| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Method of differences |
| Difficulty | Challenging +1.8 This Further Pure 3 reduction formula question requires knowledge of hyperbolic functions, the identity tanh²x = 1 - sech²x, and integration of sech²x. Part (a) demands deriving a non-standard reduction formula with a specific numerical term, while part (b) requires iterative application. The hyperbolic context and the specific form of the result make this significantly harder than standard reduction formulae, though the technique itself is methodical once the identity is recognized. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.07d Differentiate/integrate: hyperbolic functions8.06a Reduction formulae: establish, use, and evaluate recursively |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tanh^{2n} x = \tanh^{2(n-1)} x \tanh^2 x\) | B1 | — |
| \(\tanh^{2n} x = \pm\tanh^{2(n-1)} x\left(1 - \text{sech}^2 x\right)\) | M1 | — |
| \(I_n = \int_0^{\ln 2}\tanh^{2(n-1)} x\, dx - \int_0^{\ln 2}\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx\) | — | — |
| \(I_n = I_{n-1} - \left[\frac{1}{2n-1}\tanh^{2n-1} x\right]_0^{\ln 2}\) | M1A1 | M1: Correctly substitutes for \(I_{n-1}\) and obtains \(\int\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx = k\tanh^{2n-1} x\). A1: Correct expression |
| \(= I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}\) * | A1* | Correct completion with no errors (cso) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_n - I_{n-1} = \int_0^{\ln 2}\left(\tanh^{2n} x - \tanh^{2(n-1)} x\right)dx = \int_0^{\ln 2}\tanh^{2(n-1)} x\left(\tanh^2 x - 1\right)dx\) | B1 | — |
| \(= \int_0^{\ln 2}\tanh^{2(n-1)} x\left(-\text{sech}^2 x\right)dx\) | M1 | \(= \int_0^{\ln 2}\tanh^{2(n-1)} x\left(\pm\text{sech}^2 x\right)dx\) |
| \(I_n - I_{n-1} = -\left[\frac{1}{2n-1}\tanh^{2n-1} x\right]_0^{\ln 2}\) | M1A1 | M1: Obtains \(\int\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx = k\tanh^{2n-1} x\). A1: Correct expression |
| \(= I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}\) * | A1* | Correct completion with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_0 = \ln 2\) | B1 | The integration must be seen |
| \(I_2 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^3\) | M1 | Applies the reduction formula once |
| \(I_2 = I_0 - \frac{1}{1}\left(\frac{3}{5}\right)^1 - \frac{1}{3}\left(\frac{3}{5}\right)^3\) | M1A1 | M1: Second application of the reduction formula. A1: Correct expression |
| \(I_2 = \ln 2 - \frac{84}{125}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_1 = \int_0^{\ln 2}\tanh^2 x\, dx = \int_0^{\ln 2}\left(1 - \text{sech}^2 x\right)dx\) | — | — |
| \(I_1 = \left[x - \tanh x\right]_0^{\ln 2}\) | B1 | Correct integration |
| \(I_2 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^3\) | M1 | Applies the reduction formula once |
| \(I_1 = \ln 2 - \tanh(\ln 2) = \ln 2 - \frac{3}{5}\) | M1A1 | M1: Uses limits. A1: Correct expression |
| \(I_2 = \ln 2 - \frac{3}{5} - \frac{1}{3}\left(\frac{3}{5}\right)^3 = \ln 2 - \frac{84}{125}\) | A1 | — |
## Question 8:
**Given:** $I_n = \int_0^{\ln 2} \tanh^{2n} x\, dx$, $n \geq 0$
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tanh^{2n} x = \tanh^{2(n-1)} x \tanh^2 x$ | B1 | — |
| $\tanh^{2n} x = \pm\tanh^{2(n-1)} x\left(1 - \text{sech}^2 x\right)$ | M1 | — |
| $I_n = \int_0^{\ln 2}\tanh^{2(n-1)} x\, dx - \int_0^{\ln 2}\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx$ | — | — |
| $I_n = I_{n-1} - \left[\frac{1}{2n-1}\tanh^{2n-1} x\right]_0^{\ln 2}$ | M1A1 | M1: Correctly substitutes for $I_{n-1}$ and obtains $\int\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx = k\tanh^{2n-1} x$. A1: Correct expression |
| $= I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}$ * | A1* | Correct completion with no errors (cso) |
**ALT method:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_n - I_{n-1} = \int_0^{\ln 2}\left(\tanh^{2n} x - \tanh^{2(n-1)} x\right)dx = \int_0^{\ln 2}\tanh^{2(n-1)} x\left(\tanh^2 x - 1\right)dx$ | B1 | — |
| $= \int_0^{\ln 2}\tanh^{2(n-1)} x\left(-\text{sech}^2 x\right)dx$ | M1 | $= \int_0^{\ln 2}\tanh^{2(n-1)} x\left(\pm\text{sech}^2 x\right)dx$ |
| $I_n - I_{n-1} = -\left[\frac{1}{2n-1}\tanh^{2n-1} x\right]_0^{\ln 2}$ | M1A1 | M1: Obtains $\int\tanh^{2(n-1)} x\,\text{sech}^2 x\, dx = k\tanh^{2n-1} x$. A1: Correct expression |
| $= I_{n-1} - \frac{1}{2n-1}\left(\frac{3}{5}\right)^{2n-1}$ * | A1* | Correct completion with no errors |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_0 = \ln 2$ | B1 | The integration must be seen |
| $I_2 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^3$ | M1 | Applies the reduction formula once |
| $I_2 = I_0 - \frac{1}{1}\left(\frac{3}{5}\right)^1 - \frac{1}{3}\left(\frac{3}{5}\right)^3$ | M1A1 | M1: Second application of the reduction formula. A1: Correct expression |
| $I_2 = \ln 2 - \frac{84}{125}$ | A1 | cao |
**Special Case:** If $I_4$ is found, award B1 for $I_0$ or $I_1$ and M1M0A0A0
**Part (b) Alternative:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_1 = \int_0^{\ln 2}\tanh^2 x\, dx = \int_0^{\ln 2}\left(1 - \text{sech}^2 x\right)dx$ | — | — |
| $I_1 = \left[x - \tanh x\right]_0^{\ln 2}$ | B1 | Correct integration |
| $I_2 = I_1 - \frac{1}{3}\left(\frac{3}{5}\right)^3$ | M1 | Applies the reduction formula once |
| $I_1 = \ln 2 - \tanh(\ln 2) = \ln 2 - \frac{3}{5}$ | M1A1 | M1: Uses limits. A1: Correct expression |
| $I_2 = \ln 2 - \frac{3}{5} - \frac{1}{3}\left(\frac{3}{5}\right)^3 = \ln 2 - \frac{84}{125}$ | A1 | — |8.
$$I _ { n } = \int _ { 0 } ^ { \ln 2 } \tanh ^ { 2 n } x \mathrm {~d} x , \quad n \geqslant 0$$
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 1$
\item Hence show that
$$\int _ { 0 } ^ { \ln 2 } \tanh ^ { 4 } x \mathrm {~d} x = p + \ln 2$$
where $p$ is a rational number to be found.\\
8. $\quad I _ { n } = \int _ { 0 } ^ { \ln 2 } \tanh ^ { 2 n } x \mathrm {~d} x , \quad n \geqslant 0$\\
(a) Show that, for $n \geqslant 1$
$$I _ { n } = I _ { n - 1 } - \frac { 1 } { 2 n - 1 } \left( \frac { 3 } { 5 } \right) ^ { 2 n - 1 }$$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q8 [10]}}