C3 June 2005 Q4
4. \[\mathrm{f}(x) = 3\mathrm{e}^x - \tfrac{1}{2}\ln x - 2, \quad x \gt 0.\]
(a) Differentiate to find \(\mathrm{f}'(x)\). (3)
The curve with equation \(y = \mathrm{f}(x)\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
(b) Show that \(\alpha = \tfrac{1}{6}\mathrm{e}^{-\alpha}\). (2)
The iterative formula\[x_{n+1} = \tfrac{1}{6}\mathrm{e}^{-x_n}, \quad x_0 = 1,\]is used to find an approximate value for \(\alpha\).
(c) Calculate the values of \(x_1\), \(x_2\), \(x_3\) and \(x_4\), giving your answers to 4 decimal places. (2)
(d) By considering the change of sign of \(\mathrm{f}'(x)\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places. (2)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}'(x) = 3\mathrm{e}^x - \dfrac{1}{2x}\) | M1 A1 A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(3\mathrm{e}^x - \dfrac{1}{2x} = 0\) | M1 |
| \(\Rightarrow 6\alpha\mathrm{e}^{\alpha} = 1 \quad \Rightarrow \alpha = \tfrac{1}{6}\mathrm{e}^{-\alpha}\) (*) | A1 cso |
| (2) |
| Scheme | Marks |
|---|---|
| \(x_1 = 0.0613\ldots,\ x_2 = 0.1568\ldots,\ x_3 = 0.1425\ldots,\ x_4 = 0.1445\ldots\) | M1 A1 |
| (2) |
Notes
[M1 at least \(x_1\) correct, A1 all correct to 4 d.p.]
| Scheme | Marks |
|---|---|
| Using \(\mathrm{f}'(x) = 3\mathrm{e}^x - \dfrac{1}{2x}\) with suitable interval e.g. \(\mathrm{f}'(0.14425) = -0.0007\) \(\phantom{\text{e.g. }}\mathrm{f}'(0.14435) = +0.002(1)\) | M1 |
| Accuracy (change of sign and correct values) | A1 |
| (2) | |
| (9 marks) |