In the diagram, A is the mid-point of the semicircle with centre O and radius r. A circular arc with centre A meets the semicircle at B and C. The angle OAB is equal to x radians. The area of the shaded region bounded by AB, AC and the arc with centre A is equal to half the area of the semicircle. [Figure 6.1]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [B1] Correct use of trigonometry to obtain AB = 2r cos x (AG)
2. [M1] Use correct method for finding the area of the sector and the semicircle and form an equation in x (½xπr² = ½(2r cos x)² 2x)
3. [A1] Obtain x = cos⁻¹ √(π / (16x)) correctly (AG) (Via correct simplification e.g. from cos²x = π / (16x))
4. [M1] Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 (Must be working in radians. e.g. x=1 → 1.11, x=1.5 → 1.20; Accept f(1)=1.11, f(1.5)=1.20; f(x) = x-cos⁻¹√(π/(16x)): f(1)=-0.111, f(1.5)=0.3; f(x) = cos x - √(π/(16x)): f(1)=0.097, f(1.5)=-0.291; For 16x cos²x - π: f(1)=1.529, f(1.5)=-3.02. Must find values. M1 if at least one value correct)
5. [A1] Correct values and complete the argument correctly
6. [M1] Use x_(n+1) = cos⁻¹ √(π / (16x_n)) correctly at least twice (Must be working in radians. 1, 1.11173, 1.13707, 1.14225, 1.14329, 1.14349, 1.14354, 1.14354; 1.25, 1.16328, 1.14742, 1.14432, 1.14370; 1.5, 1.20060, 1.15447, 1.14570, 1.14397, 1.14363)
7. [A1] Obtain final answer 1.144
8. [A1] Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p. or show there is a sign change in the interval (1.1435, 1.1445)