Higher June 2023 Paper 1 Q11
11 The curve C has equation \(\;y = 4x^3 + x^2 - 20x\)
(a) Find \(\;\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
(b) Find the \(x\) coordinates of the points on C where the gradient is 4
Show clear algebraic working. (4)
Show clear algebraic working. (4)
| Scheme | Marks |
|---|---|
| M1 | |
| \(12x^2 + 2x - 20\) | A1 |
| (2) |
Notes
M1: for at least 2 of \(12x^2\), \(2x\), \(-20\)
| Scheme | Marks |
|---|---|
| \(12x^2 + 2x - 20 = 4\) oe | M1 |
| \(12x^2 + 2x - 24 \;(= 0)\) or \(6x^2 + x - 12 \;(= 0)\) | M1 |
eg \((6x - 8)(2x + 3) \;(= 0)\) or \((3x - 4)(2x + 3) \;(= 0)\) or \(x = \dfrac{-2 \pm \sqrt{(2)^2 - (4 \times 12 \times -24)}}{2 \times 12}\) | M1 |
Working required Answer: \(\dfrac{4}{3}, -\dfrac{3}{2}\) | A1 |
| (4) | |
| (6 marks) |
Notes
M1: ft, for equating their \(\mathrm{d}y/\mathrm{d}x\) to 4
M1: (dep on M1) ft their \(\mathrm{d}y/\mathrm{d}x\) in the form \(ax^2 + bx\,(+ c)\)
M1: for solving their three-term quadratic equation using any correct method - if factorising, allow brackets which expanded give 2 out of 3 terms correct (if using formula or completing the square allow one sign error and some simplification – allow as far as eg \(\dfrac{-2 \pm \sqrt{4 + 1152}}{24}\) oe)
A1: (dep on M2) oe, allow 1.33(3...) for \(\dfrac{4}{3}\), both values – isw any attempt to find \(y\) coordinates