| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.2 This is a structured multi-part question with clear guidance through each step. Part (a) is standard arc length formula application. Part (b) follows a prescribed hyperbolic substitution with routine algebraic manipulation and integration of cosh²θ using standard identities. While hyperbolic functions are Further Maths content (making it inherently harder than Core), the question provides significant scaffolding and uses well-established techniques, placing it moderately above average difficulty. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges4.08h Integration: inverse trig/hyperbolic substitutions8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{2}{\sqrt{x}}\) | B1 | accept \(2x^{-\frac{1}{2}}\) etc |
| \(\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\frac{4}{x}}\) | M1A1F | ft sign error in \(\frac{dy}{dx}\) |
| \(= \sqrt{\frac{x+4}{x}}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x = 4\sinh^2\theta\), \(dx = 8\sinh\theta\cosh\theta\,d\theta\) | M1A1 | M1 for any attempt at \(\frac{dx}{d\theta}\) |
| \(I = \int\sqrt{\frac{4\sinh^2\theta+4}{4\sinh^2\theta}}\cdot 8\sinh\theta\cosh\theta\,d\theta\) | M1 | |
| \(= \int\frac{2\cosh\theta}{2\sinh\theta}\cdot 8\sinh\theta\cosh\theta\,d\theta\) | m1 | use of \(\cosh^2\theta - \sinh^2\theta = 1\) |
| \(= \int 8\cosh^2\theta\,d\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Use of \(2\cosh^2\theta = 1 + \cosh 2\theta\) | M1 | allow if sign error |
| \(I = \int 4(1+\cosh 2\theta)\,d\theta\) | A1 | oe |
| \(= 4\theta + 2\sinh 2\theta\) | A1F | oe |
| Use of \(\sinh 2\theta = 2\sinh\theta\cosh\theta\) | m1 | |
| \(= 4\sinh^{-1}\frac{1}{2} + 4\times\frac{1}{2}\sqrt{1+\frac{1}{4}}\) | A1F | |
| \(= 4\sinh^{-1}\frac{1}{2} + \sqrt{5}\) | A1 | AG |
## Question 7:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{dy}{dx} = \frac{2}{\sqrt{x}}$ | B1 | accept $2x^{-\frac{1}{2}}$ etc |
| $\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\frac{4}{x}}$ | M1A1F | ft sign error in $\frac{dy}{dx}$ |
| $= \sqrt{\frac{x+4}{x}}$ | A1 | AG |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $x = 4\sinh^2\theta$, $dx = 8\sinh\theta\cosh\theta\,d\theta$ | M1A1 | M1 for any attempt at $\frac{dx}{d\theta}$ |
| $I = \int\sqrt{\frac{4\sinh^2\theta+4}{4\sinh^2\theta}}\cdot 8\sinh\theta\cosh\theta\,d\theta$ | M1 | |
| $= \int\frac{2\cosh\theta}{2\sinh\theta}\cdot 8\sinh\theta\cosh\theta\,d\theta$ | m1 | use of $\cosh^2\theta - \sinh^2\theta = 1$ |
| $= \int 8\cosh^2\theta\,d\theta$ | A1 | AG |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| Use of $2\cosh^2\theta = 1 + \cosh 2\theta$ | M1 | allow if sign error |
| $I = \int 4(1+\cosh 2\theta)\,d\theta$ | A1 | oe |
| $= 4\theta + 2\sinh 2\theta$ | A1F | oe |
| Use of $\sinh 2\theta = 2\sinh\theta\cosh\theta$ | m1 | |
| $= 4\sinh^{-1}\frac{1}{2} + 4\times\frac{1}{2}\sqrt{1+\frac{1}{4}}$ | A1F | |
| $= 4\sinh^{-1}\frac{1}{2} + \sqrt{5}$ | A1 | AG |7 A curve has equation $y = 4 \sqrt { x }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the length of arc $s$ of the curve between the points where $x = 0$ and $x = 1$ is given by
$$s = \int _ { 0 } ^ { 1 } \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x$$
\item \begin{enumerate}[label=(\roman*)]
\item Use the substitution $x = 4 \sinh ^ { 2 } \theta$ to show that
$$\int \sqrt { \frac { x + 4 } { x } } \mathrm {~d} x = \int 8 \cosh ^ { 2 } \theta \mathrm {~d} \theta$$
\item Hence show that
$$s = 4 \sinh ^ { - 1 } 0.5 + \sqrt { 5 }$$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2007 Q7 [15]}}