)g(θ) with point(s) of intersection marked.
THEN
k=196° (196.363…)k=196° (196.363…) A1
Note: The answer 166.153…166.153… is incorrect but the correct method is implicit. Award (M1)A0.
[2 marks]