4MA0 Higher June 2017 Paper 3R Q15
15 For the curve C with equation
\[y = 2x^3 - 3x^2 - 12x + 9\](a) find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
(b) Find the gradient of C at the point with coordinates (2, −11) (2)
The curve C has a gradient of −12 at the point where \(x = k\) and at the point where \(x = m\).
Given that \(k \gt m\)
(c) find the value of \(k\) and the value of \(m\). (3)
| Scheme | Marks |
|---|---|
| \(3 \times 2 \times x^2 - 3 \times 2x - 12\) | M1 |
| \(6x^2 - 6x - 12\) | A1 |
| (2) |
Notes
M1: for one of \(3 \times 2 \times x^2\) (\(=6x^2\)) or \(-3 \times 2x\) (\(=6x\)) or \(-12\)
| Scheme | Marks |
|---|---|
| \(6 \times 2^2 - 6 \times 2 - 12\) | M1 |
| 0 | A1 |
| (2) |
Notes
M1: substitute \(x = 2\) in (a)
ft from answer to (a) (must be a quadratic expression)
| Scheme | Marks |
|---|---|
| \(6x^2 - 6x - 12 = -12\) | M1 |
| \(x = 1\) , 0 | A1 |
| 1, 0 | A1 |
| (3) | |
| (7 marks) |
Notes
M1: ft from answer to (a) (must be a quadratic expression)
A1: for both correct solutions
A1: for \(k = 1\) and \(m = 0\)