FP1 January 2010 Q2
2. \[\mathrm{f}(x) = 3x^2 - \frac{11}{x^2}\]
(a) Write down, to 3 decimal places, the value of \(\mathrm{f}(1.3)\) and the value of \(\mathrm{f}(1.4)\). (1)
The equation \(\mathrm{f}(x) = 0\) has a root \(\alpha\) between 1.3 and 1.4
(b) Starting with the interval \([1.3,\ 1.4]\), use interval bisection to find an interval of width 0.025 which contains \(\alpha\). (3)
(c) Taking 1.4 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm{f}(x)\) to obtain a second approximation to \(\alpha\), giving your answer to 3 decimal places. (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1.3) = -1.439\) and \(\mathrm{f}(1.4) = 0.268\) (allow awrt) | B1 |
| (1) |
Notes
(a) Both answers required for B1. Accept anything that rounds to 3dp values above.
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(1.35) \lt 0 \ \ (-0.568\ldots) \quad \Rightarrow \quad 1.35 \lt \alpha \lt 1.4\) | M1 A1 |
| \(\mathrm{f}(1.375) \lt 0 \ \ (-0.146\ldots) \quad \Rightarrow \quad 1.375 \lt \alpha \lt 1.4\) | A1 |
| (3) |
Notes
(b) f(1.35) or awrt −0.6 M1
(f(1.35) and awrt −0.6) AND (f(1.375) and awrt −0.1) for first A1
\(1.375 \lt \alpha \lt 1.4\) or expression using brackets or equivalent in words for second A1
| Scheme | Marks |
|---|---|
| \(\mathrm{f}'(x) = 6x + 22x^{-3}\) | M1 A1 |
| \(x_1 = x_0 - \dfrac{\mathrm{f}(x_0)}{\mathrm{f}'(x_0)} = 1.4 - \dfrac{0.268}{16.417}, \qquad = 1.384\) | M1 A1, A1 |
| (5) | |
| [9] |
Notes
(c) One term correct for M1, both correct for A1
Correct formula seen or implied and attempt to substitute for M1
awrt 16.4 for second A1 which can be implied by correct final answer
awrt 1.384 correct answer only A1