Higher January 2022 Paper 1 Q24
24 The curve C has equation \(y = ax^3 + bx^2 - 12x + 6\) where \(a\) and \(b\) are constants.
The point \(A\) with coordinates (2, –6) lies on C
The gradient of the curve at \(A\) is 16
Find the \(y\) coordinate of the point on the curve whose \(x\) coordinate is 3
Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| eg \(-6 = 8a + 4b - 24 + 6\) or \(8a + 4b = 12\) oe | M1 |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 3ax^2 + 2bx - 12\) oe | M1 |
| eg \(16 = 12a + 4b - 12\) or \(12a + 4b = 28\) oe | M1ft |
| \(a = 4\) and \(b = -5\) | M1 |
| eg \(\text{``}{4}\text{''} \times 3^3 + \text{``}{-5}\text{''} \times 3^2 - 12 \times 3 + 6\) | M1ft |
| Working required Answer: 33 | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for substituting \(x = 2\) and \(y = -6\) into the equation for C
M1: at least 2 terms correct
M1ft: (dep on previous M1)
follow through their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
M1: for \(a = 4\) and \(b = -5\)
M1ft: correctly substituting their \(a\), their \(b\) and \(x = 3\) into the equation for C
A1: (dep on M3) allow (3, 33)