4MA0 Higher June 2015 Paper 4R Q21
21 \(y = x^2 - \dfrac{16}{x}\)
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) (3)
The graph shows part of the curve with equation \(y = x^2 - \dfrac{16}{x}\)
The point \(P\) is the turning point of the curve.
(b) Work out the coordinates of \(P\). (4)
| Scheme | Marks |
|---|---|
| \((-)\dfrac{16}{x} = (-)16x^{-1}\) | M1 |
| M1 | |
| \(2x + \dfrac{16}{x^2}\) | A1 |
| (3) |
Notes
M1: For \(16x^{-1}\) or \(-16x^{-1}\)
M1: For \(\pm 16x^{-2}\) or \(\pm\dfrac{16}{x^2}\) or \(2x\)
A1: Allow \(2x + 16x^{-2}\)
M2A0 for an incorrect answer that includes \(16x^{-2}\)
| Scheme | Marks |
|---|---|
| \(2x + \dfrac{16}{x^2} = 0\) | M1 |
| \(2x^3 = -16\) or \(x^3 = -8\) | M1 |
| \(x = -2\) | A1 |
| (−2, 12) | A1 |
| (4) |
Notes
M1: ft from (a)
Do not accept \(x^2 - 16x^{-1} = 0\) oe
M1: For a correct equation in \(x^3\)
NB: No ft allowed
A1: For \(x = -2\)
Accept as part of an incorrect coordinate, eg (−2, 10)
A1: Accept \(x = -2\), \(y = 12\)