June 2025 Paper 1 Q11
11 The equation of a curve is
\[x^2y + 4y^3 = 8x\]The curve has two stationary points.
(a) Use implicit differentiation to show that at the stationary points \(y = \dfrac{4}{x}\) [4 marks]
(b) Hence show that the \(x\)-coordinates of the stationary points can be written in the form \(\pm\sqrt{n}\) where \(n\) is an integer to be found. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses implicit differentiation with \(x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(Ay^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) seen | M1 | 1.1a |
| Uses the product rule to obtain \(Bxy + x^2\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 | 3.1a |
| Obtains \(2xy + x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 12y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8\) OE | A1 | 1.1b |
| Completes a reasoned argument using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and \(2xy + x^2\dfrac{\mathrm{d}y}{\mathrm{d}x} + 12y^2\dfrac{\mathrm{d}y}{\mathrm{d}x} = 8\) OE with at least one correct line of intermediate working before showing that \(y = \dfrac{4}{x}\) | R1 | 2.1 |
| (4) |
Typical solution
\[x^2y + 4y^3 = 8x\]\[2xy + x^2\frac{\mathrm{d}y}{\mathrm{d}x} + 12y^2\frac{\mathrm{d}y}{\mathrm{d}x} = 8\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = 0\]\[\Rightarrow 2xy = 8\]\[\Rightarrow y = \frac{4}{x}\]| Scheme | Marks | AO |
|---|---|---|
| Obtains \(x^2 \times \dfrac{4}{x} + 4\left(\dfrac{4}{x}\right)^3 = 8x\) OE | B1 | 3.1a |
| Rearranges their equation to obtain \(ax^4 = k\) or \(ax^{-4} = k\) OE | M1 | 2.1 |
| Deduces \(x = \pm\sqrt{8}\) CAO | R1 | 2.2a |
| (3) | ||
| (7 marks) |