Find the height of the seat when \(t = 0\).

The seat first reaches a height of 20 m after \(k\) minutes. Find \(k\).

Calculate the time needed for the seat to complete a full rotation, giving your answer correct to one decimal place.

valid approach     (M1)

eg\(\,\,\,\,\,\)\(h(0),{\text{ }} - 15\cos (1.2 \times 0) + 17,{\text{ }} - 15(1) + 17\)

\(h(0) = 2{\text{ (m)}}\)    A1     N2

[2 marks]

correct substitution into equation     (A1)

eg\(\,\,\,\,\,\)\(20 =  - 15\cos 1.2t + 17,{\text{ }} - 15\cos 1.2k = 3\)

valid attempt to solve for \(k\)     (M1)

eg\(\,\,\,\,\,\), \(\cos 1.2k =  - \frac{3}{{15}}\)

1.47679

\(k = 1.48\)     A1     N2

[3 marks]

recognize the need to find the period (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)next \(t\) value when \(h = 20\)

correct value for period     (A1)

eg\(\,\,\,\,\,\)\({\text{period}} = \frac{{2\pi }}{{1.2}},{\text{ }}5.23598,{\text{ }}6.7--1.48\)

5.2 (min) (must be 1 dp)     A1     N2

[3 marks]

Candidates did quite well at part a). Most substituted correctly but considered \(\cos 0 = 0\), obtaining an incorrect answer of 17.

Most candidates understood that they needed to solve \(h(t) = 20\), but could not do it. A considerable number of students tried to solve the equation algebraically and the most common errors were to obtain \(\cos k = \frac{{ - 0.2}}{{1,2}}\) or \(k = \frac{{ - 3}}{{15\cos 1.2}}\).

Part (c) proved difficult as many students had difficulties recognizing they needed to find the period of the function and many who could, did not round the final answer to one decimal place.