| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Infinite series convergence and sum |
| Difficulty | Challenging +1.3 This is a structured FP2 hyperbolic functions question with clear signposting. Part (i) is routine manipulation of hyperbolic definitions. Part (ii) uses standard reduction formula techniques with tanh²u = 1 - sech²u. Part (iii) requires direct integration. Part (iv) applies method of differences, a standard FP2 technique. While it involves multiple steps and FP2 content (inherently harder than single maths), the question guides students through each stage with no novel insights required, making it moderately above average difficulty. |
| Spec | 4.06b Method of differences: telescoping series4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tanh(\ln n) = \frac{e^{\ln n} - e^{-\ln n}}{e^{\ln n} + e^{-\ln n}}\) | M1 | For definition of tanh(lnn) seen; Or working with tanh(lnr) = x, definition of tanh⁻¹x seen |
| \(= \frac{n-\frac{1}{n}}{n+\frac{1}{n}} = \frac{n^2-1}{n^2+1}\) | A1 | For simplification to AG |
| Answer | Marks |
|---|---|
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n - I_{n-2} = \int_0^{\ln 2}\left(\tanh^n u - \tanh^{n-2}u\right) du\) | M1 | For factorising and replacing (\(\tanh^2 u - 1\)) by \(\pm\operatorname{sech}^2 u\) (or similarly considering \(I_n\)) |
| \(= \int_0^{\ln 2}\tanh^{n-2}u(\tanh^2 u - 1) du = -\int_0^{\ln 2}\tanh^{n-2}u\operatorname{sech}^2 u\, du\) | ||
| \(\Rightarrow I_n - I_{n-2} = -\left[\frac{1}{n-1}\tanh^{n-1}u\right]_0^{\ln 2}\) | A1 | For correct integrated term |
| \(\Rightarrow I_n - I_{n-2} = -\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1}\) | A1 | For simplification to AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_1 = \int_0^{\ln 2}\tanh u\, du = [\ln\cosh u]_0^{\ln 2}\) | M1 | For integration to \(k\ln\frac{\cosh u}{\sinh u}\) |
| \(= \ln\left(\cosh(\ln 2)\right) = \ln\frac{e^{\ln 2}+e^{-\ln 2}}{2} = \ln\frac{5}{4}\) | M1 | For simplifying \(\frac{\cosh(\ln 2)}{\sinh}\) |
| A1 | For correct value of \(I_1\) | |
| \(I_3 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^2 = -\frac{9}{50} + \ln\frac{5}{4}\) | B1ft | For correct \(I_3\), FT from \(I_1\); SC \(I_3 = -\frac{9}{50} + \ln(\cosh(\ln 2))\) M1 B1ft |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \((I_n - I_{n-2}) + (I_{n-2} - I_{n-4}) + \ldots + (I_3 - I_1)\) | M1 | For attempting to sum equations of the form of (ii) and cancelling soi |
| \(= I_n - I_1 = -\left[\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1} + \frac{1}{n-3}\left(\frac{3}{5}\right)^{n-3} + \ldots + \frac{1}{2}\left(\frac{3}{5}\right)^2\right]\) | ||
| \(\Rightarrow \frac{1}{2}\left(\frac{3}{5}\right)^2 + \frac{1}{4}\left(\frac{3}{5}\right)^4 + \frac{1}{6}\left(\frac{3}{5}\right)^6 + \ldots = I_1 - \ln\frac{5}{4}\) | A1ft | For correct answer ft from \(I_1\) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(I_n = \int_0^{\ln 2}\tanh^n u\, du = \int_0^{\ln 2}\tanh^{n-2}u \cdot \tanh^2 u\, du\) | M1 | For attempt to integrate by parts. |
| \(= \int_0^{\ln 2}\tanh^{n-2}u(1-\operatorname{sech}^2 u) du\) | ||
| \(= \int_0^{\ln 2}\tanh^{n-2}u \cdot du - \int_0^{\ln 2}\tanh^{n-2}u\operatorname{sech}^2 u\, du\) | ||
| \(\Rightarrow I_n = I_{n-2} - \left[\frac{\tanh^{n-1}u}{n-1}\right]_0^{\ln 2}\) | A1 | For correct integrated term |
| \(\Rightarrow I_n - I_{n-2} = -\frac{\tanh^{n-1}(\ln 2)}{n-1} = -\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1}\) | A1 | For simplification to AG |
