# Question 2:

## Part (i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| A: $1 + \ln x = 0 \Rightarrow \ln x = -1$ so A is $(e^{-1}, 0)$ | M1 | |
| $\Rightarrow x = e^{-1}$ | A1 | SC1 if obtained using symmetry; condone use of symmetry |
| B: $x=0,\ y = e^{0-1} = e^{-1}$ so B is $(0, e^{-1})$ | B1 | Penalise $A = e^{-1}$, $B = e^{-1}$, or co-ords wrong way round, but condone labelling errors |
| C: $f(1) = e^{1-1} = e^0 = 1$ | E1 | |
| $g(1) = 1 + \ln 1 = 1$ | E1 | |
| **[5]** | | |

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## Part (ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| *Either* by inversion: $y = e^{x-1},\ x \leftrightarrow y$ | | |
| $x = e^{y-1}$ | | |
| $\Rightarrow \ln x = y - 1$ | M1 | taking lns or exps |
| $\Rightarrow 1 + \ln x = y$ | E1 | |
| *or* by composing: $fg(x) = f(1 + \ln x) = e^{1+\ln x - 1}$ | M1 | $e^{1+\ln x - 1}$ or $1 + \ln(e^{x-1})$ |
| $= e^{\ln x} = x$ | E1 | |
| **[2]** | | |

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## Part (iii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\displaystyle\int_0^1 e^{x-1}\,dx = \left[e^{x-1}\right]_0^1$ | M1 | $\left[e^{x-1}\right]$ o.e or $u = x-1 \Rightarrow [e^u]$ |
| $= e^0 - e^{-1}$ | M1 | substituting correct limits for $x$ or $u$ |
| $= 1 - e^{-1}$ | A1cao | o.e. not $e^0$, must be exact |
| **[3]** | | |

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## Part (iv)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\displaystyle\int \ln x\,dx = \int \ln x \cdot \frac{d}{dx}(x)\,dx$ | M1 | parts: $u = \ln x$, $du/dx = 1/x$, $v = x$, $dv/dx = 1$ |
| $= x\ln x - \displaystyle\int x \cdot \frac{1}{x}\,dx$ | A1 | |
| $= x\ln x - x + c$ | A1cao | condone no '$c$' |
| $\Rightarrow \displaystyle\int_{e^{-1}}^1 g(x)\,dx = \int_{e^{-1}}^1 (1+\ln x)\,dx$ | | |
| $= \left[x + x\ln x - x\right]_{e^{-1}}^1$ | B1ft | ft their '$x\ln x - x$' (provided 'algebraic') |
| $= \left[x\ln x\right]_{e^{-1}}^1$ | DM1 | substituting limits dep B1 |
| $= 1\ln 1 - e^{-1}\ln(e^{-1})$ | | |
| $= e^{-1}\ *$ | E1 | www |
| **[6]** | | |

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## Part (v)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \displaystyle\int_0^1 f(x)\,dx - \int_{e^{-1}}^1 g(x)\,dx$ | M1 | Must have correct limits |
| $= (1-e^{-1}) - e^{-1}$ | | |
| $= 1 - 2/e$ | A1cao | 0.264 or better |
| **[2]** | | |
| *or* Area OCB $=$ area under curve $-$ triangle $= 1-e^{-1} - \frac{1}{2}\times 1 \times 1 = \frac{1}{2} - e^{-1}$ | M1 | OCA or OCB $= \frac{1}{2} - e^{-1}$ |
| *or* Area OAC $=$ triangle $-$ area under curve $= \frac{1}{2}\times 1 \times 1 - e^{-1} = \frac{1}{2} - e^{-1}$ | | |
| Total area $= 2(\frac{1}{2} - e^{-1}) = 1 - 2/e$ | A1cao | 0.264 or better |

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