| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.3 This is a structured Further Maths question on hyperbolic functions and arc length. Part (a) is routine differentiation using the chain rule with a standard inverse hyperbolic derivative. Part (b)(i) follows directly by combining derivatives. Part (b)(ii) requires setting up the arc length integral, but the algebra is carefully designed so the integrand simplifies to a standard form that integrates to ln x. While it requires multiple techniques (implicit differentiation, arc length formula, algebraic manipulation), the question is highly scaffolded and the calculations follow predictable patterns typical of FP2 exam questions. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| - \(\frac{d}{dx}\left(\cosh^{-1}\frac{1}{x}\right) = \frac{1}{\sqrt{x^2-1}} \cdot \left(-\frac{1}{x^2}\right)\) | M1A1 | M0 if \(\frac{dy}{dx} = f(y)\) and no attempt to substitute back to \(x\) |
| - \(= \frac{-1}{x\sqrt{1-x^2}}\) | A1 | 3 marks; AG |
| Answer | Marks | Guidance |
|---|---|---|
| - \(\frac{d}{dx}\left(\sqrt{1-x^2}\right) = \frac{-2x}{2\sqrt{1-x^2}}\) | B1 | For numerator |
| - | B1 | For denominator (not \((1-x^2)^{\frac{1}{2}}\)) |
| - \(\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}} + \frac{1}{x\sqrt{1-x^2}}\) | M1 | For attempt to put over a common denominator |
| - \(= \frac{1-x^2}{x\sqrt{1-x^2}} = \frac{\sqrt{1-x^2}}{x}\) | A1 | 4 marks; AG |
| Answer | Marks | Guidance |
|---|---|---|
| - \(s = \int_{\frac{1}{4}}^3 \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx = \int_{\frac{1}{4}}^3 \frac{1}{x} dx\) | M1, A1A1 | For use of \(\sqrt{1+\left(\frac{dy}{dx}\right)^2}\) |
| - \(= [\ln x]_{\frac{1}{4}}^3\) | M1 | |
| - \(= \ln 3 - \ln\frac{1}{4} = \ln 3\) | A1 | 5 marks; AG |
**(a)**
- $\frac{d}{dx}\left(\cosh^{-1}\frac{1}{x}\right) = \frac{1}{\sqrt{x^2-1}} \cdot \left(-\frac{1}{x^2}\right)$ | M1A1 | M0 if $\frac{dy}{dx} = f(y)$ and no attempt to substitute back to $x$
- $= \frac{-1}{x\sqrt{1-x^2}}$ | A1 | 3 marks; AG
- **Total: 3 marks**
**(b)(i)**
- $\frac{d}{dx}\left(\sqrt{1-x^2}\right) = \frac{-2x}{2\sqrt{1-x^2}}$ | B1 | For numerator
- | B1 | For denominator (not $(1-x^2)^{\frac{1}{2}}$)
- $\frac{dy}{dx} = \frac{-x}{\sqrt{1-x^2}} + \frac{1}{x\sqrt{1-x^2}}$ | M1 | For attempt to put over a common denominator
- $= \frac{1-x^2}{x\sqrt{1-x^2}} = \frac{\sqrt{1-x^2}}{x}$ | A1 | 4 marks; AG
- **Total: 4 marks**
**(ii)**
- $s = \int_{\frac{1}{4}}^3 \sqrt{1+\left(\frac{dy}{dx}\right)^2} dx = \int_{\frac{1}{4}}^3 \frac{1}{x} dx$ | M1, A1A1 | For use of $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$
- $= [\ln x]_{\frac{1}{4}}^3$ | M1 |
- $= \ln 3 - \ln\frac{1}{4} = \ln 3$ | A1 | 5 marks; AG
- **Total: 5 marks**
**Question 7 Total: 12 marks**
---7
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \cosh ^ { - 1 } \frac { 1 } { x } \right) = \frac { - 1 } { x \sqrt { 1 - x ^ { 2 } } }$$
(3 marks)
\item A curve has equation
$$y = \sqrt { 1 - x ^ { 2 } } - \cosh ^ { - 1 } \frac { 1 } { x } \quad ( 0 < x < 1 )$$
Show that:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 - x ^ { 2 } } } { x }$;\\
(4 marks)
\item the length of the arc of the curve from the point where $x = \frac { 1 } { 4 }$ to the point where
$$x = \frac { 3 } { 4 } \text { is } \ln 3 .$$
(5 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2009 Q7 [12]}}