Higher June 2024 Paper 1 Q18
18 A curve \(C\) has equation \(\;y = x^3 - 40x + 1\)
Find the coordinates of both the points on \(C\) at which the gradient is 8
(5)
| Scheme | Marks |
|---|---|
| \(3x^2\) or \(-40\) | M1 |
| \(3x^2 - 40\) | A1 |
| \(\text{``}{3x^2 - 40}\text{''} = 8\) | M1ft |
| \((y =)\; \text{``}{4}\text{''}^3 - 40 \times \text{``}{4}\text{''} + 1 (= -95)\) or \(y = (\text{``}{-4}\text{''})^3 - 40 \times \text{``}{-4}\text{''} + 1\; (= 97)\) | M1ft |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: (4, –95), (–4, 97) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for differentiating one of the first two terms correctly
A1: for both terms correct and no additions
M1ft: dep on M1 for equating their quadratic derivative with 8
(Derivative must be in the form \(ax^2 - 40\) or \(3x^2 - b\) where \(a \neq 0\) and \(b \neq 0\))
M1ft: dep on previous M1 for substituting at least one \(x\) value into \(y\)
NB Following through from \(ax^2 - 40 = 8\) or \(3x^2 - b = 8\), their \(x\) values must be correct
A1: both coordinates must be paired correctly