4MA0 Higher January 2015 Paper 3 Q18
18 \(y = x^3 - 4x^2 + 4x + 3\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
The diagram shows a sketch of the curve with equation \(y = x^3 - 4x^2 + 4x + 3\)
The point P is a turning point on the curve.
(b) Work out the coordinates of P.
Show clear algebraic working. (4)
Show clear algebraic working. (4)
(c) Write down the range of values of \(x\) for which the curve has a negative gradient. (2)
| Scheme | Marks |
|---|---|
| M1 | |
| \(3x^2 - 8x + 4\) | A1 |
| (2) |
Notes
M1: For any two of \(3x^2\), \(-8x\), or 4
A1: Do not ISW
| Scheme | Marks |
|---|---|
\((3x - 2)(x - 2)\ (= 0)\) or \((x =)\ \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 3 \times 4}}{2 \times 3}\) | M2 |
| \(x = 2\) | A1 |
| (2, 3) | A1 |
| (4) |
Notes
M2: For correct factors or correct substitution into the quadratic formula.
ft for a 3 term quadratic if M1 scored in (a)
Accept + instead of ±.
If M2 not scored:
M1 for (a) = 0 if M1 scored in (a)
A1: dep on M2
A1: dep on M2
| Scheme | Marks |
|---|---|
| M1 | |
| \(\dfrac{2}{3} \lt x \lt 2\) | A1 |
| (2) | |
| (8 marks) |
Notes
M1: For \(x \lt \text{“}2\text{”}\) or \(x \gt \text{“}\tfrac{2}{3}\text{”}\)
ft positive values from (b) if M1 scored in (b)
A1: ft positive values from (b) if M1 scored in (b).
Accept \(\frac{2}{3}\) – 2 or \(\frac{2}{3}\) to 2
Do not penalise \(\leqslant\) or \(\geqslant\)