## Question 2:

### Part (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 | In the 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 |

**[5]**

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### Part (ii):

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

**[6]**

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### Part (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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### Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Vol of rev $= \displaystyle\int_0^{5\ln 10} \pi y^2\,dx = \int_0^{5\ln 10} \pi(e^{x/5} + e^{-x/5})^2\,dx$ | 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 $(e^{2x/5}+e^{-2x/5})$ |
| $= \displaystyle\int_0^{5\ln 10} \pi(e^{2x/5} + 2 + e^{-2x/5})\,dx$ | A1 | including correct limits at some stage (condone 11.5 for this mark) |
| $= \pi\left[\dfrac{5}{2}e^{2x/5} + 2x - \dfrac{5}{2}e^{-2x/5}\right]_0^{5\ln 10}$ | B1 | 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 $ae^{2x/5}+be^{-2x/5}+cx$, $a,b,c \neq 0$; subtracting in correct order; condone absence of $\pi$ for M1 |
| $= 858$ | A1 | accept $273\pi$ and answers rounding to $273\pi$ or 858 |

**[5]**

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