Higher January 2019 Paper 1R Q13
13 A curve C has equation \(y = x^3 - x^2 - 8x + 12\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
The curve C has two turning points.
(b) Work out the \(x\) coordinates of the two turning points.
Show your working clearly. (3)
Show your working clearly. (3)
(c) Show that the \(x\)-axis is a tangent to the curve C. (2)
| Scheme | Marks |
|---|---|
| \(3x^2 - 2x - 8\) | B2 |
| (2) |
Notes
B2: (B1 for at least 1 correct non zero term)
| Scheme | Marks |
|---|---|
| “\(3x^2 - 2x - 8\)” = 0 | M1 |
\((3x + 4)(x - 2)\) (=0) or \(x = \dfrac{2 \pm \sqrt{100}}{2 \times 3}\) or \(x = \dfrac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-8)}}{2 \times 3}\) | M1 |
| \(-\dfrac{4}{3}\), 2 | A1 |
| (3) |
Notes
M1: Dep on at least B1, ft on M marks only dep on \(\dfrac{dy}{dx}\) being a 3 term quadratic
A1: (dep 2nd M1)
| Scheme | Marks |
|---|---|
At \(x = 2\), \(y = 2^3 - 2^2 - 8 \times 2 + 12\) (= 0) or at \(x = -\dfrac{4}{3}\), \(y = \left(-\dfrac{4}{3}\right)^3 - \left(-\dfrac{4}{3}\right)^2 - 8 \times \left(-\dfrac{4}{3}\right) + 12\) \(\left(= \dfrac{500}{27}\right)\) | M1 |
| Shown | A1 |
| (2) | |
| (7 marks) |
Notes
M1: Substitutes at least one of \(-\dfrac{4}{3}\) or 2 or their answer from (b) into \((y =)\,x^3 - x^2 - 8x + 12\)
A1: must show that (2,0) is a turning point on the curve and give concluding statement