June 2024 Paper 1 Q13
13 The curve with equation \(y = px + \dfrac{8}{x^2} + q\), where \(p\) and \(q\) are constants, has a stationary point at (2, 7).
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = p - 16x^{-3}\) | M1 | 3.1a |
| \(p - 16x^{-3} = 0\) \(p - 16 \times 2^{-3} = 0\) | M1 | 1.1 |
| \(p = 2\) | A1 | 1.1 |
| When \(x = 2, y = 7\) so \(7 = 2p + \dfrac{8}{2^2} + q\) So \(2p + q = 5\) | M1 | 3.1a |
| so \(q = 1\) | A1 | 1.1 |
| [5] |
Notes
M1: Uses negative powers to attempt to find expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
A term in \(x^{-3}\) needed for this mark.
M1: Equates their derivative to zero and attempts to solve using \(x = 2\)
M1: Uses the given coordinates to link \(p\) and \(q\), or their \(p\) and \(q\) in the Cartesian equation
A1: FT their \(p\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 48x^{-4} = \left[\dfrac{48}{x^4}\right]\) | B1 | 1.1 |
| [1] |
Notes
B1: Allow even from wrong values of \(p\) and \(q\)
| Scheme | Marks | AO |
|---|---|---|
| At (2, 7) \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 48 \times 2^{-4} = [3] \gt 0\) | M1 | 1.1 |
| So the stationary point is a minimum | A1 | 2.2a |
| [2] |
Notes
M1: Substitutes \(x = 2\) into their (b) Need not be fully evaluated
Also allow for arguing the \(48x^{-4}\) is always positive.
Do not allow for gradient evaluated on either side of (2, 7)
A1: Clear statement using the positivity of the second derivative from correct working in (c). FT their second derivative (stating maximum if their value is negative)