| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Differentiate inverse hyperbolic functions |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring fluency with hyperbolic functions, their inverses, and integration techniques. Part (a) requires solving a hyperbolic equation using exponential definitions; part (b) involves integrating a product of exponentials; part (c) requires differentiating an inverse hyperbolic function and then using integration by parts with a non-standard integrand. While these are standard FP2 techniques, the combination and the exact form requirements make this moderately challenging, placing it above average difficulty. |
| Spec | 1.08b Integrate x^n: where n != -1 and sums1.08i Integration by parts4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(\sinh x = \frac{e^x - e^{-x}}{2}\), \(\cosh x = \frac{e^x + e^{-x}}{2}\) | M1 | |
| \(\frac{e^x - e^{-x}}{2} + 4\cdot\frac{e^x + e^{-x}}{2} = 8\) | A1 | |
| \(5e^x + 3e^{-x} = 16\), so \(5e^{2x} - 16e^x + 3 = 0\) | M1 | Forming quadratic |
| \((5e^x - 1)(e^x - 3) = 0\) | A1 | |
| \(x = \ln 3\) or \(x = \ln\frac{1}{5} = -\ln 5\) | A1 A1 | Both answers required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integration by parts: \(u = \sinh x\), \(dv = e^x dx\) | M1 | |
| \(= [e^x \sinh x]_0^2 - \int_0^2 e^x \cosh x \, dx\) | A1 | |
| Second IBP or using \(e^x\sinh x = \frac{1}{2}(e^{2x}-1)\) | M1 | |
| \(\int_0^2 e^x \sinh x \, dx = \frac{1}{4}(e^4 - 1) - \frac{1}{2}(e^2 - 1)\)... | A1 | Exact simplified answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}\left[\text{arsinh}\left(\frac{2x}{3}\right)\right] = \frac{\frac{2}{3}}{\sqrt{1+\frac{4x^2}{9}}}\) | M1 | |
| \(= \frac{2}{\sqrt{9+4x^2}}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| IBP: \(u = \text{arsinh}(\frac{2}{3}x)\), \(dv = dx\) | M1 | |
| \(= \left[x\,\text{arsinh}(\frac{2}{3}x)\right]_0^2 - \int_0^2 \frac{2x}{\sqrt{9+4x^2}} dx\) | A1 | |
| \(= 2\,\text{arsinh}(\frac{4}{3}) - \left[\frac{1}{2}\sqrt{9+4x^2}\right]_0^2 \cdot \frac{1}{2}\)... | M1 A1 | Integrating second term |
| \(\text{arsinh}(\frac{4}{3}) = \ln(\frac{4}{3}+\frac{5}{3}) = \ln 3\) | M1 | |
| Full evaluation giving \(2\ln 3 - 1\) | A1 | Completion of proof |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $\sinh x = \frac{e^x - e^{-x}}{2}$, $\cosh x = \frac{e^x + e^{-x}}{2}$ | M1 | |
| $\frac{e^x - e^{-x}}{2} + 4\cdot\frac{e^x + e^{-x}}{2} = 8$ | A1 | |
| $5e^x + 3e^{-x} = 16$, so $5e^{2x} - 16e^x + 3 = 0$ | M1 | Forming quadratic |
| $(5e^x - 1)(e^x - 3) = 0$ | A1 | |
| $x = \ln 3$ or $x = \ln\frac{1}{5} = -\ln 5$ | A1 A1 | Both answers required |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integration by parts: $u = \sinh x$, $dv = e^x dx$ | M1 | |
| $= [e^x \sinh x]_0^2 - \int_0^2 e^x \cosh x \, dx$ | A1 | |
| Second IBP or using $e^x\sinh x = \frac{1}{2}(e^{2x}-1)$ | M1 | |
| $\int_0^2 e^x \sinh x \, dx = \frac{1}{4}(e^4 - 1) - \frac{1}{2}(e^2 - 1)$... | A1 | Exact simplified answer |
## Part (c)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}\left[\text{arsinh}\left(\frac{2x}{3}\right)\right] = \frac{\frac{2}{3}}{\sqrt{1+\frac{4x^2}{9}}}$ | M1 | |
| $= \frac{2}{\sqrt{9+4x^2}}$ | A1 | |
## Part (c)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| IBP: $u = \text{arsinh}(\frac{2}{3}x)$, $dv = dx$ | M1 | |
| $= \left[x\,\text{arsinh}(\frac{2}{3}x)\right]_0^2 - \int_0^2 \frac{2x}{\sqrt{9+4x^2}} dx$ | A1 | |
| $= 2\,\text{arsinh}(\frac{4}{3}) - \left[\frac{1}{2}\sqrt{9+4x^2}\right]_0^2 \cdot \frac{1}{2}$... | M1 A1 | Integrating second term |
| $\text{arsinh}(\frac{4}{3}) = \ln(\frac{4}{3}+\frac{5}{3}) = \ln 3$ | M1 | |
| Full evaluation giving $2\ln 3 - 1$ | A1 | Completion of proof |4
\begin{enumerate}[label=(\alph*)]
\item Solve the equation
$$\sinh x + 4 \cosh x = 8$$
giving the answers in an exact logarithmic form.
\item Find the exact value of $\int _ { 0 } ^ { 2 } \mathrm { e } ^ { x } \sinh x \mathrm {~d} x$.
\item \begin{enumerate}[label=(\roman*)]
\item Differentiate $\operatorname { arsinh } \left( \frac { 2 } { 3 } x \right)$ with respect to $x$.
\item Use integration by parts to show that $\int _ { 0 } ^ { 2 } \operatorname { arsinh } \left( \frac { 2 } { 3 } x \right) \mathrm { d } x = 2 \ln 3 - 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI FP2 2006 Q4 [18]}}