| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Volume of revolution with hyperbolics |
| Difficulty | Challenging +1.8 This is a substantial Further Maths question requiring multiple hyperbolic techniques: differentiation to find turning points and stationary gradients, solving hyperbolic equations, squaring and simplifying using double-angle identities, and integrating for volume of revolution. While each individual step uses standard FP2 techniques, the multi-part structure, algebraic manipulation of squared hyperbolics, and the need to evaluate definite integrals with limits involving ln(5) make this significantly harder than average A-level questions. It's challenging for Further Maths but follows predictable patterns once the techniques are known. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = 3\sinh x - 2\cosh x \Rightarrow \frac{dy}{dx} = 3\cosh x - 2\sinh x\) | B1 | |
| At TPs, \(\frac{dy}{dx} = 0 \Rightarrow \tanh x = \frac{3}{2}\), which has no (real) solutions | M1, A1(ag) | Considering \(\frac{dy}{dx} = 0\); showing no real roots www. Note: \(e^{2x} = -5\); \(e^x > 0\) and \(e^{-x} > 0\) |
| \(y = 0 \Rightarrow \tanh x = \frac{2}{3}\) | M1 | Solving \(y=0\) as far as \(e^{2x}\) or \(\tanh x\) etc. Note: \(e^{2x}=5\); \(\cosh x = \frac{3}{\sqrt{5}}\); \(\sinh x = \frac{2}{\sqrt{5}}\) |
| \(\Rightarrow x = \frac{1}{2}\ln\frac{1+\frac{2}{3}}{1-\frac{2}{3}}\) | M1 | Solving as far as \(x\). Attempt to verify: award M1 for substituting \(x = \frac{1}{2}\ln 5\) |
| \(\Rightarrow x = \frac{1}{2}\ln 5\) | A1(ag) | Completion www; M1 for clearly attempting to evaluate exactly |
| \(\frac{d^2y}{dx^2} = 3\sinh x - 2\cosh x = y\), so \(y=0 \Rightarrow \frac{d^2y}{dx^2} = 0\) | B1(ag) | \(3\sinh(\frac{1}{2}\ln 5) - 2\cosh(\frac{1}{2}\ln 5) = 0\) must be explained, e.g. connected with \(y=0\) |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Curve with: increasing, intersecting positive \(x\)-axis, \((0,-2)\) indicated, gradient increasing with large \( | x | \), one point of inflection |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3\sinh x - 2\cosh x)^2 = 9\sinh^2 x - 12\sinh x\cosh x + 4\cosh^2 x\) | B1 | |
| \(= \frac{9}{2}(\cosh 2x -1) - 6\sinh 2x + 2(\cosh 2x +1)\) | M1 | Using double angle formulae or complete alternative. Condone sign errors but need \(\frac{1}{2}\)s |
| \(= \frac{13}{2}\cosh 2x - 6\sinh 2x - \frac{5}{2}\) | A1 | Accept unsimplified. Note: \(\frac{1}{4}e^{2x} + \frac{25}{4}e^{-2x} - \frac{5}{2}\) |
| \(V = \pi\int_0^{\frac{1}{2}\ln 5} y^2\,dx\) | M1 | Attempting to integrate their \(y^2\) (ignore limits) |
| \(= \pi\left[\frac{13}{4}\sinh 2x - 3\cosh 2x - \frac{5}{2}x\right]_0^{\frac{1}{2}\ln 5}\) | A2 | Correct results and limits c.a.o., ignore omitted \(\pi\). Give A1 for one error, or for all three terms correct with incorrect limits |
| M1 | Substituting both limits | |
| \(= \pi\left[\frac{13}{4}\times\frac{12}{5} - 3\times\frac{13}{5} - \frac{5}{4}\ln 5 + 3\right]\) | M1 | Obtaining exact values of \(\sinh(\ln 5)\) and \(\cosh(\ln 5)\). Note: \(\sinh(\ln 5)=\frac{12}{5}\), \(\cosh(\ln 5)=\frac{13}{5}\) |
| OR \(= \pi\left[\frac{1}{8}e^{2x} - \frac{25}{8}e^{-2x} - \frac{5}{2}x\right]_0^{\frac{1}{2}\ln 5}\) | A2 | Correct results and limits. Give A1 for one error, or all three terms correct with incorrect limits |
| \(= \pi\left[\frac{5}{8} - \frac{5}{8} - \frac{5}{4}\ln 5 + 3\right]\) | M1, M1 | Substituting both limits; obtaining exact values of \(e^{2x}\) and \(e^{-2x}\). Note: \(e^{2x}=5\), \(e^{-2x}=\frac{1}{5}\) |
| \(= \pi\left[3 - \frac{5}{4}\ln 5\right]\) | A1(ag) | Completion www |
