| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| 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 This is a standard Further Maths hyperbolic functions question with routine steps: (a) is a bookwork proof using exponential definitions, (b) follows the standard R-formula method analogous to trigonometry, and (c) is straightforward solving once the form is established. While it requires multiple techniques, each step follows well-established procedures with no novel insight needed. |
| 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 |
| \(\frac{e^A - e^{-A}}{2} \times \frac{e^B+e^{-B}}{2} + \frac{e^A+e^{-A}}{2} \times \frac{e^B-e^{-B}}{2}\) | M1 | Replaces two of the four hyperbolic functions with correct exponential expressions. Condone poor bracketing. |
| \(= \frac{e^{A+B} - e^{B-A} + e^{A-B} - e^{-A-B} + e^{A+B} + e^{B-A} - e^{A-B} - e^{-A-B}}{4}\) | M1 | Expands numerator. Must see at least four terms. Allow sign errors only with coefficients and indices. |
| \(= \frac{2e^{A+B} - 2e^{-(A+B)}}{4} = \sinh(A+B)\) | A1* | Reaches \(\sinh(A+B)\) with no errors. "= LHS" or "= RHS" must be seen. All bracketing correct where required. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(R\sinh\alpha = 8,\ R\cosh\alpha = 10\) | B1, M1 on ePen | Equates coefficients to obtain two correct equations. \(R^2 = 10^2 - 8^2\) provided incorrect equations are not seen. |
| \(R^2(\cosh^2\alpha - \sinh^2\alpha) = 10^2 - 8^2 \Rightarrow R^2 = 36 \Rightarrow R = 6\) | 1st M1 | Complete attempt at finding positive \(R\). Correct exponential definitions must be used. |
| \(\tanh\alpha = \frac{8}{10} \Rightarrow \alpha = \text{artanh}\left(\frac{4}{5}\right) = \frac{1}{2}\ln\frac{1+\frac{4}{5}}{1-\frac{4}{5}} = \frac{1}{2}\ln 9 = \ln 3\) | 2nd M1 | Complete attempt at finding positive \(\alpha\) where \(\alpha = k\ln p\), \(k>0\), \(p>1\). Correct logarithmic form must be used. |
| \(6\sinh(x+\ln 3)\), \(R=6\) and \(\alpha = \ln 3\) (or \(p=3\)) | A1 | Correct expression. A0 for additional solutions e.g. \(6\sinh(x\pm\ln 3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \text{arsinh}(3\sqrt{7}) - \ln 3 \Rightarrow x = \ln\left(3\sqrt{7}+\sqrt{(3\sqrt{7})^2+1}\right) - \ln 3\) | M1 | Obtains \(x = \text{arsinh}\!\left(\frac{18\sqrt{7}}{\text{"6"}}\right) \pm \ln\text{"3"}\) and uses correct logarithmic form. Work must be exact, not decimals. |
| \(x = \ln\!\left(\frac{3\sqrt{7}+8}{3}\right) = \ln\!\left(\sqrt{7}+\frac{8}{3}\right)\) | A1 | Correct answer in correct form. Accept e.g. \(\ln\!\left(2\tfrac{2}{3}+\sqrt{7}\right)\). Must be fully bracketed correctly. No additional answers. |
## Question 4(a):
Prove $\sinh(A+B) = \sinh A\cosh B + \cosh A\sinh B$
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{e^A - e^{-A}}{2} \times \frac{e^B+e^{-B}}{2} + \frac{e^A+e^{-A}}{2} \times \frac{e^B-e^{-B}}{2}$ | M1 | Replaces two of the four hyperbolic functions with correct exponential expressions. Condone poor bracketing. |
| $= \frac{e^{A+B} - e^{B-A} + e^{A-B} - e^{-A-B} + e^{A+B} + e^{B-A} - e^{A-B} - e^{-A-B}}{4}$ | M1 | Expands numerator. Must see at least four terms. Allow sign errors only with coefficients and indices. |
| $= \frac{2e^{A+B} - 2e^{-(A+B)}}{4} = \sinh(A+B)$ | A1* | Reaches $\sinh(A+B)$ with no errors. "= LHS" or "= RHS" must be seen. All bracketing correct where required. |
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## Question 4(b):
$10\sinh x + 8\cosh x = R\sinh(x+\alpha)$
| Answer | Mark | Guidance |
|--------|------|----------|
| $R\sinh\alpha = 8,\ R\cosh\alpha = 10$ | B1, M1 on ePen | Equates coefficients to obtain two correct equations. $R^2 = 10^2 - 8^2$ provided incorrect equations are not seen. |
| $R^2(\cosh^2\alpha - \sinh^2\alpha) = 10^2 - 8^2 \Rightarrow R^2 = 36 \Rightarrow R = 6$ | 1st M1 | Complete attempt at finding positive $R$. Correct exponential definitions must be used. |
| $\tanh\alpha = \frac{8}{10} \Rightarrow \alpha = \text{artanh}\left(\frac{4}{5}\right) = \frac{1}{2}\ln\frac{1+\frac{4}{5}}{1-\frac{4}{5}} = \frac{1}{2}\ln 9 = \ln 3$ | 2nd M1 | Complete attempt at finding positive $\alpha$ where $\alpha = k\ln p$, $k>0$, $p>1$. Correct logarithmic form must be used. |
| $6\sinh(x+\ln 3)$, $R=6$ and $\alpha = \ln 3$ (or $p=3$) | A1 | Correct expression. A0 for additional solutions e.g. $6\sinh(x\pm\ln 3)$ |
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## Question 4(c):
$6\sinh(x+\ln 3) = 18\sqrt{7}$
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \text{arsinh}(3\sqrt{7}) - \ln 3 \Rightarrow x = \ln\left(3\sqrt{7}+\sqrt{(3\sqrt{7})^2+1}\right) - \ln 3$ | M1 | Obtains $x = \text{arsinh}\!\left(\frac{18\sqrt{7}}{\text{"6"}}\right) \pm \ln\text{"3"}$ and uses correct logarithmic form. Work must be exact, not decimals. |
| $x = \ln\!\left(\frac{3\sqrt{7}+8}{3}\right) = \ln\!\left(\sqrt{7}+\frac{8}{3}\right)$ | A1 | Correct answer in correct form. Accept e.g. $\ln\!\left(2\tfrac{2}{3}+\sqrt{7}\right)$. Must be fully bracketed correctly. No additional answers. |
---\begin{enumerate}
\item (a) Use the definitions of hyperbolic functions in terms of exponentials to show that
\end{enumerate}
$$\sinh ( A + B ) \equiv \sinh A \cosh B + \cosh A \sinh B$$
(b) Hence express $10 \sinh x + 8 \cosh x$ in the form $R \sinh ( x + \alpha )$ where $R > 0$, giving $\alpha$ in the form $\ln p$ where $p$ is an integer.\\
(c) Hence solve the equation
$$10 \sinh x + 8 \cosh x = 18 \sqrt { 7 }$$
giving your answer in the form $\ln ( \sqrt { 7 } + q )$ where $q$ is a rational number to be determined.
\hfill \mbox{\textit{Edexcel F3 2024 Q4 [9]}}