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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 $0 \leq x \leq 4 π$ .

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 $12 π$ , 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 $B$ .

[3]

Show that the area of the shaded region is $12 π$ .

[5]

Find the value of $l$ .

[3]

Hence, find the volume of the cone.

[4]

Markscheme

$6 + 6 \cos x = 0$ (or setting their $f' (x) = 0$ ) (M1)

$\cos x = - 1$ (or $\sin x = 0$ )

$x = π, x = 3 π$ A1A1

[3 marks]

attempt to integrate $\int_{π}^{3 π} (6 + 6 \cos x) d x$ (M1)

$= {[6 x + 6 \sin x]}_{π}^{3 π}$ A1A1

substitute their limits into their integrated expression and subtract (M1)

$= (18 π + 6 \sin 3 π) - (6 π + 6 \sin π)$

$= (6 (3 π) + 0) - (6 π + 0) (= 18 π - 6 π)$ A1

area $= 12 π$ AG

[5 marks]

attempt to substitute into formula for surface area (including base) (M1)

$π (2^{2}) + π (2) (l) = 12 π$ (A1)

$4 π + 2 π l = 12 π$

$2 π l = 8 π$

$l = 4$ A1

[3 marks]

valid attempt to find the height of the cone (M1)

e.g. $2^{2} + h^{2} = {(their l)}^{2}$

$h = \sqrt{12} (= 2 \sqrt{3})$ (A1)

attempt to use $V = \frac{1}{3} π r^{2} h$ with their values substituted M1

$(\frac{1}{3} π (2^{2}) (\sqrt{12}))$

$volume = \frac{4 π \sqrt{12}}{3} (= \frac{8 π \sqrt{3}}{3} = \frac{8 π}{\sqrt{3}})$ A1

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.1—3d space, volume, angles, distance, midpoints

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Topic 3— Geometry and trigonometry

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