| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of hyperbolic functions and finding/classifying stationary points. While it involves multiple steps (differentiation, solving a quadratic in cosh x, finding x values, second derivative test), the techniques are standard for FM students. The algebraic manipulation using the identity sinh²x = cosh²x - 1 is routine, and solving the resulting equation is straightforward. It's harder than average A-level due to being FM content, but represents a typical FM calculus exercise rather than requiring novel insight. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(12\cosh x \sinh x\) or \(-13\sinh x\) | M1 | Either term correct |
| \(\frac{dy}{dx} = 12\cosh x \sinh x - 13\sinh x\) | A1 | Both correct |
| \(\sinh x(12\cosh x - 13) = 0\) | M1 | Setting to zero and attempting to solve |
| \(x = 0, y = -13\) | A1 | |
| \(x = \ln\left(\frac{13}{12} + \sqrt{\left(\frac{13}{12}\right)^2 - 1}\right)\) | M1 | Correct numerical use of formula for \(\cosh^{-1}\) (could be ±) |
| \(x = \ln\left(\frac{3}{2}\right)\) | A1 | or \(x = -\ln\left(\frac{2}{3}\right)\) |
| \(y = -\frac{313}{24}\) | A1 | ft from "their" \(x\) |
| or \(x = \ln\left(\frac{2}{3}\right)\) | A1 | From the symmetry of the sinh and cosh functions (could be explicitly calculated) |
| \(y = -\frac{313}{24}\) | A1 [9] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 12\cosh^2 x + 12\sinh^2 x - 13\cosh x\) | M1 | |
| \(x = 0 \Rightarrow \frac{d^2y}{dx^2} = -1 < 0\) so maximum point | A1 | Correct values only. Must be explicit. |
| \(x = \pm \ln\frac{3}{2} \Rightarrow \frac{d^2y}{dx^2} = \frac{25}{12} > 0\) so minimum points | A1 [3] | Correct values only. Must be explicit. Must be both. |
## 6(i)
$12\cosh x \sinh x$ or $-13\sinh x$ | M1 | Either term correct
$\frac{dy}{dx} = 12\cosh x \sinh x - 13\sinh x$ | A1 | Both correct
$\sinh x(12\cosh x - 13) = 0$ | M1 | Setting to zero and attempting to solve
$x = 0, y = -13$ | A1 |
$x = \ln\left(\frac{13}{12} + \sqrt{\left(\frac{13}{12}\right)^2 - 1}\right)$ | M1 | Correct numerical use of formula for $\cosh^{-1}$ (could be ±)
$x = \ln\left(\frac{3}{2}\right)$ | A1 | or $x = -\ln\left(\frac{2}{3}\right)$ | Second solution
$y = -\frac{313}{24}$ | A1 | ft from "their" $x$
or $x = \ln\left(\frac{2}{3}\right)$ | A1 | From the symmetry of the sinh and cosh functions (could be explicitly calculated) | Third solution
$y = -\frac{313}{24}$ | A1 [9]
## 6(ii)
$\frac{d^2y}{dx^2} = 12\cosh^2 x + 12\sinh^2 x - 13\cosh x$ | M1
$x = 0 \Rightarrow \frac{d^2y}{dx^2} = -1 < 0$ so maximum point | A1 | Correct values only. Must be explicit.
$x = \pm \ln\frac{3}{2} \Rightarrow \frac{d^2y}{dx^2} = \frac{25}{12} > 0$ so minimum points | A1 [3] | Correct values only. Must be explicit. Must be both.
---In this question you must show detailed reasoning.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of all stationary points on the graph of $y = 6\sinh^2 x - 13\cosh x$, giving your answers in an exact, simplified form. [9]
\item By finding the second derivative, classify the stationary points found in part (i). [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q6 [12]}}