AS June 2018 Paper 1 Q6

(a) Matthew is finding a formula for the inverse function \(\operatorname{arsinh} x\).
He writes his steps as follows:\[\begin{gathered}\text{Let } y = \sinh x \\ y = \frac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x}) \\ 2y = \mathrm{e}^x - \mathrm{e}^{-x} \\ 0 = \mathrm{e}^x - 2y - \mathrm{e}^{-x} \\ 0 = (\mathrm{e}^x)^2 - 2y\mathrm{e}^x - 1 \\ 0 = (\mathrm{e}^x - y)^2 - y^2 - 1 \\ y^2 + 1 = (\mathrm{e}^x - y)^2 \\ \pm\sqrt{y^2 + 1} = \mathrm{e}^x - y \\ y \pm \sqrt{y^2 + 1} = \mathrm{e}^x\end{gathered}\]

To find the inverse function, swap \(x\) and \(y\): \(x \pm \sqrt{x^2 + 1} = \mathrm{e}^y\)

\[\begin{gathered}\ln\left(x \pm \sqrt{x^2 + 1}\right) = y \\ \operatorname{arsinh} x = \ln\left(x \pm \sqrt{x^2 + 1}\right)\end{gathered}\]

Identify, and explain, the error in Matthew’s proof. [2 marks]

(b) Solve \(\ln\left(x + \sqrt{x^2 + 1}\right) = 3\) [1 mark]