| Date | May 2021 | Marks available | 4 | Reference code | 21M.1.SL.TZ2.8 |
| Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
| Command term | Hence and Find | Question number | 8 | Adapted from | N/A |
Question
Consider the function f defined by f(x)=6+6 cos x, for 0≤x≤4π.
The following diagram shows the graph of y=f(x).
The graph of f touches the x-axis at points A and B, as shown. The shaded region is enclosed by the graph of y=f(x) and the x-axis, between the points A and B.
The right cone in the following diagram has a total surface area of 12π, equal to the shaded area in the previous diagram.
The cone has a base radius of 2, height h, and slant height l.
Find the x-coordinates of A and B.
Show that the area of the shaded region is 12π.
Find the value of l.
Hence, find the volume of the cone.
Markscheme
6+6 cos x=0 (or setting their f'(x)=0) (M1)
cos x=-1 (or sin x=0)
x=π, x=3π A1A1
[3 marks]
attempt to integrate 3π∫π(6+6 cos x)dx (M1)
=[6x+6 sin x]3ππ A1A1
substitute their limits into their integrated expression and subtract (M1)
=(18π+6 sin 3π)-(6π+6 sin π)
=(6(3π)+0)-(6π+0) (=18π-6π) A1
area =12π AG
[5 marks]
attempt to substitute into formula for surface area (including base) (M1)
π(22)+π(2)(l)=12π (A1)
4π+2πl=12π
2πl=8π
l=4 A1
[3 marks]
valid attempt to find the height of the cone (M1)
e.g. 22+h2=(their l)2
h=√12 (=2√3) (A1)
attempt to use V=13πr2h with their values substituted M1
(13π(22)(√12))
volume=4π√123(=8π√33=8π√3) A1
[4 marks]
