The depth, h(t) metres, of water at the entrance to a harbour at t hours after midnight on a particular day is given by

\[h(t) = 8 + 4\sin \left( {\frac{{\pi t}}{6}} \right),{\text{ }}0 \leqslant t \leqslant 24.\]

(a)     Find the maximum depth and the minimum depth of the water.

(b)     Find the values of t for which \(h(t) \geqslant 8\).

(a)     Either finding depths graphically, using \(\sin \frac{{\pi t}}{6} = \pm 1\) or solving \(h'(t) = 0\) for t     (M1)

\(h{(t)_{\max }} = 12{\text{ (m), }}h{(t)_{\min }} = 4{\text{ (m)}}\)     A1A1     N3

(b)     Attempting to solve \(8 + 4\sin \frac{{\pi t}}{6} = 8\) algebraically or graphically     (M1)

\(t \in [{\text{0}},{\text{6}}] \cup [{\text{12}},{\text{18}}] \cup \{ {\text{24}}\} \)     A1A1     N3

[6 marks]

Not as well done as expected with most successful candidates using a graphical approach. Some candidates confused t and h and subsequently stated the values of t for which the water depth was either at a maximum and a minimum. Some candidates simply gave the maximum and minimum coordinates without stating the maximum and minimum depths.

In part (b), a large number of candidates left out t = 24 from their final answer. A number of candidates experienced difficulties solving the inequality via algebraic means. A number of candidates specified incorrect intervals or only one correct interval.