Higher June 2023 Paper 2 Q22
22 An approximate value of a root of an equation, \(x\), can be found using the iterative formula
\[x_{n+1} = \sqrt[3]{5(x_n)^2 - 2x_n - 3}\]The starting value is \(\ x_1 = 4\)
(a) Work out the values of \(x_2\) and \(x_3\) [2 marks]
(b) By continuing the iteration, show that the value of \(x\) is more than 4.25 [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| \((x_2 =)\ 4.1(0\ldots)\) | B1 | |
| \((x_3 =)\) [4.176, 4.178] or 4.18 | B1ft | ft their 4.1(0…) rounded to at least 2 dp SC1 \(x_2 =\) [4.176, 4.178] or 4.18 |
Additional guidance
| Allow second B1 for \(x_3 = 4.2\) with acceptable answer seen in working | |
| \(x_2 = 7.8\) \(x_3 = 6.59\) | B0 B1ft |
| SC1 is for using \(x_0 = 4\) |
| Answer | Mark | Comments |
|---|---|---|
| \(4.25 \lt\) value \(\leqslant 4.39\) | B1 | ignore any iteration number |
Additional guidance
Ignore other values if B1 response seen