| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve using double angle formulas |
| Difficulty | Standard +0.8 This is a Further Maths question requiring manipulation of hyperbolic identities using exponential definitions, followed by solving a hyperbolic equation. Part (i) demands algebraic stamina working with exponentials and multiple angle formulas. Part (ii) requires recognizing a quadratic in cosh 2x and applying inverse hyperbolic functions. While systematic, it's more demanding than standard A-level due to the extended algebraic manipulation and being Further Maths content, placing it moderately above average difficulty. |
| 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 | Marks | Guidance |
| \(8\sinh^4 x \equiv \frac{8}{16}(e^x-e^{-x})^4\) | B1 | \(\sinh x = \frac{1}{2}(e^x-e^{-x})\) seen or implied |
| \(\equiv\frac{8}{16}(e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x})\) | M1 | For attempt to expand \((\ldots)^4\) by binomial theorem OR otherwise |
| \(\equiv\frac{1}{2}(e^{4x}+e^{-4x})-\frac{4}{2}(e^{2x}+e^{-2x})+\frac{6}{2}\) | M1 | For grouping terms for \(\cosh 4x\) or \(\cosh 2x\); OR using \(e^{4x}\) or \(e^{2x}\) expressions from RHS |
| \(\equiv\cosh 4x - 4\cosh 2x+3\) | A1 4 | For correct expression AG |
| SR may be done wholly from RHS to LHS | M1 M1 B1 A1 | Evidence of factorising required for 2nd M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Rightarrow (8\sinh^4 x + 4\cosh 2x - 3) - 3\cosh 2x + 1 = 0\) | M1 | For using (i) and \(\cosh 2x\equiv\pm1\pm2\sinh^2 x\) |
| \(\Rightarrow 8\sinh^4 x + 2\sinh^2 x - 1 = 0\) | A1 | For correct equation |
| \((4\sinh^2 x-1)(2\sinh^2 x+1)=0 \Rightarrow \sinh x=\pm\frac{1}{2}\) | M1 | For solving their quartic for \(\sinh x\) |
| A1 | For correct \(\sinh x\) (ignore other roots) | |
| \(x=\ln\left(\pm\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)=\pm\ln\left(\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)\) | A1\(\checkmark\) 5 | For correct answers (and no more) f.t. from their value(s) for \(\sinh x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Rightarrow (2\cosh^2 2x-1)-3\cosh 2x+1=0\) | M1 | For using \(\cosh 4x\equiv\pm2\cosh^2 2x\pm1\) |
| \(\Rightarrow 2\cosh^2 2x - 3\cosh 2x = 0\) | A1 | For correct equation |
| \(\cosh 2x=\frac{3}{2} \Rightarrow x=\frac{1}{2}\ln\left(\frac{3}{2}\pm\frac{1}{2}\sqrt{5}\right)\) | M1 | For solving for \(\cosh 2x\) |
| \(=\pm\frac{1}{2}\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)\) | A1, A1\(\checkmark\) | For correct \(\cosh 2x\) (ignore others); for correct answers (and no more) f.t. from value(s) for \(\cosh 2x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\Rightarrow e^{4x}-3e^{2x}+2-3e^{-2x}+e^{-4x}=0\) | M1, A1 | For changing to \(e^{\pm kx}\); for correct equation |
| \(\Rightarrow (e^{4x}+1)(e^{4x}-3e^{2x}+1)=0\) | M1, A1 | For solving for \(e^{2x}\); for correct \(e^{2x}\) (ignore others) |
