A curve is such that d²y/dx² = 24/x³ - 4. The curve has a stationary point at P where x = 2.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2014). The mark or mark type is shown in brackets.

1. [1] Evaluate d²y/dx² at x = 2: 24/(2)³ - 4 = 24/8 - 4 = 3 - 4 = -1. Since it's negative, the stationary point is a maximum. www
2. [2] Integrate d²y/dx² to find dy/dx: dy/dx = -12x⁻² - 4x + A. oe one per term
3. [1] At a stationary point, dy/dx = 0. So, when x = 2, -12(2)⁻² - 4(2) + A = 0. Attempt at the constant A after ∫n
4. [1] -12/4 - 8 + A = 0 → -3 - 8 + A = 0 → A = 11. co
5. [2] Integrate dy/dx to find y: y = 12x⁻¹ – 2x² + Ax + c. oe Doesn’t need +c, but does need a term A to give “Ax”.
6. [1] Use the point (1, 13) to find c: 13 = 12(1)⁻¹ – 2(1)² + 11(1) + c → 13 = 12 - 2 + 11 + c → 13 = 21 + c → c = -8. Attempt at c after ∫n
7. [1] The equation of the curve is y = 12x⁻¹ – 2x² + 11x - 8. At the stationary point P, x=2, so y = 12(2)⁻¹ - 2(2)² + 11(2) - 8 = 6 - 8 + 22 - 8 = 12. Coordinates of P are (2, 12). co