FP1 June 2013 (R) Q3
3. \[\mathrm{f}(x) = \frac{1}{2}x^4 - x^3 + x - 3\]
(a) Show that the equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) between \(x = 2\) and \(x = 2.5\) (2)
(b) Starting with the interval \([2, 2.5]\) use interval bisection twice to find an interval of width 0.125 which contains \(\alpha\). (3)
The equation \(\mathrm{f}(x) = 0\) has a root \(\beta\) in the interval \([-2, -1]\).
(c) Taking \(-1.5\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(\mathrm{f}(x)\) to obtain a second approximation to \(\beta\).
Give your answer to 2 decimal places. (5)
Give your answer to 2 decimal places. (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(2) = -1\) \(\mathrm{f}(2.5) = 3.40625\) Either any one of \(\mathrm{f}(2) = -1\) or \(\mathrm{f}(2.5) =\) awrt 3.4 | M1 |
| Sign change (and \(\mathrm{f}(x)\) is continuous) therefore a root \(\alpha\) exists between \(x = 2\) and \(x = 2.5\) both values correct, sign change and conclusion | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(2.25) = 0.673828125\ \left\{= \tfrac{345}{512}\right\}\ \{\Rightarrow 2 \leqslant \alpha \leqslant 2.25\}\) \(\mathrm{f}(2.25) =\) awrt 0.7 | B1 |
Attempt to find \(\mathrm{f}(2.125)\) | M1 |
| \(\mathrm{f}(2.125) = -0.2752685547\ldots\) \(\Rightarrow 2.125 \leqslant \alpha \leqslant 2.25\) \(\mathrm{f}(2.125) =\) awrt \(-0.3\) with \(2.125 \leqslant \alpha \leqslant 2.25\) or \(2.125 < \alpha < 2.25\) or \([2.125, 2.25]\) or \((2.125, 2.25)\). | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}^{\prime}(x) = 2x^3 - 3x^2 + 1\ \{+ 0\}\) At least two of the four terms differentiated correctly. Correct derivative. | M1 A1 |
| \(\mathrm{f}(-1.5) = 1.40625\ \left(= 1\tfrac{13}{32}\right)\) \(\{\mathrm{f}^{\prime}(-1.5) = -12.5\}\) \(\mathrm{f}(-1.5) =\) awrt 1.41 | B1 |
| \(\beta_2 = -1.5 - \left(\dfrac{\text{"}1.40625\text{"}}{\text{"}-12.5\text{"}}\right)\) Correct application of Newton-Raphson using their values. | M1 |
| \(= -1.3875\ \ \left(= -1\tfrac{31}{80}\right)\) \(-1.3875\) seen as answer to first iteration, award M1A1B1M1 | |
| \(= -1.39\ (2\text{ dp})\) \(-1.39\) | A1 cao |
| (5) | |
| (10 marks) |