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Date	November 2021	Marks available	2	Reference code	21N.1.AHL.TZ0.15
Level	Additional Higher Level	Paper	Paper 1	Time zone	Time zone 0
Command term	Show that	Question number	15	Adapted from	N/A

Question

The following diagram shows a frame that is made from wire. The total length of wire is equal to $15 cm$ . The frame is made up of two identical sectors of a circle that are parallel to each other. The sectors have angle $θ$ radians and radius $r cm$ . They are connected by $1 cm$ lengths of wire perpendicular to the sectors. This is shown in the diagram below.

The faces of the frame are covered by paper to enclose a volume, $V$ .

Show that $r = \frac{6}{2 + θ}$ .

[2]

Find an expression for $V$ in terms of $θ$ .

[2]

b.i.

Find the expression $\frac{d V}{d θ}$ .

[3]

b.ii.

Solve algebraically $\frac{d V}{d θ} = 0$ to find the value of $θ$ that will maximize the volume, $V$ .

[2]

b.iii.

Markscheme

$15 = 3 + 4 r + 2 r θ$ M1

$12 = 2 r (2 + θ)$ A1

Note: Award A1 for any reasonable working leading to expected result e,g, factorizing $r$ .

$r = \frac{6}{2 + θ}$ AG

[2 marks]

attempt to use sector area to find volume (M1)

volume $= \frac{1}{2} r^{2} θ \times 1$

$= \frac{1}{2} \times \frac{36}{{(2 + θ)}^{2}} \times θ (= \frac{18 θ}{{(2 + θ)}^{2}})$ A1

[2 marks]

b.i.

$\frac{d V}{d θ} = \frac{{(2 + θ)}^{2} \times 18 - 36 θ (2 + θ)}{{(2 + θ)}^{4}}$ M1A1A1

$\frac{d V}{d θ} = \frac{36 - 18 θ}{{(2 + θ)}^{3}}$

[3 marks]

b.ii.

$\frac{d V}{d θ} = \frac{36 - 18 θ}{{(2 + θ)}^{3}} = 0$ M1

Note: Award this M1 for simplified version equated to zero. The simplified version may have been seen in part (b)(ii).

$θ = 2$ A1

[2 marks]

b.iii.

Examiners report

Several candidates missed that the angle $θ$ was in radians and used arc and sector formulas with degrees instead. This aside, part (a) was often well done. Part (b)(i) was also correctly answered by many candidates, but their failure to make any attempt to simplify their answer often led to difficulties in part (b)(ii). Again, failing to simplify the result in part (b)(ii) led to yet more difficulties in part (b)(iii). Some candidates used the product rule to differentiate $\frac{18 θ}{{(2 + θ)}^{2}}$ as $18 θ {(2 + θ)}^{- 2}$ rather than the quotient rule. This was fine but made solving the equation in (b)(iii) less straightforward.

b.i.

b.ii.

b.iii.

Syllabus sections

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

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