| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| 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 FP2 question requiring expansion of exponentials to derive a triple angle formula, then solving a resulting cubic equation in cosh x. While the algebraic manipulation is systematic, it requires careful handling of exponentials, recognition that the equation reduces to a quadratic in cosh²x, and conversion to exact logarithmic form using the inverse hyperbolic function definition—more demanding than standard A-level but routine for Further Maths students. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((e^x + e^{-x})^3 = e^{3x} + 3e^x + 3e^{-x} + e^{-3x}\) | M1 | Doing the expansion |
| \(= (e^{3x} + e^{-3x}) + 3(e^x + e^{-x})\) | A1 | |
| \(\Rightarrow (2\cosh x)^3 = 2\cosh 3x + 6\cosh x\) | M1 | Relating \(\cosh 3x\) to exponentials correctly |
| \(\Rightarrow 8\cosh^3 x = 2\cosh 3x + 6\cosh x\) | A1 | |
| \(\Rightarrow \cosh 3x = 4\cosh^3 x - 3\cosh x\) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cosh 3x = 4\cosh^3 x - 3\cosh x = 6\cosh x\) | M1 | Using result of (i) |
| \(\Rightarrow 4\cosh^3 x = 9\cosh x\) | ||
| \(\Rightarrow \cosh^2 x = \frac{9}{4}\) since \(\cosh x \neq 0\) | A1 | At least one rejection needs to be stated, or \(\cosh x \geq 1\) |
| \(\Rightarrow \cosh x = (\pm)\frac{3}{2}\), \(\cosh x \neq -\frac{3}{2}\) | A1 | |
| \(\Rightarrow x = \pm\ln\!\left(\frac{3}{2} + \sqrt{\left(\frac{3}{2}\right)^2 - 1}\right)\) | A1 | A1 for each in exact form |
| \(= \pm\ln\!\left(\frac{3}{2} + \frac{1}{2}\sqrt{5}\right)\) or \(\ln\!\left(\frac{3}{2} \pm \frac{1}{2}\sqrt{5}\right)\) | A1 [5] | Deduct from 5 marks 1 mark for additional incorrect answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\cosh 3x = 6\cosh x \Rightarrow \frac{1}{2}(e^{3x}+e^{-3x}) = 3(e^x+e^{-x})\) | M1 | Using exponentials |
| \(\Rightarrow e^{3x}-6e^x-6e^{-x}+e^{-3x}=0 \Rightarrow e^{6x}-6e^{4x}-6e^{2x}+1=0\) | ||
| Let \(y=e^{2x}\); \(\Rightarrow y^3-6y^2-6y+1=0 \Rightarrow (y+1)(y^2-7y+1)=0\) | A1 | Cubic in factorised form |
| \(\Rightarrow y=-1, \frac{7\pm\sqrt{45}}{2}\) | A1 | Rejection of \(y=-1\) must be stated |
| \(e^{2x}\neq-1 \Rightarrow e^{2x}=\frac{7\pm\sqrt{45}}{2} \Rightarrow x=\frac{1}{2}\ln\left(\frac{7\pm\sqrt{45}}{2}\right)=\frac{1}{2}\ln\left(\frac{7\pm3\sqrt{5}}{2}\right)\) | A1 | oe in exact form; Deduct from 5 marks 1 mark for additional incorrect answers |
## Question 1:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(e^x + e^{-x})^3 = e^{3x} + 3e^x + 3e^{-x} + e^{-3x}$ | M1 | Doing the expansion |
| $= (e^{3x} + e^{-3x}) + 3(e^x + e^{-x})$ | A1 | |
| $\Rightarrow (2\cosh x)^3 = 2\cosh 3x + 6\cosh x$ | M1 | Relating $\cosh 3x$ to exponentials correctly |
| $\Rightarrow 8\cosh^3 x = 2\cosh 3x + 6\cosh x$ | A1 | |
| $\Rightarrow \cosh 3x = 4\cosh^3 x - 3\cosh x$ | **[4]** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cosh 3x = 4\cosh^3 x - 3\cosh x = 6\cosh x$ | M1 | Using result of (i) |
| $\Rightarrow 4\cosh^3 x = 9\cosh x$ | | |
| $\Rightarrow \cosh^2 x = \frac{9}{4}$ since $\cosh x \neq 0$ | A1 | At least one rejection needs to be stated, or $\cosh x \geq 1$ |
| $\Rightarrow \cosh x = (\pm)\frac{3}{2}$, $\cosh x \neq -\frac{3}{2}$ | A1 | |
| $\Rightarrow x = \pm\ln\!\left(\frac{3}{2} + \sqrt{\left(\frac{3}{2}\right)^2 - 1}\right)$ | A1 | A1 for each in exact form |
| $= \pm\ln\!\left(\frac{3}{2} + \frac{1}{2}\sqrt{5}\right)$ or $\ln\!\left(\frac{3}{2} \pm \frac{1}{2}\sqrt{5}\right)$ | A1 **[5]** | Deduct from 5 marks 1 mark for additional incorrect answers |
# Question 1 (Alternative Method):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\cosh 3x = 6\cosh x \Rightarrow \frac{1}{2}(e^{3x}+e^{-3x}) = 3(e^x+e^{-x})$ | M1 | Using exponentials |
| $\Rightarrow e^{3x}-6e^x-6e^{-x}+e^{-3x}=0 \Rightarrow e^{6x}-6e^{4x}-6e^{2x}+1=0$ | | |
| Let $y=e^{2x}$; $\Rightarrow y^3-6y^2-6y+1=0 \Rightarrow (y+1)(y^2-7y+1)=0$ | A1 | Cubic in factorised form |
| $\Rightarrow y=-1, \frac{7\pm\sqrt{45}}{2}$ | A1 | Rejection of $y=-1$ must be stated |
| $e^{2x}\neq-1 \Rightarrow e^{2x}=\frac{7\pm\sqrt{45}}{2} \Rightarrow x=\frac{1}{2}\ln\left(\frac{7\pm\sqrt{45}}{2}\right)=\frac{1}{2}\ln\left(\frac{7\pm3\sqrt{5}}{2}\right)$ | A1 | **oe** in exact form; Deduct from 5 marks 1 mark for additional incorrect answers |
---1 (i) By first expanding $\left( \mathrm { e } ^ { x } + \mathrm { e } ^ { - x } \right) ^ { 3 }$, or otherwise, show that $\cosh 3 x \equiv 4 \cosh ^ { 3 } x - 3 \cosh x$.\\
(ii) Solve the equation $\cosh 3 x = 6 \cosh x$, giving your answers in exact logarithmic form.
\hfill \mbox{\textit{OCR FP2 2016 Q1 [9]}}