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Date November Example questions Marks available 3 Reference code EXN.2.SL.TZ0.9
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Show that Question number 9 Adapted from N/A

Question

The temperature T°C of water t minutes after being poured into a cup can be modelled by T=T0e-kt where t0 and T0,k are positive constants.

The water is initially boiling at 100°C. When t=10, the temperature of the water is 70°C.

Show that T0=100.

[1]
a.

Show that k=110ln107.

[3]
b.

Find the temperature of the water when t=15.

[2]
c.

Sketch the graph of T versus t, clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

[4]
d.

Find the time taken for the water to have a temperature of 50°C. Give your answer correct to the nearest second.

[4]
e.

The model for the temperature of the water can also be expressed in the form T=T0at10 for t0 and a is a positive constant.

Find the exact value of a.

[3]
f.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

when t=0, T=100100=T0e0         A1

so T0=100         AG

 

[1 mark]

a.

correct substitution of t=10, T=70         M1

70=100e-10k  or  e-10k=710

 

EITHER

-10k=ln710         A1

ln710=-ln107  or  -ln710=ln107         A1

 

OR

e10k=107         A1

10k=ln107         A1

 

THEN

k=110ln107         AG

 

[3 marks]

b.

substitutes t=15 into T         (M1)

T=58.6°C         A1

 

[2 marks]

c.

a decreasing exponential         A1

starting at 0,100 labelled on the graph or stated         A1

T0 as t         A1

horizontal asymptote T=0 labelled on the graph or stated         A1

 

Note: Award A0 for stating y=0 as the horizontal asymptote.

 

[4 marks]

d.

100e-kt=50  where k=110ln107        A1

 

EITHER

uses an appropriate graph to attempt to solve for t         (M1)

 

OR

manipulates logs to attempt to solve for t e.g. ln12=-110ln107t         (M1)

t=ln2110ln107=19.433        A1

 

THEN

temperature will be 50°C after 19 minutes and 26 seconds        A1

 

[4 marks]

e.

METHOD 1

substitutes T0=100t=10 and T=70 into T=T0at10         (M1)

70=100a1010        A1

a=710        A1

 

METHOD 2

100at10=100e-kt  where k=110ln107

 

EITHER

e-k=a110a=e-10k         (M1)

 

OR

a=e-110ln107t10t         (M1)

 

THEN

a=e-ln107 =eln710        A1

a=710        A1

 

[3 marks]

f.

Examiners report

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Syllabus sections

Topic 2—Functions » SL 2.9—Exponential and logarithmic functions
Topic 2—Functions

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