By sketching a suitable pair of graphs, show that the equation cot2x = 2 sin 2x-1 has exactly one root in the interval 0 < x < π.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [1] (a) Sketch a relevant graph, e.g. y = cot 2x. Scales must be correct if seen. For y = cot 2x, correct intercept on x-axis, asymptotes at x = 0, x = (1/2)π implied. Must be continuous.
2. [1] (a) Sketch a second relevant graph, e.g. y = 2sin2x-1 and justify the given statement. For y = 2sin2x-1, correct shape and position, no incorrect curvature, correct maximum implied, intercept at (0, -1). Need to see enough of both graphs to confirm there is not a second intersection in the given interval.
3. [1] (b) Calculate the values of a relevant expression or pair of expressions at x = 0.4 and x = 0.6. E.g. for f(x) = cot(2x) - 2sin(2x) + 1, show a sign change (f(0.4)>0 and f(0.6)<0) or by comparing values like cot(2x) and 2sin(2x)-1. M0 if working in degrees.
4. [1] (b) Justify the given statement with correct calculated values. Values to 2 sf or better. Accept values with statement f (0.4)×f(0.6)<0.
5. [1] (c) Use the iterative process xn+1 = (1/2) tan⁻¹(1/(2sin2xn - 1)) correctly at least once. M0 if working in degrees. Need to get as far as a second iteration completed.
6. [1] (c) Obtain final answer 0.49.
7. [1] (c) Show sufficient iterations to 4 d.p. to justify 0.49 to 2 d.p. or show that there is a sign change in (0.485, 0.495). E.g. -0.0157 < 0, 0.355 > 0.