| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Taylor series |
| Type | Use series to approximate integral |
| Difficulty | Challenging +1.2 This is a structured Further Maths question with clear guidance. Part (a) is routine application of arc length formula with hyperbolic functions. Part (b) requires finding a Maclaurin series (standard technique) and integrating term-by-term (straightforward). The hyperbolic identity and series expansion are well-practiced topics, making this above average but not requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08a Maclaurin series: find series for function4.08b Standard Maclaurin series: e^x, sin, cos, ln(1+x), (1+x)^n8.06b Arc length and surface area: of revolution, cartesian or parametric |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dot{x} = \sinh t \quad \dot{y} = \cosh t\) | B1 | Differentiates \(x\) and \(y\) with respect to \(t\). |
| \(\dot{x}^2 + \dot{y}^2 = \sinh^2 t + \cosh^2 t = \cosh 2t\) | M1 A1 | Applies \(\cosh^2 t + \sinh^2 t = \cosh 2t\). May use exponential form or \(\left(1+\left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{2}}dx = \left(1 + \frac{\cosh^2 t}{\sinh^2 t}\right)^{\frac{1}{2}}\sinh t\, dt\) |
| \(s = \int_0^{\frac{5}{3}} \sqrt{\cosh 2t}\, dt\) | A1 | AG |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dt}(\sqrt{\cosh 2t}) = \sinh 2t \sqrt{\text{sech } 2t} = \sinh 2t(\cosh 2t)^{-\frac{1}{2}}\) | B1 | Finds first derivative |
| \(\frac{d^2}{dt^2}(\sqrt{\cosh 2t}) = 2\sqrt{\cosh 2t} - \tanh^2 2t\sqrt{\cosh 2t} = 2\sqrt{\cosh 2t} - \sinh^2 2t(\cosh 2t)^{-\frac{3}{2}}\) | B1 | Finds second derivative |
| \(s = \int_0^{\frac{3}{5}} 1 + t^2\, dt = \left[t + \frac{1}{3}t^3\right]_0^{\frac{3}{5}}\) | M1 A1 | Applies Maclaurin's series |
| \(= \frac{84}{125} = 0.672\) | A1 | CWO |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dot{x} = \sinh t \quad \dot{y} = \cosh t$ | **B1** | Differentiates $x$ and $y$ with respect to $t$. |
| $\dot{x}^2 + \dot{y}^2 = \sinh^2 t + \cosh^2 t = \cosh 2t$ | **M1 A1** | Applies $\cosh^2 t + \sinh^2 t = \cosh 2t$. May use exponential form or $\left(1+\left(\frac{dy}{dx}\right)^2\right)^{\frac{1}{2}}dx = \left(1 + \frac{\cosh^2 t}{\sinh^2 t}\right)^{\frac{1}{2}}\sinh t\, dt$ |
| $s = \int_0^{\frac{5}{3}} \sqrt{\cosh 2t}\, dt$ | **A1** | AG |
| | **Total: 4** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dt}(\sqrt{\cosh 2t}) = \sinh 2t \sqrt{\text{sech } 2t} = \sinh 2t(\cosh 2t)^{-\frac{1}{2}}$ | B1 | Finds first derivative |
| $\frac{d^2}{dt^2}(\sqrt{\cosh 2t}) = 2\sqrt{\cosh 2t} - \tanh^2 2t\sqrt{\cosh 2t} = 2\sqrt{\cosh 2t} - \sinh^2 2t(\cosh 2t)^{-\frac{3}{2}}$ | B1 | Finds second derivative |
| $s = \int_0^{\frac{3}{5}} 1 + t^2\, dt = \left[t + \frac{1}{3}t^3\right]_0^{\frac{3}{5}}$ | M1 A1 | Applies Maclaurin's series |
| $= \frac{84}{125} = 0.672$ | A1 | CWO |
---2 The curve $C$ has parametric equations
$$x = \cosh t , \quad y = \sinh t , \quad \text { for } 0 < t \leqslant \frac { 3 } { 5 }$$
The length of $C$ is denoted by $s$.
\begin{enumerate}[label=(\alph*)]
\item Show that $s = \int _ { 0 } ^ { \frac { 3 } { 5 } } \sqrt { \cosh 2 t } \mathrm {~d} t$.\\
\includegraphics[max width=\textwidth, alt={}, center]{27485e4a-cd34-43e3-aa92-767820a9f6f9-04_2714_37_143_2008}
\begin{center}
\end{center}
\item By finding the Maclaurin's series for $\sqrt { \cosh 2 t }$ up to and including the term in $t ^ { 2 }$ ,deduce an approximation to $s$ .
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q2 [9]}}