| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Surface area of revolution: Cartesian curve |
| Difficulty | Challenging +1.2 Part (a) requires applying the surface area of revolution formula with a straightforward derivative calculation and integration that simplifies nicely using exponential properties. Part (b) is routine application of standard Maclaurin series. While this is Further Maths content (inherently harder), both parts follow standard procedures without requiring novel insight or complex problem-solving, placing it slightly above average difficulty. |
| Spec | 4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{1+(e^x-\frac{1}{4}e^{-x})^2}=\sqrt{1+e^{2x}-\frac{1}{2}+\frac{1}{16}e^{-2x}}=\sqrt{e^{2x}+\frac{1}{2}+\frac{1}{16}e^{-2x}}\) | M1 | Finds \(\sqrt{1+\left(\frac{dy}{dx}\right)^2}\) |
| \(\sqrt{(e^x+\frac{1}{4}e^{-x})^2}=e^x+\frac{1}{4}e^{-x}\) | A1 | |
| \(2\pi\int_0^1(e^x+\frac{1}{4}e^{-x})(e^x+\frac{1}{4}e^{-x})\,dx\) | M1 | Correct formula for surface area with limits |
| \(2\pi\int_0^1 e^{2x}+\frac{1}{2}+\frac{1}{16}e^{-2x}\,dx\) | A1 | Expanded integrand |
| \(2\pi\left[\frac{1}{2}e^{2x}+\frac{1}{2}x-\frac{1}{32}e^{-2x}\right]_0^1\) | M1 | Integrates and substitutes limits. Integrand must be of the form \(ae^{2x}+be^{-2x}+c\), with \(a\) and \(b\) nonzero |
| \(2\pi(\frac{1}{2}e^2-\frac{1}{32}e^{-2}+\frac{1}{32})=\pi(e^2-\frac{1}{16}e^{-2}+\frac{1}{16})\) | A1 | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(1+x+\frac{x^2}{2!}+\cdots\right)+\frac{1}{4}\left(1-x+\frac{(-x)^2}{2!}+\cdots\right)\) | M1 | Uses expansion of \(e^x\) or finds first and second derivatives \(e^x-\frac{1}{4}e^{-x}\) and \(e^x+\frac{1}{4}e^{-x}\) |
| \(\frac{5}{4}+\frac{3}{4}x+\frac{5}{8}x^2\) | A1 | |
| 2 |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{1+(e^x-\frac{1}{4}e^{-x})^2}=\sqrt{1+e^{2x}-\frac{1}{2}+\frac{1}{16}e^{-2x}}=\sqrt{e^{2x}+\frac{1}{2}+\frac{1}{16}e^{-2x}}$ | **M1** | Finds $\sqrt{1+\left(\frac{dy}{dx}\right)^2}$ |
| $\sqrt{(e^x+\frac{1}{4}e^{-x})^2}=e^x+\frac{1}{4}e^{-x}$ | **A1** | |
| $2\pi\int_0^1(e^x+\frac{1}{4}e^{-x})(e^x+\frac{1}{4}e^{-x})\,dx$ | **M1** | Correct formula for surface area with limits |
| $2\pi\int_0^1 e^{2x}+\frac{1}{2}+\frac{1}{16}e^{-2x}\,dx$ | **A1** | Expanded integrand |
| $2\pi\left[\frac{1}{2}e^{2x}+\frac{1}{2}x-\frac{1}{32}e^{-2x}\right]_0^1$ | **M1** | Integrates and substitutes limits. Integrand must be of the form $ae^{2x}+be^{-2x}+c$, with $a$ and $b$ nonzero |
| $2\pi(\frac{1}{2}e^2-\frac{1}{32}e^{-2}+\frac{1}{32})=\pi(e^2-\frac{1}{16}e^{-2}+\frac{1}{16})$ | **A1** | |
| | **6** | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(1+x+\frac{x^2}{2!}+\cdots\right)+\frac{1}{4}\left(1-x+\frac{(-x)^2}{2!}+\cdots\right)$ | **M1** | Uses expansion of $e^x$ or finds first and second derivatives $e^x-\frac{1}{4}e^{-x}$ and $e^x+\frac{1}{4}e^{-x}$ |
| $\frac{5}{4}+\frac{3}{4}x+\frac{5}{8}x^2$ | **A1** | |
| | **2** | |
---3
\begin{enumerate}[label=(\alph*)]
\item A curve has equation $\mathrm { y } = \mathrm { e } ^ { \mathrm { x } } + \frac { 1 } { 4 } \mathrm { e } ^ { - \mathrm { x } }$, for $0 \leqslant x \leqslant 1$. Find, in terms of $\pi$ and e , the area of the surface generated when the curve is rotated through $2 \pi$ radians about the $x$-axis.
\item Using standard results from the list of formulae (MF19), or otherwise, find the Maclaurin's series for $\mathrm { e } ^ { x } + \frac { 1 } { 4 } \mathrm { e } ^ { - x }$ up to and including the term in $x ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q3 [8]}}