| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Integrate using hyperbolic substitution |
| Difficulty | Challenging +1.8 This is a challenging Further Maths question requiring hyperbolic substitution (non-standard technique), manipulation of hyperbolic identities, and careful handling of limits. Part (i) demands technical skill with a sophisticated substitution; part (ii) requires exact evaluation; part (iii) tests conceptual understanding of improper integrals. The multi-step nature and advanced topic place it well above average difficulty, though it follows a structured path once the substitution is executed correctly. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08d Volumes of revolution: about x and y axes4.08h Integration: inverse trig/hyperbolic substitutions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \cosh^2 u \Rightarrow du = 2\cosh u\sinh u\,du\) | B1 | For correct result |
| \(\int\sqrt{\frac{x}{x-1}}\,dx = \int\frac{\cosh u}{\sinh u}\cdot 2\cosh u\sinh u\,du\) | M1 | For substituting throughout for \(x\) |
| \(= \int 2\cosh^2 u\,du\) | A1 | For correct simplified \(u\) integral |
| \(= \int(\cosh 2u + 1)\,du = \sinh u\cosh u + u\) | M1 | For attempt to integrate \(\cosh^2 u\) |
| A1 | For correct integration | |
| \(= x^{\frac{1}{2}}(x-1)^{\frac{1}{2}} + \ln\left(x^{\frac{1}{2}} + (x-1)^{\frac{1}{2}}\right)(+c)\) | M1 | For substituting for \(u\) |
| A1 | For correct result, oe as \(f(x) + \ln(g(x))\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\sqrt{3} + \ln(2+\sqrt{3})\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = (\pi)\int_1^4\frac{x}{x-1}\,dx = (\pi)\left[x + \ln(x-1)\right]_1^4\) | M1 | For attempt to find \(\int\frac{x}{x-1}\,dx\) |
| A1 | For correct integration (ignore \(\pi\)) | |
| \(V \to \infty\) | B1 | For statement that volume is infinite (independent of M mark) |
## Question 8:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \cosh^2 u \Rightarrow du = 2\cosh u\sinh u\,du$ | B1 | For correct result |
| $\int\sqrt{\frac{x}{x-1}}\,dx = \int\frac{\cosh u}{\sinh u}\cdot 2\cosh u\sinh u\,du$ | M1 | For substituting throughout for $x$ |
| $= \int 2\cosh^2 u\,du$ | A1 | For correct simplified $u$ integral |
| $= \int(\cosh 2u + 1)\,du = \sinh u\cosh u + u$ | M1 | For attempt to integrate $\cosh^2 u$ |
| | A1 | For correct integration |
| $= x^{\frac{1}{2}}(x-1)^{\frac{1}{2}} + \ln\left(x^{\frac{1}{2}} + (x-1)^{\frac{1}{2}}\right)(+c)$ | M1 | For substituting for $u$ |
| | A1 | For correct result, oe as $f(x) + \ln(g(x))$ |
**Total: 7 marks**
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\sqrt{3} + \ln(2+\sqrt{3})$ | B1 | |
**Total: 1 mark**
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = (\pi)\int_1^4\frac{x}{x-1}\,dx = (\pi)\left[x + \ln(x-1)\right]_1^4$ | M1 | For attempt to find $\int\frac{x}{x-1}\,dx$ |
| | A1 | For correct integration (ignore $\pi$) |
| $V \to \infty$ | B1 | For statement that volume is infinite (independent of M mark) |
**Total: 3 marks**8 (i) Use the substitution $x = \cosh ^ { 2 } u$ to find $\int \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x$, giving your answer in the form $\mathrm { f } ( x ) + \ln ( \mathrm { g } ( x ) )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d25d17c4-a87c-4dcf-900c-400086af6610-3_693_1041_927_593}\\
(ii) Hence calculate the exact area of the region between the curve $y = \sqrt { \frac { x } { x - 1 } }$, the $x$-axis and the lines $x = 1$ and $x = 4$ (see diagram).\\
(iii) What can you say about the volume of the solid of revolution obtained when the region defined in part (ii) is rotated completely about the $x$-axis? Justify your answer.
\hfill \mbox{\textit{OCR FP2 2011 Q8 [11]}}