June 2019 Paper 1 Q12
12 A curve has the equation \(\quad y = x^3 + ax^2 - 7 \quad\) where \(a\) is a constant.
The gradient of the curve when \(x = 4\) is twice the gradient of the curve when \(x = -1\)
Work out the value of \(a\).
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3x^2 + 2ax\) | M1 | allow a derivative with at least one term correct and a term in \(a\) eg \(3x^2 + 2ax + 7\) or \(3x^2 + 2a\) |
| \(3(4)^2 + 2a(4)\) or \(48 + 8a\) | A1ft | |
| \(3(-1)^2 + 2a(-1)\) or \(3 - 2a\) | A1ft | |
| \(48 + 8a = 2(3 - 2a)\) | M1dep | oe ft if first M1 earned |
| \((a =)\ -3.5\) | A1 | oe |
Additional guidance
| Minimum expected working is to see the correct derivative in the first M mark. If no working seen then no marks can be awarded | |
| If the word “twice” is interpreted the wrong way round ie equation becomes \(2(48 + 8a) = 3 - 2a\) this gives an answer of \(a = -5\tfrac{1}{6}\) or \(-5.1666\ldots\) | M1, A1, A1, M0, A0 |