| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Intersection points of hyperbolic curves |
| Difficulty | Standard +0.8 This is a Further Maths question requiring (a) algebraic manipulation of exponential definitions to derive a triple angle formula, and (b) solving a cubic equation in sinh x, then finding inverse hyperbolic functions. Part (a) is routine for FM students, but part (b) requires recognizing the cubic structure, factoring, and converting to logarithmic form—more demanding than standard A-level but typical for F3. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(4\sinh^3 x + 3\sinh x = 4\left(\frac{e^x-e^{-x}}{2}\right)^3 + 3\left(\frac{e^x-e^{-x}}{2}\right)\) | M1 | Uses \(\sinh x = \frac{e^x-e^{-x}}{2}\) on both sinh terms and attempts to cube the bracket (min accepted is a linear × a quadratic bracket) |
| \(= \frac{1}{2}e^{3x} - \frac{3}{2}e^x + \frac{3}{2}e^{-x} - \frac{1}{2}e^{-3x} + \frac{3}{2}e^x - \frac{3}{2}e^{-x} = \frac{e^{3x}-e^{-3x}}{2} = \sinh 3x\) | A1* | Fully correct proof with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\sinh 3x = 19\sinh x \Rightarrow 4\sinh^3 x + 3\sinh x = 19\sinh x \Rightarrow 4\sinh^3 x - 16\sinh x = 0\) | M1 | Uses result from (a) and combines terms |
| \((\sinh x = 0 \text{ or}) \sinh^2 x = 4\) | A1 | \(\sinh^2 x = 4\) or \(\sinh x = (\pm)2\) |
| \((0, 0)\) | B1 | States the origin as one intersection |
| \(\ln(2+\sqrt{5})\) and \(-\ln(2+\sqrt{5})\) | A1 | Two correct non-zero \(x\) values (allow e.g. \(\ln(-2+\sqrt{5})\) for \(-\ln(2+\sqrt{5})\)) |
| \(\left(\ln(2+\sqrt{5}), 38\right)\) and \(\left(-\ln(2+\sqrt{5}), -38\right)\) | A1 | Two correct points (allow e.g. \(\ln(-2+\sqrt{5})\) for \(-\ln(2+\sqrt{5})\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Notes |
| \(\sinh 3x = 19\sinh x \Rightarrow \frac{e^{3x}-e^{-3x}}{2} = \frac{19(e^x-e^{-x})}{2} \Rightarrow \ldots\) | M1 | Substitutes correct exponential forms and collects terms to one side |
| \(e^{6x} - 19e^{4x} + 19e^{2x} - 1 = 0\) | A1 | Correct equation (or equivalent) |
| \((0,0)\) | B1 | States the origin as one intersection |
| \(\frac{1}{2}\ln(9+4\sqrt{5})\) or \(\frac{1}{2}\ln(9-4\sqrt{5})\) | A1 | Two correct non-zero \(x\) values (oe) |
| \(\left(\frac{1}{2}\ln(9+4\sqrt{5}), 38\right)\) and \(\left(\frac{1}{2}\ln(9-4\sqrt{5}), -38\right)\) | A1 | Two correct points (oe) |
# Question 1:
## Part (a):
| Working/Answer | Mark | Notes |
|---|---|---|
| $4\sinh^3 x + 3\sinh x = 4\left(\frac{e^x-e^{-x}}{2}\right)^3 + 3\left(\frac{e^x-e^{-x}}{2}\right)$ | M1 | Uses $\sinh x = \frac{e^x-e^{-x}}{2}$ on both sinh terms and attempts to cube the bracket (min accepted is a linear × a quadratic bracket) |
| $= \frac{1}{2}e^{3x} - \frac{3}{2}e^x + \frac{3}{2}e^{-x} - \frac{1}{2}e^{-3x} + \frac{3}{2}e^x - \frac{3}{2}e^{-x} = \frac{e^{3x}-e^{-3x}}{2} = \sinh 3x$ | A1* | Fully correct proof with no errors |
## Part (b):
| Working/Answer | Mark | Notes |
|---|---|---|
| $\sinh 3x = 19\sinh x \Rightarrow 4\sinh^3 x + 3\sinh x = 19\sinh x \Rightarrow 4\sinh^3 x - 16\sinh x = 0$ | M1 | Uses result from (a) and combines terms |
| $(\sinh x = 0 \text{ or}) \sinh^2 x = 4$ | A1 | $\sinh^2 x = 4$ or $\sinh x = (\pm)2$ |
| $(0, 0)$ | B1 | States the origin as one intersection |
| $\ln(2+\sqrt{5})$ and $-\ln(2+\sqrt{5})$ | A1 | Two correct non-zero $x$ values (allow e.g. $\ln(-2+\sqrt{5})$ for $-\ln(2+\sqrt{5})$) |
| $\left(\ln(2+\sqrt{5}), 38\right)$ and $\left(-\ln(2+\sqrt{5}), -38\right)$ | A1 | Two correct **points** (allow e.g. $\ln(-2+\sqrt{5})$ for $-\ln(2+\sqrt{5})$) |
### Alternative for (b) using exponentials:
| Working/Answer | Mark | Notes |
|---|---|---|
| $\sinh 3x = 19\sinh x \Rightarrow \frac{e^{3x}-e^{-3x}}{2} = \frac{19(e^x-e^{-x})}{2} \Rightarrow \ldots$ | M1 | Substitutes correct exponential forms and collects terms to one side |
| $e^{6x} - 19e^{4x} + 19e^{2x} - 1 = 0$ | A1 | Correct equation (or equivalent) |
| $(0,0)$ | B1 | States the origin as one intersection |
| $\frac{1}{2}\ln(9+4\sqrt{5})$ or $\frac{1}{2}\ln(9-4\sqrt{5})$ | A1 | Two correct non-zero $x$ values (oe) |
| $\left(\frac{1}{2}\ln(9+4\sqrt{5}), 38\right)$ and $\left(\frac{1}{2}\ln(9-4\sqrt{5}), -38\right)$ | A1 | Two correct **points** (oe) |
---\begin{enumerate}
\item (a) Use the definition of $\sinh x$ in terms of exponentials to show that
\end{enumerate}
$$\sinh 3 x \equiv 4 \sinh ^ { 3 } x + 3 \sinh x$$
(b) Hence determine the exact coordinates of the points of intersection of the curve with equation $y = \sinh 3 x$ and the curve with equation $y = 19 \sinh x$, giving your answers as simplified logarithms where necessary.\\
\hfill \mbox{\textit{Edexcel F3 2020 Q1 [7]}}