Higher November 2020 Paper 1R Q13
13 The curve C has equation \(\;y = 5x^3 - x^2 - 6x + 4\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
There are two points on the curve C at which the gradient of the curve is 2
(b) Find the \(x\) coordinate of each of these two points.
Show clear algebraic working. (4)
Show clear algebraic working. (4)
| Scheme | Marks |
|---|---|
| \(15x^2 - 2x - 6\) | B2 |
| (2) |
Notes
B2: for correct differentiation
(B1 for 2 of \(15x^2\), \(-2x\), \(-6\) correct)
| Scheme | Marks |
|---|---|
| e.g. “\(15x^2 - 2x - 6\)” = 2 oe | M1 |
| \(15x^2 - 2x - 8\;(= 0)\) | M1 |
e.g. \((3x + 2)(5x - 4)\;(= 0)\) \(x = \dfrac{2 \pm \sqrt{(-2)^2 - (4 \times 15 \times -8)}}{2 \times 15}\) | M1 |
Working required Answer: \(-\dfrac{2}{3},\; \dfrac{4}{5}\) | A1 |
| (4) | |
| (6 marks) |
Notes
M1: ft, for equating their \(\mathrm{d}y/\mathrm{d}x\) to 2
M1: (dep on M1) ft their three-term quadratic
M1: for solving their 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 e.g. \(\dfrac{2 \pm \sqrt{4 + 480}}{30}\) oe)
A1: oe, dep on M2 (allow −0.66 or better),
Both values – isw any attempt to find \(y\) coordinates