## Question 10:

### Part (a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \sqrt{1+\frac{x^2}{9}} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}\left(1+\frac{x^2}{9}\right)^{-\frac{1}{2}} \times \frac{2x}{9} = \frac{x}{9}\left(1+\frac{x^2}{9}\right)^{-\frac{1}{2}}$ | M1 A1 | Attempts to find $\frac{\mathrm{d}y}{\mathrm{d}x}$ achieving the form $Ax\left(1+\frac{x^2}{9}\right)^{-\frac{1}{2}}$; Correct derivative |
| $S = 2\pi\int y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x = 2\pi\int\sqrt{1+\frac{x^2}{9}}\sqrt{1+\frac{x^2}{81}\left(1+\frac{x^2}{9}\right)^{-1}}\,\mathrm{d}x$ | M1 | Uses formula surface area $= 2\pi\int y\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\,\mathrm{d}x$ |
| $= 2\pi\int\frac{1}{3}\sqrt{9+x^2}\sqrt{\frac{9(9+x^2)+x^2}{9(9+x^2)}}\,\mathrm{d}x = 2\pi\int\frac{1}{3}\sqrt{9+x^2}\sqrt{\frac{81+10x^2}{9(9+x^2)}}\,\mathrm{d}x$ **Or** $= 2\pi\int\sqrt{1+\frac{x^2}{9}+\frac{x^2}{81}}\,\mathrm{d}x = \frac{2\pi}{9}\int\sqrt{81+9x^2+x^2}\,\mathrm{d}x$ | M1 | Manipulates the integral and simplifies to the given form |
| Circular end has area $\pi \times y^2 = \pi\left(1+\frac{16}{9}\right) = \frac{25\pi}{9}$ | B1 | Correct area for a circular end found |
| So $S = \frac{2\pi}{9}\int_{-4}^{4}\sqrt{81+10x^2}\,\mathrm{d}x + \frac{50\pi}{9}$ | A1 | Achieves the correct answer with no errors seen, including the limits from the model |

### Part (b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \frac{9}{\sqrt{10}}\sinh u \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}u} = \frac{9}{\sqrt{10}}\cosh u$ | B1 | Correct derivative statement connecting $x$ and $u$ |
| So $S = "\frac{2}{9}"\pi\int\sqrt{81+81\sinh^2 u}\cdot\frac{9}{\sqrt{10}}\cosh u\,\mathrm{d}u$ | M1 | Makes a full substitution to obtain an integral in terms of $u$ only |
| $= B\pi\int\cosh^2 u\,\mathrm{d}u = B\pi\int\frac{1}{2}(\pm1\pm\cosh 2u)\,\mathrm{d}u$ | M1 | Simplifies and applies double angle formula of the form $\cosh 2u = \pm1\pm\cosh^2 u$ |
| $= "\frac{2}{9}"\times\frac{81\pi}{\sqrt{10}}\left[\frac{u}{2}+\frac{1}{4}\sinh 2u\right]$ | A1ft | For correct integration with their $p$ from (a) |
| $S = \frac{50\pi}{9} + "\frac{2}{9}"\times\frac{81\pi}{\sqrt{10}}\left[\frac{1}{2}\mathrm{arsinh}\left(\frac{4\sqrt{10}}{9}\right)+\frac{1}{4}\sinh 2\mathrm{arsinh}\left(\frac{4\sqrt{10}}{9}\right) - \left(\frac{1}{2}\mathrm{arsinh}\left(\frac{-4\sqrt{10}}{9}\right)+\frac{1}{4}\sinh 2\mathrm{arsinh}\left(\frac{-4\sqrt{10}}{9}\right)\right)\right]$ $S = \frac{50\pi}{9} + "\frac{2}{9}"\times\frac{81\pi}{\sqrt{10}}\left[(1.7827\ldots)-(-1.7827\ldots)\right] = \ldots$ | M1 | Applies appropriate limits; either $-4$ and $4$ if returning to $x$, or $\mathrm{arsinh}\left(\frac{\pm4\sqrt{10}}{9}\right)$ if using $u$ |
| Surface area is awrt $81\ \mathrm{cm}^2$ | A1 | Correct surface area |

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

If you have additional pages from the mark scheme, please share them and I'll be happy to extract and format the content for you.