| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Volume of revolution with hyperbolics |
| Difficulty | Challenging +1.2 This is a Further Maths Core Pure question involving hyperbolic functions and volumes of revolution. Part (a) requires applying the definition of arsinh and solving, which is straightforward. Part (b) requires setting up a volume integral about the y-axis using x = sinh(y), then integrating sinh²(y) using the standard identity. While it involves Further Maths content and multiple techniques, the steps are fairly standard for this topic with no novel insights required. Slightly above average difficulty due to the Further Maths context and integration technique needed. |
| Spec | 4.07e Inverse hyperbolic: definitions, domains, ranges4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\ln\left(\alpha + \sqrt{\alpha^2 + 1}\right) = \frac{1}{2}\ln 3\) | B1 | Recalls definition for \(\sinh\left(\frac{1}{2}\ln 3\right)\) or forms equation for arcsinh \(x\) |
| \(\alpha + \sqrt{\alpha^2+1} = \sqrt{3}\), so \(\sqrt{\alpha^2+1} = \sqrt{3} - \alpha\), giving \(\alpha^2 + 1 = 3 - 2\sqrt{3}\alpha + \alpha^2 \Rightarrow \alpha = \ldots\) | M1 | Uses logarithms to find value for \(\alpha\) or forms and solves correct equation without log |
| \(\alpha = \frac{\sqrt{3}}{3}\) or \(\frac{1}{\sqrt{3}}\) | A1 | Deduces correct exact value for \(\alpha\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Volume \(= \pi\int_0^{\frac{1}{2}\ln 3} \sinh^2 y \, dy\) | B1 | Correct expression including integration signs, \(dy\) and correct limits |
| \(\{\pi\}\int\left(\frac{e^y - e^{-y}}{2}\right)^2 dy = \{\pi\}\int\left(\frac{e^{2y}-2+e^{-2y}}{4}\right)dy\) or \(\{\pi\}\int \frac{1}{2}\cosh 2y - \frac{1}{2}\,dy\) | M1 | Uses exponential formula for \(\sinh y\) or identity \(\cosh 2y = \pm 1 \pm 2\sinh^2 y\) to write integrable form |
| \(\frac{1}{4}\left(\frac{1}{2}e^{2y} - 2y - \frac{1}{2}e^{-2y}\right)\) or \(\frac{1}{4}\sinh 2y - \frac{1}{2}y\) | dM1, A1 | Dependent on previous M; correct integration |
| Uses limits \(y=0\) and \(y=\frac{1}{2}\ln 3\), subtracts correct way round | M1 | Correct use of limits |
| \(\frac{\pi}{4}\left(\frac{4}{3} - \ln 3\right)\) or exact equivalent | A1 | Correct exact volume |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\ln\left(\alpha + \sqrt{\alpha^2 + 1}\right) = \frac{1}{2}\ln 3$ | B1 | Recalls definition for $\sinh\left(\frac{1}{2}\ln 3\right)$ or forms equation for arcsinh $x$ |
| $\alpha + \sqrt{\alpha^2+1} = \sqrt{3}$, so $\sqrt{\alpha^2+1} = \sqrt{3} - \alpha$, giving $\alpha^2 + 1 = 3 - 2\sqrt{3}\alpha + \alpha^2 \Rightarrow \alpha = \ldots$ | M1 | Uses logarithms to find value for $\alpha$ or forms and solves correct equation without log |
| $\alpha = \frac{\sqrt{3}}{3}$ or $\frac{1}{\sqrt{3}}$ | A1 | Deduces correct exact value for $\alpha$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Volume $= \pi\int_0^{\frac{1}{2}\ln 3} \sinh^2 y \, dy$ | B1 | Correct expression including integration signs, $dy$ and correct limits |
| $\{\pi\}\int\left(\frac{e^y - e^{-y}}{2}\right)^2 dy = \{\pi\}\int\left(\frac{e^{2y}-2+e^{-2y}}{4}\right)dy$ or $\{\pi\}\int \frac{1}{2}\cosh 2y - \frac{1}{2}\,dy$ | M1 | Uses exponential formula for $\sinh y$ or identity $\cosh 2y = \pm 1 \pm 2\sinh^2 y$ to write integrable form |
| $\frac{1}{4}\left(\frac{1}{2}e^{2y} - 2y - \frac{1}{2}e^{-2y}\right)$ or $\frac{1}{4}\sinh 2y - \frac{1}{2}y$ | dM1, A1 | Dependent on previous M; correct integration |
| Uses limits $y=0$ and $y=\frac{1}{2}\ln 3$, subtracts correct way round | M1 | Correct use of limits |
| $\frac{\pi}{4}\left(\frac{4}{3} - \ln 3\right)$ or exact equivalent | A1 | Correct exact volume |
---\begin{enumerate}
\item Solutions based entirely on graphical or numerical methods are not acceptable.
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aaf73eef-4103-48c2-865e-e8288891ae80-20_480_930_299_555}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve with equation
$$y = \operatorname { arsinh } x \quad x \geqslant 0$$
and the straight line with equation $y = \beta$\\
The line and the curve intersect at the point with coordinates $( \alpha , \beta )$\\
Given that $\beta = \frac { 1 } { 2 } \ln 3$\\
(a) show that $\alpha = \frac { 1 } { \sqrt { 3 } }$
The finite region $R$, shown shaded in Figure 1, is bounded by the curve with equation $y = \operatorname { arsinh } x$, the $y$-axis and the line with equation $y = \beta$
The region $R$ is rotated through $2 \pi$ radians about the $y$-axis.\\
(b) Use calculus to find the exact value of the volume of the solid generated.
\hfill \mbox{\textit{Edexcel CP2 2021 Q7 [9]}}