A2 June 2024 Paper 1 Q1
1 Given that \(y = \sin^{-1}(x^2)\), find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{\sqrt{(1 - (x^2)^2}}\) | B1 | 1.1 |
| \(\times 2x\) | M1 | 1.1 |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) \dfrac{2x}{\sqrt{1 - x^4}}\) | A1 | 1.1 |
| [3] |
Notes
B1: For \(\dfrac{1}{\sqrt{(1 - (x^2)^2}}\) seen.
M1: For \(2x \times \mathrm{f}(x)\) where \(\mathrm{f}(x) = \dfrac{1}{\sqrt{1 - (x^2)^2}}\) or \(\dfrac{1}{\sqrt{1 - x^2}}\) ONLY.
A1: Allow any equivalent correct form e.g. \(2x(1 - x^4)^{-0.5}\).
Must be in terms of \(x\).
Condone \((x^2)^2\) for \(x^4\).
ISW once correct answer seen.
Alternative method
| Scheme | Marks |
|---|---|
| \(\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) or \(\cos y = 2x\dfrac{\mathrm{d}x}{\mathrm{d}y}\) | B1 |
| \(\sqrt{1 - (x^2)^2}\,\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) | M1 |
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) \dfrac{2x}{\sqrt{1 - x^4}}\) | A1 |
| [3] |
B1: For correctly differentiating implicitly with respect to either \(x\) or \(y\).
M1: Replacing \(\cos y\) with \(\pm\sqrt{\pm 1 \pm (x^2)^2}\) in their derivative of the form \(\pm\cos y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x\) (or equivalent if differentiating with respect to \(y\)).
A1: Allow any equivalent correct form e.g. \(2x(1 - x^4)^{-0.5}\).
Must be in terms of \(x\).
Condone \((x^2)^2\) for \(x^4\).
ISW once correct answer seen.