June 2024 Paper 1 Q8
8
(a) Find the first three terms, in ascending powers of \(x\), in the expansion of\[(2 + kx)^5\]where \(k\) is a positive constant. [3 marks]
(b) Hence, given that the coefficient of \(x\) is four times the coefficient of \(x^2\), find the value of \(k\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains the correct constant term 32 | B1 | 1.1b |
| Obtains \(5 \times 16kx\) or \(10 \times 8(kx)^2\) OE PI by \(\dfrac{5k}{2}x\) or \(\dfrac{5 \times 4}{2!}\left(\dfrac{kx}{2}\right)^2\) | M1 | 1.1a |
| Obtains \(32 + 80kx + 80k^2x^2\ (+\ldots)\) Accept list of correct terms. No ISW If more terms are given it must be obvious which are their first three terms. | A1 | 1.1b |
| (3) |
Typical solution
\[(2 + kx)^5 = 32 + 80kx + 80k^2x^2 + \ldots\]| Scheme | Marks | AO |
|---|---|---|
| Forms the equation their \(Ak = 4 \times\) their \(Bk^2\) OE May recover if \(x\) is initially included. | M1 | 3.1a |
| Deduces \(k = \dfrac{1}{4}\) only Or their \(k =\) their \(\dfrac{A}{4B}\) Justification of rejection \(k = 0\) not required. | A1F | 2.2a |
| (2) | ||
| (5 marks) |
Typical solution
\[80k = 4 \times 80k^2\]\[k = 0 \text{ or } \frac{1}{4}\]\[k = \frac{1}{4}\]Since \(k \gt 0\)