| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple hyperbolic function manipulations, inverse hyperbolic functions, and arc length integration. Part (i) involves standard but non-trivial algebraic manipulation and solving for x. Part (ii) requires computing arc length with dy/dx involving hyperbolic functions, then integrating over specified limits—this demands careful algebraic simplification and is beyond routine A-level. However, it follows a structured path with clear sub-parts guiding the solution, preventing it from reaching the highest difficulty tier. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms8.06a Reduction formulae: establish, use, and evaluate recursively |
(i) **(a)** $\tanh x = \frac{\sinh x}{\cosh x} = \frac{\frac{1}{2}(\mathrm{e}^x - \mathrm{e}^{-x})}{\frac{1}{2}(\mathrm{e}^x + \mathrm{e}^{-x})} \times \frac{\mathrm{e}^x}{\mathrm{e}^x} = \frac{\mathrm{e}^{2x}-1}{\mathrm{e}^{2x}+1}$ **M1**
**Answer Given** so MUST justify final answer **A1** [2]
**(b)** $\tanh x = \frac{\mathrm{e}^{2x}-1}{\mathrm{e}^{2x}+1} = \frac{1}{k} \Rightarrow k\mathrm{e}^{2x} - k = \mathrm{e}^{2x} + 1$ Equated for $k$ **M1**
$\Rightarrow (k-1)\mathrm{e}^{2x} = k+1 \Rightarrow x = \frac{1}{2}\ln\left(\frac{k+1}{k-1}\right)$ **Answer Given** **A1**
$\sinh 2x = \frac{1}{2}\left(\mathrm{e}^{2x} - \mathrm{e}^{-2x}\right) = \frac{1}{2}\left(\frac{k+1}{k-1} - \frac{k-1}{k+1}\right)$ **M1**
or via $t = \tanh\left(\frac{1}{2}\text{-"angle"}\right)$ identity **A1** [4]
$= \frac{1}{2}\left(\frac{(k^2+2k+1)-(k^2-2k+1)}{k^2-1}\right) = \frac{2k}{k^2-1}$
(ii) $y = \frac{1}{2}\ln(\tanh x) \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \cdot \frac{1}{\tanh x} \cdot \mathrm{sech}^2 x$ **M1 A1**
$= \frac{1}{\sinh 2x}$ Here or later (or equivalent) **A1**
$1 + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = 1 + \frac{1}{\sinh^2 2x} = \frac{\cosh^2 2x}{\sinh^2 2x}$ **M1 A1**
M0 if no progress towards an integrable form
$L = \int_\alpha^\beta \frac{\cosh 2x}{\sinh 2x}\ \mathrm{d}x = \left[\frac{1}{2}\ln(\sinh 2x)\right]_\alpha^\beta$ For the integration **M1 M1 A1**
NB Alternative approaches lead to $\frac{1}{2}\int(\tanh x + \coth x)\ \mathrm{d}x = \frac{1}{2}[\ln(\cosh x) + \ln(\sinh x)]$
For $\beta$, $k=2 \Rightarrow \sinh 2x = \frac{4}{3}$; for $\alpha$, $k=3 \Rightarrow \sinh 2x = \frac{3}{4}$ **M1**
$\Rightarrow L = \frac{1}{2}\ln\frac{4}{3} - \frac{1}{2}\ln\frac{3}{4} = \ln\frac{4}{3}$ **A1** [10]12
\begin{enumerate}[label=(\roman*)]
\item (a) Show that $\tanh x = \frac { \mathrm { e } ^ { 2 x } - 1 } { \mathrm { e } ^ { 2 x } + 1 }$.\\
(b) Hence, or otherwise, show that, if $\tanh x = \frac { 1 } { k }$ for $k > 1$, then $x = \frac { 1 } { 2 } \ln \left( \frac { k + 1 } { k - 1 } \right)$ and find an expression in terms of $k$ for $\sinh 2 x$.
\item A curve has equation $y = \frac { 1 } { 2 } \ln ( \tanh x )$ for $\alpha \leqslant x \leqslant \beta$, where $\tanh \alpha = \frac { 1 } { 3 }$ and $\tanh \beta = \frac { 1 } { 2 }$. Find, in its simplest exact form, the arc length of this curve.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2014 Q12 [10]}}