October 2021 Paper 1 Q4
4. The curve with equation \(y = \mathrm{f}(x)\) where
\[\mathrm{f}(x) = x^2 + \ln\left(2x^2 - 4x + 5\right)\]has a single turning point at \(x = \alpha\)
The iterative formula
\[x_{n+1} = \frac{1}{7}\left(2 + 4x_n^{\,2} - 2x_n^{\,3}\right)\]is used to find an approximate value for \(\alpha\).
Starting with \(x_1 = 0.3\)
Using a suitable interval and a suitable function that should be stated,
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{f}^{\prime}(x) = 2x + \dfrac{4x - 4}{2x^2 - 4x + 5}\) | M1 A1 | 1.1b 1.1b |
| \(2x + \dfrac{4x - 4}{2x^2 - 4x + 5} = 0 \Rightarrow 2x\left(2x^2 - 4x + 5\right) + 4x - 4 = 0\) | dM1 | 1.1b |
| \(2x^3 - 4x^2 + 7x - 2 = 0\,*\) | A1* | 2.1 |
| (4) |
Notes
M1: Differentiates \(\ln\left(2x^2 - 4x + 5\right)\) to obtain \(\dfrac{\mathrm{g}(x)}{2x^2 - 4x + 5}\) where \(\mathrm{g}(x)\) could be 1
A1: For \(\mathrm{f}^{\prime}(x) = 2x + \dfrac{4x - 4}{2x^2 - 4x + 5}\)
dM1: Sets their \(\mathrm{f}^{\prime}(x) = ax + \dfrac{\mathrm{g}(x)}{2x^2 - 4x + 5} = 0\) and uses "correct" algebra, condoning slips, to obtain a cubic equation. E.g Look for \(ax\left(2x^2 - 4x + 5\right) \pm \mathrm{g}(x) = 0\) o.e. , condoning slips, followed by some attempt to simplify
A1*: Achieves \(2x^3 - 4x^2 + 7x - 2 = 0\) with no errors. (The dM1 mark must have been awarded)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(x_2 = \dfrac{1}{7}\left(2 + 4(0.3)^2 - 2(0.3)^3\right)\) | M1 | 1.1b |
| \(x_2 = 0.3294\) | A1 | 1.1b |
| (ii) \(x_4 = 0.3398\) | A1 | 1.1b |
| (3) |
Notes
(b)(i) M1: Attempts to use the iterative formula with \(x_1 = 0.3\). If no method is shown award for \(x_2 =\) awrt 0.33
A1: \(x_2 =\) awrt 0.3294 Note that \(\dfrac{1153}{3500}\) is correct
Condone an incorrect suffix if it is clear that a correct value has been found
(b)(ii) A1: \(x_4 =\) awrt 0.3398 Condone an incorrect suffix if it is clear that a correct value has been found
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{h}(x) = 2x^3 - 4x^2 + 7x - 2\) \(\mathrm{h}(0.3415) = 0.00366\ldots \qquad \mathrm{h}(0.3405) = -0.00130\ldots\) | M1 | 3.1a |
States:
| A1 | 2.4 |
| (2) | ||
| (9 marks) |
Notes
M1: Attempts to substitute \(x = 0.3415\) and \(x = 0.3405\) into a suitable function and gets one value correct (rounded or truncated to 1 sf). It is allowable to use a tighter interval that contains the root 0.340762654
Examples of suitable functions are \(2x^3 - 4x^2 + 7x - 2,\ x - \dfrac{1}{7}\left(4x^2 - 2x^3 + 2\right)\) and \(\mathrm{f}^{\prime}(x)\) as this has been found in part (a) with \(\mathrm{f}^{\prime}(0.3405) = -0.00067\ldots,\ \mathrm{f}^{\prime}(0.3415) = (+)\ 0.0018\)
There must be sufficient evidence for the function, which would be for example, a statement such as \(\mathrm{h}(x) = 2x^3 - 4x^2 + 7x - 2\) or sight of embedded values that imply the function, not just a value or values even if both are correct. Condone \(\mathrm{h}(x)\) being mislabelled as f
\(\mathrm{h}(0.3415) = 2 \times 0.3415^3 - 4 \times 0.3415^2 + 7 \times 0.3415 - 2\)
A1: Requires
- both calculations correct (rounded or truncated to 1sf)
- a statement that there is a change in sign and that the function is continuous
- a minimal conclusion e.g. ✓, proven, \(\alpha = 0.341\), root