| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove inverse hyperbolic logarithmic form |
| Difficulty | Challenging +1.2 Part (a) is a standard FP2 derivation of the inverse hyperbolic sine formula requiring algebraic manipulation of exponentials and solving a quadratic in e^x. Parts (b)(i-ii) involve differentiation of hyperbolic functions, solving equations, and integration—all core FP2 techniques. While multi-step with 14 total marks, these are well-practiced procedures for Further Maths students without requiring novel insights, placing it moderately above average difficulty. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives4.07f Inverse hyperbolic: logarithmic forms |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = \frac{1}{2}(e^x - e^{-x})\) | M1 | allow \(e^{2x} - 2ye^x - 1 = (0)\) if attempting to complete square; terms all on one side |
| \(\Rightarrow e^{2x} - 2ye^x - 1 = 0\) | ||
| \(\left(e^x\right) = \frac{2y \pm \sqrt{4y^2+4}}{2}\) | A1 | or \(e^x - y = \pm\sqrt{y^2+1}\) after completing square |
| \(e^x > 0\) so reject negative root | E1 | any correct explanation for rejection |
| \(e^x = y + \sqrt{y^2+1} \Rightarrow x = \ln\left(y + \sqrt{y^2+1}\right)\) | A1 | 4 marks |
| (b)(i) \(\frac{dy}{dx} = 6 \times 2\cosh x \sinh x\) | B1 | directly or via \(3\cosh 2x + 3\) |
| (not \(6\sinh 2x\)) | ||
| \(+ 5\cosh x\) | B1 | |
| \(\cosh x = 0\) gives no solution | E1 | Not simply cancelling \(\cosh x\) |
| (only stationary point when) | ||
| \(\sinh x = -\frac{5}{12}\) | M1 | FT "their" \(\sinh x\) from equation of form \(A\cosh x \sinh x + B\cosh x\) |
| \(x = \ln\left(-\frac{5}{12} + \sqrt{1+\frac{25}{144}}\right)\) | or M1 for using exponentials obtaining \(e^x = \frac{2}{3}\) or \(-\frac{3}{2}\) OE | |
| \(= \ln\left(\frac{2}{3}\right)\) | A1 | 5 marks |
| (b)(ii) Area \(= \int_0^{\cosh^{-1}2}(6\cosh^2 x + 5\sinh x) dx\) | B1 | |
| \(6\cosh^2 x = 3 + 3\cosh 2x\) | or \(6\cosh^2 x = \frac{3}{2}(e^{2x}+2e^{-2x})\) | |
| \(Ax + B\sinh 2x\) or \(Cx + D(e^{2x} - e^{-2x})\) | M1 | correct FT "their" \(\int 6\cosh^2 x \, dx\) |
| \(3x + \frac{3}{2}\sinh 2x + 5\cosh x\) | A1 | integration all correct (may be in \(e^x\) form) |
| \(3\cosh^{-1}2 - \frac{3}{2}\sinh(2\cosh^{-1}2) + 10 - 5\) | m1 | F(cosh\(^{-1}2\)) – F(0) correct substitution of limits into their expression |
| \((Area) = 3\cosh^{-1}2 + 6\sqrt{3} + 5\) | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| - (a) May find \(\ln\left | y \pm \sqrt{y^2+1}\right | \) and reason about not having negative ln for E1 |
**(a)** $y = \frac{1}{2}(e^x - e^{-x})$ | M1 | allow $e^{2x} - 2ye^x - 1 = (0)$ if attempting to complete square; terms all on one side
$\Rightarrow e^{2x} - 2ye^x - 1 = 0$ | |
$\left(e^x\right) = \frac{2y \pm \sqrt{4y^2+4}}{2}$ | A1 | or $e^x - y = \pm\sqrt{y^2+1}$ after completing square
