October 2021 Paper 3 Q6
6 The equation \(6\arcsin(2x - 1) - x^2 = 0\) has exactly one real root.
In order to find the root, the iterative formula
\(x_{n+1} = p + q\sin\left(rx_n^2\right),\)
with initial value \(x_0 = 0.5\), is to be used.
| Scheme | Marks | AO |
|---|---|---|
| Considers both \(\mathrm{f}(0.5)\) and \(\mathrm{f}(0.6)\) where \(\mathrm{f}(x) = \pm\left\{6\arcsin(2x - 1) - x^2\right\}\) | M1 | 1.1 |
| \(\mathrm{f}(0.5) = -0.25 \lt 0,\ \mathrm{f}(0.6) = 0.8481\ldots \gt 0\) change of sign indicates that the root lies between 0.5 and 0.6 | A1 | 2.4 |
| [2] |
Notes
M1: With at least one correct value – values should be given to at least 2 sf (rot)
Allow degrees for M1 only: \(\mathrm{f}(0.6) = 68.8617\ldots\)
A1: Correct values together with explanation in words (change of sign) and conclusion
| Scheme | Marks | AO |
|---|---|---|
| \(6\arcsin(2x - 1) - x^2 = 0 \Rightarrow \arcsin(2x - 1) = \dfrac{1}{6}x^2\) So \(2x - 1 = \sin\left(\dfrac{1}{6}x^2\right)\) | M1 | 1.1 |
| \(x = \dfrac{1}{2} + \dfrac{1}{2}\sin\left(\dfrac{1}{6}x^2\right)\) | A1 | 2.2a |
| [2] |
Notes
M1: Correct order of operations to get \(2x - 1 = \sin\left(kx^2\right)\)
\(k \neq 0\)
A1: \(p = \frac{1}{2},\ q = \frac{1}{2}\) and \(r = \frac{1}{6}\) (oe)
| Scheme | Marks | AO |
|---|---|---|
| \((x_0 = 0.5)\) \((x_1 =)\,0.5208273057\ldots\) \((x_2 =)\,0.5225973903\ldots\) \((x_3 =)\,0.5227511445\ldots\) \((x_4 =)\,0.5227645245\ldots\) | M1 | 1.1 |
| 0.5228 | A1 | 1.1 |
| [2] |
Notes
M1: Uses their iterative formula with correct starting value to produce terms up to at least \(x_2\) to at least 4 significant figures
Allow degrees for M1 only:
For reference:
\(x_1 = 0.5003636\ldots\)
\(x_2 = 0.5003641\ldots\)
\(x_3 = 0.5003641\ldots\)
A1: Must be stated to exactly 4 significant figures