June 2018 Paper 2 Q15
15 Work out the rate of change of \(y\) with respect to \(x\) at the point on the curve
\(y = x^2(x^2 - 9) \qquad\) where \(\ x = -2\)
You must show your working. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^4 - 9x^2\) | M1 | |
| \(4x^3\) or \((-)2 \times 9x\) or \((-)18x\) | M1 | differentiates at least one of their terms their term must be a function of \(x\) after an attempt to expand brackets \(4x^3 - 18x\) implies M2 |
| \(4 \times (-2)^3 - 18 \times -2\) or \(4 \times -8 - 18 \times -2\) or \(-32 + 36\) | M1dep | oe dep on 2nd M1 substitutes \(-2\) into their \(4x^3 - 18x\) their \(4x^3 - 18x\) must be two terms, each a function of \(x\) |
| 4 | A1 |
Additional guidance
| Allow recovery of brackets | |
| \(4 \times -2^3\) is allowed for \(\ 4 \times (-2)^3\) | |
| 3rd M can still be awarded even if further substitution seen eg1 \(\ 4x^3 - 18x \quad 4 \times (-2)^3 - 18 \times -2\) and \(4^4 - 9 \times 4^2 = 112\) Answer 112 eg2 \(\ 4x^3 - 18x \quad 4 \times (-2)^3 - 18 \times -2 = 4\) and \(4 \times 4^3 - 18 \times 4 = 184\) Answer 184 | M3 A0 M3 A0 |
| Only substituting \(x = -2\) into second derivative can score a maximum of M1M1M0A0 | |
| 4 followed by answer \(y = 4\) | M3A1 |
| 4 followed by answer \(y = 4x\) | M3A0 |
| Do not regard substitution of \(x = 2\) as a misread | |
| Beware that finding the gradient of the line through \((-2,\ -20)\) and \((3,\ 0)\) gives answer 4 | Zero |
| Beware that \((-)18x\) may come from wrong method eg1 \(\ 2x(2x - 9) = 4x^2 - 18x\) eg2 \(\ 2x(x^2 - 9) = 2x^3 - 18x\) | Zero Zero |