June 2019 Paper 2 Q24
24 Show that the curve \(\quad y = \dfrac{3}{5}x^5 + x^4 \quad\) has exactly two stationary points. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3x^4\) or \(4x^3\) | M1 | oe eg \(5 \times \dfrac{3}{5}x^{5 - 1}\) |
| \(3x^4 + 4x^3\) | A1 | |
| \(x^3(3x + 4)\ (= 0)\) | M1dep | allow partial factorisation of their \(3x^4 + 4x^3\) if at least \(x\) is taken as a factor ft their two terms if M1 scored |
| \(x^3(3x + 4)\ (= 0)\) and \((x =)\ 0\) and \((x =)\ -\dfrac{4}{3}\) with no other solutions | A1 | allow partial factorisation if at least \(x\) is taken as a factor |
Additional guidance
| \(3x^4 + 4x^3 = 0\) \(x = 0\) and \(x = -\dfrac{4}{3}\) | M1A1 M0A0 |
| Condone \(\ y = 3x^4 + 4x^3\) | M1A1 |
| Ignore higher derivatives | |
| Condone \((0,\ \ldots)\) and \(\left(-\dfrac{4}{3},\ldots\right)\) for \((x =)\ 0\) and \((x =)\ -\dfrac{4}{3}\) | |
| Allow \(-1.33\ldots\) for \(-\dfrac{4}{3}\) (ignore any incorrect conversion attempt after \(-\dfrac{4}{3}\) seen) |