# Question 14:

## Part (a)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to differentiate $x = 4\sin 2y$ and inverts: $\frac{dx}{dy} = 8\cos 2y \Rightarrow \frac{dy}{dx} = \frac{1}{8\cos 2y}$ | M1 | Allow $\frac{dx}{dy} = k\cos 2y \Rightarrow \frac{dy}{dx} = \frac{1}{k\cos 2y}$. Watch for $\frac{dx}{dy} = 8\cos 2x$ which is M0 A0 |
| At $(0,0)$: $\frac{dy}{dx} = \frac{1}{8}$ | A1 | Allow both marks for sight of $\frac{1}{8}$ with no incorrect working |

## Part (b)

| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) Uses $\sin 2y \approx 2y$ when $y$ is small to obtain $x \approx 8y$ | B1 | Do not allow $\sin 2y \approx 2\theta$. Double angle formula is B0 |
| (ii) The value found in (a) is the gradient of the line found in (b)(i) | B1 | Must refer to both answers; allow "gradients are the same $\left(=\frac{1}{8}\right)$" or "both have $m = \frac{1}{8}$" |

## Part (c)

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $\frac{dy}{dx}$ as function of $y$, using $\sin^2 2y + \cos^2 2y = 1$ and $x = 4\sin 2y$, attempts to write $\frac{dy}{dx}$ or $\frac{dx}{dy}$ as function of $x$ | M1 | Allow $\frac{dy}{dx} = k\frac{1}{\cos 2y} = \ldots \frac{1}{\sqrt{1-(\ldots x)^2}}$ |
| $\frac{dy}{dx} = \frac{1}{8\sqrt{1-\left(\frac{x}{4}\right)^2}}$ or $\frac{dx}{dy} = 8\sqrt{1-\left(\frac{x}{4}\right)^2}$ | A1 | Correct unsimplified answer |
| $\frac{dy}{dx} = \frac{1}{2\sqrt{16-x^2}}$ | A1 | Must see $\frac{dy}{dx}$ in correct final form in part (c) |

The image appears to be essentially blank/white, containing only the Pearson Education Limited copyright notice at the bottom and "PMT" in the top right corner. There is no mark scheme content visible on this page to extract.

This appears to be a blank back page of a mark scheme document. If you have other pages with actual mark scheme content, please share those and I'll be happy to extract and format the information for you.