Higher June 2022 Paper 1R Q22
22 The diagram shows a sketch of part of the curve with equation \(y = x^2 - \dfrac{p}{x}\) where \(p\) is a positive constant.
Diagram NOT accurately drawn
For all values of \(p\), the curve has exactly one turning point and this turning point is a minimum shown as the point \(T\) in the sketch.
For the curve where the \(x\) coordinate of \(T\) is \(-3\)
(a) find the value of \(p\) (4)
The line with equation \(y = k\) is a tangent to the curve with equation \(y = x^2 - \dfrac{16}{x}\)
(b) Find the value of \(k\) (3)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 2x + px^{-2}\) oe | M2 |
| \(2(-3) + p(-3)^{-2}\;(= 0)\) | M1 |
| 54 | A1 |
| (4) |
Notes
M2: Both terms correct
(M1 for one term correct)
M1: (dep on M1) substitute −3 into a derivative of the form \(ax + bx^{-2}\)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 2x + 16x^{-2} = 0\) | M1 |
| eg \(2x^3 + 16 = 0\) or \(2x^3 = -16\) or \(x^3 = -8\) or \(x = \sqrt[3]{-8}\) or \(x = -2\) | M1 |
| 12 | A1 |
| (3) | |
| (7 marks) |
Notes
M1: set \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\), at least one term correct
M1: rearrangement of the correct equation to remove the negative power of \(x\)