| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve mixed sinh/cosh linear combinations |
| Difficulty | Standard +0.3 This is a straightforward Further Maths hyperbolic equation requiring substitution of standard definitions (cosh x = (e^x + e^{-x})/2, sinh x = (e^x - e^{-x})/2), algebraic manipulation to form a quadratic in e^x, and solving. While it requires knowledge of hyperbolic functions and careful algebra, the method is standard and mechanical with no novel insight needed. Slightly above average difficulty due to the Further Maths content and multi-step nature, but routine for students at this level. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| \(7\cosh x + 3\sinh x = 2e^x + 7 \Rightarrow 7\left(\frac{e^x+e^{-x}}{2}\right)+3\left(\frac{e^x-e^{-x}}{2}\right)=2e^x+7\) | M1 | Substitutes at least one correct exponential form for either hyperbolic term and achieves an equation in exponentials and constants alone |
| \(\Rightarrow 7(e^{2x}+1)+3(e^{2x}-1)=4e^{2x}+14e^x\) leading to \(5e^{2x}+2=2e^{2x}+7e^x\) | M1 | Multiplies through by \(e^x\) to obtain any equation forming a 3TQ in \(e^x\) if like terms collected |
| \(\Rightarrow 6e^{2x}-14e^x+4=0 \quad \{3e^{2x}-7e^x+2=0\}\) | A1 | Correct three term quadratic in \(e^x\); could be implied by correct root even if terms not collected |
| \(\Rightarrow (3e^x-1)(e^x-2)=0 \Rightarrow e^x = \ldots\) | M1 | Solves their 3TQ by usual rules; one correct root for their quadratic if no working |
| \(x = \ln 2, \quad \ln\frac{1}{3}\) | A1 | Both correct and simplified; allow \(-\ln\frac{1}{2}\) for \(\ln 2\) and \(-\ln 3\) or \(\ln 3^{-1}\) for \(\ln\frac{1}{3}\); not isw if other answers present |
# Question 1:
$7\cosh x + 3\sinh x = 2e^x + 7 \Rightarrow 7\left(\frac{e^x+e^{-x}}{2}\right)+3\left(\frac{e^x-e^{-x}}{2}\right)=2e^x+7$ | M1 | Substitutes at least one correct exponential form for either hyperbolic term and achieves an equation in exponentials and constants alone
$\Rightarrow 7(e^{2x}+1)+3(e^{2x}-1)=4e^{2x}+14e^x$ leading to $5e^{2x}+2=2e^{2x}+7e^x$ | M1 | Multiplies through by $e^x$ to obtain any equation forming a 3TQ in $e^x$ if like terms collected
$\Rightarrow 6e^{2x}-14e^x+4=0 \quad \{3e^{2x}-7e^x+2=0\}$ | A1 | Correct three term quadratic in $e^x$; could be implied by correct root even if terms not collected
$\Rightarrow (3e^x-1)(e^x-2)=0 \Rightarrow e^x = \ldots$ | M1 | Solves their 3TQ by usual rules; one correct root for their quadratic if no working
$x = \ln 2, \quad \ln\frac{1}{3}$ | A1 | Both correct and simplified; allow $-\ln\frac{1}{2}$ for $\ln 2$ and $-\ln 3$ or $\ln 3^{-1}$ for $\ln\frac{1}{3}$; not isw if other answers present
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. Solve the equation
\end{enumerate}
$$7 \cosh x + 3 \sinh x = 2 \mathrm { e } ^ { x } + 7$$
Give your answers as simplified natural logarithms.
\hfill \mbox{\textit{Edexcel F3 2023 Q1 [5]}}