FP1 January 2011 Q3
3. \[\mathrm{f}(x) = 5x^2 - 4x^{\frac{3}{2}} - 6, \quad x \geqslant 0\]
The root \(\alpha\) of the equation \(\mathrm{f}(x) = 0\) lies in the interval \([1.6,\ 1.8]\).
(a) Use linear interpolation once on the interval \([1.6,\ 1.8]\) to find an approximation to \(\alpha\).
Give your answer to 3 decimal places. (4)
Give your answer to 3 decimal places. (4)
(b) Differentiate \(\mathrm{f}(x)\) to find \(\mathrm{f}'(x)\). (2)
(c) Taking 1.7 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 3 decimal places. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = 5x^2 - 4x^{\frac{3}{2}} - 6, \quad x \geqslant 0\) | |
| \(\mathrm{f}(1.6) = -1.29543081\ldots\) awrt −1.30 | B1 |
| \(\mathrm{f}(1.8) = 0.5401863372\ldots\) awrt 0.54 | B1 |
| \(\dfrac{\alpha - 1.6}{\text{"}1.29543081\ldots\text{"}} = \dfrac{1.8 - \alpha}{\text{"}0.5401863372\ldots\text{"}}\) \(\alpha = 1.6 + \left(\dfrac{\text{"}1.29543081\ldots\text{"}}{\text{"}0.5401863372\ldots\text{"} + \text{"}1.29543081\ldots\text{"}}\right)0.2\) Correct linear interpolation method with signs correct. Can be implied by working below. | M1 |
| \(= 1.741143899\ldots\) awrt 1.741 Correct answer seen 4/4 | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}'(x) = 10x - 6x^{\frac{1}{2}}\) At least one of \(\pm ax\) or \(\pm bx^{\frac{1}{2}}\) correct. | M1 |
| Correct differentiation. | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1.7) = -0.4161152711\ldots\) \(\mathrm{f}(1.7) = \) awrt \(-0.42\) | B1 |
| \(\mathrm{f}'(1.7) = 9.176957114\ldots\) \(\mathrm{f}'(1.7) = \) awrt 9.18 | B1 |
| \(\alpha_2 = 1.7 - \left(\dfrac{\text{"}-0.4161152711\ldots\text{"}}{\text{"}9.176957114\ldots\text{"}}\right)\) Correct application of Newton-Raphson formula using their values. | M1 |
| \(= 1.745343491\ldots\) | |
| \(= 1.745\) (3dp) 1.745 Correct answer seen 4/4 | A1 cao |
| (4) | |
| [10] |