Find the value of n when $t = 0$ .

Find the rate at which n is increasing when $t = 15$ .

$n = 800{{\rm{e}}^0}$     (A1)

$n = 800$     A1     N2

[2 marks]

evidence of using the derivative     (M1)

$n'(15) = 731$     A1     N2

[2 marks]

METHOD 1

setting up inequality (accept equation or reverse inequality)     A1

e.g. $n'(t) > 10000$

evidence of appropriate approach     M1

e.g. sketch, finding derivative

$k = 35.1226 \ldots$     (A1)

least value of k is 36     A1     N2

METHOD 2

$n'(35) = 9842$ , and $n'(36) = 11208$    A2

least value of k is 36     A2     N2

[4 marks]

This question seemed to be challenging for the great majority of the candidates.

Part (a) was generally well answered.

This question seemed to be challenging for the great majority of the candidates.

Part (a) was generally well answered but in parts (b) and (c) they did not consider that rates of change meant they needed to use differentiation. Most students completely missed or did not understand that the question was asking about the instantaneous rate of change, which resulted in the fact that most of them used the original equation. Some did attempt to find an average rate of change over the time interval, but even fewer attempted to use the derivative.

Of those who did realize to use the derivative in (b), a vast majority calculated it by hand instead of using their GDC feature to evaluate it.

This question seemed to be challenging for the great majority of the candidates.

Part (a) was generally well answered but in parts (b) and (c) they did not consider that rates of change meant they needed to use differentiation. Most students completely missed or did not understand that the question was asking about the instantaneous rate of change, which resulted in the fact that most of them used the original equation. Some did attempt to find an average rate of change over the time interval, but even fewer attempted to use the derivative.

Of those who did realize to use the derivative in (b), a vast majority calculated it by hand instead of using their GDC feature to evaluate it.

The inequality for part (c) was sometimes well solved using the original function but many failed to round their answers to the nearest integer.