| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Challenging +1.2 This is a multi-part question on hyperbolic functions requiring standard techniques: proving an identity using exponentials (routine), volume of revolution with hyperbolic integration (standard FP2 technique using cosh²x identity), and finding a stationary point by solving 2sinh(2x) + cosh(x) = 0. While it requires multiple steps and careful algebra, all techniques are standard Further Pure material with no novel insight required. The logarithmic form answer adds minor complexity but follows directly from solving the equation. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.07d Differentiate/integrate: hyperbolic functions4.08e Mean value of function: using integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\sinh x\cosh x = 2\times\frac{e^x+e^{-x}}{2}\times\frac{e^x-e^{-x}}{2}\) | ||
| \(=\frac{e^{2x}-e^{-2x}}{2}\) | M1 | Using exponential definitions and multiplying or factorising |
| \(=\sinh 2x\) | A1 (ag) | |
| Differentiating: \(2\cosh 2x = 2\cosh^2 x + 2\sinh^2 x\) | B1 | One side correct |
| \(\Rightarrow \cosh 2x = \cosh^2 x + \sinh^2 x\) | B1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sketch of \(y=\cosh x - 1\) | G1 | Correct shape and through origin |
| Volume \(= \pi\int_0^2(\cosh x-1)^2\,dx\) | M1 | \(\int(\cosh x-1)^2\,dx\) |
| \(= \pi\int_0^2\cosh^2 x - 2\cosh x + 1\,dx\) | A1 | Correct expanded integral including limits 0, 2 |
| \(= \pi\int_0^2\frac{1}{2}\cosh 2x - 2\cosh x + \frac{3}{2}\,dx\) | M1 | Attempting integrable form; Dep. on M1 above |
| \(= \pi\left[\frac{1}{4}\sinh 2x - 2\sinh x + \frac{3}{2}x\right]_0^2\) | A2 | Give A1 for two terms correct |
| \(= \pi\left[\frac{1}{4}\sinh 4 - 2\sinh 2 + 3\right]\) | ||
| \(= 8.070\) | A1 | 3 d.p. required; Condone 8.07 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \cosh 2x + \sinh x \Rightarrow \frac{dy}{dx} = 2\sinh 2x + \cosh x\) | B1 | Any correct form |
| At S.P.: \(2\sinh 2x + \cosh x = 0\) | ||
| \(\Rightarrow 4\sinh x\cosh x + \cosh x = 0\) | M1 | Setting derivative to zero and using identity |
| \(\Rightarrow \cosh x(4\sinh x+1)=0\) | M1 | Solving \(\frac{dy}{dx}=0\) to obtain value of \(\sinh x\) |
| \(\cosh x = 0\) (rejected) | A1 | Repudiating \(\cosh x=0\) |
| \(\sinh x = -\frac{1}{4}\) | A1 | |
| \(x = \ln\left(-\frac{1}{4}+\frac{\sqrt{17}}{4}\right)\) | M1 | Using log form of arsinh, or setting up and solving quadratic in \(e^x\) |
| A1 | A0 if extra "roots" quoted |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\sinh x\cosh x = 2\times\frac{e^x+e^{-x}}{2}\times\frac{e^x-e^{-x}}{2}$ | | |
| $=\frac{e^{2x}-e^{-2x}}{2}$ | M1 | Using exponential definitions and multiplying or factorising |
| $=\sinh 2x$ | A1 (ag) | |
| Differentiating: $2\cosh 2x = 2\cosh^2 x + 2\sinh^2 x$ | B1 | One side correct |
| $\Rightarrow \cosh 2x = \cosh^2 x + \sinh^2 x$ | B1 | Correct completion |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sketch of $y=\cosh x - 1$ | G1 | Correct shape and through origin |
| Volume $= \pi\int_0^2(\cosh x-1)^2\,dx$ | M1 | $\int(\cosh x-1)^2\,dx$ |
| $= \pi\int_0^2\cosh^2 x - 2\cosh x + 1\,dx$ | A1 | Correct expanded integral including limits 0, 2 |
| $= \pi\int_0^2\frac{1}{2}\cosh 2x - 2\cosh x + \frac{3}{2}\,dx$ | M1 | Attempting integrable form; Dep. on M1 above |
| $= \pi\left[\frac{1}{4}\sinh 2x - 2\sinh x + \frac{3}{2}x\right]_0^2$ | A2 | Give A1 for two terms correct |
| $= \pi\left[\frac{1}{4}\sinh 4 - 2\sinh 2 + 3\right]$ | | |
| $= 8.070$ | A1 | 3 d.p. required; Condone 8.07 |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \cosh 2x + \sinh x \Rightarrow \frac{dy}{dx} = 2\sinh 2x + \cosh x$ | B1 | Any correct form |
| At S.P.: $2\sinh 2x + \cosh x = 0$ | | |
| $\Rightarrow 4\sinh x\cosh x + \cosh x = 0$ | M1 | Setting derivative to zero and using identity |
| $\Rightarrow \cosh x(4\sinh x+1)=0$ | M1 | Solving $\frac{dy}{dx}=0$ to obtain value of $\sinh x$ |
| $\cosh x = 0$ (rejected) | A1 | Repudiating $\cosh x=0$ |
| $\sinh x = -\frac{1}{4}$ | A1 | |
| $x = \ln\left(-\frac{1}{4}+\frac{\sqrt{17}}{4}\right)$ | M1 | Using log form of arsinh, or setting up and solving quadratic in $e^x$ |
| | A1 | A0 if extra "roots" quoted |
---4 (i) Prove, using exponential functions, that
$$\sinh 2 x = 2 \sinh x \cosh x$$
Differentiate this result to obtain a formula for $\cosh 2 x$.\\
(ii) Sketch the curve with equation $y = \cosh x - 1$.
The region bounded by this curve, the $x$-axis, and the line $x = 2$ is rotated through $2 \pi$ radians about the $x$-axis. Find, correct to 3 decimal places, the volume generated. (You must show your working; numerical integration by calculator will receive no credit.)\\
(iii) Show that the curve with equation
$$y = \cosh 2 x + \sinh x$$
has exactly one stationary point.\\
Determine, in exact logarithmic form, the $x$-coordinate of the stationary point.
\hfill \mbox{\textit{OCR MEI FP2 2010 Q4 [18]}}