| [9] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 3\sinh x - 2\cosh x \Rightarrow \frac{dy}{dx} = 3\cosh x - 2\sinh x$ | B1 | |
| At TPs, $\frac{dy}{dx} = 0 \Rightarrow \tanh x = \frac{3}{2}$, which has no (real) solutions | M1, A1(ag) | Considering $\frac{dy}{dx} = 0$; showing no real roots www. Note: $e^{2x} = -5$; $e^x > 0$ and $e^{-x} > 0$ |
| $y = 0 \Rightarrow \tanh x = \frac{2}{3}$ | M1 | Solving $y=0$ as far as $e^{2x}$ or $\tanh x$ etc. Note: $e^{2x}=5$; $\cosh x = \frac{3}{\sqrt{5}}$; $\sinh x = \frac{2}{\sqrt{5}}$ |
| $\Rightarrow x = \frac{1}{2}\ln\frac{1+\frac{2}{3}}{1-\frac{2}{3}}$ | M1 | Solving as far as $x$. Attempt to verify: award M1 for substituting $x = \frac{1}{2}\ln 5$ |
| $\Rightarrow x = \frac{1}{2}\ln 5$ | A1(ag) | Completion www; M1 for clearly attempting to evaluate exactly |
| $\frac{d^2y}{dx^2} = 3\sinh x - 2\cosh x = y$, so $y=0 \Rightarrow \frac{d^2y}{dx^2} = 0$ | B1(ag) | $3\sinh(\frac{1}{2}\ln 5) - 2\cosh(\frac{1}{2}\ln 5) = 0$ must be explained, e.g. connected with $y=0$ |
| **[7]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Curve with: increasing, intersecting positive $x$-axis, $(0,-2)$ indicated, gradient increasing with large $|x|$, one point of inflection | B2 | Award B1 for a curve lacking one of these features |
| **[2]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3\sinh x - 2\cosh x)^2 = 9\sinh^2 x - 12\sinh x\cosh x + 4\cosh^2 x$ | B1 | |
| $= \frac{9}{2}(\cosh 2x -1) - 6\sinh 2x + 2(\cosh 2x +1)$ | M1 | Using double angle formulae or complete alternative. Condone sign errors but need $\frac{1}{2}$s |
| $= \frac{13}{2}\cosh 2x - 6\sinh 2x - \frac{5}{2}$ | A1 | Accept unsimplified. Note: $\frac{1}{4}e^{2x} + \frac{25}{4}e^{-2x} - \frac{5}{2}$ |
| $V = \pi\int_0^{\frac{1}{2}\ln 5} y^2\,dx$ | M1 | Attempting to integrate their $y^2$ (ignore limits) |
| $= \pi\left[\frac{13}{4}\sinh 2x - 3\cosh 2x - \frac{5}{2}x\right]_0^{\frac{1}{2}\ln 5}$ | A2 | Correct results and limits c.a.o., ignore omitted $\pi$. Give A1 for one error, or for all three terms correct with incorrect limits |
| | M1 | Substituting both limits |
| $= \pi\left[\frac{13}{4}\times\frac{12}{5} - 3\times\frac{13}{5} - \frac{5}{4}\ln 5 + 3\right]$ | M1 | Obtaining exact values of $\sinh(\ln 5)$ and $\cosh(\ln 5)$. Note: $\sinh(\ln 5)=\frac{12}{5}$, $\cosh(\ln 5)=\frac{13}{5}$ |
| **OR** $= \pi\left[\frac{1}{8}e^{2x} - \frac{25}{8}e^{-2x} - \frac{5}{2}x\right]_0^{\frac{1}{2}\ln 5}$ | A2 | Correct results and limits. Give A1 for one error, or all three terms correct with incorrect limits |
| $= \pi\left[\frac{5}{8} - \frac{5}{8} - \frac{5}{4}\ln 5 + 3\right]$ | M1, M1 | Substituting both limits; obtaining exact values of $e^{2x}$ and $e^{-2x}$. Note: $e^{2x}=5$, $e^{-2x}=\frac{1}{5}$ |
| $= \pi\left[3 - \frac{5}{4}\ln 5\right]$ | A1(ag) | Completion www |
| **[9]** | | |
---4 (i) Show that the curve with equation
$$y = 3 \sinh x - 2 \cosh x$$
has no turning points.\\
Show that the curve crosses the $x$-axis at $x = \frac { 1 } { 2 } \ln 5$. Show that this is also the point at which the gradient of the curve has a stationary value.\\
(ii) Sketch the curve.\\
(iii) Express $( 3 \sinh x - 2 \cosh x ) ^ { 2 }$ in terms of $\sinh 2 x$ and $\cosh 2 x$.
Hence or otherwise, show that the volume of the solid of revolution formed by rotating the region bounded by the curve and the axes through $360 ^ { \circ }$ about the $x$-axis is
$$\pi \left( 3 - \frac { 5 } { 4 } \ln 5 \right) .$$
Option 2: Investigation of curves
\section*{This question requires the use of a graphical calculator.}
\hfill \mbox{\textit{OCR MEI FP2 2013 Q4 [18]}}