| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2007 |
| Session | January |
| 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 Further Maths question on hyperbolic functions requiring standard techniques: (i) proving a logarithmic form (bookwork), (ii) integration using substitution/arcosh (routine application), and (iii) finding stationary points by differentiation and solving a quadratic. While it involves Further Maths content and multiple steps, each part follows standard procedures without requiring novel insight or particularly complex algebraic manipulation. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08h Integration by substitution4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(\cosh y = x\), so \(\frac{e^y+e^{-y}}{2}=x\) | M1 | |
| \(e^{2y}-2xe^y+1=0\) | M1 | Quadratic in \(e^y\) |
| \(e^y = x \pm \sqrt{x^2-1}\), take \(+\) since \(e^y>0\) | M1, A1 | |
| \(y = \ln(x+\sqrt{x^2-1})\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{1}{2}\int_{2.5}^{3.9}\frac{1}{\sqrt{x^2-\frac{9}{4}}} \, dx = \frac{1}{2}\left[\text{arcosh}\frac{2x}{3}\right]_{2.5}^{3.9}\) | M1, A1 | Standard form |
| \(= \frac{1}{2}\left[\ln\left(\frac{2x}{3}+\sqrt{\frac{4x^2}{9}-1}\right)\right]_{2.5}^{3.9}\) | M1 | |
| \(= \frac{1}{2}\ln\frac{13}{5}\) or equivalent \(a\ln b\) form | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{\sinh x(2+\sinh x)-\cosh^2 x}{(2+\sinh x)^2} = \frac{1}{9}\) | M1 | Quotient rule |
| Numerator: \(2\sinh x + \sinh^2 x - \cosh^2 x = 2\sinh x -1\) | M1, A1 | Using \(\cosh^2-\sinh^2=1\) |
| \(2\sinh x - 1 = \frac{(2+\sinh x)^2}{9}\) | M1 | Setting equal to \(\frac{1}{9}\) |
| Let \(s=\sinh x\): \(\frac{s^2-16s+13}{9}... \Rightarrow s^2-16s+... \) solve | M1 | |
| \(\sinh x = 1\) or \(\sinh x = \frac{1}{3}\)... giving two points | A1 | |
| Verify \((\ln(1+\sqrt{2}), \frac{1}{3}\sqrt{2})\) and find other point \((\ln(\frac{1}{3}+\sqrt{\frac{1}{9}+1}), ...)\) | A1, A1 |
## Question 4 (Option 1):
**Part (i):** Show $\text{arcosh}\, x = \ln(x+\sqrt{x^2-1})$
| Let $\cosh y = x$, so $\frac{e^y+e^{-y}}{2}=x$ | M1 | |
| $e^{2y}-2xe^y+1=0$ | M1 | Quadratic in $e^y$ |
| $e^y = x \pm \sqrt{x^2-1}$, take $+$ since $e^y>0$ | M1, A1 | |
| $y = \ln(x+\sqrt{x^2-1})$ | A1 | |
**Part (ii):** $\int_{2.5}^{3.9} \frac{1}{\sqrt{4x^2-9}} \, dx$
| $= \frac{1}{2}\int_{2.5}^{3.9}\frac{1}{\sqrt{x^2-\frac{9}{4}}} \, dx = \frac{1}{2}\left[\text{arcosh}\frac{2x}{3}\right]_{2.5}^{3.9}$ | M1, A1 | Standard form |
| $= \frac{1}{2}\left[\ln\left(\frac{2x}{3}+\sqrt{\frac{4x^2}{9}-1}\right)\right]_{2.5}^{3.9}$ | M1 | |
| $= \frac{1}{2}\ln\frac{13}{5}$ or equivalent $a\ln b$ form | A1, A1 | |
**Part (iii):** Gradient of $y=\frac{\cosh x}{2+\sinh x}$
| $\frac{dy}{dx} = \frac{\sinh x(2+\sinh x)-\cosh^2 x}{(2+\sinh x)^2} = \frac{1}{9}$ | M1 | Quotient rule |
| Numerator: $2\sinh x + \sinh^2 x - \cosh^2 x = 2\sinh x -1$ | M1, A1 | Using $\cosh^2-\sinh^2=1$ |
| $2\sinh x - 1 = \frac{(2+\sinh x)^2}{9}$ | M1 | Setting equal to $\frac{1}{9}$ |
| Let $s=\sinh x$: $\frac{s^2-16s+13}{9}... \Rightarrow s^2-16s+... $ solve | M1 | |
| $\sinh x = 1$ or $\sinh x = \frac{1}{3}$... giving two points | A1 | |
| Verify $(\ln(1+\sqrt{2}), \frac{1}{3}\sqrt{2})$ and find other point $(\ln(\frac{1}{3}+\sqrt{\frac{1}{9}+1}), ...)$ | A1, A1 | |
---4 (i) Show that $\operatorname { arcosh } x = \ln \left( x + \sqrt { x ^ { 2 } - 1 } \right)$.\\
(ii) Find $\int _ { 2.5 } ^ { 3.9 } \frac { 1 } { \sqrt { 4 x ^ { 2 } - 9 } } \mathrm {~d} x$, giving your answer in the form $a \ln b$, where $a$ and $b$ are rational numbers.\\
(iii) There are two points on the curve $y = \frac { \cosh x } { 2 + \sinh x }$ at which the gradient is $\frac { 1 } { 9 }$.
Show that one of these points is $\left( \ln ( 1 + \sqrt { 2 } ) , \frac { 1 } { 3 } \sqrt { 2 } \right)$, and find the coordinates of the other point, in a similar form.
\hfill \mbox{\textit{OCR MEI FP2 2007 Q4 [18]}}