C3 June 2011 Q2
2. \[\mathrm{f}(x) = 2\sin(x^2) + x - 2, \qquad 0 \leqslant x \lt 2\pi\]
(a) Show that \(\mathrm{f}(x) = 0\) has a root \(\alpha\) between \(x = 0.75\) and \(x = 0.85\) (2)
The equation \(\mathrm{f}(x) = 0\) can be written as \(x = \left[\arcsin(1 - 0.5x)\right]^{\frac{1}{2}}\).
(b) Use the iterative formula \[x_{n+1} = \left[\arcsin(1 - 0.5x_n)\right]^{\frac{1}{2}}, \quad x_0 = 0.8\] to find the values of \(x_1\), \(x_2\) and \(x_3\), giving your answers to 5 decimal places. (3)
(c) Show that \(\alpha = 0.80157\) is correct to 5 decimal places. (3)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(0.75) = -0.18\ldots\) \(\mathrm{f}(0.85) = 0.17\ldots\) | M1 |
| Change of sign, hence root between \(x = 0.75\) and \(x = 0.85\) | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| Sub \(x_0 = 0.8\) into \(x_{n+1} = \left[\arcsin(1 - 0.5x_n)\right]^{\frac{1}{2}}\) to obtain \(x_1\) | M1 |
| Awrt \(x_1 = 0.80219\) and \(x_2 = 0.80133\) | A1 |
| Awrt \(x_3 = 0.80167\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(0.801565) = -2.7\ldots \times 10^{-5}\) \(\mathrm{f}(0.801575) = +8.6\ldots \times 10^{-6}\) | M1A1 |
| Change of sign and conclusion | A1 |
| See Notes for continued iteration method | |
| (3) | |
| (8 marks) |