| \(\Rightarrow e^{2x}=\frac{1}{2}(3\pm\sqrt{5}) \Rightarrow x=\frac{1}{2}\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)\) | A1\(\checkmark\) | For correct answers (and no more) f.t. from value(s) for \(e^{2x}\) |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $8\sinh^4 x \equiv \frac{8}{16}(e^x-e^{-x})^4$ | B1 | $\sinh x = \frac{1}{2}(e^x-e^{-x})$ seen or implied |
| $\equiv\frac{8}{16}(e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x})$ | M1 | For attempt to expand $(\ldots)^4$ by binomial theorem OR otherwise |
| $\equiv\frac{1}{2}(e^{4x}+e^{-4x})-\frac{4}{2}(e^{2x}+e^{-2x})+\frac{6}{2}$ | M1 | For grouping terms for $\cosh 4x$ or $\cosh 2x$; OR using $e^{4x}$ or $e^{2x}$ expressions from RHS |
| $\equiv\cosh 4x - 4\cosh 2x+3$ | A1 **4** | For correct expression **AG** |
| SR may be done wholly from RHS to LHS | M1 M1 B1 A1 | Evidence of factorising required for 2nd M1 |
## Question 4(ii):
**Method 1:** $\cosh 4x - 3\cosh 2x + 1 = 0$
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Rightarrow (8\sinh^4 x + 4\cosh 2x - 3) - 3\cosh 2x + 1 = 0$ | M1 | For using **(i)** and $\cosh 2x\equiv\pm1\pm2\sinh^2 x$ |
| $\Rightarrow 8\sinh^4 x + 2\sinh^2 x - 1 = 0$ | A1 | For correct equation |
| $(4\sinh^2 x-1)(2\sinh^2 x+1)=0 \Rightarrow \sinh x=\pm\frac{1}{2}$ | M1 | For solving their quartic for $\sinh x$ |
| | A1 | For correct $\sinh x$ (ignore other roots) |
| $x=\ln\left(\pm\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)=\pm\ln\left(\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)$ | A1$\checkmark$ **5** | For correct answers (and no more) f.t. from their value(s) for $\sinh x$ |
**SR:** Similar scheme for $8\cosh^4 x - 14\cosh^2 x + 5 = 0 \Rightarrow \cosh x=\frac{1}{2}\sqrt{5} \Rightarrow x=\pm\ln\left(\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)$
**Method 2:** $\cosh 4x - 3\cosh 2x + 1 = 0$
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Rightarrow (2\cosh^2 2x-1)-3\cosh 2x+1=0$ | M1 | For using $\cosh 4x\equiv\pm2\cosh^2 2x\pm1$ |
| $\Rightarrow 2\cosh^2 2x - 3\cosh 2x = 0$ | A1 | For correct equation |
| $\cosh 2x=\frac{3}{2} \Rightarrow x=\frac{1}{2}\ln\left(\frac{3}{2}\pm\frac{1}{2}\sqrt{5}\right)$ | M1 | For solving for $\cosh 2x$ |
| $=\pm\frac{1}{2}\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)$ | A1, A1$\checkmark$ | For correct $\cosh 2x$ (ignore others); for correct answers (and no more) f.t. from value(s) for $\cosh 2x$ |
**Method 3:** Put all into exponentials
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\Rightarrow e^{4x}-3e^{2x}+2-3e^{-2x}+e^{-4x}=0$ | M1, A1 | For changing to $e^{\pm kx}$; for correct equation |
| $\Rightarrow (e^{4x}+1)(e^{4x}-3e^{2x}+1)=0$ | M1, A1 | For solving for $e^{2x}$; for correct $e^{2x}$ (ignore others) |
| $\Rightarrow e^{2x}=\frac{1}{2}(3\pm\sqrt{5}) \Rightarrow x=\frac{1}{2}\ln\left(\frac{3}{2}+\frac{1}{2}\sqrt{5}\right)$ | A1$\checkmark$ | For correct answers (and no more) f.t. from value(s) for $e^{2x}$ |
---4 (i) Use the definitions of hyperbolic functions in terms of exponentials to prove that
$$8 \sinh ^ { 4 } x \equiv \cosh 4 x - 4 \cosh 2 x + 3$$
(ii) Solve the equation
$$\cosh 4 x - 3 \cosh 2 x + 1 = 0$$
giving your answer(s) in logarithmic form.
\hfill \mbox{\textit{OCR FP2 2011 Q4 [9]}}