A2 October 2021 Paper 1 Q2
2 In this question you must show detailed reasoning.
Find the gradient of the curve \(y = 6\arcsin(2x)\) at the point with \(x\)-coordinate \(\frac{1}{4}\). Express the result in the form \(m\sqrt{n}\), where \(m\) and \(n\) are integers. [4]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 \times \dfrac{1}{\sqrt{1 - 4x^2}} \times 2\) | B1 M1 A1 | 1.1 1.1 1.1 |
| when \(x = \frac{1}{4}\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 \times \dfrac{1}{\sqrt{\frac{3}{4}}} \times 2 = 8\sqrt{3}\) | A1 | 1.1 |
| [4] |
Notes
B1: \(\dfrac{k}{\sqrt{1 - 4x^2}}\)
M1: Chain rule (x 2)
A1: Fully correct \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
Alternative solution
| Scheme | Marks |
|---|---|
| DR \(\sin\dfrac{y}{6} = 2x\) \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = \dfrac{1}{12}\cos\dfrac{y}{6}\) | M1 M1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{12}{\cos\frac{y}{6}}\) and \(x = \frac{1}{4}, y = \pi\) | A1 |
| \(\rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 8\sqrt{3}\) | A1 |
| [4] |
M1: For \(\frac{\mathrm{d}x}{\mathrm{d}y} = k\cos\frac{y}{6}\)
M1: Using chain rule \(\left(\times \frac{1}{6}\right)\)
A1: Or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 6 \times \dfrac{1}{\sqrt{1 - 4x^2}} \times 2\)
(Corrected from the printed mark scheme: the guidance for the first M1 is printed as \(\frac{\mathrm{d}x}{\mathrm{d}y} = k\cos\frac{\pi}{6}\); it should be \(k\cos\frac{y}{6}\), as typed above.)