The equation of a curve is such that d²y/dx² = 24/x³ . It is given that the curve has a stationary point at (-2, 19).

Mark scheme

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

1. [1] B1 May be unsimplified. dy/dx = -24x⁻² / (-2) + c or 12x⁻² + c
2. [1] M1 Substituting x = -2 into their dy/dx of the form kx⁻² + c, and setting = 0. Condone missing bracket. 12/(-2)² + c = 0 [=> c = -3]
3. [1] A1 -24/(-2) must be simplified to 12, but accept c=-3 as the final answer. dy/dx = 12x⁻² - 3 or 12/x² - 3
4. [1] B1 May be done by inspection or calculator, but must be WWW. Condone ±2. 12/x² - 3 = 0 => x² = 4 => x = ±2 => x = 2
5. [1] B1 WWW. Or other valid method. Ignore any working for x = -2. d²y/dx² = 24/(2)³ or -3 or <0 maximum
6. [1] B1FT May be unsimplified. FT their nonzero c. y = -12x⁻¹ + their cx [+d]
7. [1] M1 Substituting (-2, 19) into their integrated expression (defined by having at least one correct power) and including +d. 19 = -12/(-2) - 3(-2) + d [=> d = 7]
8. [1] A1 Accept d = 7 as final answer if y = ... seen previously. Allow 'f(x) = ' in place of 'y = '. y = -12/x - 3x + 7
9. [1] *M1 Setting their dy/dx of the form kx⁻² + c = -9/4, and correctly solving for x from their equation where x² >0. 12/x² - 3 = -9/4
10. [1] B1 Gradient of normal = 4/9
11. [1] DM1 For using their stated positive x and their stated y and a changed gradient to correctly form a line equation; condone -4/9 or 9/4. y+8/(x-4) = 4/9
12. [1] A1 OE in correct form. 4x - 9y - 88 = 0