## Question 7:

### Part (i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Both curves correct shape and labelled | B1 | Both curves of the correct shape (ignore overlaps) and labelled |
| Gradient $= 1$ at $x = 0$ stated | B1 | gradient $= 1$ at $x = 0$ stated |
| Asymptotes $y = \pm 1$ and $x = \pm 1$ (or on sketch) | B1 | For asymptotes $y = \pm 1$ and $x = \pm 1$ |
| Sketch all correct | B1 | Sketch all correct |

**Total: 4 marks**

### Part (ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int_0^k \tanh x\,dx = \left[\ln(\cosh x)\right]_0^k = \ln(\cosh k)$ | M1 | For substituting limits into $\ln\cosh x$ |
| | A1 | For correct answer |

**Total: 2 marks**

### Part (iii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Areas shown are equal: $x = \tanh k \Rightarrow y = k$ | M1 | For consideration of areas |
| | A1 | For sufficient justification |
| $\Rightarrow \int_0^{\tanh k}\tanh^{-1}x\,dx = \text{rectangle } (k \times \tanh k) - \textbf{(ii)}$ | M1 | For subtraction from rectangle |
| $= k\tanh k - \ln(\cosh k)$ | A1 | For correct answer (**answer given**) |

**Alternative:** Otherwise by parts, as $1\times\tanh^{-1}x$ OR $1\times\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$

**Total: 4 marks**

---

### Part (iii) Alternative Method 1 (By parts):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $u = \tanh^{-1}x,\quad dv = dx$ | M1 | For integrating by parts (correct way round) |
| $du = \frac{1}{1-x^2}dx,\quad v = x$ | | |
| $\Rightarrow I = \left[x\tanh^{-1}x\right]_0^{\tanh k} - \int_0^{\tanh k}\frac{x}{1-x^2}\,dx$ | A1 | For getting this far |
| | M1 | Dealing with the resulting integral |
| $= k\tanh k + \frac{1}{2}\left[\ln(1-x^2)\right]_0^{\tanh k}$ | | |
| $= k\tanh k + \frac{1}{2}\ln(1-\tanh^2 k)$ | | |
| $= k\tanh k + \frac{1}{2}\ln(\text{sech}^2 k)$ | A1 | |
| $= k\tanh k + \ln(\text{sech}\, k)$ | | |

### Part (iii) Alternative Method 2 (By substitution):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $y = \tanh^{-1}x \Rightarrow x = \tanh y \Rightarrow dx = \text{sech}^2 y\,dy$ | M1 | For substitution to obtain equivalent integral |
| When $x=0,\, y=0$; when $x = \tanh k,\, y = k$ | | |
| $\Rightarrow I = \int_0^{\tanh k}\tanh^{-1}x\,dx = \int_0^k y\,\text{sech}^2 y\,dy$ | A1 | Correct so far |
| $u = y,\quad dv = \text{sech}^2 y\,dy;\quad du = dy,\quad v = \tanh y$ | M1 | For integration by parts (correct way round) |
| $\Rightarrow I = \left[y\tanh y\right]_0^k - \int_0^k \tanh y\,dy$ | | |
| $= k\tanh k - \ln\cosh k$ | A1 | Final answer |

---