| Exam Board | CAIE |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reduction Formulae |
| Type | Hyperbolic function reduction |
| Difficulty | Challenging +1.8 This is a challenging Further Maths reduction formula question requiring integration by parts with hyperbolic functions, manipulation of hyperbolic identities (sech²x = 1 - tanh²x), careful evaluation of boundary terms at ln 3, and recursive application to find I₄. While the structure is standard for reduction formulae, the hyperbolic context and multi-step algebraic manipulation elevate it above typical A-level questions. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.08f Integrate using partial fractions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_n = \int_0^{\ln 3}\text{sech}^{n-2}x\,\text{sech}^2x\,dx\) | B1 | Separates into correct structure |
| \(I_n = \int_0^{\ln 3}\text{sech}^{n-2}x\,\text{sech}^2x\,dx = \left[\text{sech}^{n-2}x\tanh x\right]_0^{\ln 3}+(n-2)\int_0^{\ln 3}\text{sech}^{n-2}x\tanh^2x\,dx\) | *M1 | Uses integration by parts correctly |
| \(\left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)(I_{n-2}-I_n)\) | DM1 A1 | Uses \(1-\text{sech}^2x=\tanh^2x\) |
| \((1+n-2)I_n = \left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)I_{n-2} \Rightarrow (n-1)I_n=\left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)I_{n-2}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(I_2 = \frac{4}{5}\) | B1 | |
| \(3I_4 = \left(\frac{3}{5}\right)^2\!\left(\frac{4}{5}\right)+2I_2 \Rightarrow I_4 = \frac{236}{375}=0.629\) | M1 A1 | Applies reduction formula with \(n=4\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_n = \int_0^{\ln 3}\text{sech}^{n-2}x\,\text{sech}^2x\,dx$ | B1 | Separates into correct structure |
| $I_n = \int_0^{\ln 3}\text{sech}^{n-2}x\,\text{sech}^2x\,dx = \left[\text{sech}^{n-2}x\tanh x\right]_0^{\ln 3}+(n-2)\int_0^{\ln 3}\text{sech}^{n-2}x\tanh^2x\,dx$ | *M1 | Uses integration by parts correctly |
| $\left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)(I_{n-2}-I_n)$ | DM1 A1 | Uses $1-\text{sech}^2x=\tanh^2x$ |
| $(1+n-2)I_n = \left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)I_{n-2} \Rightarrow (n-1)I_n=\left(\frac{3}{5}\right)^{n-2}\!\left(\frac{4}{5}\right)+(n-2)I_{n-2}$ | A1 | AG |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $I_2 = \frac{4}{5}$ | B1 | |
| $3I_4 = \left(\frac{3}{5}\right)^2\!\left(\frac{4}{5}\right)+2I_2 \Rightarrow I_4 = \frac{236}{375}=0.629$ | M1 A1 | Applies reduction formula with $n=4$ |
---4 It is given that, for $n \geqslant 0 , \mathrm { I } _ { \mathrm { n } } = \int _ { 0 } ^ { \ln 3 } \operatorname { sech } ^ { \mathrm { n } } \mathrm { xdx }$.
\begin{enumerate}[label=(\alph*)]
\item Show that, for $n \geqslant 2$,
$$( n - 1 ) \mathrm { I } _ { n } = \left( \frac { 3 } { 5 } \right) ^ { n - 2 } \left( \frac { 4 } { 5 } \right) + ( n - 2 ) \mathrm { I } _ { n - 2 }$$
[You may use the result that $\frac { \mathrm { d } } { \mathrm { dx } } ( \operatorname { sech } x ) = - \tanh x \operatorname { sech } x$.]
\item Find the value of $I _ { 4 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 2 2024 Q4 [8]}}