A2 June 2023 Paper 2 Q5
5 In this question you must show detailed reasoning.
| Scheme | Marks | AO |
|---|---|---|
| DR \((\text{RHS} = 2\sinh x\cosh x =)\ 2 \times \dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} \times \dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\) | M1 | 2.1 |
| \(= \dfrac{1}{2}\left(\mathrm{e}^{2x} + \mathrm{e}^0 - \mathrm{e}^0 - \mathrm{e}^{-2x}\right)\) \(= \dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} = \sinh 2x = \text{LHS}\) | A1 | 2.2a |
| [2] |
Notes
M1: AG. Use of exponential definition of sinh or cosh. Must be used on LHS or RHS
A1: AG. So an intermediate step must be shown, e.g. \(2\left(\frac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{4}\right)\). LHS must be equated to RHS. Accept full reverse argument
| Scheme | Marks | AO |
|---|---|---|
| DR \(15\sinh x + 16\cosh x - 12\sinh x\cosh x = 20\) | B1 | 3.1a |
| \(12sc - 16c - 15s + 20 = 0\) \((3s - 4)(4c - 5) = 0\) | M1 | 1.1 |
| \(x = \sinh^{-1}\left(\dfrac{4}{3}\right)\) or \(x = \cosh^{-1}\left(\dfrac{5}{4}\right)\) | A1 | 1.1 |
| \(\sinh^{-1}\dfrac{4}{3} = \ln\left(\dfrac{4}{3} + \sqrt{\left(\dfrac{4}{3}\right)^2 + 1}\right) = \ln 3\) | A1 | 1.1 |
| \(\cosh^{-1}\dfrac{5}{4} = \pm\ln\left(\dfrac{5}{4} + \sqrt{\left(\dfrac{5}{4}\right)^2 - 1}\right) = \pm\ln 2\) | A1 | 3.2a |
| [5] |
Notes
B1: Use of identity in (a).
M1: Writing as \(= 0\) and factorising (where \(s = \sinh x\) and \(c = \cosh x\))
A1: (1st) Complete solution in any form (assume that \(\cosh^{-1}\) is multi-valued here)
A1: (3rd) Must show \(\pm\) explicitly (or have both \(\ln 2\) and \(\ln\frac{1}{2}\))
Alternative method
| Scheme | Marks |
|---|---|
| \(15\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} + 16\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} - 6\dfrac{\mathrm{e}^{2x} - \mathrm{e}^{-2x}}{2} = 20\) | B1 |
| \(\therefore 15\mathrm{e}^x - 15\mathrm{e}^{-x} + 16\mathrm{e}^x + 16\mathrm{e}^{-x} - 6\mathrm{e}^{2x} + 6\mathrm{e}^{-2x} = 40\) \(\therefore 15\mathrm{e}^{3x} - 15\mathrm{e}^x + 16\mathrm{e}^{3x} + 16\mathrm{e}^x - 6\mathrm{e}^{4x} + 6 = 40\mathrm{e}^{2x}\) \(\therefore 6\mathrm{e}^{4x} - 31\mathrm{e}^{3x} + 40\mathrm{e}^{2x} - \mathrm{e}^x - 6 = 0\) | *M1 |
| \(y = \mathrm{e}^x \Rightarrow 6y^4 - 31y^3 + 40y^2 - y - 6 = 0\) \(6 \times 16 - 31 \times 8 + 40 \times 4 - 2 - 6 = 256 - 256 = 0\) \(6y^3(y - 2) - 19y^2(y - 2) + 2y(y - 2) + 3(y - 2) = 0\) \((y - 2)(6y^3 - 19y^2 + 2y + 3) = 0\) | *dep*M1 |
| \(6 \times 27 - 19 \times 9 + 2 \times 3 + 3 = 171 - 171 = 0\) \(6y^3 - 19y^2 + 2y + 3 = 6y^2(y - 3) - y(y - 3) - (y - 3)\) \(= (y - 3)(6y^2 - y - 1) = (y - 3)(2y - 1)(3y + 1)\) | dep*M1 |
| \(\therefore y = \mathrm{e}^x = 2, \frac{1}{2}, 3\) or \(-\frac{1}{3}\). But \(\mathrm{e}^x \gt 0\) \(\therefore x = \ln 2, \ln\frac{1}{2}\) (or \(-\ln 2\)) or \(\ln 3\) only | A1 |
B1: Use of exponential definitions of \(\sinh x\), \(\cosh x\) and \(\sinh 2x\) in equation. Also award if starts with main method before using exponentials
*M1: Multiplying by \(\mathrm{e}^{2x}\) and collecting like terms to write as quartic equation in \(\mathrm{e}^x\). Could see a substitution, e.g. \(y = \mathrm{e}^x\) leading to \(6y^4 - 31y^3 + 40y^2 - y - 6 = 0\). Could use Pythagoras to derive quartic in sinh or cosh
*dep*M1: Using factor theorem to deduce that \(\mathrm{e}^x = 2\) (or 3 or \(\frac{1}{2}\)) is a solution and factorising; or \((6y^2 - y - 1)(y^2 - 5y + 6)\) seen
dep*M1: Using factor theorem to find another factor and fully factorising
A1: Must reject negative root explicitly for A1. ScB1 for correct solution after B1M1M0M0