a) Add the diameter to the height of the bottom of the wheel, such as \( 13 + 122 \). The maximum height above the ground of the seat is 135 metres. [2]

bi) Calculate the period by dividing the total minutes in an hour by the number of rotations, \( \frac{60}{2.4} \). The period of \( h \) is 25 minutes. [1]

bii) The exact value of \( b \) is \( \frac{2\pi}{25} \) (approximately \( 0.08\pi \)). [1]

c) Method 1: Use the difference between the maximum height and the midline, such as \( 135 – 74 = 61 \). Since the seat starts at the bottom and the cosine function is negative at the start, \( a = -61 \). Alternatively, calculate the amplitude as \( \frac{135 – 13}{2} = 61 \), and since the minimum is at the start, \( a = -61 \). [3]

Method 2: Substitute a valid point into the equation for \( h \), such as \( 135 = 74 + a\cos\left(\frac{2\pi \times 12.5}{25}\right) \), which simplifies to \( 135 = 74 + a\cos(\pi) \). Solving gives \( a = -61 \). Alternatively, use the minimum point \( 13 = 74 + a\cos(0) \), leading to \( a = -61 \). [3]

d) The graph should show a domain from 0 to 50, a range from 13 to 135, a sinusoidal shape with 2 cycles, and maximum/minimum points at approximately correct positions starting at the minimum of 13 at \( t = 0 \) and reaching 135 at the peak. [4]

e) Set up the inequality \( h \ge 105 \) and solve \( 105 = 74 + (-61)\cos bt \), which simplifies to \( \cos bt = \frac{105 – 74}{-61} = \frac{31}{61} \). The solutions for one rotation (25 minutes) occur at \( t \) values where the height is at least 105, approximately 8.371 and 16.628 minutes (and their equivalents in the cycle). The total time above 105 metres in one rotation is the difference, \( 16.628 – 8.371 = 8.257 \) minutes. The probability is \( \frac{8.257}{25} \approx 0.330 \). [5]