# Question 1:

## Part (a)

| Working/Answer | Marks | Guidance |
|---|---|---|
| $y = \tanh^{-1}(x) \Rightarrow \tanh y = x \Rightarrow x = \frac{\sinh y}{\cosh y} = \frac{e^y - e^{-y}}{e^y + e^{-y}}$ | M1 | Begins proof by expressing tanh in terms of exponentials and forms an equation in exponentials |
| Correct expression for $x$ in terms of exponentials | A1 | Allow any variables but final answer must be in terms of $x$ |
| $x(e^{2y}+1) = e^{2y}-1 \Rightarrow e^{2y}(1-x) = 1+x \Rightarrow e^{2y} = \frac{1+x}{1-x}$ | M1 | Full method to make $e^{2y}$ the subject; correct algebra, allow sign errors only |
| $e^{2y} = \frac{1+x}{1-x} \Rightarrow 2y = \ln\left(\frac{1+x}{1-x}\right) \Rightarrow y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$* | A1* | Completes proof using logs correctly with no errors; needs $\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ as conclusion |
| $k=1$ or $-1 < x < 1$ | B1 | Correct value for $k$ or correct domain |

**Way 2:**

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) \Rightarrow x = \tanh\!\left(\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\right) = \frac{e^{\ln\frac{1+x}{1-x}}-1}{e^{\ln\frac{1+x}{1-x}}+1}$ | M1, A1 | Starts with result, takes tanh of both sides, expresses in terms of exponentials; correct expression |
| $x = \frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1} = \frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1} = x$ | M1, A1 | Eliminates exponentials and logs and simplifies; correct result $x=x$ with conclusion |
| $k=1$ or $-1 < x < 1$ | B1 | |

**(5 marks)**

## Part (b)

| Working/Answer | Marks | Guidance |
|---|---|---|
| $2x = \tanh\!\left(\ln\sqrt{2-3x}\right) \Rightarrow \tanh^{-1}(2x) = \ln\sqrt{2-3x}$ | M1 | Adopts correct strategy by taking $\tanh^{-1}$ of both sides |
| $\frac{1}{2}\ln\!\left(\frac{1+2x}{1-2x}\right) = \frac{1}{2}\ln(2-3x) \Rightarrow \frac{1+2x}{1-2x} = 2-3x$ | M1 | Makes link with part (a), uses power law of logs to obtain equation with logs removed correctly |
| $6x^2 - 9x + 1 = 0$ | A1 | Correct 3TQ |
| $6x^2 - 9x + 1 = 0 \Rightarrow x = \ldots$ | M1 | Solves their 3TQ using correct method |
| $x = \frac{9-\sqrt{57}}{12}$ | A1 | Correct value with other solution rejected (accept rejection by omission); $x = \frac{9\pm\sqrt{57}}{12}$ scores A0 unless positive root is rejected |

**Alternative (first 2 marks):**

| Working/Answer | Marks | Guidance |
|---|---|---|
| $2x = \tanh\!\left(\ln\sqrt{2-3x}\right) \Rightarrow 2x = \frac{e^{2\ln\sqrt{2-3x}}-1}{e^{2\ln\sqrt{2-3x}}+1}$ | M1 | Adopts correct strategy expressing tanh in terms of exponentials |
| $\Rightarrow \frac{2-3x-1}{2-3x+1} = 2x$ | M1 | Demonstrates use of power law of logs to obtain equation with logs removed correctly |

**(5 marks)**

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