| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Solve mixed sinh/cosh linear combinations |
| Difficulty | Standard +0.3 This is a standard Further Pure hyperbolic functions question requiring routine application of definitions (cosh x = (e^x + e^(-x))/2, sinh x = (e^x - e^(-x))/2), solving a quadratic in e^k, and straightforward differentiation. All parts follow predictable patterns with no novel insight required, making it slightly easier than average for FP2 material. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(e^{2k} - e^k - 2 = 0\) | M1, A1, A1 | Allow if 2's are missing or if coshx and sinhx interchanged. AG Condone x instead of k |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = \ln 2\) | M1, E1, A1, A1F | Must state something to earn E1. Do not accept ignoring or crossing out. A1F for final answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{1}{2}\ln 2\) | M1, A1, A1F, A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Correct reason | M1, A1, E1 | Must give a reason |
| (iii) \(\frac{d^2y}{dx^2} = y = 0\) at \(\left(\frac{1}{2}\ln 2, 0\right)\) ie one point | B1F | (1 mark) |
**(a)(i)** $\frac{e^k - e^{-k}}{2} - \frac{3(e^k - e^{-k})}{2} = -1$
$-2e^k + 4e^{-k} = -2$
$e^{2k} - e^k - 2 = 0$ | M1, A1, A1 | Allow if 2's are missing or if coshx and sinhx interchanged. AG Condone x instead of k
**(ii)** $(e^k + 1)(e^k - 2) = 0$
$e^k \neq -1$
$e^k = 2$
$k = \ln 2$ | M1, E1, A1, A1F | Must state something to earn E1. Do not accept ignoring or crossing out. A1F for final answer
**(b)(i)** $\cosh x = 3\sinh x$ or in terms of $e^x$
$\tanh x = \frac{1}{3}$ or $2e^x = 4e^{-x}$
$x = \frac{1}{2}\ln\left(\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\right)$ or $e^{2x} = 2$
$x = \frac{1}{2}\ln 2$ | M1, A1, A1F, A1 | CAO
**(ii)** $\frac{dy}{dx} = \sinh x - 3\cosh x$ or $-e^x - 2e^{-x}$
$= 0$ when $\tanh x = 3$ or $e^{2x} = -2$
Correct reason | M1, A1, E1 | Must give a reason
**(iii)** $\frac{d^2y}{dx^2} = y = 0$ at $\left(\frac{1}{2}\ln 2, 0\right)$ ie one point | B1F | (1 mark)
**Total: 15 marks**
---3 The curve $C$ has equation
$$y = \cosh x - 3 \sinh x$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item The line $y = - 1$ meets $C$ at the point $( k , - 1 )$.
Show that
$$\mathrm { e } ^ { 2 k } - \mathrm { e } ^ { k } - 2 = 0$$
\item Hence find $k$, giving your answer in the form $\ln a$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinate of the point where the curve $C$ intersects the $x$-axis, giving your answer in the form $p \ln a$.
\item Show that $C$ has no stationary points.
\item Show that there is exactly one point on $C$ for which $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2006 Q3 [15]}}