| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Standard +0.3 Part (i) is a standard identity proof using exponential definitions (routine for FP2). Part (ii) requires substituting the identity to form a quadratic in cosh x, then solving using inverse hyperbolic functions—a straightforward application of techniques with no novel insight required. Slightly above average due to being Further Maths content, but this is a textbook exercise. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh x = \frac{e^x+e^{-x}}{2}\), \(\sinh x = \frac{e^x-e^{-x}}{2}\) | B1 | Correct formulae |
| \(\Rightarrow\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2\) | M1 | Dealing with squaring correctly |
| \(= \frac{1}{4}(e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}) = \frac{1}{4}\cdot 4 = 1\) | A1 | www All steps seen |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\Rightarrow \cosh^2 x - 1 = 5\cosh x - 7\) | ||
| \(\Rightarrow \cosh^2 x - 5\cosh x + 6 = 0\) | M1 | Use (i) |
| \(\Rightarrow (\cosh x - 2)(\cosh x - 3) = 0\) | M1 | Attempt to solve quadratic |
| \(\Rightarrow \cosh x = 2, 3\) | A1 | |
| \(\Rightarrow x = \cosh^{-1}2 = \pm\ln(2\pm\sqrt{3})\) | A1 | Use correct ln formula |
| and \(x = \cosh^{-1}3 = \pm\ln(3\pm\sqrt{8})\) | A1 | Use correct ln formula |
| [5] |
# Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh x = \frac{e^x+e^{-x}}{2}$, $\sinh x = \frac{e^x-e^{-x}}{2}$ | B1 | Correct formulae |
| $\Rightarrow\cosh^2 x - \sinh^2 x = \left(\frac{e^x+e^{-x}}{2}\right)^2 - \left(\frac{e^x-e^{-x}}{2}\right)^2$ | M1 | Dealing with squaring correctly | Difference of squares can be used |
| $= \frac{1}{4}(e^{2x}+2+e^{-2x}-e^{2x}+2-e^{-2x}) = \frac{1}{4}\cdot 4 = 1$ | A1 | **www** All steps seen |
| **[3]** | | |
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# Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\Rightarrow \cosh^2 x - 1 = 5\cosh x - 7$ | | |
| $\Rightarrow \cosh^2 x - 5\cosh x + 6 = 0$ | M1 | Use **(i)** |
| $\Rightarrow (\cosh x - 2)(\cosh x - 3) = 0$ | M1 | Attempt to solve quadratic | E.g. correct formula or expansion giving 2 out of 3 terms correct |
| $\Rightarrow \cosh x = 2, 3$ | A1 | |
| $\Rightarrow x = \cosh^{-1}2 = \pm\ln(2\pm\sqrt{3})$ | A1 | Use correct ln formula | Condone lack of $\pm$ |
| and $x = \cosh^{-1}3 = \pm\ln(3\pm\sqrt{8})$ | A1 | Use correct ln formula | Condone lack of $\pm$ |
| **[5]** | | |
---2 (i) Using the definitions for $\cosh x$ and $\sinh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that $\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1$.\\
(ii) Hence solve the equation $\sinh ^ { 2 } x = 5 \cosh x - 7$, giving your answers in logarithmic form.
\hfill \mbox{\textit{OCR FP2 2013 Q2 [8]}}