The curve with equation e²ˣ – 18x + y³ + y = 11 has a stationary point at (p, q).

Mark scheme

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

1. [B1] Differentiate y³ to obtain 3y² dy/dx
2. [M1] Differentiate complete equation to produce at least one term involving dy/dx using implicit differentiation.
3. [A1] Obtain 2e²ˣ−18+3y² dy/dx + dy/dx = 0
4. [A1] Substitute dy/dx = 0 to obtain either p = (1/2)ln9 or p = ln3
5. [M1] Substitute value of p in original equation and rearrange as far as y³ = ... or q³ = ... Allow in terms of ln9.
6. [A1] Obtain given result q=³√(2+18ln3-q) or y=³√(2+18ln3-y) with sufficient detail AG
7. [M1] Consider sign of q-³√(2+18ln3-q) or equivalent for 2.5 and 3.0
8. [A1] Obtain −0.18... and 0.34... with sufficient detail and justify conclusion OE
9. [M1] Use iteration process correctly at least once
10. [A1] Obtain final answer q=2.673 Answer required to exactly 4 s.f.
11. [A1] Show sufficient iterations to 6 sf to justify answer or show sign change in the interval [2.6725, 2.6735]