FP3 June 2012 Q7
7. \[\mathrm{f}(x) = 5\cosh x - 4\sinh x, \qquad x \in \mathbb{R}\]
(a) Show that \(\mathrm{f}(x) = \dfrac{1}{2}(e^{x} + 9e^{-x})\) (2)
Hence
(b) solve \(\mathrm{f}(x) = 5\) (4)
(c) show that \(\displaystyle\int_{\frac{1}{2}\ln 3}^{\ln 3} \frac{1}{5\cosh x - 4\sinh x}\,\mathrm{d}x = \frac{\pi}{18}\) (5)
| Scheme | Marks |
|---|---|
| \(\mathrm{f}(x) = 5\cosh x - 4\sinh x = 5 \times \tfrac{1}{2}(e^{x} + e^{-x}) - 4 \times \tfrac{1}{2}(e^{x} - e^{-x})\) | M1 |
| \(= \tfrac{1}{2}(e^{x} + 9e^{-x})\) * | A1cso |
| (2) |
Notes
a1M1: Replacing both \(\cosh x\) and \(\sinh x\) by terms in \(e^{x}\) and \(e^{-x}\) condone sign errors here.
a1A1: cso (answer given)
| Scheme | Marks |
|---|---|
| \(\tfrac{1}{2}(e^{x} + 9e^{-x}) = 5 \Rightarrow e^{2x} - 10e^{x} + 9 = 0\) | M1 A1 |
| So \(e^{x} = 9\) or 1 and \(x = \ln 9\) or 0 | M1 A1 |
| (4) |
Notes
b1M1: Getting a three term quadratic in \(e^{x}\)
b1A1: cao
b2M1: solving to \(x =\)
b2A1: cao need ln9 (o.e) and 0 (not ln1)
| Scheme | Marks |
|---|---|
| Integral may be written \(\displaystyle\int \dfrac{2e^{x}}{e^{2x} + 9}\,\mathrm{d}x\) | B1 |
| This is \(\dfrac{2}{3}\arctan\left(\dfrac{e^{x}}{3}\right)\) | M1 A1 |
| Uses limits to give \(\left[\tfrac{2}{3}\arctan 1 - \tfrac{2}{3}\arctan\left(\tfrac{1}{\sqrt{3}}\right)\right] = \left[\tfrac{2}{3} \times \tfrac{\pi}{4} - \tfrac{2}{3} \times \tfrac{\pi}{6}\right] = \tfrac{\pi}{18}\) * | DM1 A1cso |
| (5) | |
| (11 marks) |
Notes
c1B1: cao getting into suitable form, may substitute first.
c1M1: Integrating to give term in arctan
c1A1: cao
c2M1: Depends on previous M mark. Correct use of ln3 and ½ ln3 as limits.
c2A1: cso must see them subtracting two terms in \(\pi\).