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Date May 2021 Marks available 3 Reference code 21M.2.AHL.TZ1.12
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number 12 Adapted from N/A

Question

The function f has a derivative given by f'x=1xk-x, x, xo, xk where k is a positive constant.

Consider P, the population of a colony of ants, which has an initial value of 1200.

The rate of change of the population can be modelled by the differential equation dPdt=Pk-P5k, where t is the time measured in days, t0, and k is the upper bound for the population.

At t=10 the population of the colony has doubled in size from its initial value.

The expression for f(x) can be written in the form ax+bk-x, where a, b. Find a and b in terms of k.

[3]
a.

Hence, find an expression for f(x).

[3]
b.

By solving the differential equation, show that P=1200kk-1200e-t5+1200.

[8]
c.

Find the value of k, giving your answer correct to four significant figures.

[3]
d.

Find the value of t when the rate of change of the population is at its maximum.

[3]
e.

Markscheme

1xk-xax+bk-x

ak-x+bx=1         (A1)

attempt to compare coefficients OR substitute x=k and x=0 and solve         (M1)

a=1k and b=1k        A1

f'(x)=1kx+1kk-x

 

[3 marks]

a.

attempt to integrate their ax+bk-x         (M1)

fx1k1x+1k-xdx

=1klnx-lnk-x+c=1klnxk-x+c         A1A1

 

Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of +c.

 

[3 marks]

b.

attempt to separate variables and integrate both sides         M1

5k1Pk-PdP=1dt

5lnP-lnk-P=t+c         A1

 

Note: There are variations on this which should be accepted, such as 1klnP-lnk-P=15kt+c. Subsequent marks for these variations should be awarded as appropriate.

 

EITHER

attempt to substitute t=0, P=1200 into an equation involving c        M1

c=5ln1200-lnk-1200=5ln1200k-1200         A1

5lnP-lnk-P=t+5ln1200-lnk-1200         A1

lnPk-12001200k-P=t5

Pk-12001200k-P=et5         A1

 

OR

lnPk-P=t+c5

Pk-P=Aet5         A1

attempt to substitute t=0, P=1200        M1

1200k-1200=A         A1

Pk-P=1200et5k-1200         A1

 

THEN

attempt to rearrange and isolate P        M1

Pk-1200P=1200ket5-1200Pet5  OR  Pke-t5-1200Pe-t5 =1200k-1200P  OR  kP-1=k-12001200et5

Pk-1200+1200et5=1200ket5  OR  Pke-t5-1200e-t5+1200=1200k         A1

 

P=1200kk-1200e-t5+1200         AG

 

[8 marks]

c.

attempt to substitute t=10, P=2400         (M1)

2400=1200kk-1200e-2+1200          (A1)

k=2845.34

k=2845          A1

 

Note: Award (M1)(A1)A0 for any other value of k which rounds to 2850

 

[3 marks]

d.

attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that P=k2=1422.67         (M1)

t=1.57814

=1.58 (days)         A2

 

Note: Accept any value which rounds to 1.6.

 

[3 marks]

e.

Examiners report

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Syllabus sections

Topic 1—Number and algebra » AHL 1.11—Partial fractions
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Topic 5 —Calculus » AHL 5.15—Further derivatives and indefinite integration of these, partial fractions
Topic 1—Number and algebra
Topic 5 —Calculus

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