| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Find stationary points of hyperbolic curves |
| Difficulty | Standard +0.3 This is a straightforward application of hyperbolic function differentiation (standard formulas) followed by solving a simple equation. The derivatives of cosh and sinh are standard recall, and solving 9sinh(x) + 3cosh(x) + 7 = 0 requires only basic algebraic manipulation and knowledge that sinh(x) and cosh(x) can be expressed in exponential form. While it's a Further Maths topic, the execution is mechanical with no conceptual challenges or multi-step reasoning required. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\frac{dy}{dx} = 9\sinh x + 3\cosh x + 7\) | B1 | Correct derivative |
| \(9\frac{(e^x - e^{-x})}{2} + 3\frac{(e^x + e^{-x})}{2} + 7 = 0\) | M1 | Replaces \(\sinh x\) and \(\cosh x\) by correct exponential forms |
| \(12e^{2x} + 14e^x - 6 = 0\) oe | M1A1 | M1: Obtains a quadratic in \(e^x\); A1: Correct quadratic |
| \((3e^x - 1)(2e^x + 3) = 0 \Rightarrow e^x = \ldots\) | M1 | Solves their quadratic as far as \(e^x = \ldots\) |
| \(x = \ln\left(\frac{1}{3}\right)\) | A1 | cso (Allow \(-\ln 3\)); \(e^x = -\frac{3}{2}\) need not be seen. Extra answers, award A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Notes |
| \(\frac{dy}{dx} = 9\sinh x + 3\cosh x + 7\) | B1 | Correct derivative |
| \(9\sinh x = -3\cosh x - 7 \Rightarrow 81\sinh^2 x = 9\cosh^2 x + 42\cosh x + 49\) | ||
| \(72\cosh^2 x - 42\cosh x - 130 = 0\) | M1 | Squares and attempts quadratic in \(\cosh x\) |
| \((3\cosh x - 5)(12\cosh x + 13) = 0 \Rightarrow \cosh x = \frac{5}{3}\) | M1A1 | M1: Solves quadratic; A1: Correct value |
| \(x = \ln\left(\frac{5}{3} \pm \sqrt{\left(\frac{5}{3}\right)^2 - 1}\right)\) | M1 | Use of ln form of arcosh |
| \(x = \ln\left(\frac{1}{3}\right)\) | A1 | cso (Allow \(-\ln 3\)) |
## Question 1:
$y = 9\cosh x + 3\sinh x + 7x$
| Answer/Working | Mark | Notes |
|---|---|---|
| $\frac{dy}{dx} = 9\sinh x + 3\cosh x + 7$ | B1 | Correct derivative |
| $9\frac{(e^x - e^{-x})}{2} + 3\frac{(e^x + e^{-x})}{2} + 7 = 0$ | M1 | Replaces $\sinh x$ and $\cosh x$ by correct exponential forms |
| $12e^{2x} + 14e^x - 6 = 0$ oe | M1A1 | M1: Obtains a quadratic in $e^x$; A1: Correct quadratic |
| $(3e^x - 1)(2e^x + 3) = 0 \Rightarrow e^x = \ldots$ | M1 | Solves their quadratic as far as $e^x = \ldots$ |
| $x = \ln\left(\frac{1}{3}\right)$ | A1 | cso (Allow $-\ln 3$); $e^x = -\frac{3}{2}$ need not be seen. Extra answers, award A0 |
**Alternative:**
| Answer/Working | Mark | Notes |
|---|---|---|
| $\frac{dy}{dx} = 9\sinh x + 3\cosh x + 7$ | B1 | Correct derivative |
| $9\sinh x = -3\cosh x - 7 \Rightarrow 81\sinh^2 x = 9\cosh^2 x + 42\cosh x + 49$ | | |
| $72\cosh^2 x - 42\cosh x - 130 = 0$ | M1 | Squares and attempts quadratic in $\cosh x$ |
| $(3\cosh x - 5)(12\cosh x + 13) = 0 \Rightarrow \cosh x = \frac{5}{3}$ | M1A1 | M1: Solves quadratic; A1: Correct value |
| $x = \ln\left(\frac{5}{3} \pm \sqrt{\left(\frac{5}{3}\right)^2 - 1}\right)$ | M1 | Use of ln form of arcosh |
| $x = \ln\left(\frac{1}{3}\right)$ | A1 | cso (Allow $-\ln 3$) |
NB: Ignore any attempts to find the $y$ coordinate
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = 9 \cosh x + 3 \sinh x + 7 x$$
Use differentiation to find the exact $x$ coordinate of the stationary point of $C$, giving your answer as a natural logarithm.\\
\hfill \mbox{\textit{Edexcel F3 2016 Q1 [6]}}