| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Summation bounds using hyperbolic integrals |
| Difficulty | Challenging +1.2 This is a structured multi-part question on hyperbolic functions with clear scaffolding. Parts (a)-(c) are routine: proving a standard identity from definitions, differentiating using chain rule, and sketching a basic hyperbolic function. Part (d) requires recognizing that rectangles provide an upper Riemann sum for the integral, which is a standard technique. Part (e) is immediate once (d) is established. While this covers Further Maths content, the question guides students through each step with no novel insights required, making it moderately above average difficulty but well within reach for prepared FM students. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07c Hyperbolic identity: cosh^2(x) - sinh^2(x) = 14.08g Derivatives: inverse trig and hyperbolic functions |
| 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 | |
| \(\frac{1}{4}(e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}) = 1\) | M1 A1 | Expands, AG. Clear LHS to RHS for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d}{dx}(\tan^{-1}\sinh x) = \frac{\cosh x}{\sinh^2 x + 1}\) | M1 A1 | Applies \(\frac{d}{du}(\tan^{-1}u) = \frac{1}{u^2+1}\) |
| \(= \frac{\cosh x}{\cosh^2 x} = \text{sech}\, x\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct shape, symmetrical about \(x = 0\), bell-shaped curve | B1 | Correct shape, symmetrical about \(x = 0\) |
| \(y = 0\) (asymptote) | B1 | Accept labels on their sketch |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sum_{r=1}^{n} \text{sech}\, r < \int_0^n \text{sech}\, x \, dx\) | M1 A1 | Compares sum with integral. Limits correct for A1 |
| \(= \left[\tan^{-1}\sinh x\right]_0^n = \tan^{-1}\sinh n\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\pi\) | B1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh x = \frac{1}{2}(e^x + e^{-x})$, $\sinh x = \frac{1}{2}(e^x - e^{-x})$ | B1 | |
| $\frac{1}{4}(e^{2x} + 2 + e^{-2x} - e^{2x} + 2 - e^{-2x}) = 1$ | M1 A1 | Expands, AG. Clear LHS to RHS for A1 |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d}{dx}(\tan^{-1}\sinh x) = \frac{\cosh x}{\sinh^2 x + 1}$ | M1 A1 | Applies $\frac{d}{du}(\tan^{-1}u) = \frac{1}{u^2+1}$ |
| $= \frac{\cosh x}{\cosh^2 x} = \text{sech}\, x$ | A1 | AG |
---
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape, symmetrical about $x = 0$, bell-shaped curve | B1 | Correct shape, symmetrical about $x = 0$ |
| $y = 0$ (asymptote) | B1 | Accept labels on their sketch |
---
## Question 4(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sum_{r=1}^{n} \text{sech}\, r < \int_0^n \text{sech}\, x \, dx$ | M1 A1 | Compares sum with integral. Limits correct for A1 |
| $= \left[\tan^{-1}\sinh x\right]_0^n = \tan^{-1}\sinh n$ | A1 | AG |
---
## Question 4(e):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\pi$ | B1 | |
---4
\begin{enumerate}[label=(\alph*)]
\item Starting from the definitions of cosh and sinh in terms of exponentials, prove that
$$\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1 .$$
\item Show that $\frac { \mathrm { d } } { \mathrm { dx } } \left( \tan ^ { - 1 } ( \sinh x ) \right) = \operatorname { sech } x$.
\item Sketch the graph of $y = \operatorname { sechx }$, stating the equation of the asymptote.
\item By considering a suitable set of $n$ rectangles of unit width, use your sketch to show that
$$\sum _ { r = 1 } ^ { n } \operatorname { sechr } < \tan ^ { - 1 } ( \operatorname { sinhn } )$$
\item Hence state an upper bound, in terms of $\pi$, for $\sum _ { r = 1 } ^ { \infty }$ sech $r$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2022 Q4 [12]}}