4MA0 Higher June 2014 Paper 3R Q14
14
(a) \(y = 2x^3 + 3x^2 + 2\)
Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (2)
(b) The point \(P\) lies on the curve with equation \(y = 2x^3 + 3x^2 + 2\)
The gradient of the curve at \(P\) is \(-\dfrac{3}{2}\)
Find the coordinates of \(P\). (5)
The gradient of the curve at \(P\) is \(-\dfrac{3}{2}\)
Find the coordinates of \(P\). (5)
| Scheme | Marks |
|---|---|
| \(6x^2 + 6x\) | B2 |
| (2) |
Notes
B2: Award B1 for any two terms differentiated correctly (remember \(\frac{\mathrm{d}}{\mathrm{d}x}(2) = 0\) is a term)
| Scheme | Marks |
|---|---|
| \(\text{“}6x^2 + 6x\text{”} = -\dfrac{3}{2}\) | M1 |
| \(6x^2 + 6x + 1\frac{1}{2}\;(= 0)\) or \(4x^2 + 4x + 1 = (0)\) oe | M1 |
\((2x + 1)^2 = 0\) or \((x =)\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times 1}}{2 \times 4}\) or \(\dfrac{-4 \pm \sqrt{16 - 16}}{8}\) or \(\dfrac{-4 \pm 0}{8}\) or \(6\left(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{3}{12}\right)\;(=0)\) oe | M1 |
| \(x = -\frac{1}{2}\) | A1 |
| \(\left(-\frac{1}{2},\; 2\frac{1}{2}\right)\) | A1 |
| (5) | |
| (7 marks) |
Notes
M1: Their derivative = −1.5 oe
dep on at least B1 in (a)
M1: Correct three terms of quadratic equation on one side.
M1: Correct factors or correct use of quadratic formula or completing the square.
A1: For correctly calculating \(x\)
A1: For completely correct coordinates.