| 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 | Solve using double angle formulas |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question requiring standard manipulation of hyperbolic identities. Part (i) is routine proof by quoting given formulas, and part (ii) involves substituting the double angle formula to get a quadratic in tanh x, then using the standard inverse formula. While it's Further Maths content, the execution is mechanical with no novel insight required, placing it slightly above average difficulty. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tanh 2x=\frac{\sinh 2x}{\cosh 2x}=\frac{2\sinh x\cosh x}{\cosh^2 x+\sinh^2 x}\) | M1 | For \(\frac{\sinh 2x}{\cosh 2x}\) and use double angle formulae |
| \(\equiv\frac{2\tanh x}{1+\tanh^2 x}\) | A1 | For division by \(\cosh^2 x\) seen. N.B. \(\tanh(A+B)\) not in formula book |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{10t}{(t^2+1)}=(1+6t)\) | M1 | For using (i) to obtain equation in \(t\) |
| \(\Rightarrow 6t^3+t^2-4t+1=0\) | A1 | Correct cubic equation |
| \(\Rightarrow (t+1)(3t-1)(2t-1)=0\) | M1 | Attempt to solve cubic (calculator OK) |
| \(\Rightarrow t=(-1),\,\frac{1}{3},\,\frac{1}{2}\) | A1 | Solution. Ignore any extra values at this stage |
| \(x=\frac{1}{2}\ln\frac{1+t}{1-t} \Rightarrow x=\frac{1}{2}\ln 2,\,\frac{1}{2}\ln 3\) | M1 | For using ln form for \(\tanh^{-1}\) |
| A1 | Correct 2 values (only) or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{4x}-5e^{2x}+6=0\) | M1, A1 | Use exponentials to obtain a quadratic in \(e^{2x}\); Correct |
| \(\Rightarrow(e^{2x}-2)(e^{2x}-3)=0\) | M1 | Solve quadratic |
| \(\Rightarrow e^{2x}=2,\;3\) | A1 | Solution |
| \(\Rightarrow 2x=\ln 2,\;\ln 3\) | M1 | Take logs |
| \(\Rightarrow x=\frac{1}{2}\ln 2,\;\frac{1}{2}\ln 3\) | A1 |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tanh 2x=\frac{\sinh 2x}{\cosh 2x}=\frac{2\sinh x\cosh x}{\cosh^2 x+\sinh^2 x}$ | M1 | For $\frac{\sinh 2x}{\cosh 2x}$ and use double angle formulae |
| $\equiv\frac{2\tanh x}{1+\tanh^2 x}$ | A1 | For division by $\cosh^2 x$ seen. N.B. $\tanh(A+B)$ not in formula book |
**[2 marks]**
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{10t}{(t^2+1)}=(1+6t)$ | M1 | For using (i) to obtain equation in $t$ |
| $\Rightarrow 6t^3+t^2-4t+1=0$ | A1 | Correct cubic equation |
| $\Rightarrow (t+1)(3t-1)(2t-1)=0$ | M1 | Attempt to solve cubic (calculator OK) |
| $\Rightarrow t=(-1),\,\frac{1}{3},\,\frac{1}{2}$ | A1 | Solution. Ignore any extra values at this stage |
| $x=\frac{1}{2}\ln\frac{1+t}{1-t} \Rightarrow x=\frac{1}{2}\ln 2,\,\frac{1}{2}\ln 3$ | M1 | For using ln form for $\tanh^{-1}$ |
| | A1 | Correct 2 values (only) or equivalent |
*Alternative:*
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{4x}-5e^{2x}+6=0$ | M1, A1 | Use exponentials to obtain a quadratic in $e^{2x}$; Correct |
| $\Rightarrow(e^{2x}-2)(e^{2x}-3)=0$ | M1 | Solve quadratic |
| $\Rightarrow e^{2x}=2,\;3$ | A1 | Solution |
| $\Rightarrow 2x=\ln 2,\;\ln 3$ | M1 | Take logs |
| $\Rightarrow x=\frac{1}{2}\ln 2,\;\frac{1}{2}\ln 3$ | A1 | |
**[6 marks]**
---3 (i) By quoting results given in the List of Formulae (MF1), prove that $\tanh 2 x \equiv \frac { 2 \tanh x } { 1 + \tanh ^ { 2 } x }$.\\
(ii) Solve the equation $5 \tanh 2 x = 1 + 6 \tanh x$, giving your answers in logarithmic form.
\hfill \mbox{\textit{OCR FP2 2012 Q3 [8]}}