$e^x > 0$ so reject negative root | E1 | any correct explanation for rejection
$e^x = y + \sqrt{y^2+1} \Rightarrow x = \ln\left(y + \sqrt{y^2+1}\right)$ | A1 | **4 marks** | AG must earn previous A1
**(b)(i)** $\frac{dy}{dx} = 6 \times 2\cosh x \sinh x$ | B1 | directly or via $3\cosh 2x + 3$
| | | (not $6\sinh 2x$)
$+ 5\cosh x$ | B1 |
$\cosh x = 0$ gives no solution | E1 | Not simply cancelling $\cosh x$
(only stationary point when) | |
$\sinh x = -\frac{5}{12}$ | M1 | FT "their" $\sinh x$ from equation of form $A\cosh x \sinh x + B\cosh x$
$x = \ln\left(-\frac{5}{12} + \sqrt{1+\frac{25}{144}}\right)$ | | or M1 for using exponentials obtaining $e^x = \frac{2}{3}$ or $-\frac{3}{2}$ OE
$= \ln\left(\frac{2}{3}\right)$ | A1 | **5 marks** | accept $\ln\left(\frac{8}{12}\right)$ OE
**(b)(ii)** Area $= \int_0^{\cosh^{-1}2}(6\cosh^2 x + 5\sinh x) dx$ | B1 |
$6\cosh^2 x = 3 + 3\cosh 2x$ | | or $6\cosh^2 x = \frac{3}{2}(e^{2x}+2e^{-2x})$
$Ax + B\sinh 2x$ or $Cx + D(e^{2x} - e^{-2x})$ | M1 | correct FT "their" $\int 6\cosh^2 x \, dx$
$3x + \frac{3}{2}\sinh 2x + 5\cosh x$ | A1 | integration all correct (may be in $e^x$ form)
$3\cosh^{-1}2 - \frac{3}{2}\sinh(2\cosh^{-1}2) + 10 - 5$ | m1 | F(cosh$^{-1}2$) – F(0) correct substitution of limits into **their** expression
$(Area) = 3\cosh^{-1}2 + 6\sqrt{3} + 5$ | A1 | **5 marks**
**Total: 14 marks**
**Notes:**
- **(a)** May find $\ln\left|y \pm \sqrt{y^2+1}\right|$ and reason about not having negative ln for E1
- **Alternative:** $y = \sinh x \Rightarrow 1 + y^2 = \cosh^2 x$ M1; Rejecting minus sign since $\cosh x > 0$ E1; $\cosh x = \sqrt{1+y^2}$; $y + \sqrt{1+y^2} = \frac{1}{2}(e^x + e^{-x} + e^x + e^{-x}) = e^x$ A1 $\Rightarrow x = \ln\left(y + \sqrt{y^2+1}\right)$ A1
- **(b)(i)** If using double angle formula incorrectly, eg $\cosh^2 x = 3\cosh 2x - 3 = \frac{dy}{dx}$, then award B0 for this term but allow final A1 although FIW, since this will be penalised heavily in part (b)(ii)
- **(b)(ii)** May use $\cosh^{-1}2 = \ln(2+\sqrt{3})$ when finding F(cosh$^{-1}2$) and m1 may be implied by correct final answer
---\begin{enumerate}[label=(\alph*)]
\item Given that $y = \sinh x$, use the definition of $\sinh x$ in terms of $e^x$ and $e^{-x}$ to show that
$$x = \ln(y + \sqrt{y^2 + 1}).$$ [4 marks]
\item A curve has equation $y = 6\cosh^2 x + 5\sinh x$.
\begin{enumerate}[label=(\roman*)]
\item Show that the curve has a single stationary point and find its $x$-coordinate, giving your answer in the form $\ln p$, where $p$ is a rational number. [5 marks]
\item The curve lies entirely above the $x$-axis. The region bounded by the curve, the coordinate axes and the line $x = \cosh^{-1} 2$ has area $A$.
Show that
$$A = a\cosh^{-1} 2 + b\sqrt{3} + c$$
where $a$, $b$ and $c$ are integers. [5 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2016 Q6 [14]}}