A curve is such that d²y/dx² = 6/x⁴ - 5/x³. It is given that the curve has a stationary point at (1/2, 9).

Mark scheme

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

1. [M1] 6(1/2)⁻⁴ - 5(1/2)⁻³ Substitute x = ½ and evaluate second derivative.
2. [A1] 96-40[=56]>0 ⇒ Minimum CWO Evidence and conclusion. SC B1 for d²y/dx² >0 without sight of 96-40 or 56.
3. [B1] ∫(6x⁻⁴ - 5x⁻³)dx = -2x⁻³ + 5/2 x⁻² [+C₁] B1 for each correct { }.
4. [M1] 0 = -2(1/2)⁻³ + 5/2 (1/2)⁻² [+C₁] Substitute x = ½ into two terms of an integrated expression (at least one correct power), now with c₁, and equate to 0 to find c₁.
5. [A1] C₁ = 6
6. [M1] ∫(-2x⁻³ + 5/2 x⁻² + 6)dx = x⁻² - 5/2 x⁻¹ + 6x [+C₂] Integration of their dy/dx to produce at least two terms with correct powers, k₁,k₂ ≠ 0.
7. [M1] 9 = (1/2)⁻² - 5/2 (1/2)⁻¹ + 6(1/2) [+C₂] Substitute (½, 9) into integrated expression (at least two correct powers) to find c₂.
8. [A1] y = x⁻² - 5/2 x⁻¹ + 6x + 7 OE Condone their final answer being c₂ = 7 if a completely correct simplified expression for the equation containing c₂ has been stated previously.