June 2019 Paper 2 Q4
4 \(P\) is a point on a curve.
The curve has gradient function \(\quad \dfrac{x^5 - 17}{10}\)
The tangent to the curve at \(P\) is parallel to the line \(\quad 3x - 2y = 9\)
Work out the \(x\)-coordinate of \(P\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((y =)\ \dfrac{3}{2}x\ldots\) or \((y =)\ 1.5x\ldots\) or \(\dfrac{3}{2}\) or 1.5 | M1 | oe eg \((y =)\ \dfrac{3x - 9}{2}\) |
| \(\dfrac{x^5 - 17}{10} = \dfrac{3}{2}\) | M1dep | oe implies M2 |
| \(x^5 = \dfrac{3}{2} \times 10 + 17\) or \(\sqrt[5]{32}\) or correctly rearranges \(\dfrac{x^5 - 17}{10} = k\) to the form \(x^5 =\) (\(k\) any non-zero value) | M1 | oe eg \(x^5 = 15 + 17\) or \(x^5 = 32\) or \(\sqrt[5]{15 + 17}\) must rearrange to the form \(x^5 =\) |
| 2 | A1 |
Additional guidance
| Condone error seen in rearrangement of \(3x - 2y = 9\) if gradient is \(\dfrac{3}{2}\) May go on to score M3A1 | |
| \(\dfrac{x^5 - 17}{10} = \dfrac{3}{2}x\) | M1M0M0A0 |
| (gradient \(=\)) 3 \(\dfrac{x^5 - 17}{10} = 3\) \(x^5 = 30 + 17 \quad\) (3rd M is not dependent) 2.16 | M0M0dep M1 A0 |
| \(\dfrac{3}{2}\) \(\dfrac{x^5 - 17}{10} = -\dfrac{2}{3}\) \(x^5 = -\dfrac{2}{3} \times 10 + 17 \quad\) (3rd M is not dependent) 1.595 | M1 M0 M1 A0 |
| Condone answer (2, …) | |
| 2 embedded | M3A0 |