4MA0 Higher January 2014 Paper 4R Q15
15 For the curve with equation \(y = 4x^3 - 2x + 5\)
(i) find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
(ii) find the coordinates of the two points on the curve where the gradient of the curve is 1
(6)
| Scheme | Marks |
|---|---|
| (i) Answer: \(12x^2 - 2\) | B2 |
| (ii) “\(12x^2 - 2\)” = 1 | M1ft |
| \(x^2 = 3/12\) oe | M1ft |
| 0.5 , 4.5 and − 0.5 , 5.5 | B2 |
| (6) | |
| (6 marks) |
Notes
B2: If not B2 then B1 for \(12x^2\) or \(-2\)
M1ft: only ft \(ax^2 \pm b = 1\)
M1ft: ft \(x^2\) = constant
B2: Award B1 for both \(x\) values correct or 1 pair of \(x\) and \(y\) values correct.