## Question 3(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = 10$, $x = 5\ln 10 = 11.5$, so $OA = 5\ln 10$ | M1, A1 | Using $u = 10$ to find OA; accept 11.5 or better |
| When $u = 1$, $y = 1 + 1 = 2$ so $OB = 2$ | M1, A1 | Using $u = 1$ to find OB or $u = 10$ to find AC |
| When $u = 10$, $y = 10 + \frac{1}{10} = 10.1$, so $AC = 10.1$ | A1 | |
| **[5]** | | In case where values given as coordinates instead of OA=, OB=, AC=, give A0 on first occasion but allow subsequent As. Where coordinates followed by length e.g. B(0,2), length=2 then allow A1. |

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## Question 3(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/du}{dx/du} = \frac{1 - 1/u^2}{5/u}$ | M1 | their $dy/du \div dx/du$ |
| $= \frac{u^2 - 1}{5u}$ | A1 | Award A1 if any correct form seen at any stage including unsimplified (can isw) |
| **EITHER:** When $u = 10$, $dy/dx = 99/50 = 1.98$ | M1 | Substituting $u = 10$ in their expression |
| $\tan(90 - \theta) = 1.98 \Rightarrow \theta = 90 - 63.2$ | M1 | Or by geometry, using a triangle and the gradient of the line |
| $= 26.8°$ | A2 | 26.8°, or 0.468 radians (or better) cao. **SC** M1M0A1A0 for 63.2° (or better) or 1.103 radians (or better) |
| **OR:** When $u = 10$, $dy/dx = 99/50 = 1.98$ | M1 | Allow use of their expression for M marks |
| $\tan(90 - \theta) = 99/50 \Rightarrow \tan\theta = 50/99$ | M1 | |
| $\theta = 26.8°$ | A2 | 26.8°, or 0.468 radians (or better) cao |
| **[6]** | | |

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## Question 3(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 5\ln u \Rightarrow x/5 = \ln u$, $u = e^{x/5}$ | M1 | Need some working |
| $\Rightarrow y = u + 1/u = e^{x/5} + e^{-x/5}$ | A1 | Need some working as AG |
| **[2]** | | |

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## Question 3(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = \int_0^{5\ln 10} \pi y^2 dx = \int_0^{5\ln 10} \pi(e^{x/5} + e^{-x/5})^2 dx$ | M1 | Need $\pi(e^{x/5} + e^{-x/5})^2$ and $dx$ soi. Condone wrong limits or omission of limits for M1. Allow M1 if $y$ prematurely squared as e.g. $(e^{2x/5} + e^{-2x/5})$ |
| $= \int_0^{5\ln 10} \pi(e^{2x/5} + 2 + e^{-2x/5})dx$ | A1 | Including correct limits at some stage (condone 11.5 for this mark) |
| $= \pi\left[\frac{5}{2}e^{2x/5} + 2x - \frac{5}{2}e^{-2x/5}\right]_0^{5\ln 10}$ | B1 | $\left[\frac{5}{2}e^{2x/5} + 2x - \frac{5}{2}e^{-2x/5}\right]$ allow if no $\pi$ and/or no limits or incorrect limits |
| $= \pi(250 + 10\ln 10 - 0.025 - 0)$ | M1 | Substituting both limits (their OA and 0) in expression of correct form $ae^{2x/5} + be^{-2x/5} + cx$, $a,b,c \neq 0$ and subtracting in correct order (−0 sufficient for lower limit). Condone absence of $\pi$ for M1 |
| $= 858$ | A1 | Accept $273\pi$ and answers rounding to $273\pi$ or 858 |
| **[5]** | | NB integral can be evaluated using change of variable to $u$. For completely correct work award full marks. Partially correct solutions must include the change in $dx$. |

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