| [3] |
### (i)
$\tanh(\ln n) = \frac{e^{\ln n} - e^{-\ln n}}{e^{\ln n} + e^{-\ln n}}$ | M1 | For definition of tanh(lnn) seen; Or working with tanh(lnr) = x, definition of tanh⁻¹x seen
$= \frac{n-\frac{1}{n}}{n+\frac{1}{n}} = \frac{n^2-1}{n^2+1}$ | A1 | For simplification to AG
SCI $\tanh(\ln n) = \frac{e^{2\ln n}-1}{e^{2\ln n}+1} = \frac{n^2-1}{n^2+1}$
| [2] |
### (ii)
$I_n - I_{n-2} = \int_0^{\ln 2}\left(\tanh^n u - \tanh^{n-2}u\right) du$ | M1 | For factorising and replacing ($\tanh^2 u - 1$) by $\pm\operatorname{sech}^2 u$ (or similarly considering $I_n$)
$= \int_0^{\ln 2}\tanh^{n-2}u(\tanh^2 u - 1) du = -\int_0^{\ln 2}\tanh^{n-2}u\operatorname{sech}^2 u\, du$ | |
$\Rightarrow I_n - I_{n-2} = -\left[\frac{1}{n-1}\tanh^{n-1}u\right]_0^{\ln 2}$ | A1 | For correct integrated term
$\Rightarrow I_n - I_{n-2} = -\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1}$ | A1 | For simplification to AG
| [3] |
### (iii)
$I_1 = \int_0^{\ln 2}\tanh u\, du = [\ln\cosh u]_0^{\ln 2}$ | M1 | For integration to $k\ln\frac{\cosh u}{\sinh u}$
$= \ln\left(\cosh(\ln 2)\right) = \ln\frac{e^{\ln 2}+e^{-\ln 2}}{2} = \ln\frac{5}{4}$ | M1 | For simplifying $\frac{\cosh(\ln 2)}{\sinh}$
| A1 | For correct value of $I_1$
$I_3 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^2 = -\frac{9}{50} + \ln\frac{5}{4}$ | B1ft | For correct $I_3$, FT from $I_1$; SC $I_3 = -\frac{9}{50} + \ln(\cosh(\ln 2))$ M1 B1ft
| [4] |
### (iv)
$(I_n - I_{n-2}) + (I_{n-2} - I_{n-4}) + \ldots + (I_3 - I_1)$ | M1 | For attempting to sum equations of the form of (ii) and cancelling soi
$= I_n - I_1 = -\left[\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1} + \frac{1}{n-3}\left(\frac{3}{5}\right)^{n-3} + \ldots + \frac{1}{2}\left(\frac{3}{5}\right)^2\right]$ | |
$\Rightarrow \frac{1}{2}\left(\frac{3}{5}\right)^2 + \frac{1}{4}\left(\frac{3}{5}\right)^4 + \frac{1}{6}\left(\frac{3}{5}\right)^6 + \ldots = I_1 - \ln\frac{5}{4}$ | A1ft | For correct answer ft from $I_1$
| [2] |
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## Alternative to Q9(ii)
$I_n = \int_0^{\ln 2}\tanh^n u\, du = \int_0^{\ln 2}\tanh^{n-2}u \cdot \tanh^2 u\, du$ | M1 | For attempt to integrate by parts.
$= \int_0^{\ln 2}\tanh^{n-2}u(1-\operatorname{sech}^2 u) du$ | |
$= \int_0^{\ln 2}\tanh^{n-2}u \cdot du - \int_0^{\ln 2}\tanh^{n-2}u\operatorname{sech}^2 u\, du$ | |
$\Rightarrow I_n = I_{n-2} - \left[\frac{\tanh^{n-1}u}{n-1}\right]_0^{\ln 2}$ | A1 | For correct integrated term
$\Rightarrow I_n - I_{n-2} = -\frac{\tanh^{n-1}(\ln 2)}{n-1} = -\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1}$ | A1 | For simplification to AG
| [3] |\begin{enumerate}[label=(\roman*)]
\item Show that $\tanh(\ln n) = \frac{n^2 - 1}{n^2 + 1}$. [2]
\end{enumerate}
It is given that, for non-negative integers $n$, $I_n = \int_0^{\ln 2} \tanh^n u du$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that $I_n - I_{n-2} = -\frac{1}{n-1}\left(\frac{3}{5}\right)^{n-1}$, for $n \geq 2$. [3]
\item Find the value of $I_3$, giving your answer in the form $a + \ln b$, where $a$ and $b$ are constants. [4]
\item Use the method of differences on the result of part (ii) to find the sum of the infinite series
$$\frac{1}{2}\left(\frac{3}{5}\right)^2 + \frac{1}{4}\left(\frac{3}{5}\right)^4 + \frac{1}{6}\left(\frac{3}{5}\right)^6 + \ldots.$$ [2]
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2012 Q9 [11]}}