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 | 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=-0.1x3+ 0.8x2, 2≤x≤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 dydx.
Hence find the maximum height of the tunnel.
x=4.
x=6.
Use the trapezoidal rule, with three intervals, to estimate the cross-sectional area of the tunnel.
Write down the integral which can be used to find the cross-sectional area of the tunnel.
Hence find the cross-sectional area of the tunnel.
Markscheme
evidence of power rule (at least one correct term seen) (M1)
dydx=-0.3x2+1.6x A1
[2 marks]
-0.3x2+1.6x=0 M1
x=5.33 (5.33333…, 163) A1
y=-0.1×5.33333…3+0.8×5.33333…2 (M1)
Note: Award M1 for substituting their zero for dydx (5.333…) into y.
7.59 m (7.58519…) 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]
One correct substitution seen (M1)
6.4 m A1
[2 marks]
7.2 m A1
[1 mark]
A=12×2((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 m2 A1
[3 marks]
A=∫82-0.1x3+0.8x2 dx OR A=∫82y dx A1A1
Note: Award A1 for a correct integral, A1 for correct limits in the correct location. Award at most A0A1 if dx is omitted.
[2 marks]
A=32.4 m2 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 6×8 rectangle).
[2 marks]