| 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 Further Maths question combining integration by substitution (part a is routine), then using Riemann sums with hyperbolic functions to establish summation bounds. Part (a) is straightforward, but parts (b) and (c) require understanding the relationship between rectangular areas and integrals, evaluating the integral of cosh^{-1}(x), and manipulating the result algebraically. The conceptual leap connecting the sum to the integral inequality is non-trivial, though the techniques themselves are standard for Further Maths students who have studied hyperbolic functions and summation bounds. |
| 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 with "\(+C\)" 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 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cosh^{-1}(1+1)+\cosh^{-1}(2+1)+\ldots+\cosh^{-1}(N-1+1) < \int_1^N \cosh^{-1}(x+1)\,dx\) | M1 A1 | Compares with integral with correct limits |
| \(= (N+1)\ln\!\left(N+1+\sqrt{N^2+2N}\right) - \sqrt{N^2+2N} - 2\ln(2+\sqrt{3})+\sqrt{3}\) | 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 with "$+C$" 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 | |
**Second alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cosh^{-1}(1+1)+\cosh^{-1}(2+1)+\ldots+\cosh^{-1}(N-1+1) < \int_1^N \cosh^{-1}(x+1)\,dx$ | M1 A1 | Compares with integral with correct limits |
| $= (N+1)\ln\!\left(N+1+\sqrt{N^2+2N}\right) - \sqrt{N^2+2N} - 2\ln(2+\sqrt{3})+\sqrt{3}$ | 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]{482b2236-1f1b-4c53-a1bc-0277cf63dc62-12_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]}}