June 2024 Paper 1 Q19
19 The equation of a curve is \(\quad y = x^3 - 3x^2 + 5\)
Work out the stationary points of the curve and determine their nature.
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3x^2 - 6x\) | M1 | either term correct could have additional incorrect terms |
| Their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1dep | follow through an incorrect differentiation as long as M1 scored |
| \(x = 2\) and \(x = 0\) | A1 | these values could imply M1dep if first M mark awarded |
| \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) 6x - 6\) and \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) 6\) and/or positive (for \(x = 2\)) or \(\left(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} =\right) -6\) and/or negative (for \(x = 0\)) or \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3x^2 - 6x\) and any correct check for both sides of one correct solution to give one side with a negative gradient and one side with a positive gradient | M1dep | a correct \(x\) value assessed correctly eg \(x = -1\ \dfrac{\mathrm{d}y}{\mathrm{d}x} \gt 0\) and \(x = 1\ \dfrac{\mathrm{d}y}{\mathrm{d}x} \lt 0\) or \(x = 1\ \dfrac{\mathrm{d}y}{\mathrm{d}x} \lt 0\) and \(x = 3\ \dfrac{\mathrm{d}y}{\mathrm{d}x} \gt 0\) |
| (0, 5) Maximum (2, 1) Minimum | A1 | both points must have been assessed correctly |
Additional guidance
| Final A mark must have everything correct including both stationary points assessed correctly | |
| Just a sketch is not enough to determine the nature of the stationary points | |
| Condone incorrect writing of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) even if it’s just \(y =\) as long as it’s recovered to get the correct nature of the turning points |