AS June 2025 Paper 1 Q7
7
(a) Geraldine wants to show that\[\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\]
She writes her steps as follows:
\[\begin{aligned} &\text{Let} && y = \tanh^{-1} x \\ &\Rightarrow && \tanh y = x \\ &\Rightarrow && \frac{\mathrm{e}^y + \mathrm{e}^{-y}}{\mathrm{e}^y - \mathrm{e}^{-y}} = x \\ &\Rightarrow && \mathrm{e}^y + \mathrm{e}^{-y} = x\mathrm{e}^y - x\mathrm{e}^{-y} \\ &\Rightarrow && (1 + x)\mathrm{e}^{-y} = (x - 1)\mathrm{e}^y \\ &\Rightarrow && \mathrm{e}^{2y} = \frac{1 + x}{x - 1} \\ &\Rightarrow && 2y = \ln\left(\frac{x + 1}{x - 1}\right) \\ &\therefore && \tanh^{-1} x = \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right) \end{aligned}\]Identify and explain the error in Geraldine’s method. [2 marks]
(b) Use the correct identity to find\[\tanh^{-1}\left(-\frac{24}{25}\right)\]
Give your answer in the form \(\ln a\) where \(a\) is a rational number. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Identifies the error, eg 3rd line. | B1 | 2.3 |
| Explains that \(\tanh y\) should be \(\dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}}\) | E1 | 2.4 |
| (2) |
Typical solution
The error is in the 3rd line of working.
Geraldine uses an incorrect definition of tanh.
She should replace \(\tanh y\) with
\[\frac{\mathrm{e}^y - \mathrm{e}^{-y}}{\mathrm{e}^y + \mathrm{e}^{-y}}\]| Scheme | Marks | AO |
|---|---|---|
| Substitutes \(-\dfrac{24}{25}\) and uses the power law of logs. PI by \(-\ln 7\) | M1 | 1.1a |
| Obtains \(\ln\left(\dfrac{1}{7}\right)\) | A1 | 2.1 |
| (2) | ||
| (4 marks) |