FP1 June 2012 Q3
3. \[\mathrm{f}(x) = x^2 + \frac{3}{4\sqrt{x}} - 3x - 7, \quad x > 0\]
A root \(\alpha\) of the equation \(\mathrm{f}(x) = 0\) lies in the interval \([3,\ 5]\).
Taking 4 as a first approximation to \(\alpha\), apply the Newton-Raphson process once to \(\mathrm{f}(x)\) to obtain a second approximation to \(\alpha\). Give your answer to 2 decimal places. (6)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = x^2 + \dfrac{3}{4\sqrt{x}} - 3x - 7,\quad x > 0\) | |
| \(\mathrm{f}(x) = x^2 + \dfrac{3}{4}x^{-\frac{1}{2}} - 3x - 7\) | |
| \(\mathrm{f}'(x) = 2x - \dfrac{3}{8}x^{-\frac{3}{2}} - 3\ \{+\,0\}\) M1: \(x^n \to x^{n-1}\) on at least one term A1: Correct differentiation. | M1A1 |
| \(\mathrm{f}(4) = -2.625 = -\dfrac{21}{8} = -2\dfrac{5}{8}\) or \(4^2 + \dfrac{3}{4\sqrt{4}} - 3 \times 4 - 7\) \(\mathrm{f}(4) = -2.625\) A correct evaluation of f(4) or a correct numerical expression for f(4). This can be implied by a correct answer below but in all other cases, f(4) must be seen explicitly evaluated or as an expression. | B1 |
| \(\mathrm{f}'(4) = 4.953125 = \dfrac{317}{64} = 4\dfrac{61}{64}\) Attempt to insert \(x = 4\) into their \(\mathrm{f}'(x)\). Not dependent on the first M but must be what they think is \(\mathrm{f}'(x)\). | M1 |
| \(\alpha_2 = 4 - \left(\dfrac{\text{"}{-2.625}\text{"}}{\text{"}4.953125\text{"}}\right)\) Correct application of Newton-Raphson using their values. | M1 |
| \(= 4.529968454\ldots\ \left(= \dfrac{1436}{317} = 4\tfrac{168}{317}\right)\) | |
| \(= 4.53\ (2\text{ dp})\) 4.53 cso | A1 cao |
| [6] | |
| 6 marks |
Notes
Note that the kind of errors that are being made in differentiating are sometimes giving 4.53 but the final mark is cso and the final A1 should not be awarded in these cases.
Ignore any further iterations
A correct derivative followed by \(\alpha_2 = 4 - \dfrac{f(4)}{f'(4)} = 4.53\) can score full marks.