| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.8 This is a Further Maths arc length question requiring hyperbolic function manipulation and integration. Part (a) needs the arc length formula with careful differentiation of ln(tanh(x/2)) using chain rule and hyperbolic identities to simplify to coth x. Part (b) requires integrating coth x = cosh x/sinh x and recognizing it as ln|sinh x|, then evaluating limits using exponential definitions. While technically demanding with multiple steps, it follows a standard arc length template with well-known hyperbolic results, making it challenging but within reach for prepared FM students. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{\tanh\frac{x}{2}} \times \frac{1}{2}\text{sech}^2\frac{x}{2}\) or equivalent | M1 | Obtains expression for \(\frac{dy}{dx}\) of appropriate form. Condone sign/coefficient errors and \(\frac{x}{2}\) written as \(x\) but no "\(y\)"s. Do not condone missing "h"s unless recovered |
| \(\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx \Rightarrow \int\sqrt{1+\left(\frac{\text{sech}^2\frac{x}{2}}{2\tanh\frac{x}{2}}\right)^2}\,dx\) | M1 | Applies arc length formula with their \(\frac{dy}{dx}\). Do not condone attempts that clearly work backwards to deduce derivative is \(\text{cosech}\, x\) |
| \(\sqrt{1+\left(\frac{1}{2\sinh\frac{x}{2}\cosh\frac{x}{2}}\right)^2} \rightarrow \sqrt{1+\left(\frac{1}{\sinh x}\right)^2}\) | ddM1 | Uses identity/identities (sign errors only) to obtain \(\sqrt{1+\left(\frac{dy}{dx}\right)^2}\) in terms of \(x\) not \(\frac{x}{2}\). Must square derivative and add 1 first before converting. Requires both previous M marks |
| \(\sqrt{1+\left(\frac{1}{\sinh x}\right)^2} = \sqrt{1+\text{cosech}^2 x} \Rightarrow s = \int_1^2\coth x\,dx\)* | A1* | Obtains given answer with no errors and at least one non-trivial intermediate line. Allow without "\(s=\)" but RHS must be exactly as printed. No missing "h"s |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\coth x\,dx = \ln(\sinh x)\) | B1 | Correct integration. May see \(-\ln(\text{cosech}\,x)\). May see \(\sinh x\) in exponentials without "2" |
| \(\ln\left(\frac{e^2-e^{-2}}{2}\right) - \ln\left(\frac{e-e^{-1}}{2}\right)\) written as single logarithm | M1 | Following \(\pm\ln(\sinh x)\) etc., substitutes limits, subtracts, writes as single logarithm. Condone sign errors |
| Obtains correct ln of single fraction with no negative powers | dM1 | Correct exponential form for sinh/cosech. Options: (1) correct ln of single fraction, (2) difference of two squares to factorise numerator, (3) correct multiplier, (4) correctly replaces \(\sinh 2\) with \(2\sinh 1\cosh 1\) |
| \(s = \ln\left(\frac{(e^2+1)(e^2-1)}{e(e^2-1)}\right) = \ln\left(e+\frac{1}{e}\right)\) or equivalent | A1* | Obtains given answer, complete and correct work. Allow \(\ln(e^{-1}+e)\) |
# Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{\tanh\frac{x}{2}} \times \frac{1}{2}\text{sech}^2\frac{x}{2}$ or equivalent | M1 | Obtains expression for $\frac{dy}{dx}$ of appropriate form. Condone sign/coefficient errors and $\frac{x}{2}$ written as $x$ but no "$y$"s. Do not condone missing "h"s unless recovered |
| $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx \Rightarrow \int\sqrt{1+\left(\frac{\text{sech}^2\frac{x}{2}}{2\tanh\frac{x}{2}}\right)^2}\,dx$ | M1 | Applies arc length formula with their $\frac{dy}{dx}$. Do not condone attempts that clearly work backwards to deduce derivative is $\text{cosech}\, x$ |
| $\sqrt{1+\left(\frac{1}{2\sinh\frac{x}{2}\cosh\frac{x}{2}}\right)^2} \rightarrow \sqrt{1+\left(\frac{1}{\sinh x}\right)^2}$ | ddM1 | Uses identity/identities (sign errors only) to obtain $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ in terms of $x$ not $\frac{x}{2}$. Must square derivative and add 1 first before converting. Requires both previous M marks |
| $\sqrt{1+\left(\frac{1}{\sinh x}\right)^2} = \sqrt{1+\text{cosech}^2 x} \Rightarrow s = \int_1^2\coth x\,dx$* | A1* | Obtains given answer with no errors and at least one non-trivial intermediate line. Allow without "$s=$" but RHS must be exactly as printed. No missing "h"s |
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# Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\coth x\,dx = \ln(\sinh x)$ | B1 | Correct integration. May see $-\ln(\text{cosech}\,x)$. May see $\sinh x$ in exponentials without "2" |
| $\ln\left(\frac{e^2-e^{-2}}{2}\right) - \ln\left(\frac{e-e^{-1}}{2}\right)$ written as single logarithm | M1 | Following $\pm\ln(\sinh x)$ etc., substitutes limits, subtracts, writes as single logarithm. Condone sign errors |
| Obtains correct ln of single fraction with no negative powers | dM1 | Correct exponential form for sinh/cosech. Options: (1) correct ln of single fraction, (2) difference of two squares to factorise numerator, (3) correct multiplier, (4) correctly replaces $\sinh 2$ with $2\sinh 1\cosh 1$ |
| $s = \ln\left(\frac{(e^2+1)(e^2-1)}{e(e^2-1)}\right) = \ln\left(e+\frac{1}{e}\right)$ or equivalent | A1* | Obtains given answer, complete and correct work. Allow $\ln(e^{-1}+e)$ |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7e38e2ed-ab5f-4906-940e-4b02c6992164-22_568_1192_376_440}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve with equation
$$y = \ln \left( \tanh \frac { x } { 2 } \right) \quad 1 \leqslant x \leqslant 2$$
(a) Show that the length, $s$, of the curve is given by
$$s = \int _ { 1 } ^ { 2 } \operatorname { coth } x \mathrm {~d} x$$
(b) Hence show that
$$s = \ln \left( \mathrm { e } + \frac { 1 } { \mathrm { e } } \right)$$
\hfill \mbox{\textit{Edexcel F3 2024 Q7 [8]}}