FP1 June 2016 Q2
2. \[\mathrm{f}(x) = 3x^{\frac{3}{2}} - 25x^{-\frac{1}{2}} - 125, \qquad x > 0\]
(a) Find \(\mathrm{f}^{\prime}(x)\). (2)
The equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) in the interval \([12, 13]\).
(b) Using \(x_0 = 12.5\) as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x)\) to find a second approximation to \(\alpha\), giving your answer to 3 decimal places. (4)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}^{\prime}(x) = \tfrac{9}{2}x^{\frac{1}{2}} + \tfrac{25}{2}x^{-\frac{3}{2}}\) | M1 A1 |
| (2) |
Notes
M1: for attempting differentiation i.e. decrease a power by 1
A1 Accept equivalent expression i.e. condone equivalent fractions.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(12.5) = 0.5115\ldots\) (at least \(0.51\ldots\)) and \(\mathrm{f}^{\prime}(12.5) = 16.1927\ldots\) (at least \(16\ldots\) seen) | B1, B1 |
| \(x_1 = 12.5 - \dfrac{\mathrm{f}(12.5)}{\mathrm{f}^{\prime}(12.5)} = 12.5 - \dfrac{0.5115}{16.1927\ldots} = 12.468\) | M1 A1 |
| (4) | |
| (6 marks) |
Notes
B1: One correct, must be explicitly seen if final answer incorrect, may be implied by correct final answer.
B1: Both correct; must be explicitly seen if final answer incorrect, may be implied by correct final answer.
M1: for attempting Newton- Raphson with their values for f(12.5) and \(\mathrm{f}^{\prime}(12.5)\)
A1: cao correct to 3dp
Newton Raphson used more than once – isw.