| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question on hyperbolic functions requiring standard techniques: differentiation to find stationary points (using known derivatives of sinh and cosh), substitution to verify, and solving a hyperbolic equation. While Further Maths content is inherently harder than Core, these are routine applications of hyperbolic function properties without requiring novel insight or complex multi-step reasoning. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2\sinh x + 3\cosh x \Rightarrow \frac{dy}{dx} = 2\cosh x + 3\sinh x\) | M1 | Diffn and setting \(= 0\) |
| \(= 0\) when \(2\cosh x = -3\sinh x \Rightarrow \tanh x = -\frac{2}{3}\) | A1 | Correct value for \(\sinh x\), \(\cosh x\) or \(\tanh x\) |
| \(x = \tanh^{-1}\!\left(-\frac{2}{3}\right) = \frac{1}{2}\ln\!\left(\frac{1 - \frac{2}{3}}{1 + \frac{2}{3}}\right) = \frac{1}{2}\ln\!\left(\frac{1}{5}\right) = -\frac{1}{2}\ln 5\) | A1 | Some numerical justification must be seen ag |
| \(\sinh x = \frac{-2}{\sqrt{5}},\ \cosh x = \frac{3}{\sqrt{5}} \Rightarrow y = \frac{-4}{\sqrt{5}} + \frac{9}{\sqrt{5}} = \sqrt{5}\) | B1 | Exact answer only |
Total: 4
| Answer | Marks | Guidance |
|---|---|---|
| \(2\sinh x + 3\cosh x = 5 \Rightarrow 2\cdot\frac{e^x - e^{-x}}{2} + 3\cdot\frac{e^x + e^{-x}}{2} = 5\) | M1 | Find exponential form |
| Answer | Marks | Guidance |
|---|---|---|
| \(5e^{2x} - 10e^x + 1 = 0\) | A1 | Correct quadratic |
| \(e^x = \frac{10 \pm \sqrt{100 - 20}}{10} = \frac{10 \pm \sqrt{80}}{10}\) | M1 | Solve their 3 term quadratic |
| \(x = \ln\!\left(1 + \frac{2\sqrt{5}}{5}\right)\) and \(\ln\!\left(1 - \frac{2\sqrt{5}}{5}\right)\) | A1 | oe Single ln only |
| A1 |
Total: 5
## Question 8:
### Part (i):
$y = 2\sinh x + 3\cosh x \Rightarrow \frac{dy}{dx} = 2\cosh x + 3\sinh x$ | **M1** | Diffn and setting $= 0$
$= 0$ when $2\cosh x = -3\sinh x \Rightarrow \tanh x = -\frac{2}{3}$ | **A1** | Correct value for $\sinh x$, $\cosh x$ or $\tanh x$
$x = \tanh^{-1}\!\left(-\frac{2}{3}\right) = \frac{1}{2}\ln\!\left(\frac{1 - \frac{2}{3}}{1 + \frac{2}{3}}\right) = \frac{1}{2}\ln\!\left(\frac{1}{5}\right) = -\frac{1}{2}\ln 5$ | **A1** | Some numerical justification must be seen ag
$\sinh x = \frac{-2}{\sqrt{5}},\ \cosh x = \frac{3}{\sqrt{5}} \Rightarrow y = \frac{-4}{\sqrt{5}} + \frac{9}{\sqrt{5}} = \sqrt{5}$ | **B1** | Exact answer only
Total: **4**
### Part (ii):
$2\sinh x + 3\cosh x = 5 \Rightarrow 2\cdot\frac{e^x - e^{-x}}{2} + 3\cdot\frac{e^x + e^{-x}}{2} = 5$ | **M1** | Find exponential form
$5e^x + e^{-x} = 10$
$5e^{2x} - 10e^x + 1 = 0$ | **A1** | Correct quadratic
$e^x = \frac{10 \pm \sqrt{100 - 20}}{10} = \frac{10 \pm \sqrt{80}}{10}$ | **M1** | Solve their 3 term quadratic
$x = \ln\!\left(1 + \frac{2\sqrt{5}}{5}\right)$ and $\ln\!\left(1 - \frac{2\sqrt{5}}{5}\right)$ | **A1** | oe Single ln only
| **A1** |
Total: **5**
---8 It is given that $\mathrm { f } ( x ) = 2 \sinh x + 3 \cosh x$.\\
(i) Show that the curve $y = \mathrm { f } ( x )$ has a stationary point at $x = - \frac { 1 } { 2 } \ln 5$ and find the value of $y$ at this point.\\
(ii) Solve the equation $\mathrm { f } ( x ) = 5$, giving your answers exactly.
\section*{Question 9 begins on page 4.}
\hfill \mbox{\textit{OCR FP2 2015 Q8 [9]}}