| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring differentiation of inverse hyperbolic functions and analysis of stationary points. Part (i) involves standard differentiation formulas and algebraic manipulation to solve f'(x)=0, while part (ii) requires understanding odd/even function properties. The techniques are well-practiced in FP2, though the algebraic manipulation to reach the specific form 2ln(1+√2) requires careful work. More routine than typical FP2 proof questions but still requires multiple steps and facility with hyperbolic identities. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.07d Differentiate/integrate: hyperbolic functions4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x)=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+\frac{1}{x^2}}}\cdot\frac{-1}{x^2}\) | M1 | For attempt to differentiate using chain rule |
| B1 | First term correct | |
| \(=\frac{1}{\sqrt{1+x^2}}\left(1-\frac{1}{x}\right)\) | ||
| \(=0 \Rightarrow x=1\) | M1 | For attempt to solve their \(f'(x)=0\) |
| A1 | For correct value of \(x\) (ignore \(x=-1\)) www | |
| \(f(1)=2\sinh^{-1}1=2\ln(1+\sqrt{2})\) | A1 | For correct value obtained www AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch: correct shape in 3rd quadrant only | B1 | For correct shape in 3rd quadrant only (condone inclusion of the 1st quadrant part given) |
| \(\{f(x)\geq 2\ln(1+\sqrt{2}),\;f(x)\leq -2\ln(1+\sqrt{2})\}\) | B1 | For one part of range |
| B1 | For other part of range. SC B1 Both ranges correct but \(<\) and \(>\) used |
# Question 5:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x)=\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+\frac{1}{x^2}}}\cdot\frac{-1}{x^2}$ | M1 | For attempt to differentiate using chain rule |
| | B1 | First term correct |
| $=\frac{1}{\sqrt{1+x^2}}\left(1-\frac{1}{x}\right)$ | | |
| $=0 \Rightarrow x=1$ | M1 | For attempt to solve their $f'(x)=0$ |
| | A1 | For correct value of $x$ (ignore $x=-1$) **www** |
| $f(1)=2\sinh^{-1}1=2\ln(1+\sqrt{2})$ | A1 | For correct value obtained **www AG** |
**[5 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch: correct shape in 3rd quadrant only | B1 | For correct shape in 3rd quadrant only (condone inclusion of the 1st quadrant part given) |
| $\{f(x)\geq 2\ln(1+\sqrt{2}),\;f(x)\leq -2\ln(1+\sqrt{2})\}$ | B1 | For one part of range |
| | B1 | For other part of range. **SC** B1 Both ranges correct but $<$ and $>$ used |
**[3 marks]**5 A function is defined by $\mathrm { f } ( x ) = \sinh ^ { - 1 } x + \sinh ^ { - 1 } \left( \frac { 1 } { x } \right)$, for $x \neq 0$.\\
(i) When $x > 0$, show that the value of $\mathrm { f } ( x )$ for which $\mathrm { f } ^ { \prime } ( x ) = 0$ is $2 \ln ( 1 + \sqrt { 2 } )$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-3_497_659_520_708}
The diagram shows the graph of $y = \mathrm { f } ( x )$ for $x > 0$. Sketch the graph of $y = \mathrm { f } ( x )$ for $x < 0$ and state the range of values that $\mathrm { f } ( x )$ can take for $x \neq 0$.
\hfill \mbox{\textit{OCR FP2 2012 Q5 [8]}}