| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Standard +0.3 This is a standard Further Maths derivation requiring straightforward algebraic manipulation of exponential definitions and solving for the inverse. While it involves hyperbolic functions (a Further Maths topic), the proof follows a well-established template with no novel insight required—students practice this exact type of derivation regularly in FP3. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tanh x = \frac{\sinh x}{\cosh x} = \frac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}}\) or \(\frac{2e^x}{e^{2x}+1}\) | M1 | Substitutes the correct exponential forms. Note that \(\tanh x = \frac{\sinh x}{\cosh x}\) may be implied. |
| \(= \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x}-1}{e^{2x}+1}\) * | A1* | Correct proof with no errors or omissions or notational errors such as using sin for sinh. Question requires starting from definitions of \(\sinh x\) and \(\cosh x\) in terms of exponentials. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \text{artanh}\,\theta \Rightarrow \tanh y = \theta \Rightarrow \theta = \frac{e^{2y}-1}{e^{2y}+1}\), then \(\theta(e^{2y}+1) = e^{2y}-1 \Rightarrow e^{2y}(\theta-1) = -1-\theta \Rightarrow e^{2y} = \frac{1+\theta}{1-\theta}\) | M1 | Setting \(\theta = \frac{e^{2y}-1}{e^{2y}+1}\) or any other variables and correct processing (allow sign errors only) to make \(e^{2y}\) the subject |
| \(e^{2y} = \frac{1+\theta}{1-\theta} \Rightarrow 2y = \ln\!\left(\frac{1+\theta}{1-\theta}\right)\) | dM1 | Removes \(e\) correctly by taking ln's. Dependent on the first method mark. |
| \(y = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)\)* | A1* | Correct completion with no errors. Must be in terms of \(\theta\). Mark withheld if "tan" appears instead of "tanh" and/or missing variables. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{artanh}\,\theta = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right) \Rightarrow \theta = \tanh\!\left(\frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)\right)\), then \(\theta = \frac{e^{\ln\!\left(\frac{1+\theta}{1-\theta}\right)}-1}{e^{\ln\!\left(\frac{1+\theta}{1-\theta}\right)}+1}\) | M1 | Uses part (a) to express \(\theta\) in terms of \(e\) |
| \(= \frac{\frac{1+\theta}{1-\theta}-1}{\frac{1+\theta}{1-\theta}+1} = \frac{1+\theta-1+\theta}{1+\theta+1-\theta} = \theta\) | dM1 | Removes \(e\)'s and ln's correctly. Dependent on first method mark. |
| \(\Rightarrow \text{artanh}\,\theta = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)\)* | A1* | Obtains \(\theta = \theta\) with no errors and makes a conclusion. Must be in terms of \(\theta\). |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tanh x = \frac{\sinh x}{\cosh x} = \frac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}}$ or $\frac{2e^x}{e^{2x}+1}$ | M1 | Substitutes the correct exponential forms. Note that $\tanh x = \frac{\sinh x}{\cosh x}$ may be implied. |
| $= \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x}-1}{e^{2x}+1}$ * | A1* | Correct proof with no errors or omissions or notational errors such as using sin for sinh. Question requires starting from definitions of $\sinh x$ and $\cosh x$ in terms of exponentials. |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \text{artanh}\,\theta \Rightarrow \tanh y = \theta \Rightarrow \theta = \frac{e^{2y}-1}{e^{2y}+1}$, then $\theta(e^{2y}+1) = e^{2y}-1 \Rightarrow e^{2y}(\theta-1) = -1-\theta \Rightarrow e^{2y} = \frac{1+\theta}{1-\theta}$ | M1 | Setting $\theta = \frac{e^{2y}-1}{e^{2y}+1}$ or any other variables and correct processing (allow sign errors only) to make $e^{2y}$ the subject |
| $e^{2y} = \frac{1+\theta}{1-\theta} \Rightarrow 2y = \ln\!\left(\frac{1+\theta}{1-\theta}\right)$ | dM1 | Removes $e$ correctly by taking ln's. Dependent on the first method mark. |
| $y = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)$* | A1* | Correct completion with no errors. Must be in terms of $\theta$. Mark withheld if "tan" appears instead of "tanh" and/or missing variables. |
**Alternative for (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{artanh}\,\theta = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right) \Rightarrow \theta = \tanh\!\left(\frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)\right)$, then $\theta = \frac{e^{\ln\!\left(\frac{1+\theta}{1-\theta}\right)}-1}{e^{\ln\!\left(\frac{1+\theta}{1-\theta}\right)}+1}$ | M1 | Uses part (a) to express $\theta$ in terms of $e$ |
| $= \frac{\frac{1+\theta}{1-\theta}-1}{\frac{1+\theta}{1-\theta}+1} = \frac{1+\theta-1+\theta}{1+\theta+1-\theta} = \theta$ | dM1 | Removes $e$'s and ln's correctly. Dependent on first method mark. |
| $\Rightarrow \text{artanh}\,\theta = \frac{1}{2}\ln\!\left(\frac{1+\theta}{1-\theta}\right)$* | A1* | Obtains $\theta = \theta$ with no errors and makes a conclusion. Must be in terms of $\theta$. |
> Note: Attempts that assume $\text{artanh}\,\theta = \frac{\text{arsinh}\,\theta}{\text{arcosh}\,\theta}$ score no marks in (b).
---\begin{enumerate}
\item (a) Starting from the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials, show that, for $x \in \mathbb { R }$
\end{enumerate}
$$\tanh x = \frac { \mathrm { e } ^ { 2 x } - 1 } { \mathrm { e } ^ { 2 x } + 1 }$$
(b) Hence, given that $- 1 < \theta < 1$, prove that
$$\operatorname { artanh } \theta = \frac { 1 } { 2 } \ln \left( \frac { 1 + \theta } { 1 - \theta } \right)$$
uestion 1 continued\\
$\_\_\_\_$ 7\\
\hfill \mbox{\textit{Edexcel FP3 2018 Q1 [5]}}