June 2025 Paper 1 Q8
8
| Scheme | Marks | AO |
|---|---|---|
| \(\left(1 + \frac{1}{2}x\right)^{-3} = 1 + (-3)\left(\frac{1}{2}x\right)\) | B1 | 1.1 |
| \(+\dfrac{(-3)(-4)}{2}\left(\frac{1}{2}x\right)^2\) | M1 | 1.1 |
| A1 | 1.1 | |
| \((2 + x)^{-3} = \frac{1}{8}\left(1 - \frac{3}{2}x + \frac{3}{2}x^2\right) = \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2\) | B1FT | 1.1 |
| [4] |
Notes
B1: Correct first two terms. Possibly unsimplified
M1: Attempt third term. Must be expanding \(\left(1 + \frac{1}{2}x\right)^{-3}\). Allow \(\frac{1}{2}x^2\) for \(\left(\frac{1}{2}x\right)^2\)
A1: Obtain correct third term. Possibly unsimplified
B1FT: Multiply their three term expansion by \(\frac{1}{8}\). Bracket expanded and coefficients simplified
Terms could be listed or summed
If B1M1A1 awarded, but attempt to simplify then goes wrong, B1FT is not also awarded
ISW once correct expansion seen
| Scheme | Marks | AO |
|---|---|---|
| \((1 + 4x)^{\frac{1}{2}} = 1 + \left(\frac{1}{2}\right)(4x) + \dfrac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(4x)^2\) | M1 | 3.1a |
| \(= 1 + 2x - 2x^2\) | A1 | 1.1 |
| \(\left(1 + 2x - 2x^2\right)\left(\frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2\right)\) \(= \frac{1}{8} - \frac{3}{16}x + \frac{3}{16}x^2 + \frac{1}{4}x - \frac{3}{8}x^2 - \frac{1}{4}x^2\) | M1 | 1.1 |
| \(\frac{1}{8} + \frac{1}{16}x - \frac{7}{16}x^2\) | A1 | 1.1 |
| [4] |
Notes
M1: Attempt expansion of \((1 + 4x)^{\frac{1}{2}}\). To obtain \(1 + 2x + kx^2\) (allow unsimplified)
A1: Obtain correct, simplified, expansion. Ignore any terms beyond \(x^2\)
M1: Attempt product of their two expansions ie their answer to (a) and their attempt at \((1 + 4x)^{\frac{1}{2}}\)
‘Hence’ so M0 for division attempts
To obtain attempts at the six relevant terms (ie 1 constant term, 2 terms in \(x\) and 3 terms in \(x^2\), but not necessarily all correct)
Ignore higher powers
A1: Obtain correct three terms. Terms could be listed or summed
| Scheme | Marks | AO |
|---|---|---|
| \(\left(1 + \frac{1}{2}x\right)^{-3}\) requires \(\left|\frac{1}{2}x\right| \lt 1\) hence \(|x| \lt 2\) \((1 + 4x)^{\frac{1}{2}}\) requires \(|4x| \lt 1\) hence \(|x| \lt \frac{1}{4}\) | B1* | 2.5 |
| Select tighter condition, hence \(|x| \lt \frac{1}{4}\) | B1dep* | 2.3 |
| [2] |
Notes
B1*: Both conditions correct oe eg \(-2 \lt x \lt 2\) and \(-\frac{1}{4} \lt x \lt \frac{1}{4}\)
Allow \(\left|\frac{1}{2}x\right| \lt 1\) and/or \(|4x| \lt 1\) oe ie conditions on \(|kx|\) not \(|x|\)
\(|x| \lt \frac{1}{4}\) could also be \(|x| \leqslant \frac{1}{4}\) oe, as \(n \gt 0\)
B1dep*: Correct conclusion; reason needed that identifies that \(|x| \lt \frac{1}{4}\) is contained within \(|x| \lt 2\)
eg \(|x| \lt \frac{1}{4}\) is a subset of \(|x| \lt 2\)
eg nested interval
eg \(\frac{1}{4} \lt 2\)
Must now be \(|x| \lt \frac{1}{4}\) not \(|4x| \lt 1\)
B0 for vague statements such as ‘so that both are valid’ without any further clarification
Or \(|x| \leqslant \frac{1}{4}\)