June 2025 Paper 2 Q6
6 A curve has equation \(y = 3x^4 - 4x^3 - 6x^2 + 12x\).
| Scheme | Marks | AO |
|---|---|---|
| (i) | M1 | 1.1 |
| \(\left[\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right] 12x^3 - 12x^2 - 12x + 12\) | A1 | 2.1 |
| [2] | ||
| (ii) \(\left[\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\right] \Rightarrow 12x^3 - 12x^2 - 12x + 12 = 0\) \(\Rightarrow 12\left(x^3 - x^2 - x + 1\right) = 0\) | M1 | 1.1 |
| \(x = 1\) and \(x = -1\) | A1 | 1.1 |
| [2] |
Notes
(a)(i)
M1: Attempt \(\frac{\mathrm{d}y}{\mathrm{d}x}\), \(\geqslant 3\) terms correct
A1: All correct (ISW if candidates go on to divide by 12, but this answer oe must be seen)
(a)(ii)
M1: Setting their \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\) and attempt to solve (i.e. reach one or more values for \(x\)).
May be implied by correct answers
A1: And no others. May be done BC
Sight of \(x = 1\) and \(x = -1\) (and no other solutions) 2/2
ISW any attempt to find corresponding \(y\) values.
| Scheme | Marks | AO |
|---|---|---|
| \(\left[\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right] 36x^2 - 24x - 12 = 0\) oe | M1 | 2.1 |
| \(x = 1\) and \(x = -\frac{1}{3}\) | A1* | 1.1 |
| Only one of these \([x = 1]\) gives a stationary point oe | A1 dep* | 2.2a |
| [3] |
Notes
M1: FT their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) (\(\geqslant 2\) terms correct when differentiating their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) a second time).
May be implied by A1 but if both correct roots not seen must be set \(= 0\) for this mark.
Allow \(x^3 - x^2 - x + 1 \Rightarrow 3x^2 - 2x - 1 = 0\)
A1*: A1 for both correct roots and no others, i.e. points of inflection
A1 dep*: Statement needed. Dependent on previous A1.
Accept e.g.:
- One of these \(\left[x = -\frac{1}{3}\right]\) is not a stationary point A1
- \(x = -\frac{1}{3}\) is not a stationary point A1
- Categorising both points (i.e. \(x = 1\) is stationary and a point of inflection, \(x = -\frac{1}{3}\) is not stationary…) A1
But not “only one of these is a point of inflection” A0