(i)     Find the value of $k$.

(ii)     Interpret the meaning of the value of $k$.

Find the least number of whole years for which $\frac{{{P_t}}}{{{P_0}}} < 0.75$.

(i)     valid approach     (M1)

eg$\,\,\,\,\,$$0.9 = {{\text{e}}^{k(1)}}$

$k =  - 0.105360$

$k = \ln 0.9{\text{ (exact), }} - 0.105$    A1     N2

(ii)     correct interpretation     R1     N1

eg$\,\,\,\,\,$population is decreasing, growth rate is negative

[3 marks]

METHOD 1

valid approach (accept an equality, but do not accept 0.74)     (M1)

eg$\,\,\,\,\,$$P < 0.75{P_0},{\text{ }}{P_0}{{\text{e}}^{kt}} < 0.75{P_0},{\text{ }}0.75 = {{\text{e}}^{t\ln 0.9}}$

valid approach to solve their inequality     (M1)

eg$\,\,\,\,\,$logs, graph

$t > 2.73045{\text{ }}({\text{accept }}t = 2.73045){\text{ }}(2.73982{\text{ from }} - 0.105)$    A1

28 years     A2     N2

METHOD 2

valid approach which gives both crossover values accurate to at least 2 sf     A2

eg$\,\,\,\,\,$$\frac{{{P_{2.7}}}}{{{P_0}}} = 0.75241 \ldots ,{\text{ }}\frac{{{P_{2.8}}}}{{{P_0}}} = 0.74452 \ldots$

$t = 2.8$    (A1)

28 years     A2     N2

[5 marks]

Part (a) was generally done well, with many candidates able to find the value of $k$ correctly and to interpret its meaning. Lack of accuracy was occasionally a concern, with some candidates writing their value of $k$ to 2 significant figures or evaluating $\ln (0.9)$ incorrectly.

Few candidates were successful in part (b) with many unable to set up an inequality or equation which would allow them to find the condition on $t$. Some were able to find the value of $t$ in decades but most were unable to correctly interpret their inequality in terms of the least number of whole years. While a solution through analytic methods was readily available, very few students attempted to use their GDC to solve their initial equation or inequality.