C4 January 2010 Q1
1.
(a) Find the binomial expansion of \[\sqrt{(1 - 8x)}, \qquad |x| < \frac{1}{8},\] in ascending powers of \(x\) up to and including the term in \(x^3\), simplifying each term. (4)
(b) Show that, when \(x = \dfrac{1}{100}\), the exact value of \(\sqrt{(1 - 8x)}\) is \(\dfrac{\sqrt{23}}{5}\). (2)
(c) Substitute \(x = \dfrac{1}{100}\) into the binomial expansion in part (a) and hence obtain an approximation to \(\sqrt{23}\). Give your answer to 5 decimal places. (3)
| Scheme | Marks |
|---|---|
| \((1 - 8x)^{\frac{1}{2}} = 1 + \left(\tfrac{1}{2}\right)(-8x) + \dfrac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2}(-8x)^2 + \dfrac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}(-8x)^3 + \ldots\) | M1 A1 |
| \(= 1 - 4x - 8x^2;\ - 32x^3 - \ldots\) | A1; A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\sqrt{(1 - 8x)} = \sqrt{\left(1 - \dfrac{8}{100}\right)}\) | M1 |
| \(= \sqrt{\dfrac{92}{100}} = \sqrt{\dfrac{23}{25}} = \dfrac{\sqrt{23}}{5}\ \ \ast\) cso | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(1 - 4x - 8x^2 - 32x^3 = 1 - 4(0.01) - 8(0.01)^2 - 32(0.01)^3\) \(= 1 - 0.04 - 0.0008 - 0.000\,032 = 0.959\,168\) | M1 |
| \(\sqrt{23} = 5 \times 0.959\,168\) | M1 |
| \(= 4.795\,84\) cao | A1 |
| (3) | |
| (9 marks) |