| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a Further Maths arc length question requiring knowledge of hyperbolic function derivatives and arc length formula. While it involves multiple steps (differentiate arcosh, set up arc length integral, simplify using hyperbolic identities, integrate), the techniques are standard for F3 and the algebraic simplification follows predictable patterns. The question is harder than typical A-level due to the Further Maths content, but it's a routine application of the arc length formula with hyperbolic functions rather than requiring novel insight. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{1}{2} \times \frac{2}{\sqrt{(2x)^2-1}}\) | M1 | Attempts \(\frac{dy}{dx}\), accept form \(\frac{A}{\sqrt{(2x)^2-1}}\). Allow \(\frac{A}{\sqrt{2x^2-1}}\) (condone missing brackets) |
| \(1+\left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4x^2-1} = \frac{4x^2}{4x^2-1}\) | M1 | Attempts to find \(1+\left(\frac{dy}{dx}\right)^2\) using their \(\frac{dy}{dx}\) and attempts common denominator |
| \(\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int\sqrt{\frac{4x^2}{4x^2-1}}\,dx = 2\int\frac{x}{\sqrt{4x^2-1}}\,dx\) | A1 | Reaches correct simplified integral with \(\sqrt{x^2}\) replaced with \(x\). Allow equivalent forms e.g. \(2\int x\sqrt{\frac{1}{4x^2-1}}\,dx\), \(\frac{1}{2}\int\frac{4x}{\sqrt{(2x)^2-1}}\,dx\) |
| \(= \frac{2(4x^2-1)^{\frac{1}{2}}}{8 \times \frac{1}{2}}\) | M1 | Attempts integration, reaches form \(\alpha(\beta x^2-1)^{\frac{1}{2}}\) |
| \(s = \left[\frac{(4x^2-1)^{\frac{1}{2}}}{2}\right]_{\frac{7}{2}}^{13} = \frac{1}{2}\left(\sqrt{4\times169-1}-\sqrt{4\times\frac{49}{4}-1}\right) = \ldots\) | dM1 | Applies limits to their integral. Depends on previous 2 method marks. |
| \(= \frac{1}{2}(15\sqrt{3}-4\sqrt{3}) = \frac{11}{2}\sqrt{3}\) | A1 | cao Accept equivalents e.g. \(\frac{1}{2}\sqrt{363}\) |
| (6) |
# Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{1}{2} \times \frac{2}{\sqrt{(2x)^2-1}}$ | M1 | Attempts $\frac{dy}{dx}$, accept form $\frac{A}{\sqrt{(2x)^2-1}}$. Allow $\frac{A}{\sqrt{2x^2-1}}$ (condone missing brackets) |
| $1+\left(\frac{dy}{dx}\right)^2 = 1 + \frac{1}{4x^2-1} = \frac{4x^2}{4x^2-1}$ | M1 | Attempts to find $1+\left(\frac{dy}{dx}\right)^2$ using their $\frac{dy}{dx}$ and attempts common denominator |
| $\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int\sqrt{\frac{4x^2}{4x^2-1}}\,dx = 2\int\frac{x}{\sqrt{4x^2-1}}\,dx$ | A1 | Reaches correct simplified integral with $\sqrt{x^2}$ replaced with $x$. Allow equivalent forms e.g. $2\int x\sqrt{\frac{1}{4x^2-1}}\,dx$, $\frac{1}{2}\int\frac{4x}{\sqrt{(2x)^2-1}}\,dx$ |
| $= \frac{2(4x^2-1)^{\frac{1}{2}}}{8 \times \frac{1}{2}}$ | M1 | Attempts integration, reaches form $\alpha(\beta x^2-1)^{\frac{1}{2}}$ |
| $s = \left[\frac{(4x^2-1)^{\frac{1}{2}}}{2}\right]_{\frac{7}{2}}^{13} = \frac{1}{2}\left(\sqrt{4\times169-1}-\sqrt{4\times\frac{49}{4}-1}\right) = \ldots$ | dM1 | Applies limits to their integral. **Depends on previous 2 method marks.** |
| $= \frac{1}{2}(15\sqrt{3}-4\sqrt{3}) = \frac{11}{2}\sqrt{3}$ | A1 | cao Accept equivalents e.g. $\frac{1}{2}\sqrt{363}$ |
| | **(6)** | |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { 1 } { 2 } \operatorname { arcosh } ( 2 x ) \quad \frac { 7 } { 2 } \leqslant x \leqslant 13$$
Using calculus, determine the exact length of the curve $C$.\\
Give your answer in the form $p \sqrt { q }$, where $p$ and $q$ are constants to be found.
\hfill \mbox{\textit{Edexcel F3 2021 Q1 [6]}}