C4 June 2005 Q1
1. Use the binomial theorem to expand
\[\surd(4 - 9x), \qquad |x| \lt \frac{4}{9},\]
in ascending powers of \(x\), up to and including the term in \(x^3\), simplifying each term. (5)
| Scheme | Marks |
|---|---|
| \((4 - 9x)^{\frac{1}{2}} = 2\left(1 - \dfrac{9x}{4}\right)^{\frac{1}{2}}\) | B1 |
| \(= 2\left(1 + \dfrac{\frac{1}{2}}{1}\left(-\dfrac{9x}{4}\right) + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{1.2}\left(-\dfrac{9x}{4}\right)^2 + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{1.2.3}\left(-\dfrac{9x}{4}\right)^3 + \ldots\right)\) | M1 |
| \(= 2\left(1 - \dfrac{9}{8}x - \dfrac{81}{128}x^2 - \dfrac{729}{1024}x^3 + \ldots\right)\) | |
| \(= 2 - \dfrac{9}{4}x, - \dfrac{81}{64}x^2, - \dfrac{729}{512}x^3 + \ldots\) | A1, A1, A1 |
| (5) | |
| (5 marks) |
Notes
Note The M1 is gained for \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{1.2}(\ \ldots\ )^2\) or \(\dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{1.2.3}(\ \ldots\ )^3\)
Special Case
If the candidate reaches \(= 2\left(1 - \dfrac{9}{8}x - \dfrac{81}{128}x^2 - \dfrac{729}{1024}x^3 + \ldots\right)\) and goes no further allow A1 A0 A0