| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Prove hyperbolic identity from exponentials |
| Difficulty | Standard +0.3 Part (a) is a straightforward proof using exponential definitions—direct substitution and algebraic simplification. Part (b) is a routine substitution integral using the identity from (a). Part (c) is a standard first-order linear ODE with integrating factor cosh x, though it requires careful application of the previous parts. Overall, this is a well-structured Further Maths question testing standard techniques without requiring novel insight, making it slightly easier than average. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08f Integrate using partial fractions4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh x = \frac{1}{2}(e^x + e^{-x})\), \(\sinh x = \frac{1}{2}(e^x - e^{-x})\) | B1 | |
| \(\frac{1}{2}(e^x - e^{-x})(e^x + e^{-x}) = \frac{1}{2}(e^{2x} - e^{-2x}) = \sinh 2x\) | M1 A1 | Expands, AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u = \sinh x \Rightarrow du = \cosh x\,dx\) | B1 | |
| \(\int\sinh^2 2x\cosh x\,dx = 4\int\sinh^2 x\cosh^2 x\,dx = 4\int\sinh^2 x(\sinh^2 x + 1)\,du\) | M1 | Applies identities to find integral in terms of \(u\) |
| \(= 4\int u^2(u^2+1)\,du\) | A1 | |
| \(= 4\left(\frac{1}{5}u^5 + \frac{1}{3}u^3\right)(+C) = 4\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x\right)(+C)\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(e^{\int\tanh x\,dx} = e^{\ln\cosh x} = \cosh x\) | M1 A1 | Finds integrating factor |
| \(\frac{d}{dx}(y\cosh x) = \sinh^2 2x\cosh x\) | M1 | Correct form on LHS and attempt to integrate RHS |
| \(y\cosh x = 4\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x\right) + C\) | M1 A1 | Integrates RHS using their part (b) |
| \(4 = C\) | M1 | Substitutes initial conditions |
| \(y = 4\text{sech}\,x\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x + 1\right)\) | A1 | |
| 7 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh x = \frac{1}{2}(e^x + e^{-x})$, $\sinh x = \frac{1}{2}(e^x - e^{-x})$ | B1 | |
| $\frac{1}{2}(e^x - e^{-x})(e^x + e^{-x}) = \frac{1}{2}(e^{2x} - e^{-2x}) = \sinh 2x$ | M1 A1 | Expands, AG |
| | **3** | |
---
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = \sinh x \Rightarrow du = \cosh x\,dx$ | B1 | |
| $\int\sinh^2 2x\cosh x\,dx = 4\int\sinh^2 x\cosh^2 x\,dx = 4\int\sinh^2 x(\sinh^2 x + 1)\,du$ | M1 | Applies identities to find integral in terms of $u$ |
| $= 4\int u^2(u^2+1)\,du$ | A1 | |
| $= 4\left(\frac{1}{5}u^5 + \frac{1}{3}u^3\right)(+C) = 4\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x\right)(+C)$ | A1 | |
| | **4** | |
---
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $e^{\int\tanh x\,dx} = e^{\ln\cosh x} = \cosh x$ | M1 A1 | Finds integrating factor |
| $\frac{d}{dx}(y\cosh x) = \sinh^2 2x\cosh x$ | M1 | Correct form on LHS and attempt to integrate RHS |
| $y\cosh x = 4\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x\right) + C$ | M1 A1 | Integrates RHS using their part (b) |
| $4 = C$ | M1 | Substitutes initial conditions |
| $y = 4\text{sech}\,x\left(\frac{1}{5}\sinh^5 x + \frac{1}{3}\sinh^3 x + 1\right)$ | A1 | |
| | **7** | |
---6
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$\sinh 2 x = 2 \sinh x \cosh x$$
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_67_1550_374_347}\\
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_58_1569_475_328}\\
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_58_1569_566_328}\\
\includegraphics[max width=\textwidth, alt={}]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_54_1566_657_328} ....................................................................................................................................................... .\\
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_54_1570_840_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_53_1570_932_324}\\
\includegraphics[max width=\textwidth, alt={}, center]{dffdf588-eb26-4d08-b1a3-a0226f5e7763-10_53_1570_1023_324}
\item Using the substitution $\mathrm { u } = \sinh \mathrm { x }$, find $\int \sinh ^ { 2 } 2 x \cosh x \mathrm { dx }$.
\item Find the particular solution of the differential equation
$$\frac { d y } { d x } + y \tanh x = \sinh ^ { 2 } 2 x$$
given that $y = 4$ when $x = 0$. Give your answer in the form $y = f ( x )$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q6 [14]}}