| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hyperbolic functions |
| Type | Summation bounds using hyperbolic integrals |
| Difficulty | Challenging +1.8 This is a sophisticated Further Maths question requiring integration with substitution, understanding of inverse hyperbolic functions (cosh^{-1}x = ln(x + √(x²-1))), and Riemann sum approximations to derive summation bounds. Part (a) is routine, but parts (b) and (c) require recognizing the connection between the sum and integral, setting up appropriate inequalities using left/right rectangles, and manipulating the resulting expressions. While the techniques are standard for Further Maths, the multi-step reasoning and synthesis of hyperbolic functions with summation bounds elevates this above average difficulty. |
| Spec | 1.08g Integration as limit of sum: Riemann sums1.08h Integration by substitution4.07e Inverse hyperbolic: definitions, domains, ranges |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int \frac{x}{\sqrt{x^2-1}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{u}}\,du\) | M1 A1 | Applies given substitution |
| \(\sqrt{x^2-1} + C\) | A1 | Allow "\(+C\)" missing. Answer must be in terms of \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh^{-1} r = \ln\!\left(r + \sqrt{r^2-1}\right)\) | B1 | |
| \(\cosh^{-1}2 + \cosh^{-1}3 + \ldots + \cosh^{-1}N\) | M1 | Forms sum of the areas of the rectangles |
| \(> \int_1^N \cosh^{-1}x\,dx\) | M1 | Compares with integral with correct limits |
| \(\int_1^N \cosh^{-1}x\,dx = \left[x\cosh^{-1}x\right]_1^N - \int_1^N \frac{x}{\sqrt{x^2-1}}\,dx\) | A1 | Evaluates integral |
| \(\sum_{r=2}^{N} \ln\!\left(r+\sqrt{r^2-1}\right) > N\ln\!\left(N+\sqrt{N^2-1}\right) - \sqrt{N^2-1}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cosh^{-1}1) + \cosh^{-1}2 + \ldots + \cosh^{-1}(N-1) < \int_1^N \cosh^{-1}x\,dx\) | M1 A1 | Compares with integral with correct limits |
| \(\sum_{r=2}^{N}\ln\!\left(r+\sqrt{r^2-1}\right) < (N+1)\ln\!\left(N+\sqrt{N^2-1}\right) - \sqrt{N^2-1}\) | A1 | Adds \(\ln\!\left(N+\sqrt{N^2-1}\right)\) to both sides |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((\cosh^{-1}1)+\cosh^{-1}2+\ldots+\cosh^{-1}(N-1)+\cosh^{-1}N < \int_1^{N+1}\cosh^{-1}x\,dx\) | M1 A1 | Compares with integral with correct limits |
| \(\sum_{r=2}^{N}\ln\!\left(r+\sqrt{r^2-1}\right) < (N+1)\ln\!\left(N+1+\sqrt{N^2+2N}\right) - \sqrt{N^2+2N}\) | A1 |
# Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int \frac{x}{\sqrt{x^2-1}}\,dx = \frac{1}{2}\int \frac{1}{\sqrt{u}}\,du$ | M1 A1 | Applies given substitution |
| $\sqrt{x^2-1} + C$ | A1 | Allow "$+C$" missing. Answer must be in terms of $x$ |
---
# Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh^{-1} r = \ln\!\left(r + \sqrt{r^2-1}\right)$ | B1 | |
| $\cosh^{-1}2 + \cosh^{-1}3 + \ldots + \cosh^{-1}N$ | M1 | Forms sum of the areas of the rectangles |
| $> \int_1^N \cosh^{-1}x\,dx$ | M1 | Compares with integral with correct limits |
| $\int_1^N \cosh^{-1}x\,dx = \left[x\cosh^{-1}x\right]_1^N - \int_1^N \frac{x}{\sqrt{x^2-1}}\,dx$ | A1 | Evaluates integral |
| $\sum_{r=2}^{N} \ln\!\left(r+\sqrt{r^2-1}\right) > N\ln\!\left(N+\sqrt{N^2-1}\right) - \sqrt{N^2-1}$ | A1 | AG |
---
# Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cosh^{-1}1) + \cosh^{-1}2 + \ldots + \cosh^{-1}(N-1) < \int_1^N \cosh^{-1}x\,dx$ | M1 A1 | Compares with integral with correct limits |
| $\sum_{r=2}^{N}\ln\!\left(r+\sqrt{r^2-1}\right) < (N+1)\ln\!\left(N+\sqrt{N^2-1}\right) - \sqrt{N^2-1}$ | A1 | Adds $\ln\!\left(N+\sqrt{N^2-1}\right)$ to both sides |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(\cosh^{-1}1)+\cosh^{-1}2+\ldots+\cosh^{-1}(N-1)+\cosh^{-1}N < \int_1^{N+1}\cosh^{-1}x\,dx$ | M1 A1 | Compares with integral with correct limits |
| $\sum_{r=2}^{N}\ln\!\left(r+\sqrt{r^2-1}\right) < (N+1)\ln\!\left(N+1+\sqrt{N^2+2N}\right) - \sqrt{N^2+2N}$ | A1 | |
---7
\begin{enumerate}[label=(\alph*)]
\item Use the substitution $\mathrm { u } = \mathrm { x } ^ { 2 } - 1$ to find $\int \frac { x } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x$.\\
\includegraphics[max width=\textwidth, alt={}, center]{d3ddf5ce-4399-4438-ab67-7bdb2e1bea6e-12_778_1548_1007_296}
The diagram shows the curve with equation $\mathrm { y } = \cosh ^ { - 1 } \mathrm { x }$ together with a set of $( N - 1 )$ rectangles of unit width.
\item By considering the sum of the areas of these rectangles, show that
$$\sum _ { r = 2 } ^ { N } \ln \left( r + \sqrt { r ^ { 2 } - 1 } \right) > N \ln \left( N + \sqrt { N ^ { 2 } - 1 } \right) - \sqrt { N ^ { 2 } - 1 }$$
\item Use a similar method to find, in terms of $N$, an upper bound for $\sum _ { \mathrm { r } = 2 } ^ { \mathrm { N } } \ln \left( \mathrm { r } + \sqrt { \mathrm { r } ^ { 2 } - 1 } \right)$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2023 Q7 [11]}}