June 2023 Paper 3 Q3
3 The cubic polynomial \(\mathrm{f}(x)\) is defined by \(\mathrm{f}(x) = x^3 + px + q\), where \(p\) and \(q\) are constants.
The curve \(y = \mathrm{f}(x)\) is translated by the vector \(\begin{pmatrix} 2 \\ -3 \end{pmatrix}\).
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathrm{f}(x) = x^3 + px + q \Rightarrow \mathrm{f}'(x) = 3x^2 + p\) | M1 | 1.1 |
| \(\mathrm{f}'(2) = 13 \Rightarrow p = 1\) | A1 | 1.1 |
| [2] | ||
| (ii) \(2^3 + 2p + q = 0\) | M1 | 1.1a |
| \(q = -10\) | A1 | 1.1 |
| [2] |
Notes
(a)(i)
M1: Attempt at differentiating \(\mathrm{f}(x)\) with at least one non-zero term correct
M0 for \(\mathrm{f}'(x) = x^2 + p + \dfrac{q}{x}\)
A1: Correct value for \(p\)
(a)(ii)
M1: Substituting \(x = 2\) into \(\mathrm{f}(x)\) and equating to 0 or for correctly re-writing as \((x - 2)(x^2 + 2x + 4 + p)\) (with \(p\) or their value of \(p\) from (a)(i))
Could be in terms of \(q\) only e.g., \(2^3 + 2 \times \mathit{their}(p) + q = 0\)
Possibly seen in an attempt at long division
A1: Correct value for \(q\)
| Scheme | Marks | AO |
|---|---|---|
| \(y = (x - 2)^3 + p(x - 2) + q - 3\) | B1 B1 | 1.1 1.1 |
| \(y = x^3 + 3x^2(-2) + 3x(-2)^2 + (-2)^3 + px - 2p + q - 3\) | M1 | 1.1 |
| \(y = (x^3 - 6x^2 + 12x - 8) + x - 2 - 10 - 3\) \(y = x^3 - 6x^2 + 13x - 23\) | A1 | 2.2a |
| [4] |
Notes
B1: Substituting \((x \pm 2)\) into both \(x\) terms of \(y = \mathrm{f}(x)\)
B1: Subtracting 3 (oe) at some stage
Allow with \(p\) and \(q\) or with incorrect values of \(p\) and/or \(q\) – note that using 2 vertically and/or 3 horizontally cannot be treated as a MR
M1: Correct expansion of \((x \pm 2)^3\). Can be unsimplified for M1. If correct bracketing not seen then must be implied by later working
A1: cao – condone just the expression \(x^3 - 6x^2 + 13x - 23\) (so do not need to see \(y = \ldots\))
For reference: \(a = -6\), \(b = 13\), \(c = -23\)