FP3 June 2009 Q1
1. Solve the equation \[7\,\mathrm{sech}\,x - \tanh x = 5\]
Give your answers in the form \(\ln a\) where \(a\) is a rational number. (5)
| Scheme | Marks |
|---|---|
| \(\dfrac{7}{\cosh x} - \dfrac{\sinh x}{\cosh x} = 5 \;\Rightarrow\; \dfrac{14}{e^{x} + e^{-x}} - \dfrac{(e^{x} - e^{-x})}{e^{x} + e^{-x}} = 5\) | M1 |
| \(\therefore 14 - (e^{x} - e^{-x}) = 5(e^{x} + e^{-x}) \;\Rightarrow\; 6e^{x} - 14 + 4e^{-x} = 0\) | A1 |
| \(\therefore 3e^{2x} - 7e^{x} + 2 = 0 \;\Rightarrow\; (3e^{x} - 1)(e^{x} - 2) = 0\) | M1 |
| \(\therefore e^{x} = \dfrac{1}{3}\) or 2 | A1 |
| \(x = \ln\left(\tfrac{1}{3}\right)\) or \(\ln 2\) | B1ft |
| (5 marks) |
Notes
Alternative (i)
| Scheme | Marks |
|---|---|
| Write \(7 - \sinh x = 5\cosh x\), then use exponential substitution \(7 - \tfrac{1}{2}(e^{x} - e^{-x}) = \tfrac{5}{2}(e^{x} + e^{-x})\) Then proceed as method above. | M1 |
Alternative (ii)
| Scheme | Marks |
|---|---|
| \((7\,\mathrm{sech}\,x - 5)^2 = \tanh^2 x = 1 - \mathrm{sech}^2 x\) | M1 |
| \(50\,\mathrm{sech}^2 x - 70\,\mathrm{sech}\,x + 24 = 0\) | A1 |
| \(2(5\,\mathrm{sech}\,x - 3)(5\,\mathrm{sech}\,x - 4) = 0\) | M1 |
| \(\mathrm{sech}\,x = \tfrac{3}{5}\) or \(\mathrm{sech}\,x = \tfrac{4}{5}\) | A1 |
| \(x = \ln\left(\tfrac{1}{3}\right)\) or \(\ln 2\) | B1ft |