Question
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 \(h(t) \geqslant 8\).
▶️Answer/Explanation
Markscheme
(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]
Examiners report
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.