| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2017 |
| 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 hyperbolic equation requiring substitution of 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 involves Further Maths content, the technique is routine and mechanical with no conceptual insight required, making it slightly easier than average overall. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.06f Laws of logarithms: addition, subtraction, power rules4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(18\left(\frac{e^x + e^{-x}}{2}\right) + 14\left(\frac{e^x - e^{-x}}{2}\right) = 11 + e^x\) | M1 | Uses the correct exponential forms |
| \(9e^{2x} + 9 + 7e^{2x} - 7 = 11e^x + e^{2x}\) | ||
| \(15e^{2x} - 11e^x + 2 = 0\) or \(15e^x - 11 + 2e^{-x} = 0\) | M1A1 | M1: Collects terms to obtain a 3 term equation; A1: Correct equation in either form shown |
| \((5e^x - 2)(3e^x - 1) = 0 \Rightarrow e^x = \ldots\) or \(\left(5e^{\frac{x}{2}} - 2e^{\frac{-x}{2}}\right)\left(3e^{\frac{x}{2}} - e^{\frac{-x}{2}}\right)\) or \((5e^x - 2)(3 - e^{-x})\) | dM1 | Attempt to solve their 3TQ; depends on the second M mark |
| \(x = \ln\frac{2}{5},\ \ln\frac{1}{3}\) | A1 | Both; \(\ln\frac{2}{5}\) or \(\ln 0.4\); \(\ln\frac{1}{3}\) or \(\ln 0.3\text{rec}\); \(-\ln 3\) scores A0 |
## Question 1: $18\cosh x + 14\sinh x = 11 + e^x$
| Working/Answer | Mark | Guidance |
|---|---|---|
| $18\left(\frac{e^x + e^{-x}}{2}\right) + 14\left(\frac{e^x - e^{-x}}{2}\right) = 11 + e^x$ | M1 | Uses the correct exponential forms |
| $9e^{2x} + 9 + 7e^{2x} - 7 = 11e^x + e^{2x}$ | | |
| $15e^{2x} - 11e^x + 2 = 0$ or $15e^x - 11 + 2e^{-x} = 0$ | M1A1 | M1: Collects terms to obtain a 3 term equation; A1: Correct equation in either form shown |
| $(5e^x - 2)(3e^x - 1) = 0 \Rightarrow e^x = \ldots$ or $\left(5e^{\frac{x}{2}} - 2e^{\frac{-x}{2}}\right)\left(3e^{\frac{x}{2}} - e^{\frac{-x}{2}}\right)$ or $(5e^x - 2)(3 - e^{-x})$ | dM1 | Attempt to solve their 3TQ; depends on the second M mark |
| $x = \ln\frac{2}{5},\ \ln\frac{1}{3}$ | A1 | Both; $\ln\frac{2}{5}$ or $\ln 0.4$; $\ln\frac{1}{3}$ or $\ln 0.3\text{rec}$; $-\ln 3$ scores A0 |
**Total: 5 marks**
---\begin{enumerate}
\item Solve the equation
\end{enumerate}
$$18 \cosh x + 14 \sinh x = 11 + \mathrm { e } ^ { x }$$
Give your answers in the form $\ln a$, where $a$ is rational.\\
\hfill \mbox{\textit{Edexcel F3 2017 Q1 [5]}}