A2 June 2024 Paper 2 Q7
7
A function is defined by \(\mathrm{f}(x) = \dfrac{1}{\sqrt{17\cosh x - 15\sinh x}}\). The region bounded by the curve \(y = \mathrm{f}(x)\), the \(x\)-axis, the \(y\)-axis and the line \(x = \ln 3\) is rotated by \(2\pi\) radians about the \(x\)-axis to form a solid of revolution \(S\).
Use a suitable substitution, together with known results from the formula book, to show that the volume of \(S\) is given by \(k\pi\tan^{-1}q\) where \(k\) and \(q\) are rational numbers to be determined. [7]
| Scheme | Marks | AO |
|---|---|---|
| \(17\cosh x - 15\sinh x = 17\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} - 15\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\) | M1 | 3.1a |
| \(= \dfrac{17 - 15}{2}\mathrm{e}^x + \dfrac{17 + 15}{2}\mathrm{e}^{-x}\ \left(= \mathrm{e}^x + 16\mathrm{e}^{-x}\right)\) | M1 | 1.1 |
| \(= \mathrm{e}^{-x}\left(\mathrm{e}^{2x} + 16\right)\) (so \(a = 1\), \(b = 2\) and \(c = 16\)) www | A1 | 2.2a |
| [3] |
Notes
M1: Substituting correct exponential definitions of hyperbolic functions into expression
M1: Collecting terms
eg \(s\mathrm{e}^x + t\mathrm{e}^{-x}\) where \(s\), \(t\) are non-zero constants, possibly unsimplified.
A1: Answer can be embedded. ISW.
| Scheme | Marks | AO |
|---|---|---|
| \(V = \pi\displaystyle\int_0^{\ln 3}\left(\frac{1}{\sqrt{17\cosh x - 15\sinh x}}\right)^2\mathrm{d}x\) | M1 | 1.1 |
| \(= \pi\displaystyle\int_0^{\ln 3}\frac{1}{\mathrm{e}^{-x}\left(\mathrm{e}^{2x} + 16\right)}\,\mathrm{d}x\) or \(\pi\displaystyle\int_0^{\ln 3}\frac{\mathrm{e}^x}{\mathrm{e}^{2x} + 16}\,\mathrm{d}x\) | M1 | 1.1 |
| \(u = \mathrm{e}^x\) | M1 | 3.1a |
| \(\mathrm{d}u = \mathrm{e}^x\,\mathrm{d}x \Rightarrow \therefore V = \pi\displaystyle\int_{\ldots}^{\ldots}\frac{\mathrm{e}^x}{\mathrm{e}^{2x} + 16}\,\mathrm{d}x = \pi\int_{\ldots}^{\ldots}\frac{1}{u^2 + 16}\,\mathrm{d}u\) | M1* | 1.1 |
| \(\displaystyle\int_{\ldots}^{\ldots}\frac{1}{u^2 + 16}\,\mathrm{d}u = \left[\frac{1}{4}\tan^{-1}\frac{u}{4}\right]_{\ldots}^{\ldots}\) | B1FT | 1.1 |
| \(x = 0 \Rightarrow u = \mathrm{e}^0 = 1,\ x = \ln 3 \Rightarrow u = \mathrm{e}^{\ln 3} = 3\) \(\therefore V = \dfrac{\pi}{4}\left[\tan^{-1}\dfrac{u}{4}\right]_1^3 = \dfrac{\pi}{4}\left(\tan^{-1}\dfrac{3}{4} - \tan^{-1}\dfrac{1}{4}\right)\) | depM1* | 1.1 |
| \(\tan\left(\tan^{-1}\dfrac{3}{4} - \tan^{-1}\dfrac{1}{4}\right) = \dfrac{\frac{3}{4} - \frac{1}{4}}{1 + \frac{3}{4} \times \frac{1}{4}} = \dfrac{12 - 4}{16 + 3} = \dfrac{8}{19}\) \(\therefore \tan^{-1}\dfrac{3}{4} - \tan^{-1}\dfrac{1}{4} = \tan^{-1}\dfrac{8}{19}\) \(\therefore V = \dfrac{1}{4}\pi\tan^{-1}\dfrac{8}{19}\ \left(\text{so } k = \dfrac{1}{4} \text{ and } q = \dfrac{8}{19}\right)\) | A1 | 3.1a |
| [7] |
Notes
M1: DR. Formula for VoR correctly used in solution (limits and/or \(\pi\) may come later) soi. \(\mathrm{d}x\) must be seen here but can be implied later.
M1: Squaring out and using their part (a) formula
M1: Useful substitution stated or used
eg \(u = \mathrm{e}^{-x}\) or \(\mathrm{e}^x = 4\tan u\) etc
M1*: Making the substitution (including \(\mathrm{d}x\)) to reduce to integrable form. Ignore limits here.
\(\mathrm{e}^x\,\mathrm{d}x = 4\sec^2 u\,\mathrm{d}u\) and \(V = \frac{\pi}{4}\int\mathrm{d}u\) if using \(\mathrm{e}^x = 4\tan u\)
B1FT: Correct arctan integration (ignore limits etc). FT their 16 and \(\sqrt{16}\)
Could be implicit in the substitution.
(ie \(\int\mathrm{d}u = [u]\) if using \(\mathrm{e}^x = 4\tan u\))
depM1*: Dealing with limits correctly (either converting to \(u\)-space or substituting back to \(x\)-space – if the latter ignore missing “\(x =\)” in the limits if correct at resubstitution)
Could see \(\dfrac{\pi}{4}\left[\tan^{-1}\dfrac{\mathrm{e}^x}{4}\right]_0^{\ln 3}\)
A1: Formula for \(\tan(A - B)\) must be used.
(NB It is possible to find 8/19 using tan and \(\tan^{-1}\) on a calculator)
0.3129987969... alone or used to derive \(k\) and/or \(q\) gets A0