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Date	May 2021	Marks available	2	Reference code	21M.2.AHL.TZ1.2
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Hence and Find	Question number	2	Adapted from	N/A

Question

The cross-sectional view of a tunnel is shown on the axes below. The line $[AB]$ represents a vertical wall located at the left side of the tunnel. The height, in metres, of the tunnel above the horizontal ground is modelled by $y = - 0.1 x^{3} + 0.8 x^{2}, 2 \leq x \leq 8$ , relative to an origin $O$ .

Point $A$ has coordinates $(2, 0)$ , point $B$ has coordinates $(2, 2.4)$ , and point $C$ has coordinates $(8, 0)$ .

Find the height of the tunnel when

Find $\frac{d y}{d x}$ .

[2]

a.i.

Hence find the maximum height of the tunnel.

[4]

a.ii.

$x = 4$ .

[2]

b.i.

$x = 6$ .

[1]

b.ii.

Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.

[3]

c.

Write down the integral which can be used to find the cross-sectional area of the tunnel.

[2]

d.i.

Hence find the cross-sectional area of the tunnel.

[2]

d.ii.

Markscheme

evidence of power rule (at least one correct term seen) (M1)

$\frac{d y}{d x} = - 0.3 x^{2} + 1.6 x$ A1

[2 marks]

a.i.

$- 0.3 x^{2} + 1.6 x = 0$ M1

$x = 5.33 (5.33333 \dots, \frac{16}{3})$ A1

$y = - 0.1 \times 5.33333 \dots^{3} + 0.8 \times 5.33333 \dots^{2}$ (M1)

Note: Award M1 for substituting their zero for $\frac{d y}{d x} (5.333 \dots)$ into $y$ .

$7.59 m (7.58519 \dots)$ A1

Note: Award M0A0M0A0 for an unsupported $7.59$ .
Award at most M0A0M1A0 if only the last two lines in the solution are seen.
Award at most M1A0M1A1 if their $x = 5.33$ is not seen.

[6 marks]

a.ii.

One correct substitution seen (M1)

$6.4 m$ A1

[2 marks]

b.i.

$7.2 m$ A1

[1 mark]

b.ii.

$A = \frac{1}{2} \times 2 ((2.4 + 0) + 2 (6.4 + 7.2))$ (A1)(M1)

Note: Award A1 for $h = 2$ seen. Award M1 for correct substitution into the trapezoidal rule (the zero can be omitted in working).

$= 29.6 m^{2}$ A1

[3 marks]

c.

$A = \int_{2}^{8} - 0.1 x^{3} + 0.8 x^{2} d x$ OR $A = \int_{2}^{8} y d x$ A1A1

Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if $d x$ is omitted.

[2 marks]

d.i.

$A = 32.4 m^{2}$ A2

Note: As per the marking instructions, FT from their integral in part (d)(i). Award at most A1FTA0 if their area is $> 48$ , this is outside the constraints of the question (a $6 \times 8$ rectangle).

[2 marks]

d.ii.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

Topic 5—Calculus » SL 5.5—Integration introduction, areas between curve and x axis

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