A2 June 2024 Paper 1 Q9
9
(a) It is given that\[p = \ln\left(r + \sqrt{r^2 + 1}\right)\]
Starting from the exponential definition of the sinh function, show that \(\sinh p = r\) [4 marks]
(b) Solve the equation\[\cosh^2 x = 2\sinh x + 16\]
Give your answers in logarithmic form. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes down \(\dfrac{\mathrm{e}^p - \mathrm{e}^{-p}}{2}\) OE | B1 | 1.2 |
| Substitutes \(p\) with the given expression and removes e and ln to obtain \(r + \sqrt{r^2 + 1}\) or \(\dfrac{1}{r + \sqrt{r^2 + 1}}\) | M1 | 3.1a |
| Collects over common denominator \(r + \sqrt{r^2 + 1}\) or Multiplies \(\dfrac{1}{r + \sqrt{r^2 + 1}}\) by \(\dfrac{r - \sqrt{r^2 + 1}}{r - \sqrt{r^2 + 1}}\) | M1 | 1.1a |
| Completes reasoned argument to obtain \(\sinh p = r\) AG | R1 | 2.1 |
| (4) |
Typical solution
\[\begin{aligned}\sinh p &= \frac{\mathrm{e}^p - \mathrm{e}^{-p}}{2} \\ &= \frac{\mathrm{e}^{\ln\left(r + \sqrt{r^2 + 1}\right)} - \mathrm{e}^{-\ln\left(r + \sqrt{r^2 + 1}\right)}}{2} \\ &= \frac{1}{2}\left(r + \sqrt{r^2 + 1} - \frac{1}{r + \sqrt{r^2 + 1}}\right) \\ &= \frac{1}{2}\left(\frac{\left(r + \sqrt{r^2 + 1}\right)^2 - 1}{r + \sqrt{r^2 + 1}}\right) \\ &= \frac{1}{2}\left(\frac{\left(r^2 + 2r\sqrt{r^2 + 1} + r^2 + 1\right) - 1}{r + \sqrt{r^2 + 1}}\right) \\ &= \frac{1}{2}\left(\frac{2r^2 + 2r\sqrt{r^2 + 1}}{r + \sqrt{r^2 + 1}}\right) \\ &= \frac{1}{2}\left(\frac{2r\left(r + \sqrt{r^2 + 1}\right)}{r + \sqrt{r^2 + 1}}\right) \\ &= r\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Uses \(\cosh^2 x = 1 + \sinh^2 x\) to obtain \(1 + \sinh^2 x = 2\sinh x + 16\) or Substitutes in correct exponential form to obtain \(\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 = 2\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) + 16\) | M1 | 3.1a |
| Obtains \((\sinh x =)\ -3, 5\) or Obtains \((\mathrm{e}^x =)\ 10.099\ldots, 0.162\ldots\) | A1 | 1.1b |
| Uses \(\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)\) FT their \(-3\) or 5 | M1 | 1.1a |
| Obtains \(\ln\left(-3 + \sqrt{10}\right)\), \(\ln\left(5 + \sqrt{26}\right)\) and no other solutions | A1 | 1.1b |
| (4) | ||
| (8 marks) |
Typical solution
Let \(s = \sinh x\)
\[s^2 + 1 = 2s + 16\]\[s^2 - 2s - 15 = 0\]\[s = -3, 5\]\[x = \ln\left(-3 + \sqrt{10}\right),\quad x = \ln\left(5 + \sqrt{26}\right)\]