AS June 2020 Paper 1 Q8
8
(a) Prove that\[\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\]
[5 marks]
(b) Prove that the graphs of\[y = \sinh x \quad \text{and} \quad y = \cosh x\]
do not intersect. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes \(\tanh x\) or \(\tanh y\) in exponential form. | B1 | 1.2 |
| Selects a method by forming an equation of the form \(x = \tanh y\). | M1 | 3.1a |
| Forms an equation in \(\mathrm{e}^{2y}\). | M1 | 1.1a |
| Isolates the terms in \(\mathrm{e}^{2y}\). | M1 | 1.1a |
| Completes a rigorous argument to show that \(\tanh^{-1} x = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\) | R1 | 2.1 |
Typical solution
Let \(y = \tanh^{-1} x\)
\[x = \tanh y\]\[x = \frac{\left(\dfrac{\mathrm{e}^y - \mathrm{e}^{-y}}{2}\right)}{\left(\dfrac{\mathrm{e}^y + \mathrm{e}^{-y}}{2}\right)} = \frac{\mathrm{e}^{2y} - 1}{\mathrm{e}^{2y} + 1}\]\[\begin{aligned}x(\mathrm{e}^{2y} + 1) &= \mathrm{e}^{2y} - 1 \\ x\mathrm{e}^{2y} + x &= \mathrm{e}^{2y} - 1 \\ 1 + x &= \mathrm{e}^{2y} - x\mathrm{e}^{2y} \\ 1 + x &= \mathrm{e}^{2y}(1 - x) \\ \frac{1 + x}{1 - x} &= \mathrm{e}^{2y} \\ 2y &= \ln\left(\frac{1 + x}{1 - x}\right) \\ y &= \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\end{aligned}\]\[\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)\]Notes
(corrected from the printed mark scheme: the typical solution prints the denominator of the compound fraction for \(x\) as \(\dfrac{(\mathrm{e}^y - \mathrm{e}^{-y})}{2}\); it should be \(\dfrac{(\mathrm{e}^y + \mathrm{e}^{-y})}{2}\).)
| Scheme | Marks | AO |
|---|---|---|
| Selects a method by assuming, for contradiction, that the graphs do meet and forms the equation \(\sinh x = \cosh x\). | M1 | 3.1a |
| Deduces that \(\tanh x = 1\) or that \(2\mathrm{e}^{-x} = 0\). | A1 | 2.2a |
| Completes a rigorous argument to prove the required result that the graphs of \(y = \sinh x\) and \(y = \cosh x\) do not intersect. | R1 | 2.1 |
| (8 marks) |
Typical solution
If the graphs meet, then \(\sinh x = \cosh x\)
\[\begin{aligned}\tfrac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x}) &= \tfrac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x}) \\ \mathrm{e}^x - \mathrm{e}^{-x} &= \mathrm{e}^x + \mathrm{e}^{-x} \\ 0 &= 2\mathrm{e}^{-x}\end{aligned}\]but \(\mathrm{e}^{-x} \gt 0\)
\(\therefore\) the graphs of \(y = \sinh x\) and \(y = \cosh x\) do not intersect