June 2022 Paper 2 Q5
5 In this question you must show detailed reasoning.
A curve has equation \(y = x^3 - 3x^2 + 4x\).
(a) Show that the curve has no stationary points. [2]
(b) Show that the curve has exactly one point of inflection. [2]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x + 4 = 0\) \(b^2 - 4ac = -12\) or \(\mathrm{D} = -12\) or \(3(x - 1)^2 + 1 = 0\) oe | M1 | 3.1a |
| No (real) roots or no value of \(x\), or can’t \(\sqrt{\text{negative}}\) or gradient always +ve. | A1 | 1.1 |
| [2] |
Notes
M1: Differentiate & equate to 0. May be implied by calc of D
or \(x = \dfrac{6 \pm \sqrt{36 - 48}}{6}\) or \(x = \dfrac{6 \pm \mathrm{i}\sqrt{12}}{6}\) oe
A1: Must see justification as line above, no errors, & statement
Other correct forms of the quadratic equation and justification may be seen.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x - 6 = 0\) | M1 | 1.1 |
| \(x = 1\) gives a point of inflection or \(x = 1\) & show that, either side of this point, gradient does not change sign or second derivative does change sign | A1 | 2.2a |
| [2] |
Notes
M1: Differentiate their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) and = 0. Can be implied by \(x = 1\)
A1: Statement “\(x = 1\) gives a point of inflection” is enough.
or This equation has one root. (so curve has one inflection)
Not just “\(x = 1\)”
Ignore \(y\)-coordinate