The equation cot (x/2) = 3x has one root in the interval 0 < x < π, denoted by α.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. M1 Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 1 [Need to evaluate at both points, but M1 still available if one value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using g(x) - f(x), there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient.]
2. A1 Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. [e.g. 3.92 > 1.5, 1.83 < 3 or 2.42 > 0, -1.17 < 0. Can all be in symbols – an explanation in words is not required.]
3. M1 State x = (1/3)(x + 4 tan⁻¹(1/(3x))) [Or rearrange cot(x/2) = 3x as far as x/2 = tan⁻¹(1/(3x))]
4. A1 Rearrange to the given equation cot(x/2) = 3x [Or continue rearrangement to x = (1/3)(x + 4 tan⁻¹(1/(3x))) and state iterative formula of x_(n+1) = (1/3)(x_n + 4 tan⁻¹(1/(3x_n))) AG]
5. M1 Use the iterative process correctly at least once [Obtain one value and substitute that back in to obtain a second value. Working in degrees is M0.]
6. A1 Obtain final answer 0.79 [Must be to 2 d.p.]
7. A1 Show sufficient iterations to at least 4 d.p. to justify 0.79 to 2 d.p. or show there is a sign change in the interval (0.785, 0.795) [e.g. 1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945 or 0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945 or 0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945 . Condone truncation. Allow recovery. Condone minor differences in the final d.p. If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c).]