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Date May 2022 Marks available 2 Reference code 22M.2.AHL.TZ1.10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Sketch Question number 10 Adapted from N/A

Question

Consider the function fx=x2-1, where 1x2.

The curve y=f(x) is rotated 2π about the y-axis to form a solid of revolution that is used to model a water container.

At t=0, the container is empty. Water is then added to the container at a constant rate of 0.4m3s-1.

Sketch the curve y=fx, clearly indicating the coordinates of the endpoints.

[2]
a.

Show that the inverse function of f is given by f-1x=x2+1.

[3]
b.i.

State the domain and range of f-1.

[2]
b.ii.

Show that the volume, Vm3, of water in the container when it is filled to a height of h metres is given by V=π13h3+h.

[3]
c.i.

Hence, determine the maximum volume of the container.

[2]
c.ii.

Find the time it takes to fill the container to its maximum volume.

[2]
d.

Find the rate of change of the height of the water when the container is filled to half its maximum volume.

[6]
e.

Markscheme

correct shape (concave down) within the given domain 1x2             A1

1,0 and 2,3=2,1.73             A1

 

Note: The coordinates of endpoints may be seen on the graph or marked on the axes.

 

[2 marks]

a.

interchanging x and y (seen anywhere)             M1

x=y2-1

x2=y2-1             A1

y=x2+1             A1

f-1x=x2+1             AG

 

[3 marks]

b.i.

0x3  OR domain 0,3=0,1.73             A1

1y2  OR  1f-1x2  OR  range 1,2             A1

 

[2 marks]

b.ii.

attempt to substitute x=y2+1 into the correct volume formula             (M1)

V=π0hy2+12dy =π0hy2+1dy             A1

=π13y3+y0h             A1

=π13h3+h             AG


Note:
Award marks as appropriate for correct work using a different variable e.g. π0hx2+12dx


[3 marks]

c.i.

attempt to substitute h=3  =1.732 into V             (M1)

V=10.8828

V=10.9m3  =23πm3             A1

 

[2 marks]

c.ii.

time =10.88280.4=23π0.4             (M1)

=27.207

=27.2=53πs             A1

 

[2 marks]

d.

attempt to find the height of the tank when V=5.4414 =3π             (M1)

π13h3+h=5.4414  =3π

h=1.1818             (A1)

attempt to use the chain rule or differentiate V=π13h3+h with respect to t             (M1)

dhdt=dhdV×dVdt=1πh2+1×dVdt  OR  dVdt=πh2+1dhdt             (A1)

attempt to substitute their h and dVdt=0.4             (M1)

dhdt=0.4π1.18182+1=0.053124

=0.0531m s-1             A1

 

[6 marks]

e.

Examiners report

Part a) was generally well done, with the most common errors being to use an incorrect domain or not to give the coordinates of the endpoints. Some graphs appeared to be straight lines; some candidates drew sketches which were too small which made it more difficult for them to show the curvature.

Most candidates were able to show the steps to find an inverse function in part b), although occasionally a candidate did not explicitly swop the x and y variables before writing down the inverse function, which was given in the question. Many candidates struggled to identify the domain and range of the inverse, despite having a correct graph.

Part c) required a rotation around the y-axis, but a number of candidates attempted to rotate around the x-axis or failed to include limits. In the same vein, many substituted 2 into the formula instead of the square root of 3 when answering the second part. Many subsequently gained follow through marks on part d).

There were a number of good attempts at related rates in part e), with the majority differentiating V with respect to t, using implicit differentiation. However, many did not find the value of h which corresponded to halving the volume, and a number did not differentiate with respect to t, only with respect to h.

a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
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d.
[N/A]
e.

Syllabus sections

Topic 2—Functions » SL 2.3—Graphing
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