| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Arc length with hyperbolic curves |
| Difficulty | Challenging +1.2 This is a structured multi-part Further Maths question on hyperbolic functions. Parts (a)-(c) involve standard techniques: sketching coth x, proving a standard identity from definitions, and differentiation using chain rule. Part (d) requires arc length formula application and solving cosh a = 2, which are bookwork-level for Further Maths students. While it spans multiple techniques, each step follows predictable patterns without requiring novel insight. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07b Hyperbolic graphs: sketch and properties4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct graph (decreasing curve, positive \(x\) region) | B1 | Correct shape and position, not too truncated |
| \(x = 0\), \(y = 1\) | B1 | States equations of asymptotes |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\coth x = \dfrac{e^x + e^{-x}}{e^x - e^{-x}}\), \(\operatorname{cosech} x = \dfrac{2}{e^x - e^{-x}}\) | B1 | |
| \(\left(\dfrac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2 - \dfrac{4}{\left(e^x - e^{-x}\right)^2} = \dfrac{e^{2x} + e^{-2x} - 2}{\left(e^x - e^{-x}\right)^2} = 1\) | M1 A1 | Writes over common denominator, AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\dfrac{dy}{dx} = -\dfrac{\operatorname{sech}^2\!\left(\frac{1}{2}x\right)}{2\tanh\!\left(\frac{1}{2}x\right)} = -\dfrac{1}{2\sinh\!\left(\frac{1}{2}x\right)\cosh\!\left(\frac{1}{2}x\right)}\) | M1 A1 | Uses chain rule |
| Or \(\dfrac{dy}{dx} = -\dfrac{\operatorname{cosech}^2\!\left(\frac{1}{2}x\right)}{2\coth\!\left(\frac{1}{2}x\right)} = -\dfrac{1}{2\sinh\!\left(\frac{1}{2}x\right)\cosh\!\left(\frac{1}{2}x\right)}\) | ||
| \(= -\dfrac{1}{\sinh x} = -\operatorname{cosech} x\) | A1 | AG |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_a^{2a} \sqrt{1 + \operatorname{cosech}^2 x}\, dx\) | M1 | Forms correct integral |
| \(\int_a^{2a} \sqrt{\coth^2 x}\, dx = \int_a^{2a} \coth x\, dx\) | M1 A1 | Uses \(\coth^2 x - \operatorname{cosech}^2 x = 1\) |
| \(= \bigl[\ln\sinh x\bigr]_a^{2a} = \ln\sinh 2a - \ln\sinh a\) | M1 | Integrates and substitutes limits |
| \(= \ln\dfrac{\sinh 2a}{\sinh a} = \ln(2\cosh a)\) | M1 | Combines logarithms and uses double angle formula |
| \(\ln(2\cosh a) = \ln 4 \Rightarrow \cosh a = 2\) | A1 | AG |
| \(a = \ln\!\left(2 + \sqrt{2^2 - 1}\right) = \ln\!\left(2 + \sqrt{3}\right)\) | A1 | Must reject \(\ln\!\left(2 - \sqrt{3}\right)\) |
| 7 |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct graph (decreasing curve, positive $x$ region) | **B1** | Correct shape and position, not too truncated |
| $x = 0$, $y = 1$ | **B1** | States equations of asymptotes |
| | **2** | |
## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\coth x = \dfrac{e^x + e^{-x}}{e^x - e^{-x}}$, $\operatorname{cosech} x = \dfrac{2}{e^x - e^{-x}}$ | **B1** | |
| $\left(\dfrac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2 - \dfrac{4}{\left(e^x - e^{-x}\right)^2} = \dfrac{e^{2x} + e^{-2x} - 2}{\left(e^x - e^{-x}\right)^2} = 1$ | **M1 A1** | Writes over common denominator, AG |
| | **3** | |
## Question 8(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{dy}{dx} = -\dfrac{\operatorname{sech}^2\!\left(\frac{1}{2}x\right)}{2\tanh\!\left(\frac{1}{2}x\right)} = -\dfrac{1}{2\sinh\!\left(\frac{1}{2}x\right)\cosh\!\left(\frac{1}{2}x\right)}$ | **M1 A1** | Uses chain rule |
| Or $\dfrac{dy}{dx} = -\dfrac{\operatorname{cosech}^2\!\left(\frac{1}{2}x\right)}{2\coth\!\left(\frac{1}{2}x\right)} = -\dfrac{1}{2\sinh\!\left(\frac{1}{2}x\right)\cosh\!\left(\frac{1}{2}x\right)}$ | | |
| $= -\dfrac{1}{\sinh x} = -\operatorname{cosech} x$ | **A1** | AG |
| | **3** | |
## Question 8(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_a^{2a} \sqrt{1 + \operatorname{cosech}^2 x}\, dx$ | **M1** | Forms correct integral |
| $\int_a^{2a} \sqrt{\coth^2 x}\, dx = \int_a^{2a} \coth x\, dx$ | **M1 A1** | Uses $\coth^2 x - \operatorname{cosech}^2 x = 1$ |
| $= \bigl[\ln\sinh x\bigr]_a^{2a} = \ln\sinh 2a - \ln\sinh a$ | **M1** | Integrates and substitutes limits |
| $= \ln\dfrac{\sinh 2a}{\sinh a} = \ln(2\cosh a)$ | **M1** | Combines logarithms and uses double angle formula |
| $\ln(2\cosh a) = \ln 4 \Rightarrow \cosh a = 2$ | **A1** | AG |
| $a = \ln\!\left(2 + \sqrt{2^2 - 1}\right) = \ln\!\left(2 + \sqrt{3}\right)$ | **A1** | Must reject $\ln\!\left(2 - \sqrt{3}\right)$ |
| | **7** | |8
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $\mathrm { y } = \operatorname { coth } \mathrm { x }$ for $x > 0$ and state the equations of the asymptotes.
\item Starting from the definitions of coth and cosech in terms of exponentials, prove that
$$\operatorname { coth } ^ { 2 } x - \operatorname { cosech } ^ { 2 } x = 1$$
The curve $C$ has equation $\mathrm { y } = \ln \operatorname { coth } \left( \frac { 1 } { 2 } \mathrm { x } \right)$ for $x > 0$.
\item Show that $\frac { \mathrm { dy } } { \mathrm { dx } } = - \operatorname { cosechx }$.
\item It is given that the arc length of $C$ from $\mathrm { x } = \mathrm { a }$ to $\mathrm { x } = 2 \mathrm { a }$ is $\ln 4$, where $a$ is a positive constant.
Show that $\cosh a = 2$ and find, in logarithmic form, the exact value of $a$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2020 Q8 [15]}}