| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Express in form R cosh(x±α) or R sinh(x±α) |
| Difficulty | Standard +0.3 Part (a) is a standard bookwork proof using exponential definitions of hyperbolic functions. Part (b) requires applying the identity from (a), expanding using definitions, and solving a straightforward equation. This is a typical Further Maths exercise with clear scaffolding and standard techniques, making it slightly easier than average for A-level overall but routine for FM 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/Working | Mark | Guidance |
| \((\cosh A \cosh B + \sinh A \sinh B) = \left(\frac{e^A + e^{-A}}{2}\right)\left(\frac{e^B + e^{-B}}{2}\right) + \left(\frac{e^A - e^{-A}}{2}\right)\left(\frac{e^B - e^{-B}}{2}\right)\) leading to \(= \frac{e^{A+B} + e^{A-B} + e^{B-A} + e^{-A-B} + e^{A+B} - e^{A-B} - e^{B-A} + e^{-A-B}}{4}\) | M1 | Expresses lhs in terms of exponentials correctly, combines terms and combines fractions with common denominator |
| \(= \frac{2e^{A+B} + 2e^{-(A+B)}}{4} = \frac{e^{A+B} + e^{-(A+B)}}{2} = \cosh(A+B)\) | A1* | Fully correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cosh(x + \ln 2) = \cosh x \cosh(\ln 2) + \sinh x \sinh(\ln 2) = \left(\frac{2 + \frac{1}{2}}{2}\right)\cosh x + \left(\frac{2 - \frac{1}{2}}{2}\right)\sinh x\) | M1 | Applies result from part (a) and evaluates both \(\cosh(\ln 2)\) and \(\sinh(\ln 2)\). Use of (a) must be seen |
| \(\frac{5}{4}\cosh x + \frac{3}{4}\sinh x = 5\sinh x \Rightarrow \frac{5}{4}\cosh x = \frac{17}{4}\sinh x\) | dM1 | Collects terms and reaches \(a\cosh x = b\sinh x\). Depends on first M mark |
| \(5\cosh x = 17\sinh x\) | A1 | Correct equation |
| \(x = \frac{1}{2}\ln\left(\frac{1 + \frac{5}{17}}{1 - \frac{5}{17}}\right)\) or \(\frac{e^{2x}-1}{e^{2x}+1} = \frac{5}{17} \Rightarrow x = \ldots\) | ddM1 | Moves to \(\tanh x\) and uses correct logarithmic form for \(\text{artanh}\, x\) or reverts to exponential forms and solves for \(x\). Depends on both M marks |
| \(x = \frac{1}{2}\ln\left(\frac{11}{6}\right)\) | A1 | Cao. Accept integer multiples of \(\frac{11}{6}\) |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cosh A \cosh B + \sinh A \sinh B) = \left(\frac{e^A + e^{-A}}{2}\right)\left(\frac{e^B + e^{-B}}{2}\right) + \left(\frac{e^A - e^{-A}}{2}\right)\left(\frac{e^B - e^{-B}}{2}\right)$ leading to $= \frac{e^{A+B} + e^{A-B} + e^{B-A} + e^{-A-B} + e^{A+B} - e^{A-B} - e^{B-A} + e^{-A-B}}{4}$ | M1 | Expresses lhs in terms of exponentials correctly, combines terms and combines fractions with common denominator |
| $= \frac{2e^{A+B} + 2e^{-(A+B)}}{4} = \frac{e^{A+B} + e^{-(A+B)}}{2} = \cosh(A+B)$ | A1* | Fully correct proof with no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cosh(x + \ln 2) = \cosh x \cosh(\ln 2) + \sinh x \sinh(\ln 2) = \left(\frac{2 + \frac{1}{2}}{2}\right)\cosh x + \left(\frac{2 - \frac{1}{2}}{2}\right)\sinh x$ | M1 | Applies result from part (a) and evaluates both $\cosh(\ln 2)$ and $\sinh(\ln 2)$. Use of (a) must be seen |
| $\frac{5}{4}\cosh x + \frac{3}{4}\sinh x = 5\sinh x \Rightarrow \frac{5}{4}\cosh x = \frac{17}{4}\sinh x$ | dM1 | Collects terms and reaches $a\cosh x = b\sinh x$. Depends on first M mark |
| $5\cosh x = 17\sinh x$ | A1 | Correct equation |
| $x = \frac{1}{2}\ln\left(\frac{1 + \frac{5}{17}}{1 - \frac{5}{17}}\right)$ or $\frac{e^{2x}-1}{e^{2x}+1} = \frac{5}{17} \Rightarrow x = \ldots$ | ddM1 | Moves to $\tanh x$ and uses correct logarithmic form for $\text{artanh}\, x$ or reverts to exponential forms and solves for $x$. Depends on both M marks |
| $x = \frac{1}{2}\ln\left(\frac{11}{6}\right)$ | A1 | Cao. Accept integer multiples of $\frac{11}{6}$ |
---\begin{enumerate}
\item (a) Use the definitions of $\sinh x$ and $\cosh x$ in terms of exponentials to show that
\end{enumerate}
$$\cosh A \cosh B + \sinh A \sinh B \equiv \cosh ( A + B )$$
(b) Hence find the value of $x$ for which
$$\cosh ( x + \ln 2 ) = 5 \sinh x$$
giving your answer in the form $\frac { 1 } { 2 } \ln k$, where $k$ is a rational number to be determined.\\
(5)
\hfill \mbox{\textit{Edexcel F3 2022 Q1 [7]}}