June 2023 Paper 1 Q5
5 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -2x + 32x^{-3} = 0\) \(x^4 = 16\) | M1 | 1.1a |
| So \(x = \pm 2\) | A1 | 1.1b |
| When \(x = \pm 2,\ y = 7\) [So the points are \((-2, 7)\) and \((2, 7)\)] | A1 | 1.1b |
| [3] |
Notes
M1: Attempt to differentiate and equate to zero soi
A1: Both \(x\)-values and no others.
A1: FT their \(x\)-coordinate(s) Do not FT \(x = 0\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2 - 96x^{-4}\) When \(x = 2,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2 - \dfrac{96}{16} < 0\) When \(x = -2,\ \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -2 - \dfrac{96}{16} < 0\) | M1 | 2.1 |
| So both points are maximum points | E1 | 2.1 |
| [2] |
Notes
M1: Attempts to find the second derivative. FT their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Or convincing statement that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\) for any \(x\ [\neq 0]\) because \(x^{-4}\) is always positive
E1: AG Complete argument required from correct second derivative and their \(x \neq 0\). ISW if \(-2 - \frac{96}{x^4}\) wrongly evaluated. Also allow for one point established and an argument from symmetry
Alternative method
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} > 0\) for \(0 < x < 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} < 0\) for \(x > 2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} > 0\) for \(x < -2\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} < 0\) for \(-2 < x < 0\) | M1 |
| So both points are maximum points | E1 |
M1: Evaluating gradient for suitable values of \(x\) on either side of each turning point
Also allow for \(y\)-coordinates in these ranges.
E1: AG Complete argument required from correct first derivative and their \(x \neq 0\). Also allow for one point established and an argument from symmetry