October 2020 Paper 1 Q7
7 In this question you must show detailed reasoning.
A curve has equation \(y = 4x^3 - 6x^2 - 9x + 4\).
(a) Sketch the gradient function for this curve, clearly indicating the points where the gradient is zero. [4]
(b) Find the set of values of \(x\) for which the gradient function is decreasing. Give your answer using set notation. [2]
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 12x - 9\) | M1 | 1.1a |
| When \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 - 12x - 9 = 0\) | M1 (dep) | 1.1a |
| \(3(2x + 1)(2x - 3) = 0\) so \(x = -0.5,\ 1.5\) | A1 | 1.1a |
 | B1 | 1.1 |
| [4] |
Notes
M1: Attempt to differentiate seen
M1: Attempt to solve their \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)
A1: Both values seen – may be indicated on the graph
B1: Correct shape through (0, −9)
SC For cubic graph of the function drawn with M0M0A0 allow SC1 for correct shape with minimum when \(x = 1.5\), and maximum when \(x = -0.5\)
| Scheme | Marks | AO |
|---|---|---|
| DR | ||
| Min point of gradient function when \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 24x - 12 = 0\) so \(x = \dfrac{1}{2}\) | M1 | 3.1a |
| Gradient is decreasing for \(\left\{x : x < \dfrac{1}{2}\right\}\) | A1 | 2.5 |
| [2] |
Notes
M1: Attempt to find the vertex (including completing the square or symmetry argument)
A1: Inequality correctly formed and expressed as a set.
Allow either \(<\) or \(\leqslant\)