| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.8 Part (i) is a standard FP2 proof of the logarithmic form of inverse sinh, requiring routine manipulation of exponentials. Part (ii) is more demanding: students must apply logarithmic forms, manipulate the resulting equation to isolate surds, then square carefully to solve—this requires multiple algebraic steps and careful handling of square roots. The question is moderately challenging for FP2 but follows established patterns for this topic. |
| Spec | 4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \sinh y = \frac{e^y - e^{-y}}{2}\) | M1 | For correct expression for sinh y and attempt to obtain quadratic |
| \(\Rightarrow e^{2y} - 2xe^y - 1 = 0 \Rightarrow e^y = x \pm \sqrt{x^2+1}\) | A1 | For correct solution(s) for \(e^y\) |
| reject – sign as \(e^y > 0 \Rightarrow y = \ln\left(x+\sqrt{x^2+1}\right)\) | A1 | For justification of + sign to AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\ln\left(x+\sqrt{x^2+1}\right) - \ln\left(x+\sqrt{x^2-1}\right) = \ln 2\) | M1 | For stating both ln expressions and attempting to exponentiate |
| \(\Rightarrow \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}} = 2\) | A1 | For correct equation AG |
| \(\Rightarrow \sqrt{x^2+1} - 2\sqrt{x^2-1} = x\) | M1 | For attempting to square once |
| \(\Rightarrow 4x^2 - 3 = 4\sqrt{x^4-1}\) | A1 | For a correct equation with \(\sqrt{}\) as subject |
| \(\Rightarrow 24x^2 = 25 \Rightarrow x = \frac{5}{\sqrt{24}}\left(= \frac{5}{2\sqrt{6}}\right)\) | A1 | For correct x and no others isw |
| [5] |
### (i)
$x = \sinh y = \frac{e^y - e^{-y}}{2}$ | M1 | For correct expression for sinh y and attempt to obtain quadratic
$\Rightarrow e^{2y} - 2xe^y - 1 = 0 \Rightarrow e^y = x \pm \sqrt{x^2+1}$ | A1 | For correct solution(s) for $e^y$
reject – sign as $e^y > 0 \Rightarrow y = \ln\left(x+\sqrt{x^2+1}\right)$ | A1 | For justification of + sign to AG
| [3] |
**Alt:** $\sinh y + \cosh y = e^y$; $\sinh y = x \Rightarrow \cosh y = \pm\sqrt{x^2+1}$; reject –ve sign as $e^y > 0 \Rightarrow e^y = x + \sqrt{x^2+1} \Rightarrow y = \ln\left(x+\sqrt{x^2+1}\right)$
### (ii)
$\ln\left(x+\sqrt{x^2+1}\right) - \ln\left(x+\sqrt{x^2-1}\right) = \ln 2$ | M1 | For stating both ln expressions and attempting to exponentiate
$\Rightarrow \frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}} = 2$ | A1 | For correct equation AG
$\Rightarrow \sqrt{x^2+1} - 2\sqrt{x^2-1} = x$ | M1 | For attempting to square once
$\Rightarrow 4x^2 - 3 = 4\sqrt{x^4-1}$ | A1 | For a correct equation with $\sqrt{}$ as subject
$\Rightarrow 24x^2 = 25 \Rightarrow x = \frac{5}{\sqrt{24}}\left(= \frac{5}{2\sqrt{6}}\right)$ | A1 | For correct x and no others isw
| [5] |
---\begin{enumerate}[label=(\roman*)]
\item Given that $y = \sinh^{-1} x$, prove that $y = \ln\left(x + \sqrt{x^2 + 1}\right)$. [3]
\item It is given that $x$ satisfies the equation $\sinh^{-1} x - \cosh^{-1} x = \ln 2$. Use the logarithmic forms for $\sinh^{-1} x$ and $\cosh^{-1} x$ to show that
$$\sqrt{x^2 + 1} - 2\sqrt{x^2 - 1} = x.$$
Hence, by squaring this equation, find the exact value of $x$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2012 Q7 [8]}}