A2 June 2025 Paper 1 Q14
14
(a) Using the exponential definitions of \(\sinh x\) and \(\cosh x\), prove that\[\coth^{-1}(x) = \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right)\] [3 marks]
(b) Hence, solve the equation\[\coth^{-1}(x) = -\ln 5\]
Give your answer in an exact form. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses definitions of sinh and cosh. PI \(x = \dfrac{\mathrm{e}^y + \mathrm{e}^{-y}}{\mathrm{e}^y - \mathrm{e}^{-y}}\) or \(x = \dfrac{\mathrm{e}^{2y} + 1}{\mathrm{e}^{2y} - 1}\) | M1 | 1.1a |
| Rearranges to make \(\mathrm{e}^{2y}\) or \(\mathrm{e}^{2x}\) the subject. or Uses the quadratic equation formula | M1 | 3.1a |
| Completes a reasoned argument, starting from the full definitions of sinh and cosh, to obtain \(\coth^{-1}(x) = \dfrac{1}{2}\ln\left(\dfrac{x + 1}{x - 1}\right)\) AG | R1 | 2.1 |
| (3) |
Typical solution
Let \(y = \coth^{-1}(x)\)
Then \(x = \coth y\)
\[x = \frac{\frac{1}{2}(\mathrm{e}^y + \mathrm{e}^{-y})}{\frac{1}{2}(\mathrm{e}^y - \mathrm{e}^{-y})} = \frac{\mathrm{e}^{2y} + 1}{\mathrm{e}^{2y} - 1}\]\[x(\mathrm{e}^{2y} - 1) = \mathrm{e}^{2y} + 1\]\[x\mathrm{e}^{2y} - x = \mathrm{e}^{2y} + 1\]\[\mathrm{e}^{2y}(x - 1) = x + 1\]\[\mathrm{e}^{2y} = \frac{x + 1}{x - 1}\]\[2y = \ln\left(\frac{x + 1}{x - 1}\right)\]\[y = \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right)\]\[\coth^{-1}(x) = \frac{1}{2}\ln\left(\frac{x + 1}{x - 1}\right)\]| Scheme | Marks | AO |
|---|---|---|
| Uses the result from part (a) to form an equation and applies at least one law of logs correctly | M1 | 3.1a |
| Obtains \(\dfrac{x + 1}{x - 1} = \dfrac{1}{25}\) OE | A1 | 1.1b |
| Obtains \(-\dfrac{13}{12}\) OE | A1 | 1.1b |
| (3) | ||
| (6 marks) |