FP3 June 2014 Q4
4. Using the definitions of hyperbolic functions in terms of exponentials,
(a) show that \[\mathrm{sech}^2 x = 1 - \tanh^2 x\] (3)
(b) solve the equation \[4\sinh x - 3\cosh x = 3\] (4)
| Scheme | Marks |
|---|---|
| \(\tanh x = \dfrac{\frac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}}{\frac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}}\) or \(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\) or \(\dfrac{\mathrm{e}^{2x} - 1}{\mathrm{e}^{2x} + 1}\) Use of the correct exponential form of \(\tanh x\) | M1 |
| \(1 - \tanh^2 x = 1 - \left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}}\right)^2 = \dfrac{(\mathrm{e}^{2x} + \mathrm{e}^{-2x} + 2) - (\mathrm{e}^{2x} + \mathrm{e}^{-2x} - 2)}{(\mathrm{e}^x + \mathrm{e}^{-x})^2}\) Attempts \(1 - \tanh^2 x\) with their \(\tanh x\), obtains a common denominator and expands the numerator correctly – three terms from \((a + b)^2\) at least once | dM1 |
| \(= \dfrac{2\mathrm{e}^x \cdot 2\mathrm{e}^{-x}}{(\mathrm{e}^x + \mathrm{e}^{-x})^2}\) | |
| \(= \dfrac{4}{(\mathrm{e}^x + \mathrm{e}^{-x})^2} = \mathrm{sech}^2 x\ ^*\) Correct completion with no errors | A1* |
| (3) |
Notes
Allow candidates to process both sides and ‘meet in the middle’
Note that it is possible to start from \(\mathrm{sech}^2 x\) and obtain \(1 - \tanh^2 x\) by reversing the above work
| Scheme | Marks |
|---|---|
| Ignore any imaginary solutions in (b) | |
| \(4\left(\dfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2}\right) - 3\left(\dfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2}\right) = 3\) Substitutes the correct exponential forms for \(\sinh x\) and \(\cosh x\) | M1 |
| \(\mathrm{e}^x - 7\mathrm{e}^{-x} = 6\) | |
| \(\mathrm{e}^{2x} - 6\mathrm{e}^x - 7 = 0\) Obtains a quadratic in \(\mathrm{e}^x\) | M1 |
| \((\mathrm{e}^x + 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ in \(\mathrm{e}^x\) as far as \(\mathrm{e}^x = \ldots\) | M1 |
| \(x = \ln 7\) or awrt 1.95 | A1 |
| (4) | |
| (7 marks) |
Notes
Alternatives for (b)
| Scheme | Marks |
|---|---|
| \(4\sinh x - 3\cosh x = 3 \Rightarrow 4\sinh x = 3 + 3\cosh x\) \(\Rightarrow 7\cosh^2 x - 18\cosh x - 25 = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\cosh x\) | M1 |
| \(7\cosh^2 x - 18\cosh x - 25 = 0 \Rightarrow (7\cosh x - 25)(\cosh x + 1) = 0\) | |
| \(\cosh x = \tfrac{25}{7} \Rightarrow \tfrac{\mathrm{e}^x + \mathrm{e}^{-x}}{2} = \tfrac{25}{7} \Rightarrow 7\mathrm{e}^{2x} - 50\mathrm{e}^x + 7 = 0\) Uses the correct form of \(\cosh x\) in terms of exponentials to obtain a 3TQ in \(\mathrm{e}^x\) | M1 |
| \(7\mathrm{e}^{2x} - 50\mathrm{e}^x + 7 = 0 \Rightarrow (7\mathrm{e}^x - 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ as far as \(\mathrm{e}^x = \ldots\) | M1 |
| \(x = \ln 7\) or awrt 1.95 No other values | A1 |
| Scheme | Marks |
|---|---|
| \(4\sinh x - 3\cosh x = 3 \Rightarrow 4\sinh x - 3 = 3\cosh x\) \(\Rightarrow 7\sinh^2 x - 24\sinh x = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\sinh x\) | M1 |
| \(7\sinh^2 x - 24\sinh x = 0 \Rightarrow \sinh x(7\sinh x - 24) = 0\) | |
| \(\sinh x = \tfrac{24}{7} \Rightarrow \tfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{2} = \tfrac{24}{7} \Rightarrow 7\mathrm{e}^{2x} - 48\mathrm{e}^x - 7 = 0\) Uses the correct form of \(\sinh x\) in terms of exponentials to obtain a 3TQ in \(\mathrm{e}^x\) | M1 |
| \(7\mathrm{e}^{2x} - 48\mathrm{e}^x - 7 = 0 \Rightarrow (7\mathrm{e}^x + 1)(\mathrm{e}^x - 7) = 0 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 3TQ as far as \(\mathrm{e}^x = \ldots\) | M1 |
| \(x = \ln 7\) or awrt 1.95 No other values | A1 |
| Scheme | Marks |
|---|---|
| \(4\sinh x - 3\cosh x = 3 \Rightarrow 4\tanh x - 3 = 3\,\mathrm{sech}\,x\) \(\Rightarrow 25\tanh^2 x - 24\tanh x = 0\) M1: Attempt to square correctly and obtains a quadratic in \(\tanh x\) | M1 |
| \(25\tanh^2 x - 24\tanh x = 0 \Rightarrow \tanh x(25\tanh x - 24) = 0\) | |
| \(\tanh x = \tfrac{24}{25} \Rightarrow \tfrac{\mathrm{e}^x - \mathrm{e}^{-x}}{\mathrm{e}^x + \mathrm{e}^{-x}} = \tfrac{24}{25} \Rightarrow \mathrm{e}^{2x} = 49\) Uses the correct form of \(\tanh x\) in terms of exponentials to obtain a 2TQ in \(\mathrm{e}^x\) | M1 |
| \(\mathrm{e}^{2x} = 49 \Rightarrow \mathrm{e}^x = \ldots\) Attempt to solve their 2TQ as far as \(\mathrm{e}^x = \ldots\) | M1 |
| \(x = \ln 7\) or awrt 1.95 No other values | A1 |
(corrected from the printed mark scheme: in the first two alternatives the guidance is printed the other way round, “a quadratic in \(\sinh x\)” for the \(\cosh x\) quadratic and “a quadratic in \(\cosh x\)” for the \(\sinh x\) quadratic)