November 2021 Paper 2 Q13
13 \(P\) is the point on the curve \(\quad y = ax^3 + 10x^2 \quad\) where \(\quad x = 2\)
The gradient of the normal to the curve at \(P\) is \(-\dfrac{1}{4}\)
Work out the value of \(a\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(3ax^2\) or \(20x\) | M1 | oe eg \(3 \times ax^{3-1}\) or \(2 \times 10x^{2-1}\) |
| \(3a \times 2^2 + 20 \times 2\) or \(12a + 40\) | M1 | ft substitution of \(x = 2\) into their derivative must have attempted differentiation and have two terms with one involving \(a\) may be seen in a denominator |
| their \((12a + 40) = -1 \div -\dfrac{1}{4}\) or their \((12a + 40) = 4\) | M1dep | oe eg \(-\dfrac{1}{\text{their } (12a + 40)} = -\dfrac{1}{4}\) dep on 2nd M1 |
| \(-3\) | A1 |
Additional guidance
| Only substituting \(x = 2\) into \(y\) | Zero |
| \(ax^2 + 10x\) \(4a + 20 = 4\) | M0 M1M1 |
| \(3x^2 + 20x\) \(12 + 20\) | M1 M0M0 |