June 2025 Paper 2 Q20
20 Here is a sketch of the curve with equation \(\quad y = 4 + 2x - x^2\)
The tangent to the curve at the turning point \(P\) is shown.
The normal to the curve at \(Q\) meets this tangent at \(R\).
Work out the \(x\)-coordinate of \(R\). [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(2 - 2x\) | M1 | oe eg \(2x^0 - 2x^{2 - 1}\) may be implied eg \(2 - 2 \times 3\) or \(-4\) or \(-\dfrac{1}{2 - 2x}\) |
| Substitutes \(x = 3\) into \(-\dfrac{1}{\text{their } (2 - 2x)}\) | M1 | oe eg \(-1 \div -4\) their \((2 - 2x)\) must be a linear expression in \(x\) |
| (gradient of \(QR =\)) \(\dfrac{1}{4}\) or 0.25 | A1 | oe value may be seen in an equation eg \(y = \dfrac{1}{4}x + c\) |
| Substitutes \(y = 5\) into correct equation of \(QR\) | M1dep | eg \(5 - 1 = \dfrac{1}{4}(x - 3)\) or \(\dfrac{5 - 1}{x - 3} = \dfrac{1}{4}\) or \(5 = \dfrac{1}{4}x + \dfrac{1}{4}\) dep on M2A1 |
| 19 | A1 | implied by (19, 5) |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| \(y = 2 - 2x\) is awarded 1st M1 and may be awarded up to full marks with further correct working | |
| \(2 + 2x\) \(-\dfrac{1}{2 + 2 \times 3} = -\dfrac{1}{8}\) | M0 M1A0 |
| Escalate responses using similar triangles |