| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using substitution u = cosh x or u = sinh x |
| Difficulty | Challenging +1.2 Part (i) is a standard identity proof using exponential definitions requiring algebraic manipulation but following a clear path. Part (ii) applies the proven identity to solve a cubic, requiring the insight to recognize 4(5u³) - 3(5u) = 5(4u³ - 3u) = 5cosh(3x) = 13, then using inverse hyperbolic functions. While this requires connecting multiple ideas, it's a fairly standard Further Maths technique with moderate algebraic complexity. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use correct definition of \(\cosh x\) | B1 | |
| Attempt to cube their definition involving \(e^x\) and \(e^{-x}\) (or \(e^{2x}\) and \(e^x\)) | M1 | Must be 4 terms |
| Put their 4 terms into LHS and attempt to simplify | M1 | |
| Clearly get A.G. | A1 | |
| SC: Allow one B1 for correct derivation from \(\cosh 3x = \cosh(2x+x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rewrite as \(k\cosh 3x = 13\) | M1 | |
| Use ln equivalent on \(13/k\) | M1 | Allow \(\pm\ln\) or \(\ln(13/k \pm \sqrt{(13/k)^2-1})\) for their \(k\), or attempt to set up and solve quadratic via exponentials |
| Get \(x = \pm\frac{1}{3}\ln 5\) | A1 | |
| Replace in \(\cosh x\) for \(u\) | M1 | |
| Use \(e^{a\ln b} = b^a\) at least once | M1 | |
| Get \(\frac{1}{2}(5^{\frac{1}{3}} + 5^{-\frac{1}{3}})\) | A1 |
# Question 8:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use correct definition of $\cosh x$ | B1 | |
| Attempt to cube their definition involving $e^x$ and $e^{-x}$ (or $e^{2x}$ and $e^x$) | M1 | Must be 4 terms |
| Put their 4 terms into LHS and attempt to simplify | M1 | |
| Clearly get A.G. | A1 | |
| SC: Allow one B1 for correct derivation from $\cosh 3x = \cosh(2x+x)$ | | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rewrite as $k\cosh 3x = 13$ | M1 | |
| Use ln equivalent on $13/k$ | M1 | Allow $\pm\ln$ or $\ln(13/k \pm \sqrt{(13/k)^2-1})$ for their $k$, or attempt to set up and solve quadratic via exponentials |
| Get $x = \pm\frac{1}{3}\ln 5$ | A1 | |
| Replace in $\cosh x$ for $u$ | M1 | |
| Use $e^{a\ln b} = b^a$ at least once | M1 | |
| Get $\frac{1}{2}(5^{\frac{1}{3}} + 5^{-\frac{1}{3}})$ | A1 | |
---8 (i) Using the definition of $\cosh x$ in terms of $\mathrm { e } ^ { x }$ and $\mathrm { e } ^ { - x }$, show that
$$4 \cosh ^ { 3 } x - 3 \cosh x \equiv \cosh 3 x$$
(ii) Use the substitution $u = \cosh x$ to find, in terms of $5 ^ { \frac { 1 } { 3 } }$, the real root of the equation
$$20 u ^ { 3 } - 15 u - 13 = 0 .$$
\hfill \mbox{\textit{OCR FP2 2010 Q8 [10]}}