(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.