June 2024 Paper 3 Q3
3
Determine the value of the constant \(a\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \((3 - 2x)^{-2} = \frac{1}{9}(1 + \ldots)^{-2}\) | B1 | 1.1a |
| \((1 + kx)^{-2} = 1 + (-2)(kx) + \ldots\) | B1FT | 1.1 |
| \(\ldots + \dfrac{(-2)(-3)}{2!}(kx)^2\) | B1FT | 1.1 |
| \((3 - 2x)^{-2} = \frac{1}{9}\left(1 + \frac{4}{3}x + \frac{4}{3}x^2 + \ldots\right)\) | B1 | 1.1 |
| [4] |
Notes
B1: For reference: \(\frac{1}{9}\left(1 - \frac{2}{3}x\right)^{-2}\) - soi
or for \(3^{-2}(1 + \ldots)^{-2}\)
B1FT: Correct first two terms follow through their \(k\) – allow un-simplified
\(k \ne \pm 1, \pm 2\) - if correct \(k = -\frac{2}{3}\)
B1FT: Correct third term following through their \(k\) – allow un-simplified but must imply that the third term contains their \(k^2\) - for correct \(k\) condone \(\frac{2 \times 3}{2!}\left(\frac{2}{3}x\right)^2\) (or similar for their \(k\) if negative)
\(k \ne \pm 1, \pm 2\)
Condone \(\frac{2 \times 3}{2!}(kx)^2\) and 2 for 2!
B1: Or correct equivalent e.g. \(\frac{1}{27}\left(3 + 4x + 4x^2\right)\), \(\frac{1}{9} + \frac{4}{27}x + \frac{4}{27}x^2\), etc.
ISW after correct expansion seen
Ignore higher order terms if found – a correct answer scores all 4 marks www
| Scheme | Marks | AO |
|---|---|---|
| \(|x| \lt \frac{3}{2}\) | B1 | 2.5 |
| [1] |
Notes
B1: oe, for example, \(-\frac{3}{2} \lt x \lt \frac{3}{2}\) - allow \(-\frac{3}{2} \leqslant x \lt \frac{3}{2}\) but not \(-\frac{3}{2} \leqslant x \leqslant \frac{3}{2}\) (or any inequality that includes the \(\frac{3}{2}\)) - ISW once correct inequality seen. Allow \(\left[-\frac{3}{2}, \frac{3}{2}\right)\) or \(\left(-\frac{3}{2}, \frac{3}{2}\right)\) oe but not \(\left[0, \frac{3}{2}\right)\) (or equivalents in set notation)
\(-\frac{3}{2} \lt |x| \lt \frac{3}{2}\) is B0 but \(0 \leqslant |x| \lt \frac{3}{2}\) is B1
Note that \(|2x| \lt 3\) only is B0 (must be in terms of \(x\))
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{a + x}{(3 - 2x)^2} = (a + x)\left(\frac{1}{9} + \frac{4}{27}x + \ldots\right)\) \(= \ldots + \left(\frac{1}{9} + \frac{4}{27}a\right)x + \ldots\) | B1FT | 3.1a |
| \(\frac{4}{3}a + 1 = 0 \Rightarrow a = -\frac{3}{4}\) | B1FT | 2.2a |
| [2] |
Notes
B1FT: Finding correct coefficient of \(x\) or the \(x\) term for their \((p + qx + \ldots)(a + x)\) - FT their \(p\) and \(q\) from part (a) (so their \(x\)-coefficient must be \(p + aq\)). Allow embedded in an expansion e.g. \(= \frac{1}{9}\left(\ldots + \left(\frac{4}{3}a + 1\right)x + \ldots\right)\) or \(= \frac{1}{9}\left(\ldots + \frac{4}{3}ax + x + \ldots\right)\)
This mark can be implied by the correct answer for \(a\) (or on the FT as detailed in the next mark)
B1FT: Follow through \(-\dfrac{\textit{their}\text{ constant term}}{\textit{their}\text{ coefficient of }x}\) from part (a)