FP1 January 2012 Q2
2.
(a) Show that \(\mathrm{f}(x) = x^4 + x - 1\) has a real root \(\alpha\) in the interval \([0.5,\ 1.0]\). (2)
(b) Starting with the interval \([0.5,\ 1.0]\), use interval bisection twice to find an interval of width 0.125 which contains \(\alpha\). (3)
(c) Taking 0.75 as a first approximation, apply the Newton Raphson process twice to \(\mathrm{f}(x)\) to obtain an approximate value of \(\alpha\). Give your answer to 3 decimal places. (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = x^4 + x - 1\) | |
| \(\mathrm{f}(0.5) = -0.4375\ \left(-\dfrac{7}{16}\right)\) \(\mathrm{f}(1) = 1\) Either any one of \(\mathrm{f}(0.5) = \) awrt \(-0.4\) or \(\mathrm{f}(1) = 1\) | M1 |
| Sign change (positive, negative) (and \(\mathrm{f}(x)\) is continuous) therefore (a root) \(\alpha\) is between \(x = 0.5\) and \(x = 1.0\) \(\mathrm{f}(0.5) = \) awrt \(-0.4\) and \(\mathrm{f}(1) = 1\), sign change and conclusion | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(0.75) = 0.06640625\ \left(\dfrac{17}{256}\right)\) Attempt \(\mathrm{f}(0.75)\) | M1 |
| \(\mathrm{f}(0.625) = -0.222412109375\ \left(-\dfrac{911}{4096}\right)\) \(\mathrm{f}(0.75) = \) awrt 0.07 and \(\mathrm{f}(0.625) = \) awrt \(-0.2\) | A1 |
| \(0.625 \leqslant \alpha \leqslant 0.75\) \(0.625 \leqslant \alpha \leqslant 0.75\) or \(0.625 < \alpha < 0.75\) or \([0.625,\ 0.75]\) or \((0.625,\ 0.75)\). or equivalent in words. | A1 |
| (3) |
Notes
In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}'(x) = 4x^3 + 1\) Correct derivative (May be implied later by e.g. \(4(0.75)^3 + 1\)) | B1 |
| \(x_1 = 0.75\) | |
| \(x_2 = 0.75 - \dfrac{f(0.75)}{f'(0.75)} = 0.75 - \dfrac{0.06640625}{2.6875\,(43/16)}\) Attempt Newton-Raphson | M1 |
| \(x_2 = 0.72529(06976\ldots) = \dfrac{499}{688}\) Correct first application – a correct numerical expression e.g. \(0.75 - \dfrac{17/256}{43/16}\) or awrt 0.725 (may be implied) | A1 |
| \(x_3 = 0.724493\left(\dfrac{499}{688} - \dfrac{0.002015718978}{2.562146811}\right)\) Awrt 0.724 | A1 |
| \((\alpha) = 0.724\) cao | A1 |
| (5) | |
| (10 marks) |
Notes
A final answer of 0.724 with evidence of NR applied twice with no incorrect work should score 5/5