The diagram shows the curve y = x³cos2x for 0 ≤ x ≤ ¼π. The curve has a maximum point at M, where x = p. [Figure 7.1]

Mark scheme

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

1. [M1] Use correct product rule 3x² cos(2x) + x³ (d/dx(cos2x)) OE
2. [A1] Obtain derivative 3x² cos(2x) – 2x³ sin(2x) OE
3. [A1 AG] Equate correct derivative to zero and obtain p = ½ tan⁻¹(3/(2p)) after full and correct working Must reach tan 2p = ... or tan 2x = ... before given answer.
4. [M1] Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 0.7 Must have at least 1 of 2 values correct or 3 of 4 values correct f(p) = p - ½ tan⁻¹(3/(2p)) f(0.5) = -0.124... <0, f(0.7) = 0.132.. > 0 or comparing p and ½ tan⁻¹(3/(2p)) 0.5 < ½ tan⁻¹(3/(2*0.5)) = 0.624... => 0.5 < 0.624... 0.7 > ½ tan⁻¹(3/(2*0.7)) = 0.567... => 0.7 > 0.567
5. [A1] Complete the argument correctly with correct calculated values
6. [M1] Use the iterative formula correctly at least twice
7. [A1] Obtain final answer 0.596
8. [A1] Show sufficient iterations to at least 5 decimal places to justify 0.596 to 3 decimal places or show there is a sign change in the interval (0.5955, 0.5965) E.g. 0.5, 0.62452, 0.58814, 0.59856, 0.59556, 0.59642 OR 0.6, 0.59514, 0.59654, 0.59614, 0.59626 OR 0.7, 0.56708, 0.60467, 0.59380, 0.59693, 0.59603, 0.59629 Can recover from wrong values.