A2 June 2020 Paper 1 Q12
12
(a) Use the definition of the cosh function to prove that\[\cosh^{-1}\left(\frac{x}{a}\right) = \ln\left(\frac{x + \sqrt{x^2 - a^2}}{a}\right) \qquad \text{for } a \gt 0\] [6 marks]
(b) The formulae booklet gives the integral of \(\dfrac{1}{\sqrt{x^2 - a^2}}\) as\[\cosh^{-1}\left(\frac{x}{a}\right) \quad \text{or} \quad \ln\left(x + \sqrt{x^2 - a^2}\right) + c\]
Ronald says that this contradicts the result given in part (a).
Explain why Ronald is wrong. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Expresses \(x\) in terms of \(y\) | M1 | 3.1a |
| Recalls the exponential form of cosh. | B1 | 1.1b |
| Forms a quadratic equation in \(\mathrm{e}^y\) | M1 | 1.1a |
| Solves their quadratic equation in \(\mathrm{e}^y\), to obtain two solutions. | A1F | 1.1b |
| Explains that inverse cosh is defined to be non-negative or explains that \(\dfrac{x - \sqrt{x^2 - a^2}}{a}\) is \(\lt 1\) or shows that \(\dfrac{x - \sqrt{x^2 - a^2}}{a} = \dfrac{1}{\frac{x + \sqrt{x^2 - a^2}}{a}}\) | M1 | 2.4 |
| Completes a rigorous argument to show the required result, with reference to the larger root of the quadratic being the valid one with a clear reason. | R1 | 2.1 |
Typical solution
Let \(y = \cosh^{-1}\left(\frac{x}{a}\right)\); then \(x = a\cosh y\)
\[x = \frac{a}{2}\left(\mathrm{e}^y + \mathrm{e}^{-y}\right)\]\[\frac{2x}{a} = \mathrm{e}^y + \mathrm{e}^{-y}\]\(\times\,\mathrm{e}^y\):
\[\mathrm{e}^{2y} - \frac{2x}{a}\mathrm{e}^y + 1 = 0\]\[\mathrm{e}^y = \frac{\frac{2x}{a} \pm \sqrt{\frac{4x^2}{a^2} - 4}}{2}\]\[\mathrm{e}^y = \frac{x \pm \sqrt{x^2 - a^2}}{a}\]Product of roots = 1, so one root is greater than 1 and the other is less than 1.
\(y \geqslant 0\) by definition of \(\cosh^{-1} \therefore \mathrm{e}^y \geqslant 1\)
So we choose the larger root.
\[\mathrm{e}^y = \frac{x + \sqrt{x^2 - a^2}}{a}\]and
\[\cosh^{-1}\left(\frac{x}{a}\right) = \ln\left(\frac{x + \sqrt{x^2 - a^2}}{a}\right) \text{ as required.}\]| Scheme | Marks | AO |
|---|---|---|
| States \(\ln\left(\dfrac{x + \sqrt{x^2 - a^2}}{a}\right) = \ln\left(x + \sqrt{x^2 - a^2}\right) - \ln(a)\) | E1 | 2.4 |
| States \(c = -\ln(a)\). | E1 | 2.3 |
| (8 marks) |