21N.1.SL.TZ0.3

pestleMathematicsAISLPaper 121N· sl-2-5-modelling-functionssource ↗

Natasha carries out an experiment on the growth of mould. She believes that the growth can be modelled by an exponential function

$P (t) = A e^{k t}$ ,

where $P$ is the area covered by mould in ${mm}^{2}$ , $t$ is the time in days since the start of the experiment and $A$ and $k$ are constants.

The area covered by mould is $112 {mm}^{2}$ at the start of the experiment and $360 {mm}^{2}$ after $5$ days.

Write down the value of $A$ .

[1]

Find the value of $k$ .

[3]

Markscheme / solution

$(A =) 112$ A1

[1 mark]

$112 e^{5 k} = 360$ (M1)

Note: Award (M1) for their correct equation.

EITHER

graph of $y = 112 e^{5 k}$ and $y = 360$ with indication of point of intersection (M1)

$(k =) \frac{1}{5} \ln (\frac{360}{112})$ (M1)

Note: Award (M1) for correct rearranging and use of $\log$ .

THEN

$(k =) 0.234 (0.233521 \dots)$ A1

Note: Award (M1)(M1)(A0) for $0.233$ .

[3 marks]

Examiners’ report

In part (a), there were some problems for a few candidates to identify the value of $A$ . Many answers were left as $\frac{112}{e^{k (0)}}$ and thus scored no marks. Those candidates who could identify the value of $A$ were generally able to find a correct solution. Most candidates were able to substitute into the given formula, and many were able to find a correct solution to the resulting equation. The exponential function did not seem to have put candidates off. In several responses the use of logs was seen or implied in the candidates' work; this topic is off syllabus and candidates are expected to use technology (and not logs) to solve such problems. However, very few candidates showed workings between the substitution and the final answer, which was to their detriment in the awarding of marks for their method whenever an incorrect answer was seen. A few candidates did not seem to understand the function notation $P (t)$ . In part (b) a few candidates wrote $360 (5)$ for $P (t)$ and multiplied the two values.