C3 January 2008 Q3
3.\[\mathrm{f}(x) = \ln(x + 2) - x + 1, \qquad x > -2,\ x \in \mathbb{R}.\]
(a) Show that there is a root of \(\mathrm{f}(x) = 0\) in the interval \(2 < x < 3\). (2)
(b) Use the iterative formula\[x_{n+1} = \ln(x_n + 2) + 1, \quad x_0 = 2.5\]to calculate the values of \(x_1\), \(x_2\) and \(x_3\) giving your answers to 5 decimal places. (3)
(c) Show that \(x = 2.505\) is a root of \(\mathrm{f}(x) = 0\) correct to 3 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(2) = 0.38\ \ldots\) \(\mathrm{f}(3) = -0.39\ \ldots\) | M1 |
| Change of sign (and continuity) \(\Rightarrow\) root in \((2, 3)\ \ast\) cso | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(x_1 = \ln 4.5 + 1 \approx 2.50408\) | M1 |
| \(x_2 \approx 2.50498\) | A1 |
| \(x_3 \approx 2.50518\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Selecting \([2.5045, 2.5055]\), or appropriate tighter range, and evaluating at both ends. \(\mathrm{f}(2.5045) \approx 6\times 10^{-4}\) \(\mathrm{f}(2.5055) \approx -2\times 10^{-4}\) | M1 |
| Change of sign (and continuity) \(\Rightarrow\) root \(\in (2.5045, 2.5055)\) \(\Rightarrow\) root \(= 2.505\) to 3 dp \(\ast\) cso | A1 |
| (2) | |
| (7 marks) |
Notes
Note: The root, correct to 5 dp, is 2.50524