A2 October 2021 Paper 1 Q8
8 You are given that \(\mathrm{f}(x) = 4\sinh x + 3\cosh x\).
(a) Show that the curve \(y = \mathrm{f}(x)\) has no turning points. [3]
(b) Determine the exact solution of the equation \(\mathrm{f}(x) = 5\). [5]
| Scheme | Marks | AO |
|---|---|---|
| \(y = 4\sinh x + 3\cosh x\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\cosh x + 3\sinh x\) | M1 | 1.1 |
| \(= 0\) when \(4\cosh x + 3\sinh x = 0\) \(\Rightarrow 4\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) + 3\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) = 0\) | M1 | 2.1 |
| \(\Rightarrow \mathrm{e}^{2x} = -\dfrac{1}{7}\) which is not possible as \(\mathrm{e}^{2x} \gt 0\) so no turning points | A1 | 2.4 |
| [3] |
Notes
M1: Diffn (Hyperbolics or exponentials)
M1: Set = 0 and use exponential forms – can change to exponentials before diffn.
A1: Conclusion with justification
Alternative method
| Scheme | Marks |
|---|---|
| \(y = 4\sinh x + 3\cosh x\) \(\Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = 4\cosh x + 3\sinh x\) | M1 |
| \(= 0\) when \(\tanh x = -\dfrac{4}{3}\) | M1 |
| But \(|\tanh x| \lt 1\) for all \(x\). So there are no values of \(x\) for which \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) So no turning points | A1 |
| [3] |
M1: Differentiate
M1: Set = 0 and use formula for tanh
A1: Conclusion with justification
| Scheme | Marks | AO |
|---|---|---|
| \(y = 4\sinh x + 3\cosh x = 5\) \(\Rightarrow 4\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) + 3\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) = 5\) | M1 | 3.1a |
| \(\Rightarrow 7\mathrm{e}^x - \mathrm{e}^{-x} = 10 \Rightarrow 7\mathrm{e}^{2x} - 10\mathrm{e}^x - 1 = 0\) | M1 | 3.1a |
| \(\mathrm{e}^x = \dfrac{10 \pm \sqrt{100 + 28}}{14} = \dfrac{5 + \sqrt{32}}{7}\) or \(\dfrac{5 - \sqrt{32}}{7}\) | A1 | 1.1 |
| But \(\mathrm{e}^x \gt 0\) so cannot \(= \dfrac{5 - \sqrt{32}}{7}\) So the only root is \(\mathrm{e}^x = \dfrac{5 + \sqrt{32}}{7}\) | A1 | 2.3 |
| \(\Rightarrow x = \ln\left(\dfrac{5 + 4\sqrt{2}}{7}\right)\) | A1 | 1.1 |
| [5] |
Notes
M1: Use of exponentials
M1: equation of the form \(a\mathrm{e}^{2x} + b\mathrm{e}^x + c = 0\) (for non-zero \(a\), \(b\) and \(c\))
A1: Two roots for \(\mathrm{e}^x\)
A1: One rejected plus reason
A1: Ignore inclusion of 2nd root
Alternative method
| Scheme | Marks |
|---|---|
| \(4\sinh x + 3\cosh x = 5 \Rightarrow 4\sinh x = 5 - 3\cosh x\) \(\therefore 16\sinh^2 x = 16\left(\cosh^2 x - 1\right) = 25 - 30\cosh x + 9\cosh^2 x\) | M1 |
| \(7\cosh^2 x + 30\cosh x - 41 = 0\) | M1 |
| \(\cosh x = \dfrac{-30 \pm \sqrt{30^2 - 4 \times 7 \times -41}}{2 \times 7} = \dfrac{-30 \pm \sqrt{2048}}{14}\) \(\cosh x \geqslant 1 \Rightarrow \cosh x = \dfrac{-30 + 32\sqrt{2}}{14} = \dfrac{-15 + 16\sqrt{2}}{7}\) | A1 |
| \(\Rightarrow x = \cosh^{-1}\dfrac{-15 + 16\sqrt{2}}{7} = \pm\ln\left(\dfrac{-15 + 16\sqrt{2}}{7} + \sqrt{\left(\dfrac{-15 + 16\sqrt{2}}{7}\right)^2 - 1}\right)\) \(= \pm\ln\left(\dfrac{-15 + 16\sqrt{2}}{7} + \sqrt{\dfrac{225 + 512 - 480\sqrt{2}}{49} - \dfrac{49}{49}}\right)\) \(= \pm\ln\left(\dfrac{-15 + 16\sqrt{2}}{7} + \sqrt{\dfrac{688 - 480\sqrt{2}}{49}}\right) = \pm\ln\left(\dfrac{-15 + 16\sqrt{2} + 4\sqrt{43 - 30\sqrt{2}}}{7}\right)\) But the negative root does not work in the original equation since LHS would be negative while RHS would be positive (but equal when squared). | A1 |
| \(\therefore x = \ln\left(\dfrac{-15 + 16\sqrt{2} + 4\sqrt{43 - 30\sqrt{2}}}{7}\right)\) NB \(\left(5 - 3\sqrt{2}\right)^2 = 25 + 18 - 30\sqrt{2} = 43 - 30\sqrt{2}\) and \(5 - 3\sqrt{2} \gt 0\) \(\therefore x = \ln\left(\dfrac{-15 + 16\sqrt{2} + 4\left(5 - 3\sqrt{2}\right)}{7}\right) = \ln\left(\dfrac{5 + 4\sqrt{2}}{7}\right)\) | A1 |
| [5] |
M1: Use Pythagoras
M1: Quadratic in cosh (or sinh)
A1: Two roots
A1: One rejected plus reason
(Full working of this alternative method is taken from the appendix of the mark scheme.)