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Date	May 2014	Marks available	1	Reference code	14M.2.sl.TZ1.5
Level	SL only	Paper	2	Time zone	TZ1
Command term	Interpret	Question number	5	Adapted from	N/A

Question

The population of deer in an enclosed game reserve is modelled by the function $P(t) = 210\sin (0.5t - 2.6) + 990$ , where $t$ is in months, and $t = 1$ corresponds to 1 January 2014.

Find the number of deer in the reserve on 1 May 2014.

[3]

Find the rate of change of the deer population on 1 May 2014.

[2]

b(i).

Interpret the answer to part (i) with reference to the deer population size on 1 May 2014.

[1]

b(ii).

Markscheme

$t = 5$ (A1)

correct substitution into formula (A1)

eg $210\sin (0.5 \times 5 - 2.6) + 990,{\text{ }}P(5)$

$969.034982 \ldots$

969 (deer) (must be an integer) A1 N3

[3 marks]

evidence of considering derivative (M1)

eg $P'$

$104.475$

$104$ (deer per month) A1 N2

[2 marks]

b(i).

(the deer population size is) increasing A1 N1

[1 mark]

b(ii).

Examiners report

[N/A]

b(i).

[N/A]

b(ii).

Syllabus sections

Topic 6 - Calculus » 6.1 » Derivative interpreted as gradient function and as rate of change.

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