A2 June 2022 Paper 2 Q5
5
(a) By using the exponential definitions of \(\sinh x\) and \(\cosh x\), prove the identity \(\cosh 2x \equiv \cosh^2 x + \sinh^2 x\). [2]
(b) Hence find an expression for \(\cosh 2x\) in terms of \(\cosh x\). [1]
(c) Determine the solutions of the equation \(5\cosh 2x = 16\cosh x + 21\), giving your answers in exact logarithmic form. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle \begin{aligned} \text{RHS} &= \cosh^2 x + \sinh^2 x \\[4pt] &= \left(\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right)^2 + \left(\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right)^2 \\[4pt] &= \frac{1}{4}\left(\mathrm{e}^{2x} + \mathrm{e}^{-2x} + 2 + \mathrm{e}^{2x} + \mathrm{e}^{-2x} - 2\right) \\[4pt] &= \frac{1}{4}\left(2\mathrm{e}^{2x} + 2\mathrm{e}^{-2x}\right) = \frac{1}{2}\left(\mathrm{e}^{2x} + \mathrm{e}^{-2x}\right) \end{aligned}\) | M1 | 2.1 |
| \(= \cosh 2x = \text{LHS}\) | A1 | 1.1 |
| [2] |
Notes
M1: Using definitions of \(\cosh x\) and \(\sinh x\).
A1: AG. Intermediate working must be seen.
| Scheme | Marks | AO |
|---|---|---|
| \(\cosh^2 x + \sinh^2 x = \cosh 2x\) & \(\cosh^2 x - \sinh^2 x = 1\) \(\Rightarrow 2\cosh^2 x = \cosh 2x + 1\) \(\Rightarrow \cosh 2x = 2\cosh^2 x - 1\) | B1 | 2.2a |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \(10\cosh^2 x - 5 = 16\cosh x + 21\) \(\Rightarrow 10c^2 - 16c - 26 = 0 \Rightarrow 5c^2 - 8c - 13 = 0\) | M1 | 1.1 |
| \(c = -1\) rejected since \(\cosh x \geqslant 1\) (or \(\geqslant 0\) oe) or \(c = 13/5\) | A1 | 2.3 |
| \(\cosh^{-1}\dfrac{13}{5} = \ln\left(\dfrac{13}{5} + \sqrt{\left(\dfrac{13}{5}\right)^2 - 1}\right) = \ln 5\) | M1 | 1.1 |
| \(\therefore x = \pm\ln 5\) | A1 | 2.2a |
| [4] |
Notes
M1: Using the identity from (b) to reduce equation to 3 term quadratic
A1: Both solutions found… …and \(-1\) rejected explicitly with valid reason. E.g “\(-1\) is outside the range of \(\cosh x\)”
Could be BC
M1: Use of formula for \(\cosh^{-1}\) (or by solving quadratic in \(\mathrm{e}^x\)).
A1: \(x = \ln 5\) or \(\ln(1/5)\). Must have both solutions.
Alternative method
| Scheme | Marks |
|---|---|
| \(\dfrac{5}{2}\left(\mathrm{e}^{2x} + \mathrm{e}^{-2x}\right) = 8\left(\mathrm{e}^x + \mathrm{e}^{-x}\right) + 21\) \(5\mathrm{e}^{4x} - 16\mathrm{e}^{3x} - 42\mathrm{e}^{2x} - 16\mathrm{e}^x + 5 = 0\) | M1 |
| \(\mathrm{e}^x = \mathrm{e} \Rightarrow\) \(5\mathrm{e}^3(\mathrm{e} - 5) + 9\mathrm{e}^2(\mathrm{e} - 5) + 3\mathrm{e}(\mathrm{e} - 5) - (\mathrm{e} - 5) = 0\) \((\mathrm{e} - 5)(5\mathrm{e}^3 + 9\mathrm{e}^2 + 3\mathrm{e} - 1) = 0\) \((\mathrm{e} - 5)(\mathrm{e}^2(5\mathrm{e} - 1) + 2\mathrm{e}(5\mathrm{e} - 1) + (5\mathrm{e} - 1)) = 0\) \((\mathrm{e} - 5)(5\mathrm{e} - 1)(\mathrm{e}^2 + 2\mathrm{e} + 1) = 0\) \((\mathrm{e} - 5)(5\mathrm{e} - 1)(\mathrm{e} + 1)^2 = 0\) \(\therefore \mathrm{e} = -1,\ \dfrac{1}{5}\) or \(5\) | M1 |
| \(\mathrm{e} = -1 \Rightarrow \mathrm{e}^x = -1\) which is not possible since \(\mathrm{e}^x \gt 0\) for all (real) \(x\). | A1 |
| So \(\mathrm{e}^x = 5\) or \(1/5 \Rightarrow x = \ln 5\) or \(\ln(1/5)\) | A1 |
| [4] |
M1: Using the exponential definition of cosh to reduce the given equation to a quartic equation in \(\mathrm{e}^x\).
M1: Factorising or using quartic solver.
A1: Negative solution must be rejected and a valid reason given.
A1: \(x = \pm\ln 5\). Must have both solutions.